Symmetries of Particles and Fields

Table of Contents

Metric tensor

  1. $g_{\mu \nu}$ has real eigenvalues $\lambda_i$:

    \begin{equation*} g_{\mu \nu} = \text{diag}(\lambda_1, \dots, \lambda_n) \end{equation*}

  2. Rescale sucht that $\lambda_i \in \left\{ -1, 0, 1 \right\}$ keeping $x^2 = g_{\mu \nu}x^{\mu} x^{\nu}$ fixed. Rescaling follows

    \begin{equation*} \begin{split} x^{\mu} &\mapsto s x^{\mu} \\ g_{\mu \nu} & \mapsto s^{-2} g_{\mu \nu}, \quad s^2 > 0 \end{split} \end{equation*}

    which we can do separately for each $\lambda_i$, hence

    \begin{equation*} g_{\mu \nu} = \text{diag} \big( +1, +1, \dots, 0, 0, \dots, -1, -1 \big) \end{equation*}

    which we call the canonical basis.

Example: space time

\begin{equation*} g_{\mu \nu} = \text{diag}(\underbrace{1, 1, 1}_{\text{space}}, \underbrace{- 1}_{time}) \end{equation*}

such that

\begin{equation*} x^2 = x_1^2 + x_2^2 + x_3^2 - x_4^2 \end{equation*}

Usually, we use the opposite signature metric, using $x^4 = x^0$ and $x4 = x0:

\begin{equation*} g_{\mu \nu} = \text{diag}\big(1, -1, -1, -1 \big) \end{equation*}

which gives

\begin{equation*} x^2 = x_0^2 - \big( x_1^2 + x_2^2 + x_3^2 \big) \end{equation*}

Raising and lowering indices using the metric tensor

$\det g_{\mu \nu} \ne 0$ if and only if $g_{\mu \nu}$ is invertible.

Can use metric to raise and lower indices

\begin{equation*} x_{\mu} \equiv g_{\mu \nu} x^{\nu} \end{equation*}
\begin{equation*} x^2 = x_{\mu} x^{\mu} = x^{\mu} x_{\mu} = g_{\mu \nu} x^{\mu} x^{\nu} = \tensor{\delta}{_{\mu}^{\nu}} x^{\mu} x_{\nu} \end{equation*}

Can raise indices of metric

\begin{equation*} \begin{split} g^{\mu \nu} &= g^{\mu \alpha} g^{\nu \beta} g_{\alpha \beta} \\ &= g^{\mu \alpha} \tensor{\delta}{_{\alpha}^{\nu}} \end{split} \end{equation*}

were we've contracted over $\beta$. Then

\begin{equation*} g^{\nu \beta} g_{\alpha \beta} = \tensor{\delta}{_{\alpha}^\nu} \end{equation*}

which implies $g^{\mu \nu}$ is the /inverse of $g_{\mu \nu}$.

Same with Levi-Civita symbols
\begin{equation*} \varepsilon^{\mu \nu \dots \alpha \beta} = g^{^{\mu \mu'}} \tensor{g}{^\nu \nu'} \dots \tensor{g}{\alpha \alpha'} \tensor{g}{^{\beta \beta'}} \end{equation*}

and derivatives wrt. $xμ

\begin{equation*} \partial_{\mu} \equiv \frac{\partial }{\partial x^{\mu}} = g_{\mu \nu} \frac{\partial }{\partial x_{\nu}} = g_{\mu \nu} \partial^{\nu} \end{equation*}
If metric is definite

If metris is definite, the we can choose basis such that $g_{\mu \nu} = \delta_{\mu \nu}$, i.e. can ignore distinction between upper and lower indices!

Matrix groups

Lorentz group

\begin{equation*} g_{\mu \nu} = \big( 1, 1, 1, -1 \big) \end{equation*}

is called $\text{SO}(3, 1)$ which has $\det D = + 1$.

1D groups

All 1D Lie groups are such that the parameter $a$ can be chosen such that

\begin{equation*} g(a) g(b) = g(a + b) \end{equation*}

Furthermore, all 1D Lie groups are Abelian.

$U(1)$ and $SO(2)$ are isomorphic.

Let $G$ be a $n$ dimensional compact Abelian Lie group. Then

\begin{equation*} G \cong U(1) \otimes U(1) \otimes \dots \otimes U(1) \end{equation*}

If $G$ is compact, then for every representation there exists an equivalent unitary representation.

Lie algebras

Generators

Consider group $G = \left\{ g(\alpha) \right\}$ where we choose $\alpha_a$ with $a \in [1, \dim G]$ such that

\begin{equation*} g(0) = 1 \end{equation*}

Taylors thm expand $g(\alpha)$ about $g(\alpha = 0)$, so we can write

\begin{equation*} g(\alpha) = 1 + \alpha^a \frac{\partial g(\alpha)}{\partial \alpha^a} \bigg|_{\alpha = 0} + \mathcal{O}(\alpha^2) \end{equation*}

for "small" $\alpha$.

Define

\begin{equation*} T_a := \frac{1}{i} \frac{\partial g(\alpha)}{\partial \alpha^a} \bigg|_{\alpha = 0} \end{equation*}

Matrices $\{ T_a \}$ are the generators of the group.

The $\frac{1}{i}$ is put in to make $T^a$ hermitian (in a unitary representation).

Examples

Example: $SU(N)$
\begin{equation*} g^\dagger g = g g^\dagger = 1 , \quad \det g = 1 \end{equation*}

which implies

\begin{equation*} T_a^{\dagger} = T_a \text{ (i.e. hermitian)} \quad \text{ and } \quad \tr T_a = 0 \end{equation*}

Thus, the generators are hermitian and traceless.

Example: $SU(2)$

We generally choose Pauli matrices as generators:

\begin{equation*} \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{equation*}
Example: $SU(3)$

  • Generators are Gell-Mann

    \begin{equation*} \left\{ \lambda_a : a = 1, 2, \dots, 8 \right\} \end{equation*}
Example: $U(N)$
  • Don't have constraint $\det D = 1$.
  • Generators of $SU(N)$, $\{ T_a \}$ are also the part a subset of the generators of $U(N)$, and "including the trace" we have $\{ 1, T_a \}$ as the generators for $U(N)$
Example: $SO(N)$
\begin{equation*} g^T g = g g^T = 1, \quad \det g = 1 \end{equation*}

for $SO(N)$, $g$ is real, where we generally omit the $i$ in the def. of generators.

Taking Taylor expansion, as for $SU(N)$, we have

\begin{equation*} T_a = - T_a^T \end{equation*}

i.e. $T_a$ are anti-symmetric.

Example: $SO(2)$ and $SO(3)$

For $SO(2)$, generally choose:

\begin{equation*} T = T_a = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad \alpha = 1 \end{equation*}

For $SO(3)$, generally choose

\begin{equation*} \left\{ T_a \right\} = \left\{ \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} \right\} \end{equation*}

Additional constraint $\det g = 1$ is automatically satisfied (why?).

Commutations

Consider $f = g h g^{-1} h^{-1}$ with $g, h \in G$.

Expanding each of these using their Taylor expansion up to and including 2nd order terms, then take the commutation relation between $\alpha^a T_a$ and $\beta_b T^b$, the expansion of $g$ and $h$, respectively, we have

\begin{equation*} \comm{\alpha^a T_a}{\beta^b T_b} = - \frac{i}{2} \gamma^c T_c \end{equation*}

For this to be true $\forall \alpha^a, \beta^b$ we must have

\begin{equation*} \gamma^c = - 2 \tensor{C}{^c_a_b} \alpha^a \beta^b \end{equation*}

for some numbers $\tensor{C}{^c_a_b}$.

Which implies

\begin{equation*} \comm{T_a}{T_b} = -2 \tensor{C}{^c_{ab}} \end{equation*}

where $\tensor{C}{^c_{ab}}$ are called the structure constants.

Exponentiation

Consider 1D subgroup $G_1 \subset G$, $\{ g(t) \} = \left\{ g \big(\alpha(t)\big) \right\}$ labeled by the single parameter $t$.

Chose $t \in \mathbb{R}$ such that

\begin{equation*} g(s) g(t) = g(t + s), \quad g(0) = 1 \implies \alpha_a(0) = 0 \end{equation*}

Diff. wrt. $s$

\begin{equation*} g'(s) g(t) = g'(t + s) \end{equation*}

and set $s= 0$, which gives us a differential equation for all $g(t)$:

\begin{equation*} \frac{d}{dt} g(t) = g'(0) g(t) \end{equation*}

which has the solution

\begin{equation*} g(t) = \exp \Big( t \ g'(0) \Big) \end{equation*}

Or, if you consider this in the familiar notation used in Diff. Geom. for the exponential map, we can write this as

\begin{equation*} \gamma(t) = \exp \Big( t X \Big) \end{equation*}

where $X \in T_e G$, s.t.

\begin{equation*} \gamma(1) = \exp (X) \end{equation*}

where $\exp: T_e G \to G$, i.e. maps from the Lie algebra to the Lie group.

Going back to the earier notation of $g(t)$, we can write

\begin{equation*} \begin{split} g'(0) &= \frac{\partial g}{\partial \alpha^a}\bigg|_{\alpha = 0} \frac{d \alpha^a}{dt} \bigg|_{t = 0} \\ &= i T_a \gamma^a \end{split} \end{equation*}

where

\begin{equation*} \gamma^a := \frac{d \alpha^a}{dt} \bigg|_{t = 0} \end{equation*}

which implies

\begin{equation*} g(t) = \exp \Big( i (t \gamma^a) T_a \Big) \end{equation*}

Adjoint representation

Can define an adjoint representation:

\begin{equation*} \tensor{\big( A_a \big)}{^b_c} = i \tensor{C}{^b_{ac}} \end{equation*}

then

\begin{equation*} \comm{A_a}{A_b} = i A_c \tensor{C}{^c _{ab}} \end{equation*}
Examples
  • $SU(2)$ and $SO(3)$

    • $SU(2)$ take generators $\{ \frac{1}{2} \sigma_j \}$, with

      \begin{equation*} \comm{\frac{1}{2} \sigma_i}{\frac{1}{2} \sigma_j} = i \epsilon_{ijk}\ \frac{1}{2} \sigma_k \end{equation*}

      and so the structure constants are $\epsilon_{ijk}$.

    • Consider

      \begin{equation*} g = 1 + \frac{i}{2} \boldsymbol{\omega} \cdot \boldsymbol{\sigma} + \mathcal{O}(\omega^2) \end{equation*}

      Then

      \begin{equation*} \begin{split} g \sigma_k g^{-1} &= \bigg( 1 + \frac{i}{2} \omega_i \sigma_i \bigg) \sigma_k \bigg( 1 - \frac{i}{2} \omega_i \sigma_i \bigg) \\ &= \sigma_k + \omega_i \epsilon_{kim} \sigma_m \end{split} \end{equation*}

      So if we write $\boldsymbol{\sigma} = \big( \sigma_1, \sigma_2 ,\sigma_3 \big)$, then we can write

      \begin{equation*} g \boldsymbol{\sigma} g^{-1} = \boldsymbol{\sigma} + \boldsymbol{\omega} \times \boldsymbol{\sigma} \end{equation*}

      which is just rotation in $\mathbb{R}$!

    • Therefore the adjoint representation of $SU(2)$ is the defining or fundamental representation of $SO(3)$. hence

      \begin{equation*} SU(2) \cong SO(3) \end{equation*}

      for elements "close to the identity".

    In general, the adjoint representation is not the fundamental representation of some other group…

Killing form

The Killing form

\begin{equation*} g_{ab} := \text{tr}(A_a A_b) \end{equation*}

real symmetric $\dim G \times \dim G$ matrix.

Usually we'll choose diagonal basis such that

\begin{equation*} g_{ab} := \big( 1, 1, 1, \dots, 1, 0, 0, 0, -1, -1, \dots, -1 \big) \end{equation*}

Examples

$SU(2)$
\begin{equation*} g_{ij} = - \epsilon_{ikl} \epsilon_{j l k} = 2 \delta_{ij} \end{equation*}

So we rescale generators such that

\begin{equation*} - \frac{1}{\sqrt{2}} \epsilon_{ijk} : g_{ij} = \delta_{ij} \end{equation*}

Application of killing form

  • Can use Killing form to lower and raise indices

  • E.g. kan define structure consts with all lower indices

    \begin{equation*} C_{abc} := g_{ad} \tensor{C}{^d_{bc}} \end{equation*}

    • This is an invariant tensor → can be written as a product of generators

      \begin{equation*} \begin{split} \text{tr} \Big( A_{a} \comm{A_b}{A_c} \Big) &= i \text{tr} \big( A_a A_d \big) \tensor{C}{^d_{bc}} \\ &= i g_{ad} \tensor{C}{^d_{bc}} \\ &= i \tensor{C}{_{abc}} \end{split} \end{equation*}

Invariant Tensors

Let $G$ be a Lie group.

An invariant tensor

\begin{equation*} d_{a_1, a_2, \dots, a_n} \end{equation*}

is a tensor that remains invariant under the actions of group $G$, i.e.

\begin{equation*} d_{a_1, a_2, \dots, a_n} = d_{a_1, a_2, \dots, a_n} = d_{a_1', \dots, a_n'} \tensor{D}{} \end{equation*}