Geometry

Table of Contents

Notation

  • $C^k$ denotes the space of all functions which have continuous derivatives to the k-th order
  • smooth means $C^\infty$, i.e. infitively differentiable, more specificely, $C^\infty(D)$ means all infitively differentiable functions with domain $D$
  • Maps $f : D \subseteq \mathbb{R}^n$ are assumed to be smooth unless stated otherwise, i.e. partial derivatives of every order exist and are continuous on $D$
  • Euclidean space $\mathbb{E}^n$ as the set $\mathbb{R}^n$ together with its natural vector space operations and the standard inner product
  • sub-scripts (e.g. basis vectors $\mathbf{e}_i$ for $V$ and coeffs. $b_i$ for $V^*$ ) are co-variant
  • super-scripts (e.g. basis vectors of $\mathbf{f}^i$ for $V^*$ and coeffs $a^i$ for $V$ ) are contra-variant
  • $M \overset{\sim}{\to} N$ where $M, N$ are manifolds, uses the $\sim$ to refer to a linear map from $M$ to $N$

Stuff

Curves

Examples

Helix

The helix in $\mathbb{E}^3$ is defined by

\begin{equation*}
\mathbf{x}(s) = \frac{1}{\sqrt{2}} \begin{pmatrix} \cos s \\ \sin s \\ s \end{pmatrix}
\end{equation*}

Constructing a sphere in Euclidean space

I suggest having a look at this page in the notes

Surface of a sphere in the Euclidean space is defiend as:

\begin{equation*}
S^2 = \{ \mathbf{x} \in \mathbb{E}^3 \ | \ |\mathbf{x}| = 1 \} \subset \mathbb{E}^3
\end{equation*}

As it turns out, $S^2$ is not a vector space. How do you define vectors on this spherical surface $S^2$ ?

Defining vectors on spherical surface

At each point on $S^2$, construct a whole plan which is tangent to the sphere, called the tangent plane.

This plane $\mathbb{R}^2$ is the two dimensional vector space of lines tangent to the sphere at the given point, called tangent vectors.

Each point on the sphere $S^2$ defines a different tangent plane. This leads to the notion of a vector field which is: a rule for smoothly assigning a tangent vector to each point on $S^2$.

The above description of a vector space on $S^2$ is valid everywhere, and so we refer to it as a global description.

Usually, we don't have this luxury. Then we we parametrise a number of "patches" of the surface using coordinates, in such a way that the patches cover the whole surface. We refer to this as a local description.

Motivation

The tangent-space $T_p D$ at some point $p$ on the 2-sphere is a function of the point $p$.

The issue with the 2-sphere is that we cannot obtain a (smooth) basis for the surface. We therefore want to think about the operations which do not depend on having a basis. $T_p D$ gives a way of doing this, since each of the derivatives are linear inpendent.

Ricci calculus and Einstein summation

This is reason why we're using superscript to index our coordinates, $x^i$.

Suppose that I have a vector space $V$ and dual space $V^*$. A choice of basis $\mathbf{e}_1, \dots, \mathbf{e}_n$ for $V$ induces a dual basis $\mathbf{f}^1, \dots, \mathbf{f}^n$ on the dual vector space $V^*$, determined by the rule

\begin{equation*}
\mathbf{f}^i(\mathbf{e}_j) = \delta_j^i
\end{equation*}

where $\delta_j^i$ is the Kronecker delta.

Any element $\mathbf{e}$ of $V$ or $\mathbf{f}$ of $V^*$ can be written as lin. comb. of these basis vectors:

\begin{equation*}
\mathbf{e} = \sum_{i} a^i \mathbf{e}_i \text{ and } \mathbf{f} = \sum_{j} b_j \mathbf{f}^j
\end{equation*}

and we have $\mathbf{f}(\mathbf{e}) = \sum_{i} \mathbf{a}^i \mathbf{b}_i$.

If we do a change of basis of $V$, which induces a change of basis for $V^*$, then the coefficients of a vector in $V$ transform in the same way as the basis of vectors $V^*$ and vice versa, the coefficients of a vector $V^*$ transform in the same way as the basis vectors of $V$.

Suppose a new basis for $V$ is given by $\overset{\sim}{\mathbf{e}}_1, \dots, \overset{\sim}{\mathbf{e}}_n$, with

\begin{equation*}
\overset{\sim}{\mathbf{v}}_i = \sum_{j} h_i^j \mathbf{v}_j
\end{equation*}

where the $h_i^j$ are the coefficients of the invertible change-of-basis matrix, and $g_l^k$ are the coefficients of its inverse (i.e. $\sum_j h_i^j g_j^l = \delta_i^l$). If we denote the new induced dual basis for $V^*$ by $\widetilde{\mathbf{f}}^1, \dots, \widetilde{\mathbf{f}}^n$, we have

\begin{equation*}
\widetilde{\mathbf{f}}^i = \sum_j g_j^i \mathbf{f}^j
\end{equation*}

Moreover, for any elements of $\mathbf{e}$ of $V$ and $\mathbf{f}$ of $V^*$ which we can write as

\begin{equation*}
\mathbf{e} = \sum_i a^i \mathbf{e}_i = \sum_i \widetilde{a}^i \widetilde{\mathbf{e}}_i \text{ and } \mathbf{f} = \sum_j b_j \mathbf{f}^i = \sum_j \widetilde{b}_j \widetilde{\mathbf{f}}^j
\end{equation*}

we have

\begin{equation*}
\widetilde{a}^i = \sum_j g_j^i a^j \text{ and } \widetilde{b}_i = \sum_j h_i^j b_j
\end{equation*}

See how the order of the indices are different?

The entities $b_i$ and $\mathbf{e}_i$ are co-variant .

The entities $a^i$ and $\mathbf{f}^i$ are contra-variant .

One-forms are sometimes referred to as co-vectors , because their coefficients transform in a co-variant way.

The notation then goes:

  • sub-scripts (e.g. basis vectors $\mathbf{e}_i$ for $V$ and coeffs. $b_i$ for $V^*$ ) are co-variant
  • super-scripts (e.g. basis vectors of $\mathbf{f}^i$ for $V^*$ and coeffs $a^i$ for $V$ ) are contra-variant

Very important: "super-script indicies in the denominator" are understood to be lower indices, i.e. co-variant in denominator equals contravariant.

Now, consider this notation for our definition of tangent space and dual space:

If you choose coordinates $(x^1, \dots, x^n)$ on an open set $D \subset \mathbb{R}^n$ containing a point $p$, then you get a basis for the tangent space $T_p D$ at $p$

\begin{equation*}
  \Big\{ \frac{\partial}{\partial x^1} \big|_p, \dots,     \frac{\partial}{\partial x^n} \big|_p \Big\}
\end{equation*}

which have super-script in denominator, indicating a co-variant entity (see note).

Similarily we get a basis for the cotangent space $T_p^* D$ at $p$

\begin{equation*}
\{ dx^1 |_p, \dots dx^n|_p \}
\end{equation*}

which have super-script indices, indicating a contra-variant entity.

Why did we decide the first case is the co-variant (co- and contra- are of course relative)?

Because in differential geometry the co-variant entities transform like the coordinates do, and we choose the coordinates to be our "relative thingy".

Differential forms

Differential forms are an approach to multivariable calculus which is independent of coordinates.

Surfaces

Notation

  • $D \subseteq \mathbb{R}^2$ is the domain in the plane whose Cartesian coordinates will be denoted $(u^1, u^2)$ unless otherwise stated
  • $(x^1, x^2, x^3) \in \mathbb{E}^3$, unless otherwise stated
  • $\mathbf{x}(D)$ denotes the image of the smooth, injective map $\mathbf{x}: D \to \mathbb{E}^3$

Regular surfaces

A local surface in $\mathbb{E}^3$ is smooth, injective map $\mathbf{x}: D \to \mathbb{E}^3$ with a continuous inverse of $\mathbf{x}^{-1}: \mathbf{x}(D) \to D$. Sometimes we denote the image $\mathbf{x}(D)$ by $S$.

The assumptation that $\mathbf{x}$ is injective means that points in the image $\mathbf{x}(D)$ are uniquely labelled by points in $D$.

Given a local surface we define

\begin{equation*}
\mathbf{x}_{u^1} = \frac{\partial x^i}{\partial u^1} \frac{\partial }{\partial x^i}, \quad \mathbf{x}_{u^2} = \frac{\partial x^i}{\partial u^2} \frac{\partial}{\partial x^i}, \quad \mathbf{x}_{u^1}, \mathbf{x}_{u^2} \in T_{\mathbf{x}(p)} \mathbb{E}^3
\end{equation*}

For every point $p \in D$, these are vectors in $T_{\mathbf{x}(p)} \mathbb{E}^3$, which we will identify with $\mathbb{E}^3$ itself. We say that a local surface $\mathbf{x}$ is regular at $p \in D$ if $\mathbf{x}_{u^1}(p)$ and $\mathbf{x}_{u^2}(p)$ are linearly independent. A local surface is regular if it is regular at $p$ for all $p \in D$.

This gives rise to the differential form $d \mathbf{x}: T_p D \to T_{\mathbf{x}(p)} \mathbb{E}^3$:

\begin{equation*}
d \mathbf{x} = \mathbf{x}_{u^1} d u^1 + \mathbf{x}_{u^2} d u^2
\end{equation*}

Here is a quick example of evaluating the differential form induced by the definition of a regular local surface:

\begin{equation*}
\begin{split}
  d \mathbf{x}(\mathbf{v}) &= \Big( \mathbf{x}_{u^1} d u^1 + \mathbf{x}_{u^2} du^2 \Big) \Big( a \frac{\partial}{\partial u^1} + b \frac{\partial}{\partial u^2} \Big) \\
  &= a \mathbf{x}_{u^1} + b \mathbf{x}_{u^2} \\
  &= a \frac{\partial x^i}{\partial u^1} + b \frac{\partial x^i}{\partial u^2}
\end{split}
\end{equation*}

$\Sigma \subset \mathbb{E}^3$ is a regular surface if for each $p \in \Sigma$ there exists a regular local surface $\mathbf{x}: D \to \mathbb{E}^3$ such that $p \in \mathbf{x}(D)$ and $\mathbf{x}(D) = U \cap \Sigma$ for some open set $U \subset \mathbb{R}^3$.

In other words, if for each point $p$ on the surface we can construct a regular local surface, then the entire surface is said to be regular.

A map $\mathbf{x}$ defines a local surface which is part of some surface $\Sigma$, is sometimes called a coordinate chart on $\Sigma$.

Thus, if the surface $\Sigma$ is a regular surface (not just locally regular) we can "define" $\Sigma$ from a set of all these coordinate charts .

At a regular point on a local surface, the plane spanned by $\mathbf{x}_{u^1}(p)$ and $x_{v^1}(p)$ is the tangent plane to the surface at $\mathbf{x}(p)$, which we denote by $T_{\mathbf{x}(p)}S$. At a regular point, the unit normal to the surface is

\begin{equation*}
\mathbf{N}(p) = \frac{\mathbf{x}_{u^1}(p) \times \mathbf{x}_{u^2}(p)}{|\mathbf{x}_{u^1}(p) \times \mathbf{x}_{u^2}(p)|}
\end{equation*}

Clearly, $\mathbf{N}(p)$ is orthogonal to the tangent plane $T_{\mathbf{x}(p)}S$.

Given a local surface the map $\mathbf{N} : D \to \mathbb{E}^3$ is a smooth function whose image lies in a unit sphere $S^2 \subset \mathbb{E}^3$. The map $\mathbf{N}$ is called the local Gauss map.

Standard Surfaces

Let $g: D \to \mathbb{R}$ be a smooth function. The graph of $g$ is the local surface defined by

\begin{equation*}
\mathbf{x}(u^1, u^2) = \begin{pmatrix}u^1 \\ u^2 \\ g(u^1, u^2) \end{pmatrix}
\end{equation*}

An implicitly defined surface $\Sigma$ is zero set of a smooth function $f: \mathbb{E}^3 \to \mathbb{R}$, i.e.

\begin{equation*}
\Sigma = f^{-1}(0)
\end{equation*}

Note that $f$ is a mapping from $\mathbb{E}^3$, and we're saying that the inverse of this function defines a surface, where it's also important to note the smooth requirement, as this implies that $f$ is differentiable.

An implicitly defined surface $\Sigma = f^{-1}(0)$, such that $d f \ne 0$ everywhere on $\Sigma$, is a regular surface.

This is due to the fact that if there is a point $p \in \Sigma$ such that $df|_p = 0$, then that implies that $\mathbf{x}_u$ and $\mathbf{x}_v$ are linearly dependent, hence not a regular surface.

A surface of revolution with profile curve $f(u)$ is a local surface of the form

\begin{equation*}
\mathbf{x}(u, \phi) = \begin{pmatrix}
  f(u) \cos \phi \\
  f(u) \sin \phi \\
  u
\end{pmatrix}
\end{equation*}

A surface of revolution can be constructed by rotation a curve $x_1 = f(x_3)$ around the $x_3$ axis in $\mathbb{R}^3$. It thus has cylindrical symmetry.

A ruled surface is a surface of the form

\begin{equation*}
\mathbf{x}(u, v) = z(u) + v \mathbf{p}(u)
\end{equation*}

Notice that curves of constant $u$ are straight lines in $\mathbb{E}^3$ through $\mathbf{z}(u)$ in the direction $\mathbf{p}(u)$.

Examples of surfaces

Quadratic surfaces

Quadratic surfaces are the graphs of any equation that can be put into the general form:

\begin{equation*}
(ax + by + cz + d)^2 = 0
\end{equation*}

The general equation for a cone

\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0
\end{equation*}

The general equation for a hyperboloid of one sheet

\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1
\end{equation*}

hyperboloid-one-sheet.gif

The general equation for a hyperboloid of two sheets

\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = -1
\end{equation*}

hyperboloid-two-sheet.gif

The general equation for an ellipsoid

\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
\end{equation*}

with $a = b = c = 1$ being a sphere.

General equation for an elliptic paraboloid

\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z}{c}
\end{equation*}

elliptic-paraboloid.gif

General equation for an hyperbolic paraboloid

\begin{equation*}
\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{z}{c}
\end{equation*}

hyperbolic-paraboloid.gif

Fundamental forms

Symmetric tensors

Notation
  • $D \subseteq \mathbb{R}^n$
  • $(x^1, x^2, \dots, x^n) \in D$ are coordinates
Definitions

A (Riemannian) metric on $D$ is a symmetric tensor $g = g_{ij} dx^i dx^j$ which is positive definite at each point; $g(v, v) \ge 0, \quad \forall v \in T_p D$, with equality if and only if $v = 0$.

Equivalently, it is a choice for each $p \in D$ of an inner product on $T_p D$

First fundamental form

Notation
  • $D \subseteq \mathbb{R}^2$ and $(u^1, u^2) \in D$ are our coordinates
Stuff

Consider a regular local surface defined by $\mathbf{x} : D \to \mathbb{E}^3$. The linear map

\begin{equation*}
d \mathbf{x} |_p : T_p D \to T_{\mathbf{x}(p)} S
\end{equation*}

is a bijection.

This bijectivity can be used to give a coordinate free definition of regularity of a local surface.

Given a regular local surface $\mathbf{x}: D \to \mathbb{E}^3$, the first fundamental form is defined by

\begin{equation*}
I = d \mathbf{x} \cdot d \mathbf{x}
\end{equation*}

where we have introduced the notation $d \mathbf{x} \cdot d \mathbf{x} = (dx^1)^2 + (dx^2)^2 + (dx^3)^2$.

The first fundamental form of a local surface $\mathbf{x}: D \to \mathbb{E}^3$ is

\begin{equation*}
I = E \ (du)^2 + 2 F  \ du \  dv +  G \ (dv)^2
\end{equation*}

where $E, F, G$ are functions on $D$ given by

\begin{equation*}
E = \mathbf{x}_u \cdot \mathbf{x}_u, \quad F = \mathbf{x}_u \cdot \mathbf{x}_v, \quad G = \mathbf{x}_v \cdot \mathbf{x}_v
\end{equation*}

The first fundamental form is a metric on $D$.

Problems
  • Prove bijectivity of linear map from regular local surface

    Let $w$ be a vector field on $D$,

    \begin{equation*}
w = a \frac{\partial}{\partial u^1} + b \frac{\partial}{\partial u^2}
\end{equation*}

    Also, we have the $\mathbb{E}^3$ valued 1-form on $D$:

    \begin{equation*}
d \mathbf{x} = \mathbf{x}_{u^1} du^1 + \mathbf{x}_{u^2} du^2
\end{equation*}

    Then,

    \begin{equation*}
d \mathbf{x}(w) = a \mathbf{x}_{u^1} + b \mathbf{x}_{u^2}
\end{equation*}

    Evaluating this at each $p \in D$ it is clear that this is onto $T_{\mathbf{x}(p)} S$. Further, since the surface is $d \mathbf{x}(w) = 0$ implies $w = 0$, so the map is one-to-one.

Second Fundamental form

Notation
  • $D \subseteq \mathbb{R}^2$ and $(u^1, u^2) \in D$ are our coordinates
  • $\mathbf{N}$ is the normal of the surface (if my understanding is correct)
Stuff

Given a local surface $\mathbf{x}: D \to \mathbb{E}^3$, the second fundamental form is defined by

\begin{equation*}
II = - d \mathbf{x} \cdot d \mathbf{N}
\end{equation*}

with the dot product interpreted as usual.

The $\mathbb{E}^3$ valued 1-form $d \mathbf{N}$ is linear map which may have a non-trivial kernel. It is convenient to use the isomorphism $d \mathbf{x}$ to rewrite the map $d \mathbf{N}$ as a symmetric bilinear form.

Since $\mathbf{N}$ is unit normalised it follows that $\mathbf{N} \cdot \mathbf{N}_{u^1} = 0$ (by differentiating $\mathbf{N} \cdot \mathbf{N} = 1$ by $u^1$ and $u^2$ respectively).

Hence, $\mathbf{N}_{u^1}(p)$ and $\mathbf{N}_{u^2}(p)$ must belong to the tangent plane $T_{\mathbf{x}(p)} S$. In other words, $d \mathbf{N}|_p : T_p D \to T_{\mathbf{x}(p)} S$.

The second fundamental form is given by

\begin{equation*}
II = l \ (du)^2 + 2 m \ du \ dv + n \ (dv)^2
\end{equation*}

where $l, m, n$ are continuous functions on $D$ given by

\begin{equation*}
l = - \mathbf{x}_u \cdot \mathbf{N}_u, \quad m = - \mathbf{x}_u \cdot \mathbf{N}_v = - \mathbf{x}_v \cdot \mathbf{N}_u, \quad n = - \mathbf{x}_v \cdot \mathbf{N}_v
\end{equation*}

Which can also be written as

\begin{equation*}
l = \mathbf{x}_{uu} \cdot \mathbf{N}, \quad m = \mathbf{x}_{uv} \cdot \mathbf{N}, \quad n = \mathbf{x}_{vv} \cdot \mathbf{N}
\end{equation*}
Q & A
  • DONE What do we mean by a 1-form having a "non-trivial kernel"?

    In Group-theory we have the following definition of a kernel :

    \begin{equation*}
\ker (\phi) = \{ x \in G: \phi(x) = e_H \}
\end{equation*}

    where $\phi: G \to H$ is a homomorphism.

    When we say the mapping $\phi$ has a non-trivial kernel, we mean that there are more elements in $G$ than just the identity element which is being mapped to the identity-element in $H$, i.e.

    \begin{equation*}
\ker (\phi) = \{ x \in G: \phi(x) = e_H \} \ne \{e_G\}
\end{equation*}

    Hence, in the case of the some 1-form $d \mathbf{N}$, we have mean

    \begin{equation*}
\ker (d \mathbf{N} ) \ne \{ \mathbf{0} \}
\end{equation*}

    i.e. non-trivial kernel refers to the 1-form mapping more than just the zero-vector to the zero-vector in the target vector-space.

  • DONE What do we mean when we write dx from TpD to Tx(p) S?

    What do we mean when we write the following:

    \begin{equation*}
d \mathbf{x} |_p : T_p D \to T_{\mathbf{x}(p)} S
\end{equation*}

    where:

    • $S \subset \mathbb{E}^3$ is some surface in $\mathbb{E}^3$
    • $D \subseteq \mathbb{R}^2$ is the domain of our "coordinates"
    • $\mathbf{x}: D \to \mathbb{E}^3$ is a smooth map
    • Answer

      We're saying that the differential 1-form $d \mathbf{x}$ maps from the vector-fields defined on $D$ at $p$ to the vector-fields defined on the point $\mathbf{x}(p)$ on the surface $S$.

Curvature

Notation

Bilinear algebra

The eigenvalues of $B$ wrt. $A$ are roots of the polynomial

\begin{equation*}
\det (B - \lambda A) = 0
\end{equation*}

where $A, B$ are represented by symmetric $n \times n$ matrices.

If $A$ is positive definite (i.e. $\mathbf{v}^T \mathbf{A} \mathbf{w}$ defines an inner product) there exists a basis $\{e_1, \dots, e_n\}$ of $V$ such that:

  1. $\{e_1, \dots, e_n\}$ is orthonormal wrt. $A$
  2. each $e_k$ is an eigenvector of $B$ wrt. $A$ with a real eigenvalue

Gauss and mean curvatures

$\forall \mathbf{p} \in \Sigma$ have 2 symmetric bilinear forms on $T_{\mathbf{p}} \Sigma$, $I$ and $II$ look for eigenvalues & eigenvectors of $I$ and $II$.

The eigenvalues $k_1, k_2$ of $II$ wrt. $I$ are the principal curvatures of the surface. The corresponding eigenvectors are the principal directions of the surface. Hence the principal curvatures are the roots of the polynomial $\det (II - \lambda I) = 0$.

The principal curvatures may vary with position and so are (smooth) functions on $D$.

The product of the principal curvatures is the Gauss curvature :

\begin{equation*}
K = k_1 k_2
\end{equation*}

Average of the principal curvatures is the Mean curvature :

\begin{equation*}
H = \frac{1}{2} (k_1 + k_2)
\end{equation*}

If $k_1 = k_2$ we have that all directions are principal.

\begin{equation*}
K = \frac{ln - m^2}{EG - F^2}, \qquad H = \frac{l G + n E - 2mF}{2 (EG - F^2)}
\end{equation*}

where all variables are as given by the first and second fundamental forms.

We get the elegant basis independent expressions

\begin{equation*}
K = \frac{\det II}{\det I}, \quad H = \tr \Big( I^{-1} (II) \Big)
\end{equation*}

Thus, the Gauss curvature is positive if and only if $II$ is positive definite.

Meaning of curvature

Notation

Curves on surfaces

The composition

\begin{equation*}
\mathbf{x} \circ c : [a, b] \to \mathbb{E}^3, \quad t \mapsto \mathbf{x}\big(c(t)\big)
\end{equation*}

describes a curve in $\mathbb{E}^3$ lying on the surface.

\begin{equation*}
\mathbf{x}' = \frac{d}{dt} \mathbf{x}\big( c(t) \big)
\end{equation*}

and

\begin{equation*}
\mathbf{N}' = \frac{d}{dt} \mathbf{N} \big( c(t) \big)
\end{equation*}

The arclength of the curve $\mathbf{x}\big( c(t) \big), \quad t \in [a, b]$, lying on the surface is

\begin{equation*}
s = \int_a^b \sqrt{I(c', c')} \ dt
\end{equation*}

For a curve lying on a surface,

\begin{equation*}
\mathbf{N} \cdot \frac{d^2}{dt^2} \mathbf{x} \big( c(t) \big) = II(c', c')
\end{equation*}

where $II$ is the second fundamental form of the surface.

Invariance under Euclidean motions

Let $\mathbf{x} : D \to \mathbb{E}^3$ and $\hat{\mathbf{x}} : D \to \mathbb{E}^3$ be two surfaces related by a Euclidean motion, so

\begin{equation*}
\hat{\mathbf{x}}(u, v) = \mathbf{A} \mathbf{x}(u, v) + \mathbf{a}, \quad \forall u, v \in D
\end{equation*}

where $\mathbf{A}$ is a orthogonal matrix with $\det \mathbf{A} = 1$ and $\mathbf{a} \in \mathbb{E}^3$.

Then,

\begin{equation*}
\hat{I} = I, \quad \hat{II} = II
\end{equation*}

and hence, in particular,

\begin{equation*}
\hat{H} = H, \quad \hat{K} = K
\end{equation*}

The first fundamental form and second fundamental form determine the surface (up to Euclidean motions).

Taylor series

Let $\mathbf{p} \in \mathbf{x}(D)$ be a point on a regular local surface. By Euclidean motion, choose $\mathbf{p}$ to be at the origin, and the unit normal at that point to be along the positive $x_3$ axis so $T_{\mathbf{p}}S$ is the $(x_1, x_2)$ plane.

Near $\mathbf{p}$ we can parametrise the surface as a graph:

\begin{equation*}
\mathbf{x}(u, v) = \begin{bmatrix}
                     u \\ v \\ f(u, v)
                   \end{bmatrix}
\end{equation*}

where at the origin

\begin{equation*}
\begin{split}
  I_{(0, 0)} &= (du)^2 + (dv)^2 \\
  II_{(0, 0)} &= f_{uu}(0,0) (du)^2 + 2 f_{uv}(0, 0) \ du \ dv + f_{vv}(0, 0) (dv)^2
\end{split}
\end{equation*}

Using the above parametrization, and observing that $\mathbf{x}_u(0, 0)$ and $\mathbf{x}_v(0, 0)$ span $T_\mathbf{p} S$, which is the plane orthogonal to $\mathbf{N}(0, 0)$, we see that $f_u(0, 0) = 0$ and $f_v(0, 0) = 0$.

Further, supposing the $x_1, x_2$ axes correspond to the principal directions, then the Taylor series of the surface near the origin is

\begin{equation*}
f(u, v) = \frac{k_1}{2} u^2 + \frac{k_2}{2} v^2 + \text{higher order terms}
\end{equation*}

where $k_1, k_2$ are the principal curvatures at $\mathbf{p}$.

Umbilical points

Let $\mathbf{x} : D \subset \mathbb{R} \to \mathbb{E}^3$ be a regular local surface.

We then say a point $p$ is a umbilical if and only if

\begin{equation*}
II_p = \lambda I_p, \qquad \lambda \ne 0
\end{equation*}

or equivalently,

\begin{equation*}
k_1 = k_2
\end{equation*}

i.e. all directions are pricipal directions.

An umbilical point is part of a sphere.

We can see the "being a part of a sphere" from the fact that a point $p$ on a sphere can be written as

\begin{equation*}
\mathbf{x} - \mathbf{x}_0 = - \frac{1}{\lambda} \mathbf{N}
\end{equation*}

where $\lambda > 0$ corresponds to $\mathbf{N}$ pointing inwards, while $\lambda < 0$ is $\mathbf{N}$ pointing outwards. In this case, we have

\begin{equation*}
d \mathbf{N} = - \lambda d \mathbf{x}
\end{equation*}

hence,

\begin{equation*}
II = \lambda I
\end{equation*}

Conversely, if $II = \lambda I$ then

\begin{equation*}
- d \mathbf{x} \cdot d \mathbf{N} = \lambda \mathbf{x} \cdot \mathbf{x}
\end{equation*}

Which tells us that

\begin{equation*}
N_{u^i} = \lambda \mathbf{x}_{u^i}, \quad \forall i
\end{equation*}

Thus,

\begin{equation*}
\mathbf{N} + \lambda \mathbf{x} = \text{constant} = \lambda \mathbf{x}_0
\end{equation*}

where $\mathbf{x}_0$ is just some constant. Then,

\begin{equation*}
\mathbf{N} = - \lambda ( \mathbf{x} - \mathbf{x}_0 )
\end{equation*}

A regular local surface has $II = 0$ if and only if it is (a piece of) a plane.

The statement that $\mathbf{N} = \text{constant}$ or $\mathbf{N}_{u^i} = 0$ is equivalent of saying that $\mathbf{x}$ is part of a plane, since the tangents of the map $\mathbf{x}$ are perpendicular to the normal.

Every point is umbilical if and only if the surface is a plane or a sphere.

If $II_p = \lambda(p) I_p$ for some smooth function $\lambda : D \to \mathbb{R}$, then

\begin{equation*}
d \mathbf{N} = - \lambda d \mathbf{x} \implies 0 = d (d \mathbf{N}) = d ( - \lambda d \mathbf{x} ) = - d \lambda \wedge d \mathbf{x}
\end{equation*}

(here we have $\lambda$ as a function, thus the exterior derivative of $\lambda$ gives us a 1-form).

And since

\begin{equation*}
0 = d \lambda \wedge d \mathbf{x} = (d\lambda \wedge d u) \mathbf{x}_u + (d \lambda + dv) \mathbf{x}_v
\end{equation*}

and hence by regularity of the surface $d \lambda \wedge d u = d \lambda \wedge d v = 0$. Thus $\lambda$ is a constant function on $D$ which implies $S \subset \text{plane or sphere}$.

This is because we've already stated that if $\lambda = 0$ $S$ is part of a plane (thm:second-fundamental-form-zero-everywhere-on-surface) and if $\lambda \ne 0$ and constant we have $S$ to be part of a sphere (thm:all-points-umbilical-surface-is-sphere-or-plane).

Problems

Lemma 9.1 (in the notes)
\begin{equation*}
\mathbf{x}' = d \mathbf{x} (c') \quad \text{and} \quad \mathbf{N}' = d \mathbf{N}(c')
\end{equation*}
\begin{equation*}
\begin{split}
  \frac{d}{dt} \mathbf{x} \big( c(t) \big) &= \frac{\partial x^i}{\partial t} \frac{\partial}{\partial x^i} \\
  &= \bigg( \frac{\partial x^i}{\partial u} \frac{\partial u}{\partial t} \bigg) \frac{\partial}{\partial x^i} + \bigg( \frac{\partial x^i}{\partial v} \frac{\partial v}{\partial t} \bigg) \frac{\partial}{\partial x^i} \qquad \text{(by chain rule)} \\
  &= \mathbf{x}_u u'(t) + \mathbf{x}_v v'(t) \qquad \qquad \qquad \qquad \qquad \bigg( \frac{du}{dt} = u'(t) \bigg) \\
  &= \mathbf{x}_u u'(t) \bigg( du \frac{\partial}{\partial u} \bigg) + \mathbf{x}_v v'(t) \bigg( dv \frac{\partial}{\partial v} \bigg), \qquad \bigg( dv \frac{\partial}{\partial v} = 1 \bigg) \\
  &= d \mathbf{x} \Bigg[ u'(t) \frac{\partial}{\partial u} + v'(t) \frac{\partial}{\partial v} \Bigg] \\
  &= d \mathbf{x} \big( c'(t) \big)
\end{split}
\end{equation*}
\begin{align*}
  \frac{d}{dt} \mathbf{x} \big( c(t) \big) &= \frac{\partial x^i}{\partial t} \frac{\partial}{\partial x^i} \\
  &= \bigg( \frac{\partial x^i}{\partial u} \frac{\partial u}{\partial t} \bigg) \frac{\partial}{\partial x^i} + \bigg( \frac{\partial x^i}{\partial v} \frac{\partial v}{\partial t} \bigg) \frac{\partial}{\partial x^i}, & \text{(by chain rule)} \\
  &= \mathbf{x}_u u'(t) + \mathbf{x}_v v'(t),  & \bigg( \frac{du}{dt} = u'(t) \bigg) \\
  &= \mathbf{x}_u u'(t) \bigg( du \frac{\partial}{\partial u} \bigg) + \mathbf{x}_v v'(t) \bigg( dv \frac{\partial}{\partial v} \bigg), & \bigg( dv \frac{\partial}{\partial v} = 1 \bigg) \\
  &= d \mathbf{x} \Bigg[ u'(t) \frac{\partial}{\partial u} + v'(t) \frac{\partial}{\partial v} \Bigg] \\
  &= d \mathbf{x} \big( c'(t) \big)
\end{align*}

Moving frames in Euclidean space

Notation

  • $\mathbf{x} : D \to \mathbb{E}^m$ is a smooth map
  • $D \subseteq \mathbb{R}^n$
  • $u = (u^1, \dots, u^n)$ denotes the coordinates on $D$
  • $\mathbf{x}(u) = \mathbf{x}(u^1, \dots, u^n)$
  • moving frame denotes a collection of maps $\mathbf{e}_i : D \to \mathbb{E}^3$ for $i = 1, 2, 3$ such that these $\mathbf{e}_i$ form a oriented orthonormal basis of $\mathbb{E}^3$
  • oriented means that $\mathbf{e}_3(u) = \mathbf{e}_1(u) \times \mathbf{e}_2(u)$
  • $\mathbf{E} = \begin{pmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \end{pmatrix}$, which, because the frame is oriented, we have $\det \mathbf{E} = (\mathbf{e}_1 \times \mathbf{e}_2) \cdot \mathbf{e}_3 = + 1$, i.e. it's a rotation matrix

Stuff

A moving frame for $\mathbb{E}^3$ on $D$ is a collection of maps $\mathbf{e}_i : D \to \mathbb{E}^3$ for $i = 1, 2, 3$ such that for all $u \in D$ the $\mathbf{e}_i(u)$ form an oriented orthonormal basis of $\mathbb{E}^3$.

Oriented means that $\mathbf{e}_3(u) = \mathbf{e}_1(u) \times \mathbf{e}_2(u)$.

This definition uses the notation of orientedness in three dimensions. For general $m$ there is a different definition of a oriented frame.

If $\mathbf{v} : D \to \mathbb{E}^3$, given by

\begin{equation*}
\mathbf{v}(u) = \mathbf{v}(u^1, \dots, u^n) = \begin{pmatrix}
  v^1(u^1, \dots, u^n) \\
  v^2(u^1, \dots, u^n) \\
  v^3(u^1, \dots, u^n)
\end{pmatrix}
\end{equation*}

we write $d \mathbf{v}$ for its entry by entry exterior derivative:

\begin{equation*}
d \mathbf{v} = \begin{pmatrix}
                 dv^1 \\
                 dv^2 \\
                 dv^3
               \end{pmatrix}
\end{equation*}

Thus, $d \mathbf{v}$ takes vector fields in $D$ and spits out vectors in $\mathbb{E}^3$.

Connection forms and the structure equations

Since $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ is an orthonormal basis for $\mathbb{E}^3$, any vector $\mathbf{v}$ can be expanded as $\mathbf{v} = (\mathbf{e}_i \cdot \mathbf{v}) \mathbf{e}_i$ in the moving frame, and the same applies to a vector-valued 1-form, e.g. $d \mathbf{x}$.

Therefore we define 1-forms $\theta_1, \theta_2, \theta_3$ by

\begin{equation*}
\theta_i = \mathbf{e}_i \cdot d \mathbf{x} \ \in \Omega^1(D) \quad \text{so that} \quad d \mathbf{x} = \sum_{i=1}^{3} \theta_i \mathbf{e}_i
\end{equation*}

The 1-forms $w_{ij} = \mathbf{e}_i \cdot d \mathbf{e}_j \in \Omega^1(D)$ are called the connection 1-forms and by definition satisfy

\begin{equation*}
d \mathbf{e}_i = \sum_{j=1}^{3} \mathbf{e}_j w_{ji}
\end{equation*}

Each $w_{ij}$ are in this case a 1-form.

The connection 1-forms $w_{ij}$ are related by the antisymmetry property:

\begin{equation*}
w_{ij} = - w_{ji}
\end{equation*}

for all $i, j$. In particular $w_{ii} = 0$ for all $i$.

We can now write the structure equations for a surface using matrix-notation:

\begin{equation*}
\boldsymbol{\omega} = \begin{pmatrix}
                    0 & \omega_{12} & \omega_{13} \\
                    - \omega_{12} & 0 & \omega_{23} \\
                    - \omega_{13} & -\omega_{23} & 0
                  \end{pmatrix}
\end{equation*}

We can also write

\begin{equation*}
d \mathbf{E} = \mathbf{E} \boldsymbol{\omega}
\end{equation*}

We will also write

\begin{equation*}
d \boldsymbol{\theta} = \begin{pmatrix} d \theta_1 & d \theta_2 & d \theta_3 \end{pmatrix}
\end{equation*}

The first structure equations are

\begin{equation*}
d \boldsymbol{\theta} + \boldsymbol{\omega} \wedge \boldsymbol{\theta} = 0
\end{equation*}

where the wedge product between the vectors are taken as

\begin{equation*}
\Big( \boldsymbol{\omega} \wedge \boldsymbol{\theta} \Big)_i = \sum_{j=1}^{3} \omega_{ij} \wedge \theta_j
\end{equation*}

The second structure equations are

\begin{equation*}
d \boldsymbol{\omega} + \boldsymbol{\omega} \wedge \boldsymbol{\omega} = 0
\end{equation*}

Definition of connection 1-forms and second structure equations only requires the existence of a moving frame and not a map $\mathbf{x}: D \to \mathbb{E}^3$.

The structure equations exist in the more general context of Riemannian geometry, where $\mathbf{R} = d \boldsymbol{\omega} + \omega \wedge \omega$ is the Riemann curvature, which in general is non-vanishing. In our case it's zero because our moving frame is in $\mathbb{E}^m$.

Structure equations for surfaces

Notation

  • $\mathbf{x} : D \to \mathbb{E}^3$
  • $\theta_i = \mathbf{e}_i \cdot d \mathbf{x}$ are 1-forms
  • $\omega_{ij} = \mathbf{e}_i \cdot d \mathbf{e}_j$ are "connection" 1-forms

Adapted frames and the structure equations

A moving frame $(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$ for $\mathbb{E}^3$ on $D$ is said to be adapted to the surface if $\mathbf{e}_3 = \mathbf{N}$.

I.e. it's adapted to the surface if we orient the basis such that $\mathbf{e}_3$ corresponds to the normal of the surface.

The first and second structure equations for a local surface wrt. to an adapted frame, give the structure equations for a surface:

First structure equations:

\begin{equation*}
\begin{split}
  d \theta_1 + \omega_{12} \wedge \theta_2 &= 0 \\
  d \theta_2 - \omega_{12} \wedge \theta_1 &= 0
\end{split}
\end{equation*}

Symmetry equation:

\begin{equation*}
\omega_{13} \wedge \theta_1 + \omega_{23} \wedge \theta_2 = 0
\end{equation*}

Gauss equation:

\begin{equation*}
d \omega_{12} - \omega_{13} \wedge \omega_{23} = 0
\end{equation*}

Codazzi equations:

\begin{equation*}
\begin{split}
  d \omega_{13} + \omega_{12} \wedge \omega_{23} &= 0 \\
  d \omega_{23} - \omega_{12} \wedge \omega_{13} &= 0
\end{split}
\end{equation*}

Notice how $\theta_3$ has just vanished if you compared to in a moving frame, which comes from the fact that in an adapted moving frame we have $\mathbf{e}_3 = \mathbf{N}$.

The Gauss equation above is equivalent to

\begin{equation*}
d \omega_{12} = K \theta_1 \wedge \theta_2
\end{equation*}

This shows that the Gauss curvature can be computed simply from a knowledge of $\theta_1$ and $\theta_2$ without reference to the local description of the surface $\mathbf{x}: D \to \mathbb{E}^3$.

Let $\mathbf{x}: D \to \mathbb{E}^3$ be a local surface with the first fundamental form $I$ and $\theta_1, \theta_2$ be the 1-forms on $D$ such that

\begin{equation*}
I = \theta_1^2 + \theta_2^2
\end{equation*}

Then there exists a unique adapted frame such that $\theta_1 = \mathbf{e}_1 \cdot d \mathbf{x}$ and $\theta_2 = \mathbf{e}_2 \cdot d \mathbf{x}$.

We say a 1-form is degenerate if wrt. any basis , the matrix representing the 1-form has $\det = 0$.

Two local surfaces $\mathbf{x} : D \to \mathbb{E}^3$ and $\tilde{\mathbf{x}} : D \to \mathbb{E}^3$ are isometric if and only if $I = \tilde{I}$.

Isometric surfaces have the same Gauss curvature. More specifically,

If $\mathbf{x}, \tilde{\mathbf{x}} : D \to \mathbb{E}^3$ are two isometric surfaces, then

\begin{equation*}
K = \tilde{K}
\end{equation*}

The Guass curvature is an instrinsic invariant of a surface!

The first fundamental form $I$ of a surface actually then turns out to determine the following properties:

  • distance
  • angles
  • area

Geodesics

Notation

  • $\mathbf{x}: [a, b] \to \mathbb{E}^n$ which defines the map $t \to \mathbf{x}(t)$, and has unit speed $|\mathbf{x}'| = 1$ joining two points $\mathbf{x}(a), \mathbf{x}(b) \in \mathbb{E}^n$

Stuff

Consider a 1-parameter family of nearby curves

\begin{equation*}
\mathbf{x}_{\varepsilon}(t) = \mathbf{x}(t) + \varepsilon \mathbf{y}(t)
\end{equation*}

where $\mathbf{x}'(t) \cdot \mathbf{y}(t) = 0$ and $\mathbf{y}(a) = \mathbf{y}(b) = 0$ so that all curves in the family join $\mathbf{x}(a)$ to $\mathbf{x}(b)$. We refer to $\mathbf{y}$ as a connecting vector.

It's very important that $\mathbf{x}'(t) \cdot \mathbf{y}(t) = 0$, because if $\mathbf{y}$ has a component along $\mathbf{x}'$ we could remove the shared component by reparametrising $\mathbf{x}(t)$.

We say a unit speed curve $\mathbf{x}(t)$ as above has stationary length if the length of the nearby curves $s_\varepsilon = \int_{b}^{a} | \mathbf{x}_\varepsilon'| \dd t$ satisfies

\begin{equation*}
\frac{\dd s_\varepsilon}{\dd \varepsilon} \Big|_{\varepsilon = 0} = 0
\end{equation*}

for all connecting vector $\mathbf{y}(t)$.

A unit speed curve $\mathbf{x}(t)$ in Euclidean space has stationary length if and only if it is the straight line joining the two points.

Let $\mathbf{x}(t)$ be a unit speed curve in Euclidean space. We then have to prove the following:

  1. $\mathbf{x}(t)$ is a straight line, then it has stationary length
  2. $\mathbf{x}(t)$ has stationary length then it's a straight line

Remember, stationary length is equivalent of

\begin{equation*}
\frac{\dd}{\dd \varepsilon}\Big|_{\varepsilon=0} s_\varepsilon = 0, \qquad s_\varepsilon = \int_{b}^{a} | \mathbf{x}_\varepsilon'| \dd t
\end{equation*}

First, suppose that $\mathbf{x}(t)$ is in fact a straight line, then

\begin{equation*}
\begin{split}
  |\mathbf{x}_\varepsilon'|^2 &= \frac{\dd}{\dd t} \Big( \mathbf{x}_\varepsilon' \cdot \mathbf{x}_\varepsilon' \Big) \\
  &= (\mathbf{x}' + \varepsilon \mathbf{y}') \cdot (\mathbf{x}' + \varepsilon \mathbf{y}') \\
  &= | \mathbf{x}' |^2 + 2 \varepsilon \ \mathbf{x}' \cdot \mathbf{y}' + \varepsilon^2 | \mathbf{y}' |^2
\end{split}
\end{equation*}

Now, taking the square root and the derivative wrt. $\varepsilon$ we have

\begin{equation*}
\begin{split}
  \frac{\dd }{\dd \varepsilon} \Big|_{\varepsilon = 0} | \mathbf{x}_\varepsilon' | &= \frac{1}{\sqrt{| \mathbf{x}' |^2 + 2 \varepsilon \ \mathbf{x}' \cdot \mathbf{y}' + \varepsilon^2 | \mathbf{y}' |^2}} \Bigg|_{\varepsilon = 0} \Big( 2 \mathbf{x}' \cdot \mathbf{y}' + 2 \varepsilon | \mathbf{y}' |^2 \Big) \Bigg|_{\varepsilon=0} = \frac{1}{|\mathbf{x}'|^2} (2 \mathbf{x}' \cdot \mathbf{y}')
\end{split}
\end{equation*}

Remembering that $\mathbf{x}(t)$ is a unit-speed curve, i.e. $|\mathbf{x}'| = 1$, thus

\begin{equation*}
\frac{\dd}{\dd \varepsilon}\Big|_{\varepsilon=0} | \mathbf{x}_\varepsilon'| = 2 \mathbf{x}' \cdot \mathbf{y}' 
\end{equation*}

Now, substituting this into the expression for $s_\varepsilon$, and observing that interchanging the integral wrt. $t$ and derivative wrt. $\varepsilon$ is alright to do, we get

\begin{equation*}
\begin{split}
  \frac{\dd}{\dd \varepsilon} \Bigg|_{\varepsilon=0} s_\varepsilon &= \int_{a}^{b} 2 \mathbf{x}' \cdot \mathbf{y}' \ \dd t \\
  &= 2 \Big( \mathbf{x}' \cdot \mathbf{y} \Big|_{a}^b - \int_{a}^{b} \mathbf{x}'' \cdot \mathbf{y} \ \dd t \Big) \\
  &= - 2 \int_{a}^{b} \mathbf{x}'' \cdot \mathbf{y} \ \dd t
\end{split}
\end{equation*}

since $\mathbf{y}(a) = \mathbf{y}(b) = 0$ by definition of connecting vectors. The final integral is zero if and only if $\mathbf{x}''(t) = 0$, which is equivalent of saying that $\mathbf{x}(t)$ is linear in $t$ and thus is a straigt line, concluding the first part of our proof.

Now, for the second part, we suppose that $\mathbf{x}(t)$ has stationary length

We again perform exactly the same computation and end up with the same integral as we got previously (since we did not use any of our assumptions until the very end), i.e.

\begin{equation*}
\frac{\dd}{\dd \varepsilon} \Bigg|_{\varepsilon=0} s_\varepsilon = - 2 \int_{a}^{b} \mathbf{x}'' \cdot \mathbf{y} \ \dd t
\end{equation*}

And since $\mathbf{x}(t)$ is assumed to have stationary length,

\begin{equation*}
\int_{a}^{b} \mathbf{x}'' \cdot \mathbf{y} \ \dd t = 0
\end{equation*}

which is true if and only if $\mathbf{x}'' = 0$, hence by the same argument as above, $\mathbf{x}(t)$ is the straight line between the two points $a$ and $b$.

Notice the "calculus of variations" spirit of the proof! Marvelous, innit?!

Geodesics on surfaces

A unit-speed curve $\mathbf{x} \big( c(t) \big)$ lying in a surface is a geodesic if its acceleration is everywhere normal to the surface, that is,

\begin{equation*}
\frac{\dd^2}{\dd t^2} \mathbf{x} \big( c(t) \big) = A(t) \mathbf{N} \big( c(t) \big)
\end{equation*}

where $\mathbf{N}$ is the unit normal to the surface and $A$ is some function along the curve.

This means that for a geodesic the acceleration in the direction tangent to the surface vanishes thus generalising the concept of a straight line in a plane.

You can see this from looking at the proof of stationary length in Euclidean space being equivalent to the curve being the straight line: in the final integral we have a dot-product between $\mathbf{x}''$ and $\mathbf{y}$,

\begin{equation*}
\int_{a}^{b} \mathbf{x}'' \cdot \mathbf{y} \ \dd t \overset{?}{=} 0
\end{equation*}

But, all $\mathbf{x}_\varepsilon$ defined in the definition of a connecting vector / nearby curves also lies on the surface, hence $\mathbf{x}_\varepsilon$ cannot have a component in the direction perpendicular to surface. Neither can $\mathbf{x}(t)$ since this is also on the surface, which implies $\mathbf{y}(t)$ also cannot have a component normal to the surface. Thus,

\begin{equation*}
\mathbf{N} \cdot \mathbf{y} = 0 \quad \implies \quad \mathbf{x}'' \cdot \mathbf{y} = 0
\end{equation*}

Finally implying

\begin{equation*}
\frac{\dd}{\dd \varepsilon} \Bigg|_{\varepsilon=0} s_\varepsilon = 0 \quad \iff \quad \mathbf{x}(t) \text{ is \textbf{geodesic}}
\end{equation*}

A curve lying in a surface has stationary length (among nearby curves on the surface joining the same endpoints) if and only if it's a geodesic.

A curve $\mathbf{x} \big( c(t) \big)$ lying in a surface is a geodesic if and only if, in an adapted moving frame it obeys the geodesic equations

\begin{equation*}
\begin{split}
  \frac{\dd}{\dd t} \big( \theta_1 (c') \big) + \omega_{12} (c') \theta_2(c') &= 0 \\
  \frac{\dd}{\dd t} \big( \theta_2 (c') \big) - \omega_{12} (c') \theta_1(c') &= 0
\end{split}
\end{equation*}

and the energy equation

\begin{equation*}
\big( \theta_1 (c ') \big)^2 + \big( \theta_2 (c') \big)^2 = 1
\end{equation*}

Given a point $\mathbf{p}$ on a surface and a unit tangent vector $\mathbf{v}$ to the surface at $\mathbf{p}$, there exists a unique geodesic on the surface $t \mapsto \mathbf{x}\big( c(t) \big)$ for $|t| < \varepsilon$ (with $\varepsilon$ sufficiently small), such that $\mathbf{x}(0) = \mathbf{p}$ and $\mathbf{x}'(0) = \mathbf{v}$.

The geodesic equations only depend on the first fundamental form of a surface. Hence they are partof the intrinsic geometry of a surface and isometric surefaces have the same geodesics!

Two-dimensional hyperbolic space is the upper half plane

\begin{equation*}
H = \{ (x, y) \mid y > 0 \} \subset \mathbb{R}^2
\end{equation*}

equipped with the first fundamental form given by

\begin{equation*}
I = \frac{(dx)^2 + (dy)^2}{y^2}
\end{equation*}

Integration over surfaces

Notation

  • $x: D \to \mathbb{R}^n$ defines a local map , where we drop the bold-face notation due to not anymore using the Euclidean structure
  • $x^* \alpha$ denotes the pull-back of $\alpha$ by the map $x$

Integration of 2-forms over surfaces

Let $x: D \to \mathbb{R}^3$ define a local surface

\begin{equation*}
x(u, v) = \Big( x_1(u, v), x_2(u, v), x_3(u, v) \Big)
\end{equation*}

Note we do not write the map defining the surface in bold here, to emphasise we are not going to use the Euclidean structure).

Let

\begin{equation*}
\alpha = \alpha_{12} dx_1 \wedge dx_2 + \alpha_{13} d x_1 \wedge dx_3 + \alpha_{23} dx_2 \wedge dx_3
\end{equation*}

be a 2-form on $\mathbb{R}^3$. We define the pull-back $x^* \alpha$ of $\alpha$ by the map $x$ to be the 2-form on $D$ given by

\begin{equation*}
x^* \alpha = \alpha_{12}(x) dx_1 \wedge dx_2 + \alpha_{13}(x) dx_1 \wedge dx_3 + \alpha_{23}(x) dx_2 \wedge dx_3
\end{equation*}

IMPORTANT: where here $dx_k$ is the exterior derivative of $x_k: D \to \mathbb{R}$, i.e.

\begin{equation*}
dx_k = \frac{\partial x_k}{\partial u} du + \frac{\partial x_k}{\partial v} dv
\end{equation*}

Let $x: D \to \mathbb{R}^3$ be a local surface and let $\alpha$ be a 2-form on $\mathbb{R}^3$. We define the integral of $\alpha$ over the local surface to be

\begin{equation*}
\int_{x(D)} \alpha = \int_{D} x^* \alpha
\end{equation*}

So, we're defining the integral of the 2-form $\alpha$ over the map $x: D \to \mathbb{R}^3$ as the integral over the pull-back of $\alpha$ over the domain $D$.

Why is this useful? It's useful because we can integrate some 2-form in the "target" manifold $\mathbb{R}^3$ over the "input" domain $D$.

Let $\Sigma$ be a k-dimensional oriented closed and bounded submanifold in $\mathbb{R}^n$ with boundary $\partial \Sigma$ given the induced orientation and $\alpha \in \Omega^{k - 1}(\Sigma)$. Then

\begin{equation*}
\int_{\Sigma} d \alpha = \int_{\partial \Sigma} \alpha
\end{equation*}

The Stokes' and divergence of vector calculus are the $k = 2$ and $k = 3$ special cases respectively.

Integration of functions over surfaces

For a local surface, we have

\begin{equation*}
$| \mathbf{x}_u \times \mathbf{x}_v | = \sqrt{\det I}$
\end{equation*}

Hence, we obtain an alternate expression for the area

\begin{equation*}
A = \int_{D} \sqrt{\det I} \ \dd u \ \dd v
\end{equation*}

Thus the are depends only on $I$, hence it's an intrinsic property of the surface.

For a local surface $x: D \to \mathbb{E}^3$ with an adapted frame,

\begin{equation*}
\theta_1 \wedge \theta_2 = | \mathbf{x}_u \times \mathbf{x}_v | du \wedge dv
\end{equation*}

Let $\mathbf{x}: D \to \mathbb{E}^3$ be a local surface and $f: \mathbb{E}^3 \to \mathbb{R}$ be a function.

Then the integral over $f$ over the surface is given by

\begin{equation*}
\int_{\mathbf{x}(D)} f = \int_{D} f \big( \mathbf{x}(u, v) \big) \ \theta_1 \wedge \theta_2
\end{equation*}

In particular,

\begin{equation*}
A = \int_{D} \theta_1 \wedge \theta_2
\end{equation*}

gives the are of the local surface. The 2-form $\theta_1 \wedge \theta_2$ is called the area form.

Definitions

Words

space-curves
curves in $\mathbb{E}^3$
plane curves
curves in $\mathbb{E}^2$
canonically
"independent of the choice"
rigid motion / euclidean motion
motion which does not change the "structure", i.e. translation or rotation

Regular curves

A curve $\mathbf{x}(t)$ is regular if its velocity (or tangent) vector $\mathbf{x}'(t) \ne 0, \forall t \in (a, b)$.

The tangent line to a regular curve $\mathbf{x}(t)$ at $\mathbf{x}(t^0)$ is the line $\{ \mathbf{x}(t^0) + \lambda \mathbf{x}'(t^0) \ | \ \lambda \in \mathbb{R} \}$.

A unit-speed curve $\mathbf{x}(s)$ is biregular if $\kappa(s) \ne 0, \forall s$, where $\kappa$ denotes the curvature.

(Note that a unit-speed curve is necessarily regular.)

The principal normal along a unit-speed biregular curve $\mathbf{x}(s)$ is

\begin{equation*}
\mathbf{N}(s) = \frac{\mathbf{T}'}{| \mathbf{T}'(s) |} = \frac{\mathbf{T}'(s)}{\kappa}
\end{equation*}

The binormal vector field along $\mathbf{x}(s)$ is

\begin{equation*}
\mathbf{B} = \mathbf{T} \times \mathbf{N}
\end{equation*}

The norm of the velocity

\begin{equation*}
v(t) = \sqrt{\mathbf{x}'(t) \cdot \mathbf{x}'(t)}
\end{equation*}

is the speed of th curve at $\mathbf{x}(t)$.

A parametrisation of a regular curve $\mathbf{x}(t)$ s.t. $v(t) = 1, \forall t$ is called a unit-speed parametrisation.

Level set

The level set of a real-valued function $f$ of $n$ variables is the set of the form

\begin{equation*}
L_c(f) = \{(x_1, \dots, x_n) \ | \ f(x_1, \dots, x_n) = c \}
\end{equation*}

Arc-length

The arg-length of a regular curve $\mathbf{x} \ : \ (a, b) \to \mathbb{E}^n$ from $\mathbf{x}(t^0)$ to $\mathbf{x}(t)$ is

\begin{equation*}
s(t) = \int_{t^0}^t v(t) dt
\end{equation*}

For a unit-speed parametrisation we have $s = t - t^0$, hence it is also called an arc-length parametrisation.

As we can see in the notes, there's a theorem which says that for any regular curve, there exists a reparametrisation of which is unit-speed.

Most reparametrisations are difficult to compute, and thus it's mostly used as a theoretical tool.

Example: Helix

The helix in $\mathbb{E}^3$ is defined by

\begin{equation*}
\mathbf{x}(s) = \frac{1}{\sqrt{2}} \begin{pmatrix}
\cos x \\ \sin x \\ s
\end{pmatrix}
\end{equation*}

which is an arc-length parametrisation

Curvature

The unit tangent vector field along a regular curve is $\mathbf{x}(t)$ is

\begin{equation*}
\mathbf{T}(t) = \frac{\mathbf{x}'(t)}{v(t)}
\end{equation*}

Thus, for a unit-speed curve $\mathbf{x}(s)$ it is simply $\mathbf{T}(s) = \mathbf{x}'(s)$.

For a unit-speed curve $\mathbf{x}(s)$ the curvature $\kappa(s)$ is defined by

\begin{equation*}
\kappa(s) = |\mathbf{T}'(s)| = | \mathbf{x}''(s) |
\end{equation*}

Torsion

The torsion of a biregular unit-speed curve $\mathbf{x}(s)$ is defined by

\begin{equation*}
\mathbf{B}' = - \tau \mathbf{N}
\end{equation*}

or equivalently $\tau = - \mathbf{B}' \cdot \mathbf{N}$.

The oscillating plane at a point on a curve is the plane spanned by $\mathbf{T}$ and $\mathbf{N}$. The torsion measure how fast the curve is twisting out of this plane.

Isometry

An isometry of $\mathbb{E}^3$ is a map $\mathbb{E}^3 \to \mathbb{E}^3$ given by

\begin{equation*}
\mathbf{x} \mapsto A \mathbf{x} + \mathbf{b}
\end{equation*}

where $A$ is an orthogonal matrix and $\mathbf{b}$ is a fixed vector.

If $\det A = 1$, so that $A$ is a rotation matrix, then the isometry is said to be Euclidean motion or a rigid motion.

If $\det A = -1$ the isometry is orientation-reversing.

By definition, an isometry preserves the Euclidean distance between two points $|\mathbf{x} - \mathbf{y}|$.

Tangent spaces

$\forall p \in D$ we define the tangent space $T_p D$ to $D$ at $p$ as the set of all derivative operators at $p$, called tangent vectors at $p \in D$

\begin{equation*}
\begin{split}
  T_p D &= \Big\langle \frac{\partial}{\partial x^1} \big|_p, \dots,     \frac{\partial}{\partial x^n} \big|_p \Big\rangle \\
  &= \Big\{ X : C^\infty(D) \to \mathbb{R} \ | \ X(f \circ g) = X(f) g(p) + f(p) X(g) \Big\} \\
  &\subset \{ \text{linear maps s.t. } C^\infty (D) \to \mathbb{R} \}
\end{split}
\end{equation*}

and thus we have

\begin{equation*}
\frac{\partial}{\partial x^i} \Big|_p (f) = \frac{\partial f}{\partial x_i} \Big|_p
\end{equation*}

in the notation we love so much.

Vector fields are directional derivatives.

A vector field is defined by the tangent at each point $p \in D$ for all $p$ in the domain of the vector field.

It's important to remember that these $x^i$ are curves which are parametrised arbitrarily, and thus describe any potential curve not just the $x^i$ you are "used" to seeing.

In words

  • Tangent space of a manifold facilitiates the generalization of vectors from affine spaces to general manifolds

Tangent vector

There are different ways to view a tangent vector:

  • embedded, i.e. with the manifold where we want to define the tangent vector embedded in a surrounding space, so that we can refer to the tangent vector as "sticking out" of the manifold
  • intrinsically, i.e. without having to refer to some surrounding space
Physists view

Basically considers the tangent vector as a directional derivative

A tangent vector to $D$ at $p$ is determined by an n-tuple

\begin{equation*}
(v^1, \dots, v^n) \in \mathbb{R}^n
\end{equation*}

for each choice of coordinates $(x^1, \dots, x^n)$ at $p$, such that, $f(\overset{\sim}{x}^1, \dots, \overset{\sim}{x}^n)$ is the set of coordinates, we have

\begin{equation*}
\overset{\sim}{v}^i = \sum_{j = 1}^n \frac{\partial \overset{\sim}{x}^i}{\partial x^j} v^j
\end{equation*}

In your "normal" vector spaces we're used to thinking about direction and derivatives as two different concepts (which they are) which can exist independently of each other.

Now, in differential geometry, we only consider these concepts together ! That is, the direction is defined by the basis which the tangent vectors ("derivative" operators) defines.

"Geometric" view

This is a more "intuitive" way of looking at tangent vectors, which directly generalises the concept used in Euclidean space.

A (regular) curve in $D$ is a (smooth) map $c : (a, b) \to D$, given by

\begin{equation*}
c(t) = \Big( x^1(t), x^2(t), \dots, x^n(t) \Big)
\end{equation*}

where each $x^i : (a, b) \to \mathbb{R}$ is a smooth function, such that its velocity

\begin{equation*}
c'(t) = \Big( (x^1)'(t), (x^2)'(t), \dots, (x^n)'(t) \Big)
\end{equation*}

is non-vanishing, $(x^i)'(t) \ne 0, \forall t$, (as an element of $\mathbb{R}^n$) for all $t \in (a, b)$. We say that a curve $c$ passes through $p \in D$ if, say $c(0) = p$ (without loss of generality one can always take the parameter value at $p$ to be 0).

$(a, b) \to \mathbb{R}$ means a map from the open range $(a, b)$ to $\mathbb{R}$, NOT a map which "takes two arguments", duh…

Let $c : (a, b) \to D$ be a curve that passes through $p$. There exists a unique $c_p' \in T_p D$ such that for any smooth function $f : D \to \mathbb{R}$

\begin{equation*}
c_p'(f) = \frac{d}{dt}\Big|_{t=0} f(c(t))
\end{equation*}

There is a one-to-one correspondence between velocities of curves that pass through $p \in D$ and tangent vectors in $T_p D$. By (standard) abuse of notation sometimes we denote $c_p'$ by the corresponding velocity $c'(0)$.

Tangent vector of smooth curves

This approach is quite similar to the geometric view of tangent vectors described above, but I prefer this one.

As of right now, you should have a look at the section about Tangent space and manifolds, as I'm not entirely sure whether or not this can be confusing together with the different notation and all. Nonetheless, the other section is more interesting as it's talking about tangent vectors and general manifolds rather than the more "specific" cases we've been looking at above.

Let $\gamma: I \to M$ be a smooth curve and $\gamma(0) = p \in M$ (wlog).

The tangent vector to curve $\gamma$ at $p$ is a linear map

\begin{equation*}
\begin{split}
  X_p: \quad & C^{\infty}(M) \to \mathbb{R} \\
  & f \mapsto \dv{}{t}\bigg|_{t = 0} f \big( \gamma(t) \big)
\end{split}
\end{equation*}

where

\begin{equation*}
\begin{split}
  \dv{}{t}\bigg|_{t = 0} f \big( \gamma(t) \big) &:= \big( f \circ \gamma \big)'(0) \\
  &= \big( f \circ \mathbf{x}^{-1} \circ \mathbf{x} \circ \gamma \big)'(0) \\
  &= \partial_b \big( f \circ \mathbf{x}^{-1} \big) \big( \mathbf{x}(p) \big) \cdot \big( x^b \circ \gamma \big)'(0) \\
  &= \big( x^b \circ \gamma \big)'(0) \cdot  \bigg( \pdv{}{x^b} \bigg|_{p} f \bigg)
\end{split}
\end{equation*}

where $\mathbf{x}: U \to \mathbb{R}^n$ is a chart map.

Often denote $X_p$ by $\dot{\gamma}_p$.

Tangent as the dual-space of the cotangent space

This section introduces the tanget space as the dual of the cotangent space. Furthermore, we construct the cotangent space in quite a "axiomatic" manner: defining the cotangent space as a quotient space of real-valued smooth functions on the manifold $M$. It is almost an exact duplicate of the lecture notes provided by Prof. José Miguel Figueroa-O'Farrill in the course Differentiable Manifolds taught at University of Edinburgh in 2019.

  • Notation
    • Zero-derivative vector subspace

      \begin{equation*}
Z_p := \left\{ f \in C^{\infty}(M): D \big( f \circ \varphi^{-1} \big) \big|_{\varphi(p)} = 0 \text{ for some chart } U \ni p \text{ with chart-map } \varphi \right\}
\end{equation*}
  • Stuff

    The cotangent space at some point $p \in M$ is the quotient vector space

    \begin{equation*}
T_p^*M := C^{\infty}(M) / Z_p
\end{equation*}

    where

    \begin{equation*}
Z_p := \left\{ f \in C^{\infty}(M): D \big( f \circ \varphi^{-1} \big) \big|_{\varphi(p)} = 0 \text{ for some chart } U \ni p \text{ with chart-map } \varphi \right\}
\end{equation*}

    i.e. all those functions which have vanishing derivative when composed with the inverse of some chart map $\varphi$.

    The derivative of $f \in C^{\infty}(M)$ at $p \in M$ is the image of $f$ under the surjective linear map

    \begin{equation*}
\begin{split}
  d: \quad & C^{\infty}(M) \to T_p^* M \\
  & f \mapsto df \big|_{p} = f \mod Z_p
\end{split}
\end{equation*}

    which is simply the canonical map arising from the original space $C^{\infty}(M)$ to the quotient space $C^{\infty}(M) / Z_p$.

    Observe that $f \circ \varphi^{-1}: \mathbb{R}^d \to \mathbb{R}$, and so we can indeed take the derivative.

    $Z_p$ as defined in the definition of the cotangent space forms a vector subspace.

    If $\psi$ is a smooth function in a neighborhood of $p \in M$, we can multiply $\psi$ by a some bump-function $k$ to construct $f \in C^{\infty}(M)$.

    For any choice of bump function $k$, $f$ agrees with $\psi$ in some neighborhood of $p$. Therefore its derivative at $p$ is independent of the bump function chosen. Thus we can define the derivative at $p$ of $C^{\infty}(M)$ functions which are only defined in a neighborhood of $p$, e.g. the coordinate functions!

    Let $M$ be an n-dimensional manifold. Then

    1. $T_p^*M$ is an n-dimensional vector space $\forall p \in M$
    2. If $\big( U, \varphi \big)$ is a coordinate chart around $p$ with local coordinates $\big( x^1, \dots, x^n \big)$ then $\dd{x^1} \big|_{p}, \dots, \dd{x^n} \big|_{p}$ are a basis for $T_p^*M$.
    3. If $f \in C^{\infty}(M)$, then

      \begin{equation*}
\dd{f} \big|_{p} = \partial_i \big( f \circ \varphi^{-1} \big) \big|_{\varphi(p)} \dd{x^i} \big|_{p}
\end{equation*}

    If $f \in C^{\infty}(M)$ then letting $\phi: f \circ \varphi^{-1}$ means that

    \begin{equation*}
f - \partial_i \phi \big|_{\varphi(p)} x^i
\end{equation*}

    is a (locally defined) smooth function whose derivative vanish at $p$. This is seen by considering the composition with $\varphi^{-1}$:

    \begin{equation*}
\begin{split}
  f \circ \varphi^{-1} - \partial_i \phi \big|_{\varphi(p)} x^i \circ \varphi^{-1} &= \phi - \partial_i \phi \big|_{\varphi(p)} \chi^i
\end{split}
\end{equation*}

    where $\chi^i$ denotes the Euclidean coordinates, i.e. $\chi^i := x^i \circ \varphi^{-1}$. This implies that

    \begin{equation*}
\pdv{}{\chi^j} \bigg|_{\varphi(p)} \bigg( \phi - \pdv{\phi}{\chi^i} \bigg|_{\varphi(p)} \chi^i \bigg) = 0
\end{equation*}

    (since the partial derivative wrt. $\chi^i$ "pass through $\phi$). Therefore,

    \begin{equation*}
\dd{f} \big|_{p} = \partial_i \phi \big|_{\varphi(p)} \dd{x^i} \big|_{p}
\end{equation*}

    and hence $\dd{x^1} \big|_{p}, \dots, \dd{x^n} \big|_{p}$ span $T_p^*M$.

    Now we just need to show that $\dd{x^1} \big|_{p}, \dots, \dd{x^n} \big|_{p}$ are also linearly independent. Suppose

    \begin{equation*}
a_i \dd{x^i} \big|_{p} = 0, \quad a_i \in \mathbb{R}
\end{equation*}

    Then the function $a_i x^i$ has vanishing derivative at $p$, and so $a_i \chi^i$ has vanishing derivative at $\varphi(p)$. But $a_i \chi^i$ is a linear function and so the derivative at any point vanish if and only if it is the zero function. Therefore $a_i = 0$ for all $i$, and so $\dd{x^1} \big|_{p}, \dots, \dd{x^n} \big|_{p}$ are also linearly independent, and hence form a basis of $T_p^* M$.

    The tangent space $T_p M$ at $p \in M$ is the dual of the cotangent vector space. $T_p^*M$.

    This is reasonable for finite-dimensional spaces since in these cases $\big( V^* \big)^* = V$ for vector space $V$, i.e. dual of dual is original vector space.

    If $\big( x^1, \dots, x^n \big)$ is the local coordinate at $p \in M$ and $\dd{x^1} \big|_{p}, \dots, \dd{x^n} \big|_{p}$ is a basis of $T_p^* M$, the canonical basis for $T_p M$ is denoted

    \begin{equation*}
\pdv{x^1}{} \bigg|_{p}, \dots, \pdv{}{x^n} \bigg|_{p}
\end{equation*}

    To relate the tangent space to a more intuitive notion, we introduce the directional derivative.

    A directional derivative at $p \in M$ is a linear map

    \begin{equation*}
\begin{split}
  X_p: \quad & C^{\infty}(M) \to \mathbb{R} \\
  & f \mapsto X_p(f)
\end{split}
\end{equation*}

    s.t.

    \begin{equation*}
X_p(fg) = f(p) X_p g + g(p) X_p f
\end{equation*}

    Observe that if $\xi \in T_p M = \big( T_p^* M \big)^*$ it defines a linear map

    \begin{equation*}
\begin{split}
  \xi: \quad & C^{\infty}(M) \to \mathbb{R} \\
  & f \mapsto \xi \big( \dd{f} \big|_{p} \big)
\end{split}
\end{equation*}

    and from the formula for $\dd{f} \big|_{p}$,

    \begin{equation*}
\dd{(fg)} \big|_{p} = f(p) \dd{g} \big|_{p} + g(p) \dd{f} \big|_{p}
\end{equation*}

    Therefore

    \begin{equation*}
X_p f := \xi \big( \dd{f} \big|_{p} \big)
\end{equation*}

    is a directional derivative at $p$.

    All tangent vectors are of this form.

    An example of a directional derivative is if $f: S^2 \to \mathbb{R}$, then for any tangent direction $\mathbf{u}$ to $S^2$ at $p \in S^2$ we can define the derivative of $f$ at $p$ along $\mathbf{u}$ to be the real number

    \begin{equation*}
\mathbf{u} \cdot \boldsymbol{\nabla} f \big|_{p} = D f \big|_{p}(\mathbf{u})
\end{equation*}

    Let $X_p$ be a directional derivative at $p$ and let $f \in Z_p$. Then

    \begin{equation*}
X_p f = 0
\end{equation*}

    Use a coordinate chart near $p$. By the FTC,

    \begin{equation*}
\begin{split}
  f(x) - f(p) &= \int_{0}^{1} \pdv{}{t} f \big( p + t (x - a) \big) \dd{t} \\
  &= \underbrace{\big( x^i - p^i \big)}_{=: h^i(x)} \int_{0}^{1} \underbrace{\pdv{f}{x^i} \bigg|_{a + t(x - a)}}_{=: g_i(x)} \dd{t}
\end{split}
\end{equation*}

    Using a bump function we can extend $g_i$ and $h^i$ from a neighborhood of $p$ to $C^{\infty}(M)$.

    Notice that $h^i(p) = 0$ and if $\dd{f} \big|_{p} = 0$ then $g_i(p) = 0$ as well. Therefore

    \begin{equation*}
f = f(p) + h^i(x) g_i(x) \quad \text{and} \quad h^i(p) = g_i(p) = 0
\end{equation*}

    By the Leibniz rule,

    \begin{equation*}
X_p(1) = X_p(1 \cdot 1) = 2 X_p(1) \implies X_p(1) = 0
\end{equation*}

    and by linearity, $X_p(\text{constant}) = 0$. Therefore

    \begin{equation*}
\begin{split}
  X_p f &= X_p \bigg( \sum_{i}^{} h^i g_i \bigg) \\
  &= \sum_{i}^{} X_p (h^i g_i) \\
  &= \sum_{i}^{} \Big( h^i(p) X_p g_i + g_i(p) X_p h^i \Big) \\ 
  &= 0
\end{split}
\end{equation*}

    Therefore $X_p$ kills $Z_p$ and descends to a linear map $C^{\infty}(M) / Z_p \to \mathbb{R}$, i.e.

    \begin{equation*}
X_p \in T_p M
\end{equation*}

    Therefore, as a result of the above lemma, we get

    \begin{equation*}
T_p M = \left\{ \text{directional derivatives at } p \right\}
\end{equation*}

    we can also write

    \begin{equation*}
\begin{split}
  T_p M &:= \big( T_p^* M \big)^* \\
  &= \left\{ \xi: C^{\infty}(M) \overset{\sim}{\to} \mathbb{R} \mid \xi \text{ annihilates } Z_p \right\} \\
  &= \left\{ \xi: C^{\infty}(M) \overset{\sim}{\to} \mathbb{R} \mid \xi(fg) = f(p) \xi g + g(p) \xi f \right\}
\end{split}
\end{equation*}

    Relative to local coordinates,

    \begin{equation*}
X_p = c^i \pdv{}{x^i} \bigg|_{p}
\end{equation*}

    and

    \begin{equation*}
X_p f = c^i \pdv{f}{x^i} \bigg|_{p}
\end{equation*}

Tangent bundle

The tangent bundle of a differential manifold $M$ is a manifold $TM$, which assembles all the tangent vectors in $M$. As a set it's given by the disjoint union of the tangent spaces of $M$, i.e.

\begin{equation*}
TM = \{(x, y) \ | \ x \in M, y \in T_x M \}
\end{equation*}

Thus, an element in $TM$ can be thought of as a pair $(x, y)$, where $x$ is a point in the manifold $M$ and $y$ is a tangent vector to $M$ at the point $x$.

Let $M$ be a smooth manifold. Then the tangent bundle is the set

\begin{equation*}
\text{TM} := \bigcup_{p \in M} T_p M
\end{equation*}

and further we define the bundle projection:

\begin{equation*}
\begin{split}
  \pi: \quad & \text{TM} \to M \\
  & X \mapsto p
\end{split}
\end{equation*}

where $p$ is the point for which $X \in T_p M$. This gives us a set bundle; now we just have to show that the fibres are indeed isomorphic, and thus we've obtained a fibre bundle.

Idea: construct a smooth atlas on $\text{TM}$ from a given smooth atlas on $M$.

  • Take some chart $(U, x) \in \mathcal{A}_M$
  • Construct

    \begin{equation*}
  \big( \text{preim}_{\pi}(U),  \xi\big)
\end{equation*}

    where we define $\xi$ as

    \begin{equation*}
\begin{split}
  \xi: \quad & \text{preim}_{\pi}(U) \to \xi \Big( \text{preim}_{\pi}(U) \Big) \\
  & \textcolor{green}{X} \mapsto \bigg( x^1 \Big( \underbrace{\pi(\textcolor{green}{X})}_{\in U} \Big), \dots, x^{\dim M} \Big( \underbrace{\pi(\textcolor{green}{X})}_{\in U} \Big), \textcolor{red}{X^1}, \dots, \textcolor{red}{X^{\dim M}} \bigg)
\end{split}
\end{equation*}

    where

    • First $\dim M$ coordinates we observe is projecting the tangent at some point $X_p$ onto the point itself $p$, i.e. $\pi(X_p) = p \in U$ (we don't write $X_p$ in the above because we can do this for any point in the manifold)
    • Second $\dim M$ coordinates account of the direction and magnitude of the tangent $X$, i.e. we choose the coefficients of $X$ in the tangent space at that point!

      \begin{equation*}
  X = \underbrace{\textcolor{red}{X^a}}_{\in \mathbb{R}} \bigg( \frac{\partial}{\partial x^a} \bigg)_{\pi(X)}
\end{equation*}
  • Finally we need to ensure that this map $\xi$ is indeed smooth : We start by considering the total space, which is the space of all sections $\sigma$, i.e.

    \begin{equation*}
  \Gamma(TM) := \{ \sigma: M \to TM \mid \pi \circ \sigma = \text{id}_M  \}
\end{equation*}

    equipped with the two operations:

    \begin{equation*}
\begin{split}
  \oplus: \quad & \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM) \\
  & \big( \sigma, \tau \big) \to \sigma \oplus \tau \\
  \forall p \in M: \quad & \big( \sigma \oplus \tau \big)(p) := \sigma(p) \underbrace{+}_{\text{in } T_p M} \tau(p)
\end{split}
\end{equation*}

    and multiplication:

    \begin{equation*}
  \begin{split}
    \otimes : \quad & C^{\infty}(M) \to \Gamma(TM) \to \Gamma(TM) \\
    & \big( f, \sigma \big) \mapsto f \odot \sigma \\
    \forall p \in M \quad & \big( f \odot \sigma \big) := f(p) \cdot \sigma(p), \quad (\ \cdot \text{ is s-mult. in } T_p M )
  \end{split}
\end{equation*}

Cotangent bundle

Let $M$ be n-dimensional and let

\begin{equation*}
T^* M := \bigsqcup_{a \in U} T_a^* M
\end{equation*}

denote the disjoint union of all the cotangent spaces of $M$.

If $\big( U, \varphi \big)$ is a chart of $M$, then the map

\begin{equation*}
\begin{split}
  \psi: \quad & U \times \mathbb{R}^n \to T^* U \\
  & \big(a, p_1, \dots, p_n\big) \mapsto p_i \dd{x}^i \big|_{a}
\end{split}
\end{equation*}

defines a bijection, and allows us to define

\begin{equation*}
\begin{split}
  \Phi: \quad & T^*U \to \varphi(U) \times \mathbb{R}^n \\
  & p_i \dd{x}^i \big|_{a} \mapsto \big( (\varphi \times \id) \circ \psi^{-1} \big) \big( p_i \dd{x}^i \big|_{a} \big) = \big( x^1(a), \dots, x^n(a), p_1, \dots, p_n \big) \in \mathbb{R}^{2n}
\end{split}
\end{equation*}

It then follows that $\Phi$ is a bjection from $T^* U$ to an open subsets of $\mathbb{R}^{2n}$, hence $\big( T^* U, \Phi \big)$ defines a chart of $T^* M$. In this way we can bring the charts of $M$ up to $T^* M$, thus $T^* M$ is a manifold.

Let $\left\{ U_{\alpha}, \varphi_{\alpha} \right\}_{\alpha \in \mathcal{A}}$ be an atlas of $M$.

Then $\left\{ T^* U_{\alpha}, \Phi_{\alpha} \right\}_{\alpha \in \mathcal{A}}$ is an atlas of $T^* M$, where

\begin{equation*}
\Phi_{\alpha} = (\varphi_{\alpha} \times \id) \circ \psi^{-1}
\end{equation*}

Since $M = \bigcup_{\alpha \in \mathcal{A}} U_{\alpha}$ we have that $T^* M = \bigcup_{\alpha \in \mathcal{A}}^{} T^* U_{\alpha}$. Therefore we only need to check that indeed the transition maps

\begin{equation*}
\big( \Phi_{\alpha} \circ \Phi_{\beta}^{-1} \big): \Phi_{\beta} \big( T^* U_{\alpha} \cap T^* U_{\beta} \big) \to \Phi_{\alpha} \big( T^* U_{\alpha} \cap T^* U_{\beta} \big) \subseteq \mathbb{R}^{2n}
\end{equation*}

First observe that

\begin{equation*}
\Phi_{\beta} \big( T^* U_{\alpha} \cap T^* U_{\beta} \big) = \varphi_{\beta} \big( U_{\alpha} \cap U_{\beta} \big) \times \mathbb{R}^n
\end{equation*}

which is an open subset of $\mathbb{R}^{2n}$. So the transition map is open. To see smoothness, let $\big( x^i \big)$ be local coordinates of $U_{\alpha}$ and $\big( y^i \big)$ be local coordinates of $U_{\beta}$. Then

\begin{equation*}
p_i \dd{y}^i \in \Phi_{\beta}^{-1} \Big( \Phi_{\beta} \big( T^* U_{\alpha} \cap T^* U_{\beta} \big) \Big) = T^* U_{\alpha} \cap T^* U_{\beta}
\end{equation*}

and

\begin{equation*}
\dd{y}^i = \pdv{y^i}{x^j} \dd{x}^j
\end{equation*}

so

\begin{equation*}
p_i \pdv{y^i}{x^j} \dd{x}^j \in T^* U_{\alpha} \cap T^* U_{\beta}
\end{equation*}

Therefore

\begin{equation*}
\begin{split}
  \big( \Phi_{\alpha} \circ \Phi_{\beta}^{-1} \big) \big( \underbrace{y^1, \dots, y^n, p_1, \dots, p_n }_{\in \varphi_{\beta} (U_{\alpha} \cap U_{\beta}) \times \mathbb{R}^n} \big) &= \Phi_{\alpha} \big( p_i \dd{y}^i  \big) \\
  &= \Phi_{\alpha} \bigg( \underbrace{p_i \pdv{y^i}{x^j}}_{=: q_j} \dd{x}^j \bigg) \\
  &= \Phi_{\alpha} \big( q_j \dd{x}^j \big) \\
  &= \big( x^1, \dots, x^n, q_1, \dots, q_n \big) \in \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) \times \mathbb{R}^{n}
\end{split}
\end{equation*}

Since $\big( \varphi_{\alpha} \circ \varphi_{\beta}^{-1} \big) \big|_{U_{\alpha} \cap U_{\beta}}$ is a diffeomorphism $\big( \Phi_{\alpha} \circ \Phi_{\beta} \big) \big|_{T^* U_{\alpha} \cap T^* U_{\beta}}$ is smooth in the first $n$ components. Furthermore, $p_i \mapsto p_i \pdv{y^i}{x^j} =: q_j$ is smooth since the derivative of smooth functions are smooth and $q_j$ depends linearly on $p_i$.

Hence, $\big( \Phi_{\alpha} \circ \Phi_{\beta} \big) \big|_{T^* U_{\alpha} \cap T^* U_{\beta}}$ is smooth for all $\alpha, \beta \in \mathcal{A}$, and so $\left\{ \big( T^* U_{\alpha}, \Phi_{\alpha} \big) \right\}_{\alpha \in \mathcal{A}}$ defines an atlas for $T^* M$.

Let

\begin{equation*}
\begin{split}
  \pi: \quad & T^* M \to M \\
  & \omega \mapsto a \in M, \quad \forall \omega \in T_a^* M
\end{split}
\end{equation*}

i.e. in local coords we have $\pi(\omega_i \dd{x}^i \big|_{a}) = a$.

$\pi$ is smooth.

$\pi$ being "smooth" means that

\begin{equation*}
\big( \varphi \circ \pi \circ \Phi^{-1} \big): \varphi(U) \times \mathbb{R}^{n} \to \varphi(U) 
\end{equation*}

is a smooth map.

Let $\big( x^i \big)$ be local coordinates of $U_{\alpha}$. Then

\begin{equation*}
\begin{split}
  \big( \varphi \circ \pi \circ \Phi^{-1} \big) \big( x^1, \dots, x^n, p_1, \dots, p_n \big) 
  &= \big( \varphi \circ \pi \big) \big( p_i \dd{x}^i \big) \\
  &= (x^1, \dots, x^n)
\end{split}
\end{equation*}

since $\pi(p_i \dd{x}^i) \in U$. More concretely,

\begin{equation*}
\big( \varphi \circ \pi \big) \big( p_i \dd{x}^i \big|_{a} \big) = \varphi (a) = \big( x^1(a), \dots, x^n(a) \big)
\end{equation*}

Hence,

\begin{equation*}
\big( \varphi \circ \pi \circ \Phi^{-1} \big) \big( x^1, \dots, x^n, p_1, \dots, p_n \big) = \big( x^1, \dots, x^n \big)
\end{equation*}

which is clearly smooth since $x^i$ are all smooth maps.

Second-countability follows directly from the fact that $M = \bigcup_{\alpha \in \mathcal{A}}^{} U_{\alpha}$ is second-countable, and so $T^* M = \bigcup_{\alpha \in \mathcal{A}}^{} T^* U_{\alpha}$ is second-countable.

In what follows we are considering $\omega_1, \omega_2 \in T^* M$ as points, i.e. $\omega_i \in T_a^* M = \preim_{\pi}(\left\{ a \right\})$ for some $a \in M$.

Let $\omega_1 \in T_a^* M$ and $\omega_2 \in T_b^* M$, then we have the following two cases:

  1. $a \ne b$: since $M$ is Hausdorff, there exists two sets $U, V \subset M$ such that $\preim_{\pi}(U) \cap \preim_{\pi}(V) = \emptyset$, and so we're good.
  2. $a = b$: we have chart $(T^* U, \Phi)$, and so $\Phi(T^* U)$ is homeomorphic to some subset of $\mathbb{R}^{2n}$. $\mathbb{R}^{2n}$ is Hausdorff, therefore $\exists U_{\mathbb{R}}, V_{\mathbb{R}} \subset \Phi(T^* U)$ such that $\preim_{\Phi}(U_{\mathbb{R}}) \ni \omega_1$ and $\preim_{\Phi}(V_{\mathbb{R}}) \ni \omega_2$.

Hence $T^* M$ is also Hausdorff.

In the above we are talking about open subsets of $T^* M$ but as we have seen before, since chart $U \subseteq M$ induces chart $T^* U \subseteq T^* M$, any set $U^* \in T^* M$ open is equivalent to saying that the intersection $U^* \cap T^* U_{\alpha}$ is open for $\forall \alpha \in \mathcal{A}$. This in turn means that there exist some $U \subseteq M$ such that $U^* = T^* U$, therefore we can equivalently consider this open set $T^* U$.

Dual space

Let $V$ be a vectorspace over $\mathbb{R}$. Then the dual space of $V$ denoted as $V^*$, is given by

\begin{equation*}
V^* = \{ \varphi : V \to \mathbb{R} \text{ linear } \}
\end{equation*}

Properties

Dual Basis

Honestly, "automatically" is a bit weird. What is actually happening as follows:

Suppose that we have a basis in $V$ defined by the set of vectors $\{ \mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n \}$. then we can construct a basis in the dual space $V^*$, called the dual basis. This dual basis is defined by the set $\{ \mathbf{e}^1, \mathbf{e}^2, \dots, \mathbf{e}^n \}$ of linear functions / 1-forms on $V$, defined by the relation

\begin{equation*}
\mathbf{e}^i ( c^1 \mathbf{e}_1 + c^2 \mathbf{e}_2 + \dots + c^n \mathbf{e}_n ) = c^i, \quad i = 1, \dots, n
\end{equation*}

for any choice of coefficients $c^i$ in the field we're working in (which is usally $\mathbb{R}$).

In particular, letting each of these coefficients be equal to 1 and the rest equal zero, we get the following set of equations

\begin{equation*}
\mathbf{e}^i ( \mathbf{e}_j ) = \delta_j^i = \begin{cases}
  0 \quad \text{if } i \ne j \\
  1 \quad \text{if } i = j
\end{cases}
\end{equation*}

which defines a basis.

If $\mathbf{v}_1, \dots, \mathbf{v}_n$ is a basis for $V$, we automatically get a dual basis $\mathbf{w}_1, \dots, \mathbf{w}_n$ for $V^*$, defined by

\begin{equation*}
\mathbf{w}_i \Big( \sum_{j=1} a^j \mathbf{v}_j \Big) = a^i
\end{equation*}

If $\dim V < \infty$ (is finite), then $\dim(V) = \dim(V^*)$

Dual of the dual

If $\dim V < \infty$

\begin{equation*}
\Big( V^* \Big)^* = V
\end{equation*}
Map between duals

If $f : V \to W$ in a linear map between (dual) vector spaces $\implies$ get canonically a dual map :

\begin{equation*}
\begin{split}
f^* : W^* \to V^* \ | \ f^* (\mathbf{w}^*) (\mathbf{v}) = \mathbf{w}^* \Big( f(\mathbf{v}) \Big), \text{ where } \mathbf{w} \in W^*, \ \mathbf{v} \in V  
\end{split}
\end{equation*}

1-forms

Aight, so this is the proper definition of a one-form.

A (differential) one-form is a smooth section $\omega: M \to T^* M$ on the cotangent bundle, i.e. satisfying

\begin{equation*}
\pi \circ \omega = \id_M
\end{equation*}

We denote the space of one-forms as $\Omega^1(M)$.

$\Omega^1(M)$ is a $C^{\infty}(M) \text{-module}$.x

Let

  • $f \in C^{\infty}(M)$
  • $\big( U, \varphi \big)$ be a chart of $M$ with local coordinates $\big( x^i \big)$
  • $\big( T^* U, \Phi \big)$ be a chart of $T^* M$ as defined in def:cotangent-bundle
  • Define

    \begin{equation*}
\begin{split}
  (f \omega): \quad & M \to T^* M \\
  & a \mapsto f(a) \omega(a), \quad \forall a \in M 
\end{split}
\end{equation*}

    To see that $(f \omega) \in \Omega^1(M)$ we need the map

    \begin{equation*}
\Phi \circ (f \omega) \circ \varphi^{-1}: \varphi(U) \to \varphi(U) \times \mathbb{R}^n
\end{equation*}

    to be smooth. Writing the the map out explicitly, we have

    \begin{equation*}
\Big( \Phi \circ (f \omega) \circ \varphi^{-1} \Big) \big( x^1, \dots, x^n \big) = \Phi \big( (f \omega_i) \dd{x}^i \big) = \bigg( x^1, \dots, x^n, (f \omega_1)(x), \dots, (f \omega_n)(x) \bigg)
\end{equation*}

    where $x = \big( x^1, \dots, x^n \big)$ (i.e. it's really just $\varphi$), which is smooth by smoothness of $f, \omega_i \in C^{\infty}(M)$.

  • Define

    \begin{equation*}
\begin{split}
  \omega_1 + \omega_2: \quad & M \to T^* M \\
  & a \mapsto \big( \omega_1 + \omega_2 \big)(a) := \omega_1(a) + \omega_2(a), \quad \forall a \in M
\end{split}
\end{equation*}

    Again we require smoothness of the corresponding composition with the charts:

    \begin{equation*}
\begin{split}
  \Big( \Phi \circ (\omega + \tilde{\omega}) \circ \varphi^{-1} \Big): \quad & \varphi(U) \to \varphi(U) \times \mathbb{R}^n \\
  & (x^1, \dots, x^n) \mapsto \bigg( x^1, \dots, x^n, \omega_1(x) + \tilde{\omega}_1(x), \dots, \omega_n(x) + \tilde{\omega}_n(x) \bigg)
\end{split}
\end{equation*}

    which again is smooth since $\omega, \tilde{\omega}: M \to T^* M$ are smooth sections, i.e. $\omega_i, \tilde{\omega}_j \in C^{\infty}(M)$.

Hence, $\Omega^1(M)$ is closed under (scalar) $C^{\infty} \text{-scalar multiplication}$ and addition, i.e. defines a module.

Let $\omega \in \Omega^1(M)$ (one-form) and $X \in \mathfrak{X}(M)$ (vector field).

We define the

\begin{equation*}
\begin{split}
  \omega(X): \quad & M \to \mathbb{R} \\
  & a \mapsto \omega(X)(a) := \omega(a) X(a), \quad \forall a \in M
\end{split}
\end{equation*}

Then

\begin{equation*}
\begin{split}
  \Omega^1(M) \times \mathfrak{X}(M) & \to C^{\infty}(M) \\
  (\omega, X) & \mapsto \omega(X)
\end{split}
\end{equation*}

defines a $C^{\infty} \text{-bilinear}$ and non-degenerate pairing.

  • $C^{\infty} \text{-bilinear}$: follows directly from the fact that both $\Omega^1(M)$ and $\mathfrak{X}(M)$ are $C^{\infty} \text{-modules}$, and $\omega(X)(a) = \omega(a) X(a)$.
  • Non-degenerate: suppose $\omega \in \Omega^1(M)$ is non-zero and

    \begin{equation*}
\omega(X)(a) = \omega(a) X(a) = 0, \quad \forall X \in \mathfrak{X}(M), \quad \forall a \in M
\end{equation*}

    Then either $\omega = 0$ or $X(a) = 0$ forall $a \in M$. The non-degeneracy for $X \in \mathfrak{X}(M)$ follows by an almost identical argument.

[DEPRECATED] Old definition

A 1-form at $p \in D$ is a linear map $\alpha : T_p D \to \mathbb{R}$. This means, for all $v, w \in T_p D$ and $a \in \mathbb{R}$,

\begin{equation*}
\alpha (v + w) = \alpha(v) + \alpha(w), \quad \alpha(av) = a \alpha(v)
\end{equation*}

1-forms is equivalent to linear functionals

The set of 1-forms at $p \in D$, denoted by $T_p^* D$, is called the dual vector space of $T_p D$

We define 1-forms $dx^j$ at each $p \in D$ by their action on the basis $\{ \frac{\partial}{\partial x^i} \}$:

\begin{equation*}
dx^j \Bigg( \frac{\partial}{\partial x^k} \Bigg) = \begin{cases}
  &1, \quad \text{if } j = k \\
  &0, \quad \text{otherwise}
\end{cases}
\end{equation*}

Or equivalently, $dx^i$ are defined by their action on an arbitrary tangent vector $v = \sum_{i=1}^{n} v^i \frac{\partial}{\partial x^i}$ :

\begin{equation*}
dx^i(v) = v^j
\end{equation*}

Differential 1-form

A differential 1-form on $D$ is a smooth map $\alpha$ which assigns to each $p \in D$ a 1-form in $T_p^* D$ ; it can be written as:

\begin{equation*}
\alpha = \sum_{i=1}^{n} \alpha_i dx^i
\end{equation*}

where $\alpha_i : D \to \mathbb{R}$ are smooth functions.

Line integrals

Let $c : [a, b] \to D$ be a curve (the end points are included to ensure the integrals exist) and $\alpha$ on the 1-form on $D$. The integral of $\alpha$ over the curve $c$ is

\begin{equation*}
\int_c \alpha = \int_a^b \alpha(c'(t)) \ dt
\end{equation*}

where $c'(t)$ is the tangent vector field to the curve.

Working in coordinates, the result of applying the 1-form $\alpha$ on $c'(t)$ gives the expression

\begin{equation*}
\alpha(c') = \sum_{i=1}^{n}(x^i)'(t) \ \alpha_i(c(t))
\end{equation*}

i.e. the derivative of $x^i$ wrt. $t$ times the evaluation of $\alpha_i$ at $c(t)$, where $\alpha_i$ denotes the evaluation of $\alpha$ along $x^i$.

Example
  • Question

    Consider the parametrized graph of a function $f$:

    \begin{equation*}
c(t) = \big( t, f(t) \big), \qquad t \in [a, b]
\end{equation*}

    as a curve in the plane. Show that $\int_c x^2 \ dx^1$ is just the usual integral $\int_{a}^{b} f \big( x^1 \big) \ dx^1$.

  • Answer
    \begin{equation*}
\int_c \alpha = \int_{a}^{b} \alpha \big( c'(t) \big) \ dt
\end{equation*}

    for any 1-form $\alpha$ over the curve $c$. We then simply let

    \begin{equation*}
\alpha = x^2 dx^1
\end{equation*}

    Then,

    \begin{equation*}
\alpha \big( c'(t) \big) = x^2(t) dx^1 \bigg( \frac{\partial}{\partial x^1} + f'(t) \frac{\partial }{\partial x^2} \bigg) = x^2(t)
\end{equation*}

    Finally giving us the integral

    \begin{equation*}
\int_c x^2 \ dx^1 = \int_{a}^{b} x^2(t) \ dt = \int_{a}^{b} f(t) \ dt
\end{equation*}

    Where we can obtain the wanted form by noting that $x^1 = t$.

k-form

A 2-form at $p \in D$ is a map $\alpha : T_p D \times T_p D \to \mathbb{R}$ which is linear in each argument and alternating

\begin{equation*}
\alpha(v, w) = - \alpha(w, v), \forall v, w \in T_p D
\end{equation*}

More generally, a k-form at $p \in D$ is a map of $k$ vectors in $T_p D$ to $\mathbb{R}$ which is multilinear (linear in each argument) and alternating (changes sign under a swap of any two arguments).

And even more general, on the vector space $V$ with $\dim V = d$, a k-form ($0 \le k \le d$) is a $T_k^0V$ tensor that is anti-symmetric, e.g. for a 2-form $\omega \in T_2^0 V$

\begin{equation*}
\omega(u, v) = - \omega(v, u)
\end{equation*}

In the case of a k-form, if $k = d$, where $d = \dim V$, then $\omega, \omega'$ are top forms, both non-vanishing:

\begin{equation*}
\exists c \in K: \omega' = c \omega
\end{equation*}

i.e. any two top-forms are equal up to a constant factor.

Further, the definition of a volume on some d-dimensional vector space, completely depends on your choice of top-form.

Wedge product

The wedge product or exterior product $\alpha \wedge \beta$ of 1-forms $\alpha$ and $\beta$ is a 2-form defined by the following bilinear (linear in both arguments) and alternating map

\begin{equation*}
(\alpha \wedge \beta)(v, w) = \alpha(v) \beta(w) - \alpha(w) \beta(v) = \begin{vmatrix}
\alpha(v) & \alpha(w) \\
\beta(v) & \beta(w) 
\end{vmatrix}
\end{equation*}

More generally, the wedge product of $k$ 1-forms, $\alpha_1, \alpha_2, \dots, \alpha_k$ can be defined as a map acting on $k$ vectors $v_1, v_2, \dots, v_k$

\begin{equation*}
(\alpha^1 \wedge \dots \wedge \alpha^k) (v_1, \dots, v_k) = \begin{vmatrix}
\alpha^1(v_1) & \dots & \alpha^1(v_k) \\
\vdots & \ddots & \vdots \\
\alpha^k(v_1) & \dots & \alpha^k(v_k)
\end{vmatrix}
\end{equation*}

From the properties of the determinant it follows that the resulting map is linear in each vector sperarately an changes sign if any pair of vectors is exchanged (this corresponds to exchanging two columns in the determinant). Hence it defines a k-form.

Wedge product between different forms

We extend $\wedge$ linearly in order to define the wedge product of a $k$ -form $\alpha$ and an $\ell$ -form $\beta$. Explicitly,

\begin{equation*}
(\alpha_I dx^I) \wedge (\beta_J dx^J) = \alpha_I \beta_J dx^I \wedge dx^J
\end{equation*}

Here the sum is happening over all multi-indices $I$ and $J$ with $|I| = k$ and $|J| = \ell$.

Now two things can happen:

  • $I \cap J \ne \emptyset$, in which case $dx^I \wedge dx^J = 0$ since there will be a repeated index
  • $I \cap J = \emptyset$, in which chase $dx^I \wedge dx^J = \pm dx^K$, for some muli-index K of length $|K| = k + \ell$. The sign is due to having to reorder them to be increasing.

Therefore, the wedge product defines a (bilinear) map

\begin{equation*}
\wedge : \Omega^k(D) \times \Omega^\ell(D) \to \Omega^{k + \ell} (D)
\end{equation*}

Multi-index

Useful as more "compact" notation.

By a multi-index $I$ of length $|I| = k$ we shall mean an increasing sequence $I = (i_1, \dots, i_k)$ of integers $1 \le 1 < i_1 < i_2 < \dots < i_k \le n$. We will write

\begin{equation*}
dx^I = dx^{i_1} \wedge \dots \wedge dx^{i_k}
\end{equation*}

The set of k-forms at $p$ is a vector space of dimension ${n \choose k}$ for $0 \le k \le n$ with basis $\{dx^I |_p : |I| = k\}$.

Here $n$ denotes the maximum number of dimensions. So we're just saying that we're taking the wedge-product between some $k$ indicies of the 1-forms we're considering.

Differential k-form

A differential k-form or a differential form of degree k on $D$ is a smooth map $\alpha$ which assigns to each $p \in D$ a k-form at $p$; it can be written as

\begin{equation*}
\alpha = \alpha_I dx^I
\end{equation*}

where $\alpha_I : D \to \mathbb{R}$ are smooth functions, and the sum happens over all multi-indices $I$ with $|I| = k$.

Given two differential k-forms $\alpha, \beta$ and a function $f$ the differential k-forms $\alpha + \beta$ and $f \alpha$ are

\begin{equation*}
\alpha + \beta = (\alpha_I + \beta_I) dx^I, \quad f \alpha = f \alpha_I dx^I
\end{equation*}

The set of k-forms on $D$ is denoted $\Omega^k(D)$.

By convention, a zero-form is a function. If $k > n$ then $\Omega^k(D) = \emptyset$ (for every form has a repeated index).

To make the notation used a bit more apparent, we can expand $\alpha$ for $|I| = 2$ in for a vector-space in $\mathbb{R}^3$, i.e. $\alpha \in \Omega^2(\mathbb{R}^3)$, defined above as follows:

\begin{equation*}
\begin{split}
\alpha &= \alpha_I dx^I \\
  & = \alpha_1 \alpha_2 (dx^1 \wedge dx^2) + \alpha_1 \alpha_3 (dx^1 \wedge dx^3) + \alpha_2 \alpha_1 (dx^2 \wedge dx^1) + \alpha_2 \alpha_3 (dx^2 \wedge dx^1) + \alpha_3 \alpha_1 (dx^3 \wedge dx^1) + \alpha_3 \alpha_2 (dx^3 \wedge dx^2) \\
  &= (\alpha_1 \alpha_2 - \alpha_2 \alpha_1) (dx^1 \wedge dx^2) + (\alpha_1 \alpha_3 - \alpha_3 \alpha_1) (dx^1 \wedge dx^3) + (\alpha_2 \alpha_3 - \alpha_3 \alpha_2) (dx^2 \wedge dx^3)
\end{split}
\end{equation*}

where we've used the fact that $dx^i \wedge dx^j = - (dx^j \wedge dx^i)$. and just combined the "common" wedge-products. It's very important to remember that the $\alpha_i$ here represents a 0-form / smooth function. The actual definition of $\alpha$ is as a sum of all possible $dx^i \wedge dx^j$ but the above definition is just encoding the fact that $dx^i \wedge dx^i = 0$.

A form $\alpha \in \Omega^k(D)$ is said to be closed if $d \alpha = 0$.

A form $\alpha \in \Omega(D)$ is said to be exact if

\begin{equation*}
\alpha = d \beta
\end{equation*}

for some $\beta \in \Omega^{k - 1}(D)$.

If a k-form is closed on $\mathbb{R}^m$, then it is also exact.

Exterior derivative

Let $f \in C^{\infty}(M)$ (i.e. it's a 0-form), and define

\begin{equation*}
\begin{split}
  \dd{f}: \quad & M \to T^* M \\
  & a \mapsto \dd{f} \big|_{a}
\end{split}
\end{equation*}

where $\dd{f}$ denotes the exterior derivative (or the "differential") of $f \in C^{\infty}(M)$.

Then $\dd{f}$ is a one-form, i.e.

\begin{equation*}
\dd{f} \in \Omega^1(M)
\end{equation*}

First recall that for a chart $\big( U, \varphi \big)$ with local coordinates $\big( x^i \big)$,

\begin{equation*}
\dd{f} = \pdv{f}{x^i} \dd{x}^i
\end{equation*}
  • $a \mapsto \dd{f} \big|_{a}$ is smooth: Let $\big( T^*U, \Phi \big)$ be a chart defined as in def:cotangent-bundle, then the map is smooth iff

    \begin{equation*}
\Big( \Phi \circ \big(a \mapsto \dd{f} \big|_{a} \big) \circ \varphi^{-1} \Big): \varphi(U) \to \varphi(U) \times \mathbb{R}^n
\end{equation*}

    is smooth. Writing out this map we simply have

    \begin{equation*}
\big(x^1, \dots, x^n\big) \mapsto \bigg( x^1, \dots, x^n, \pdv{f}{x^1}, \dots, \pdv{f}{x^n} \bigg)
\end{equation*}

    REMINDER: recall what $\pdv{f}{x^i}$ actually means.

    \begin{equation*}
\pdv{f}{x^i} = \partial_i \big( f \circ \varphi^{-1} \big)
\end{equation*}

    where $\partial_i$ denotes the partial derivative wrt. the i-th component of the map.

  • $\pi \circ (a \mapsto \dd{f} \big|_{a}) = \id_M$ is clear

Given a smooth function $f$ on $D$, its exterior derivative (or differential) is the 1-form $df$ defined by

\begin{equation*}
(df)(v) = v(f)
\end{equation*}

for any vector field $v$. Equivalently

\begin{equation*}
df = \sum_{i=1}^{n} \frac{\partial f}{\partial x^i} dx^i
\end{equation*}

Let $f: M \to \mathbb{R}$ be a smooth function, i.e. $f \in \Omega^0(M)$.

As it turns out, in this particular case, the push-forward of $f$, denoted $f_{*p}: T_p M \to T_{f(p)} \mathbb{R}$ is equivalent to the exterior derivative!

If $\alpha = \alpha_I dx^I \in \Omega$, then its exterior derivative $d\alpha \in \Omega^{k + 1}(D)$ is

\begin{equation*}
d \alpha = d \alpha_I \wedge dx^I
\end{equation*}

where $d \alpha_I$ denotes the exterior derivative of the function $\alpha_I$ (which we defined earlier).

More explicitly, take the example of the exterior derivative of a 1-form, i.e. $\alpha \in \Omega^1$:

\begin{equation*}
\begin{split}
  d \alpha &= \sum_{j=1}^{n} d f_j \wedge d x^j \\
  &= \sum_{j=1}^{n} \frac{\partial f_j}{\partial x^i} \big( d x^i \wedge d x^j \big)
\end{split}
\end{equation*}

from the the definition of $df$ where $f$ is a function (0-form), and $\alpha = \sum_{j=1}^{n} f_j \ d x^j$.

Theorems

The exterior derivative $d: \Omega^k(D) \to \Omega^{k + 1}(D)$ is a linear map satisfying the following properites

  1. $d$ obeys the graded derivation property, for any $\alpha \in \Omega^k(D)$
\begin{equation*}
d ( \alpha \wedge \beta) = d \alpha \wedge \beta + (-1)^k \alpha \wedge d\beta
\end{equation*}
  1. $d (d\alpha) = 0$ for any $\alpha \in \Omega^k(D)$, or more compactly, $d^2 = 0$

Example problems

Handin 2

Let $c : [0, 2 \pi] \to \mathbb{R}^3$ be the helix $c(t) = (\cos t, \sin t, t)$ and consider the 1-form on $\mathbb{R}^3$

\begin{equation*}
\alpha = 2 x^1 x^2 \ dx^1 + (x^1)^2 x^2 \ dx^2 + x^3 \ dx^3
\end{equation*}
  1. Find the tangent $c'(t)$ at each point along curve. Hence evaluate the line integral of the 1-form $\alpha$ along the curve $c$.

    \begin{equation*}
c'(\tau) = - \sin \tau \frac{\partial}{\partial x^1} \big|_{t = \tau} + \cos \tau \frac{\partial }{\partial x^2} \big|_{t = \tau} + \frac{\partial}{\partial x^3} \big|_{t = \tau}
\end{equation*}
1

    Hence the integral is

    The tangent plane at some point $p = c(\tau)$ along the curve $c(t)$ for a specified $t = \tau$ is given by

    \begin{equation*}
c'(\tau) = \sum_{j=1}^{3} (x^j)'(\tau) \frac{\partial}{\partial t} \big|_{t = \tau}
\end{equation*}
2

    which in this case is equivalent of

    \begin{equation*}
c'(\tau) = - \sin \tau \frac{\partial}{\partial x^1} \big|_{t = \tau} + \cos \tau \frac{\partial }{\partial x^2} \big|_{t = \tau} + \frac{\partial}{\partial x^3} \big|_{t = \tau}
\end{equation*}
3

    Concluding the first part of the claim.

    For the integral, we know that

    \begin{equation*}
\int_c \alpha = \int_a^b \alpha(c'(t)) \ dt
\end{equation*}
4

    for the boundaries $t = a$ and $t = b$. Computing $\alpha(c'(t))$ we get

    \begin{equation*}
\alpha(c'(t)) = 2 \cos t \sin t \ dx^1 + \cos^2 t \sin t \ dx^2 + t \ dx^3
\end{equation*}
5
    \begin{equation*}
\begin{split}
  \therfore \int_c \alpha &= \int_0^{2 \pi} 2 \cos t \sin t \ dx^1 + \cos^2 t \sin t \ dx^2 + t \ dx^3 \\
  &= \int_0^{2 \pi} 2 \cos t \sin t \ dx^1 + \int_0^{2 \pi} \cos^2 t \sin t \ dx^2 + \int_0^{2 \pi} t \ dt^
\end{split}
\end{equation*}
6
  2. Show that $d \alpha = 0$. Now find a smooth function $f: \mathbb{R}^3 \to \mathbb{R}$ such that $\alpha = df$. Hence evaluate the above line integral without explicit integration.

Integration in Rn

The standard orientation (which we always assume) is defined by

\begin{equation*}
dx^1 \wedge dx^2 \wedge \dots \wedge dx^n
\end{equation*}

Coordinates $(y^1, \dots, y^n)$ (an ordered set) are said to be oriented on $D$ if and only if $dy^1 \wedge \dots \wedge dy^n$ is a positive multiple of $dx^1 \wedge \dots \wedge dx^n$ for all $x \in D \subseteq \mathbb{R}^n$.

Observe that this induces an orientation on $\mathbb{R}^n$, since we simply apply $dx^1 \wedge \dots \wedge dx^n$ to the coordinates $(x^1, \dots, x^n)$, thus returning a $-1$ or $+1$ dependening on whether or not the surface is oriented.

Let $(x^1, \dots, x^n)$ be oriented coordinates for $\mathbb{R}^n$. Let $(y^1, \dots, y^n)$ be smooth functions on $\mathbb{R}^n$. Then

\begin{equation*}
dy^1 \wedge \dots \wedge dy^n = \frac{\partial (y^1, \dots, y^n)}{\partial (x^1, \dots, x^n)} dx^1 \wedge \dots \wedge dx^n
\end{equation*}

where the factor on the RHS is hte Jacobian of the coordinate transformation (i.e. the determinant of the matrix whose $ij$ component is $\frac{\partial y^i}{\partial x^i}$.

Let $(x^1, \dots, x^n)$ be oriented coordinates on $D \subseteq \mathbb{R}^n$ and write

\begin{equation*}
\omega = f(x^1, \dots, x^n) dx^1 \wedge \dots \wedge dx^n \in \Omega^n(D)
\end{equation*}

Then the integral of $\omega$ over $D$ is defined by

\begin{equation*}
\int_D \omega = \int_D f(x^1, \dots, x^n) \ dx^1 \cdots dx^n
\end{equation*}

where the RHS is now the usual multi-integral of several variable caculus (provided it exists).

Topological space

A topological space may be defined as a set of points, along with a set of neighbourhoods for each point, satisfying a set of axioms relating points and neighbourhoods.

Or more rigorously, let $X$ be a set. A topology on $X$ is a collection $\mathcal{T}$ of subsets of $X$, called open subsets, satisfying:

  • $X$ and $\emptyset$ are open
  • The union of any family of open subsets is open
  • The intersection of any finite family of open subsets is open

A topological space is then a pair $(X, \mathcal{T})$ consisting of a set $X$ together with a topology $\mathcal{T}$ on $X$.

The definition of a topological space relies only upon set theory and is the most general notion of mathematical space that allows for the definition of concepts such as:

  • continuity
  • connectedness
  • convergence

A topology is a way of constructing a set of subsets of $X$ such that theese subsets are open and satisfy the properties described above.

Atlases & coordinate charts

A chart for a topological space $M$ (also called a coordinate chart, coordinate patch, coordinate map, or local frame ) is a homeomorphism $\varphi: U \to \mathbb{E}^n$, where $U$ is an open subset of $M$. The chart is traditionally denoted as the ordered pair $(U, \varphi)$.

An atlas for a topological space $M$ is a collection $\{ (U_{\alpha}, \varphi_{\alpha}) \mid \alpha \in A\}$, indexed by the set $A$, of charts on $M$ s.t. $\bigcup_{\alpha  \in A} U_\alpha = M$.

If the codomain of each chart is the n-dimensional Euclidean space, then $M$ is said to be n-dimensional manifold.

Two atlases $\left\{ \big( U_{\alpha}, \varphi_{\alpha} \big) \right\}_{\alpha \in A}$ and $\left\{ \big( V_{\beta}, \psi_{\beta} \big) \right\}_{\beta \in B}$ on $X$ are compatible if their union is also an atlas.

So we need to check the following properties:

  1. The following are open in $\mathbb{R}^n$ for all $\alpha \in A$ and all $\beta \in B$

    \begin{equation*}
\begin{cases}
  \varphi_{\alpha}( U_{\alpha} \cap V_{\beta} ) \\
  \psi_{\beta} (U_{\alpha} \cap V_{\beta})
\end{cases}
\end{equation*}
  2. $\varphi_{\alpha} \circ \psi_{\beta}^{-1}$ and $\psi_{\beta} \circ \varphi_{\alpha}^{-1}$ are $C^{\infty}$ for all $\alpha \in A$ and all $\beta \in B$.

Compatibility of atlases define a equivalence relation of atlases.

A differentiable structure on $X$ is an equivalence class of compatible atlases.

Often one defines differentiable structure with a "maximal atlas" instead of an equivalence class. The "maximal atlas" is obtained by simply taking the union of all atlases in the equivalence class.

A transition map is a composition of one chart with the inverse of another chart, which defines a homeomorphism of an open subset of the $\mathbb{E}^n$ onto another open subset of the $\mathbb{E}^n$..

Suppose we have the following two charts on some manifold $M$

\begin{equation*}
  (U_{\alpha}, \varphi_{\alpha}), \quad (U_{\beta}, \varphi_{\beta})
\end{equation*}

such that

\begin{equation*}
U_{\alpha} \cap U_{\beta} \ne \emptyset
\end{equation*}

The transition map is defined

\begin{equation*}
\begin{split}
  \tau_{\alpha, \beta}: \quad & \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) \to \varphi_{\beta}(U_{\alpha} \cap U_{\beta}) \\
  & \tau_{\alpha, \beta} = \varphi_{\beta} \circ \varphi_{\alpha}^{-1} \big|_{\varphi_{\alpha}(U_{\alpha} \cap U_{\beta})}
\end{split}
\end{equation*}

where we've used the notation

\begin{equation*}
f \big|_{\varphi_{\alpha}(U_{\alpha} \cap U_{\beta})}
\end{equation*}

to denote that the function $f$ is restricted to the domain $\varphi_{\alpha}(U_{\alpha} \cap U_{\beta})$, i.e. the statement is only true on that domain.

A differentiable manifold is a topological manifold equipped with an equivalence class of atlases whose transition map are all differentiable.

More generally, a $C^k$ manifold is a topological manifold for which all the transition maps are all k-times differentiable.

A smooth manifold or $C^\infty$ manifold is a differentiable manifold for which all the transition map are smooth.

To prove $X$ is a smooth manifold if suffices to find one atlas due to the compatibility of atlases being an equivalence relation.

A complex manifold is a topological space modeled on the Euclidean space over the complex field and for which all the transition maps are holomorphic.

When talking about "some-property-manifold", it's important to remember that the "some-property" part is specifying properties of the atlas which we have equipped the manifold with.

$f: M \to \mathbb{R}$ is smooth if for any chart $\big( U, \varphi \big)$, the function

\begin{equation*}
f \circ \varphi^{-1}: \varphi(U) \subset \mathbb{R}^n \to \mathbb{R}
\end{equation*}

is smooth.

Observe that if $f \circ \varphi_{\alpha}^{-1}$ is smooth for a chart $\big( U_{\alpha}, \varphi_{\alpha} \big)$, then we can transition between patches to get a smooth map everywhere.

Let $f: M \to N$ with $\dim M = m$ and $\dim N = n$.

Then $f$ is smooth if for every $a \in M$ and charts $\big( U, \varphi \big)$ with $a \in U$ and $\big( V, \psi \big)$ with $f(a) \in V$, we have

\begin{equation*}
\psi \circ f \circ \varphi^{-1} \in C^{\infty}
\end{equation*}

That is,

\begin{equation*}
f \in C^{\infty}(M; N) \quad \iff \quad \psi \circ f \circ \varphi^{-1} \in C^{\infty}
\end{equation*}

where $C^{\infty}$ denotes smooth functions on $\mathbb{R}^d$.

Examples

Real projective space
\begin{equation*}
\mathbb{R}P^n = \left\{ \text{1d vector subspaces of } \mathbb{R}^{n + 1} \right\}
\end{equation*}

If $0 \ne v \in \mathbb{R}^{n + 1}$ spans a 1d subspace (up to multiplication by real numbers). So, for each $i = 1, \dots, n + 1$, we let

\begin{equation*}
U_i = \left\{ \text{1d subspaces spanned by } v \text{ where } v^i = 1 \right\}
\end{equation*}

Then

\begin{equation*}
\mathbb{R} P^n = \bigcup_{i}^{} U_i
\end{equation*}

and we further let

\begin{equation*}
\begin{split}
  \varphi_i: \quad & U_i \to \mathbb{R}^n \\
  & \ell \mapsto \left\{ v \in \mathbb{R}^{n + 1} \mid v^i = 1 \right\}, \quad \forall \ell \in \mathbb{R}P^n
\end{split}
\end{equation*}
Real with global charts
\begin{equation*}
\begin{split}
  M = \mathbb{R} \quad \text{with} & \quad \big( \mathbb{R}, \varphi \big), \quad \varphi(x) = x \\
  N = \mathbb{R} \quad \text{with} & \quad \big( \mathbb{R}, \psi \big), \quad \psi(x) = x^3
\end{split}
\end{equation*}

then

\begin{equation*}
\big( \varphi \circ \psi^{-1} \big)(x) = \varphi (x^{1 / 3}) = x^{1 / 3}
\end{equation*}

is not diff. at $x = 0$.

$M \ne N$ as a manifold (which can be seen by noticing that the identity map is not $C^{\infty}$).

But $M$ and $N$ are diffeomorphic:

\begin{equation*}
\begin{split}
  f: \quad & N \to M \\
  & x \mapsto x^3
\end{split}
\end{equation*}

Then

\begin{equation*}
\begin{split}
  \varphi \circ f \circ \psi^{-1}: \quad & \mathbb{R} \to \mathbb{R} \\
  & x \mapsto x
\end{split}
\end{equation*}

is $C^{\infty}$ and invertible with $C^{\infty}$ inverse.

Manifolds

A topological space that locally resembles the Euclidean space near each point.

More precisely, each n-dimensional manifold has a neighbourhood that is homomorphic to the Euclidean space of dimension $n$ .

Immersed and embedded submanifolds

An immersed submanifold in a manifold $M$ is a subset $N \subset M$ with a structure of a manifold (not necessarily the one inherited from $M$!) such that the inclusion map $i: N \hookrightarrow{}  M$ is an immersion.

Note that the manifold structure on $N$ s part of the data, thus, in general, it is not unique.

Note that for any point $p \in N$, the tangent space to $N$ is naturally a subspace of the tangent space to $M$, i.e. $T_p N \subset T_p M$.

An embedded submanifold $N \subset M$ is an immersed manifold such that the inclusion map $i: N \hookrightarrow{}  M$ is a homeomorphism, i.e. $i$ is an embedding.

In this case the smooth structure on $N$ is uniquely determined by the smooth structure on $M$.

In words, the inclusion map being a homeomorphism means the manifold topology of $M$ agrees with the subspace topology of $\phi(M) \subseteq N$.

Let $\phi: M \to N$ be smooth. Then $c \in N$ if $\phi_{*_p}$ is surjective on $\preim_{\phi}(\left\{ c \right\}) := \left\{ p \in M \mid F(p) = c \right\}$, i.e.

\begin{equation*}
\phi_{*_p} \text{ is surjective } \forall p \in \preim_{\phi} ( \left\{ c \right\} )
\end{equation*}

or equiv,

\begin{equation*}
\im \big( \phi_{*_p} \big) = T_{\phi(p)} N, \quad \forall p \in \preim_{\phi} ( \left\{ c \right\} )
\end{equation*}

then we say that $c$ is a regular value (for $\phi$).

Let

Then $\preim_{\phi} ( \left\{ c \right\})$ is an embedded submanifold of $M$ of dimension $n$, and

\begin{equation*}
T_p \big( \preim_{\phi} ( \left\{ c \right\} ) \big) = \text{ker}(\phi_{*_p})
\end{equation*}

From Rank-Nullity theorem applied to $\phi_{*_p}$, using the fact that $\dim \big( \im(\phi_{*_p})  \big)= m$ since it's surjective on $\preim_{\phi}(\left\{ c \right\})$.

Let

\begin{equation*}
\begin{split}
  \iota: \quad & \preim_{\phi} (\left\{ c \right\}) \to M \\
  & p \mapsto \iota(p) = p
\end{split}
\end{equation*}

be an embedding. Then $\phi \circ \iota$ is the constant map $\preim_{\phi}(\left\{ c \right\}) \to \left\{ c \right\}$.

\begin{equation*}
p \in \preim_{\phi} (\left\{ c \right\}) \implies 0 = (\phi \circ \iota)_{*_p} = \phi_{*_p} \circ \iota_{*_p} \implies \text{im}(\iota_{*_p}) \subseteq \text{ker}(\phi_{*_p})
\end{equation*}

The rest follows from the Rank-Nullity theorem.

Examples
  • Figure 8 loop in $\mathbb{R}^2$
    • It is immersed via the map

      \begin{equation*}
t \mapsto \big( \cos(\pi / 2 + t), \sin(2t) \big), \quad 0 < t < 2 \pi
\end{equation*}
    • This immersion of $S^1$ in $\mathbb{R}^2$ fails to be an embedding at the crossing point in the middle of the figure 8 (though the map itself is indeed injective)
    • Thus, $S^1$ is not homeomorphic to its image in the subspace / induced topology.

Riemannian manifold

A (smooth) Riemannian manifold or (smooth) Riemannian space $(M, g)$ is a real smooth manifold $M$ equipped with an inner product $g_p$ on the tangent space $T_p M$ at each point $p$ that varies smoothly from point to point in the sense that if $X$ and $Y$ are vector fields on the space $M$, then:

\begin{equation*}
p \mapsto g_p \Big(X(p), Y(p) \Big)
\end{equation*}

is a smooth function .

The family $g_p$ of inner products is called a Riemannian metric (tensor).

The metric "locally looks linear":)

The Riemannian metric (tensor) makes it possible to define various geometric notions on Riemannian manifold, such as:

  • angles
  • lengths of curves
  • areas (or volumes)
  • curvature
  • gradients of functions and divergence of vector fields

Euclidean space is a subset of Riemannian manifold

Resolving some questions
  • Why do we need to map the point $p$ to two vector-spaces $X, Y$ before applying the metric $g_p$ ? Because it's the tangent space $T_p M$ which is equipped with the metric $g_p$, not the manifold $M$ itself, and since vector spaces are defined by a basis in $T_p M$ we need to map $p$ into this space before applying $g_p$.
  • What do we really mean by the map $p \mapsto g_p \Big( X(p), Y(p) \Big)$ being smooth ? This means that this maps varies smoothly wrt. the point $p$.
  • The Riemannian metric $g_p$ is dependent on $p$, why is that? Same reason as the first question: the inner product is equipped on the tangent space, not the manifold itself, and since we have a different tangent space at each point $p$, the inner product itself depends on the point chosen.

Differential manifold

A differentiable manifold is a type of manifold that is locally similar enough to a linear space to allow one to do calculus.

Homeomorphism

A homeomorphism or topological isomorphism is a continuous function between topological spaces that has a continuous inverse function.

Diffeomorphism

A diffeomorphism is an isomorphism of smooth manifolds. It is an invertible function that maps one differentiable manifold to another such that both the function and its inverse are smooth.

That is, a smooth $f: M \to N$ with a smooth inverse $f^{-1}: N \to M$.

It's worth noting that the term diffeomorphism is also sometimes used to only mean once-differentiable rather than smooth!

Isometric

An isometry or isometric map is a distance-preserving transformation between metric-spaces, usually assumed to be bijective.

Algebra

A K-vector space $(V, +, \cdot)$ equipped by a product, i.e. a bilinear map

\begin{equation*}
\bullet: V \times V \overset{\sim}{\to} V
\end{equation*}

is called an algebra $(V, +, \cdot, \bullet)$

Example: Algebra over differentiable functions

On some manifold $M$, we have the vector-space $(\mathcal{C}^\infty(M), +, \cdot)$ which is a $\mathbb{R}$ -vector space, and we define the product $\bullet$ as

\begin{equation*}
\bullet: \mathcal{C}^\infty(M) \times \mathcal{C}^\infty(M) \to \mathcal{C}^\infty(M)
\end{equation*}

by the map, for some $f, g \in \mathcal{C}^\infty(M)$,

\begin{equation*}
(f \bullet g) (p) = f(p) \cdot g(p)
\end{equation*}

where $p \in M$ and the product on the RHS is just the s-multiplication in $\mathbb{R}$.

Equations / Theorems

Frenet-Serret frame

The vector fields $\{\mathbf{T}(s), \mathbf{N}(s), \mathbf{B}(s) \}$ along a biregular curve $\mathbf{x}(s)$ are an orthonormal basis for $\mathbb{E}^3$ for each $s$.

This is called the Frenet-Serret frame of $\mathbf{x}(s)$.

By definition of the unit tangent, $\mathbf{T}(s) \cdot \mathbf{T}(s) = 1$. Differentiate this wrt. $s$ to find $\mathbf{T}(s) \cdot \mathbf{T}'(s) = 0, \forall s$. Thus, the principal normal satisfies $\mathbf{N} \cdot \mathbf{N} = 1$ and $\mathbf{T} \cdot \mathbf{N} = \mathbf{T} \cdot \mathbf{T}' / \kappa = 0$.

By definition of the binormal we also have $\mathbf{B} \cdot \mathbf{B} = 1, \mathbf{B} \cdot \mathbf{T} = 0$ and $\mathbf{B} \cdot \mathbf{N} = 0$. Hence, $\mathbf{T}, \mathbf{N}, \mathbf{B}$ form an orthonormal basis.

Structure equations

Let $\mathbf{x}$ be a unit-speed biregular curve in $\mathbb{E}^3$. The Frenet-Serret frame $\mathbf{T}(s), \mathbf{N}(s), \mathbf{B}(s)$ along $\mathbf{x}(s)$ satisfies:

\begin{equation*}
\begin{align}
  \mathbf{T}' &= \kappa \mathbf{N} \\
  \mathbf{N}' &= - \kappa \mathbf{T} + \tau \mathbf{B} \\
  \mathbf{B}' &= - \tau \mathbf{N}
\end{align}
\end{equation*}

These are called the structure equations for unit-speed space curve, or sometimes the "Frenet-Serret equations".

See p. 9 in the notes for a proof.

For a general parametrisation $\mathbf{x}(t)$ of a biregular space curve the structure equations become

\begin{equation*}
\begin{align}
  \mathbf{T}' &= v \kappa \mathbf{N} \\
  \mathbf{N}' &= - v \kappa \mathbf{T} + v \tau \mathbf{B} \\
  \mathbf{B}' &= - v \tau \mathbf{N}
\end{align}
\end{equation*}

where $v$ is the speed of the curve.

Extras

A biregular curve $\mathbf{x}(t)$ is a plane curve if and only if $\tau = 0$ everywhere.

If $\mathbf{x}(t)$ lies in a place, then $\mathbf{T}$ and $\mathbf{N}$ are tangent to the plane and so $\mathbf{B}$ must be a unit normal to this plane and hence constant.

The structure equations then imply $\tau = 0$.

The curvature and torsion of a biregular space curve in any parametrisation can be computed by

\begin{equation*}
v^3 \kappa = | \mathbf{x}' \times \mathbf{x}'' |, \quad v^6 \kappa^2 \tau = (\mathbf{x}' \times \mathbf{x}'') \cdot \mathbf{x}'''
\end{equation*}

Matrix formulation

The structure equations can also be expressed in matrix form:

\begin{equation*}
\frac{d}{ds} \begin{pmatrix} \mathbf{T} & \mathbf{N} & \mathbf{B} \end{pmatrix} = \begin{pmatrix} \mathbf{T} & \mathbf{N} & \mathbf{B} \end{pmatrix}
\begin{pmatrix}
  0 & -\kappa & 0 \\
  \kappa & 0 & -\tau \\
  0 & \tau & 0
\end{pmatrix}
\end{equation*}

By ODE theory, for given $\kappa(s)$ and $\tau(s)$ and initial conditions

\begin{equation*}
\begin{pmatrix} \mathbf{T}(0) & \mathbf{N}(0) & \mathbf{B}(0) \end{pmatrix}
\end{equation*}

there exists a unique solution

\begin{equation*}
\begin{pmatrix} \mathbf{T}(s) & \mathbf{N}(s) & \mathbf{B}(s) \end{pmatrix}
\end{equation*}

to the ODE system, and hence it must conicide with the Frenet-Serret frame.

There is then a unique curve $\mathbf{x}(s)$ satisfying

\begin{equation*}
\mathbf{x}' = \mathbf{T}, \quad \mathbf{x}(0) = \mathbf{b}
\end{equation*}

Equivalence problem

The equivalence problem is the problem of classifying all curves up to rigid motions.

Uniqueness of biregular curve

Let $\kappa(s)$ and $\tau(s)$ be given, with $\kappa(s)$ everywhere positive. Then there exists a unique unit-speed biregular curve $\mathbf{x}(s)$ with these as curvature and torsion such that $\mathbf{x}(0) = \mathbf{b}$ and $\{\mathbf{T}(0), \mathbf{N}(0), \mathbf{B}(0) \}$ is any fixed oriented orthonormal basis in $\mathbb{E}^3$.

Fundamental Theorem of Curves

If two biregular space curves have the same curvature $\kappa$ and torsion $\tau$ then they differ at most by a Euclidean motion.

Tangent spaces

Orthogonality

\begin{equation*}
\frac{\partial}{\partial x^i} \big|_p (x^j) = \delta_{ij} \implies \frac{\partial}{\partial x^i} \big|_p \text{ are lin. indep.}
\end{equation*}

Change of basis

Suppose we have two different bases for a space:

\begin{equation*}
(x_1, x_2, \dots, x_n) \text{ and } (\overset{\sim}{x}_1, \overset{\sim}{x}_2, \dots, \overset{\sim}{x}_n)
\end{equation*}

we have the following relationship

\begin{equation*}
\frac{\partial}{\partial x^i} \Big|_p = \sum_{j=1}^{n} \frac{\partial \overset{\sim}{x}^j}{\partial x^i} \frac{\partial}{\partial \overset{\sim}{x}^j} \Big|_p
\end{equation*}

and for the dual-space

Implicit Function Theorem

Let $n, k \in \mathbb{Z}^+$, where $n$ is the base and $k$ is the "extra" dimension.

Let $U$ be an open subset of $\mathbb{R}^n \times \mathbb{R}^k$, and let $x^1, \dots, x^n, y^1, \dots, y^k$ denote standard coordinates in $\mathbb{R}^n \times \mathbb{R}^k$.

Suppose $\Phi: U \to \mathbb{R}^k$ is smooth, $(a, b) \in U$ with $a \in \mathbb{R}^n$ and $b \in \mathbb{R}^k$, and

\begin{equation*}
c = \Phi(a, b) \in \mathbb{R}^k
\end{equation*}

If the $k \times k$ matrix

\begin{equation*}
\begin{bmatrix}
  \frac{\partial \Phi^i}{\partial y^j} (a, b)
\end{bmatrix}
\end{equation*}

is invertible, then there exists neighbourhoods $V_0 \subset \mathbb{R}^n$ of $a$ and $W_0 \subset \mathbb{R}^k$ of $b$ and a smooth function

\begin{equation*}
F: V_0 \to W_0
\end{equation*}

such that:

\begin{equation*}
\Phi^{-1}(c) \cap \big( V_0 \times W_0 \big)
\end{equation*}

is the graph of $F$.

Or, equivalently, such that

\begin{equation*}
\Phi(x, y) = c \quad \text{for} \quad (x, y) \in V_0 \times W_0 \quad \iff \quad y = F(x)
\end{equation*}

I like to view it like this:

Suppose we have some $N$ dimensional space, and we split it up into two subspaces of dimension $k, n$ such that

\begin{equation*}
N = k + n
\end{equation*}

Then, using Implicit Function Theorem, we can simply check the invertibility of the Jacobian of $\Phi$, as described in the theorem, to find out if there exists a function $F: V_0 \to W_0$ where $V_0 \subset \mathbb{R}^n$ and $W_0 \subset \mathbb{R}^k$.

Where $\Phi: U \to \mathbb{R}^k$, where $U \subseteq \mathbb{R}^n \times \mathbb{R}^k$, is a map "projecting" some neighbourhood / open set of $\mathbb{R}^n \times \mathbb{R}^k$ to $\mathbb{R}^k$.

Example

Consider $n = 2$ and $k = 1$. Ket

\begin{equation*}
\Phi(x^1, x^2, y) = \big( x^1 \big)^2 + \big( x^2 \big)^2 + y^2
\end{equation*}

Then,

\begin{equation*}
\frac{\partial \Phi}{\partial y} = 2 y
\end{equation*}

Consider $\mathbf{x}_0 = \big( x_0^1, x_0^2 \big) = (0, 0)$ and $y_0 = 1$, and $c = 1 = \Phi(x_0^1, x_0^2, y_0)$.

Thus,

\begin{equation*}
\frac{\partial \Phi}{\partial y} = 2 \cdot 1 = 2 \ne 0 \quad \implies \text{invertible}
\end{equation*}

Thus, in the neighbourhood of $(0, 0)$ in $\mathbb{R}^2$ we can consider the level set $\Phi(x^1, x^2, y) = 1$ and locally solve $y$ as a function of $(x^1, x^2)$, i.e. there exists a function $F$ such that $y = F(x^1, x^2)$.

Inverse Function Theorem

Let $\phi: M \to N$ be a differentiable map, i.e. $F \in C^1(M, N)$. If

\begin{equation*}
\dd{\phi}\big|_p: T_p M \to T_{\phi(p)} N
\end{equation*}

is a linear isomorphism at a point $p$ in $M$, then there exists an open neighborhood $U \ni p$ such that

\begin{equation*}
\phi \big|_{U} : U \to \phi(U)
\end{equation*}

is a diffeomorphism.

Note that this implies that $M$ and $N$ must have the same dimension at $p$.

If the above holds $\forall p \in M$, then $\phi$ is a local diffeomorphism.

Gauss-Bonnet

Let $\Sigma$ be an oriented closed and bounded surface with no boundary. Then

\begin{equation*}
\chi(\Sigma) = \frac{1}{2 \pi} \int_{\Sigma} K
\end{equation*}

where $\chi(\Sigma)$ is the Euler characteristic of the surface $\Sigma$, defined as

\begin{equation*}
\chi(\Sigma) = V - E + F
\end{equation*}

where $V$ denotes the vertices, $E$ the edges, and $F$ the faces obtained by dissecting the surface $\Sigma$ into polygons (this turns out to be independent of choice of dissection).

Change of basis

Notation

  • Einstein summation notation
  • $V$ is a K-vector space, i.e. vector space over some field $K$
  • $e_b$ denotes the b-th basis-vector of some basis in $V$
  • $\cong$ means isomorphic to
  • $T_q^p$ denotes a tensor of order $(p, q)$

Stuff

Suppose we have two different bases in some K-vector space $V$:

\begin{equation*}
\tilde{e}_a = A_a^b e_b, \quad e_a = B_a^m \tilde{e}_m
\end{equation*}

Now, how does the this affect the 1-forms / covectors (contra-variant) change?

k-forms / covectors

Let $\omega$ be a covector:

\begin{equation*}
\omega_a := \omega(e_a) = \omega (B_a^m \tilde{e}_m) = B_a^m \omega(\tilde{e}_m) = \tilde{\omega}_m
\end{equation*}

where $\tilde{\omega}_m$ denotes the m-th new basis!

\begin{equation*}
\omega (B_a^m \tilde{e}_m) = B_a^m \omega(\tilde{e}_m)
\end{equation*}

is true since $\omega$ is a linear map by definition.

vectors

\begin{equation*}
v^a := v(\epsilon^a) = \epsilon^a(v) = \epsilon^a( \tilde{v}^b \tilde{e}_b) = \epsilon^a (\tilde{v}^b A_b^m e_m ) = \tilde{v}^b A_m^b \epsilon^a ( e_m ) = A_b^a \tilde{v}^b
\end{equation*}

where the only thing which might seem a bit weird is the

\begin{equation*}
v(\epsilon^a) = \epsilon^a(v)
\end{equation*}

which relies on

\begin{equation*}
T_0^1 V \cong V
\end{equation*}

where

\begin{equation*}
T_0^1 V = \{ \varphi: V^* \to K \mid \varphi \text{ is linear} \} = \big( V^* \big)^*
\end{equation*}

i.e. $V$ is isomorphic to the the dual of the dual space, which is only true for a finite basis!

Apparently this can be shown using a "constructive proof", i.e. you build up the notation mentioned above and then show that it does indeed define an isomorphism between the vector-space $V$ and the dual of the dual space.

Determinants

Notation

Stuff

Problem with matrix-representation

A matrix is a $(1, 1)$ tensor, and we can thus write it as

\begin{equation*}
\phi \in T_1^1 V  : \quad \phi_b^a \sim \begin{pmatrix}
                                  \phi_1^1 & \dots & \phi_d^1 \\
                                  \vdots & \ddots & \vdots \\
                                  \phi_1^d & \dots & \phi_d^d
                                \end{pmatrix}
\end{equation*}

where $\phi_b^a \sim \dots$ means that we can write out an exhaustive representation of all the $a$ and $b$ entries in such a way (in this case a matrix).

Also, it turns out that we can write a bilinear map as

\begin{equation*}
g \in T_2^0 V : \quad g_{ab} \sim \begin{pmatrix}
                                    g_{11} & \dots & g_{1d} \\
                                    \vdots & \ddots & \vdots \\
                                    g_{d1} & \dots & g_{dd}
                                  \end{pmatrix}
\end{equation*}

See?! We can represent both as a matrix, but the way the change with the basis are completely different!

The usual matrix representation that we're used to (the one with the normal matrix-multiplication, etc.) is the $T_1^1 V$ tensor, and it's an endomorphism on $V$, i.e. homomorphism which takes a vector to a vector.

Definition

Let $\phi \in \text{End}(V) \cong T_1^1 V \cong \{ \varphi: V \to V \mid \varphi \text{ is linear} \}$. Then

\begin{equation*}
\det \phi := \frac{\omega \big( \phi(e_1), \dots, \phi(e_d) \big)}{\omega(e_1, \dots, e_d)}
\end{equation*}

for some volume form / top-form $\omega$, and for some basis $e_1, \dots, e_d$ of $V$, i.e. it's completely independent from the choice of basis and top-form.

Due to the top-forms being equal up to some constant, we see that any constant in the above expression would cancel.

Tangent space and manifolds

Notation

  • $\mathbf{x} : M \overset{\sim}{\to} \mathbb{R}^n$, that is a linear map from the manifold $M$ and $\mathbb{R}^n$, often called a chart map (as it is related to a chart $(U, \mathbf{x})$)
  • $f: M \to \mathbb{R}$
  • $\mathcal{C}^{\infty}(\mathbb{R}^n, \mathbb{R}) = \{ g: \mathbb{R}^n \to \mathbb{R} \mid g \ \text{is smooth} \}$
  • $\gamma: \mathbb{R} \to M$ is a smooth curve, i.e. a smooth mapping taking in a single parameter and mapping it to a point on the manifold $M$
  • $\partial_a \Big|_{\mathbf{x}(p)} : \mathcal{C}^{\infty}(\mathbb{R}^n, \mathbb{R}) \to \mathbb{R}$ is the partial derivative, which at each point $\mathbf{x}(p) \in \mathbb{R}^n$ (since $p \in M$) for some function $g \in \mathcal{C}^{\infty}(\mathbb{R}^n, \mathbb{R})$ we take the partial derivative of $g$ wrt. a-th entry of the Cartesian product $\mathbb{R} \times \dots \times \mathbb{R}$ (n times)

Stuff

We define a new symbol

\begin{equation*}
\pdv{x^a}\Big|_p f := \partial_a \Big( f \circ \mathbf{x}^{-1} \Big) \big( \mathbf{x}(p) \big), \quad \text{where } f: M \to \mathbb{R}
\end{equation*}

that is

\begin{equation*}
\pdv{x^a} \Big|_p : \mathcal{C}^{\infty}(M) \to \mathbb{R}
\end{equation*}

Why is this all necessary? It's pretty neat because we're first using the chart-map $\mathbf{x}$ to map the point $p \in M$ to Euclidean space.

Then, the composite function $f \circ \mathbf{x}^{-1}: \mathbb{R}^n \to \mathbb{R}$

The tangent space $T_p M$ is an n-dimensional (real) vector space.

Addition structure: Consider two curves $\gamma, \mu$ in $M$ s.t.

\begin{equation*}
\gamma(0) = \mu(0) = p
\end{equation*}

with tangent vectors $X_p$ and $Y_p$. We let

\begin{equation*}
\big( \alpha X_p + \beta Y_p \big)(f) := \alpha X_p(f) + \beta Y_p(f), \quad \alpha, \beta \in \mathbb{R}
\end{equation*}

Need to show that $\exists$ curve $\nu$, s.t.

\begin{equation*}
\nu(0) = p, \quad \dot{\nu}_p = \alpha X_p + \beta Y_p
\end{equation*}

Let $\big( U, \phi \big)$ be a chart, $p \in U$. Then we define $\nu$ by

\begin{equation*}
\phi \big( \nu(t) \big) = \alpha \big[ \phi \big( \gamma(t) \big) - \phi(p) \big] + \beta \big[ \phi \big( \mu(t) \big) - \phi(p) \big] + \phi(p)
\end{equation*}

so $\nu(0) = p$ and

\begin{equation*}
x^i \big( \nu(t) \big) = \alpha \Big( x^i \big( \gamma(t) \big) - x^i(p) \Big) + \dots
\end{equation*}

where $x^i$ are the components of the chart. Tangent vector $Z_p$ to $\nu$ at $p$:

\begin{equation*}
\begin{split}
  Z_p(f) &= \dv{}{t} \bigg|_{t = 0} f \big( \nu(t) \big) \\
  &= \partial_i \big( f \circ \phi^{-1} \big) \big( \phi(p) \big) \dv{}{t} \bigg|_{t = 0} x^i \big( \nu(t) \big) \\
  &= \alpha \dv{}{t} \bigg|_{t = 0} f \big( \gamma(t) \big) + \beta \dv{}{t} \bigg|_{t = 0} f \big( \mu(t) \big) \\
  &= \alpha X_p(f) + \beta Y_p(f)
\end{split}
\end{equation*}

where we have used the fact that

\begin{equation*}
\dv{}{t} \bigg|_{t = 0} x^i \big( \nu(t) \big) = \alpha \dv{}{t} \bigg|_{t = 0} x^i \big( \gamma(t) \big) + \beta \dv{}{t} \bigg|_{t = 0} x^i \big( \mu(t) \big)
\end{equation*}

N-dimensional follows from Theorem thm:tangent-vectors-form-a-basis: They form the basis

\begin{equation*}
\left\{ \pdv{}{x^i}\bigg|_{p}: i = 1, \dots, n \right\}
\end{equation*}

thus $\dim T_p M = n$.

Construting a basis

From above, we can construct vectors

\begin{equation*}
e_a := \pdv{x^a} \Big|_p
\end{equation*}

which are the tangent vectors to the chart-induced curves $\gamma_{(a)}$.

Any $X \in T_p M$ ca be written as

\begin{equation*}
X = X^a \cdot \pdv{x^a} \Big|_p, \quad X^a \in \mathbb{R}
\end{equation*}

where we're using Einstein summation and $\cdot$ refers to the s-multiplication of a in the vector space.

Further,

\begin{equation*}
\Big\{ \pdv{x^1} \Big|_p, \dots, \pdv{x^n} \Big|_p \Big\}
\end{equation*}

form a basis for $T_pM$.

$\exists \mu : \mathbb{R} \to M$, $\mu$ is a smooth curve through $p$

\begin{equation*}
p = \mu(0)
\end{equation*}

Then the map $X_{\mu, p}: \mathcal{C}(M) \to \mathbb{R}$, which is the tangent vector of the curve $\mu$ at point $p$ given by

\begin{equation*}
\begin{split}
  X_{\mu, p} f &= \big( f \circ \mu \big)' (0) \\
  &= \big( f \circ \mathbf{x}^{-1} \circ \mathbf{x} \circ \mu \big)' (0)
\end{split}
\end{equation*}

Then by the chain rule, we have:

\begin{equation*}
\begin{split}
  &= \partial_b \big( f \circ \mathbf{x}^{-1} \big) \big( \mathbf{x}(p) \big) \cdot \big( x^b \circ \mu \big)' (0) \\
  &= \big( x^b \circ \mu \big)'(0) \Big( \pdv{x^b} \Big|_p f \Big) \qquad \Bigg( \text{by def. of } \pdv{x^b} \Big|_p f \Bigg)
\end{split}
\end{equation*}

where we've used the fact that $\big( x^b \circ \mu \big)'$ is just a real number, allowing us to move it to the front.

Hence, for any smooth curve $\mu$ in $M$ at some point $p$ we can "generate" the tangent of this curve $\mu$ at point $p$ from the set

\begin{equation*}
\left\{ \pdv{x^1} \Big|_p, \dots, \pdv{x^n} \Big|_p \right\}
\end{equation*}

which we say to be a generating system of $T_p M$.

Now, all we need to prove is that they are also linearly independent, that is

\begin{equation*}
\lambda^a \pdv{x^a} \Big|_p x^b = 0 \quad \implies \quad \lambda^b = 0 \quad \text{or } x^b = 0
\end{equation*}

This is just the definition of basis, that if a vector is zero in this basis, then either all the coefficients are zero or the vector itself is the zero-vector.

One really, really important thing to notice in this proof is the usage of

\begin{equation*}
\begin{split}
  X_{\mu, p} f &= \big( f \circ \mu \big)' (0) \\
  &= \big( f \circ \mathbf{x}^{-1} \circ \mathbf{x} \circ \mu \big)' (0)
\end{split}
\end{equation*}

where we just "insert" the $\mathbf{x}^{-1} \circ \mathbf{x}$ since itself is just an identity operation, but which allows us to "work" in Euclidean space by mapping the point in the manifold to the Euclidean space, and then mapping it back to the manifold for $f$ to finally act on it!

Push-forward and pullback

Let $\phi: M \to N$ be a smooth map between smooth manifolds $M$ and $N$.

Then the push-forward $\phi_*$ at the point $p$ is the linear map

\begin{equation*}
\phi_{*_p}: T_p M \overset{\sim}{\to} T_{\phi(p)} N
\end{equation*}

which defines the map

\begin{equation*}
X \mapsto \phi_{*_p}(X)
\end{equation*}

as

\begin{equation*}
\phi_{*_p}(X) f := X ( f \circ \phi )
\end{equation*}

where:

  • $f: N \to \mathbb{R}$ is a smooth function
  • $X \in T_p M$
  • $\phi_{*_p}(X) \in T_{\phi(p)} N$
  • Elements in $T_p M$ and $T_{\phi(p)} N$ define maps of functions, hence we need to apply it to some function to define it's operation

A couple of remarks:

  • $\phi_{*_p}$ defined as above, is the only linear map from $T_p M$ to $T_{\phi(x)} N$ one can actually define!
  • $\phi_{*_p}$ is often referred to as the derivative of $f$ at the point $p$
  • The tangent vector $X_{\gamma, p}$ of the curve $\gamma$ at the point $p$ in the manifold $M$ is pushed forward to the tangent vector of the curve $\phi \circ \gamma$ at the point $\phi(p)$ in the manifold $N$; i.e. for a curve $\gamma$ we map the tangent vector at some point $p$ to the tangent vector of the "new" curve $\phi \circ \gamma$ in the target manifold at the resulting point $\phi(p)$
    • Or, pull-back (on functions) the curve by composition $\phi^* f = f \circ \phi$, and then let the tangent act on it

Let $\phi: M \to N$ be a smooth map between smooth manifolds $M$ and $N$.

Then the pull-back $\phi^*$ of $\phi$ at the point $\phi(p) \in N$ is the linear map

\begin{equation*}
\phi_p^*: T_{\phi(p)}^* N \to T_p^* M 
\end{equation*}

i.e. a linear map from the cotangent space at the target TO the cotagent space of the originating manifold at point $p$!

We define the map

\begin{equation*}
\omega \mapsto \phi_p^*(\omega)
\end{equation*}

as, acting on $X \in T_p M$,

\begin{equation*}
\phi_p^*(\omega) (X) := \omega \Big( \phi_{*_p} (X) \Big)
\end{equation*}

which is linear since $\phi_{*_p}$ and $\omega$ are linear, where $\phi_{*_p}$ is the push-forward.

Let

  • $\phi: M \to N$ be a smooth map
  • $\omega \in \Omega^1(M)$, i.e. $\omega: M \to T^* M$ s.t.

    \begin{equation*}
\pi \circ \omega = \id_M
\end{equation*}

Then

\begin{equation*}
\phi^* \omega \in \Omega^1(N)
\end{equation*}

We need to show that $\phi^* \omega$ is indeed smooth.

Let

  • $\big( U, x \big)$ be a chart in $M$ (with $a \in U$) and $\big( V, y \big)$ be a local chart in $N$ (with $\phi(a) \in V$)
  • $\omega \in \Omega^1(M)$
  • $Y \in \mathfrak{X}(N)$

Then in $U$ we can write $\omega = \omega_i \dd{x}^i$ and $Y = Y^i \pdv{}{y^i}$ with $\omega_i \in C^{\infty}(M)$ and $Y^i \in C^{\infty}(N)$. Since $\phi^* \omega: M \to T^* M$, we have $(\phi^* \omega)(a) \in T_a M$, therefore it's sufficient to consider how $(\phi^* \omega)(a)$ acts on a basis vector of $T_a M$, e.g. $\pdv{}{x^i}$:

\begin{equation*}
(\phi^* \omega) \pdv{}{x^i} = \omega \bigg( \phi_* \pdv{}{x^i} \bigg)
\end{equation*}

Recall that

\begin{equation*}
\bigg( \phi_* \pdv{}{x^i} \bigg) \bigg|_{\phi(a)} f = \pdv{}{x^i} (f \circ \phi^{-1}) \bigg|_{\phi(a)} = \pdv{(f \circ \phi^{-1})}{y^j} \bigg|_{\phi(a)} \pdv{y^j}{x^i} \bigg|_{a}
\end{equation*}

since $\big( y^j \big)$ are the local coords of $\phi(a)$, which means $\pdv{}{y^j}$ can act on $C^{\infty}(N)$; observe that $(f \circ \phi^{-1}): N \to \mathbb{R}$. Therefore, going back to our original expression for $\phi^* \omega$:

\begin{equation*}
\begin{split}
  \big( \phi^* \omega \big) \pdv{}{x^i} \bigg|_{a} &= \omega \bigg( \varphi_* \pdv{}{x^i} \bigg|_{a} \bigg) \\
  &= \omega \bigg( \pdv{y^j}{x^i} \bigg|_{a} \pdv{}{y^j} \bigg|_{\phi(a)} \bigg) \\
  &= \pdv{y^j}{x^i} \bigg|_{a} \omega \bigg( \pdv{\cdot \circ \phi^{-1}}{y^j} \bigg|_{\phi(a)} \bigg) \\
  &= \pdv{y^j}{x^i} \bigg|_{a} \omega_j \Big( (y^j \circ x \circ)(\phi^{-1}(a)) \big) \Big)
\end{split}
\end{equation*}
\begin{equation*}
\big( F^* \omega \big)(a) = \omega_i(a) \ \big( F^* \dd{x}^i \big)(a)
\end{equation*}
Equivalence with Jacobian in some chart

Let $\phi: M \to N$ be a smooth math between manifolds.

Relative to local coordinates $x^i$ near $p$ and $y^i$ near $\phi(p)$

\begin{equation*}
\begin{split}
  \phi_{*_p} \bigg( \pdv{}{x^i}\bigg|_{p} \bigg) (f) &= \pdv{}{x^i} \bigg|_{p} \big( f \circ F \big) \\
  &= \pdv{\phi^j}{x^i}\bigg|_{p} \pdv{f}{y^j}\bigg|_{\phi(p)} \\
  \implies \quad \phi_{*_p} \bigg( \pdv{}{x^i}\bigg|_{p} \bigg) &= \pdv{\phi^j}{x^i}\bigg|_{p} \pdv{}{y^j}\bigg|_{\phi(p)}
\end{split}
\end{equation*}
Comments on push-forward and exterior derivative

First I was introduced to the exterior derivative in the Geomtry course I was doing, and afterwards I was, through the lectures by Schuller, introduced to the concept of the push-forward, and the pull-back defined using the push-forward. Afterwards, in certain cotexts (e.g. in Geometry they defined the pull-back of a 2-form on $\matbb{E}^3$ involving the exterior derivative), I kept thinking "Hmm, there seems to be some connection between the exterior derivative and the push-forward!

Then I read this stachexchange answer, where you'll find the following snippet:

Except in one special situation (described below), there is essentially no relationship between the exterior derivative of a differential form and the differential (or pushforward) of a smooth map between manifolds, other than the facts that they are both computed locally by taking derivatives and are both commonly denoted by the symbol $d$.

And the special case he's referring to is; when the function is a smooth map $f: M \to \mathbb{R}$, where the two are equivalent.

Immersion, submersion and embedding

A submersion is a differentiable map between differential manifolds whose differential is surjective everywhere.

Let $M$ and $N$ be differentiable manifolds and $f: M \to N$ be a differentiable map between them. The map $f$ is a submersion at the point $p \in M$ if its differential

\begin{equation*}
D f_p : T_p M \to T_{f(p)} N
\end{equation*}

is a surjective linear map.

Let $\phi: M \overset{\sim}{\to} N$ be a smooth map on manifolds $M$ to $N$.

We say $\phi$ is an immersion if and only if the derivative / push-forward $\phi_{*_p}$ is injective for each point $p$, or equivalently $\rank \phi_* = \dim M$.

Remember the push-forward is a map from the tangent space of $M$ at point $p$ to the tangent space of $N$ at $\phi(p)$: $\phi_{*_p} : T_p M \to T_{\phi(p)} N$. We do NOT require the map $\phi$ itself to be injective!

Let $\phi: M \overset{\sim}{\to} N$ be a smooth map on manifolds $M$ and $N$.

We say $\phi$ is an embedding of $M$ in $N$ if and only if:

  1. $\phi$ is an immersion
  2. $\phi(M) \cong_{\text{topo}} N$, where $\cong_{\text{topo}}$ means a homeomorphism / topological isomorphism

To summarize: A smooth map $\phi: M \to N$ is called a

  • submersion if $\phi_{*_p}: T_p M \to T_{\phi(p)} N$ is surjective for for all $p \in M$ (so $\dim M \ge \dim N)$
  • immersion if $\phi_{*_p}$ is injective for all $p \in M$ (so $\dim M \le \dim N$)
  • embedding if $\phi_{*_p}$ is an immersion and if the manifold topology of $M$ agrees with the subspace topology of $\phi(M) \subseteq N$

Any smooth manifold $M$ can be:

  • embedded in $R^{2d}$
  • immersed in $R^{2d - 1}$

Where $d = \dim M$.

This is of course "worst-case scenarios", i.e. there exists manifolds which can be embedded / immersed in lower-dimensional manifolds than the rules mentioned here.

There exists even stronger / better lower bounds for a lot of target manifolds, which requires slightly more restrictions on the manifold.

Let $M$ and $N$ me differentiable manifolds. A function

\begin{equation*}
f: M \to N
\end{equation*}

is a local diffeomorphism, if, for each point $x \in M$, there exists an open set $U$ such that $x \in U$, and the image $f(U)$ is open and

\begin{equation*}
f|_U : U \to f(U)
\end{equation*}

is a diffeomorphism.

A local diffeomorphism is then a special case of an immersion $f$ from $M$ to $N$ , where the image $f(U)$ of $U$ under $f$ locally has the differentiable structure of a submanifold of $N$.

Example: 2D Klein bottle in 3D

The klein bottle is a 2D surface, as we can see below: klein_bottle.png

But due to the self-intersecting nature of the Klein bottle, it is not a manifold when it "sits" in $\mathbb{R}^3$. Nonetheless, the mapping of the Klein bottle as shown in the picture, does in fact have a injective puh-forward! That is, we can injectively map each tangent vector at a point in such a manner that no two tangent vectors are mapped to the same tangent vector on the Klein bottle 2D surface in $\mathbb{R}^3$.

Hence, the Klein bottle can be immersed in $\mathbb{R}^3$ but NOT embedded, as "predicted" by Whitney Theorem. And the same theorem tells us that we can in fact embed the Klein bottle in $\mathbb{R}^4$.

Tensor Fields and Modules

Notation

  • $\Big( \Gamma(TM), \oplus, \odot \Big)$ $\quad \Gamma(TM) = \{ \sigma \in C^{\infty}(M; TM) : \pi \circ \sigma = \text{id}_M \}$ where $TM$ denotes the tangent bundle of the manifold $M$.
  • $R \oplus \dots \oplus R$ is taking the Cartesian product and equipping it with addition
  • Homomorphisms

    \begin{equation*}
\text{Hom}_R(P, Q) := \{ \varphi: P \overset{\sim}{\to} Q \mid \text{R-linear} \}
\end{equation*}

    where $R \text{-linear}$ refers to the fact that if $r \in R$ and $v \in P$ we have

    \begin{equation*}
\varphi(r v) = r \varphi(v)
\end{equation*}
  • Let $T \in T_q^p$ be a (p, q)-tensor, then
  • Symmetric summation

    \begin{equation*}
T_{(ab)} = \frac{1}{2} \big( T_{ab} + T_{ba} \big)
\end{equation*}
  • Anti-symmetric summation

    \begin{equation*}
T_{[a b]} = \frac{1}{2} \Big( T_{ab} - T_{ba} \Big)
\end{equation*}
  • Vertical bars denote arguments which are excluded

    \begin{equation*}
\begin{split}
  T_{(a |b| c)} &= \frac{1}{2} \Big( T_{abc} - T_{cba} \Big) \\
  T_{[a |bc| d]} &= \frac{1}{2} \Big( T_{abcd} - T_{dbca} \Big)
\end{split}
\end{equation*}

Stuff

Let

A vector field is a smooth section $\sigma$ of $TM$, i.e.

  • $\sigma: M \to TM$ is smooth
  • $\pi: TM \to M$ is smooth
  • $\pi \circ \sigma = \text{id}_M$

Informally, a vector field on a manifold $M$ can be defined as a function that inputs a point $p \in M$ and outputs an element of the tangent space $X \in T_pM$. Equivalently, a vector field is a section of the tangent bundle.

Defined as a function on $M$, a vector field $X = X^i \pdv{}{x^i}$ is smooth if $X^i$ are all smooth functions of $p$.

Let $f \in C^{\infty}(M)$ and $X$ be a vector field on $M$.

The function

\begin{equation*}
\begin{split}
  X(f): \quad & M \to \mathbb{R} \\
  & p \mapsto X_p(f)
\end{split}
\end{equation*}

is smooth if and only if $X$ is smooth.

If chart $\pdv{}{x^i}(f) = p \maspto \pdv{}{x^i}\bigg|_{p} f = \pdv{f \circ \phi^{-1}}{x^i}\bigg|_{\phi(p)}$ are smooth because $f \circ \phi^{-1}$ is on the chart.

Conversely, if $X(f) \in C^{\infty}(M)$, we choose $f = x^i$, thus $X^i = X(x^i)$ is smooth.

Module

We say $(M, \oplus, \odot)$ is a R-module, $R$ being a ring, if

\begin{equation*}
\begin{split}
  \oplus:& M \times M \to M \\
  \odot:& R \times M \to M
\end{split}
\end{equation*}

satisfying

\begin{equation*}
C^{\oplus} \ A^{\oplus} \ N^{\oplus} \ I^{\oplus} \quad A \ D \ D \ U
\end{equation*}

Thus, we can view it as a "vector space" over a ring, but because it behaves wildly different from a vector space over a field, we give this space a special name: .

Important: $M$ denotes a module here, NOT manifold as usual.

If $D$ is a division ring, then $D$ has a basis.

This is not a $\iff$ but simply says that we guarantee the existence of a basis if $D$ is a division ring.

First we require the Axiom of Choice, in the incarnation of Zorn's lemma, which is just the Axiom of Choice restated, given that we already have all the other axioms of Zermelo-Fraenkel set theory.

Zorn's Lemma: A partially ordered set $P$ whose every totally ordered subset $T$ has an upper bound in $P$ contains a maximal element.

where:

partially ordered

Every module $V$ over a divison ring has a basis

Notation
Theorem

Every module $V$ over a division ring has a basis.

  1. Let $S \in V$ be a generating system of $V$, i.e.

    \begin{equation*}
\forall v \in V, \quad \exists e_1, \dots, e_N \in S, \quad \exists v^1, \dots, v^N \in D : v = v^a e_a
\end{equation*}

    Observe that $S$, the generating system, always exists since we can simply have $S = V$

  2. Define a partially ordered set $(P, \le)$ by

    \begin{equation*}
 P := \{ u \in \mathcal{P}(S) \mid  u \text{ is lineraly indep.} \}
\end{equation*}

    where $\mathcal{P}(S)$ denotes the powerset of $S$. We partially order it by inclusion:

    \begin{equation*}
 \le \equiv \subseteq
\end{equation*}

    i.e. if a set is a subset of another, then the it's smaller than the other subset.

  3. Let $T$ be any totally ordered subset of $P$, then

    \begin{equation*}
 \bigcup T \quad \text{upper bound to T}
\end{equation*}

    and it is a lin. indep. subset of $S$. Thus, by Zorn's Lemma, $P$ has a maximal element, one of which we call $B$. By construction, $B$ is a maximal lin. indep. subset of $S$.

  4. Claim: $S = \text{span}_D ( B ) = \{ \text{any finite D-lin. comb. of elements of } B \}$ Proof: Let $v \in S$. Since $B$ is maximal lin. indep. subset, we have $B \cup \{ v \}$ is linearly dependent. That is,

    \begin{equation*}
 \begin{split}
   \exists e_1, \dots, e_N \in B, \ & \exists a^1, \dots, a^N \in D, \ \exists a \in D: \\
   a^i e_i + a v &= 0
 \end{split}
\end{equation*}

    and not all of $a^1, \dots, a^N$, $a$ vanish, i.e. $a^1, \dots, a^N, a \ne 0$. Now it is clear that $a \ne 0$, because

    \begin{equation*}
 a = 0 \quad \implies \quad a^i e_i = 0 \quad \text{and} \quad a^1, \dots, a^N = 0
\end{equation*}

    but this is a contradiction to $e_1, \dots, e_N$ being linearly independent, as assumed previously. Hence we consider $a \ne 0$; then, since $0 \ne a \in D$ (remembering that $D$ is a division ring)

    \begin{equation*}
 \implies \exists a^{-1} \in D : a^{-1} a = 1_D
\end{equation*}

    Thus, if we multiply the equation above with the inverse of $a$, we get

    \begin{equation*}
 v = - (a^{-1} a^i) e_i
\end{equation*}

    for the finite subsets $\{ e_1, \dots, e_N \}$ of B. Thus,

    \begin{equation*}
 S = \text{span}_D(B)
\end{equation*}

    Hence, we have existence of a linear indepndent subset of $B$ which also spans $B$.

As we see above, here we're making use of the fact that $D$ is a division ring, when we're using the inverse of $a \in D$.

Observe that $C^\infty(M)$ is not a division ring, hence $\Gamma(TM)$ consider as a $C^\infty(M)$ module is not guaranteed to have a basis.

Definition of $\Gamma(TM)$ can be found here.

Examples

$C^\infty$ module

One simply example of a module is

\begin{equation*}
\Big( \Gamma(TM), \oplus, \odot \Big), \quad \Gamma(TM) = \{ \sigma \in C^{\infty}(M) : \pi \circ \sigma = \text{id}_M \}
\end{equation*}

where:

  • $M$ is a manifold
  • $\pi: TM \to M$ is the projection
  • $\Gamma(TM)$ denotes the set of all sections of $TM$, i.e. the total space of the bundle

which is a $C^\infty(M)$ module.

Terminology

A module over a ring is called free if it has a basis.

Examples:

\begin{equation*}
\begin{split}
  \Gamma(T \mathbb{R}^2) \quad &\text{is free} \\
  \Gamma(T \mathbb{S}^2) \quad &\text{is NOT free}
\end{split}
\end{equation*}

A module $\Gamma$ over a ring $R$ is called projective if it is a directed summand of a free R-module $F$:

\begin{equation*}
\Gamma \oplus Q = F
\end{equation*}

where $Q$ is the R-module.

Remark: free $\implies$ projective

Theorems

\begin{equation*}
\{ \text{smooth sections of a vector fibre fundle } (E, \pi, M) \text{ over a smooth manifold M} \}
\end{equation*}

is a finitely projective $C^\infty(M)$ module $\Gamma(E)$.

From this have the following corollary:

\begin{equation*}
\Gamma(E) \oplus Q = F
\end{equation*}

where $Q$ is the $C^\infty(M)$ module and $F$ is a free module.

Thus, $Q$ "quantifies" how much $\Gamma(E)$ fails to have a basis, since $Q$ is how much we have to add to $\Gamma(E)$ to make it free.

Let $P, Q$ be finitely generated projective modules over a commutative ring $R$.

Then,

\begin{equation*}
\Bigg( \text{Hom}_R(P, Q) := \{ \varphi: P \overset{\sim}{\to} Q \mid \text{R-linear} \}, \oplus, \odot \Bigg)
\end{equation*}

is again a finitely generated projective module.

This falls out of the commutativity of the ring $R$.

In particular:

\begin{equation*}
\text{Hom}_{C^\infty(M)} \bigg( \Gamma(TM), C^\infty(M) \bigg) =: \Gamma(TM)^* = \Gamma(T^*M)
\end{equation*}

where the equality $\Gamma(TM)^* = \Gamma(T^*M)$ can be shown (but we haven't done that here).

Finally, this gives us the "standard textbook definition" of a tensor-field:

A $(r, s)$ tensor field $t$ on a smooth manifold $M$ is a $C^\infty(M)$ multilinear map

\begin{equation*}
t: \underbrace{\Gamma(T^*M) \times \dots \times \Gamma(T^*M)}_r \times \underbrace{\Gamma(TM) \times \dots \times \Gamma(TM)}_s \overset{\sim}{\to} C^\infty(M)
\end{equation*}

We can then view

\begin{equation*}
\underbrace{\Gamma(T^*M) \times \dots \times \Gamma(T^*M)}_r \times \underbrace{\Gamma(TM) \times \dots \times \Gamma(TM)}_s
\end{equation*}

as the space of all tensor-fields on $C^\infty(M)$, which again, forms a module!

Hence, we when we talk about the mapping $t$ (see def:tensor-field for def) being multilinear, of course it must be multilinear to the underlying ring-structure of the module, i.e. $C^\infty(M)$ multilinear!

Some textbooks give the above definition, and then note that $t$ is not $\mathbb{R}$ linear, but $C^\infty(M)$ linear, and that this needs to be checked. But because we're aware of the commutative ring that "pops out" we know that of course it has to be multilinear wrt. the underlying ring-structure, which in this case $C^\infty(M)$.

Metric tensor

A metric gensor $g$ on $V$ is a $(0, 2) \text{-tensor}$ over $V$ s.t.

  1. symmetric:

    \begin{equation*}
g(v, w) = g(w, v), \quad \forall v, w \in V
\end{equation*}
  2. non-degenerate:

    \begin{equation*}
g(v, w) = 0 \quad \forall w \in V \quad \implies v = 0
\end{equation*}

Unlike an inner product, a metric tensor does NOT have to be positive-definite!

A standard example of this is the Lorentz metric.

Let $g$ be a metric tensor on $V$. The inverse metric is a $(2, 0) \text{-tensor}$ with the components $g^{ab}$ given by the matrix inverse of $\big( g_{ab} \big)$, i.e.

\begin{equation*}
\tensor{g}{^{ac}_{}} \tensor{g}{^{}_{cb}} = \tensor{\delta}{^{a}_{b}}
\end{equation*}

Raising and lowering of indices

Let $V$ be a vector space with $\dim V < \infty$.

A metric tensor $g$ on $V$ defines an isomorphism $V \cong V^*$ by the map

\begin{equation*}
\begin{split}
  & V \to V^* \\
  & X \mapsto \lambda_X = g(\cdot, X)
\end{split}
\end{equation*}

then

\begin{equation*}
\lambda_X(Y) = g(Y, X) \in \mathbb{K}
\end{equation*}

where $\mathbb{K}$ is the underlying field.

Therefore, we often write

\begin{equation*}
\begin{split}
  \lambda_a &= g_{ab} X^b \\
  X^a &= \tensor{g}{^{ab}_{}} \lambda_b
\end{split}
\end{equation*}

and we often surpress the difference between vector $X$ and covector $\lambda$, by simply writing $X_a$ for the vector corresponding to $\lambda_a$, i.e.

\begin{equation*}
X_a = g_{ab} X^b \quad \text{and} \quad X^b = g^{ab} X_a
\end{equation*}

which defines the raising and lowering operation often seen.

More generally, this works similarily for arbitrary tensors, e.g. $(r, s) \text{-tensor} \mapsto (r - 1, s + 1) \text{-tensor}$ by the "lowering operation". Consider the example of a $(0, 2) \text{-tensor}$ $T_{ab}$:

\begin{equation*}
\tensor{T}{^{a}_{b}} = \tensor{g}{^{ac}_{}} \tensor{T}{^{}_{cb}}, \quad \tensor{T}{_{a}^{b}} = \tensor{g}{^{bc}_{}} \tensor{T}{^{}_{ac}}, \quad \tensor{T}{^{ab}_{}} = \tensor{g}{^{ac}_{}} \tensor{g}{^{bd}_{}} \tensor{T}{^{}_{cd}}
\end{equation*}

Signature of tensor

The signature $(p, q)$ of metric $g$ on $V$ is the number of negative values $p$ and the number of positive values $q$, and $p + q = n$.

If $p = 0$, i.e. $\text{sig}(g) = (0, n)$ for some $q \in \mathbb{N}$, then the metric is positive-definite and thus defines an inner product!

The signature of a tensor is basis independent.

Lorentzian metrics

Metrics with signature $(1, q)$ are known as Lorentzian metrics, i.e. have one negative component.

First I wrote $(1, n - 1)$, but this is not true! One can also have coefficients which are zero. So we don't necessarily have to have $n$ non-zero elements.

Let $v \in V$ and $g$ be a Lorentzian metric. Then we say $v$ is

  • timelike if $g(v, v) < 0$
  • spacelike if $g(v, v) > 0$
  • null if $g(v, v) = 0$

Null vectors form a "double cone" in the sense that if $v$ is null, then so is $tv$ for all $t \in \mathbb{R} \setminus \left\{ 0 \right\}$.

In relativity, this is called the light-cone.

Future-directed vectors lie inside the future light-cone, and past-directed vectors lie inside the past light-cone.

Let $\big( M, g \big)$ be a Riemannian manifold.

If there exists a nowhere vanishing vector field $Z$, then $M$ admits a Lorentzian metric given by

\begin{equation*}
\hat{g}(X, Y) = g(X, Y) - 2 \frac{g(X, Z) g(Y, Z)}{g(Z, Z)}
\end{equation*}

Let $\left\{ e_a: a = 0, 1, \dots, n \right\}$ such that

\begin{equation*}
e_0 = \frac{1}{\sqrt{\hat{g}(Z, Z)}} Z
\end{equation*}

so that $e_0$ has unit norm. From definition of $g$, we have

\begin{equation*}
\begin{split}
  g(e_0, e_0 ) &= \hat{g}(e_0, e_0) - 2 \frac{\hat{g}(e_0, Z) \hat{g}(e_0, Z)}{\hat{g}(Z, Z)} \\
  &= 1 - 2 \frac{\hat{g}(Z, Z) \hat{g}(Z, Z) / \hat{g}(Z, Z)}{\hat{g}(Z, Z)} \\
  &= 1 - 2 \\
  &= - 1
\end{split}
\end{equation*}

And

\begin{equation*}
g(e_0, e_a) = 0, \quad \forall a \ne 0
\end{equation*}

and

\begin{equation*}
g(e_a, e_b) = \delta_{ab}, \quad \forall a, b = 1, \dots, n
\end{equation*}

Therefore, $g$ in an orthonormal basis clearly has Lorentzian signature!

Psuedo-Riemannian manifolds

A metric tensor $g$ on a smooth manifold $M$ is a (0, 2)-tensor field which is symmetric and non-degenerate at every point $m \in M$.

The pair $\big( M, g \big)$ is called a psuedo-Riemannian manifold.

So, for a psuedo-Riemannian manifold $(M, g)$, the metric tensor has a particular signature at each point. Now, since the signature is integer-valued, by continuity, it must be constant from point to point. There are two types of signature $p$ which are of particular importance:

  1. $p = 0$: $g$ is positive-definite, then $(M, g)$ is a Riemannian manifold
  2. $p = 1$: $g$ is Lorentzian, then $(M, g)$ is a Lorentzian manifold

Let $(M, g)$ be a psuedo-Riemannian manifold. If $g$ is Lorentzian, i.e. has signature $p = 1$, then we say $(M, g)$ is a Lorentzian manifold.

In a coordinate basis, a metric tensor takes the form

\begin{equation*}
g = g_{ij} \dd{x^i} \otimes \dd{x^j}
\end{equation*}

It is conventional to denote the symmetric tensor product of covectors by

\begin{equation*}
\dd{x^i} \dd{x^j} = \frac{1}{2} \big( \dd{x^i} \otimes \dd{x^j} + \dd{x^j} \otimes \dd{x^i} \big)
\end{equation*}

so we can write

\begin{equation*}
g = \tensor{g}{^{}_{ij}} \dd{x^i} \dd{x^j}
\end{equation*}
Examples
  1. Consider $\mathbb{R}^n$ with Cartesian coordinates $\big( x^1, \dots, x^n \big)$. The Euclidean metric is

    \begin{equation*}
g = \big( \dd{x^1} \big)^2 + \cdots + \big( \dd{x^n} \big)^2
\end{equation*}

    and $\big( \mathbb{R}^n, g \big)$ is the Euclidean space.

  2. Consider $\mathbb{R}^4$ with coordinates $\big( x^0, x^1, x^2, x^3 \big)$. The Minkowski metric is

    \begin{equation*}
\eta = - \big( \dd{x^0} \big)^2 + \big( \dd{x^1} \big)^2 + \big( \dd{x^2} \big)^2 + \big( \dd{x^3} \big)^2
\end{equation*}

    and $\big( \mathbb{R}^4, \eta \big)$ is the Minkowski spacetime.

  3. Consider $S^2$. In polar coordinates $\big( \theta, \phi \big)$ the uni round sphere is

    \begin{equation*}
g = \big( \dd{\theta} \big)^2 + \sin^2 (\theta) \big( \dd{\phi} \big)^2
\end{equation*}

    This is positive definite for all $\theta \in (0, \pi)$; this corresponds to everywhere this chart is defined. However, the chart does not cover $S^2$, so this does not fully define $g$ on $S^2$. To fully define $g$ on $S^2$, we need to consider an atlas on $S^2$ and ensure that $g$ is a smooth tensor field.

Length on Riemannian manifolds

Let

The length of the curve is

\begin{equation*}
L[\gamma] = \int_{a}^{b} \sqrt{g \big( \dot{\gamma}, \dot{\gamma} \big) \big|_{\gamma(t)}} \dd{t}
\end{equation*}

where $\dot{\gamma}$ is the tangent vector to the curve.

It may be checked that this definition is invariant under reparametrization of $\gamma(t)$. You can see this by noting that

\begin{equation*}
g \big( \dot{\gamma}, \dot{\gamma} \big) \big|_{\gamma(t)} = \tensor{g}{^{}_{ij}} \big( (x \circ \gamma)(t) \big) \dv{x^i}{t} \dv{x^j}{t}
\end{equation*}

Because of this, the metric is often written as the "line element"

\begin{equation*}
\dd{s^2} = \tensor{g}{^{}_{ij}} \dd{x^i} \dd{x^j}
\end{equation*}

Consider defining a metric on $(M, g)$ by defining

\begin{equation*}
d(p, q) = \inf_{\gamma} L[\gamma]
\end{equation*}

i.e. the distance between two points $p, q \in M$ is defined to be the shortest path between the two points (wrt. length on the Riemannian manifold defined by $g$).

Clearly $d$ is non-negative, and vanish if and only if $a = b$, since the term inside the integrand is positive for all values. It's also symmetric by the symmetry of the term inside the integrand.

Finally, one can show that the topology induced by $d$ is indeed the same as the topology defined on the manifold $M$!

Let's define the open ball of radius $r$ centered at $p \in M$ as

\begin{equation*}
B_r(p) = \left\{ q \in M : d(p, q) = \inf_{\gamma} L[\gamma] < r \right\}
\end{equation*}

Observe then that if $p, q \in U \subseteq M$ for some chart $(U, x)$, then the shortest line is simply the straight line in the Euclidean plane, i.e. directly corresponds with the open ball notion in the standard topology in $\mathbb{R}^n$. Therefore we only need to consider the case where $B_r(p)$ is not fully contained in a single chart, but instead requires, say, two charts $\big( U_x, x \big)$ and $\big( U_y, y \big)$.

But, we know that the value $g$ takes on is basis-independent, and so we know the value will be exactly the same irregardless of the chart. Therefore it follows for multiple charts imediately as a consequence of the case where $B_r(p)$ is fully contained in a single chart.

Or, we consider the determinant of the Jacobian when moving from $U_x$ to $U_y$, and the explicitly write out the corresponding integral.

Defining a notion of length is not so straight-forward for Lorentzian manifolds.

A smooth curve $\gamma: I \to M$ is said to be timelike, spacelike, or null if the tangent vector $\dot{\gamma}(t)$ is the corresponding term for all $t$!

Observe that most curves do not fit into this definition; the "type" of tangent vector can change along the curve, i.e. we can cross the boundaries of the "light-cones".

Thus, for a spacelike curve $\gamma(t)$ we have direct correspondance with the length defined on a Riemannian manifold, since $g \big( \dot{\gamma}, \dot{\gamma} \big) > 0$ along $\gamma(t)$. For the case of $\gamma(t)$ being timelike, we need a new notion.

Let

The proper time from $p$ along the curve is

\begin{equation*}
\tau = \int_{0}^{t} \sqrt{- g \big( \dot{\gamma}, \dot{\gamma} \big) \big|_{\gamma(t)}} \dd{t}
\end{equation*}

Again, this is invariant under reparametrization of $\gamma(t)$.

Consider a timelike curve $\gamma(t)$ parametrized by proper time.

It's tangent vector satisfies

\begin{equation*}
\tensor{g}{^{}_{ab}} X^a X^b = -1
\end{equation*}

The tangent vector $X$ of a reparametrized curve $\tilde{\gamma} \big( \tau(t) \big) = \gamma(t)$ is given by

\begin{equation*}
X = \dv{t}{\tau} \dot{\gamma}(t)
\end{equation*}

by the chain rule. Hence

\begin{equation*}
g(X, X) = \bigg( \dv{t}{\tau} \bigg)^2 g \big( \dot{\gamma}, \dot{\gamma} \big)
\end{equation*}

If $\tau$ is the proper time, then

\begin{equation*}
\dv{\tau}{t} = \sqrt{- g \big( \dot{\gamma}, \dot{\gamma} \big)}
\end{equation*}

Hence

\begin{equation*}
g(X, X) = g_{ab} X^a X^b = -1
\end{equation*}

Grassman algebra and deRham cohomology

Notation

  • $X_i \in \Gamma(TM) = \{ \sigma \in C^\infty(M) : \text{proj} \circ \sigma = \text{id}_M \}$, i.e. maps in $C^\infty$ s.t. composed with the projection from the total space to the base space of the tangent bundle form the identity. That is, $\sigma$ maps a point in $M$ to it's fibre in $T_pM$, known as sections
  • $\pi$ denotes a permutation in what follows section, NOT a projection as seen earlier
  • $\text{Perm}(n)$ denotes the set of permutations on $n$ letters / digits
  • $\otimes$ is the tensor product
  • $(\circ)_{\textcolor{red}{[}m_1, \dots, m_n \textcolor{red}{]}} := \frac{1}{n!} \sum_{\pi \in \text{Perm}(n)} \text{sgn}(\pi) (\circ)_{\pi(m_1), \dots, \pi(m_n)}$ is called the anti-symmetric bracket notation, where $(\circ)$ denotes some "indexable object"
  • $(\circ)_{\textcolor{red}{(}m_1, \dots, m_n \textcolor{red}{)}}$ is the same as anti-symmetric bracket notation, but dropping the $\text{sgn}(\pi)$, we call symmetric bracket notation
  • \begin{equation*}
\begin{split}
  B^n &= \text{Im}(\textcolor{red}{d}) \subseteq \Omega^n(M) \\
  Z^n &= \text{Ker}(\textcolor{green}{d}) \subseteq \Omega^n(M)
\end{split}
\end{equation*}

    where $B^n$ is the exact forms and $Z^n$ is the closed forms, where $\textcolor{red}{d}$ denotes the previous and $\textcolor{green}{d}$ the next

Grassman Algebra

The set of all n-forms is denoted

\begin{equation*}
\Omega^n(M)
\end{equation*}

which naturally is a $C^\infty(M)$ module since:

  • sum of two n-forms is a n-form
  • $C^\infty(M)$ multiple of some $\omega \in \Omega^n$ is again in $\Omega^n$

We have a problem though: taking the tensor product of forms does not yield a form!

I.e. the space is not closed.

In what follows we're slowly building up towards a way of defining a product in such a way that we do indeed have the space of forms being closed under some additional operation (other than $+$ and $\cdot$), which is called the Grassman algebra.

We define the wedge product as follows:

\begin{equation*}
\wedge : \Omega^n(M) \times \Omega^m(M) \to \Omega^{n + m}(M)
\end{equation*}
\begin{equation*}
(\omega, \sigma) \mapsto \omega \wedge \sigma
\end{equation*}

defined by

\begin{equation*}
(\omega \wedge \sigma)(X_1, \dots, X_{n + m}) = \frac{1}{n!} \frac{1}{m!} \sum_{\pi \in \text{Perm}(m + n)} \text{sign}(\pi) (\omega \otimes \sigma) \big( X_{\pi(1)}, \dots, X_{\pi(n)} \big)
\end{equation*}

e.g.

\begin{equation*}
\omega \wedge \sigma = \omega \otimes \sigma - \sigma \otimes \omega
\end{equation*}

where the tensor product $\otimes$ is just defined as

\begin{equation*}
(\omega \otimes \sigma) (X_1, \dots, X_{n + m}) = \omega(X_1, \dots, X_n) \sigma(X_{n + 1}, \dots, X_{n + m})
\end{equation*}

as usual.

Further, this allows us to construct the pull-back for some arbitrary $n$ form:

Let $\omega \in \Omega^n(M)$ and $h: M \to N$ be a smooth mapping between the manifolds.

This induces the pull-back:

\begin{equation*}
h_p^*: T_{h(p)}^* N \to T_p^* M
\end{equation*}

can be used to define

\begin{equation*}
h^*: \Omega^1(N) \overset{\sim}{\to} \Omega^1(M)
\end{equation*}

where $\Gamma(T^* M) = \Omega^1(N)$, which is the pull-back of the entire space rather than at a specific point $p$.

Then,

\begin{equation*}
\big( h^* \omega \big)\big( X_1, \dots, X_n \big) := \omega \big( h_*(X_1), \dots, h_*(X_n) \big)
\end{equation*}

where $h_*$ is the push forward of a vector field.

The pull-back distributes over the wedge product:

\begin{equation*}
h^*(\omega \wedge \sigma) := h^*(\omega) \wedge h^*(\sigma)
\end{equation*}

The $C^\infty(M)$ module defined

\begin{equation*}
\text{Gr}(M) \equiv \Omega(M) := \Omega^0(M) \oplus \Omega^1(M) \oplus \Omega^2(M) \oplus \dots \oplus \Omega^n(M)
\end{equation*}

(where we've seen $\Omega^0(M) = C^\infty(M)$ and $\Omega^1(M) = \Gamma(T^*M)$ before!)

Then $(\Omega(M), +, \cdot, \wedge)$ defines the Grassman algebra / exterior algebra of $M$, with the $\wedge$ being a bilinear map

\begin{equation*}
\wedge : \Omega(M) \times \Omega(M) \to \Omega(M)
\end{equation*}

is defined by linear contiuation of $\wedge$ (wedge product for forms), which means that for example if we haven

\begin{equation*}
\psi = \omega + \sigma, \quad \psi \in \Omega(M)
\end{equation*}

where $\omega \in \Omega^1(M)$ and $\sigma \in \Omega^3(M)$, and another $\varphi \in \Omega^n(M) \subseteq \Omega(M)$, then

\begin{equation*}
\psi \wedge \varphi = (\omega + \sigma) \wedge \varphi = \omega \wedge \varphi + \sigma \wedge \varphi
\end{equation*}

Now, as it turns out, we cannot define a differentiable structure on tensors on some manifold $M$. But do not despair! We can in fact do this on anti-symmetric tensors, i.e. forms, we can indeed define a differentiable structure without any further restrictions on the manifold!

The exterior derivative operator

\begin{equation*}
d: \Omega^n(M) \to \Omega^{n + 1}(M)
\end{equation*}

where

\begin{equation*}
\begin{split}
  \big( d \omega \big) \big( X_1, \dots, X_{n + 1} \big) := &\sum_{i=1}^{n + 1} (-1)^{i + 1} X_i \Big( \omega \big( X_1, \dots, \cancel{X_i}, \dots, X_{n + 1} \big) \Big) \\
  &+ \sum_{i < j} (-1)^{i + j} \omega \big( [X_i, X_j], X_1, \dots, \cancel{X_i}, \dots \cancel{X_j}, \dots X_{n + 1} \big)
\end{split}
\end{equation*}

i.e. since $\omega$ takes $n$ entries we leave out the i-th entry, and $[X_i, X_j]$ is the commutator.

Equivalently, this can also be defined as

\begin{equation*}
\dd{\omega} = \sum_{\left| I \right| = k}^{} \dd{\omega}_I \wedge \dd{x}^I
\end{equation*}

where we hare using multi-indexing, and

\begin{equation*}
\omega = \sum_{\left| I \right| = k}^{} \omega_{I} \dd{x}^I
\end{equation*}

with $\omega_i \in C^{\infty}(M)$, and so

\begin{equation*}
\dd{\omega_i} = \pdv{\omega_i}{x^i} \dd{x}^i
\end{equation*}

Commutator of two vector fields $X, Y \in \Gamma(TM)$ is given by

\begin{equation*}
[X, Y]f := X \big( Yf \big) - Y \big( Xf \big)
\end{equation*}

for $f \in C^\infty(M)$.

A geometrical interpretation of $\comm{X}{Y}$..

Let $r$ be the endpoint of concatenating $\gamma^X$ with $\gamma^Y$, and $r'$ be the endpoint of starting with $\gamma^Y$ and concatenating with $\gamma^X$. Then one can show that

\begin{equation*}
x^i(r) - x^i(r') = ts \comm{X}{Y}_p^i + \dots
\end{equation*}

where $x^i$ is the i-th component of some chart. Here we have Taylor expanded in which case the first-order terms turn out to cancel, and we get the above.

Let $\omega \in \Omega^n(M)$ and $\varphi \in \Omega^m(M)$.

\begin{equation*}
d \big( \omega \wedge \varphi \big) = dw \wedge \varphi + (- 1)^n \omega \wedge d \varphi
\end{equation*}

If we have the smooth map $h: M \to N$, then

\begin{equation*}
h^*(d \omega) = d(h^* \omega)
\end{equation*}

where we observe that the $d$ are "different":

  • LHS: $d: \Omega^n(N) \to \Omega^{n + 1}(N)$
  • RHS: $d: \Omega^n(M) \to \Omega^{m + 1}(N)$

which is why we use the word "commute" rather than that they are "the same" (cuz they ain't)

Further, action of $d$ extends by linear continuation to $\Omega(M)$:

\begin{equation*}
d: \Omega(M) \to \Omega(M)
\end{equation*}

where $\Omega(M)$ denotes the Grassman algebra.

Exterior power

Let $V$ be a vector space with $\dim(V) = n$.

Then the exterior power is defined as $\bigwedge^k V$, a subspace of the Grassman algebra,

\begin{equation*}
\bigwedge^k 
\end{equation*}

In the case where $V = TM$, we have

\begin{equation*}
\bigwedge^k (T^* M) = \mathrm{span}\left(\left\{ \dd{x}^1 \wedge \cdots \wedge \dd{x}^k \right\} \right)
\end{equation*}

Then we have

\begin{equation*}
\dim \big( \bigwedge^k V^* \big) = {n \choose k}
\end{equation*}

Physical examples

Maxwell electrodynamics

Let $F$ be the field strength, i.e. the Lorentz force:

\begin{equation*}
F = |v \times B + q E|
\end{equation*}

then $F$ is a two-form (since it maps both $B$ and $E$ to $\mathbb{R}$), and we require

\begin{equation*}
d F = 0, \qquad F \in \Omega^3(M)
\end{equation*}

which is called the homogenous Maxwell equations.

Since $F$ is a two-form on the reals, we know from Poincaré lemma that if a n-form is closed on $\mathbb{R}$ and thus we must exact, thus

\begin{equation*}
F = d A
\end{equation*}

for some $A \in \Omega^2(M)$. $A$ is called the gauge potential.

de Rham Cohomology

The following theorem has already been stated before, but I'll restate it due to it's importance in this section:

\begin{equation*}
d^2 = 0
\end{equation*}

where $d$ is the exterior derivative.

\begin{equation*}
d^2 = d \circ d
\end{equation*}

and

\begin{equation*}
\omega \in \Omega^n(M): d^2 \omega = d \big( d \omega \big)
\end{equation*}

in local coords:

\begin{equation*}
\begin{split}
  dw &= \big( \partial_b \omega_{a_1, \dots, a_n} \big) dx^b \wedge dx^{a_1} \wedge \dots \wedge dx^{a_n} \\
  &= \big( \partial_{\textcolor{red}{[} b} \omega_{a_1, \dots, a_n \textcolor{red}{]}}  \big) dx^b \wedge dx^{a_1} \wedge \dots \wedge dx^{a_n}
\end{split}
\end{equation*}

(remember we're using Einstein summation), where we've used the fact that $A_{ab} B^{\textcolor{red}{[}a, b \textcolor{red}{]}} = A_{\textcolor{red}{[}a, b \textcolor{red}{]}} B^{ab}$, which gives us

\begin{equation*}
\begin{split}
  d \big( d \omega \big) &= \big( \partial_{\textcolor{red}{[} c} \partial_b \omega_{a_1, \dots, a_n \textcolor{red}{]}} \Big) dx^c \wedge dx^b \wedge dx^{a_1} \wedge \dots dx^{a_n} \\
  &= 0
\end{split}
\end{equation*}

where the last equality is due to Schwartz ($\frac{\partial f}{\partial x_i \partial x_j} = \frac{\partial f}{\partial x_j \partial x_i}$ under certain conditions).

$d^2 = 0$ implies that there exists a sequence of maps such that

\begin{equation*}
0 \overset{d}{\longrightarrow} \Omega^0(M) \overset{d}{\longrightarrow} \Omega^1(M) \overset{d}{\longrightarrow} \Omega^2(M) \overset{d}{\longrightarrow} \cdots \overset{d}{\longrightarrow} \Omega^n(M) \overset{d}{\longrightarrow} \Omega^{n + 1}(M) \overset{d}{\longrightarrow} \cdots \overset{d}{\longrightarrow} \Omega^{\text{dim } M}(M) \overset{d}{\longrightarrow} 0
\end{equation*}

We then observe that:

\begin{equation*}
\begin{split}
  \text{Ker}(\textcolor{green}{d}) &\subseteq \Omega^n(M) \\
  \text{Im}(\textcolor{red}{d}) &\subseteq \Omega^n(M)
\end{split}
\end{equation*}

where the above theorem tells us that:

\begin{equation*}
d^2 = 0 \quad \iff \quad \text{Im}(\textcolor{red}{d}) \subseteq \text{Ker}(\textcolor{green}{d})
\end{equation*}

Now, we introduce some terminology:

\begin{equation*}
\dots \longrightarrow \Omega^{n - 1}(M) \overset{\textcolor{red}{d}}{\longrightarrow} \Omega^n(M) \overset{\textcolor{green}{d}}{\longrightarrow}\Omega^{n + 1}(M) \longrightarrow \dots
\end{equation*}

We then say that $\omega \in \Omega^n(M)$ is called:

  • exact if $\omega \in \text{Im}(\textcolor{red}{d})$
  • closed if $\omega \in \text{Ker}(\textcolor{green}{d})$

which is equivalent to the exact and closed definitions that we've seen before, since $d \omega = \beta$ for some $\beta \in \Omega^{n + 1}(M)$ is exact, i.e. $\omega \in \text{Im}(\textcolor{red}{d})$. Observe that we here consider $d$ as a mapping $\Omega(M) \to \Omega(M)$ rather than $\Omega^n(M) \to \Omega^{n + 1}(M)$, thus the different colored mappings are sort of the same but sort of not :)

As we know from Poincaré lemma, there are cases where

\begin{equation*}
Z^n = B^n
\end{equation*}

but then you might wonder, if it's not the case: how would one quantify the difference between $Z^n$ and $B^n$?

The n-th deRham cohomology group is the quotient vector space

\begin{equation*}
H^n(M) := Z^n / B^n = \frac{\left\{ \alpha \in \Omega^k(M) \mid \dd{\alpha} = 0 \right\}}{\left\{ \alpha = \dd{\beta} \mid \beta \in \Omega^{k - 1}(M) \right\}}
\end{equation*}

where on $Z^n$ we have equivalence relation:

\begin{equation*}
\omega \sim_{B^n} \sigma \quad \iff \quad \omega - \sigma \in B^n
\end{equation*}

and we write (this is just notation)

\begin{equation*}
Z^n / B^n \equiv Z^n / \sim_{B^n}
\end{equation*}
\begin{equation*}
H_{dR} (M) = \bigoplus_{k = 0}^n H_{dR}^k(M)
\end{equation*}

defines the deRham cohomology.

This we can equip with a "wedge"

\begin{equation*}
\begin{split}
  \wedge: \quad & H_{dR}^k(M) \times H_{dR}^l(M) \overset{\wedge}{\longrightarrow} H_{dR}^{k + l}(M) \\
  & (\alpha, \beta) \mapsto [\alpha] \wedge [\beta] := [\alpha \wedge \beta]
\end{split}
\end{equation*}

which we can check is well-defined.

The idea of de Rham cohomology is to classify the different types of closed forms on a manifold.

One performs this classification by saying that two closed forms $\omega, \sigma \in \Omega^n(M)$ are cohomologous if they differ by an exact form, i.e. $\omega - \sigma$ is a exact:

\begin{equation*}
\omega \sim \sigma \quad \iff \quad \omega - \sigma \in B_n
\end{equation*}

or equivalently,

\begin{equation*}
\omega \sim \omega + \dd{\beta}, \quad \forall \beta \in \Omega^{k - 1}(M)
\end{equation*}

where $B_n$ is the set of exact forms.

Thus, the definition of n-th de Rham cohomology group is the set of equivalence classes with the equiv. relation described above; that is, the set of closed forms in $\Omega^n(M)$ modulo the exact forms.

Further, framing it slightly different, it might become a bit more apparent what we're saying here.

Observe that

\begin{equation*}
\begin{split}
  Z^n &:= \text{Ker}(d) = \{ \beta \in \Omega^n(M) : d \beta = 0 \} \quad \text{(set of \textbf{closed} forms)} \\
  B^n &:= \text{Im}(d) = \{ \alpha \in \Omega^n(M) : \alpha = d \beta \} \quad \text{(set of \textbf{exact} forms)}
\end{split}
\end{equation*}

Since $d^2 \omega = 0, \quad \forall \omega \in \Omega^n(M)$, then clearly

\begin{equation*}
Z^n \subseteq B^n
\end{equation*}

And further, it turns out that by partitioning all the closed forms by taking the "modulo" the exact forms, we get a set of unique and disjoint partitions (due to this being an equiv. relation).

That is,

\begin{equation*}
\omega \sim_{B^n} \sigma \iff \omega - \sigma \in B^n \iff \exists \alpha \in B^n \quad \text{such that} \quad \omega = \alpha + \sigma
\end{equation*}

$H^n(M)$ only depends on the global topology of the manifold $M$

This is quite a remarkable result, since all our "previous" work depends on the exterior derivative of the local structure, and then it turns out that $H^n(M)$ only depends no the actual topology of the manifold! Woah dude!

We have the following example:

\begin{equation*}
H^0(M) \cong \mathbb{R}^{\text{\# of connected pieces of M)}
\end{equation*}

Since $\Omega^{-1}(M) = 0$ then $B^0 = 0$, therefore

\begin{equation*}
H_{dR}^0(M) = Z^0 = \left\{ f \in \Omega^0(M) \mid \underbrace{\dd{f} = 0}_{\text{locally constant}} \right\}
\end{equation*}

If $M$ is connected, then

\begin{equation*}
H_{dR}^0(M) \cong \mathbb{R}
\end{equation*}

with the bijection $[f] \mapsto f(a)$ for some $a$ (which is arbitrary since $M$ is connected).

Assume $c: [0, 1] \to M$ is smooth, then FTC implies

\begin{equation*}
\int_{0}^{1} \dv{}{t} \underbrace{f \big( c(t) \big)}_{= \dd{f}(c'(t)) = 0} \dd{t} \quad \implies \quad f \big( c(1) \big) = f \big( c(0) \big)
\end{equation*}

Let $F: N \to M$ be $C^{\infty}$. then it induces

\begin{equation*}
\begin{split}
  F^*: \quad & H_{dR}(M) \to H_{dR}(N) \\
  & [\alpha] \mapsto F^*[\alpha] := \big[ F^* \alpha \big]
\end{split}
\end{equation*}
\begin{equation*}
F^* (\dd{\alpha}) = \dd{\big( F^* \alpha \big)} \quad \text{and} \quad F^* (\alpha \wedge \beta) = F^* \alpha \wedge F^* \beta
\end{equation*}

and so

\begin{equation*}
\begin{split}
  \dd{\alpha} &= 0 \quad \implies \dd{\big( F^* \alpha \big)} = F^* \dd{\alpha} = 0 \\
  \alpha &= \dd{\beta} \quad \implies F^* \alpha = F^* \dd{\beta} = \dd{\big( F^* \beta \big)}
\end{split}
\end{equation*}

shows that it's well-defined.

And it is a homomorphism:

\begin{equation*}
F^* \big( [\alpha] \wedge [\beta] \big) = F^* [\alpha \wedge \beta]  = [F(\alpha \wedge \beta) ] = [F^* \alpha \wedge F^* \beta] = [F^* \alpha] \wedge [F^* \beta] = F^*[\alpha] \wedge F^*[\beta]
\end{equation*}

Let

\begin{equation*}
\begin{split}
  F: \quad & N \times [0, 1] \to M \\
  & (a, t) \mapsto F_t(a) := F(a, t)
\end{split}
\end{equation*}

be "smooth" in the sense that $F$ is smooth on $M \times (- \varepsilon, 1 + \varepsilon)$ for any $\varepsilon > 0$. Then $[\alpha] \in H_{dR}^k(M)$ then $F^* \alpha \in \Omega^K \big( N \times [0, 1] \big)$, and let

\begin{equation*}
F^* \alpha := \beta + \dd{t} \wedge \gamma =:\underbrace{ F_t^* \alpha}_{= \alpha} + \dd{t} \wedge \underbrace{\iota_{\partial_t} F^* \alpha}_{= \gamma}
\end{equation*}

so $\beta \in \Omega^k(M)$ and $\gamma \in \Omega^{k - 1}(M)$. Then

\begin{equation*}
0 = F^* \big( \dd{\alpha} \big) = \dd{\big( F^* \alpha \big)} = \dd{\big( \beta + \dd{t} \wedge \big)} = \dd_N{\beta} + \dd{t} \wedge \pdv{\beta}{t} - \dd{t} \wedge \dd_N{\gamma}
\end{equation*}

which implies

\begin{equation*}
\pdv{}{t} F_t^* \alpha = \pdv{\beta}{t} = \dd_M{\gamma}
\end{equation*}

Integrating

\begin{equation*}
F_1^* \alpha - F_0^* \alpha \overset{\matrm{FTC}}{=} \int_{0}^{1} \bigg( \pdv{}{t} F_t^* \alpha \bigg) \dd{t} = \int_{0}^{1} \dd_N{\gamma} \dd{t} = \dd_N{\bigg( \int_{0}^{1} \gamma \dd{t} \bigg)}
\end{equation*}
\begin{equation*}
\therefore \quad \big[ F_1^* \alpha \big] = \big[ F_0^* \alpha \big], \quad \forall \alpha
\end{equation*}

hence

\begin{equation*}
F_0^* = F_1^*
\end{equation*}

From this we can restate Poincaré Lemma using the deRham cohomology.

\begin{equation*}
H_{dR}^k(\mathbb{R}^n) \cong 
\begin{cases}
  \mathbb{R}, \quad &k = 0 \\
  0, \quad & k > 0
\end{cases}
\end{equation*}
\begin{equation*}
\begin{split}
  F: \quad & \mathbb{R}^n \times [0, 1] \to \mathbb{R}^n \\
  & (x, t) \mapsto tx
\end{split}
\end{equation*}

Then

\begin{equation*}
\begin{split}
  F_1(x) &= 0 \implies F_1^* = \id \\
  F_0(x) &= 0 \implies F_0^* = 0
\end{split}
\end{equation*}

for $k > 0$.

So $[\alpha] \in H_{dR}^{k > 0}(\mathbb{R}^n)$, therefore

\begin{equation*}
[\alpha] = F_1^* [\alpha] = F_0^*[\alpha] = 0 \quad \implies \quad H_{dR}^{k > 0}(\mathbb{R}^n) = 0
\end{equation*}

Summary

  • From $d^2 = 0$ we get the sequence with inclusions where always the images are included in the kernels of the next map.n
  • Then if we want to quantify how much the images diviate from the kernel, we can quantify by "modding out" the closed forms, $B^n$, from the exact forms, $Z^n$.
  • We then learn purely topological invariants, $H^n(M)$

$H_{dR}^k (M \times \mathbb{R}) \cong H_{dR}^k (M), \quad \forall k$

So now we want to show that this is

\begin{equation*}
\pi^* \circ s^* \overset{?}{=} \id_{H_{dR}^k(M)
\end{equation*}
\begin{equation*}
H_{dR}^k (M \times \mathbb{R}) \cong H_{dR}^k (M), \quad \forall k = 0, \dots, n
\end{equation*}

First we show that

\begin{equation*}
s^* \circ \pi^* = \id_{H_{dR}^k(M)
\end{equation*}

and then we need to show that also

\begin{equation*}
\pi^* \circ s^* = \id_{H_{dR}^k(M)
\end{equation*}

We have

\begin{equation*}
M \times \mathbb{R} \overset{\overset{\pi}{\longrightarrow}}{\underset{s}{\longleftarrow}} M
\end{equation*}

with

  • $\pi(a, t) = a$
  • $s(a) = (a, 0)$
  • $\pi \circ s = \id_M$
  • $(s \circ \pi)(a, t) = (a, 0) \ne (a, t)$
  • $\pi \circ s = \id_M$

Therefore,

\begin{equation*}
\begin{split}
  \pi^*& : \Omega^k(M) \longrightarrow \Omega^k(M \times \mathbb{R}) \\
  s^* & : \Omega^k(M \times \mathbb{R}) \longrightarrow \Omega^k(M)
\end{split}
\end{equation*}

Then we have

\begin{equation*}
\pi \circ s = \id_M \quad \implies \quad s^* \circ \pi^* = \id_{\Omega^k(M)} \quad \implies s^* \pi^* [\alpha] = \big[ s^* \pi^* \alpha \big] = [\alpha]
\end{equation*}

which implies

\begin{equation*}
s^* \circ \pi^* = \id_{H_{dR}^k(M)
\end{equation*}

Now we show

\begin{equation*}
\pi^* \circ s^* = \id_{H_{dR}^k(M)
\end{equation*}

It's sufficient to show that on $\Omega^k(M \times \mathbb{R})$, $\exists K: \Omega^{l}(M \times \mathbb{R}) \to \Omega^{l - 1}(M \times \mathbb{R})$

\begin{equation*}
\pi^* \circ s^* = \id_{\Omega^k(M \times \mathbb{R})} \pm \big( K \dd{} \pm \dd{K} \big)
\end{equation*}

(such a $K$ is called a chain homotopy)

If we have such a $K$, then

\begin{equation*}
\begin{split}
  \pi^* \circ s^* [\alpha] &= \big[ \pi^* s^* \alpha \big] \\
  &= \big[ \alpha \pm (K \dd{} + \dd{K}) \alpha \big] \\
  &= \big[ \alpha \pm K \underbrace{\dd{\alpha}}_{= 0} \pm \underbrace{\dd{K \alpha}}_{\text{exact}} \big] \\
  &= [\alpha]
\end{split}
\end{equation*}

Construction of $K$: "integration along the fibre"

\begin{equation*}
K: \Omega^k(M \times \mathbb{R}) \longrightarrow \Omega^{k - 1}(M \times \mathbb{R})
\end{equation*}

Let

  • $\phi \in \Omega^k(M)$
  • $f \in C^{\infty}(M \times \mathbb{R})$
  • $\beta \in \Omega^{k - 1}(M)$

In $\Omega^k(M \times \mathbb{R})$ we have two kinds of forms:

  • $\pi^* \phi f$
  • $\pi^* \beta \wedge f \dd{t}$

So we define $K$ to

\begin{equation*}
\begin{split}
  K( \pi^* \phi f) &= 0 \\
  K ( \pi^* \beta \wedge f \dd{t}) &= \pi^* \beta \int_{0}^{t} f
\end{split}
\end{equation*}

where

\begin{equation*}
\int_{0}^{t} f = \int_{0}^{t} f(a, z) \dd{z}
\end{equation*}
\begin{equation*}
\begin{split}
  \big( \id - \pi^* \circ s^* \big) \big( \pi^* \phi f \big) &= \pi^* \phi f(a, t) - \pi^* s^* \big( \pi^* \phi f \big) \\
  &= \pi^* \phi f(a, t) - \pi^* \big( \underbrace{s^* \pi^*}_{= \id} \phi \underbrace{s^* f}_{= f \circ s = f(a, 0)} \big) \\
  &= \pi^* \phi f(a, t) - \pi^* \phi f(a, 0) \\
  &= \pi^* \phi \Big( f(a, t) - f(a, 0) \Big)
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
  \big( \dd{K} - K \dd{} \big) \big( \pi^* \phi f \big) &= - K \Bigg( \dd{\big( \pi^* \phi f \big)} + (-1)^k \pi^* \phi \wedge \bigg( \dd{f}_M + \pdv{f}{t} \dd{t} \bigg) \Bigg) \\
      &= - (-1)^k K \bigg( \pi^* \phi \wedge \pdv{f}{t} \dd{t} \bigg) \\
      &= (-1)^{k + 1} \pi^* \phi \underbrace{\int_{0}^{t} \pdv{f}{t}}_{\overset{FTC}{=} f(a, t) - f(a, 0)}
\end{split}
\end{equation*}

where on the LHS initially $\dd{K}$ vanish, since

\begin{equation*}
\dd{K} \pi^* \phi f = \dd{\big( \underbrace{K \pi^* \phi f}_{= 0} \big)} = 0
\end{equation*}

Upshot:

\begin{equation*}
\big( \id - \pi^* s^* \big) \big( \pi^* \phi f \big) = (-1)^{k + 1} \big( \dd{K} - K \dd{} \big) \big( \pi^* \phi f \big)
\end{equation*}

which is an identity for "forms of the form" $\pi^* \phi f$; now we need for $\pi^* \beta \wedge f \dd{t}$ also.

\begin{equation*}
\begin{split}
  \big( \id - \pi^* \circ s^* \big) \big( \pi^* \beta \wedge f \dd{t} \big) &= \pi^* \beta \wedge f \dd{t} - \pi^* \Big( s^* \pi^* \beta \wedge s^* (f \dd{t}) \Big) \\
  &= \pi^* \beta \wedge f \dd{t}
\end{split}
\end{equation*}

where we have used the fact that

\begin{equation*}
s^* (f \dd{t}) = \big( s^* f \big) +  \big( s^* \dd{t} \big) = \big( f \circ s \big) + \underbrace{\big( \dd{s^*(t)} \big)}_{= 0}
\end{equation*}
\begin{equation*}
\begin{split}
  \dd{K} \big( \pi^* \beta \wedge f \dd{t} \big) &= \dd{\bigg( \pi^* \beta \int_{0}^{t} f \bigg)} \\
  &= \pi^* \dd{\beta} \int_{0}^{t} f + (-1)^{k - 1} \pi^* \beta \bigg( \dd_M{\Big( \int_{0}^{t} f \Big)} + f(a, t) \dd{t} \bigg)
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
  K \dd{\big( \pi^* \beta \wedge f \dd{t} \big)} &= K \Big( \pi^* \dd{\beta} \wedge f \dd{t} + (-1)^{k - 1} \pi^* \beta \wedge \dd_M{f} \wedge \dd{t} \Big) \\
  &= \pi^* \dd{\beta} \int_{0}^{t} f + (-1)^{k - 1} \pi^* \beta \dd_M{\bigg( \int_{0}^{t} f \bigg)}
\end{split}
\end{equation*}

Comparing terms in $\dd{K} \big( \pi^* \beta \wedge f \dd{t} \big)$ and $K \dd{\big( \pi^* \beta \wedge f \dd{t} \big)}$, we see

\begin{equation*}
\big( \dd{K} - K \dd{} \big) \big( \pi^* \beta \wedge f \dd{t} \big) = (-1)^{k - 1} \pi^* \beta \wedge f \dd{t}
\end{equation*}

Upshot:

\begin{equation*}
\big( \id - \pi^* s^* \big) \big( \pi^* \beta \wedge f \dd{t} \big) = (-1)^{k - 1} \big( \dd{K} - K \dd{} \big) \big( \pi^* \beta \wedge f \dd{t} \big)
\end{equation*}

Thus we have such a $K: \Omega^k(M \times \mathbb{R}) \longrightarrow \Omega^{k - 1}(M \times \mathbb{R})$, and hence

\begin{equation*}
\pi^* \circ s^* = \id_{H_{dR}^k(M)}
\end{equation*}

which concludes our proof.

Lie Theory

Notation

  • $L$ often used as short-hand notation for the K-vectorspace $(L, +, \cdot)$ when this vector space is further equipped with the Lie brackets $[\![ \cdot, \cdot ]\!]$, i.e. when writing

    \begin{equation*}
(L, [\![ \cdot, \cdot ]\!])  
\end{equation*}

    $L$ refers the underlying K-vectorspace.

  • $L(G)$ is the set of left-invariant vector fields in the Lie group $G$
  • $\cong_\text{vec}$ refers to two vector spaces being isomorphic (which is not the same as, say, a group isomorphism)
  • $[\![ \cdot, \cdot ]\!]$ denotes the abstract Lie-brackets, i.e. any function which takes two arguments as satisfy the properties of a Lie bracket.
  • $\comm{\cdot}{\cdot}$ denotes the particular instance of a Lie bracket defined by

    \begin{equation*}
  \comm{a}{b} = ab - ba
\end{equation*}

    known as the commutation relation

  • $\pi_0$ refers to the 0th fundmantal group / path components
  • $\pi_1$ refers to the 1st fundamental group
  • $G^0$ denotes the connected component of the identity
  • Submanifold refers to embedded submanifold
  • $T^2 = S^1 \times S^1 = \mathbb{R}^2 / \mathbb{Z}^2$ denotes the 2-torus
  • $\text{Vect}(M)$ denotes the space of vector fields
  • $G \lcirclearrowright M$ denotes that $G$ acts on $M$ as a group
  • $\text{Lie}(G)$ denotes the Lie algebra generated by the Lie group $G$
  • $G$ denotes a Lie group, and $\mathfrak{g} = \Lie(G)$ denotes the corresponding Lie algebra
  • $\Hom_G(V, W)$ and $\Hom_{\mathfrak{g}}(V, W)$ denotes the space of morphisms of Lie group and Lie algebra reps
  • Sometimes you might see $g \cdot v$ or something for vector fields $v \in \mathrm{Vect}(G)$ (though I try to avoid it) which really means $\big( L_g \big)_* v$ and similarily $v \cdot g = \big( R_{g^{-1}} \big)_* v$

Stuff

A Lie group $(G, \bullet)$ is

  • a group with group operation $\bullet$:

    \begin{equation*}
  (C^\bullet) \ A^{\bullet} \ N^{\bullet} \ I^{\bullet}
\end{equation*}

    where $(C^\bullet)$ denotes that the group could be commutative, but is not necessarily so.

  • $G$ is a smooth manifold and the maps:

    \begin{equation*}
\begin{split}
  \mu: &G \times G \to G \\
  &(g_1, g_2) \mapsto g_1 \bullet g_2
\end{split}
\end{equation*}

    where $G \times G$ inherits smooth atlas from $G$, thus $\mu$ is a map between smooth manifolds.

    \begin{equation*}
\begin{split}
  i: &G \to G \\
  & g \mapsto g^{-1}
\end{split}
\end{equation*}

    are both smooth maps.

Let $(G, \bullet)$ be a Lie group.

Then for any $g \in G$, there exists a map

\begin{equation*}
\begin{split}
  \ell_g :& G \to G \\
  & h \mapsto \ell_g(h) := g \bullet h, \quad h \in G
\end{split}
\end{equation*}

called the left translation wrt. $g$.

Each left translation of a Lie group $\ell_g$ is an isomorphism but NOT a group isomorphism.

It is also a diffeomorphism on $G$, by the definition of a Lie group.

Let $G_1$ and $G_2$ be Lie groups

If $\varphi: G_1 \to G_2$ is a smooth / analytic map preserving group structure, i.e.

\begin{equation*}
\varphi(gh) = \varphi(g) \varphi(h), \quad \forall g, h \in G_1
\end{equation*}

then $\varphi$ is a morphism of Lie groups.

  1. Lie groups do not have to be connected, neither simply-connected
  2. Discrete groups are Lie groups

Let

  • $G$ be a Lie group
  • $G^0$ be the connected component (which always exists around identity) of $\text{id} \in G$
  • Then $G^0$ is a normal subgroup in $G$ and $G^{0}$ is a Lie group itself.
  • $G \setminus G^0$ is a discrete group
  1. First we show that $G^0$ is indeed a Lie group. By definition of a Lie group, the inversion map

    \begin{equation*}
\begin{split}
  i: \quad & G \to G \\
  & g \mapsto g^{-1}
\end{split}
\end{equation*}

    is continuous. The image of a connected topological space under a continous map is connected, hence $i$ takes $G^o$ to a connected comp of $G$ containing the identity, since

    \begin{equation*}
i(e) = e \implies i \big|_{G^0}: G^0 \to G^0
\end{equation*}

    Similar argument for $G^0 \times G^0 \to G^0$. Hence $G^0$ is a Lie group. At the same time, conjugation

    \begin{equation*}
h \mapsto g h g^{-1} 
\end{equation*}

    is cont. in $h$ for all $g \in G$. Thus $g G^0 g^{-1}$ is a conn. comp. of $G$ which contains $e \in G^0$ since $g e g^{-1} = e$.

    \begin{equation*}
\implies \quad g G^0 g^{-1} = G^0        , \quad \forall g \in G
\end{equation*}
  2. Let $q: G \to G / G^0$ be the quotient map. $q$ is an open map (i.e. maps open to open) since $G / G^0$ is equipped with the quotient topology. This implies that for every $g \in G$ we have

    \begin{equation*}
q(g G^0) = q(g) \in \mathcal{O}_{~}
\end{equation*}

    i.e. it's open. This implies that every element of $G / G^0$ is an open subset, hence the union of all elements in $G / G^0$ cover $G / G^0$ and each of them open, i.e. we have an open covering in which every open subset contains exactly one element of $G / G^0$ (which is the definition of a discrete topological group).

Let

FINISH IT!

  1. Show that $\tilde{G}$ is a Lie group. Let $M, N$ be a connected manifolds, $m \in M$, $n \in N$, and

    \begin{equation*}
\varphi: M \to N, \quad \varphi(m) = n
\end{equation*}

    be cont. $\tilde{M}, \tilde{N}$ be universal covers, $\tilde{m} \in \tilde{M}$ and $\tilde{n} \in \tilde{N}$ with

    \begin{equation*}
p(\tilde{m}) = m, \quad p(\tilde{n}) = n
\end{equation*}

    Then $\varphi$ lifts to $\tilde{\varphi}: \tilde{M} \to \tilde{N}$ s.t.

    \begin{equation*}
\tilde{\varphi}(\tilde{m}) = \tilde{n}
\end{equation*}

    Choose $\tilde{e} \in \tilde{G}$ s.t. $\rho(\tilde{e}) = e$ implies that $i: G \to G$ lifts in a unique way to $\tilde{i}: \tilde{G} \to \tilde{G}$ taking $\tilde{e} \mapsto \tilde{e}$. Same trick works for $\tilde{m}: \tilde{G} \times \tilde{G} \to \tilde{G}$.

  2. $\text{ker}(\rho)$ is discrete and central

Lie subgroups

A closed Lie subgroup $H$ of a Lie group $G$ is a (embedded) submanifold which is also a subgroup.

A Lie subgroup $H$ of a Lie group $G$ is an immersed (as opposed to embedded) submanifold which is also a subgroup.

  1. Any closed Lie subgroup is closed (as a submanifold)
  2. Any subgroup of a Lie group which is a closed subset is a closed Lie group.
  1. $G$ connected Lie group, $U$ neighborhood of $e \in G$, then $U$ generates $G$.
  2. $\varphi: G_1 \to G_2$ is a morphism of Lie groups, and $G_2$ is connected. If $\varphi_*: T_e G_1 \to T_e G_2$ is surjective, then $\varphi$ is surjective.
  1. $H$ subgroup generated by $U$, then $H$ is open in $G$ because $\forall h \in H$, we have $h \cdot U$ is open neighborhood of $h$ in $G$. Then
  2. Inverse function theorem says that $\varphi_*$ is surjective onto some neighborhood $U \ni e$, Since an image of a group morphism is a subgroup, and $U$ generates $G_2$, $\phi$ is surjective.
Example

$H = \mathbb{R}$ and $G = T^2 = \mathbb{R}^2 / \mathbb{Z}^2$ with

\begin{equation*}
\begin{split}
  f: \quad & \mathbb{R} \to T^2 \\
  & x \mapsto (x \mod \mathbb{Z}\ , \alpha x \mod \mathbb{Z}), \quad \alpha \in \mathbb{R} / \mathbb{Q}
\end{split}
\end{equation*}

Then it is well-known (apparently) that the image of this map is everywhere dense in $T^2$, and is often called the irrational or dense winding of $T^2$, and the map is open "one way" but the "other way".

This is an example of a Lie subgroup which is NOT a closed Lie subgroup. The image of the map $\mathbb{R} \to T^2$ is a Lie subgroup which is not closed. It can be shown that if a Lie subgroup is closed in $G$, then it is automatically a closed Lie subgroup. We do not get a proof of that though, apparently.

Factor groups

  • As in for discrete groups, given a closed Lie subgroup $H \subset G$, we can define notion of cosets and define $G / H$ as the set of equivalence classes.
  • Following theorem shows that the coset space is actually a manifold

Let

Then $G / H$ is a submanifold of $\dim(G / H) = n - k$ and there exists a fibre bundle with $G \overset{\rho}{\to} G / H$, where $\rho$ is the canonical map, with $H$ as it's fibre. The tangent space is given by

\begin{equation*}
\begin{split}
  T_e (G / H) &= (T_e G) / (T_e H) \\
  \overline{e} &= \rho(e)
\end{split}
\end{equation*}

Further, if $H$ is a normal closed Lie subgroup then $G / H$ has a canonical structure of a Lie group (i.e. transition maps are smooth and the smooth structure does not depend on the choice of $g$ and $M$ (see proof).

Let

  • $\rho: G \to G/ H$ be the canonical map
  • $g \in G$ and $\bar{g} = \rho(g) \in G / H$

Then $g \cdot H$ is a (embedded) submanifold in $G$ as it's an image of $H$ under diffeomorphism $x \mapsto g x$. Choose a submanifold $M \subseteq G$ such that $g \in M$ and $M$ is traversal to the manifold $g \cdot H$, i.e.

\begin{equation*}
T_g G = \big( T_g (g \cdot H) \big) \otimes T_g M
\end{equation*}

which implies that $\dim M = \dim G - \dim H$.

Let $U \subset M$ be a sufficently small neighborhood of $g$ in $M$. Then the set

\begin{equation*}
UH = \left\{ uh \mid u \in U, h \in H \right\}
\end{equation*}

is open in $G$. This follows from the IFT applied to the map $U \times H \to G$.

Consider $\bar{U} = \rho(U)$. Since $\rho^{-1}(\bar{U}) = UH$ is open, $\bar{U}$ is an open neighborhood of $\bar{g}$ in $G / H$ and the map $\rho \big|_U: U \to \bar{U}$ is a homeomorphism. This gives a local chart for $G / H$ by $\big( \bar{U}, \varphi \circ (\rho \big|_{U})^{-1} \big)$, where $\varphi$ denotes a chart map for $G$. At the same time this shows that $G \overset{\rho}{\to} G /H$ is a fibre bundle with fibre $H$.

GET CONFIRMATION ABOUT THIS. With the the atlas $\left\{ \big( \bar{U}_{\alpha}, \varphi_{\alpha} \circ (\rho \big|_{U})^{-1} \big) \right\}$ we see that the transition maps are smooth by the smoothness of $\rho$ and $\varphi_{\alpha}$. Further, observe that choosing any other $g$ and $M$ does not alter the proof, since $\dim M = \dim G - \dim H$ still holds, and therefore ???

The above argument also shows that the push-forward of $\rho$, i.e. $\rho_*: T_g G \to T_{\bar{g}}(G / H)$ has the kernel

\begin{equation*}
\text{ker}(\rho_*) = T_g (g H)
\end{equation*}

In particular, $g = e$ gives the isomorphism (since $\rho_*$ is an isomorphism)

\begin{equation*}
T_{\bar{e}} (G / H) = T_e G / T_1 H
\end{equation*}

as wanted.

REMINDER: If $G \overset{\rho}{\to} G / H$ is a fibre bundle with fibre $H$, then there exists a long exact sequence of homotopy groups

\begin{equation*}
\dots \longrightarrow \pi_2(G / H) \longrightarrow \pi_1(G) \longrightarrow\pi_1(G / H) \longrightarrow\pi_0(H) \longrightarrow\pi_0(G) \longrightarrow \pi_0(G / H) \to 0
\end{equation*}

Exact means that $\overset{\beta}{\longrightarrow} \circ \overset{\alpha}{\longrightarrow}$ with $\text{Im}(\beta) = \text{ker}(\alpha)$.

Let $H$ be a closed Lie subgroup of a Lie group $G$.

  1. $H$ connected $\implies$ $\pi_0(G) = \pi_0(G / H)$ where $\left| \pi_0(G) \right| = \text{number of components}$. In paricular, if $H$ and $G / H$ are both connected, then so is $G$.
  2. $H$ connected, $G$ connected $\implies$

    \begin{equation*}
\dots \longrightarrow \pi_2(G / H) \longrightarrow \pi_1(H) \longrightarrow \pi_1(G) \longrightarrow\pi_1(G / H) \longrightarrow \left\{ e \right\}
\end{equation*}

Push-forward on fields

What does this mean? It means that on a Lie group we can in fact construct a diffeomorphism using the left translations, and thus a push-forward from $G$ to $G$, i.e. we can map vector fields to vector fields in the group $G$!

We can push forward on vector field $X$ on $G$ to another vector field, defined

\begin{equation*}
\big( \ell_{g * } X \big) := \ell_{g * } \big( X_h \big)
\end{equation*}

where $X_h \in T_h G$, $h \in G$.

Let $(G, \bullet)$ be a Lie group, and $X$ a vector field on $G$, then $X$ is called left invariant vector field, if for any $g \in G$

\begin{equation*}
\ell_{g * } X = X
\end{equation*}

Alternatively, one can write this as

\begin{equation*}
\forall h \in G: \quad \ell_{g * } \big( X_h \big) := X_{gh}
\end{equation*}

where we write the map pointwise on the vector field.

Alternatively, again, $\forall h \in G$ and $\forall f \in \mathcal{C}^\infty(G)$

\begin{equation*}
\begin{split}
  \big( \ell_{g * } X_h \big) f &= X_{gh} f \\
  X_h(f \circ \ell_g) &= \big( X f \big) (gh) \\
  \big[ X \big( f \circ \ell_g \big) \big](h) &= \big[ \big( Xf \big) \circ \ell_g \big] (h) \\
  X \big( f \circ \ell_g \big) &= \big( Xf \big) \circ \ell_g
\end{split}
\end{equation*}

Similarily we can define right invariant and bi-invariant (both left- and right-invariant).

The set of left-invariant vector fields of a Lie group $G$ can be denoted

\begin{equation*}
L (G) \subset \Gamma(TG)
\end{equation*}

where $\Gamma(TG)$ is a $\mathcal{C}^\infty(G)$ module.

We observe that the following is true:

\begin{equation*}
\begin{split}
  + &: L(G) \times L(G) \to L(G) \\
  \cdot &: \mathcal{C}^\infty(G) \times L(G) \to L(G) \\
\end{split}
\end{equation*}

which implies that $L(G)$ is a $\mathcal{C}^\infty(G)$ module.

Example: $G$ acting on space of vector fields $\text{Vect}(M)$

$G \lcirclearrowright M$ yields $G$ acts on $\text{Vect}(M)$

\begin{equation*}
\begin{split}
  g: \quad & M \to M \\
  & g^{-1} \cdot m \mapsto m
\end{split}
\end{equation*}

Then push-forward

\begin{equation*}
g_* = \dd{}_{g^{-1} \cdot m} g: T_{g^{-1} \cdot m} M \to T_m M
\end{equation*}

If $g \in G$ and $v \in \text{Vect}(M)$, we let

\begin{equation*}
\big( gv \big)(m) = \big( g_* \big) v \big( g^{-1} \cdot m \big)
\end{equation*}

If $g \cdot m = m$ $\forall g \in G$ then $G$ acts on $T_m M$.

Example: $G$ acting on dual space of $\text{Vect}(M)$

Let $\big( T_m M \big)^*$ be a dual space to $T_m M$, i.e. cotangent space.

\begin{equation*}
T^*M = \coprod_{m \in M} T_m^* M
\end{equation*}

is again a vector bundle called the cotangent bundle.

\begin{equation*}
\bigwedge^K T^* M = \coprod_{m \in M} \bigwedge^K \big( T_m^* M \big)
\end{equation*}

is a vector bundle as well and it sections called k-forms.

$G \lcirclearrowright M$ yields that $\bigwedge^k T^*M$ is a $G \text{-rep}$.

Lie Algebra

A one-parameter subgroup corresponds to an map $x \in \mathfrak{g} = T_e G$ is a morphism of Lie groups

\begin{equation*}
\gamma_x: \mathbb{K} \to G \quad \text{s.t.} \quad \dot{\gamma}_x(0) = x
\end{equation*}

$\forall m \in M$ $\exists \varepsilon > 0$ s.t. integral curve $\gamma^m(t) = \big( - \varepsilon, \varepsilon \big) \to M$ with initial condition $\gamma^m(0) = m$ exists and is unique.

Moreover, the map (called the flow)

\begin{equation*}
\begin{split}
  \Phi: \quad & (-\varepsilon, \varepsilon) \times M \to M \\
  & (t, m) \mapsto \Phi^t(m) := \gamma^m(t)
\end{split}
\end{equation*}

is smooth.

$\forall A \in \mathfrak{g}$ there exists a unique one-parameter subgroup corresponding to $A$.

Choose $A \in T_e G =: \mathfrak{g}$ and let $X^A$ be the corresponding left-invariant vector field, $\Phi^t: M \to M$ be the time $t$ flow of $X^A$. Then

\begin{equation*}
\begin{split}
  \Phi^t(g_1 \cdot g_2) &= g_1 \Phi^t(g_2) \\
  \Phi^t(g_2) &= \gamma^{g_2}(t)
\end{split}
\end{equation*}

satsifies

\begin{equation*}
\dv{t} \gamma^{g_2}(t) = X^A\big(\gamma^{g_2}(t) \big) \quad \text{and} \quad \gamma^{g_2}(0) = g_2
\end{equation*}

Furthermore, $g_1 \Phi^t(g_2)$ satisfies

\begin{equation*}
\dv{}{t} \Big( g_1 \cdot \Phi^t(g_2) \Big) = \big( L g_1 \big)_{*} X^A \big( \gamma^{g_2}(t) \big)
\end{equation*}

by the chain rule. Then $g_1\gamma^{g_2}(0) = g_1 g_2$.

In other words, the RHS and LHS satisfy the same diff. eqn. but solutions are unique by thm:uniqueness-of-ODE-solution-and-flows.

For it to be a morphism, we observe that it preserves the group operation. Letting $\gamma_A(t) = \Phi^t(e)$, we have

\begin{equation*}
\begin{split}
  \gamma_A(t + s) &= \Phi^{t + s}(e) \\
  &= \Phi^{t} \big( \Phi^s(e) \big) \\
  &= \Phi^t \big( \gamma_A(s) \cdot e \big) \\
  &= \gamma_A(s) \Phi^t(e) \\
  &= \gamma_A(s) \gamma_A(t)
\end{split}
\end{equation*}

For $\mathbb{K} = \mathbb{R}$, let $\gamma: \mathbb{R} \to G$ be a one-param. subgroup..

Identify $T_0 \mathbb{R} \cong \mathbb{T}_t \mathbb{R}$, then $\dot{\gamma}(t): T_t \mathbb{R} \to T_{\gamma(t)} G$ is a composition

\begin{equation*}
\dot{\gamma}(t) = \big( L_{\gamma(t)} \big)_* \dot{\gamma}(0)
\end{equation*}

Therefore $\gamma(t)$ is an integral curve for a left-invariant vector field corresponding to $\dot{\gamma}(0) \in T_eG$.

Hence, $\gamma(t)$ is unique.

An abstract Lie algebra $(L, +, \cdot, [\![ \cdot, \cdot ]\!])$ is a K-vectorspace $(L, +, \cdot)$ equipped with an abstract lie bracket $[\![ \cdot, \cdot ]\!]$ that satisfies:

  • bilinear (in $K$): $[\![ \cdot, \cdot ]\!]: L \times L \to L$
  • anti-symmetric: $[\![ x, y ]\!] = - [\![ x, y ]\!]$
  • Jacobi identity:

    \begin{equation*}
 [\![ x, [\![ y, z ]\!] ]\!] + [\![ y, [\![ z, x ]\!] ]\!] + [\![ z, [\![ x, y ]\!] ]\!] = 0
\end{equation*}

One might wonder why we bother with these weird brackets, or Lie algebras at all; as we'll see, there is a correspondance between Lie groups, which are geometrical objects, and these Lie algebras, which are linear objects.

\begin{equation*}
L(G) \cong_\text{vec} T_e G, \quad e \text{ is the identity in } G
\end{equation*}

Equivalently, one could do the same as in Thm. thm:left-invariant-vector-fields-isomorphic-to-tangent-identity for right-invariant vector fields $R(G)$, thus making $R(G) \cong_{\text{vec}} L(G)$.

Further, for the bi-invariant vector fields defined

\begin{equation*}
T_e G^{\Ad_{G}} := \left\{ X \in T_e G \mid \Ad_{g}(X) = X, \forall g \in G \right\}
\end{equation*}

we also have being isomorphic to the left- and right-invariant vector fields:

\begin{equation*}
T_e G^{\Ad_{G}} \cong_{\text{vec}} L(G) \cong_{\text{vec}} R(G)
\end{equation*}

We start with the following corollary

\begin{equation*}
\begin{split}
  \dim L(G) &= \dim (T_e G) \\
  &= \dim (G)
\end{split}
\end{equation*}

We need to construct a linear isomorphism

\begin{equation*}
\begin{split}
  j : T_e G &\overset{\sim}{\to} L(G) \\
  A \mapsto j(A)
\end{split}
\end{equation*}

where

\begin{equation*}
j(A)_g := \ell_{g^*} A \in T_{ge} G = T_g G \quad \forall g \in G
\end{equation*}

where $j(A)_g$ denotes at the point $g$.

That is, we push forward the vector-field at the point $e$ for every point $g \in G$, thus creating a vector at every point.

  1. We now prove that it's left-invariant vector field:

    \begin{equation*}
\begin{split}
  \ell_{h^*} \big( j(A)_g \big) &= \ell_{h^*} \big( \ell_{g^*} A \big) \\
  &= \ell_{(hg)^*} A = j(A)_{hg}
\end{split}
\end{equation*}
  2. It's clearly linear, since $j(A)_g := \ell_{g^*} A$ and the push-forward on a vector-field at the point $g$, $\ell_{g^*}$, is by definition linear.
  3. $j$ is injective:

    \begin{equation*}
\begin{split}
  j(A) = j(B) \quad \iff \quad \forall g \in G: j(A)_g = j(B)_g
\end{split}
\end{equation*}

    which can be seen from

    \begin{equation*}
 j(A)_e = j(B)_e \quad \iff \quad \ell_{0^*} A = \ell_{0^*} B \quad \iff \quad A = B
\end{equation*}
  4. $j$ is surjective: Let $X \in L(G)$, i.e. $X$ is a left-invariant vector field. Then we let $A^X$ be some vector field $A^X$ associated with $X$, defined by

    \begin{equation*}
 A^X = X_e \in T_e G
\end{equation*}

    Consider:

    \begin{equation*}
\begin{split}
  j(A^X)_g &= \ell_{g^*} A^X \\
  &= \ell_{g^*}(X_e) = X_{ge} = X_g
\end{split}
\end{equation*}

    which implies $j(A^X) = X$

Hence, as claimed,

\begin{equation*}
T_e G \cong_\text{vec} L(G)
\end{equation*}

Which means that as a vector space we can work with $T_e G$ to prove properties of the vector-field of left-invariant vector field!

Only problem is that we do not have the same algebra, i.e.

\begin{equation*}
\Big( L(G), \comm{\cdot}{\cdot} \Big) \subseteq \Big( \Gamma(TG) , \comm{\cdot}{\cdot} \Big)
\end{equation*}

and we would really like for the following to be the case

\begin{equation*}
\Big( T_e G, [\![ \cdot, \cdot ]\!] \Big) \cong_\text{alg} \Big( L(G), \comm{\cdot}{\cdot} \Big)
\end{equation*}

that is, we want some bilinear map $[\![ \cdot, \cdot ]\!]:  T_e G \times T_e G \to T_e G$ s.t.

\begin{equation*}
j \Big( \textcolor{red}{[\![ A, B ]\!]} \Big) = \comm{j(\textcolor{red}{A})}{j(\textcolor{red}{B})}
\end{equation*}

Thus, we simply define the commutation brackets on $T_e G$, $[\![ \cdot, \cdot ]\!]$ such that

\begin{equation*}
\underbrace{\textcolor{red}{[\![ A, B ]\!]}}_{\in T_e G} = \underbrace{j^{-1} \bigg( \comm{j(\textcolor{red}{A})}{j(\textcolor{red}{B})} \bigg)}_{\in L(G)}
\end{equation*}

as desired we get

\begin{equation*}
\Big( T_e G, [\![ \cdot, \cdot ]\!] \Big) \cong_\text{alg} \Big( L(G), \comm{\cdot}{\cdot} \Big)
\end{equation*}
Example of Lie Algebra

Let $L = \Gamma(TM)$ is a $\mathbb{R}$ vectorspace. Then

\begin{equation*}
 \Big( \Gamma(TM), +, \cdot, [\![ \cdot,  \cdot]\!] \Big)
\end{equation*}

is an infinite-dimensional (abstract) Lie algebra.

Examples of Lie groups

Unit circle

\begin{equation*}
G := \{ z \in \mathbb{C} \mid |z| = 1 \} = S^1
\end{equation*}

where we let the group operation $\bullet = \cdot_\mathbb{C}$, i.e. multiplication in $\mathbb{C}$.

Whenever we multiply a two complex numbers which are both unit length, then we still end up on the unit-circle.

General linear group

\begin{equation*}
G := \{ \varphi: \mathbb{R}^n \overset{\sim}{\to} \mathbb{R} \mid \det \varphi \ne 0 \} = \text{GL}(n, \mathbb{R})
\end{equation*}

equipped with the $\circ$ operation, i.e. composition. Due to the nature of linear maps, this group is clearly satisfies $A N I$ (but not $C$), hence it's a Lie group.

Why is GL a manifold?

$\text{GL}(, \mathbb{R})$ can be represented in $\mathbb{R}^{n^2}$ (as matrices), and due to the $\det \varphi \ne 0$ the set is also open.

Thus we have an open set of on $\mathbb{R}^{n^2}$ which we can represent as

Relativistic Spin Group

In the definition of the relativistic spin groups we make use of the very useful method for constructing a topology over some set by inheriting a topology from some other space.

  1. Define topology on the "components" of the larger set
  2. Take product topology
  3. Take induced subset-topology
Proof / derivation

To define the relativistic spin group $\mathrm{SL}(2, \mathbb{C})$ we start with the set

\begin{equation*}
\mathrm{SL}(2, \mathbb{C}) = \left\{ 
  \begin{pmatrix}
    a & b \\
    c & d
  \end{pmatrix}
  \in \mathbb{C}^4 \ \bigg| \ ad - bc = 1
 \right\}
 \subseteq \mathbb{C}^4
\end{equation*}
  • As a group

    We make $\mathrm{SL}(2, \mathbb{C})$ into a group $\Big( \mathrm{SL}(2, \mathbb{C}), \cdot \Big)$:

    \begin{equation*}
\begin{split}
  \bullet: \quad & \mathrm{SL}(2, \mathbb{C}) \to \mathrm{SL}(2, \mathbb{C}) \\
  & \begin{pmatrix} a & b \\ c & d \end{pmatrix}
  \bullet
  \begin{pmatrix} e & f \\ g & h \end{pmatrix}
   :=
  \begin{pmatrix}
    ae + bg & af + bh \\
    ce + dg & cf + dh
  \end{pmatrix}
\end{split}
\end{equation*}

    i.e. matrix multiplication, which we know is ANI (but not commutative):

    • Associative
    • Exists neutral element
    • Invertible (since we recognize $\det( \cdot ) = 1$)
  • As a topological space

    From this group, we can create a topological space $\Big( \mathrm{SL}(2, \mathbb{C}), \mathcal{O} \Big)$:

    1. Define topology $\mathcal{O}_{\mathbb{C}}$ on $\mathbb{C}$ by virtue of defining "open balls":

      \begin{equation*}
B_r(z) := \left\{ y \in \mathbb{C} : |y - z| < r \right\}
\end{equation*}

      which is the same as we do for the standard topology in $\mathbb{R}^d$.

    2. Take the product topology:

      \begin{equation*}
\mathcal{O}_{\mathbb{C} \times \mathbb{C} \times \dots \times \mathbb{C}}
\end{equation*}
    3. Equip $\mathrm{SL}(2, \mathbb{C}) \subset \mathbb{C}^4$ with the induced subset topology of the product topology over $\mathbb{C}^4$, i.e.

      \begin{equation*}
\mathcal{O} := \mathcal{O}_{\mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C}}\big|_{\mathrm{SL}(2, \mathbb{C})} := \left\{ U \cap \mathrm{SL}(2, \mathbb{C}) \mid U \in \mathcal{O}_{\mathbb{C}^4} \right\}
\end{equation*}

    Verify that we have $\Big( \mathrm{SL}(2, \mathbb{C}), \mathcal{O} \Big)$, with $\mathcal{O}$ as given above, is a topological manifold. We do this by explicitly constructing the charts which together fully covers $\mathrm{SL}(2, \mathbb{C})$, i.e. defines an atlas of $\mathrm{SL}(2, \mathbb{C})$:

    1. First chart $(U, x)$:

      \begin{equation*}
U := \left\{ 
\begin{pmatrix}
  a & b \\ c & d
\end{pmatrix}
\in \mathrm{SL}(2, \mathbb{C})
\ \bigg| \ \\
a \ne 0
\right\}
\end{equation*}

      and the map

      \begin{equation*}
\begin{split}
  x: \quad & U \to x(U) \subseteq \mathbb{C}^3 \\
  & \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto (a, b, c)
\end{split}
\end{equation*}

      which is continuous and invertible, with the inverse:

      \begin{equation*}
\begin{split}
  x^{-1}: \quad & x(U) \to U \\
  & (a, b, c) \mapsto \begin{pmatrix} a & b \\ c & \frac{1 + bc}{a} \end{pmatrix}
\end{split}
\end{equation*}

      hence $x: U \to x(U)$ is a homeomorphism, and thus $(U, x)$ is a coordinate chart of $\mathrm{SL}(2, \mathbb{C})$.

    2. Second chart $(V, y)$:

      \begin{equation*}
V := \left\{ 
\begin{pmatrix}
  a & b \\ c & d
\end{pmatrix}
\in \mathrm{SL}(2, \mathbb{C})
\ \bigg| \ \\
d \ne 0
\right\}
\end{equation*}

      and the map

      \begin{equation*}
\begin{split}
  y: \quad & U \to x(U) \subseteq \mathbb{C}^3 \\
  & \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto (b, c, d)
\end{split}
\end{equation*}

      which is continuous and invertible, with the inverse:

      \begin{equation*}
\begin{split}
  y^{-1}: \quad & y(V) \to V \\
  & (b, c, d) \mapsto \begin{pmatrix} \frac{1 + bc}{d} & b \\ c & d \end{pmatrix}
\end{split}
\end{equation*}
    3. Third chart $(W, z)$:

      \begin{equation*}
W := \left\{ 
\begin{pmatrix}
  a & b \\ c & d
\end{pmatrix}
\in \mathrm{SL}(2, \mathbb{C})
\ \bigg| \ \\
b \ne 0
\right\}
\end{equation*}

      and the map

      \begin{equation*}
\begin{split}
  z: \quad & W \to z(W) \subseteq \mathbb{C}^3 \\
  & \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto (a, b, d)
\end{split}
\end{equation*}

      which is continuous and invertible, with the inverse:

      \begin{equation*}
\begin{split}
  z^{-1}: \quad & z(W) \to W \\
  & (a, b, d) \mapsto \begin{pmatrix} a & b \\ \frac{ad - 1}{b} & d \end{pmatrix}
\end{split}
\end{equation*}

    Then we have an atlast in $\{ (U, x), (V, y), (W, z) \}$, since these cover all of $\mathrm{SL}(2, \mathbb{C})$, since the only case we're missing is when all $a = b = c = d = 0$, which does not have determinant $1$, therefore is not in $\mathrm{SL}(2, \mathbb{C})$. Hence, $\Big( \mathrm{SL}(2, \mathbb{C}), \mathcal{O} \Big)$ is a complex topological manifold, with

    \begin{equation*}
\dim \mathrm{SL}(2, \mathbb{C}) = 3 
\end{equation*}
  • As a differentiable manifold

    Now we need to check if $\Big( \mathrm{SL}(2, \mathbb{C}), \mathcal{A} \Big)$ is differentiable manifold (specifically, a $\mathbb{C} \text{-differentiable}$); that is, we need the transition maps to be "$C^?$ compatible", where ? specifies the order of differentiability.

    One can show that the atlast $\mathcal{A}$ defined above is differentiable to arbitrary degree. We therefore let $\mathcal{A}$ be the maximal atlast with differentiability to arbitrary degree, containing the atlast we constructed above. This is just to ensure that in the case we late realize we need some other chart with these properties, then we don't have to redefine our atlas to also contain this new chart. By using the maximum atlas, we're implicitly including all these possible charts, which is convenient.

    One can show that $U$ above defines open subsets, by observing that the subset where $a = 0$ is closed, hence the complement ($a \ne 0$) is open.

  • As a Lie group

    As seen above, we have the group $\Big( \mathrm{SL}(2, \mathbb{C}), \bullet \Big)$, where $\mathrm{SL}(2, \mathbb{C})$ is a $\mathbb{C} \text{-differentiable}$ manifold to arbitrary degree, with the maximum atlas $\mathcal{A}$ containing $\{ (U, x), (W, z) \}$ as defined previously.

    To prove that this is indeed a Lie group, we need to show that both the following maps:

    \begin{equation*}
\begin{split}
  \mu: \quad & \mathrm{SL}(2, \mathbb{C}) \times \mathrm{SL}(2, \mathbb{C}) \to \mathrm{SL}(2, \mathbb{C}) \\
  & \mu \left( \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \begin{pmatrix} e & f \\ g & h \end{pmatrix} \right) :=
  \begin{pmatrix}
    a & b \\ c & d
  \end{pmatrix}
  \bullet
  \begin{pmatrix}
    e & f \\ g & h
  \end{pmatrix}
\end{split}
\end{equation*}

    and

    \begin{equation*}
\begin{split}
  i: \quad & \mathrm{SL}(2, \mathbb{C}) \to \mathrm{SL}(2, \mathbb{C}) \\
  & i \begin{pmatrix} a & b \\ c & d \end{pmatrix} := \frac{1}{ad - bc} \begin{pmatrix} d & - b \\ -c & a \end{pmatrix}
\end{split}
\end{equation*}

    are both smooth.

    Differentiability is a rather strong notion in the complex case, and so one needs to be careful in checking this. Nonetheless, we can check this to the arbitrary degree.

    relativistic-spin-group-smoothness-of-inverse-map.png

    We observe that the Fig. fig:commutation-diagram-inverse-relativistic-spin-group-charts is the case, since the inverse map restricted to $U$, where $a \ne 0$, is then mapped to $d \ne 0$, i.e. the image is $V$.

    We cannot talk about differentiablility on the manifold itself, hence we say $i$ is differentiable if and only if the map $y \circ i \circ x^{-1}$ is differentiable (since we already know $x$ and $y$ are differentiable). We observe that

    \begin{equation*}
(y \circ i \circ x^{-1}) \left( 
\begin{pmatrix}
  a & b \\ c & d
\end{pmatrix}
\right) =
(y \circ i) \left(
\begin{pmatrix}
  a & b \\ c & \frac{1 + bc}{a}
\end{pmatrix}
\right) = 
y \left(
\begin{pmatrix}
  \frac{1 + bc}{a} & -b \\ -c & a
\end{pmatrix}
\right) =
\big( -b, -c, a \big)
\end{equation*}

    which is most certainly a differentiable map. We've used the fact that all these matrices have $\det (\cdot) =1$ in the inverse above.

    Performing the same verification for the other charts, we find the same behavior. Hence, we say that $i$, on the manifold-level, is differentiable.

    For $\mu$ we can simply let the product-space $\mathrm{SL}(2, \mathbb{C}) \times \mathrm{SL}(2, \mathbb{C})$ inherit the smooth atlast $\mathcal{A}$ on $\mathrm{SL}(2, \mathbb{C})$, hence $\mu$ is also smooth.

    That is, the composition map $\mu$ and the inverse map $i$ are both smooth for the group $\Big( \mathrm{SL}(2, \mathbb{C}), \bullet \Big)$, hence $\Big( \mathrm{SL}(2, \mathbb{C}), \bullet \Big)$ is a complex 3-dimensional Lie group!

  • TODO As a Lie algebra

    In this section our aim is to construct the Lie algebra $L \big( \mathrm{SL}(2, \mathbb{C}) \big)$ of the Lie group $\Big( \mathrm{SL}(2, \mathbb{C}), \bullet \Big)$.

    We will use the standard notation of

    \begin{equation*}
\mathfrak{sl} \big( 2, \mathbb{C} \big) = L \big( \mathrm{SL}(2, \mathbb{C}) \big)
\end{equation*}

    Recall

    \begin{equation*}
L(G) := \left\{ \underbrace{X \in \Gamma(TG)}_{\text{vector field on} G} \ \bigg| \  g \in G: \ \Big( \ell_{g^*} \big( X_h \big) \Big)_{gh} = X_{gh}, \forall h \in G \right\}
\end{equation*}

    i.e. it's the set of left-invariant vector fields.

    Further, recall

    \begin{equation*}
T_{\text{id}}
\mathrm{SL}(2, \mathbb{C})
\cong_{\text{vec}}
L \Big( \mathrm{SL}(2, \mathbb{C}) \Big)
\end{equation*}

    where

    \begin{equation*}
\text{id} = 
\begin{pmatrix}
  1 & 0 \\ 0 & 1
\end{pmatrix}
\end{equation*}

    Now, we need to equip $\mathrm{SL}(2, \mathbb{C})$ with the Lie brackets:

    \begin{equation*}
\begin{split}
  [\![ \cdot, \cdot ]\!]: \quad & T_{\mathrm{id}} \ \mathrm{SL}(2, \mathbb{C}) \times T_{\mathrm{id}} \ \mathrm{SL}(2, \mathbb{C}) \to T_{\mathrm{id}} \ \mathrm{SL}(2, \mathbb{C}) \\
  & (A, B) \mapsto [\![ A, B ]\!] := \ell_{(g^{-1})^*} \comm{\ell_{g^*} A}{\ell_{g^*} B}
\end{split}
\end{equation*}

    where

    \begin{equation*}
g = 
\begin{pmatrix}
  a & b \\ c & d
\end{pmatrix}
\end{equation*}

    To explicitly write out this $[\![ \cdot, \cdot ]\!]$, we use the chart $(U, x)$ since $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \in U$. For any $f \in \mathcal{C}^{\infty} \Big( \mathrm{SL}(2, \mathbb{C}), \mathbb{C} \Big)$, we have

    \begin{equation*}
\Bigg( \ell_{g^*} \bigg( \frac{\partial }{\partial x^i} \bigg)_{\mathrm{id}} \Bigg)_{g} f 
\end{equation*}

$\mathrm{SU}(2)$

\begin{equation*}
\begin{split}
  \mathrm{SU}(2) &= \left\{ A \in \mathrm{GL}( \mathbb{C}, 2) \mid A \bar{A}^T = \text{id}, \ \det A = 1 \right\}\\
  &= \left\{ \begin{pmatrix} \alpha & \beta \\ - \bar{\beta} & \bar{\alpha} \end{pmatrix} : \alpha, \beta \in \mathbb{C}, \quad \left| \alpha^2 \right| + \left| \beta \right|^2 = 1 \right\}
\end{split}
\end{equation*}

Observe that if we write

\begin{equation*}
\alpha = x_1 + i x_2 \quad \text{and} \quad \beta = x_3 + i x_4
\end{equation*}

then we observe that $\mathrm{SU}(2)$ is diffeomorphic to $S^3 = \left\{ x_1^2 + x_4^2 = 1 \right\} \subset \mathbb{R}^4$.

Classification of Lie Algebras

Notation

Similar stuff

If a representation $V$ is reducible, then it has a non-trivial subrepresentation $W$ and thus, $V$ can be included in a short exact sequence

\begin{equation*}
0 \longrightarrow W \overset{i}{\longrightarrow} V \overset{\mathrm{can}}{\longrightarrow} V / W \longrightarrow 0
\end{equation*}

where $i: W \to V$ is the inclusion mapping, and $\can: V \to V / W$ is the canonical mapping.

Apparently, the natural question is whether the above sequence splits, i.e. whether

\begin{equation*}
V \simeq W \oplus V / W
\end{equation*}

If so, one can iterate the process to decompose $V$ into a direct sums of irreducibles.

Example of non-semisimple representation

Let $G = \mathbb{R}$ and so $\mathfrak{g} = \mathbb{R}$.

Note: I find the below confusing. I think a better approach is to note that $\mathbb{R}$ is commutative and so the rep must also be commutative. The only matrices which are commutative are the diagonal matrices, hence we have full reducibility.

The a representation of $\mathfrak{g}$ is the same as a vector space $V$ with a linear map $\mathbb{R} \to \End(V)$. Every such map is of the form

\begin{equation*}
t \mapsto t A
\end{equation*}

for some arbitrary $A \in \End(V)$.

The corresponding representation of the group $\mathbb{R}$ is given by

\begin{equation*}
t \mapsto \exp(t A)
\end{equation*}

Thus, classiying representations of $\mathbb{R}$ is equivalent to classifying linear maps $V \to V$ up to a change of basis.

Such a classification is known (it's the Jordan normal form!), but is non-trivial.

Now consider the case where $v$ is an eigenvector of $A$, then the one-dimensional space $\mathbb{C} v \subset V$ is invariant under $A$, and thus is a subrerepresentation in $V$. Since every linear operator in a complex vector space has an eigenvector, this shows that every representation of $\mathbb{R}$ is reducible, unless it is one-dimensional. We just saw above that the one-dimensional rep is indeed irreducible, hence the only irreducible representations of $\mathbb{R}$ are one-dimensional.

One can see that writing a representation given by

\begin{equation*}
t \mapsto \exp(t A)
\end{equation*}

as a direct sum of irreducible representations is equivalent to diagonalizing $A$; recall that the direct sum of two vector subspaces $U$ and $W$ is the sum of the subspaces in which $U$ and $W$ only have $\left\{ 0 \right\}$ in common, and the direct sum of matrices is

\begin{equation*}
\bigoplus_{i = 1}^n A_i = \diag \big( A_1, \dots, A_n \big) = 
   \left(
   \begin{array}{c|c|c|c}
     A_1 & 0 & 0 & 0 \\
     \hline
     0 & A_2 & 0 & 0 \\
     \hline
     0 & 0 & \ddots & 0 \\
     \hline
     0 & 0 & 0 & A_n
   \end{array}
   \right)
\end{equation*}

Therefore, a representation is completely reducible if and only if $A$ is diagonalizable. Since not every linear operator is diagonalizable, not every representation is completely reducible.

More stuff

Let $\rho: G \to \GL(V)$ be a representation of $G$ (respectively $\mathfrak{g}$) and $A: V \to V$ be a diagonalizable intertwining operator.

Then each $\lambda \text{-eigenspace}$ $V_{\lambda}$ is a subrepresentation (of Lie group), so

\begin{equation*}
V = \otimes_{\lambda} V_{\lambda}
\end{equation*}

In particular, for any $z \in Z(G)$ (the center of $G$) such that $\rho(z)$ is diagonalizable, $V$ decomposes into a direct sum of $\rho(z) \text{-eigenspaces}$.

Stuff

Every finite-dim. complex Lie algebra $(L, \comm{\cdot}{\cdot})$ can be decomposed as

\begin{equation*}
L = R \oplus_{s, \text{Lie}} \big( L_1 \oplus_\text{Lie} \dots \oplus_\text{Lie} L_n \big)
\end{equation*}

where:

  1. $a$ is a Lie sub-algebra of $L$ which is solvable , i.e.:

    \begin{equation*}
\{ \comm{r_1}{r_2} \mid r_1, r_2 \in R \} = \comm{R}{R} \supseteq \comm{\comm{R}{R}}{\comm{R}{R}} \supseteq [ \ \cdot \ , \ \cdot \ ] \supset \dots \overset{!}{=} \{ 0 \}
\end{equation*}
  2. $L_1, \dots, L_n$ are simple Lie algeabras, i.e.
    • $L_i$ is non-abelian
    • $_i$ contains no non-trivial ideals, where:
      • An ideal means some sub-vector space $I \subseteq L$ s.t. $\comm{I}{L}  \subseteq I$, i.e. if you bracket the sub vector spaces from the outside, then you're still in the ideal $I$.
      • $\{ 0 \}$ is clearly ideal since $\comm{\{ 0 \}}{L} = \{ 0 \}$
  3. The direct sum $\oplus_\text{Lie}$ between Lie algebras is defined as:

    \begin{equation*}
 \begin{split}
   L_1 \oplus_\text{Lie} L_2 & := L_1 \oplus L_2 \\
   \text{plus the condition: } & \comm{L_1}{L_2} = 0
 \end{split}
\end{equation*}
  4. semi-direct sum $\oplus_{s, \text{Lie}}$

    \begin{equation*}
 \begin{split}
   R \oplus_{s, \text{Lie}} L_1 & := R \oplus L_1 \\
   \text{plus the condition: } & \comm{R}{L} \subseteq R
 \end{split}
\end{equation*}

Which saying that we can always decompose a complex Lie algeabra in a solvable $R$, semi-direct sum, and a direct sum between simple Lie algebras.

Every Lie algebra can be decomposed into a semi-direct sum $\oplus_{s, \text{Lie}}$ between a solvable Lie algebra $R$ and a direct sum $\oplus_\text{Lie}$ between simple Lie algebras

An example of solvable Lie algebra is the upper-triangular matrices.

An example of un-solvable Lie algebra is $\mathfrak{sl}(n, \mathbb{C})$ since taking the commutator gives you the space itself, i.e.

\begin{equation*}
\comm{\mathfrak{sl}(n, \mathbb{C})}{\mathfrak{sl}(n, \mathbb{C})} = \mathfrak{sl}(n, \mathbb{C})
\end{equation*}

A Lie algebra that has no solvable part, i.e. $L = L_1 \oplus_\text{Lie} \dots \oplus_\text{Lie} L_n$ is called semi-simple.

A Lie algebra is called simple if there are no ideals other than $0$ and itself, and it is not abelian.

If $\mathfrak{g}$ is simple, then no solvable ideals other than (possibly) $\mathfrak{g}$. Therefore

\begin{equation*}
\comm{\mathfrak{g}}{\mathfrak{g}} \subset \mathfrak{g} \quad \implies \quad \comm{\mathfrak{g}}{\mathfrak{g}} = 0 \quad \implies \mathfrak{g} \text{ abelian}
\end{equation*}

since otherwise we would not be able to repeatedly apply commutators and eventually reach $0$. This is a contradiction, concluding our proof.

We say the radical of $\mathfrak{g}$ is the maximal sovlable ideal which contains every other solvable ideal.

We denote the radical of $\mathfrak{g}$ a $\mathrm{rad}(\mathfrak{g})$.

A representation $V$ is called semi-simple or completely reducible if it is isomorphic to a direct sum of irreducible representations, i.e.

\begin{equation*}
V \simeq \bigoplus_{i \in I} V_i
\end{equation*}

In this case, it is common to group isomorphic direct summands, and write

\begin{equation*}
V \simeq \bigoplus_{i \in I} n_i V_i
\end{equation*}

the number $n_i$ is called the multiplicity of the subrepresentation $V_i$.

It turns out it's quite hard to classify the solvable Lie algebras $R$, and simpler to classify the semi-simple Lie algeabras. Thus we put our focus towards classifying the semi-simple Lie algebras and then using these as building blocks to classify the full Lie algebra of interest.

$(L \comm{\cdot}{\cdot})$ is a complex Lie algebra and $h \in L$, then define

\begin{equation*}
\begin{split}
  \ell \mapsto \ad(h) (\ell) &:= \comm{h}{\ell} \\
  \text{or simply: } \ad(h) &:= \comm{h}{\cdot}
\end{split}
\end{equation*}

is the adjoint map wrt. $h \in L$.

The bililinear map

\begin{equation*}
\begin{split}
  K: L \times L  &\overset{\sim}{\to} \mathbb{C} \\
  K(a, b) & := \tr \Big( \underbrace{\ad(a) \circ \ad(b)}_{L \overset{\sim}{\to} L} \Big)
\end{split}
\end{equation*}

is called the killing "form" (it's a symmetric map, so not the kind of form you're used to).

And we make the following remarks about the Killing form:

  • $L$ is finite-dim. thus the $\tr$ is cyclic, hence $K(a, b) = K(b, a)$
  • $L$ is solvable if and only if

    \begin{equation*}
K \big( [L, L], L \big) = 0
\end{equation*}
  • $L$ semi-simple (i.e. no solvable part) if and only if $K$ is non-degenerate:

    \begin{equation*}
  \forall a \in L : K(a, b) = 0 \implies b = 0
\end{equation*}
  • $G$ invariant since

    \begin{equation*}
\begin{split}
  K \big( g x g^{-1}, g y g^{-1} \big) &= \tr \Big( \ad_{g x g^{-1}} \circ \ad_{g y g^{-1}} \Big) \\
  &= \tr \big( g \ad_{x} g^{-1} g \ad_{y} g^{-1} \big) \\
  &= k(x, y)
\end{split}
\end{equation*}

Now, how would we then compute these simple forms?

Now consider, for actual calculations, components of $\ad(h)$ and $K$ wrt. a basis:

\begin{equation*}
\begin{split}
  E_1, \dots, E_{\dim{L}} & \quad \text{of } L \\
  \varepsilon^1, \dots, \varepsilon^{\dim{L}} & \quad \text{of } L^*
\end{split}
\end{equation*}

Then

\begin{equation*}
\begin{split}
  \Big( \ad(E_i) \Big)_k^j & := \varepsilon^j \Big( \ad(E_i)(E_k) \Big) \quad \text{(def. by how it maps the basis)} \\
  &= \varepsilon^j \underbrace{\Big( \comm{E_i}{E_k} \Big)}_{c_{ik}^m E_m} \qquad \ \  \text{(expand result in basis)} \\
  &= \varepsilon^j \Big( c_{ik}^m E_m \Big) \\
  &= c_{ik}^m \varepsilon^j \big( E_m \big) \\
  &= c_{ik}^j
\end{split}
\end{equation*}

where $c_{ik}^m \in \mathbb{C}$ are just coefficients of expanding the commutation in the space of the complex numbers, which we clearly can do since $\comm{\cdot}{\cdot}: L \times L \overset{\sim}{\to} L$. These coefficients are called the structure constants of $L$ wrt. chosen basis.

The killing form in components is:

\begin{equation*}
\begin{split}
  K_{ij} = K(E_i, E_j) & = \tr \Big( \ad(E_i) \circ \ad(E_j) \Big) \\
  &= \underbrace{c_{im}^l c_{jl}^m}_{\big( \ad(E_i) \circ \ad(E_j) \big)_l^l}
\end{split}
\end{equation*}

Where the last bit is taking the $\tr$. Thus, each component of the killing form is given by

\begin{equation*}
K_{ij} = c_{im}^n c_{jn}^m
\end{equation*}

We then empasize that $L$ is semi-simple if and only if $K$ is a psuedo-inner product (i.e. inner product but instead of being positive-definite, we only require it to be non-degenerate).

One can check that $\ad(h)$ is anti-symmetric wrt. killing form $K$ a (for a simple Lie algeabra, which implies semi-simple).

A Cartau subalgebra $H$ of a Lie algebra $(L, [\cdot, \cdot])$ is:

  • $H \subset L$ as a vector subspace
  • maximal subalgeabra of $L$ and that there exists a basis:

    \begin{equation*}
  h_1, \dots, h_m
\end{equation*}

    of $H$ that can be extended to a basis of $L$

    \begin{equation*}
 h_1, \dots, h_m, e_1, \dots, e_{d - m}, \quad d = \dim L
\end{equation*}

    such that the extension vectors $e_1, \dots, e_{d - m}$ are eigenvectors for any $\ad(h)$ where $h \in H$ :

    \begin{equation*}
 \ad(h) e_\alpha = \underbrace{\lambda_\alpha(h)}_{\in \mathbb{C}} e_\alpha, \qquad \alpha = 1, \dots, d - m
\end{equation*}

    where the eigenvalue $\lambda(h)$ depends on $h$ since with any other $h$ we have a different map $\ad(h)$

Now, one might wonder, does such a Cartau subalgebra exists?!

  1. Any finite-dimensional Lie algebra posses a Cartau subalgebra
  2. If $L$ is simple Lie algebra then $H$ is abelian, i.e. $[H, H] = 0$
\begin{equation*}
\comm{h}{e_\alpha} = \lambda_\alpha(h) e_\alpha
\end{equation*}

is linear in $h \in H$. Thus,

\begin{equation*}
\lambda_\alpha : H \overset{\sim}{\to} \mathbb{C}
\end{equation*}

I.e. we can either view $\lambda_\alpha$ as a linear map, OR as a specific value $\lambda_\alpha(h)$. This is simply equivalent of saying that

\begin{equation*}
\lambda_\alpha \in H^*
\end{equation*}

The $\lambda_1, \dots, \lambda_{d - m} \in H^*$ are the roots of the Lie algebra, and we call

\begin{equation*}
\Phi := \{ \lambda_1, \dots, \lambda_{d - m} \} \subseteq H^*
\end{equation*}

the root set.

Since $\ad(h)$ is anti-symmetric wrt. killing form, then

\begin{equation*}
\lambda \in \Phi \implies - \lambda \Phi
\end{equation*}

Also, $\lambda$ are not linearly independent.

A set of fundamental roots $\Pi \subset \Phi$ such that

  1. $\Pi$ linearly independent, $\Pi = \{ \pi_1, \dots, \pi_f \}$
  2. Then $\forall \lambda \in \Phi$ such that

    \begin{equation*}
 \exists n_1, \dots, n_f \in \mathbb{N}, \quad \exists \varepsilon \{ -1, +1 \} : \lambda = \varepsilon \sum_{i=1}^{f} n_i \pi_i
\end{equation*}

Observe that the $\varepsilon$ makes it so that we're basically choosing either to take all the positive $\lambda$ or all the negative $\lambda$, since they are linearly independent, AND this is different from saying that $n_i \in \mathbb{Z}$!!! Since we could then have some negative and some positive. We still need to be able to produce $- \lambda$ from $\lambda$, therefore we need the $\varepsilon$.

And as it turns out, such a $\Pi \subset \Phi$ can always be found!

The fundamental roots of $\Phi$ span the Cartau subalgebra $H$

\begin{equation*}
\text{span}_\mathbb{C} (\Pi) = H^*
\end{equation*}

buuut note that $\Pi$ is not unique (which is apparent by the fact that we can choose $\varepsilon \in \left\{ -1, 1 \right\}$ for the expressing the roots, see definition of fundamental roots).

If we let $H_\mathbb{R}^* := \text{span}_\mathbb{R} \Pi$, then we have

\begin{equation*}
\Pi \subset \underbrace{\Phi}_{ \subseteq \text{span}_{\pm \mathbb{N}} (\Pi)} \subset \underbrace{H_{\mathbb{R}}}_{= \text{span}_{\mathbb{R}}(\Pi)} \subset \underbrace{H^*}_{= \text{span}_{\mathbb{C}}(\Pi)}
\end{equation*}

We define the dual of the Killing form as $K^*: H^* \times H^* \to \mathbb{C}$ defined by

\begin{equation*}
K^*(\mu, \nu) := K \big( i^{-1}(\mu), i^{-1}(\nu) \big)
\end{equation*}

where we define

\begin{equation*}
\begin{split}
  i: & H \overset{\sim}{\to} H^* \\
  i(h) & := K(h, \cdot)
\end{split}
\end{equation*}

where $i^{-1}$ exists if $L$ is a semi-simple Lie algebra (and thus of course if it's a simple Lie algebra).

If we restrict the dual of the Killing form $K^*$ to $H_{\mathbb{R}}^*$ (as opposed to $H_{\mathbb{C}}^*$), that is,

\begin{equation*}
K^* : H_{\mathbb{R}}^* \times H_{\mathbb{R}} \overset{\sim}{\to} \mathbb{R}
\end{equation*}

and

\begin{equation*}
K^* (\alpha, \alpha) \ge 0
\end{equation*}

with equality if and only if $\alpha = 0$.

Then, on $H_{\mathbb{R}}^* = \text{span}_{\mathbb{R}}(\Pi)$ we can calculate lengths and angles.

In particular, one can calculate lengths and angles of the fundamental roots of $L$ (all roots are spanned by fundamental roots => can calculate such on all roots).

Now, we wonder, can we recover precisely the set $\Phi$ from the set $\Pi$?

For any $\lambda \in \Phi$ define

\begin{equation*}
\begin{split}
  s_\lambda :& H_{\mathbb{R}}^* \overset{\sim}{\to} H_{\mathbb{R}}^* \\
  s_\lambda(h) & = \mu - 2 \frac{K^*(\lambda, \mu)}{K^*(\lambda, \lambda)} \lambda
\end{split}
\end{equation*}

which is:

  • linear in $\mu$
  • non-linear in $\lambda$

and such $s_\lambda$ is called a Weyl transformation, and

\begin{equation*}
W := \{ s_\lambda \mid \lambda \in \Phi \}
\end{equation*}

called Weyl group with the group operation being the composition of maps.

  1. The Weyl group is generated by the fundamental roots in $\Pi$:

    \begin{equation*}
\forall w \in W, \exists \pi_1, \dots, \pi_n \in \Pi : w = s_{\pi_1} \circ s_{\pi_2} \circ \dots \circ s_{\pi_n}   
\end{equation*}
  2. Every root $\lambda \in \Phi$ can be produced from a fundamental root $\pi \in \Pi$ by action of the Weyl group $W$:

    \begin{equation*}
\forall \lambda \in \Phi, \exists w \in W, \pi \in \Pi : \lambda = w (\pi)
\end{equation*}
  3. The Weyl group merely permutates the roots:

    \begin{equation*}
\forall w \in W, \forall \lambda \in \Phi : w(\lambda) \in \Phi
\end{equation*}

Thus, if we know the fundamental roots $\Pi$, we can, by 1., find the entire Weyl group, and thus, by 2., we can find all the roots $\Phi$!

Conclusion

Consider: for any fundamental roots $\pi_i, \pi_j \in \Pi$, by definition of $s_{\lambda}$ we have

\begin{equation*}
S_{\pi_i} (\pi_j) = \pi_j - 2 \frac{K^* (\pi_i, \pi_j)}{K^*(\pi_i, \pi_i)} \pi_i
\end{equation*}

And

\begin{equation*}
s_{\pi_i}(\pi_j) \in \Phi = \{ \varepsilon \sum n_i \pi_i \}
\end{equation*}

Which means that both terms on LHS of $s_{\pi_i}(\pi_j)$ must be the same sign, and further because it's an element in $\Phi$, we know the coefficient must be in the integers (for $i \ne j$):

\begin{equation*}
C_{ij} = 2 \frac{K^*(\pi_i, \pi_j)}{K^*(\pi_i, \pi_j)}
\end{equation*}

where, for $i \ne j$, we have $- C_{ij} \in \mathbb{N}$.

Observe, $C_{ij}$ is not symmetric.

We call the matrix defined by $C_{ij}$ the Cartau matrix, and observe that $C_{ii} = 2$ while every other entry is some non-positive number!

Now we define the bond number:

\begin{equation*}
\begin{split}
  n_{ij} & := C_{ij} C_{ji} \\
  &= 4 \frac{K^*(\pi_i, \pi_j)}{K^*(\pi_i, \pi_i)} \frac{K^*(\pi_j, \pi_i)}{K^*(\pi_j, \pi_j)} \\
  &= 4 \cos^2 \big( \sphericalangle (\pi_i, \pi_j) \big)
\end{split}
\end{equation*}

which implies

\begin{equation*}
0 \le n_{ij} = C_{ij} C_{ji} < 4, \quad \forall i \ne j
\end{equation*}

where $C_{ij}$ and $C_{ji}$ are non-positive numbers, hence:

\begin{equation*}
n_{ij} \in \{ 0, 1, 2, 3 \}
\end{equation*}

Therefore:

$C_{ij}$ $C_{ji}$ $n_{ij}$
0 0 0
-1 -1 1
-1 -2 2
-2 -1 2
-1 -3 3
-3 -1 3

Which further implies that

\begin{equation*}
K^*(\pi_j, \pi_i) < K^*(\pi_j, \pi_j) \quad \text{OR} \quad K^*(\pi_i, \pi_i) > K^*(\pi_j, \pi_j)
\end{equation*}
Dynkin diagrams

We draw these diagrams as follows:

  1. for every fundamental root draw circle:

    fundamental_roots.png

  2. if two circles represent $\pi_i, \pi_j \in \Pi$, draw $n_{ij}$ lines between them:

    fundamental_roots_connected.png

  3. if there are 2 or 3 lines between two roots, use the $<$ sign on the lines between to indicate which is the greatest root

Any fininte-dimensional simple $\mathbb{C}$ - Lie algebra can be reconstructed from the set $\Pi$ of fundamental roots and the latter only comes in the following forms:

dynkin-diagrams.png

Taken from [[https:/commons.wikimedia.org/wiki/]] /

Representation Theory of Lie groups and Lie algebras

Representation of Lie Algebras

Let $\big( L, [\![ \cdot, \cdot ]\!] \big)$ be a Lie algebra.

Then a representation $\varrho$ of this Lie algebra is:

\begin{equation*}
\varrho: L \overset{\sim}{\to} \text{End}(V)
\end{equation*}

(in the Interactions of Algebra, Geometry and Topology course we specific that it should be $\text{Aut}(V)$ rather than $\text{End}(V)$, since $\text{Aut}(V)$ contains inverses)

s.t.

\begin{equation*}
\varrho \big( [\![ a, b ]\!] \big) = \comm{\varrho(a)}{\varrho(b)} = \varrho(a) \circ \varrho(b) - \varrho(b) \circ \varrho(a)
\end{equation*}

where the vector space $V$ (a finite-dimensional vector space) is called the representation space.

A morphism between representations $V$ and $W$ of a Lie group $G$ is a linear map $f: V \to W$ such that

\begin{equation*}
f \circ \rho_V(g) = \rho_W(g) \circ f, \quad \forall g \in G
\end{equation*}

We denote the space of morphisms between Lie group representations $V$ and $W$ by

\begin{equation*}
\Hom_G(V, W)
\end{equation*}

A morphism between representations $V$ and $W$ of a Lie algebra $\mathfrak{g}$ is a linear map $f: V \to W$ such that

\begin{equation*}
f \circ \rho_V(x) = \rho_W(x) \circ f, \quad \forall x \in \mathfrak{g}
\end{equation*}

We denote the space of morphisms between Lie algebra representations $V$ and $W$ by

\begin{equation*}
\Hom_{\mathfrak{g}}(V, W)
\end{equation*}

A representation $V$ of a Lie group $G$ contains the same "data" as an action of the Lie group on a vector space $V$.

Representations of Lie algebra $\mathfrak{g}$ are often called modules over $\mathfrak{g}$ or $\mathfrak{g} \text{-modules}$.

Morphisms between representations are often referred to as intertwining operators.

An example of a representation is $G$ acts on $M$, then

\begin{equation*}
V = C^{\infty}(M)
\end{equation*}

is a representation of $G$ via

\begin{equation*}
\big( \rho_V(g) \circ f \big)(m) = f \big( g^{-1} \triangleright m \big)
\end{equation*}

In general, $v \in T_m M$ is an equiv. class of curves where $\gamma_1 \sim \gamma_2$ if $\gamma_1(0) = \gamma_2(0) = m$ and $\gamma_1'(0) = \gamma_2'(0)$.

$M, N$ are manifolds, $\varphi: M \to N$ with $\varphi(m) = n$. Then $\varphi$ takes curves through $m$ to curves through $n$, and

\begin{equation*}
\gamma: \mathbb{R} \to M, \quad \gamma(0) = m \quad \implies \quad \varphi \circ \gamma: \mathbb{R} \to N \quad \varphi \big( \gamma(0) \big) = \varphi(m) = n
\end{equation*}

And $\varphi$ takes equiv. classes to equiv. classes (CHECK THIS). Thus differential of $\varphi$, $\varphi_*: T_m M \to T_n N$.

A representation $\varrho: L \overset{\sim}{\to} \text{End}(V)$ is called reducible if there exists a vector subspace $U \subseteq V$ s.t.

\begin{equation*}
\forall a \in L, \forall u \in U : \varrho(a)(u) \in U, \quad U \ne \{ 0 \} \quad \text{and} \quad U \ne L
\end{equation*}

in other words, the representation map $\varrho$ restricts to

\begin{equation*}
\varrho |_u : L \overset{\sim}{\to} \text{End}(U)
\end{equation*}

Otherwise, $\varrho: L \to \text{End}(V)$ is called irreducible or simple. In other words, if $V$ has no non-trivial subrepresentations (i.e. other than 0 or itself).

Example of irreducible representation

The vector representation $\mathbb{C}^n$ of $\SL(n , \mathbb{C})$ is irreducible.

Irreducible representations
Finite-dimensional reps. of $\mathfrak{sl}(2, \mathbb{C})$ [EXAM IMP]

First we make the following observation:

  1. Complex reps. of $\mathfrak{sl}(2, \mathbb{C})$ are isomorphic to those of its real form $\mathfrak{su}(2)$
  2. Reps. of $\mathfrak{su}(2)$ are the same as the reps. of $\SU(2)$
  3. $\SU(2)$ is compact, thus it's reps. are completely reducible.
  4. Hence reps. of $\mathfrak{sl}(2, \mathbb{C})$ are iso. to completely reducible reps. $\implies$ reps. are completely reducible, as claimed.

Recall that the generators of $\mathfrak{sl}(2, \mathbb{C})$ satisfy the relations

\begin{equation*}
\comm{h}{e} = 2 e, \quad \comm{h}{f} = - 2f, \quad \comm{e}{f} = h
\end{equation*}

Let $V$ be a representation of $\mathfrak{sl}(2, \mathbb{C})$ with rep. $\varrho_V: \mathfrak{sl}(2, \mathbb{C}) \longrightarrow V$ .

A vector $v \in V$ is said to be a vector of weight $\lambda$ if

\begin{equation*}
\varrho_V(h) v = \lambda v
\end{equation*}

i.e. $\lambda$ is an eigenvalue of $\varrho_V(h)$.

We denote $V[\lambda]$ the subspace of vectors of weight $\lambda$.

Note that

\begin{equation*}
\varrho_V(e) V[\lambda] \subset V[\lambda + 2] \quad \text{and} \quad \varrho_V(f) V[\lambda] \subset V[\lambda - 2]
\end{equation*}

To see this, let $v \in V[\lambda]$ and observe that

\begin{equation*}
\varrho_V(h) \Big( \varrho_V(e) v \Big) = \Big( \underbrace{\comm{\varrho_V(h)}{\varrho_V(e)}}_{= 2 \varrho_V(e)} + \varrho_V(e) \varrho_V(h) \Big) v = \underbrace{(2 + \lambda) \big( \varrho_V(e) v \big)}_{ \in V[\lambda + 2]}
\end{equation*}

since $v \in V[\lambda]$. Observe then that this means that $\varrho_V(e) v \in V[\lambda + 2]$.

Similarily we also find that $\varrho_V(f) V[\lambda] \subset V[\lambda - 2]$, as claimed above.

Every finite-dimensional rep. of $\mathfrak{sl}(2, \mathbb{C})$ admits decomposition into the weight subspaces:

\begin{equation*}
V \simeq \bigoplus_{\lambda_i} V[\lambda_i]
\end{equation*}

Let $V$ be a rep. of $\mathfrak{sl}(2, \mathbb{C})$.

Since reps. of $\mathfrak{sl}(2, \mathbb{C})$ are semi-simple, we can assume $V$ to be irreducible.

Let $W \subset V$ be the subset spanned by eigenvectors of $\varrho_V(h)$. By Remark remark:generators-of-sl-2-C-takes-us-from-weight-space-to-weight-space we know that this space is stable under $\varrho_V(e)$ and $\varrho_V(f)$. Hence $W$ is a subrepresentation, contradicting the irreducibility of $V$. Therefore $W = V$.

Let $V$ be a rep. of $\mathfrak{sl}(2, \mathbb{C})$ with $\dim(V) < \infty$.

A weight $\lambda$ is said to be a highest weight of $V$ if, for any other weight $\tilde{\lambda}$, one has

\begin{equation*}
\Re(\lambda) \ge \Re(\tilde{\lambda})
\end{equation*}

Then $v \in V[\lambda]$ are called highest weight vectors.

Let $v \in V[\lambda]$ be a highest-weight vector in a finite-dim. rep. of $\mathfrak{sl}(2, \mathbb{C})$.

Then

  1. $\varrho_V(e) v= 0$
  2. For all $k \ge 0$, define

    \begin{equation*}
v^k := \frac{1}{k!} \varrho_V(f)^k v
\end{equation*}

    Then we have

    \begin{equation*}
\begin{split}
  \varrho_V(h) v^k &= \big( \lambda - 2k \big)v^k \\
  \varrho_V(f) v^k &= \big(  k + 1 \big) v^{k + 1} \\
  \varrho_V(e) v^k &= \big( \lambda - k + 1 \big) v^{k - 1}
\end{split}
\end{equation*}
  1. Follows immediately from the fact that

    \begin{equation*}
e V[\lambda] \subset V[\lambda + 2] = 0
\end{equation*}

    since $V$ is finite-dimensional and $\lambda$ is highest-weight.

  2. We have the following:
    • Action of $f$ on $v^k$ follows immediately from the def. of $v^k$.
    • For $h$ it follows from the relation $\comm{h}{f} = - 2f$.
    • For $e$ we use induction:
      1. Base step:

        \begin{equation*}
\varrho_V(e) v^1 = \varrho_V(e) \varrho_V(f) v^0 = \varrho_V(f) \varrho_V(e) v + \varrho_V(h) v = \lambda v
\end{equation*}
      2. General case:

        \begin{equation*}
\begin{split}
  \varrho_V(e) v^{k + 1} &= \frac{1}{k + 1} \Big( \varrho_V(f) \varrho_{V}(e) v^k + \varrho_V(h) v^k \Big) \\
  &= \frac{1}{k + 1} \big( \varrho(f) (\lambda - k + 1) v^{k - 1} + (\lambda - 2k) v^k \big) \\
  &= \frac{1}{k + 1} \big( (\lambda - k + 1)k v^{k} + (\lambda - 2k) v^k \big) \\
  &= (\lambda - k) v^k
\end{split}
\end{equation*}

For any $n > 0$, let $V_n$ be the $(n + 1)$ dimensional space with basis $v^0, v^1, \dots, v^n$.

Define the action of $\mathfrak{sl}(2, \mathbb{C})$ on $V$ by

\begin{equation*}
\begin{split}
  \varrho_V(h) v^k &= \big( n - 2k \big) v^k \\
  \varrho_V(f) v^k &= 
  \begin{cases}
    \big( k + 1 \big)v^{k + 1} \quad & \text{if } k < n \\
    0 & \text{if } k = n
  \end{cases} \\
  \varrho_V(e) v^k &= 
  \begin{cases}
    \big( n - k + 1 \big)v^{k - 1}  \quad & \text{if } k > 1 \\
    0 & \text{if } k = 0
  \end{cases}
\end{split}
\end{equation*}

where $v \in V[n]$ is a highest-weight vector and

\begin{equation*}
v^k := \frac{1}{k!} \varrho_V(f)^k v
\end{equation*}
  1. Then $V_n$ is an irreducible rep. of $\mathfrak{sl}(2, \mathbb{C})$, it is referred to as the irreducible rep. with highest-weight $n$.
  2. For any $n \ne m$ we have $V_n \not\simeq V_m$
  3. Every finite-dimensional irreducible representation of $\mathfrak{sl}(2, \mathbb{C})$ is isomorphic to $V_n$ for some $n \in \mathbb{N}$.
  1. For any $v \in V$, we can find an integer $k$ such that

    \begin{equation*}
\varrho_V(e)^k v = 0 \quad \text{and} \quad \varrho_V(e)^{k - 1} v \ne 0
\end{equation*}

    Then $\varrho_V(e)^{k - 1} v$ is proportional to $v^n$. Since $v^n$ generates the whole of $V$ under action of $\varrho_V(f)$, we see that $V_n$ is irreducible.

  2. Since rep. $V_m$ and $V_n$ are different dimensions and therefore cannot possibly be isomorphic.
  3. Let $\lambda \in \mathbb{C}$ and consider an infinite-dimensional representation of $M_{\lambda}$ with the basis $v^0, v^1, \dots$ and the action of $\mathfrak{sl}(2, \mathbb{C})$ defined as before. Any irreducible finite-dim. rep. which contains highest weight $\lambda$ is a quotient of $M_{\lambda}$ by certain subrepresentation. Let $V$ be such a finite-dim. subrep. of $\mathfrak{sl}(2, \mathbb{C})$. Note, that only finitely many $v^k$ are non-zero in $V$, since all of them are linearly independent. Let $n$ be the maximal integer such that $v^n \ne 0$. Then

    \begin{equation*}
\varrho_V(e)v^{n + 1} = (\lambda - n) v^n = 0
\end{equation*}

    since $v^{ n + 1} = 0$, from which we derive that $\lambda = n$. Now consider a subspace $\tilde{M}$ spanned by vectors $\left\{ v^k \right\}$ with $k \ge n + 1$. Then $\tilde{M}$ is closed under the action of $\mathfrak{sl}(2, \mathbb{C})$, and hence $\tilde{M} \subset M$ is a subrep. Now it is easy to see that $\big(M_{\lambda} / \tilde{M} \big) \simeq V_n$.

Then we can consider an infinite dimensional rep. $M$, and quotient by some subrep $W$

In the quotient, there is a "last" vector $v^n$, which is not killed by $/ W$.

Take $v^{n + 1}, v^{n + 2}, \dots$, then

\begin{equation*}
\implies \underbrace{e v^{n + 1}}_{=  0} = \big( \lambda + - (n + 1) \big) v^n \quad \implies \quad \lambda = n
\end{equation*}
\begin{equation*}
V = \bigoplus_k n_k V_k = \bigoplus_r V [r]
\end{equation*}

where $V[r]$ denotes eigenspace of "weight" $r$ (with weight referring to the eigenvalue). This implies that

\begin{equation*}
V = \left\{ v^0, \dots, v^n \right\} = : V_n = \underbrace{V_n[n]}_{= \mathbb{C}(v^0)} \oplus \underbrace{V_n[n - 2]}_{\mathbb{C}(v^1)} \oplus \cdots \oplus \underbrace{V_n[ - n]}_{ = \mathbb{C}(v^n)}
\end{equation*}

where $V_n[k]$ is the h-eigenspace (in $V_n$) of weight $k$ (which is same if we drop the subscript $n$)

\begin{equation*}
\dim \big( V[k] \big) = \dim \big( V[-k] \big)
\end{equation*}

and we get isomorphisms

\begin{equation*}
V[k] \overset{\overset{f^k}{\longrightarrow}}{\underset{e^k}{\longleftarrow}} V[-k]
\end{equation*}

$V_1$ is the std. 2-dimensional $\mathfrak{sl}_2 \text{-module}$, then

\begin{equation*}
S^n V_1 = V_n
\end{equation*}

where $S^n V_1$ is irreducible of dim. $n + 1$.

Let

\begin{equation*}
t = \exp(h) \in \SL(2, \mathbb{C})
\end{equation*}

then the character is given by

\begin{equation*}
\chi_V(t) = \sum_{k}^{} \dim \big( V[k] \big) z(t)^k
\end{equation*}

where we let $z(t) = e$. For some $V_n$ we have

\begin{equation*}
\chi_{V_n} = \underbrace{z^n + z^{n - 2} + \cdots + z^{-n}}_{n + 1} = \frac{z^{n + 1} - z^{- (n + 1)}}{z - z^{-1}}
\end{equation*}

Finite-dimensional rep. theory of $\mathfrak{sl}(2, \mathbb{C})$ is controlled by symmetric functions in $\big( z_1, z_2 \big)$ modulo $z_1 z_2 = 1$.

In fact: Finite-dimensional rep. theory of $\mathfrak{sl}(k, \mathbb{C})$ is controlled by symmetric functions in $\big( z_1, z_2, \dots, z_k \big)$ modulo $z_1 z_2 \cdots z_k = 1$.

This follows from Lemma lemma:properties-of-character, by taking the tensor products.

\begin{equation*}
\begin{split}
  \chi_{V_2 \otimes V_4} &= \chi_{V_2} \cdot \chi_{V_4} \\
  &= \big( z^2 + 1 + z^{-2} \big) \big( z^4 +  z^2 + 1 + z^{-2} + z^4 \big) \\
  &= z^6 + 2 z^4 + 3z^2 + 3 + 3 z^{-2} + 2 z^{-4} + z^{-6}
\end{split}
\end{equation*}

Sanity check at this point: make sure the coefficients sum to the dimensionality of the space!

\begin{equation*}
\begin{split}
  \chi_{V_2 \otimes V_4} &= \chi_{V_2} \cdot \chi_{V_4} \\
  &= \big( z^2 + 1 + z^{-2} \big) \big( z^4 +  z^2 + 1 + z^{-2} + z^4 \big) \\
  &= z^6 + 2 z^4 + 3z^2 + 3 + 3 z^{-2} + 2 z^{-4} + z^{-6} \\
  &= \chi_{V_6} + \chi_{V_4} + \chi_{V_2}
\end{split}
\end{equation*}

Hence

\begin{equation*}
V_2 \otimes V_4 \simeq V_6 \oplus V_4 \oplus V_2
\end{equation*}

Say $g \in G$, with eigenvalues in a complex representation are $\lambda_1, \dots, \lambda_n$. Eigenvalues of $g$ in $S^2 V$ ($\bigwedge^2 V$) are products $\lambda_i \lambda_j$ for $i \le j$ ($i < j$)

\begin{equation*}
\chi_{S^2 V}(g) = \sum_{i \le j}^{} \lambda_i \lambda_j \quad \text{and} \quad \chi_{\bigwedge^2 V}(g)  = \sum_{i < j}^{} \lambda_i \lambda_j
\end{equation*}

Then

\begin{equation*}
\chi_{S^2 V}(g) + \chi_{\bigwedge^2 V}(g) = \big( \lambda_1 + \dots + \lambda_n \big)^2 = \big( \chi_V(g) \big)^2
\end{equation*}

and

\begin{equation*}
\chi_{S^2 V}(g) - \chi_{\bigwedge^2 V}(g) = \lambda_1^2 + \dots + \lambda_n^2 = \chi_V(g^2)
\end{equation*}
\begin{equation*}
\chi_{S^2 V}(g) = \frac{\chi_v(g)^2 + \chi_V(g^2)}{2} \quad \text{and} \quad \chi_{\bigwedge^2 V}(g) = \frac{\chi_V(g)^2  - \chi_V(g^2)}{2}
\end{equation*}

Using the above, we have

\begin{equation*}
\begin{split}
  \chi_{S^2 V_2} &= \frac{1}{2} \Big( (z^2 + 1 + z^{-2})^2 + (z^4 + 1 + z^{-4}) \Big) \\
  &= \frac{1}{2} \Big( z^4 + 1 + z^{-4} + 2 z^2 + 2 + 2 z^{-2} + z^4 + 1 + z^{-4} \Big) \\
  &= z^4 + z^2+  2 + z^{-2} + z^{-4} \\
  &= \chi_{V_4} + \chi_{V_0}
\end{split}
\end{equation*}

Hence

\begin{equation*}
S^2 V_2 \simeq V_4 \oplus \underbrace{V_0}_{= \mathbb{C}}
\end{equation*}

Adjoint representation

Let $\mathfrak{g}$ be a Lie algebra over a field $K$.

Given an element $x$ of a Lie algebra $\mathfrak{g}$, one defines the adjoint action of $x$ on $\mathfrak{g}$ is given by the adjoint map.

Then there is a linear mapping

\begin{equation*}
\begin{split}
  \ad: \quad & \mathfrak{g} \to \text{End}(\mathfrak{g}) \\
  & x \mapsto \ad_x
\end{split}
\end{equation*}

Within $\text{End}(\mathfrak{g})$, the Lie bracket is, by definition, given by the commutator of the two operators:

\begin{equation*}
\comm{\ad_x}{\ad_y} = \ad_x \circ \ad_y - \ad_y \circ \ad_x
\end{equation*}

Using the above definition of the Lie bracket, the Jacobi idenity

\begin{equation*}
\comm{x}{\comm{y}{z}} + \comm{y}{\comm{z}{x}} + \comm{z}{\comm{x}{y}} = 0
\end{equation*}

takes the form

\begin{equation*}
\big( \comm{\ad_x}{\ad_y} \big)(z) = \big( \ad_{\comm{x}{y}} \big)(z)
\end{equation*}

where $x$, $y$, and $z$ are arbitrary elements of $\mathfrak{g}$.

This last identity says that $\ad$ is a Lie algebra homomorphism; i.e. a linear mapping that takes brackets to brackets. Hence $\ad$ is a representation of a Lie algebra and is called the adjoint representation of the algebra $\mathfrak{g}$.

Casamir operator

Let $\rho: L \overset{\sim}{\to} \text{End}(V)$ be the representation of complex Lie algebra $L$.

We define the ρ-Killing form on $L$ as

\begin{equation*}
\begin{split}
  \kappa_{\rho}: \quad & L \times L \overset{\sim}{\to} \mathbb{C} \\
  & (x, y) \mapsto \text{tr} \Big( \rho(x) \circ \rho(y) \Big)
\end{split}
\end{equation*}

/Note that this is not the same as the "standard" Killing form, where we only consider $\kappa_{\ad}$.

Let $\rho: L \overset{\sim}{\to} \text{End}(V)$ be a faithful representation of a complex semi-simple Lie algebra $L$.

Then $\kappa_{\rho}$ is non-degenerate.

Hence $\kappa_{\rho}$ induces an isomorphism $L \overset{\sim}{\to} L^*$ by

\begin{equation*}
x \mapsto \kappa_{\rho}(x, \cdot) \in L^*, \quad \forall x \in L
\end{equation*}

Recall that if $\{ X_1, \dots, X_{\dim L} \}$ is a basis of $L$, then the dual basis $\{ \tilde{X}^1, \dots, \tilde{X}^{\dim L} \}$ of $L^*$ which is defined by

\begin{equation*}
\tilde{X}^i \big( X_j \big) = \tensor{\delta}{^i_j}
\end{equation*}

By using the isomorphism induced by $\kappa_{\rho}$ (when $\rho$ is a faithful representation, by proposition:ρ-killing-form-induces-isomorphism-for-complex-semi-simple-algebras), we can find some $\xi_1, \dots, \xi_{\dim L} \in L$ such that we have

\begin{equation*}
\kappa_{\rho}(\xi_i, \cdot) = \tilde{X}^i
\end{equation*}

or equivalently,

\begin{equation*}
\kappa_{\rho}(x, \xi_i) = \tilde{X}^i(x), \quad \forall x \in L
\end{equation*}

We thus have,

\begin{equation*}
\kappa_{\rho}(X_i, \xi_j) = \delta_{ij} := 
\begin{cases}
  1 & \text{if } i \ne j \\
  0 & \text{otherwise}
\end{cases}
\end{equation*}

Let $\{ X_i \}$ and $\{ \xi_j \}$ be defined as above. Then

\begin{equation*}
\comm{X_j}{\xi_k} = \sum_{m = 1}^{\dim L} \tensor{C}{^k _{mj}} \xi_m
\end{equation*}

where $\tensor{C}{^k_{mj}}$ are the structure constatns wrt. $\{ X_i \}$.

Let $\rho: L \overset{\sim}{\to} \text{End}(V)$ be a faithful representation of a complex (compact) Lie algebra $L$ and let $\{ X_1, \dots, X_{\dim L} \}$ be a basis of $L$.

The Casimir operator associated to the representation $\rho$ is the endomorphisms $\Omega_{\rho}: V \overset{\sim}{\to} V$

\begin{equation*}
\Omega_{\rho} := \sum_{i=1}^{\dim L} \rho(X_i) \circ \rho(\xi_i)
\end{equation*}

Let $\Omega_{\rho}$ be the Casimir operator of a representation $\rho: L \overset{\sim}{\to} \text{End}(V)$.

Then

\begin{equation*}
\forall x \in L:  \comm{\Omega_{\rho}}{\rho(x)} = 0
\end{equation*}

that is, $\Omega_{\rho}$ commutes with every endormorphism in $\text{im}_{\rho}(L)$.

If $\rho: L \overset{\sim}{\to} \text{End}(V)$ is irreducible, then any operator $S$ which commutes with every endomorphism in $\text{im}_{\rho}(L)$, i.e.

\begin{equation*}
\comm{S}{\phi} = 0, \quad \forall \phi \in \text{im}_{\rho}(L)
\end{equation*}

has the form

\begin{equation*}
S = c_{\rho} \text{id}_V
\end{equation*}

for some constant $c_{\rho} \in \mathbb{C}$ (or $\mathbb{R}$, if $L$ is a real Lie algebra).

Or equivalently, if $V$ and $W$ are two irreducible complex representations of $G$, then

\begin{equation*}
\Hom_G(V, W) =
\begin{cases}
  \mathbb{C} \id \quad & \text{if } V \simeq W \\
  0 & \text{otherwise}
\end{cases}
\end{equation*}

Observe that if we let $W = V$ in the second, we recover the first statement.

For any $A \in \Hom_G(V, W)$, one has that

\begin{equation*}
\text{ker}(A) \subset V
\end{equation*}

and

\begin{equation*}
\text{im}(A) \subset W
\end{equation*}

are subrepresentations.

Since both $V$ and $W$ are irreducible, we conclude that either

\begin{equation*}
\im(A) = 0 \quad \text{or} \quad \Big( \text{ker}(A) = 0 \text{ and } \im(A) = W \Big)
\end{equation*}

In other words, either $A = 0$ or $A$ is an isomorphism.

Thus, for $V \not\simeq W$ we have $\Hom_G(V, W) = 0$.

Now, if $V \simeq V$, the operator $A$ is an isomorphism and thus invertible. Let $\lambda$ be an eigenvalue of $A$. On one hand, $A - \lambda \id$ is still an intertwining operator, on the other it is not invertible anymore and thus $A - \lambda \id = 0$. Hence $A = \lambda \id$.

The Casimir operator of $\rho: L \overset{\sim}{\to} \text{End}(V)$ is

\begin{equation*}
\Omega_{\rho} = c_{\rho} \text{id}_V
\end{equation*}

where

\begin{equation*}
c_{\rho} = \frac{\dim L}{\dim V}
\end{equation*}

The first part follows from Schurs lemma and Thm. thm:casimir-operator-representation-of-lie-algebra.

The following follows immediately from Schurs lemma.

Let $V$ be a completely reducible representation of Lie group $G$ or Lie algebra $\mathfrak{g}$.

Then

  1. $V \simeq \bigoplus_i n_i V_i = \bigoplus C^{n_i} \otimes V_i$ where $V_i$ are irreducible pairwise non-isomorphic representations.
  2. Every intertwining operator $A: V \to V$ has the form

    \begin{equation*}
A = \bigoplus_i \big( A_i \otimes \id_i \big)
\end{equation*}

    for some $A_i \in \End(\mathbb{C}^{n_i})$

For 1), notice that any operator $\Phi: V \to V$ can be written in block form

\begin{equation*}
\Phi = \bigoplus_{i, j} \Phi_{ij}
\end{equation*}

with $\Phi_{ij}: V_i \to V_j$. By Schur's lemma, $\Phi_{ij} = 0$ for $i \ne j$ and $\Phi_{ii} = \lambda_i \id_{V_i}$.

Part 2) is proven similarily.

Note that Corollary corollary:4.25-kirillov can be quite useful:

  • if we can decompose a representation $V$ into irreducible representations, this gives us a very effective way of analysing intertwining operators
  • For example, if $V = \bigoplus_i V_i$ with $V_i \nsimeq V_j$, then $\Phi = \bigoplus \lambda_i \id_{V_i}$ so one can find $\lambda_i$ by computing $\Phi(v)$ for just one vector in $V_i$.
  • It can also show that each eigenvalue $\lambda_i$ will appear with multiplicity equal to $\dim (V_i)$

If $G$ is a commutative group, then any irreducible complex representation of $G$ is one-dimensional.

Similarily, if $\mathfrak{g}$ is a commutative Lie algebra, then any irreducible complex representation of $\mathfrak{g}$ is one-dimensional.

Complete reducibility of unitary representations

Notation
  • $\hat{G}$ denotes the set of isomorphism classes of irreducible representations of $G$
  • Matrix coefficients of a representation

    \begin{equation*}
\big( \rho_V(g) \big)^i_j =  \left\langle f^i , \rho_V(g) v_j \right\rangle
\end{equation*}
Unitary representations

A unitary representation $V$ of a real Lie group $G$ (resp. Lie algebra $\mathfrak{g}$) is a complex vector space $V$ together with a homomorphism $\rho: G \to U(V)$ (resp. $\rho: \mathfrak{g} \to \mathfrak{u}(v)$.

Equivalently, $V$ is unitary if it has a $G \text{-invariant}$ (resp. $\mathfrak{g} \text{-invariant}$) inner product.

If $V$ is irreducible, we're done.

If $V$ has a subrepresentation $W$, let $W^{\perp}$ be the orthogonal complement of the latter. Then

\begin{equation*}
V = W \oplus W^{\perp}
\end{equation*}

and $W^{\perp}$ is a subrep. as well.

Indeed, if $\left\langle v, w \right\rangle = 0$ for any $w \in W$, then

\begin{equation*}
\left\langle v ,\rho(g) w \right\rangle = \left\langle \rho(g)^{-1} v, w \right\rangle = 0
\end{equation*}

We simple iterate the above argument until we're done, i.e. reached $\dim(V)$.

Every representation of a finite group is unitary.

Let $V$ be a rep. of a finite group $G$, and let $B: V \times V \to \mathbb{R}$ be some inner product on $V$.

If $B$ is $G \text{-invariant}$, we are done.

If not, we "average" $B$ over $G$ in the following fashion: define

\begin{equation*}
\tilde{B}(v, w) = \frac{1}{\left| G \right|} \sum_{g \in G}^{} B(g \cdot v, g \cdot w)
\end{equation*}

Thus This defined $\tilde{B}$ is positive-definite Hermitian form (because it is a sum of such forms) and is clearly $G \text{-invariant}$:

\begin{equation*}
\tilde{B}(h \cdot v, h \cdot w) = \frac{1}{\left| G \right|} \sum_{g \in G}^{} B(gh \cdot v, gh \cdot w) = \frac{1}{\left| G \right|} \sum_{g \in G}^{} B(g' \cdot v, g' \cdot w)
\end{equation*}

Every rep. of a finite group is completely reducible.

Haar measure on compact Lie groups

A (right) Haar measure on a real Lie group $G$ is a Borel measure $\dd{g}$ which is invariant under the right action of $G$ on itself.

Similarily, one defines a left Haar measure.

Let $V$ be a one-dimensional real representation of a compact Lie group $G$. Then

\begin{equation*}
\left| \rho(g) \right| = 1, \quad \forall g \in G
\end{equation*}

Suppose for the sake of contradiction that $\left| \rho(g) \right| < 1$, then

\begin{equation*}
\rho(g^n) \to 0 \quad \text{as} \quad n \to \infty
\end{equation*}

On the other hand, $\rho(G)$ is a compact subset of $\mathbb{R}^{\times} = \mathbb{R} \setminus \left\{ 0 \right\}$, therefore cannot contain a sequence tending to 0 (since $0 \not\in \mathbb{R}^{\times}$ and a compact subspace contains all its limit points).

Similar argument shows that $\left| \rho(g) \right| > 1$ leads to a contradiction as well (just consider inverse of $g$).

Let $G$ be a real Lie group. Then

  1. $G$ is orientable, and the orientation can be chosen so that it is preserved by the right action of $G$ on itself.
  2. If $G$ is compact, then for a fixed choice of right-invariant orientation on $G$, there exists a unique right-invariant top-form $\omega$ such that

    \begin{equation*}
\int_G \omega = 1
\end{equation*}
  3. The form $\omega$ is also left-invariant and satisfies

    \begin{equation*}
i^* \omega = \big( -1 \big)^{\dim G} \omega \quad \text{where} \quad i(g) = g^{-1}
\end{equation*}
  1. Choose a nonzero element in $\Lambda^n \mathfrak{g}^*$ where $n = \dim G$. Then, it can be uniquely extended to right-invariant top form $\tilde{\omega}$ on $G$. Since this form is non-vanishing on $G$, it shows that $G$ is orientable.
  2. If $G$ is compact, we have a finite integral, i.e.

    \begin{equation*}
\int_G \omega < \infty
\end{equation*}

    Defining $\omega := \frac{\tilde{\omega}}{I}$ we obtain a right-invariant top form satisfying $\int_G \omega = 1$. The uniqueness follows from the identification of the space of right-invariant top forms with $\Lambda^n \mathfrak{g}^*$ and the fact that the latter is 1-dimensional. Therefore, any right-invariant form $\omega' = c \omega$ for some $c$ and the condition $\int_G \omega' = 1$ forces $c = 1$.

  3. To prove that $\omega$ is also left-invariant, it is sufficient to show that it is $\text{Ad}_{}^* \text{-invariant}$. The result then follows from the fact that $\Lambda^n \mathfrak{g}^*$ is one-dimensional and Lemma lemma:4.35-kirillov-real-rep-of-compact-Lie-group-has-unit-length. Finally, let us notice that since $\omega$ is left-invariant, the form $i^* \omega$ is right-invariant, and therefore,

    \begin{equation*}
i^* \omega = c \omega
\end{equation*}

    It is clear that $i_*: \mathfrak{g} \to \mathfrak{g}$ is given by $x \mapsto -x$, thus on $\Lambda^n \mathfrak{g}$, we have

    \begin{equation*}
i_* = - 1^n \quad \implies \quad i^* \omega = \big( -1 \big)^n \omega
\end{equation*}

Choosing the orientation on $B$ and the bi-invariant volume form $\omega$ as in Theorem thm:4.34-kirillov, there exists a unique Borel measure $\dd{g}$ on $G$ such that for any continuous function $f$ on $G$, one has

\begin{equation*}
\int_G f \dd{g} = \int_G f \omega
\end{equation*}

Let $G$ be a compact real Lie group.

Then it has a canonical Borel measure, which is both left- and right-invariant under the involution $i: g \mapsto g^{-1}$, and satisfies

\begin{equation*}
\int_G \dd{g} = 1
\end{equation*}

This measure is called the Haar measure on $G$.

Any finite-dimensional representation of a a compact Lie group $G$ is unitary and thus completely reducible.

Consider a positive-definite inner product $B(\cdot, \cdot)$ on $V$ and "average" it over $G$:

\begin{equation*}
\tilde{B}(v, w) = \int_G B(g \cdot v, g \cdot w) \dd{g}
\end{equation*}

wher $\dd{g}$ is the Haar measure on $G$. Then $\tilde{B}(v, v )> 0$ as an integral of a positive function, and $B(h \cdot v, h \cdot w) = B(v, w)$ since the Haar measure is right-invariant. Therefore, any finite dimensional representation of a compact Lie group is unitary.

Characters
  • Now know that any finite-dim rep. $V$ of a compact Lie group is completely reducible, that is

    \begin{equation*}
V = \bigoplus_i n_i V_i
\end{equation*}

    for some irreducible reps. $V_i$ and positive integers $n_i$, let's consider how we can find the multiplicities $n_i$

  • For this we need character theory
  • Fix a compact Lie group $G$ together with a Haar measure $\dd{g}$
  1. Let $V, W$ be non-isomorphic irreducible reps. of $G$ and $f: V \overset{\sim}{\to} W$. Then

    \begin{equation*}
\int_G g f g^{-1} \dd{g} = 0
\end{equation*}
  2. If $V$ is an irreducible repr. and $f: V \overset{\sim}{\to} V$, then

    \begin{equation*}
\int g f g^{-1} \dd{g} = \frac{\tr(f)}{\dim V} \ \id
\end{equation*}

Let

\begin{equation*}
\tilde{f} = \int_G g f g^{-1} \dd{g}
\end{equation*}

Then $\tilde{f}$ commutes with the action of $G$, i.e.

\begin{equation*}
h \tilde{f} h^{-1} = \int_G \big( hg \big)f \big( hg \big)^{-1} \dd{g} = \tilde{f}
\end{equation*}

By Schur's lemma, we have

  • $\tilde{f} = 0$ if $W \not\simeq V$
  • $\tilde{f} = \lambda \id$ for $W \simeq V$. Since

    \begin{equation*}
\tr( \tilde{f} ) = \tr (f)
\end{equation*}

    and so

    \begin{equation*}
\lambda = \frac{\tr(f)}{\dim V}
\end{equation*}

Let $V$ be a representation of $G$ with a basis $\left\{ v_i \right\}$ and $\left\{ f^i \right\}$ be a dual basis of $V^*$; that is,

\begin{equation*}
\left\langle f^i, v_j \right\rangle = \delta^i_j
\end{equation*}

Matrix coefficients are functions on $G$ of the form

\begin{equation*}
\tensor{\rho}{^{i}_{j}}(g) = \left\langle f^i, \rho(g) v_j \right\rangle
\end{equation*}

If one writes $\rho(g): V \to V$ as a matrix, $\tensor{\rho}{^{i}_{j}}$ represents $(i, j) \text{-th}$ entry; hence the name.

  1. Let $V, W$ be non-isomorphic irreducible reps. of $G$. Choose bases $\left\{ v_j \right\}_{j = 1}^n$ for $V$ and $\left\{ w_a \right\}_{a = 1}^{m}$ for $W$. Then for any $i, j, a, b$ the matrix coeffs. $\tensor{\big( \rho_V \big)}{^{i}_{j}}$ and $\tensor{\big( \rho^W \big)}{^{a}_{b}}$ are orthogonal:

    \begin{equation*}
\Big( \tensor{\big( \rho^V \big)}{^{i}_{j}}, \tensor{\big( \rho^W \big)}{^{a}_{b}} \Big) = 0
\end{equation*}

    where $(\cdot, \cdot)$ is an inner product on $C^{\infty}(G, \mathbb{C})$ given by

    \begin{equation*}
\Big( \psi_1, \psi_2 \Big) = \int_G \psi_1(g) \overline{\psi_2(g)} \ddc{g}, \quad \forall \psi_1, \psi_2 \in C^{\infty}(G, \mathbb{C})
\end{equation*}
  2. Let $V$ be an irreducible rep. of $G$ and let $V_i \in V$ be an orthonormal basis wrt. $G \text{-invariant}$ inner product (which exists, as seen before). Then

    \begin{equation*}
\Big( \tensor{\big( \rho^V \big)}{^{i}_{j}}, \tensor{\big( \rho^W \big)}{^{k}_{l}} \Big) = \frac{\delta_{jl} \delta^{ik}}{\dim V}
\end{equation*}
  1. Choose a pair of indices $i, a$ and apply Lemma lemma:4.42-kirillov to the mapping

    \begin{equation*}
\begin{split}
  E_{ai}: \quad & V \to W \\
  & v_j \mapsto \delta_{ij} w_a
\end{split}
\end{equation*}

    Then, we have

    \begin{equation*}
\int_G \rho_W(g) E_{ai} \rho_V \big( g^{-1} \big) \dd{g} = 0
\end{equation*}

    Rewriting the above equality in matrix form and using the unitarity of $\rho(g)$, $\rho \big( g^{-1} \big) = \overline{\rho(g)^T}$, we obtain for any $b, j$

    \begin{equation*}
\int_G \tensor{\big( \rho_W \big)}{^{b}_{a}}(g) \overline{\tensor{\big( \rho_V \big)}{^{j}_{i}}(g)} \dd{g} = 0
\end{equation*}

    This proves it for orthonormal bases; the general case follows immediately.

  2. Apply Lemma lemma:4.42-kirillov to the matrix unit $E_{lj}}: V \to V$ to obtain

    \begin{equation*}
\sum_{l, j}^{} E_{lj} \int_G \tensor{\big( \rho_W \big)}{^{k}_{l}}(g) \overline{\tensor{\big( \rho_V \big)}{^{i}_{j}}(g)} \dd{g} = \frac{\tr(E^{ki})}{\dim V} \id
\end{equation*}

    concluding our proof.

A character of a representation $V$ is a function on the group defined by

\begin{equation*}
\chi_V(g) = \tr_V \Big( \rho(g) \Big)
\end{equation*}
  1. Let $V = \mathbb{C}$ be a trivial representation. Then

    \begin{equation*}
\chi_V = 1
\end{equation*}
  2. Character of $V \oplus W$:

    \begin{equation*}
\chi_{V \oplus W} = \chi_V + \chi_W
\end{equation*}
  3. Character of tensor product

    \begin{equation*}
\chi_{V \otimes W} = \chi_V \chi_W
\end{equation*}
  4. Characters are class functions:

    \begin{equation*}
\chi_V \big( g h g^{-1} \big) = \chi_V(h)
\end{equation*}
  5. Let $V^*$ be the dual representation of $V$. Then

    \begin{equation*}
\chi_{V^*} = \overline{\chi_V}
\end{equation*}

Let $V, W$ be irreducible representations of a real Lie group $G$. Then

\begin{equation*}
\big( \chi_V, \chi_W \big) = 
\begin{cases}
  0 & \text{if } V \not\simeq W \\
  1 & \text{if } V \simeq W
\end{cases}
\end{equation*}

In other words, if $\hat{G}$ denotes the set of isomorphism classes of irreducible representations of $G$, then $\left\{ \chi_V \mid V \in \hat{G} \right\}$ is an orthonormal family of functions on $G$.

A representation $V$ of $G$ is irreducible if and only if

\begin{equation*}
\big( \chi_V, \chi_V \big) = 1
\end{equation*}

On the other hand, if $V$ is semi-simple, and its decomposition into irreducibles reads $V = \bigoplus_i n_i V_i$, then the multiplicities can be extracted by the following formula

\begin{equation*}
n_i = \big( \chi_{V_i}, \chi_V \big)
\end{equation*}

From def:matrix-coefficients-Lie-theory, the matrix coefficients admits a basis-independent definition, namely for $v \in V$, $f \in V^*$, we set

\begin{equation*}
\rho_{f, g}^V(g) = \left\langle f, \rho_V(g) v \right\rangle
\end{equation*}

then

\begin{equation*}
\tensor{\rho}{^{i}_{j}} = \rho_{f^i, v_j}^V
\end{equation*}

where $\left\{ v_j, f^i \right\}_{i, j = 1}^n$ are dual bases in $V$ and $V^*$. Therefore, for any rep. $V$ of $G$ we obtain a map

\begin{equation*}
\begin{split}
  m(f, v): \quad & V^* \times V \to C^{\infty}(G, \mathbb{C}) \\
  & g \mapsto m(f, v)(g) := \left\langle f, \rho_V(g) v \right\rangle
\end{split}
\end{equation*}

$V^* \otimes V$ has additional structure:

  1. It is a $G \text{-bimodule}$, i.e. admits two commuting actions on $G$, those on two tensor factors
  2. If $V$ is unitary, then the inner product on $V$ defines an inner product on $V^*$

We then define an inner product on $V^* \otimes V$ by

\begin{equation*}
\big( f^1 \otimes v_2, f^2 \otimes v_2 \big) = \frac{1}{\dim(V)} \big( f^1, f^2 \big) \big( v_1, v_2 \big)
\end{equation*}

Let $\hat{G}$ be the set of isomorphism classes of irreducible representations of $G$.

Define the map

\begin{equation*}
\begin{split}
  m: \quad & \bigoplus_{V_i \in \hat{G}} V_i^* \times V_i \to C^{\infty}(G, \mathbb{C}) \\
  & f \otimes v \mapsto m(f \otimes v)(g) := \left\langle f, \rho_V(g) v \right\rangle, \quad \forall g \in G
\end{split}
\end{equation*}

where $\oplus$ is the space of finite linear combinations. Then

  1. The map $G$ is a morphism of $G \text{-bimodules}$:

    \begin{equation*}
\begin{split}
  m \Big( (g \cdot f) \otimes v \Big) &= L_g \big( m(f \otimes v \big) \\
  m \Big( f \otimes (g \cdot v) \Big) &= R_g \big( m( f \otimes v) \big)
\end{split}
\end{equation*}

    where $L_g$ and $R_g$ are the left- and right-actions of $G$ on $C^{\infty}(G, \mathbb{C})$, respectively.

  2. The map $m$ respects the inner product, if it is defined by

    \begin{equation*}
 \big( f^1 \otimes v_2, f^2 \otimes v_2 \big) = \frac{1}{\dim(V)} \big( f^1, f^2 \big) \big( v_1, v_2 \big)
 \end{equation*}

    on $V_i^* \otimes V_i$, and on the $C^{\infty}(G)$ by the formula from the previous lecture:

    \begin{equation*}
(f^1, f^2) = \int_G f^1(g) \overline{f^2(g)} \dd{g}
\end{equation*}
  1. Follow immediately from the orthogonality theorem from before.
  2. Follows by direct computation:

    \begin{equation*}
\begin{split}
  L_g \big( m(f \otimes v) \big)(h) &= m(f \otimes v)\big( g^{-1} h \big) = \left\langle f, \rho_V \big( g^{-1} h \big) v \right\rangle = \left\langle f, \rho_V \big( g^{-1} \big) \rho_V \big( h \big) v \right\rangle \\
  &= \left\langle \underbrace{\big( \rho(g)^{-1} \big)^*}_{= \rho_{V^*}(g)} f, \rho_V(h) v \right\rangle = m \Big( (g \cdot f) \otimes v \Big)
\end{split}
\end{equation*}

    and

    \begin{equation*}
\begin{split}
  R_g \big( m(f \otimes v) \big)(h) &= m(f \otimes v) (hg) = \left\langle f, \rho_V(hg) v \right\rangle = \left\langle f, \rho_V(h) \rho_V(g) v \right\rangle \\
  &= m \big( f \otimes (g \cdot v) \big)
\end{split}
\end{equation*}

    as wanted.

$m$ is injective.

This follows immediately from the respecting the inner product.

The map $m$ in Theorem thm:120-lectures defines an isomorphism

\begin{equation*}
m: \widehat{\bigoplus}_{V_i \in \hat{G}} \big( V_i^* \otimes V_i \big) \to L^2(G, \dd{g})
\end{equation*}

where

  • $\widehat{\bigoplus}$ is the Hilbert space direct sum, i.e. the completion of the algebraic direct sum wrt. the metric given by inner product

    \begin{equation*}
\big( f^1 \otimes v_2, f^2 \otimes v_2 \big) = \frac{1}{\dim(V)} \big( f^1, f^2 \big) \big( v_1, v_2 \big)
\end{equation*}
  • $L^2(G, \dd{g})$ is the Hilbert space of complex-valued square-integrable functions on $G$ wrt. the Haar measure, wrt. inner product

    \begin{equation*}
(f^1, f^2) = \int_G f^1(g) \overline{f^2(g)} \dd{g}
\end{equation*}

The set of characters $\left\{ \chi_V, V \in \hat{G} \right\}$ is an orthonormal basis (in the sense of Hilbert space) of the space $\big( L^2(G, \dd{g}) \big)^G$ of conjugation-invariant functions on $G$.

Example: $G = S^1 = \mathbb{R} / \mathbb{Z}$
  • The Haar measure on $G$ is given by $\dd{x}$ and the irreducible representations are parametrized by $\mathbb{Z}$
  • Therefore, the orthogonality relation is given by

    \begin{equation*}
\int_{0}^{1} e^{2 \pi i k a} \overline{e^{2 \pi i ka}} \dd{x} = \delta_{kl}
\end{equation*}

    which is the usual ortogonality relation for exponents.

  • Peter-Weyl theorem in this cases just says that the exponents $e^{2 \pi i k x}$ for $k \in \mathbb{Z}$ form an orthonormal basis of $L^2(S^1, \dd{x})$, which is one of the main statements of the theory of Fourier series!
    • Every $L^2$ function $f(x)$ on $S^1$ can be written as a series

      \begin{equation*}
f(x) = \sum_{k \in \mathbb{Z}}^{} c_k e^{2 \pi i k x}
\end{equation*}

      which converges in the $L^2$ metric.

    • For this reason, the study of the structure of $L^2(G)$ can be considered as a generalization of harmonic analysis!

Pretty neat stuff, if I might say so myself.

Representation of Lie groups

A representation of a Lie group $\big( G, \bullet \big)$ is a Lie group homomorphism

\begin{equation*}
R: G \to \text{GL}(V)
\end{equation*}

for some finite-dimensional vector space $V$.

Recall that $R: G \to \text{GL}(V)$ is a Lie group homomorphism if it is smooth and

\begin{equation*}
R(g_1 \bullet g_2) = R(g_1) \circ R(g_2), \quad \forall g_1, g_2 \in G
\end{equation*}

Let $G$ be a Lie group. For each $g \in G$, we define the Adjoint map:

\begin{equation*}
\begin{split}
  \Ad_{_g}: \quad & G \to G \\
  & h \mapsto g h g^{-1}
\end{split}
\end{equation*}

Notice the capital "A" here to distinguish from the adjoint map of a Lie algebra.

Since $\mathrm{Ad}_{g}$ is a composition of a Lie group, then it's a smooth map. Further,

\begin{equation*}
\big( \mathrm{Ad}_{g_*} \big)_e: T_e G \overset{\sim}{\to} T_{\mathrm{Ad}_{g}(e)} G = T_e G
\end{equation*}

Thus, $\big( \Ad_{g_*} \big)_e$ yields a representation of $G$ with $\rho(g) = \big( \Ad_{g} \big)_*$. Thus, $\big( \Ad_{g_*} \big)_e \in \text{End}(T_e G)$, i.e. it yields a representation of $G$ on $\text{Vect}(G)$.

Consider the following map

\begin{equation*}
\begin{split}
  \Ad_{}: \quad & G \to \text{End}(T_e G) \\
  & g \mapsto \Ad_{g_*}
\end{split}
\end{equation*}

where $\Ad_{g_*} = \big( \Ad_{g_*} \big)_e$ since we are in $T_e G$.

Then

\begin{equation*}
\Ad_{*} = \ad
\end{equation*}

as operators on $T_e G$, where $\ad_{}$ is the adjoint map of the Lie algebra. To make it explicit, this means

\begin{equation*}
\ad(X)(Y) = \comm{X}{Y} = \Ad_{*} (X)(Y) = \comm{X}{Y}, \quad \forall X, Y \in T_e G
\end{equation*}

By definition of $\Ad_{}$ we have

\begin{equation*}
\Ad(g)(X) = \dv{}{t} \bigg|_{t = 0} \big( g \exp(t X) g^{-1} \big) = g X g^{-1}
\end{equation*}

Thus, $\Ad_{*}$ is defined as

\begin{equation*}
\begin{split}
  \Ad_{*}(X)(Y)
  &= \dv{}{s} \bigg|_{s = 0} \dv{}{t} \bigg|_{t = 0} \exp(s X) \exp(t Y) \exp (- s X) \\
  &= \dv{}{s} \bigg|_{s = 0} \dv{}{t} \bigg|_{t = 0} \exp \Big(tY + ts (XY - YX) + \dots \Big) \\
  &= XY - YX \\
  &= \comm{X}{Y} \\
  &= \ad_{X}(Y)
\end{split}
\end{equation*}

$\text{Ad}_{}: G \to \Aut(\mathfrak{g})$ is a group homomorphism.

Let $X \in \mathfrak{g}$ and $g_1, g_2 \in G$, then

\begin{equation*}
\text{Ad}_{g_1} \big( \text{Ad}_{g_2} (X) \big) = \text{Ad}_{g_1} \big( g_2 X g_2^{-1} \big) = g_1 g_2 X g_2^{-1} g_1^{-1} = \text{Ad}_{g_1 g_2} (X)
\end{equation*}

Moreover, the image of $\text{Ad}_{}$ is invertible endomorphisms, i.e. automorphisms since if $g \in G$, then

\begin{equation*}
\text{Ad}_{g^{-1}} \big( \text{Ad}_{g}(X) \big) = X
\end{equation*}

which holds for all $g \in G$.

Lie algebra of vector fields

Stabilizers and centers

Let

Then

Furhermore, if $\text{im}(f)$ is a (embedded) submanifold adn thus a closed Lie subgroup, we have a Lie group isomorphism

\begin{equation*}
\text{im}(f) \simeq G_1 / \text{ker}(f)
\end{equation*}

Let $V$ be a represention of Lie group $G$, and $v \in V$.

Then the stabilizer

\begin{equation*}
G_v := \left\{ g \in G : g \cdot v = v \right\}
\end{equation*}

is a closed Lie subgroup in $G$ with Lie algebra

\begin{equation*}
\text{Lie}(G_v) = \left\{ x \in \mathfrak{g} \mid x \cdot v = 0 \right\}
\end{equation*}

Example 3.32 in Kirillov: WRITE IT OUT

Let $\mathfrak{g}$ be a Lie algebra. The center of $\mathfrak{g}$ is defined by

\begin{equation*}
\mathfrak{z}(\mathfrak{g}) = \left\{ x \in \mathfrak{g} \mid \comm{x}{y} = 0, \quad \forall y \in \mathfrak{g} \right\}
\end{equation*}

$\mathfrak{z}(\mathfrak{g})$ is an ideal in $\mathfrak{g}$.

Let $G$ be a connected Lie group. Then its center $Z(G)$ is a closed Lie group with Lie algebra $\mathfrak{z}(\mathfrak{g})$.

If $G$ is not connected, then $Z(G)$ is still a closed Lie subgroup; however, its Lie algebra might be "smaller" than $\mathfrak{z}(\mathfrak{g})$.

If $g \in G$ and $x \in \mathfrak{g}$, then

\begin{equation*}
\Ad_{g} \big( \exp(tx) \big) = \exp \Big( \ad_{g}(tx) \Big)
\end{equation*}

Since $G$ is connected, then by $\exp(tx)$

\begin{equation*}
g \in Z(G) \iff g \in \text{ker}(\Ad_{}), \quad \Ad_{}: G \to \text{GL}(\mathfrak{g})
\end{equation*}

implies

\begin{equation*}
Z(G) = \text{ker}(\Ad_{})
\end{equation*}

and

\begin{equation*}
\text{Lie} \big( Z(G) \big) = \text{ker}(\ad_{}) = Z(g)
\end{equation*}

Let $G$ be connected.

The Adjoint group

\begin{equation*}
\Ad_{} G := G / Z(G)
\end{equation*}

Examples:

\begin{equation*}
\begin{split}
  \text{PSL}(n, \mathbb{K}) &= \text{SL}(n, \mathbb{K}) / Z \\
  \text{PGL}(n, \mathbb{K}) &= \text{GL}(n, \mathbb{K}) / Z
\end{split}
\end{equation*}
TODO When $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$, when $\text{PGL} \simeq \text{PSL}$?

Fundmental Theorems of Lie Theory

  1. All Lie groups $G$ has a canonical structure of a Lie algebra on $T_e G = \mathfrak{g}$.
    • For every morphism of Lie groups $f: G_1 \to G_2$, we have $f_*: \mathfrak{g}_1 \to \mathfrak{g}_2$.
    • Moreover, if $G_1$ is connected then $\text{Hom}(G_1, G_2) \to \text{Hom}(\mathfrak{g}_1, \mathfrak{g}_2)$ is injective.
    • Q: When is it also surjective?
      • $G = S^1 = \mathbb{R} / \mathbb{Z}$ and $G_2 = \mathbb{R}$ then

        \begin{equation*}
\mathfrak{g}_1 = \mathfrak{g}_2 = \mathbb{R}
\end{equation*}

        Consider the identity map $\text{id}: \mathfrak{g}_1 \overset{\sim}{\longrightarrow} \mathfrak{g}_2$. Is there a morphism $\mathbb{R} / \mathbb{Z}$ which is locally $\text{id}$?

        • A: NO! Suppose it exists, then it should be given by $\theta \mapsto \theta$; on the other hand, it must also satisfy $f(\mathbb{Z}) = \left\{ 0 \right\}$. This, this morphism of Lie algebras can not be lifted to a morphism of Lie groups.
  2. Every Lie subgroup $H \subset G$ defines a Lie subalgebra $\mathfrak{h} \subset \mathfrak{g}$.
    • Q: Does every Lie subalgebra come as $\text{Lie}(H)$ for some subgroup $H \subset G$?
      • A: For any Lie group $G$ there is a bijection between connected Lie subgroups and Lie subalgebras.
  3. Given a topology of a Lie group, its group law can be recovered from commutator.

Any real or complex Lie algebra, is isomorphic to an algebra of some Lie group.

For any real or complex finite-dimensional Lie algebra $\mathfrak{g}$, there is a unique (up to isomorphism) connected, simply-connected Lie group $G$ (respectively, real or complex) with

\begin{equation*}
\text{Lie}(G) = \mathfrak{g}
\end{equation*}

Moreover, any other connected Lie group $G'$ with $\text{Lie}(G') = \mathfrak{g}$, i.e. same Lie algebra, must be of the form $G / Z$ for some discrete central subgroup $Z \subset G$, and $G$ is a universal cover of $G'$.

By Lie's 3rd theorem, $\exists$ a Lie group $G$ s.t. $\text{Lie}(G) = \mathfrak{g}$. Then just let $\tilde{G}$ be the universal cover of the connected component of $G$ containing $e$. Suppose $G'$ is connected, simply-connected, and $\text{Lie}(G') = \mathfrak{g}$. Then the morphism $\text{id}: \mathfrak{g} \overset{\sim}{\longrightarrow} \mathfrak{g}$ lifts to $f: G \to G'$ which is locally an identity $\implies$ $\text{ker}(f)$ is discrete $\implies$ $G$ is a covering space of $G'$ so

\begin{equation*}
\text{ker}(f) = \pi_1 (G') \text{ is commutative}
\end{equation*}

Complexification / Real forms of Lie algebras and Lie groups

Complexification

Let $\mathfrak{g}$ be a real Lie algebra.

The complexification of $\mathfrak{g}$ is the complex Lie algebra

\begin{equation*}
\mathfrak{g}_{\mathbb{C}} := \mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C} = \mathfrak{g} \oplus i \mathfrak{g}
\end{equation*}

where $i^2 = -1$ as usual.

The bracket on $\mathfrak{g}_{\mathbb{C}}$ induced by $\comm{\cdot}{\cdot}_{\mathfrak{g}$ by $\mathbb{C} \text{-linearity}$

\begin{equation*}
\comm{x + iy}{\tilde{x} + i \tilde{y}} = \underbrace{\bigg( \comm{x}{\tilde{x}} - \comm{y}{\tilde{y}} \bigg)}_{= \mathfrak{g}} + \underbrace{\bigg( \comm{x}{\tilde{y}} + \comm{\tilde{x}}{y} \bigg)}_{= \mathfrak{g}}
\end{equation*}

Conversely, we say that $\mathfrak{g}$ is a real form of $\mathfrak{g}_{\mathbb{C}}$.

  • Examples
    • $\mathfrak{gl}(n, \mathbb{C}) = \mathfrak{gl}(, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C}$
    • $\mathfrak{gl}(n, \mathbb{C}) = \mathfrak{u}(n) \otimes_{\mathbb{R}} \mathbb{C} = \mathfrak{u} \oplus i \mathfrak{u}(n)$
      • Recall that $Y \in \mathfrak{u}(n) \oplus i \mathfrak{u}(n)$ then $Y + \overline{Y}^T = 0$
      • To see $\mathfrak{gl}(n, \mathbb{C}) \subseteq \mathfrak{u}(n) \oplus i \mathfrak{u}(n)$:
        • $X \in \mathfrak{gl}(, \mathbb{C})$ then $X - \overline{X}^T \in \mathfrak{u}(n)$
        • Similarily, $i \big( X + \overline{X}^T \big) \in \mathfrak{u}(n)$
        • So we can write

          \begin{equation*}
X = \frac{X - \overline{X}^T}{2} + i \bigg( - \frac{i \big( X + \overline{X}^T \big)}{2} \bigg)
\end{equation*}

          which shows $\mathfrak{gl}(n, \mathbb{C}) \subseteq \mathfrak{u}(n) \oplus i \mathfrak{u}(n)$

      • For $\mathfrak{u}(n) \oplus i \mathfrak{u}(n) \subseteq \mathfrak{gl}(n, \mathbb{C})$ we ?
Real form

Let $G$ be a connected complex Lie group with $\mathfrak{g} = \text{Lie}(G)$.

A real form of $G$ is a closed real Lie subgroup $K \subseteq G$ s.t.

\begin{equation*}
\mathfrak{k} = \text{Lie}(K)
\end{equation*}

is a real form of $\mathfrak{g}$.

Let $G$ be connected and simply-connected, with $\mathfrak{g} = \text{Lie}(G)$.

Then for every real form $\mathfrak{k} \subseteq \mathfrak{g}$ there exists a real form $K \subseteq G$ s.t.

\begin{equation*}
\mathfrak{k} = \text{Lie}(K)
\end{equation*}
  • Examples
    • $\text{SU}(n)$ is compact real form $\text{SL}(n, \mathbb{C})$
  • Lie algebras of $\mathfrak{so}(3, \mathbb{R})$, $\mathfrak{su}(2)$ and $\mathfrak{sl}(2, \mathbb{c})$

    $\mathfrak{so}(3, \mathbb{R})$ has a basis

    \begin{equation*}
J_x = 
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & -1 \\
  0 & 1 & 0
\end{pmatrix}, \quad
J_y = 
\begin{pmatrix}
  0 & 0 & -1 \\
  0 & 0 & 0 \\
  1 & 0 & 0
\end{pmatrix}, \quad
J_z =
\begin{pmatrix}
  0 & -1 & 0 \\
  1 & 0 & 0 \\
  0 & 0 & 0
\end{pmatrix}
\end{equation*}

    with commutation relations

    \begin{equation*}
\comm{J_x}{J_y} = J_z, \quad \comm{J_y}{J_z} = J_z, \quad \comm{J_z}{J_x} = J_y
\end{equation*}

    $\mathfrak{su}(2)$ have basis

    \begin{equation*}
i \sigma_1 = 
\begin{pmatrix}
  0 & i  \\ i & 0
\end{pmatrix}, \quad
i \sigma_2 = 
\begin{pmatrix}
  0 & 1 \\ -1 & 0
\end{pmatrix}, \quad
i \sigma_3 = 
\begin{pmatrix}
  i & 0 \\ 0 & -i
\end{pmatrix}
\end{equation*}

    which commutation relations

    \begin{equation*}
\comm{i \sigma_1}{i \sigma_2} = - 2 i \sigma_3, \quad \comm{i \sigma_2}{i \sigma_3} = - 2 i \sigma_1, \quad \comm{i \sigma_3}{i \sigma_1} = - 2 i \sigma_2
\end{equation*}

    $\mathfrak{sl}(2, \mathbb{C})$ has basis

    \begin{equation*}
e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad f = \begin{pmatrix} 0 & 0 \\\ 1 & 0 \end{pmatrix}, \quad h = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix}
\end{equation*}

    with commutation relations

    \begin{equation*}
\comm{e}{f} = h, \quad \comm{h}{e} = + 2e, \quad \comm{h}{f} = - 2f
\end{equation*}

    The isomorphism $\mathfrak{su}(2) \longrightarrow \mathfrak{so}(3, \mathbb{R})$ is defined by

    \begin{equation*}
i \sigma_1 \mapsto - 2 J_x, \quad i \sigma_2 \mapsto - 2 J_y, \quad i \sigma_3 \mapsto - 2 J_z
\end{equation*}

    Since we saw earlier that $\mathfrak{gl}(n, \mathbb{C}) = \mathfrak{u}(n) \oplus i \mathfrak{u}(n)$, we can argue from this that $\mathfrak{sl}(2, \mathbb{C})$ is a complexification of $\mathfrak{su}(2)$, i.e. $\mathfrak{sl}(2, \mathbb{C}) \simeq \mathfrak{su}(2) \otimes_{\mathbb{R}} \mathbb{C}$, where we have the additional constraint of the matrices being traceless. The explicit isomorphism $\mathfrak{su}(2) \longrightarrow \mathfrak{sl}(2, \mathbb{C})$ is defined

    \begin{equation*}
i \sigma_1 \mapsto i \big( e + f \big), \quad i \sigma_2 \mapsto e - f, \quad i \sigma_3 \mapsto i h
\end{equation*}

    Finally, the isomorphism $\mathfrak{so}(3, \mathbb{R}) \longrightarrow \mathfrak{sl}(2, \mathbb{C})$ is given by

    \begin{equation*}
J_x \mapsto - \frac{i}{2} (e + f), \quad J_y \mapsto \frac{1}{2} \big( f - e \big), \quad J_z \mapsto - \frac{i}{2} h
\end{equation*}
Representations

Let $V$ be some $\mathbb{K} \text{-vector space}$ with $\mathbb{K} \in \left\{ \mathbb{R}, \mathbb{C} \right\}$.

\begin{equation*}
\begin{split}
  \rho_V : & G \underset{\text{map of Lie gp}}{\longrightarrow} \text{GL}(V) = \text{Aut}_{\mathbb{K}}(V) \\
  \pi_V: & \mathfrak{g} \underset{\text{map of Lie alg}}{\longrightarrow} \mathfrak{gl}(V) = \text{End}_{\mathbb{K}}(V)
\end{split}
\end{equation*}

A representation is a pair $\big( V, \pi_V \big)$.

And given two representations $\big( V, \pi_V \big)$ and $\big( W, \pi_W \big)$ a morphism of representations is a linear map $f: V \to W$ s.t. the following diagram commutes:

morphism-of-representation.png

$G$ is Lie group with $\mathfrak{g} = \text{Lie}(G)$

  1. Every $G \text{-rep}$ $\rho: G \to \text{GL}(V)$ lifts to a $\mathfrak{g} \text{-rep}$ by $\rho_*: \mathfrak{g} \to \mathfrak{gl}(V)$. Moreover,

    \begin{equation*}
\text{Hom}_{G}(V, W) \subseteq \text{Hom}_{g}(V, W)
\end{equation*}
  2. If $G$ is connected and simply-connected, then

    \begin{equation*}
\text{Hom}_{G}(V, W) = \text{Hom}_{g}(V, W)
\end{equation*}

    which is equivalent to saying that the category of G-representations is equal to the category of $\mathfrak{g}$ representations.

Let $\mathfrak{g}$ be a real Lie algebra, and $\mathfrak{g}_{\mathbb{C}}$ be its complexification. The any complex representation of $\mathfrak{g}$ has a unique structure of representation of $\mathfrak{g}_{\mathbb{C}}$, and

\begin{equation*}
\Hom_{\mathfrak{g}}(V, W) \simeq \Hom_{\mathfrak{g}_{\mathbb{C}}}(V, W)
\end{equation*}

In other words, categories of complex representations of $\mathfrak{g}$ and $\mathfrak{g}_{\mathbb{C}}$ are equivalent.

This tells us that if you want to work with complex vector spaces, then there is no reason to work with a real form rather than using the complex version, since they are equal (and over a complex vector space, it's easier to work with complex Lie algebra).

  • Examples
    • Trivial representation: $V = \mathbb{C}$
      • Group rep.: $\rho(g) = \id$ for any $g \in G$
      • Alg. rep.: $\rho(x) = 0$ for any $x \in \mathfrak{g}$
    • Adjoint rep: $V = \mathfrak{g}$
      • Group rep.: $\rho(g) = \text{Ad}_{g}$
      • Alg. rep.: $\rho(x) = \text{ad}_{x}$

Operations on representations

Let $V$ be a representation of a Lie group $G$.

A subspace $W \subset V$ is called a suprepresentation of $G$ if it's stable under the action by $\rho_V(g)$, i.e.

\begin{equation*}
\rho_V(g) W \subset W, \quad \forall g \in G
\end{equation*}

Let $V$ be a representation of a Lie algebra $\mathfrak{g}$.

A subspace $W \subset V$ is called a subrepresentation of $\mathfrak{g}$ if it's stable under the action by $\rho_V(x)$, i.e.

\begin{equation*}
\rho_V(x) W \subset W, \quad \forall x \in \mathfrak{g}
\end{equation*}

Let $V$ be a representation (of group or algebra) and $W \subset V$ a subrepresentation (of group or algebra).

Then the quotient space $V / W$ carries a structure of a representation and is called the factor representation or the quotient representation.

Let $G$ be a connected Lie group with a representation $V$.

Then $W \subset V$ is a subrep. for $G$ if and only if $W$ is a subrep. for $\mathfrak{g}$

Let $V, W$ be representations of $G$ (respectively, $\mathfrak{g}$).

Then there is a canoncial structure of a representation on

  1. $V^*$: let $A^*: V^* \to V^*$ denote the adjoint of an operator $A: V \to V$, then we have the following reps.:

    \begin{equation*}
\begin{split}
  \rho_{V^*}(g) &= \big( \rho_V(g)^{-1} \big)^*, \quad \forall g \in G \\
  \rho_{V^*}(x) &= - \big( \rho_V(x) \big)^*, \quad \forall x \in \mathfrak{g}
\end{split}
\end{equation*}
  2. $V \oplus W$:

    \begin{equation*}
\begin{split}
  \rho_{V \oplus W}(g) (v \oplus w) &= \big( \rho_V(g) v \big) \oplus \big( \rho_W(g) w \big), \quad \forall g \in G \\
  \rho_{V \oplus W}(x) &= \big( \rho_V(x) v \big) \oplus \big( \rho_W(x) w \big), \quad \forall x \in \mathfrak{g} 
\end{split}
\end{equation*}
  3. $V \otimes W$:

    \begin{equation*}
\begin{split}
  \rho_{V \otimes W}(g) (v \otimes w) &= \big( \rho_V(g) v \big) \otimes \big( \rho_W(g) w \big), \quad \forall g \in G \\
  \rho_{V \otimes W}(x) (v \otimes w) &= \big( \rho_V(x) v \big) \otimes w + v \otimes \big( \rho_W(x) w \big), \quad x \in \mathfrak{g}
\end{split}
\end{equation*}

Let $g \in G$ and $\gamma: \mathbb{R} \to G$ be a "curve" (one-parameter subgroup) such that $\dot{\gamma}(0) = x \in \mathfrak{g}$.

  1. Let

    \begin{equation*}
\rho_{V^*}(g) = \big( \rho_V(g)^{-1} \big)^*
\end{equation*}

    where for an operator $A: V \to V$ we denote the adjoint $A^*: V^* \to V^*$. Observe then that this preserves the natural pairing between $v \in V$ and $\varphi_v \in V^*$:

    \begin{equation*}
\left\langle \rho_V(g) v, \rho_{V^*}(g) \varphi_{v} \right\rangle = \left\langle \rho_V(g)^{-1} \rho_V(g) v, \varphi_v \right\rangle = \left\langle v, \varphi_v \right\rangle
\end{equation*}

    The corresponding map for the Lie algebra is obtained by simply taking th derivative of some integral curve:

    \begin{equation*}
\begin{split}
  0 = \dv{}{t}\bigg|_{t = 0}  \left\langle v, \varphi_v \right\rangle &= \dv{}{t}\bigg|_{t = 0} \left\langle \rho_V \big( \gamma(t) \big) v, \rho_{V^*} \big( \gamma(t) \big) \varphi_v \right\rangle \\
  &= \left\langle \rho_V(x) v, \varphi_{v} \right\rangle + \left\langle v, \rho_{V^*}(x) \varphi_v \right\rangle
\end{split}
\end{equation*}

    by the product rule. Therefore,

    \begin{equation*}
\left\langle \rho_V(x) v, \varphi_v \right\rangle = - \left\langle v, \rho_{V^*}(x) \varphi_v \right\rangle
\end{equation*}

    Which implies that

    \begin{equation*}
\rho_{V^*}(x) = - \rho_V(x)^*
\end{equation*}

    i.e. the adjoint of $\rho_V(x)$.

  2. Since $\rho_V(g) \in \GL(V)$, we simply define the representation on $V \oplus W$ by the linearisation of the action on $V$ and $W$ respectively. That is, if $v \in V$ and $w \in W$,

    \begin{equation*}
\rho_{V \oplus W}(g) \big( v \oplus w \big) = \rho_V(g) v \oplus \rho_{W}(g) w
\end{equation*}

    Due to the linearity of $\oplus$, the Lie algebra rep. is identitical (since the derivative-operator is linear):

    \begin{equation*}
\rho_{V \oplus W}(x)(v \oplus w) = \rho_V(x) v \oplus \rho_W(x) w
\end{equation*}
  3. The group rep. on $V \otimes W$ is simply

    \begin{equation*}
\rho_{V \otimes W}(g)(v \otimes w) = \big( \rho_V(g) v \big) \otimes \big( \rho_W(g) w \big)
\end{equation*}

    But the algebra rep. is slightly more tricky. A naive approach would be to define it similarily as the group rep. but this does in fact not define a reresentation (it is not even linear in $x$!). We again consider the derivative of $\gamma(t) \in G$:

    \begin{equation*}
\begin{split}
  \rho_{V \otimes W}(x)(v \otimes w) &= \dv{}{t}\bigg|_{t = 0} \rho_{V \otimes W} \big( \gamma(t) \big) (v \otimes w) \\
  &= \dv{}{t}\bigg|_{t = 0} \Big( \rho_V \big( \gamma(t) \big) v \otimes \rho_W \big( \gamma(t) \big) w \Big) \\
  &= \bigg( \dv{}{t}\bigg|_{t = 0} \rho_V \big( \gamma(t) \big) v \bigg) \otimes w + v \otimes \bigg( \dv{}{t}\bigg|_{t = 0} \rho_W \big( \gamma(t) \big) w \bigg) \\
  &= \big( \rho_V(x) v  \big) \otimes w + v \otimes \big( \rho_W(x) w \big)
\end{split}
\end{equation*}

    where we made use of the Leibniz rule.

Following Lemma lemma:4.10-kirillov-canonical-representations-of-dual-direct-sum-and-tensor-product we immediately have a canonical representation on the tensor space

\begin{equation*}
V^{\otimes k} \otimes \big( V^* \big)^{\otimes l}
\end{equation*}

The coadjoint representation of a Lie group $G$ and its Lie algebra $\mathfrak{g}$ is the dual vector space $\mathfrak{g}^*$ together with the following actions of $g$ and $\mathfrak{g}$ respectively:

\begin{equation*}
\begin{split}
  \left\langle \Ad_{g}^*(\varphi), y \right\rangle &= \left\langle \varphi, \Ad_{g}^{-1}(y) \right\rangle  \quad \text{for any} \quad g \in G, \ y \in \mathfrak{g}, \ \varphi \in \mathfrak{g}^* \\
  \left\langle \ad_{x}^*(\varphi), y \right\rangle &= - \left\langle \varphi, \ad_{x}(y) \right\rangle \quad \text{for any} \quad x, y \in \mathfrak{g}, \ \varphi \in \mathfrak{g}^* 
\end{split}
\end{equation*}

Let $V, W$ be a representation of $G$ or $\mathfrak{g}$.

The space $\Hom(V, W) \simeq W \oplus V^*$ is naturally a representation of $G$ and $\mathfrak{g}$, respectively. More precisely, we have

\begin{equation*}
\begin{split}
  \rho(g) A = \rho_{W}(g) A \rho_V(g)^{-1} \quad \text{for any} \quad g \in G, \ A \in \Hom_G(V, W) \\
  \rho_(x) A = \rho_{W}(x) A - A \rho_V(x) \quad \text{for any} \quad x \in \mathfrak{g}, \ A \in \Hom_{\mathfrak{g}}(V, W)
\end{split}
\end{equation*}

which easily follows from the derivative of a one-parameter subgroup $\gamma(t)$.

Let $V$ be a representation of $G$ or $\mathfrak{g}$, respectively.

Then the space of bilinear forms on $V$ is a representation as well:

\begin{alignat*}{2}
  \rho(g) B(v, w) &= B \big( \rho_V(g)^{-1} v, \rho_V(g)^{-1} w \big), \qquad & g \in G \\
  \rho(x) B(v, w) &= - \bigg( B \big( \rho_V(x), w \big) + B \big( v, \rho_V(x) w \big) \bigg), \quad & x \in \mathfrak{g}
\end{alignat*}

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and representation $V$.

Then a vector $v \in V$ is called $G \text{-invariant}$ if

\begin{equation*}
\rho(g) v = v, \quad \forall g \in G
\end{equation*}

The subspace of all $G \text{-invariant}$ vectors is denoted $V^G$.

Similarily, $v$ is called $\mathfrak{g} \text{-invariant}$ if

\begin{equation*}
\rho(x) v = 0, \quad \forall x \in \mathfrak{g}
\end{equation*}

The subspace of all $\mathfrak{g} \text{-invariant}$ vectors is denoted $V^{\mathfrak{g}}$.

If $G$ is a connected Lie group, then the the $G \text{-invariant}$ and $\mathfrak{g} \text{-invariant}$ vector spaces are equal, i.e.

\begin{equation*}
V^G = V^{\mathfrak{g}}
\end{equation*}

In general, one can easily see that $V^{\mathfrak{g}} \subseteq V^G$, since

\begin{equation*}
\dv{}{t}\bigg|_{t = 0} \underbrace{\rho \big( \gamma(t) \big) v}_{= v} = \dv{}{t}\bigg|_{t = 0} v = 0 \quad \implies \quad v \in V^{\mathfrak{g}}
\end{equation*}

since $v$ is just a constant…

Let $V, W$ be $G \text{-reps.}$, then

\begin{equation*}
\big( \Hom(V, W) \big)^G = \Hom_G(V, W)
\end{equation*}

In particular, considering $\mathbb{C}$ as a trivial $G \text{-rep}$ one gets

\begin{equation*}
V^G = \Hom(\mathbb{C}, V)^G = \Hom_G(\mathbb{C}, V)
\end{equation*}

Reconstruction of Lie group from it's Lie algebra

Notation

  • $\exp |_V \to \exp(V) \subseteq G$ denotes the $\exp$ map restricted to the vector space $V$

Stuff

$M$ is a smooth manifold and $Y$ be is a smooth vector field on $M$.

Then a smooth curve

\begin{equation*}
\gamma : (a, b) \to M
\end{equation*}

is called an integral curve if

\begin{equation*}
\forall \lambda \in (a, b) : X_{\gamma, \gamma(\lambda)} = Y_{\gamma(\lambda)}
\end{equation*}

There is a unique integral curve of $Y$ through each point of the manifold $M$.

This follows from the existence and uniqueness of solutions to ordinary differential equations.

Given any $Y \in \Gamma(TM)$, and any $p \in M$, there exists $\varepsilon > 0$ and a smooth curve $\gamma: (-\varepsilon, \varepsilon) \to M$ with

\begin{equation*}
\gamma(0) = p
\end{equation*}

The maximal integral curve of $Y \in \Gamma(TM)$ through $p \in M$ is the unique integral curve $\gamma: I_{\mathrm{max}}^p \to M$ of $Y$ through $p$, where

\begin{equation*}
I_{\mathrm{max}}^p := \bigcup \left\{ I \subseteq \mathbb{R} \mid \text{there exists an integral curve } \gamma: I \to M \text{ of } Y \text{ through } p \right\}
\end{equation*}

So we're just taking the "largest" interval for which there exists integral curves "locally" for $Y$.

In general, $I_{\mathrm{max}}^p$ differ from point to point for a given vector field $Y$.

An integral curve $\gamma$ of a vector field $Y$ is called complete if its domain can be extended to $(a, b) = R$.

Using the definition of maximal integral curve, we say a vector field $Y$ is complete if $I_{\mathrm{max}}^p = \mathbb{R}$ for all $p \in M$.

On a compact manifold, every vector field is complete.

Every left-invariant vector field on a Lie group $G$ is complete.

Let $A \in T_e G$ and define the thus uniquely determined left-invariant vector field $X^A$:

\begin{equation*}
X_g^A := \ell_{g^*} A
\end{equation*}

Then let $\gamma^A : \mathbb{R} \to G$ be the integral curve of $X^A$ (which we can due to this theorem) through the point

\begin{equation*}
\gamma(0) = e
\end{equation*}

This defines the so-called exponential map

\begin{equation*}
\begin{split}
  \exp: \quad & T_e G \to G \\
  & \exp(A) := \gamma^A(1)
\end{split}
\end{equation*}

It might be a bit clearer if one writes

\begin{equation*}
\exp(t A) \big|_{t = 0} := \gamma^A(t) \big|_{t = 0}
\end{equation*}
  1. Derivative

    \begin{equation*}
\dv{}{t} \bigg|_{t = 0} \gamma_A(\lambda A) \quad \implies \quad \gamma_A(\lambda t ) = \gamma_{\lambda A}(t)
\end{equation*}

    which implies

    \begin{equation*}
\exp(t A ) = \gamma_{t A}(1) = \gamma_{A}(t)
\end{equation*}
  2. If $X^A \in \Gamma(TM)$ is a left-invariant vector field implies that time $t$ flow along $X^A$ is given by

    \begin{equation*}
g \mapsto g \exp (tA)
\end{equation*}

    (and if we replace left-invariant with right-invariant we have $g \mapsto \exp(t A) g$)

Some places you might see people talking about using infinitesimally small generators, say $T$. Then they will write something like "we can generate the group from the generator $T$ by expanding about the identity:

\begin{equation*}
\exp(i \alpha T) \approx 1 + i \alpha T
\end{equation*}

where $1 + i \alpha T$ is our Lie group element generated from the Lie group element $T$!

Now, there are several things to notice here:

  • Factor of $i$; this is just convention, and might be convenient in some cases
  • $\alpha$ is just what we refer to as $t$
  • This is really not a rigorous way to do things…

Finally, and this is the interesting bit IMO, one thing that this notation might make a bit clearer than our definition of $\exp$ is:

  • We don't need an expression for $\gamma^A(t)$; $\gamma^A(t)$ is a smooth curve, and so we can
    1. Taylor expand $\gamma^A(t)$ to obtain new Lie group element
    2. $\exp$ new Lie group element
    3. Goto 1. and repeat until generated entire group
  • Matching with the expression above, only considering first-order expansion, we see that

    \begin{equation*}
\alpha = t \quad \text{and} \quad T = A
\end{equation*}

    where

    \begin{equation*}
\exp(t A) \approx e + t \exp'(s A) \big|_{s = 0} = e + t \underbrace{\big( \gamma^A \big)'(s) \big|_{s = 0}}_{= A} = e + t A
\end{equation*}
  • Neat, innit?!
  1. $\exp$ is a local diffeomorphism:

    \begin{equation*}
 \begin{split}
   \exists V \subseteq T_e G & , 0 \in V \\ 
   \exp |_V : V \to & \exp(V) \subseteq G
 \end{split}
\end{equation*}
    1. This restricted map is bijective
    2. $\exp|_V$ and $\exp|_V^{-1}$ are smooth
    3. Follows from IFT
  2. If $G$ is compact then $\exp$ is surjective: $\exp(T_e G) = G$. This is super-nice, as it means we can recover the entire Lie group from the Lie algebra!!! (due to this theorem) However:
    • $T_e G$ is non-compact
    • $G$ is compact
    • Hence $\exp$ cannot be injective!
  3. If $G$ is non-compact: it may be not surjective, may be injective and may be bijective. Just saying it can be whatever if $G$ is non-compact.

Letting $\mathfrak{g} := T_e G$.

  1. $\exp(0) = e$ and $\exp_*(0) = \text{id}$ by definition
  2. $\exp$ is a diffeo (as stated above) from a neighborhood of $0$ in $g$ to a neighborhoud of $e \in G$.
    • By IFT $\exp$ has a local inverse, which we call $\log$
  3. $\ep \big( (t + s) A \big) = \exp (t A ) \exp (s A)$
  4. For any morphism of Lie groups $\varphi: G_1 \to G_2$, we have

    \begin{equation*}
\exp \big( \varphi_* (A) \big) = \varphi \big( \exp(A) \big), \quad \forall A \in \mathfrak{g}_1
\end{equation*}
    • This follows from $\varphi \big( \exp(t A) \big)$ being a one-param subgroup in $G_2$, giving us

      \begin{equation*}
\dv{}{t} \bigg|_{t = 0} \varphi_* \big( \exp_*(A) \big) = \varphi_*(A)
\end{equation*}

      has to coincide with $\exp \big( t \varphi_*(A) \big)$

  5. Follows directly from (4):

    \begin{equation*}
\Ad_{g} \big( \exp(A) \big) = \exp \Big( \big( \Ad_{g} \big)_* A \Big)
\end{equation*}

The ordinary exponential function is a special case of the exponential map when $G$ is the multiplicative group of positive real numbers (whose Lie algebra is the additive group of all real numbers).

The exponential map of a Lie group satisfies many properties analogous to those of the ordinary exponential function, however, it also differs in many important respects.

Let $G_1$ be connected, and $\varphi: G_1 \to G_2$ a morphism of Lie groups.

Then $\varphi$ is determined by $\varphi_*$

$G_1$ connected, hence generated by a neighbourhoud of $e$, i.e. generated by elements $\exp(X), \quad X \in \mathfrak{g}$. But

\begin{equation*}
\varphi \big( \exp(X) \big) = \exp \big( \varphi_*(X) \big)
\end{equation*}

Letting $\mathfrak{g} = T_e G$, then if $x, y \in \mathfrak{g}$ are small enough (can expand any analytic function in a sufficiently small neighborhood) we have

\begin{equation*}
\exp(x) \cdot \exp(y) = \exp \big( \mu(x, y) \big)
\end{equation*}

because $\exp(x) \cdot \exp(y)$ is close enough to $e \in G$.

Can write

\begin{equation*}
\mu(x, y) = \underbrace{\alpha_1(x) + \alpha_2(y)}_{\text{linear}} + \underbrace{Q_1(x)}_{\text{quadratic}} + \underbrace{\lambda(x, y)}_{\text{bilinear}} + \underbrace{Q_2(y)}_{\text{quadratic}} + \dots
\end{equation*}

Observe that for the above equation, we let $x = 0$

\begin{equation*}
\exp(0) \cdot \exp(y) = \exp(y) = \exp \big( \mu(x, y) \big) \quad \implies \quad \alpha_2(x) = y, \quad Q_2(x) = 0
\end{equation*}

since $\alpha_1(0) = 0$ and $Q_1(0) = 0$. Similarily for $y = 0$. Thus we see that

\begin{equation*}
\mu(x, y) = x + y + \lambda(x, y) + \dots
\end{equation*}

And finally, letting $x = y$, we see that

\begin{equation*}
\lambda(x, x) = 0 \quad \implies \quad \lambda \text{ is skew-symmetric}
\end{equation*}

From Prop. proposition:expansion-of-exponential-map-in-neighbourhoood-of-identity we can write the commutation in the definition of adjoint map $\ad_{X}(Y) = \comm{X}{Y}$ by

\begin{equation*}
\comm{x}{y} := 2 \lambda(x, y)
\end{equation*}

$\varphi: G_1 \to G_2$ is a Lie group morphism. Then

  1. $\varphi_* \big( \comm{X}{Y} \big) = \comm{\varphi_*(x)}{\varphi_*(y)}$ for all $x, y \in \mathfrak{g}$
  2. Directly from 1): $\Ad_{g} \big( \comm{x}{y} \big) = \comm{\Ad_{g}(x)}{\Ad_{g}(y)}$ for all $x, y \in \mathfrak{g}$ and all $g \in G$
  3. Exponential

    \begin{equation*}
\exp(x) \exp(y) \exp(-x) \exp(-y) = \exp \big( \comm{x}{y} + \dots \big), \quad \forall x, y \in \mathfrak{g}
\end{equation*}

If $G$ is commutative, then so is $\mathfrak{g}$, i.e.

\begin{equation*}
\comm{x}{y} = 0, \quad \forall x, y \in \mathfrak{g}
\end{equation*}
\begin{equation*}
\mu(x, y) = x + y + \frac{1}{2} \comm{x}{y} + \frac{1}{12} \bigg( \comm{x}{\comm{x}{y}} + \comm{y}{\comm{y}{x}} \bigg) + \underbrace{\dots}_{\text{iterated commutators}}
\end{equation*}

Due to Theorem thm:baker-campbell-hausdorff we observe that commutators defines multiplication locally.

If $G$ is connected, then

  1. $\comm{\cdot}{\cdot}$ defines $m: G \times G \to G$
  2. $G$ is commutative $\iff$ $\mathfrak{g}$ is commutative.

Example

$G = \text{SO}(3, \mathbb{R})$ with Lie algebra $\mathfrak{g} = \mathfrak{so}(3, \mathbb{R})$ generated by

\begin{equation*}
J_x = 
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & -1 \\
  0 & 1 & 0
\end{pmatrix}, \quad
J_y = 
\begin{pmatrix}
  0 & 0 & -1 \\
  0 & 0 & 0 \\
  1 & 0 & 0
\end{pmatrix}, \quad
J_z =
\begin{pmatrix}
  0 & -1 & 0 \\
  1 & 0 & 0 \\
  0 & 0 & 0
\end{pmatrix}
\end{equation*}

and

\begin{equation*}
\exp(t J_x) = 
\begin{pmatrix}
  1 & 0 & 0 \\
  0 & \cos(t) & - \sin(t) \\
  0 & \sin(t) & \cos(t)
\end{pmatrix}
\end{equation*}

These one-param subgroups generates a neighbourhood of $e \in \text{SO}(3, \mathbb{R})$, and thus by connectedness of $\text{SO}(3, \mathbb{R})$, it generates the whole of $\text{SO}(3, \mathbb{R})$.

$J_x, J_y, J_z$ are called "infinitesimal generators".

Flow

A one-parameter subgroup of a Lie group $G$ is a Lie group homomorphism

\begin{equation*}
\xi: \mathbb{R} \to G
\end{equation*}

with $\mathbb{R}$ understood as a Lie group under ordinary addition.

Let $M$ be a smooth manifold and let $Y \in \Gamma(TM)$ be a complete vector field.

The flow of $Y$ is a smooth map

\begin{equation*}
\begin{split}
  \Theta: \quad & \mathbb{R} \times M \to M \\
  & (t, p) \mapsto \Theta_t(p) := \gamma_p(t)
\end{split}
\end{equation*}

where $\gamma_p$ is the maximal integral curve of $Y$ through $p$. For a fixed $p$, we have

\begin{equation*}
\Theta_0 = \id_M, \quad \Theta_{t_1} \circ \Theta_{t_2} = \Theta_{t_1 + t_2}, \quad \Theta_{-t} = \Theta_t^{-1}
\end{equation*}

For each $t \in \mathbb{R}$ the map $\Theta_t$ is a diffeomorphism $M \to M$. Denoting by $\Diff(M)$ the group (under composition) of the diffeomorphisms $M \to M$, we have that the map

\begin{equation*}
\begin{split}
  \xi: \quad & \mathbb{R} \to \Diff(M)  \\
  & t \mapsto \Theta_t
\end{split}
\end{equation*}

is a one-parameter subgroup of $\Diff(M )$.

Lie group action, on a manifold

Notation

  • Unless otherwise specified, assume maps in this section to be continuous
  • $[\varepsilon]$ is sometimes used to the denote a specific equivalence class related to the group-element $\varepsilon$
  • $d = \dim M$
  • $O_m$ denotes the orbit of $G$ of the element $m \in M$, i.e. $O_m := \Orb_G(m)$
  • $G_m$ denotes the stabilizer of $G$ of the element $m \in M$, i.e. $G_m := \Stab_G(m)$

Preparation

Let $(G, \bullet)$ be a Lie group, and $M$ be a smooth manifold.

Then a smooth map

\begin{equation*}
\triangleright : G \times M \to M
\end{equation*}

satisfying

  1. $e \triangleright p = p, \quad \forall p \in M$
  2. $g_2 \triangleright (g_1 \triangleright p) = (g_2 \bullet g_1) \triangleright p$

is called a left G-action on the manifold $M$.

Similarily, we define a right action:

\begin{equation*}
\triangleleft : M \times G \to M
\end{equation*}

s.t.

  1. $p \triangleleft e = p, \quad \forall p \in M$
  2. $(p \triangleleft g_1) \triangleleft g_2 = p \triangleleft (g_1 \bullet g_2)$

Observe that we can define the right action $\textcolor{green}{\triangleleft}$ using the left action $\textcolor{red}{\triangleright}$:

\begin{equation*}
p \ \textcolor{green}{\triangleleft} \ g := g^{-1} \ \textcolor{red}{\triangleleft} \ p
\end{equation*}

In the literature you might often see the following maps defined as the left action and right action:

\begin{equation*}
\begin{split}
  L_g: \quad & G \to G \\
  & h \mapsto gh
\end{split}
\end{equation*}

and

\begin{equation*}
\begin{split}
  R_g: \quad & G \to G \\
  & h \mapsto h g^{-1}
\end{split}
\end{equation*}

These actions define representations of $G$ on vector fields and k-forms on $G$. In particular, for $X \in \text{Vect}(G)$ and $g, m \in G$ we have

\begin{equation*}
\begin{split}
  \big( gX \big)(m) &= \big( L_g \big)_* X \big( g^{-1}m \big) \in T_e G \\
  \big( Xg \big)(m) &= \big( R_{g^{-1}} \big)_* X(mg) \in T_e G
\end{split}
\end{equation*}

where we've written $mg$ for $R_{g^{-1}}(m)$.

\begin{equation*}
\tilde{e}_a = A_a^m e_m
\end{equation*}

and

\begin{equation*}
\tilde{X}^a = \big( A^{-1} \big)_m^a X^m
\end{equation*}

can be understood as a right action of $\text{GL}(d)$ on the basis $e_m$ and a left action of $\text{GL}(d)$ on the components $X^a$.

Let:

  • two Lie group $(G, \bullet)$ and $(H, \blacksquare)$ and a Lie group homomorphism $\rho: G \to H$.
  • $M$ and $N$ be two smooth manifolds
  • two left actions

    \begin{equation*}
\begin{split}
  \triangleright &: G \times M \to M \\
  \blacktriangleright &: H \times N \to N
\end{split}
\end{equation*}
  • $f: M \to N$ be a smooth map

Then $f$ is called $\rho$ equivariant if the following diagram commutes:

equivariant-lie-group-action.png

where $\rho \times f: G \times M \to H \times N$ is a function where $\rho$ takes the first entry and maps to $H$, and $f$ takes the second entry and maps to $N$.

Let $\triangleright : G \times M \to M$ be a left action.

  1. For any $p \in M$ we define it's orbit under the action as the set

    \begin{equation*}
\text{orb}(p) := \{ q \in M \mid \exists g \in G : g \triangleright p = q \} \subseteq M
\end{equation*}
  2. Let

    \begin{equation*}
 p \sim q \quad \iff \quad \exists g \in G : q = g \triangleright p
\end{equation*}

    which defines a equivalence relation, thus defining a partition of $M$ called orbit space

    \begin{equation*}
 \big( M / {\sim} \big) \cong \ M / \text{orb}(p)
\end{equation*}
  3. For any $p \in M$ we define the stabilizer

    \begin{equation*}
  \text{stab}(p) := \{ g \in G \mid g \triangleright p = p \} \subseteq G
\end{equation*}

An action $\triangleright$ is called free if and only if

\begin{equation*}
\forall p \in M: \quad \text{stab}(p) = \{ e \}
\end{equation*}

where $e$ is the identity of the group.

Examples of Lie group actions

  1. $\text{GL}(n, \mathbb{R})$ acts on $\mathbb{R}^n$ and is a Lie group
  2. $O(n, \mathbb{R})$ acts on $S^{n - 1} = \left\{ v \in \mathbb{R}^n \mid \norm{v} = 1  \right\}$ since

    \begin{equation*}
\norm{v} = 1 \iff v^T v = 1
\end{equation*}

    and thus $g \in O(n ,\mathbb{R})$ we have

    \begin{equation*}
\big( gv \big)^T gv = v^T g^T gv = v^T v = 1
\end{equation*}
    • Of course, $SO(n, \mathbb{R})$ also acts on $S^{n - 1}$
  3. $U(n)$ acts on $S^{2n - 1} \subset \mathbb{R}^{2n} \cong \mathbb{C}$
Classical groups

Let

\begin{equation*}
J_{2n} = \begin{pmatrix} 0_n & I_n \\ - I_n & 0 \end{pmatrix} \quad \text{and} \quad I_{p, q} = \begin{pmatrix} I_p & 0 \\ 0 & - I_q \end{pmatrix}
\end{equation*}

We use the term classical groups to refer to the following groups:

  1. General linear:

    \begin{equation*}
\text{GL}(n, \mathbb{K}) := \left\{ A \in \text{Mat}(n, \mathbb{K}) \mid \det(A) \ne 0 \right\}
\end{equation*}
  2. Special linear:

    \begin{equation*}
\text{SL}(n, \mathbb{K}) := \left\{ A \in \text{Mat}(n, \mathbb{K}) \mid \det(A) = 1 \right\}
\end{equation*}
  3. Orthogonal:

    \begin{equation*}
O(n, \mathbb{K}) := \left\{ A \in \text{GL}(n, \mathbb{K}) \mid A A^T = I_n \right\}
\end{equation*}
  4. Special orthogonal:

    \begin{equation*}
\text{SO}(n, \mathbb{K}) := O(n, \mathbb{K}) \cap \text{SL}(n, \mathbb{K})
\end{equation*}
  5. Indefinite orthogonal:

    \begin{equation*}
O(p, q; \mathbb{R}) = \left\{ A \in \text{GL}(n, \mathbb{R}) \mid A I_{p, q} A^T = I_{p, q} \right\}
\end{equation*}
  6. Indefinite special orthogonal:

    \begin{equation*}
\text{SO}(p, q; \mathbb{R}) = O(p, q; \mathbb{R}) \cap \text{SL}(n, \mathbb{R})
\end{equation*}

    This group can equivalently be said to be the group of transformations under which a metric tensor with signature $(p, q)$ is invariant!

    • Ths is why in relativity we consider the group $\SO(1, 3)$ (or $\SO(3, 1)$ depending on the sign-convention being used)!
  7. Symplectic:

    \begin{equation*}
\text{Sp}(2n, \mathbb{K}) = \left\{ A \in \text{GL}(2n, \mathbb{K}) \mid A J_{2n} A^T = J_{2n} \right\}
\end{equation*}

    or equivalently

    \begin{equation*}
\Sp(2n, \mathbb{K}) = \left\{ A: \mathbb{K}^{2n} \to \mathbb{K}^{2n} \mid \omega \big( Ax, Ay) = \omega(x, y) \big) \right\}
\end{equation*}

    where $\omega$ is the skew-symmetric bilinear form

    \begin{equation*}
\omega(x, y) = \sum_{i=1}^{n} x_i y_{i + n} - y_i x_{i + n}
\end{equation*}

    (which, up to a change of basis, is the unique nondegenerate skew-symmetric bilinear form on $\mathbb{K}^{2n}$). The equivalence of the two above comes from the fact that $\omega(x, y) = \left\langle J x, y \right\rangle$.

  8. Unitary:

    \begin{equation*}
U(n) = \left\{ A \in \text{GL}(n, \mathbb{C}) \mid A \overline{A}^T = I_n \right\}
\end{equation*}
  9. Special unitary:

    \begin{equation*}
\text{SU}(n) = U(n) \cap \text{SL}(n, \mathbb{C})
\end{equation*}
  10. Unitary quaternionic or compact Symplectic:

    \begin{equation*}
\text{Sp}(n) = \text{Sp}(2n, \mathbb{C}) \cap \text{SU}(2n)
\end{equation*}

Consider $\exp$ for matrices, then $\exp(x)$ converges for all $x \in \text{Mat}(\mathbb{K}, n \times n)$, defined

\begin{equation*}
\exp: \text{Mat}(\mathbb{K}, n \times n) \to \text{Mat}(\mathbb{K}, n \times n)
\end{equation*}

Furthermore, in some neighborhood $U$ of $I_n \in \text{Mat}(\mathbb{K}, n \times n)$ we have the following series be convergent

\begin{equation*}
\log(x) = \sum_{n \ge 1}^{} \frac{\big( -1 \big)^{n + 1} \big( x - 1 \big)^n}{n}
\end{equation*}

with $\log: U \to \text{Mat}(\mathbb{K}, n \times n)$.

  1. $\exp$ and $\log$ are inverse of each other whenever defined.
  2. $\exp(0) = I_n$ and $d \exp(0) = \exp_*(0) = \text{id}$
  3. $xy = yx$ implies $\exp(x + y) = \exp(x) \exp(y)$ $XY = YX$ implies $\log(XY) = \log(X) + \log(Y)$ with $X, Y$ around $I_n$
  4. $\forall x \in \text{Mat}(\mathbb{K}, n \times n)$ the map

    \begin{equation*}
\begin{split}
  \varphi: \quad & \mathbb{K} \to \text{GL}_n \\
  & t \mapsto \exp(t x)
\end{split}
\end{equation*}

    is a morphism of Lie groups

  5. We have

    \begin{equation*}
\exp \Big( \big( \mathrm{Ad}_{g} \big)_* (x) \Big) = \mathrm{Ad}_{g} \big( \exp(x) \big)
\end{equation*}

    and

    \begin{equation*}
\exp \big( x^t \big) = \big( \exp(x) \big)^t
\end{equation*}

$T_e G$ then consists of curves whose entries are smooth (or holomorphic) functions in $t$, satisfying $A(0) = e$ and $\det \big( A(t) \big) \ne 0$ for all $t \in \mathbb{K}$, i.e. regular curves. Due to the smoothness, we can write

\begin{equation*}
A(t) = 1 + A_e t + \mathcal{O}(t) \quad \text{as} \quad t \to 0
\end{equation*}

passing through $e$. This implies that

\begin{equation*}
T_e \text{GL}_n \subset \text{Mat}_{n}
\end{equation*}

On the other hand, $\exp(tx)$ is a curve in $\text{GL}_{n}$ for any $x \in \text{Mat}_{n}$, and the corresponding tangent vector is $x$.

\begin{equation*}
T_e \text{GL}_n = \text{Mat}_{n}
\end{equation*}

From now on, we let

\begin{equation*}
\mathfrak{gl}_n := T_e \text{GL}_{n}
\end{equation*}

Exercise: For $G = \text{GL}_{n}$ we have

\begin{equation*}
\big( \mathrm{Ad}_{g} \big)_*(x) = g x g^{-1}, \quad g \in \text{GL}_{n}, x \in \text{Mat}(\mathbb{K}; n \times n)
\end{equation*}

where multiplication is matrix multiplication. NOTE: this does not make sense in general (i.e. multiplying a group element $x$ with elements of tangent space $g$), but in this case it does, which you can see be looking at the curves, etc.

For all classical $G \subset \text{GL}_{n}$ there exists a vector space $\mathfrak{g} \subset \mathfrak{gl}_n$ s.t. for a neighbourhood $U$ of $e \in \text{GL}_{n}$ and a neighbourhood $u$ of $e \in \mathfrak{gl}_n$ where them maps

\begin{equation*}
\begin{split}
  \exp:& u \cap \mathfrak{g} \to U \cap G \\
  \log:& U \cap G \to u \cap \mathfrak{g}
\end{split}
\end{equation*}

are mutually inverse.

Each classical group $G$ is a Lie group with the tanget space at identity being $T_e G = \mathfrak{g}$, where $\mathfrak{g}$ is as in Theorem thm:classical-groups-exp-and-log-mutually-inverse.

If $X = \exp(x) \in G$ and $x \in \mathfrak{g}$ with

\begin{equation*}
X X^T = X^T X
\end{equation*}

i.e. $X$ and $X^T$ commute, then we also have

\begin{equation*}
x^T x = x x^T
\end{equation*}

i.e. $x$ and $x^T$ also commute.

\begin{equation*}
X X^T = X^T X \quad \iff \quad \bigg(I + \sum_{k_1 = 1}^{\infty} \frac{1}{k_1 !} x^{k_1} \bigg) \bigg(I + \sum_{k_2 = 1}^{\infty} \frac{1}{k_2 !} (x^T)^{k_2} \bigg) = \bigg(I + \sum_{s_1 = 1}^{\infty} \frac{1}{s_1 !} x^{s_1} \bigg) \bigg(I + \sum_{s_2 = 1}^{\infty} \frac{1}{s_2 !} (x^T)^{s_2} \bigg)
\end{equation*}

Substracting and combining powers,

\begin{equation*}
\sum_{k_1, k_2 = 1}^{\infty} \frac{1}{k_1 ! \ k_2 !} \big( x^{k_1} (x^T)^{k_2} - (x^T)^{k_1} x^{k_2} \big) = 0
\end{equation*}

Suppose now that there exists terms which are non-vanishing and let $m_1, m_2$ be the first indices where we have

\begin{equation*}
x^{m_1} (x^T)^{m_2} - (x^T)^{m_1} x^{m_2} \ne 0
\end{equation*}

Then

\begin{equation*}
x \Big( x^{m_1 - 1} (x^T)^{m_2} \Big) - x^T \big( (x^T)^{m_1 - 1} x^{m_2} \big) \ne 0
\end{equation*}

so we must have one of the following:

  1. $x \ne x^T$ and $x^{m_1 - 1} (x^T)^{m_2} = (x^T)^{m_1 - 1} x^{m_2}$
  2. $x = x^T$ and $x^{m_1 - 1} (x^T)^{m_2} \ne (x^T)^{m_1 - 1} x^{m_2}$

But if 2. is true, then clearly we have all terms vanish, so this cannot be the case. Therefore we must have 1. We can then iterate this argument for both $m_1$ and $m_2$ until we have $m_1 = m_2 = 1$, i.e. the term

\begin{equation*}
x x^T - x^T x \ne 0
\end{equation*}

as the first non-vanishing term. But this is under the assumption that all higher order terms are indeed vanishing, the series is simply

\begin{equation*}
\sum_{k_1, k_2 = 1}^{\infty} \frac{1}{k_1 ! \ k_2 !} \big( x^{k_1} (x^T)^{k_2} - (x^T)^{k_1} x^{k_2} \big) = x x^T - x^T x
\end{equation*}

But the LHS vanish, and so we must have

\begin{equation*}
x x^T = x^T x
\end{equation*}

as wanted.

G-homogenous space

Let $G$ act on $M$, then $\forall m \in M$

  1. $G_m$ is a closed Lie subgroup of $G$
  2. $g \mapsto g \cdot m$ is an injective immersion $G / (G_m) \hookrightarrow{} G$

In particular, $O_m$ is an immersed submanifold in $M$ with tangent space

\begin{equation*}
T_m O_m = (T_e G) / (T_1 G_m)
\end{equation*}

Moreover, if $O_m$ is an (embedded) submanifold, then $g \mapsto g \cdot m$ is a diffeomorphism $G / G_m \to M$.

  1. It's sufficient to prove that in some neighbourhood $U \ni e$ contained in $G$, the intersection $U \cap G_m$ is a submanifold with the tangent space $T_e G_m = \mathfrak{h}$.
    • Recall that the commutator on vector fields on $M$, denoted $\mathscr{X}(M)$, can be defined for $X, Y \in \mathscr{X}(M)$

      \begin{equation*}
\comm{X}{Y}(f) = X(Y(f)) - Y(X(f)), \quad \forall f \in C^{\infty}(M)
\end{equation*}

      Thus the space

      \begin{equation*}
\mathfrak{h} := \left\{ x \in \mathfrak{g} \mid \rho_{ * }(x) (m) = 0 \right\}
\end{equation*}

      is closed under the Lie bracket since $\rho_{ * }: T_e G_m \to T_e M$ is a Lie algebra morphism (where we know $T_e G$ is in direct correspondance with left-invariant vector fields on $M$)

    • Also, since the vector field $\xi = \rho_* (x)$ vanishes at $m \in M$, we have

      \begin{equation*}
\rho \big( \exp(tx) \big)(m) = \Phi_{\xi}^t(m) = m, \quad \forall t \in (- \varepsilon, \varepsilon)
\end{equation*}

      where $\Phi_{\xi}^t(m)$ corresponds to the flow of $\xi$ with $\Phi_{\xi}^0(m) = m$. Therefore

      \begin{equation*}
\exp(t x) \in G_m \quad \forall x \in \mathfrak{h}
\end{equation*}

      That is, every element of $\mathfrak{h}$ exponentiates to $G_m$.

    • Choose a vector subspace $\mathfrak{u} \subset \mathfrak{g}$ complementary to $\mathfrak{h}$, i.e.

      \begin{equation*}
\mathfrak{g} = \mathfrak{u} \oplus \mathfrak{h}
\end{equation*}

      Since $\mathfrak{h} = \ker(\rho_{ * })$, by First Isomorphism Theorem of vector spaces, we know that $\rho_{ * } \big|_{\mathfrak{u}}$ is injective. Therefore the map

      \begin{equation*}
\begin{split}
  & \mathfrak{u} \to M \\
  & y \mapsto \rho \big( \exp(y) \big) m 
\end{split}
\end{equation*}

      is also injective for "sufficiently small $y$" (i.e. $\exists V \ni y$ near $e \in G$ s.t. that the above holds); this is just $\Phi_{Y}^1(m)$. Therefore

      \begin{equation*}
\exp(y) \in G_m \quad \iff \quad y = 0, \quad \forall y \in \mathfrak{u}
\end{equation*}
    • By IFT, any element $g$ in a sufficiently small neighbourhood $U \ni e$ in $G$ can be written as

      \begin{equation*}
g = \exp(y) \exp(x), \quad \text{for} \quad x \in \mathfrak{h} \text{ and } y \in \mathfrak{u}
\end{equation*}

      On the other hand,

      \begin{equation*}
\begin{split}
  \rho \big( \exp(y) \exp(x) \big) m &= \rho \big( \exp(y) \big) \underbrace{\rho \big( \exp(x) \big) m}_{= m } \\
  &= \rho \big( \exp(y) \big) m
\end{split}
\end{equation*}

      thus

      \begin{equation*}
g \in G_m \quad \iff \quad g \in \exp (\mathfrak{h})
\end{equation*}
    • Since $\exp(\mathfrak{h})$ is a submanifold in a neighbourhood of $e \in G$, we see that $G_m$ must also be a submanifold (which means that it's a closed Lie subgroup, since when we say submanifold, we mean an embedded submanifold)
  2. The proof of 1) also tells us that there is an isomorphism

    \begin{equation*}
T_e(G / G_m) = \mathfrak{g} / \mathfrak{h} \cong \mathfrak{u}
\end{equation*}

    so the injectivity of the map $\rho_{ * }: \mathfrak{u} \to T_m M$ shows that the map $G / G_m \to M$ given by $g \mapsto \rho(g) m$ is an immersion.

A G-homogenous space is a manifold $M$ with transitive action of $G$, i.e. $\forall m_1, m_2 \in M$

\begin{equation*}
\exists g \in G: \quad g m_1 = m_2
\end{equation*}

Why is this notation of G-homogenous so useful? Because if $G$ acts transitively on $M$, then $G / G_m$ for some $m \in M$ also acts freely on $M$. From this we then have the bijection $[g] \mapsto g \cdot m$! Moreover, this map is diffeomorphic

If $M$ is a G-homogenous space, then $\forall m \in M$ there is a fibre bundle $G \overset{p}{\longrightarrow} M = (G / G_m)$ with fibre $G_m$, where

\begin{equation*}
p(g) \mapsto g \cdot m
\end{equation*}

The proof follows from the fact that we have a diffeomorphism between $G / G_m$ and $M$ following thm:stabilizer-is-closed-lie-group-and-action-diffeomorphism-to-manifold , and we already know that $G \overset{p}{\longrightarrow} (G / G_m)$ is a fibre bundle with fibre $G_m$ from earlier.

Let's clarify the above corollary.

If we were to construct a fibration from a group $G$ and its action on some G-homogenous manifold $M$ we could approach it as follows.

  1. Let $\pi: G \to M$ be the projection map we are interested in
  2. Consider some point $m \in M$, then the corresponding fibre is $\pi^{-1}(m)$
  3. Naturally, we must then have

    \begin{equation*}
 \text{ker}(\pi) = \underbrace{\Stab_G(m)}_{= G_m}
\end{equation*}

    since the action by $\Stab_G(m)$ on $m$ is identity.

  4. Since $M$ is G-homogenous, i.e. $G$ acts transitively, at every point $a \in M$ we therefore have

    \begin{equation*}
\pi^{-1}(a) \cong \Stab_G(m)
\end{equation*}
  5. Results in a fibre bundle $G \overset{\pi}{\longrightarrow} M$ with fibre $\Stab_G(m)$ (so $m$ is a particularly chosen point)
  6. Since $G$ is transitive, $G / G_m$ is free, and therefore we have the bijective mapping

    \begin{equation*}
\begin{split}
  \varphi_m: \quad & (G / G_m) \to M \\
  & [g] \mapsto g \cdot m
\end{split}
\end{equation*}

    where we write $\varphi_m$ to remind ourselves that this bijection depends on the point $m \in M$.

  7. Need $\varphi_m$ to be diffeo. Since $G$ is a Lie group and $G_m$
Example of application of G-homogenous space
\begin{equation*}
SU(n)  = \left\{ A \in GL(n, \mathbb{C}) \mid A \bar{A}^T = \text{id}, \ \det(A) = 1 \right\}
\end{equation*}

$SU(n)$ acts on $S^{2n - 1}$ with

\begin{equation*}
x = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \in \mathbb{C}^n \simeq \mathbb{R}^{2n}
\end{equation*}

So $SU(n)_x$ has $x$ on the first column, and then zeros on the rest of the first row, and then the rest of the matrix is $SU(n - 1)$. Recall that $\SU(n)$ acts as rotation, and so in a basis containing $x$ we have $\SU(n - 1)$ acting by rotation around the axis $x$.

We get the fibre bundle $SU(n) \longrightarrow S^{2n - 1}$ with fibre $SU(n - 1)$. We then have the exact sequence

\begin{equation*}
\dots \longrightarrow \pi_2(S^{2n - 1}) \longrightarrow \pi_1 \big( SU(n - 1) \big) \longrightarrow \pi_1 \big( SU(n) \big) \longrightarrow \pi_1 \big( S^{2n - 1} \big) \longrightarrow \pi_0 \big( SU(n - 1) \big) \longrightarrow \pi_0 \big( SU(n) \big) \longrightarrow \pi_0 \big( S^{2n - 1} \big)
\end{equation*}

If $k > 0$ and $n \le k$ then

\begin{equation*}
\pi_n(S^k) =
\begin{cases}
  0 & \text{if } n < k \\
  \mathbb{Z} & \text{if } n = k
\end{cases}
\end{equation*}

If $n \ge 2$, we have $\pi_2(S^{2n - 1} = \pi_1(S^{2n - 1}) = \pi_0(S^{2n - 1}) = 0$. This implies

\begin{equation*}
\begin{cases}
  \pi_1 \big( SU(n - 1) \big) &= \pi_1 \big( SU(n) \big) \\
  \pi_0 \big( SU(n - 1) \big) &= \pi_0 \big( SU(n) \big)
\end{cases}
\quad \implies \quad
\begin{cases}
  \pi_1 \big( SU(n) \big) &= \pi_1 \big( SU(1) \big) \\
  \pi_0 \big( SU(n) \big) &= \pi_0 \big( SU(1) \big)
\end{cases}
\end{equation*}

$SU(1)$ is a point, hence connected and simply connected, hence so is $SU(n)$ for all $n \ge 1$!

Principal fibre bundles

A bundle $(E, \pi, M)$ is called a principal G-bundle if

  1. $E$ is a right G-space, i.e. equipped with a right G-action
  2. $\triangleleft$ is free
  3. $(E, \pi, M)$ and $(E, \rho, E / G)$ are isomorphic as bundles, where
    • $\rho: E \to E / G$, takes a point $\varepsilon$ to it's orbit / equivalence class
    • Since $\triangleleft$ is free:

      \begin{equation*}
\rho^{-1} ([\varepsilon]) \cong G
\end{equation*}

Suppose we have two principal G-bundles:

Then a principal bundle morphism or map needs to satisfy the following commutation relations:

principal-bundle-morphism.png

and a further restriction is also that there exists some Lie group homomorphism $\rho: G \to G'$, i.e. $\rho$ has to be a smooth map satisfying:

\begin{equation*}
\rho(g_1 \bullet_G g_2) = \rho(g_1) \bullet_{G'} \rho(g_2)
\end{equation*}

A principal bundle map is a diffeomorphism.

A principal G-bundle $(P, \pi, M)$ under the action $\triangleleft$ by $G$ is called trivial if it is diffeomorphic to the principal G-bundle $(M \times G, \pi_1, M)$ equipped with

\begin{equation*}
(x, g) \triangleleft g' := (x, g \bullet g')
\end{equation*}

and

\begin{equation*}
\pi(x, g) := x
\end{equation*}

That is, $(P, \pi, M)$ is trivial if and only if it's diffeomorphic to the bundle where the total space is the mfd. $M$ with $G$ attached as the fibre to each point, or equivalently (due to this theorem) if there exists a principal bundle map between these bundles.

trivial-principal-bundle.png

A principal G-bundle $(P, \pi, M)$ is trivial if and only if

\begin{equation*}
\exists \sigma: M \to P: \quad \pi \circ \sigma = \text{id}_M, \quad \sigma \text{ is smooth}
\end{equation*}

i.e. there exists a smooth global section from $M$ to $P$.

Let $M$ be a manifold. First observe that

\begin{equation*}
m \in M : \quad L_m M := \{ (e_1, \dots, e_{\dim M} ) \mid e_1, \dots, e_{\dim M} \text{ basis for } T_m M \} \cong \mathrm{GL}(\dim M, \mathbb{R})
\end{equation*}

i.e. the bundle is simply the set of all possible bases of $T_m M$.

The frame bundle is then

\begin{equation*}
\mathrm{LM} = \dot{\bigcup_{m \in M}} L_m M
\end{equation*}

where $\dot{\bigcup}$ denotes the unique union.

Equip $\mathrm{LM}$ with a smooth atlas inherited from $M$.

Further we define the projection $\pi: \text{LM} \to M$ by

\begin{equation*}
(e_1, \dots, e_{\dim M}) \mapsto m, \quad \exists! m \in M : (e_1, \dots, e_{\dim M}) \in L_m M
\end{equation*}

which implies that $(\mathrm{LM}, \pi, M)$ is a smooth bundle.

Now, to get a principal bundle, we need to establish a right action $\mathrm{GL}(\dim M, \mathbb{R})$ $\triangleleft$ on $\mathrm{LM}$, which we define to be

\begin{equation*}
(e_1, \dots, e_{\dim M}) \triangleleft g := \big( g_1^m e_m, g_2^m e_m, \dots, g_{\dim M}^m e_m \big)
\end{equation*}

which is just change of basis, and $\triangleleft$ is therefore free.

Checking that this in fact a principal $\mathrm{GL}(\dim M, \mathbb{R})$ bundle by verifying that this bundle is in fact isomorphic to $(\mathrm{LM}, \tilde{\pi}, \mathrm{LM} / G)$.

Observe that the Frame bundle allows us to represent a choice of basis by a choice of section $\sigma$ in each neighborhood $U \subseteq M$, with $m \in U$.

That is, any $p \in L_m U$ is a choice of basis for the tangent space at $m \in U$. This is then equipped with the general linear group $\mathrm{GL}(\dim M, \mathbb{R})$, i.e invertible transformations, which just happens to be used for to construct change of bases!!!

Ya'll see what this Frame bundle is all about now?

Associated bundles

Given a G-principal bundle $(P, \pi, M)$ (where the total space $P$ is equipped with $p \triangleleft g$ for $g \in G$) and a smooth manifold $F$ on which we can have a left G-action:

\begin{equation*}
\triangleright: G \times F \to F
\end{equation*}

we define the associated bundle

\begin{equation*}
P_F \overset{\pi_F}{\longrightarrow} M
\end{equation*}

by:

  1. let $\sim_G$ be the equivalence relation:

    \begin{equation*}
(p, f) \sim_g (p', f') \quad \iff \quad \exists g \in G: p' = p \triangleleft g, \quad f' = g^{-1} \triangleright f   
\end{equation*}

    Thus, consider the quotient space:

    \begin{equation*}
P_F := (P \times F) / \sim_G
\end{equation*}

    In other words, the elements of $P_F$ are the equivalence classes $[(p, f)]$ (short-hand sometimes used $[p, f]$) where $p \in P$, $f \in F$.

  2. Define $\pi_F : P_F \to M$ by

    \begin{equation*}
[p, f] \mapsto \pi(p)   
\end{equation*}

    which is well-defined since

    \begin{equation*}
\pi_F \big( [ p \triangleleft g, g^{-1} \triangleright f ] \big) = \pi (p \triangleleft g ) = \pi(p) = \pi_F \big( [p, f] \big)
\end{equation*}

This defines a fibre bundle with typical fibre $F$:

\begin{equation*}
P_F \overset{\pi_F}{\longrightarrow} M
\end{equation*}
Example of associated bundle: tangent bundle

That is, if we change the frame by the right-action (which represents a change of basis), then we must change the components of the tangent space (if we let $F = \mathbb{R}^d$ be the tangent space).

Then $\text{LM}_{\mathbb{R}^d} \overset{\pi_{\mathbb{R}^d}}{\longrightarrow} M$ is the associated bundle of the frame bundle.

Example: tensor associated bundle
  • $P = \text{LM}$
  • $G = \text{GL}(d, \mathbb{R})$
  • $F = \big( \mathbb{R}^d \big)^{\times p} \times \big( \mathbb{R}^d \big)^{\times q}$

With the left action of $G$ on $F$:

\begin{equation*}
\big( g \triangleright f \big)_{j_1, \dots, j_q}^{i_1, \dots, i_p} := f_{\tilde{j}_1, \dots, \tilde{j}_p}^{\tilde{i}_1, \dots, \tilde{i}_q} g_{\tilde{i}_1}^{\tilde{i}_1} \dots g_{\tilde{i}_p}^{\tilde{i}_p} \big( g^{-1} \big)_{j_1}^{\tilde{j}_1} \dots \big( g^{-1} \big)_{j_q}^{\tilde{j}_q}
\end{equation*}

Which defines the tensor bundle wrt. some frame bundle $LM \overset{\pi}{\longrightarrow} M$.

Observe that we can easily obtain $T^*M$ from $TM$ using the notion of a dual bundle!

So what we observe here is that "changes of basis" in the frame bundle, which is the principal bundle for the associated bundles tangent bundle and tensor bundle, corresponds to the changes to the tangent and tensors as we are familiar with!

That is, upon having the group $G = \text{GL}(d, \mathbb{R})$ act on the frame bundle, thus changing bases, we also have this same group act on the associated bundles!

Example of associated bundle: tensor densitites
  • $P = \text{LM}$
  • $G = \text{GL}(d, \mathbb{R})$
  • $F = \big( \mathbb{R}^d \big)^{\times p} \times \big( \mathbb{R}^d \big)^{\times q}$

But now, left action of $G$ on $F$:

\begin{equation*}
\big( g \triangleright f \big)_{j_1, \dots, j_q}^{i_1, \dots, i_p} := \big( \det g^{-1} \big)^{\omega} f_{\tilde{j}_1, \dots, \tilde{j}_p}^{\tilde{i}_1, \dots, \tilde{i}_q} g_{\tilde{i}_1}^{\tilde{i}_1} \dots g_{\tilde{i}_p}^{\tilde{i}_p} \big( g^{-1} \big)_{j_1}^{\tilde{j}_1} \dots \big( g^{-1} \big)_{j_q}^{\tilde{j}_q}
\end{equation*}

for some $\omega \in \mathbb{Z}$.

Then $\text{LM}_F \overset{\pi_F}{\longrightarrow} M$ is called the (p, q)-tensor $\omega$ density bundle over $M$.

Observe that this is the same as the tensor bundle, but with the factor of $\det g^{-1}$ in front, thus if we had instead used $G = \text{O}(d, \mathbb{R})$, i.e. orthogonal group, then $\det g^{-1} = 1$, hence it would be exactly the same as the tensor bundle!

Associated bundle map

An associated bundle map between two associated bundles $(\tilde{u}, \tilde{h})$ (sharing the same fibre, but being associated to arbitrarily different respective G-principal bundle $P, P'$)

associated-bundle-maps.png

is a bundle map $(u, h)$ (structure-preserving map of bundles) which can be costructed from a principal bundle map between the underlying principal bundles,

principal-bundle-map-in-associated-bundle-map-definition.png

where

\begin{equation*}
\begin{split}
  u(p \triangleleft g) &= u(p) \triangleleft' g \\
  \pi' \circ u &= h \circ \pi
\end{split}
\end{equation*}

as

\begin{equation*}
\begin{split}
  \tilde{u} \big( [p, f] \big) &:= \big[ u(p), f \big] \\
  \tilde{h}(m) &:= h(m) \\
  \forall p \in P, \ f &\in F, \ m \in M
\end{split}
\end{equation*}
Restricted Associative bundles

Let

If there exists a bundle morphism (NOT principal) such that

principal-bundle-map-in-restricted-association-bundle-map-definition.png

with:

  • $\pi' \circ u = f \circ \pi$
  • $u(p \triangleleft h) = u(p) \triangleleft h, \quad \forall h \in H \subseteq G, \forall p \in M$

Then

  • $P \overset{\pi}{\longrightarrow} M$ is called a G-extension of the H-principal bundle $P' \overset{\pi'}{\longrightarrow} M$
  • $P' \overset{\pi'}{\longrightarrow} M$ is called an H-restriction of the /G-principal bundle $P \overset{\pi}{\longrightarrow} M$

i.e. if one is an extension of the other, then the other is a restriction of the one.

Example: vector bundle

A real vector bundle of rank $m$ on a manifold $M$ (the base space), is a space $E$ (the total space) and a $C^{\infty}$ projection map $\pi: E \to M$ such that

  1. For every $a \in M$, the fibre $F_p = \preim_{\pi} \big( \left\{ a \right\} \big)$ is a real vector space of $\dim F_p = m$
  2. Each $a \in M$ has a neighborhood $U \ni a$ and a diffeo.

    \begin{equation*}
\psi_U: \quad \preim_{\pi}(U) \overset{\cong}{\longrightarrow} U \times \mathbb{R}^m
\end{equation*}

    i.e. a local trivialization, such that $\psi_U$ maps the vector space $\preim_{\pi} \big( \left\{ a \right\} \big)$ isomorphically to the vector space $\left\{ a \right\} \times \mathbb{R}^m$

  3. On $U \cap V$, the composition

    \begin{equation*}
\psi_U \circ \psi_V^{-1}: (U \cap V) \times \mathbb{R}^m \to (U \cap V) \times \mathbb{R}^m
\end{equation*}

    takes the form

    \begin{equation*}
\big(a, v \big) \mapsto \big( a, g_{uv}(a) v \big), \quad a \in M, \quad v \in \mathbb{R}^m
\end{equation*}

    for a smooth transition map $g_{uv}: U \cap V \to \GL(m, \mathbb{R})$.

Let $\big( E, \pi, M \big)$ be a real rank $m$ vector bundle.

We say $\left\{ \big( U_{\alpha}, \psi_{\alpha} \big) \right\}_{\alpha \in A}$ is a trivializing cover with transition maps $\left\{ g_{\alpha \beta} : U_{\alpha} \cap U_{\beta} \to \GL(m, \mathbb{R}) \right\}$ if

\begin{equation*}
\bigcup_{\alpha}^{} U_{\alpha} = M
\end{equation*}

and $\big( U_{\alpha}, \psi_{\alpha} \big)$ are local trivializations.

The cocycle conditions are that for all $\alpha, \beta, \gamma \in A$, we have

  1. $g_{\alpha \alpha}(a) = I$ for all $a \in U_{\alpha}$
  2. $g_{\alpha \beta}(a) g_{\beta \alpha}(a) = I$ for all $a \in U_{\alpha} \cap U_{\beta}$
  3. $g_{\alpha \beta}(a) g_{\beta \gamma}(a) g_{\gamma \alpha}(a) = I$ for all $a \in U_{\alpha} \cap U_{\beta} \cap U_{\gamma}$

One can show that a vector bundle satisfy these conditions.

Let

Then there exists a vector bundle $E$ with $C^{\infty}$ projection map $\pi: E \to M$ and transition maps $\left\{ g_{\alpha \beta} \right\}$.

Define

\begin{equation*}
\tilde{E} := \bigsqcup_{\alpha \in A} \big( U_{\alpha} \times \mathbb{R}^m \big)
\end{equation*}

and the equivalence relation

\begin{equation*}
(a, v) \sim (b, w) \quad \iff \quad a = b \quad \text{and} \quad v = g_{\beta \alpha}(a) w
\end{equation*}

for all $(a, v) \in U_{\alpha} \times \mathbb{R}^m$ and $(b, w) \in U_{\beta}$.

Let's check that this is indeed a equivalence relation:

  1. Clearly $(a, v) \sim (a, v)$ since $g_{\alpha \alpha}(a) = I$ since $\left\{ g_{\alpha \beta} \right\}$ satisfy the cocycle conditions
  2. If $(a, v) \sim (b, w)$ then $\exists g_{\beta \alpha}: U_{\beta} \cap U_{\alpha} \to \GL(m ,\mathbb{R})$ such that

    \begin{equation*}
v = g_{\beta \alpha}(a) w \quad \implies \quad g_{\alpha \beta}(b) v = g_{\alpha \beta}(b) g_{\beta \alpha}(a) w \quad \implies g_{\alpha \beta}(b) v = w
\end{equation*}

    since $a = b$ and $g_{\alpha \beta}$ satisfy the cocycle conditions.

  3. Finally, if $(a, v) \sim (b, w)$ and $(b, w) \sim (c, u)$ then by satisfying the third condition of the cocycle conditions, we find that this is also satsified.

Hence it is indeed an equivalence relation.

Then we define the quotient space $E = \tilde{E} / \sim$ with the equivalence relation from above, and we define the projection map $\pi$ by

\begin{equation*}
\begin{split}
  \pi: \quad & E \to M \\
  & [(a, v)] \mapsto \pi \big( [(a, v)] \big) = a, \quad \forall a \in M, v \in \mathbb{R}^m
\end{split}
\end{equation*}

where $[(a, v)]$ denotes the equivalence class of $(a, v) \in U_{\alpha} \times \mathbb{R}^m$ for some $\alpha$. Then

\begin{equation*}
\preim_{\pi} \big( U_{\alpha} \big) = \left\{ [(a, v)] \in E \mid a \in U_{\alpha}, v \in \mathbb{R}^m \right\}
\end{equation*}

and we define the map $\psi_{\alpha}$

\begin{equation*}
\begin{split}
  \psi_{\alpha}: \quad & \preim_{\pi}(U_{\alpha}) \to U_{\alpha} \times \mathbb{R}^m \\
  & [(a, v)] \mapsto (a, v)
\end{split}
\end{equation*}

Refining (if necessary) the open cover $\left\{ U_{\alpha} \right\}_{\alpha \in A}$, we can assume that the $U_{\alpha}$ are charts, with corresponding chart maps $\varphi_{\alpha}: U_{\alpha} \to \mathbb{R}^n$ (which are also bijections). Then the map

\begin{equation*}
\Phi_{\alpha} := \big( \varphi_{\alpha} \times \id \big) \circ \psi_{\alpha} \quad \implies \quad \Phi_{\alpha}: \preim_{\pi}(U_{\alpha}) \to \varphi( U_{\alpha} ) \subseteq \mathbb{R}^{n + m}
\end{equation*}

Finally, we observe that $\left\{ \big( \preim_{\pi}(U_{\alpha}), \Phi_{\alpha} \big) \right\}$ is indeed an atlas for $E$. The charts clearly form a open cover of $E$ by definition of $\tilde{E}$ and the quotient topology, terefore we only need to check that $\Phi_{\alpha}$ are (smooth ?) homeomorphisms to ϕ(Uα)$.

Using Theorem thm:existence-of-vector-bundle-of-manifold we can construct new vector bundles from old ones by using the "functional" constructions $\oplus$, $\Hom$, and $\otimes$ on vector spaces. The Whitney sum is an example of this.

Let $E_1 \to M$ and $E_2 \to M$ be vector bundles of rank $l$ and $l$, respectively.

Then we define $E_1 \oplus E_2$ to be the vector bundle with fibres

\begin{equation*}
\big( E_1 \oplus E_2 \big)_a = \big( E_1 \big)_a \oplus \big( E_2 \big)_a
\end{equation*}

and transition maps

\begin{equation*}
\begin{split}
  g_{\alpha \beta}^{E_1 \otimes E_2}: \quad & U_{\alpha} \cap U_{\beta} \to \GL(k + l, \mathbb{R}) \\
  & a \mapsto \left[
\begin{array}{@{}c|c@{}}
   g_{\alpha \beta}^E(a) &  \\
    \hline  
    & g_{\alpha \beta}^F(a)
\end{array}
\right]
\end{split}
\end{equation*}

where we have used a common refinement $\left\{ U_{\alpha} \right\}$, since we wlog. we can take intersections of the of covers of the two bundles to obtain a new open trivializing cover which is common to both bundles.

One can then see that $g_{\alpha \beta}^{E_1 \otimes E_2}$ are smooth and obey the cocycle conditions as follows. The map

\begin{equation*}
\begin{split}
  \varphi: \quad & \GL(k, \mathbb{R}) \times \GL(l, \mathbb{R}) \to \GL(k + l, \mathbb{R}) \\
  & (A, B) \mapsto \left[
\begin{array}{@{}c|c@{}}
   A &  \\
    \hline  
    & B
\end{array}
\right]
\end{split}
\end{equation*}

is a smooth map on $\mathbb{R}^{k^2} \times \mathbb{R}^{l^2}$ to $\mathbb{R}^{(k + l)^2}$. Furthermore, it's a group homomorphism, i.e. it satisfies

\begin{equation*}
\varphi( A_1 A_2, B_1 B_2 ) = \varphi(A_1, B_1) \varphi(A_2, B_2)
\end{equation*}

Following the example of the Whitney sum, we can obtain the generalized tangent bundle:

Using the Whitney sum we can construct the generalized tangent bundle:

\begin{equation*}
TM \oplus T^*M \longrightarrow M
\end{equation*}

Let $E \to M$ be a $m$ rank real vector bundle.

Then consider $E^* \to M$, with fibres

\begin{equation*}
\big( E^* \big)_a = \Hom(E_a, \mathbb{R}) = \big( E_a \big)^*
\end{equation*}

with transition maps $g_{\alpha \beta}^{E^*}$ are the inverse transpose of $g_{\alpha \beta}^{E}$. The group homomorphism underlying this is

\begin{equation*}
\begin{split}
  \GL(m, \mathbb{R}) & \to \GL(m, \mathbb{R}) \\
  A & \mapsto \big( A^T \big)^{-1}
\end{split}
\end{equation*}

This is a homomorphism since the entries of $\big( A^T \big)^{-1}$ are rational functions of entries of $A$ and the denominators are $\det(A)$, which is non-zero in $\GL(m ,\mathbb{R})$!

Observe that we can easily obtain $T^*M$ from $TM$ using the notion of a dual bundle!

Let

Then define $E \otimes F \longrightarrow M$ is denotes a vector bundle with the typical fibre

\begin{equation*}
\big( E \otimes F \big)_a = E_a \otimes F_a
\end{equation*}

with the transition maps

\begin{equation*}
g_{\alpha \beta}^{E \otimes F}(a) = g_{\alpha \beta}^{E}(a) \otimes g_{\alpha \beta}^{F}(a)
\end{equation*}

and if $v \otimes w \in E_a \otimes F_a$, then

\begin{equation*}
\big( g_{\alpha \beta}^{E \otimes F}(a) \big) \big(v \otimes w) = \big( g_{\alpha \beta}^{E}(a) \big) v \otimes \big( g_{\alpha \beta}^{F}(a) \big) w
\end{equation*}

defines a $C^{\infty}$ map

\begin{equation*}
\begin{split}
  \GL(k, \mathbb{R}) \times \GL(l, \mathbb{R}) & \to \GL(kl, \mathbb{R}) \\
  (A, B) & \mapsto A \otimes B
\end{split}
\end{equation*}

(whose entrires are polynomial in entries of $A$ and $B$).

We can iterate this construction to arrive a $E^{\otimes p}$. In particular, we can start with $TM$ and arrive at the $(r, s)$ tensor bundle:

\begin{equation*}
T_s^r(M) := \big( TM \big)^{\otimes r} \otimes \big( T^* M \big)^{\otimes s}
\end{equation*}
  • Examples of vector bundles
    • For the tangent bundle $TM$, the $g_{uv}$ is the Jacobian matrix of a change of coordinates
    • For the cotangent bundle $T^*M$, the $\big( (g_{uv})^T \big)^{-1}$ is the inverse of the transpose Jacobian!

Connections

Let $P \overset{\pi}{\longrightarrow} M$ be a principal G-bundle.

Then each $A \in T_e G$ induces a vector field on $P$

\begin{equation*}
\forall p \in P : X_{p}^A f := f \Big( p \triangleleft \exp \big( t A \big) \Big)' (0)
\end{equation*}

It is useful to define the map

\begin{equation*}
\begin{split}
  i: \quad & T_e G \longrightarrow \Gamma(TP) \\
  & A \mapsto X^A
\end{split}
\end{equation*}

which can be shown to be a Lie algebra homomorphism

\begin{equation*}
i \Big( [\![ A, B,  ]\!] \Big) = \comm{i(A)}{i(B)}
\end{equation*}

where:

  • $[\![ \cdot, \cdot ]\!]$ is the Lie bracket on $T_e G$
  • $\comm{\cdot}{\cdot}$ is the commutation bracket on $\Gamma(TP)$

Let $p \in P$. Then

\begin{equation*}
\begin{split}
  T_p P \supset V_p P & := \text{ker}(\pi_*) \\
  &= \{ X \in T_p P \mid \pi_* (X) = 0 \}
\end{split}
\end{equation*}

where:

$V_p P$ is called the vertical subspace at point $p$.

Idea of a connection is to make a choice of how to "connect" the individual points of "neighboring" fibres in a principal bundle.

Let's take a moment to think about this. For a principal G-bundle we know the fibre at each point $m \in M$ is isomorphic to the group $G$. Consider two points in $p_1, p_2 \in \pi^{-1}(\left\{ m \right\})$, and $X_1 \in T_{p_1}P$, $X_2 \in T_{p_2} P$. One then wonder how these vectors are pushed down to the manifold at $m$, i.e. what is $\pi_*(X_1)$ and $\pi_*(X_2)$, both of which lie in $T_m M$. We are then defining $V_p$, the vertical space, as those $X_1$ and $X_2$, in possibly different tangent spaces, are pushed down to the zero-tangent (i.e. the tangent correspondig to the constant path through $m \in M$).

A connection on a principal G-bundle $P \overset{\pi}{\longrightarrow} M$ is an assignment where every for $p \in P$ a vector subspace $H_p P$ of $T_p P$ is chosen such that

  1. $H_p \oplus V_p P = T_p p$ where $V_p$ is the vertical subspace
  2. The push-forward by the right-action of $G$ satisfy:

    \begin{equation*}
(\triangleleft g)_* \big(H_p P \big) = H_{p \triangleleft g} P
\end{equation*}
  3. The unique decomposition:

    \begin{equation*}
X_p = \underbrace{\text{hor}(X_p)}_{\text{by def. } \in H_p P} + \underbrace{\text{ver}(X_p)}_{\text{by def. } \in V_p P}
\end{equation*}

    leads, for every smooth vector field $X \in \Gamma(TP)$, to two smooth vector fields $\text{hor}(X)$ and $\text{ver}(X)$.

Some principal G-bundle $P \overset{\pi}{\longrightarrow} M$

The choice of horizontal subspace $H_p P$ at each $p \in P$, which is required to provide a connection, is conveniently "encoded" in the thus induced Lie-algebra-valued one-form $\omega$, which is defined as follows:

\begin{equation*}
\begin{split}
  \omega_p: \quad & T_p P \overset{\sim}{\longrightarrow} T_e G \\
  & X_p \mapsto \omega_p \big( X_p \big) := i_p^{-1} \Big( \underbrace{\text{ver}(X_p)}_{\in V_p P} \Big)
\end{split}
\end{equation*}

where we need to remember that $\text{ver}(X_p)$ depends on the choice of the horizontal subspace $H_p P$!

Recall that $i_p^{-1}$ is the map defined in the beginning of this section:

\begin{equation*}
\begin{split}
  i_p: \quad & T_e G \to V_p P \\
  & A \mapsto X_p^A
\end{split}
\end{equation*}

where $\omega$ is called the connection 1-form wrt. the connection.

That is, the connection is a choice of horizontal subspaces $H_p P$ of the fibre of the principal bundle, and once we have such a space, we can define this connection 1-form.

Therefore one might wonder "can we go the other way around?":

Yes, we can!

\begin{equation*}
H_p P = \text{ker}(\omega_p) := \left\{ X \in T_p P \mid \omega_p(X) = 0 \right\}
\end{equation*}

A connection 1-form $\omega$ wrt. a given connection has the properties

  1. $\omega_p \big( X_p^A \big) = A$
  2. Pull-back

    \begin{equation*}
\Big( (\triangleleft g)^* \omega \Big)_p \big( X_p \big) = \Big( \Ad_{g^{-1}} \Big)_* \Big( \omega_p (X_p) \Big)
\end{equation*}

    where we recall

    \begin{equation*}
\begin{split}
\Ad_g : \quad & G \to G \\
& h \mapsto g h g^{-1} \\
\Ad_g_* : & T_e G \to T_e G
\end{split}
\end{equation*}
  3. Smooth, since

    \begin{equation*}
\omega = i^{-1} \circ \text{ver}
\end{equation*}

    where $i^{-1}$ is smooth since the exponential map is smooth.

Different approach to connections

This approach is the one used by Schuller in the International Winter School on Gravity and Light 2015.

A connection (or covariant derivative) $\nabla$ on a smooth manifold $\big( M, \mathcal{O}, \mathcal{A} \big)$ is a map which takes a pair $\big( X, T \big)$ consisting of a vector field $X$ and a (p, q)-tensor field $T$ and sends this to a (p, q)-tensor field $\nabla_X T$, satisfying the following properties:

  1. $\nabla_X f = X f, \quad \forall f \in \mathcal{C}^{\infty}(M)$
  2. $\nabla_X \big( T + S \big) = \nabla_X T + \nabla_X S$ for $T, S \in T_q^p$
  3. The Leibnitz rule for $T \in T_1^1$:

    \begin{equation*}
\nabla_X \Big( T(\omega, Y) \Big) = \Big( \nabla_X T \Big) (\omega, Y) + T \Big( \nabla_X \omega, Y \Big) + T \Big( \omega, \nabla_X Y \Big)
\end{equation*}

    and the generalization to a (p, q)-tensor follows from including further terms corresponding to $\nabla_X$ of the arguments. It's worth noting that this is actually the definition obtained from

    \begin{equation*}
\nabla_X (T \otimes S) = \Big( \nabla_X T \Big) \otimes S + T \otimes \Big( \nabla_X S \Big)
\end{equation*}

    which is the more familiar form of the Leibnitz rule.

  4. $C^{\infty}\text{-linear}$ (in $X$)

    \begin{equation*}
\nabla_{f X + Z} T = f \cdot \nabla_X T + \nabla_Z T, \quad f \in \mathcal{C}^{\infty}(M)
\end{equation*}

Consider vector fields $X, Y \in \Gamma(TM)$. Then

\begin{equation*}
\begin{split}
  \nabla_X Y &= \nabla_{X^i \frac{\partial }{\partial x^i}} \bigg( Y^m \frac{\partial }{\partial x^m} \bigg) \\
  &= X^i \bigg( \nabla_{\frac{\partial }{\partial x^i}} Y^m \bigg) \cdot \frac{\partial }{\partial x^m} + X^i Y^m \cdot \nabla_{\frac{\partial }{\partial x^i}} \bigg( \frac{\partial }{\partial x^m} \bigg) \\
  &= X^i \bigg( \frac{\partial }{\partial x^i} Y^m \bigg) \frac{\partial }{\partial x^m} + X^i Y^m \tensor{\Gamma}{^i_{kl}}
\end{split}
\end{equation*}
  1. A vector field $X$ on $M$ is said to be parallely transported along a smooth curve $\gamma: \mathbb{R} \to M$ with tangent vector $v_{\gamma}$ if

    \begin{equation*}
\nabla_{v_{\gamma}} X = 0
\end{equation*}

    i.e.

    \begin{equation*}
\Big( \nabla_{v_{\gamma, \gamma(\lambda)}} X \Big)_{\gamma(\lambda)} = 0, \quad \forall \lambda \in \mathbb{R}
\end{equation*}
  2. A weaker notion is

    \begin{equation*}
\Big( \nabla_{v_{\gamma, \gamma(\lambda)}} X \Big)_{\gamma(\lambda)} = \mu(\lambda) \cdot X_{\gamma(\lambda)}
\end{equation*}

    for some $\mu \in \mathbb{R} \to \mathbb{R}$, i.e. it's "parallel".

Local representations of connections the base manifold: "Yang-Mills" fields

In practice, e.g. for computational purposes, one whishes to restrict attention to some $U \subseteq M$:

  • Choose local section $\sigma: U \to P$, thus $\pi \circ \sigma = \text{id}_M$

Such a local section induces:

  1. "Yang-Mills" field, $\omega^U$:

    \begin{equation*}
\begin{split}
  \omega^U : \quad  & \Gamma(TU) \to T_e G \\
  & \omega^U = \sigma^* \omega
\end{split}
\end{equation*}

    i.e. a "local" version of the connection 1-form on the principle fiber $\omega$, defined through the pull-back of the chosen section $\sigma$.

  2. Local trivialization , $h$, of the principal bundle

    \begin{equation*}
\begin{split}
  h : \quad & U \times G \to P \\
  & (m, g) \mapsto \sigma(m) \triangleleft g
\end{split}
\end{equation*}

    Then we can define the local representation of $\omega$

    \begin{equation*}
h^* \omega: \quad T_{(m, g)} (U \times G) \cong T_m U \oplus T_g G \to \text{TP}   
\end{equation*}

local-trivialization-yang-mills.png

Suppose we have chosen a local section $\sigma: U \to P$.

The Yang-Mills field $\omega^U$, i.e. the connection 1-form restricted to the subspace $U$, is then defined

\begin{equation*}
\begin{split}
  \omega^U : \quad  & \Gamma(TU) \to T_e G \\
  & \omega^U = \sigma^* \omega
\end{split}
\end{equation*}

Thus, this is a "Lie algebra"-valued 1-form.

Choosing the local trivialization $h$:

\begin{equation*}
\begin{split}
  h : \quad & U \times G \to P \\
  & (m, g) \mapsto \sigma(m) \triangleleft g
\end{split}
\end{equation*}

Then we can define the local representation of the global connection $\omega$:

\begin{equation*}
h^* \omega: \quad & T_{(m, g)} (U \times G) \cong T_m U \oplus T_g G \to \text{TP}   
\end{equation*}

given by

\begin{equation*}
\big( h^* \omega \big)_{(m, g)} \big( v, \gamma \big) = \Big( \Ad_{g^{-1}} \Big)_* \Big( \omega^U(v) \Big) + \Xi_g (\gamma)
\end{equation*}

where

local-trivialization-yang-mills.png

\begin{equation*}
\begin{split}
  \Xi_g : \quad & T_g G \longrightarrow T_e G \\
  & L_g^A \mapsto A
\end{split}
\end{equation*}

where we have

  • $L_g^A := \ell_{g^*} A$

Examples of Yang-Mills field:

Gauge map
  • Can we use the "local understading" / restriction to $U$, to construct a global connection 1-form?
  • Can do so by defining the Yang-Mills field for different, overlapping subspaces of the base-manifold!
    • Need to be able to map between the intersection of these subspaces; introduce gauge map

Suppose we have two subspaces of the base manifold $U^{(1)}, U^{(2)} \subseteq M$, such that $U^{(1)} \cap U^{(2)} \ne \emptyset$ for which we also have chosen two corresponding sections $\sigma^{(1)}, \sigma^{(2)}$, such that

gauge-map-definition.png

We then define the gauge map as

\begin{equation*}
\Omega: \quad U^{(1)} \cap U^{(2)} \longrightarrow G
\end{equation*}

where $G$ is the underlying Lie group (on $P$), defined on the unique $\Omega(m)$ for all $m \in U^{(1)} \cap U^{(2)}$:

\begin{equation*}
\sigma^{(2)}(m) = \sigma^{(1)}(m) \triangleleft \Omega(m)
\end{equation*}

where $\Xi$ is the Maurer-Cartan form.

From the definition of gauge map, get

\begin{equation*}
\omega^{U^{(2)}}_{\mu}(m) = \Ad_{\Omega^{-1}(m)}^* \omega_{\mu}^{U^{(1)}} + \Big( \Omega^* \Xi_{m, \mu} \Big)
\end{equation*}

where:

  • $m$ denotes a point in the base-manifold
  • $\mu$ denotes the the component (which $m$ is NOT, hence the comma)

This theorem gives us the relationship between the Yang-Mill fields on the two different subspaces $U^{(1)}, U^{(2)}$ of $M$!

Example: Frame bundle

Recall that in the case of the Frame budle we have

\begin{equation*}
P = LM, \quad  G = \mathrm{GL}(\dim M , \mathbb{R})
\end{equation*}

Then a particular choice of section $\sigma: U \to \mathbb{R}^{\dim M}$ for some chart $\big( U, x \big)$ is equivalent of a specific choice of coordinates:

\begin{equation*}
\sigma(m) = \Bigg( \frac{\partial }{\partial x^1}, \dots, \frac{\partial }{\partial x^{\dim M }} \Bigg)
\end{equation*}

Let's first consider this as an instance of a Yang-Mills field:

\begin{equation*}
\omega^U = \sigma^* \omega
\end{equation*}

is then a Lie-algebra valued one-form on $U$, with components

\begin{equation*}
\tensor{\big( \omega^U \big)}{^i_{j \mu}}
\end{equation*}

where

  • $\mu = 1, \dots, \dim M$ comes from being a one-form on $M$, hence $\dim M$ components
  • $i, j = 1, \dots, \dim M$ from $G = \mathrm{GL}(\dim M, \mathbb{R})$

In fact, these are the Christoffel symbols from GR!

\begin{equation*}
\tensor{\Gamma}{^{i}_{j \mu}} := \tensor{\big( \omega^U \big)}{^{i}_{j \mu}}
\end{equation*}

In GR we pretend that all these indices $i, j, \mu$ are related to the manifold $M$, but really the indices $i, j$ are related to the Lie algebra (resulting from the Lie group $G$) while the index $\mu$ is an proper component of the one-form!

We can obtain the gauge map for the Frame bundle. For this we first need to compute the Maurer-Cartan form $\Xi$ in $G = \mathrm{GL}(d, \mathbb{R})$. We do this as follows:

  1. Choose coords an open set $\mathrm{GL}^+$ on $\mathrm{GL}(d, \mathbb{R})$ containing $\mathrm{id}_G$:

    \begin{equation*}
\mathrm{GL}^+(d, \mathbb{R}) \overset{\tensor{x}{^i_j}}{\longrightarrow} R^{d \cdot d}
\end{equation*}

    where

    \begin{equation*}
\tensor{x}{^i_j}(g) =: \tensor{g}{^i_j}
\end{equation*}

    which are the "matrix entries".

  2. We then consider

    \begin{equation*}
\Big( L^A \tensor{x}{^i_j} \Big)_g =  \Big( \tensor{x}{^i_j} \circ \gamma^A \Big)'(0) =  \tensor{x}{^i_j} \Big( \underbrace{g \cdot \exp \big( t A \big)}_{\gamma(t) \in G: \ \gamma(0) = g} \Big)'(0)
\end{equation*}

    since

    • $\exp \big( tA \big)$ is the unique integral curve of $L^A$
    • by def. of vector-field acting on a map

This can then be written

\begin{equation*}
\begin{split}
  \Big( L^A \tensor{x}{^i_j} \Big)_g &= \Big( \tensor{x}{^i_j} \big( g \cdot e^{t A} \big) \Big) \\
  &= \Big( \tensor{g}{^i_k} \tensor{\big( e^{tA} \big)}{^k_j} \Big)'(0) \\
  &= \tensor{g}{^i_k} \tensor{A}{^k_j}
\end{split}
\end{equation*}

Hence,

\begin{equation*}
L_g^A = \tensor{g}{^i_k} \tensor{A}{^k_j} \frac{\partial }{\partial \tensor{x}{^i_j}}\Big|_g
\end{equation*}

Since

\begin{equation*}
\begin{split}
  \Xi_g: & T_g P \to T_e P \\
  & L_g^A \mapsto A
\end{split}
\end{equation*}

we, in this case, have

\begin{equation*}
\tensor{\big( \Xi_g \big)}{^i_j} = \tensor{\big( g^{-1} \big)}{^i_k} \big( \tensor{dx}{^k_j} \big)_g
\end{equation*}

Then,

\begin{equation*}
\begin{split}
  \tensor{\big( \Xi_g \big)}{^i_j} \Big( L_g^A \Big)
  &= \bigg[ \tensor{\big( g^{-1} \big)}{^i_k} \tensor{dx}{^k_j}\big|_g \bigg] \bigg[ \tensor{g}{^p_r} \tensor{A}{^r_q} \frac{\partial }{\partial \tensor{x}{^p_q}}\bigg|_g \bigg] \\
  &= \tensor{\big( g^{-1} \big)}{^i_k} \tensor{g}{^p_r} \tensor{A}{^r_q} \tensor{dx}{^k_j}\big|_g \frac{\partial }{\partial \tensor{x}{^p_q}}\bigg|_g \\
  &= \tensor{\big( g^{-1} \big)}{^i_k} \tensor{g}{^p_r} \tensor{A}{^r_q} \tensor{\delta}{^k_p} \tensor{\delta}{^q_j} \\
  &= \tensor{\big( g^{-1} \big)}{^i_k} \tensor{g}{^k_r} \tensor{A}{^r_j} \\
  &= \tensor{A}{^i_j}
\end{split}
\end{equation*}

as we wanted!

Recalling the definition of the gauge map, we now want to compute

\begin{equation*}
\omega^{U^{(2)}}_{\mu} = \Ad_{\Omega(m)^{-1}^*} \omega^{U^{(1)}}_{\mu} + \Big( \Omega^* \Xi_{m} \Big)_{\mu}
\end{equation*}

where $\mu$ is the index of the components. To summarize, we're interested in

\begin{equation*}
\begin{split}
  G = \mathrm{GL}(d, \mathbb{R}) \qquad & \Omega: U^{(1)} \cap U^{(2)} \\
  P = \mathrm{LM} \qquad \qquad & \Xi_g : T_g G \to T_e G \\
  & \Omega^* \Xi: T U^{(1)} \cap U^{(2)} \to T_e G
\end{split}
\end{equation*}

Let $p \in U^{(1)} \cap U^{(2)}$, then

\begin{equation*}
\begin{split}
  \tensor{\Big( \Omega^* \Xi \Big)}{_p^i_j} \bigg( \frac{\partial }{\partial x^{\mu}}\bigg|_p \bigg)
  &= \underbrace{\tensor{\Xi}{_{\Omega(p)}^i_j} \Bigg( \Omega_* \frac{\partial }{\partial x^{\mu}}\bigg|_p \Bigg)}_{\text{def. of pull-back}} \\
  &= \tensor{\Big( \Omega^{-1}(p) \Big)}{^i_k} \tensor{dx}{^k_j}\big|_{\Omega(p)} \Bigg( \Omega_* \frac{\partial }{\partial x^{\mu}}\bigg|_p \Bigg) \\
  &= \tensor{\Big( \Omega^{-1}(p) \Big)}{^i_k} \underbrace{\Omega_* \Bigg( \frac{\partial }{\partial x^{\mu}}\bigg|_p \tensor{x}{^i_j}\Bigg)}_{\text{by def. of } \dd{x}} \\
  &= \tensor{\Big( \Omega^{-1}(p) \Big)}{^i_k} \frac{\partial }{\partial x^{\mu}}\bigg|_p \Big( \tensor{x}{^k_j} \circ \Omega \Big)(p) \\
  &= \tensor{\Big( \Omega^{-1}(p) \Big)}{^i_k} \frac{\partial }{\partial x^{\mu}}\bigg|_p \tensor{\Omega(p)}{^k_j}
\end{split}
\end{equation*}

Hence, considering the components of this

\begin{equation*}
\begin{split}
  \tensor{\big( \Omega^* \Xi \big)}{_p^i_j} 
  &= \tensor{\Big( \Omega^{-1}(p) \Big)}{^i_k} \frac{\partial }{\partial x^{\mu}} \Big( \tensor{\Omega(p)}{^k_j} \Big) \dd{x^{\mu}} \\
  &= \tensor{\big( \Omega^{-1}(p) \big)}{^i_k} \tensor{\dd{\Omega (p)}}{^k_j} \\
  &= \boldsymbol{\Omega}^{-1} \dd{ \boldsymbol{\Omega}}
\end{split}
\end{equation*}

where $\boldsymbol{\Omega}$ denotes the corresponding matrix.

Now, we need to compute the second term:

\begin{equation*}
\Ad_{\Omega^{-1}(p)^*} \omega^{U^{(2)}}
\end{equation*}

Observe that

\begin{equation*}
\begin{split}
  \Ad_{g}:\quad & G \to G \\
  \Ad_{g^*}: \quad & T_e G \to T_{\Ad_{g}(e)} = T_{e} \quad \text{since} \quad \Ad_{g}(e) = g e g^{-1} = e
\end{split}
\end{equation*}

Thus,

\begin{equation*}
\Ad_{g^*} A = g A g^{-1}
\end{equation*}

The above can be seen from:

\begin{equation*}
\Big( \Ad_{g} \circ \exp(A t) \Big)'(0) = \Big( g \exp(A t) g^{-1} \Big)'(0) = g A g^{-1}
\end{equation*}

Finally, this gives us the transition between the two Yang-Mills fields

\begin{equation*}
\begin{split}
  \tensor{\Big( \omega^{U^{(2)}} \Big)}{^i_{j \mu}} &= \tensor{\Big( \sigma^* \omega^{U^{(1)}} \Big)}{^i_{j \mu}} \\
  &= \underbrace{\tensor{\Big( \Omega^{-1} \Big)}{^i_k} \tensor{\Big( \omega^{U^{(1)}} \Big)}{^k_{l \mu}} \tensor{\Omega}{^l_{j \mu}}}_{= \Ad_{\Omega^{-1}(m)^*} \omega^{U^{(1)}}_{\mu}} + \underbrace{\tensor{\Big( \Omega^{-1} \Big)}{^i_k} \partial_{\mu} \tensor{\Omega}{^k_l}}_{= \big( \Omega^* \Xi_m \big)_{\mu}}
\end{split}
\end{equation*}

That is, we got it yo!

Structure theory of Lie algebras

Notation

  • $\mathfrak{g}$ denotes finite-dimensional Lie algebra over ground field $\mathbb{K} \in \left\{ \mathbb{R}, \mathbb{C} \right\}$

Universal enveloping algebra

The universal enveloping algebra of $\mathfrak{g}$, denoted by $U \mathfrak{g}$, is the associative algebra wih unit over $\mathbb{K}$ with generators $\left\{ i(x) : x \in \mathfrak{g} \right\}$ subject to relations

\begin{equation*}
\begin{split}
  i(x + y) &= i(x) + i(y) \\
  i(cx) &= c i(x), \quad \forall c \in \mathbb{K} \\
  i(x) i(y) - i(y) i(x) &= i \big( \comm{x}{y} \big)
\end{split}
\end{equation*}

For simplified notation, one will often write $x \in U \mathfrak{g}$ instead of $i(x)$, which will be justified later when we show that $i: \mathfrak{g} \to U \mathfrak{g}$ is injective (and so we can consider $\mathfrak{g}$ as a subspace in $U \mathfrak{g}$).

If we dropped the relation

\begin{equation*}
i(x) i(y) - i(y) i(x) = i \big( \comm{x}{y} \big)
\end{equation*}

from universal enveloping algebra, we would get the associative algebra generated by elements $x \in \mathfrak{g}$ with no relations other than linearity and associativity, i.e. tensor algebra of $\mathfrak{g}$:

\begin{equation*}
T \mathfrak{g} = \bigoplus_{n \ge 0} \mathfrak{g}^{\otimes n}
\end{equation*}

Therefore, we can alternatively describe the universal enveloping algebra as a quotient of the tensor algebra:

\begin{equation*}
U \mathfrak{g} = T \mathfrak{g} / \big( xy - yx - \comm{x}{y} \big), \quad x, y \in \mathfrak{g}
\end{equation*}

Even when $\mathfrak{g} \subseteq \mathfrak{gl}(n, \mathbb{K})$ is a matrix algebra, multiplication in $U \mathfrak{g}$ is not necessarily std. multiplication of matrices.

E.g. let $e \in \mathfrak{sl}(2, \mathbb{C})$, then $e^2 = 0$ as $2 \times 2$ matrices, but $e^2 \ne 0$ in $U \mathfrak{g}$; there are many representations of $\mathfrak{sl}(2, \mathbb{C})$ in which $\rho(e)^2 \ne 0$.

Let

  • $A$ be an associative algebra with unit over $\mathbb{K}$
  • $\rho: \mathfrak{g} \to A$ be a linear map such that

    \begin{equation*}
\rho(x) \rho(y) - \rho(y) \rho(x) = \rho ( \comm{x}{y})
\end{equation*}

Then $\rho$ can be uniquely extended to a morphism of associative algebras $U \mathfrak{g} \to A$.

This is the reason why we call $U \mathfrak{g}$ the universal associative algebra.

Any rep. of $\mathfrak{g}$ (not necessarily finite-dim) has a canonical structure of a $U \mathfrak{g}$ module.

Conversely, every $U \mathfrak{g}$ has a canoncial structure of a representation of $\mathfrak{g}$.

In other words, categories of reps. of $\mathfrak{g}$ and $U \mathfrak{g}$ modules are equivalent.

Example: commutative Lie algebra

Letting $\mathfrak{g}$ be a commutative Lie algebra, then $U \mathfrak{g}$ is generated by elements $x \in \mathfrak{g}$ with relations $xy = yx$, i.e.

\begin{equation*}
U \mathfrak{g} = S \mathfrak{g}
\end{equation*}

is the symmetric algebra of $\mathfrak{g}$, which can alos be described as the algebra of polynomial functions n $\mathfrak{g}^*$. Choosing a basis $\left\{ x_i \right\}$ in $\mathfrak{g}$, we see that

\begin{equation*}
U \mathfrak{g} = S \mathfrak{g} = \mathbb{K}[x_1, \dots, x_n]
\end{equation*}
Example: $\mathfrak{sl}(2, \mathbb{C})$

Universal enveloping algebra of $\mathfrak{sl}(2, \mathbb{C})$ is the associative algebra over $\mathbb{C}$ generated by elements $e$, $f$, and $h$ with relations

\begin{equation*}
\begin{split}
  ef - fe &= h \\
  he - eh &= 2e \\
  hf - fh &= - 2f
\end{split}
\end{equation*}
\begin{equation*}
f(\comm{x}{y}) = \comm{f(x)}{f(y)}
\end{equation*}

So if say $x \in \ker(f)$, then

\begin{equation*}
f ( \comm{x}{y} ) = \comm{f(x)}{f(y)} = \comm{0}{f(y)} = 0 \quad \implies \quad \comm{x}{y} \in \ker(f), \quad \forall y \in \mathfrak{g}
\end{equation*}

Hence $\ker(f)$ is an ideal.

Parallel Transport

This is the proper way of introducing the concept of Parallel transport.

The idea behind parallel transport is to make a choice of a curve $\gamma^{\uparrow}: [0, 1] \to P$ from the curve in the manifold $\gamma: [0, 1] \to M$, with

\begin{equation*}
\gamma(0) = a, \quad \gamma(1) = b, \quad a, b \in M
\end{equation*}

Since we have connections between the fibres, one might think of constructing "curves" in the total space $P$ by connecting a chosen point $g \in F_p$ for $p \in M$ to some other chosen point $g' \in F_{p'}$ for $p' \in M$, with $p$ and $p'$ being "near" each other.

Then the unique curve

\begin{equation*}
\gamma^{\uparrow}: [0, 1] \to P
\end{equation*}

through a point $\gamma^{\uparrow}(0) =: p \in \pi^{-1} \big( \left\{ a \right\} \big)$ which satisfies:

  1. $\pi \circ \gamma^{\uparrow} = \gamma$
  2. $\text{ver} \big( X_{\gamma^{\uparrow}, \gamma^{\uparrow}(\lambda)} \big) = 0$, for all $\lambda \in [0, 1]$
  3. $\pi_* \big( X_{\gamma^{\uparrow}, \gamma^{\uparrow}(\lambda)} \big) = X_{\gamma, \gamma(\lambda)}$, for all $\lambda \in [0, 1]$

is called the lift of $\gamma$ through $p$.

My initial thoughts for how to approach this would be to make a choice of section at each $p \in \text{im}(\gamma)$, such that

\begin{equation*}
(\sigma \circ \gamma)(\gamma) = g \in P
\end{equation*}

where $g$ would be a choice for every $\lambda$. Just by looking at this it becomes apparent that this is not a very good way of going about this, since we would have to choose these sections for each $U \subseteq M$ such that it's all well-defined. But this seems like a very "intricate" way of going about this.

The idea is to take some such curve as a "starting curve", denoted $\delta(\gamma)$, and then construct every other $\gamma^{\uparrow}$ from this $\delta$ by acting on it at each point by elements of $G$, or rather, choosing a curve $g: [0, 1] \to G$ in the Lie-group, such that

\begin{equation*}
\gamma^{\uparrow}(\lambda) = \delta(\lambda) \triangleright g(\lambda)
\end{equation*}

i.e. we can generate any lift of $\gamma$ simply by composing the arbitrary curve $\delta$ with some curve in the Lie-group.

It turns out that the choice of $g: [0, 1] \to G$ is the solution to an ODE with the initial condition

\begin{equation*}
g(0) = g_0
\end{equation*}

where $g_0$ is the unique group element such that

\begin{equation*}
\delta(0) \triangleright g_0 = p \in P
\end{equation*}

The ODE for $g: [0, 1] \to G$ is

\begin{equation*}
\Ad_{g^{-1}^*} \Big( \omega_{\delta(\lambda)} \big( X_{\delta, \delta(\lambda)} \big) \Big) + \Xi_{g(\lambda)} \Big( X_{g, g(\lambda)} \Big) = 0
\end{equation*}

with initial condition

\begin{equation*}
g(0) = g_0
\end{equation*}

such that

\begin{equation*}
\delta(0) \triangleright g_0 = p \in P
\end{equation*}

Worth noting that this is a first-order ODE.

Horizontal lift

Let

Then the horizontal lift of $\gamma$ to the associated bundle $P_F$ through the point $[p, f] \in P_F$ is the curve

\begin{equation*}
\begin{split}
  \overset{\overset{P_F}{\uparrow}}{\gamma}_{[p, f]}: \quad & [0, 1] \to P_F \\
  & \lambda \mapsto \big[ \overset{\uparrow}{\gamma}_p(\lambda),  \ f \big]
\end{split}
\end{equation*}

Curvature and torsion (on principle G-bundles)

Curvature

Let $P \overset{\lefttriangle G}{\longrightarrow} P \overset{\pi}{\longrightarrow} M$ be a principle G-bundle with a connection 1-form $\omega$.

Then let $\phi$ be a A-valued (e.g. Lie-algebra valued) k-form, then

\begin{equation*}
\begin{split}
  D \phi: \quad & \Gamma(T_0^K P) \to A \\
  & \big( X_1, \dots, X_k \big) \mapsto D \phi \big( X_1, \dots, X_k \big) := \dd{\phi} \Big( \text{hor}(X_1), \dots, \text{hor}(X_{k + 1}) \Big)
\end{split}
\end{equation*}

is called the covariant exterior derivative of $\phi$.

Let $P \overset{\pi}{\longrightarrow} M$ be a principal G-bundle with the connection 1-form $\omega$.

Then the curvature (of the connection 1-form) is the Lie-algebra-valued 2-form on $P$

\begin{equation*}
\begin{split}
  \Omega: \quad & \Gamma(T_0^T P) \to T_e G \\
  & (X_1, X_2) \mapsto \Omega(X_1, X_2) := (D \omega)(X_1, X_2)
\end{split}
\end{equation*}

Curvature (of 1-form) can also be written

\begin{equation*}
\Omega = \dd{\omega} + \omega \doublewedge \omega
\end{equation*}

where in the case Lie-algebras we have

\begin{equation*}
(\omega \doublewedge \omega)(X, Y) := [\![ \omega(X), \omega(Y) ]\!]
\end{equation*}

but the $\doublewedge$ is used to not restrict ourselves to Lie-algebra-valued forms, in which case $\doublewedge$ becomes different.

First observe that $\Omega$ is $C^{\infty}(P)$ bilinear.

We do this on a case-by-case basis.

  1. $X$ and $Y$ are both vertical, i.e. $\text{hor}(X) = \text{hor}(Y) = 0$. Then

    \begin{equation*}
\exists A, B \in T_e G: \quad X = X^A, \quad Y = X^B
\end{equation*}

    Then

    \begin{equation*}
\text{LHS} = \Omega(X^A, X^B) = \big( D \omega \big)(X^A, X^B) = \dd{\omega} \big( \underbrace{\text{hor}(X^A)}_{= 0}, \underbrace{\text{hor}(X^B)}_{= 0} \big) = 0
\end{equation*}

    And

    \begin{equation*}
\begin{split}
  \text{RHS} &= \big( \dd{\omega} \big)(X^A, X^B) + \big( \omega \doublewedge \omega \big)(X^A, X^B) \\
   &= X^A \Big( \underbrace{\omega(X^B)}_{ = B} \Big) - X^B \Big( \underbrace{\omega(X^A)}_{= A} \Big) - \omega \Big( \underbrace{\comm{X^A}{X^B}}_{\overset{(1)}{=} X^{[\![ A, B ]\!]}} \Big) + [\![ \underbrace{\omega(X^A)}_{ = A}, \underbrace{\omega(X^B)}_{ = B} ]\!] \\
   &= \underbrace{X^A(B)}_{ = 0} + \underbrace{X^B(A)}_{= 0} - \underbrace{\omega \Big( X^{[\![ A, B ]\!]} \Big)}_{= [\![ A, B ]\!]} + [\![ A, B ]\!] \\
   &= 0
\end{split}
\end{equation*}

    where at $(1)$ we've used the the fact that the map

    \begin{equation*}
\begin{split}
  & T_eG \to \Gamma(TP) \\
  & A \mapsto X^A
\end{split}
\end{equation*}

    defines a Lie-algebra homomorphism.

  2. $X$ and $Y$ are both horizontal, i.e. $\text{ver}(X) = \text{ver}(Y) = 0$. Then

    \begin{equation*}
LHS = \Omega(X, Y) = \big( D \omega \big)(X, Y) = \dd{\omega} \big( \underbrace{\text{hor}(X)}_{= X}, \underbrace{\text{hor}(Y)}_{= Y} \big) = \dd{\omega}(X, Y)
\end{equation*}

    and

    \begin{equation*}
\begin{split}
  \text{RHS} &= \dd{\omega}(X, Y) + \big( \omega \doublewedge \omega \big)(X, Y) \\
  &= \dd{\omega}(X, Y) + [\![ \underbrace{\omega(X)}_{= 0}, \underbrace{\omega(Y)}_{= 0} ]\!] \\
  &= \dd{\omega}(X, Y)
\end{split}
\end{equation*}

    since from Remark remark:horizontal-space-as-kernel-of-connection-1-form we know that

    \begin{equation*}
\text{hor}(X) = X \quad \implies \quad X \in \text{ker}(\omega) \quad \implies \quad \omega(X) = 0
\end{equation*}
  3. (wlog) $X$ is horizontal, and $Y$ is vertical, i.e.

    \begin{equation*}
\text{ver}(X) = 0 \quad \text{and} \quad \text{hor}(Y) = 0 \implies \exists A \in T_e G: \quad Y = X^A
\end{equation*}

    Since any tangent can be separated into horizontal and vertical component, we know showing these cases is sufficient. Furthermore, since both sides of the expression we want to show is anti-symmetric in the arguments, if $X$ and $Y$ were swapped we'd pick up a minus sign on both sides which cancel each other out. Then