# Algebra

## Notation

• denotes a group
• CANI stands for:
• Commutative
• Associative
• Neutral element (e.g. 0 for addition)
• Inverse

## Terminology

almost all
is an abbreviation meaning "all but finitely many"

## Definitions

### Homomorphisms

A homomorphism is a structure-preserving map between the groups and , i.e.

A endomorphism is a homomorphism from the group to itself, i.e.

An automorphism is an invertible homomorphism, denoted .

In the case of a vector space , we have

### Linear isomorphism

Two vector spaces and are said to be isomorphic if and only if there exists a linear bijection .

### Field

An (algebraic) field is a set and the maps

that satisfy

### Vector space

A vector space over a field is a pair consisting of an abelian group and a mapping

such that for all and we have A D D U:

• Associativity:
• Distributivity over field-multiplication:
• Uint:

### Ring

A ring over a set with the maps:

which satisfy the same as a field, but without C and I for , i.e. multiplication.

A division ring is a ring where the multiplicative operation allows an inverse, i.e. it's

### Equivalence relations

A equivalence relation on some set is defined as some relation betweeen and , denoted , such that the relation is:

• reflexive:
• symmetric:
• transistive

Why do we care about these?

• Partitions the set it's defined on into unique and disjoint subsets

### Unitary transformations

A unitary transformation is a transformation which preserves the inner product.

More precisely, a unitary transformation is an isomorphism between two Hilbert spaces.

## Groups

### Notation

• or denotes the number of left cosets of a subgroupd of , and is called the index

### Definitions

The symmetric group of a finite set of symbols is the group whose elements are all the permutation operations that can be performed on the distinct symbols, and whose group operation is the composition of such permutation operations, which are defined as bijective functions from the set of symbols to itself.

An action of a group is a formal way of interpreting the manner in which the elements of the group correspond to transformations of some space in a way that preserves the structure of the space.

If is a group and is a set, then a (left) group action of on is a function

that satisfies the following two axioms (where we denote as ):

• identity: for all ( denotes the identity element of )
• compatibility: for all and all

where denotes the result of first applying to and then applying to the result.

From these two axioms, it follows that for every , the function which maps to is a bijective map from to . Therefore, one may alternatively define a group action of on as a group homomorphism from into the symmetric group of all bijections from to .

The action of on is called transistive if is non-empty and if:

Faithful (or effective ) if

That is, in a faithful group action, different elements of induce different permutations of .

In algebraic terms, a group acts faithfully on if and only if the corresponding homomorphism to the symmetric group, , has a trivial kernel.

If does not act faithfully on , one can easily modify the group to obtain a faithful action. If we define:

then is the normal subgroup of ; indeed, it is the kernel of the homomorphism . The factor group acts faithfully on by setting .

We say a group action is free (or semiregular or fixed point free ) if, given ,

We say a group action is regular if and only if it's both transitive and free; that is equivalent to saying that for every two there exists precisely one s.t. .

Consider a group acting on a set . The orbit of an element in is the set of elements in to which can be moved by the elements of . The orbit of is denoted by :

An abelian group, or commutative group, is simply a group where the group operation is commutative!

A monoid is an algebraic structure with a single associative binary operation and an identity element, i.e. it's a semi-group with a binary operation.

#### Cosets

Let be a group and be a subgroup of . Let . The set

of products of with elements of , with on the left is called a left coset of in .

The number of left cosets of a subgroup of is the index of in and is denoted by or . That is,

#### Center

Given a group , the center of , denoted is defined as the set of elements which commute with every element of the group, i.e.

We say that a subgroup of is central if it lies inside , i.e. .

#### Abelianisation

Given a group , define a abelianisation of to be the quotient group

with is the normal subgroup generated by the commutators for , i.e.

### Theorems

Let be a group of order and let be a prime divison of .

Then has an element of order .

#### Fermat's Little Theorem

Let be a prime number.

Then for , the number is an integer multiple of . That is

#### Isomorphism theorems

Let

Then is a normal subgroup of , and . Furthermore, there is an isomorphim

In particular, if is surjective, then

Le be a group, and .

1. First so clearly . Let and . We then want to show that . Observe that

since . So so that ,

And now we check that the

2. Let , . Wan to show that .

3. Let , . Then since , so , for all .
4. Want to show that . First Isomorphism Theorem tells us that

Therefore, letting

where we simply factor out the from every element in .

So maps into , but we need it to be surjective, i.e.

An element of is a coset for , which is clearly . And finally,

Hence, by the First Isomorphism Theorem,

Notice . In fact, .

As an example, . Then

by the First Isomorphism Theorem and since . And, we can also write

since .

Want to show that both are isom. to . We do this by constructing map:

and just take the coset. And,

Hence by 1st Isom. Thm.

### Kernels and Normal subgroups

Arising from an action of a group on a set , is the homomorphism

the permutation of corresponding to .

The kernel of this homomorphism is also called the kernel of the action and is denoted .

Remember,

is the permutation . Therefore, if and only if for all , and so

consists of those elements which stabilize every element of .

Let be a subgroup of a group , and let .

The conjugate of by , written , is the set

of all conjugate of elements of by .

This is the image of under the conjugation homomorphism where .

Hence, is a subgroup of .

Let be a subgroup of .

If for all , then

A subgroup of a group, , is called a normal subgroup if it is invariant under conjugation; that is, the conjugation of an element of by an element of is still in :

. Then if and only if for all .

Let and be groups.

The kernel of any homomorphism is normal in .

Hence, the kernel of any group action of is normal in .

Let be a group acting on the set of left cosets of a subgroup of .

1. The stabilizer of each left coset is the conjugate
2. If for all then

### Factor / Quotient groups

Let be a group acting on a set and be the kernel of the action.

The set of cosets of in is a group with binary operation

which defines the factor group or quotient group of by and is denoted

But I find the following way of defining a quotient group more "understandable":

Let be a group homomorphism such that

That is, maps all distinct which are equivalent under to the same element in , but still preserves the group structure by being a homomorphism of groups.

There is a function with

For we have that

thus is a homomorphism.

Then, clearly

is called the natural homomorphism from to .

The order of a factor group is the number of distinct cosets of in .

If is finite, then

### Group presentations

#### Notation

• or refers to the group generated by such that
• denote is the free group as generated by
• In general: , e.g.

where the "unit-condition" simply specifies that the group is commutative

#### Free groups

is the free group generated by .

Elements are symbols in , subject to

• group axioms
• " and all logical consequences :) "

Let .

The group with presentation

is the group generated by subject to

• group axioms
• " and all logical consequences :) "

There's no algorithm for deciding whether is the trivial group.

Let and let , where is a group.

Then there is a unique homomorphism:

And the image of is the subgroup of generated by .

### Central Extensions of Groups

Let

• be an abelian group
• be an arbitrary group

An extension of by the group is given an exact sequence of group homomorphisms

The extension is called central if is abelian and its image , where denotes the center of , i.e.

For a group acting on another group by a homomorphism , the semi-direct product group is the set with the multiplication given by the formula

This is a special case of a group extension with and :

### Exact sequence

An exact sequence of groups is given by

of groups and group homomorphisms, where exact refers to the fact that

## Linear Algebra

### Notation

• denotes the set of all matrices on the field
• denotes the representing matrix of the mapping wrt. bases and , where is ordered basis for and ordered basis for :

• and where denotes the "identity-mapping" from elements represented in the basis to the representation in .
• denotes the n-dimensional standard basis
• , i.e. the set of non-zero elements of

• is a set
• is a field

### Basis

A subset of a vector space is called a generating set of the vector space if its span is all of the vector space.

A vector space that has a finite generating set is said to be finitely generated.

is called linearly independent if for all pairwise different vectors and arbitrary scalars ,

A basis of a vector space is a linearly independent generating set in .

The following are equivalent for a subset of for a bector space :

1. is a basis, i.e. linearly independent generating set
2. is minimal among all generating sets, i.e. does not generate for any
3. is maximal among all linearly independent subsets, i.e. is not lineraly independent for any .

"Minimal" and "maximal" refers to the inclusion and exclusion.

Let be a vector space containing vector subspaces . Then

#### Linear mappings

Let be vector spaces over a field . A mapping is called linear or more precisely linear if

This is also a homomorphism of vector spaces.

Let be a linear mapping between vector spaces. Then,

where usually we use the terminology:

• rank of is
• nullity of is

### Linear Mappings and Matrices

Let be a field and let .

There exists a bijection and set of matrices with rows and columns:

Which attaches to each linear mapping , its representing matrix , defined

i.e. the matrix-representation of is a defined by how maps the basis of the target space.

Observe that the matrix product between two matrices and ,

An elementary matrix is any square matrix which differs from the identity matrix in at most one entry.

Any matrix whose only non-zero entries lie on the diagonal, and which has the first 1's along the diagonal and then 0's elsewhere, is said to be in Smith Normal Form

there exists invertible matrices and s.t. is a matrix in Smith Normal Form.

A linear mapping is injective if and only if

Let , then

Hence, if

as claimed.

Let be square matrices over some commutative ring are conjugate if

for an invertible P ∈ (n; R).

Further, conjugacy is an equivalence relation on .

#### Trace of linear map

The trace of a matrix is defined

The trace of a finite product of matrices is independent of the order of the product (given that the products are valid). In other words, trace is invariant under cyclic permutations.

To see that trace is a invariant under cyclic permutations, we observe

This case of two matrices can easily be generalized to case of products of multiple matrices.

### Rings and modules

A ring is a set with two operatiors that satisfy:

1. is an abelian group
2. is a monoid
3. The distributive laws hold, meaning that :

Important: in some places, e.g. earlier in your notes, they use a slightly less restrictive definition of a ring, and in that case we'd call this definition a unitary ring.

#### Polynomials

A field is algebraically closed if each non-constant polynomial with coefficients in has a root in .

E.g. is algebraically closed, while is not.

If a field is algebraically closed, then every non-zero polynomial decomposes into linear factors

with , and .

This decomposition is unique up to reordering of the factors.

Let be an integral domain and let with monic.

Then there exists unique such that

and or .

#### Ideals and Subrings

Let and be rings. A linear map is a ring homomorphism if the following hold for all :

A subset of a ring is an ideal, written , if the following hold:

1. is closed under subtraction
2. for all and , i.e. closed under multiplication by elements of
• I.e. we stay in even when multiplied by elements from outside of

Ideals are sort of like normal subgroups for rings!

Let be a commutative ring and let .

Then the ideal of generated by is the set

together with the zero element in the case .

If , a finite set, we will often write

Let be a commutative ring. An ideal of is called a principal ideal if

for some .

Let be a subset of a ring . Then is a subring if and only if

1. has a multiplicative identity
2. is closed under subtraction:
3. is closed under multiplication

It's important to note that and does not necessarily have the same identity element, even though is a subring of !

Let be a ring. An element is a called a unit if it's invertible in or in other words has a multiplicative inverse in , i.e.

We will use the notation for the group of units of a ring .

In a ring a non-zero element is called a zero-divisor or divisor of zero if

An integral domain is a non-zero commutative ring that has no zero-divisors, i.e.

1. If then or
2. and then

#### Factor Rings

Let be an ideal in a ring . The set

is a coset of in or the coset of wrt. in .

Let be a ring and and ideal of .

The mapping

Has the following properties:

1. is surjective
2. If is a ring homomorphism with so that

then there is a unique ring homomorphism:

Where the second point states that factorizes uniquely through the canonical mapping to the factor whenever the ideal is sent to zero.

Let and be rings.

Then every ring homomorphism induces a ring isomorphism

#### Modules

We say is a R-module, being a ring, if

satisfying

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: module.

Important: denotes a module here, NOT manifold as usual.

A unitary module is in the case where we also have , i.e. the ring is a unitary ring and contains a multiplicative identity-element.

Let be a ring and let be an R-module. A subset of is a submodule if and only if

Let be a ring, let and be R-modules and let be an R-homomorphism.

Then is injective if and only if

Let . Then is the smallest submoduel of that contains .

The intersection of any collection of submodules of is a submodule of .

Let and be submodules of . Then

is a submodule of .

Let be a ring, and R-module and submodule of .

For ever the coset of wrt. in is

It is a coset of in the abelian group and so is an equivalence class for the equivalence relation

The factor of by or quotient of by is the set

of all cosests of in .

for all and .

The R-module is the factor module of by submodule .

Let be a ring, let and be R-modules, and a submodule of .

1. The mapping defined by

is surjective.

2. If is an R-homomorphism with

so that , then there is a unique homomorphism such that .

Let be a ring and let and be R-module. Then every R-homomorphism induces an R-isomorphism

Let be submodules of a R-module .

Then is a submodule of and is a submodule of .

Also,

Let be submodules of an R-module , where .

Then is a submodule of .

Also,

### Determinants and Eigenvalue Reduction

#### Definitions

An inversion of a permutation is a pair such that and .

The number of inversions of the permutation is called the length of and writtein .

The sign of a permutation is defined to be the parity of the number of inversions of :

where is the length of the permutation.

For , the set of even permutations in forms a subgroup of because it is the kernel of the group homomorphism .

This group is the alternating group and is denoted .

(XÌƒ / ker(p))

Let be a ring.

The determinant is a mapping given by

Let be a commutative ring, then

Let for some ring , and let be the defined by removing the i-th row and j-th column of .

Then the cofactor matrix of is defined (component-wise)

The adjugate matrix of is defined

where is the cofactor matrix.

The reason for this definition is so that we have the following identity

To see this, recall the "standard" approach to computing the determinant of a matrix , where we do so by choosing some row and some column , cut out the rest of the matrix, and write the determinant as a expansion in these coefficients. The cofactor matrix has exactly the determinant of that matrix as a entries, hence the above expression is simply going to give us the same expression as the "expansion-method" for computing the determinant.

Let be an matrix with entries from a commutative ring .

For a fixed the i-th row expansion of the determinant is

and for a fixed the j-th column expansion of the determinant is

where is the cofactor of

where is the matrix obtained from deleting the i-th row and j-th column.

#### Cayley-Hamilton Theorem

Let be a square matrix with entries in a commutative ring .

Then evaluating its characteristic polynomial at the matrix gives zero.

Let

where denotes the adjugate matrix of and .

Observe then that

by the propoerty of the adjugate matrix.

Since is a matrix whose entries are a polynomial, we can write as

for some matrices .

Therefore,

Now, letting

we obtain the following relations

Multiplying the above relations by for each we get the following

Substituting back into the above series, we have

we observe that the LHS vanishes, since we can group terms which cancel! Hence

#### Eigenvalues and Eigenvectors

##### Theorems

Each endomorphism of non-zero finite dimensional vector space over an algebraically closed field has an eigenvalue.

### Inner Product Spaces

#### Definitions

##### Inner product

An inner product is a (anti-)bilinear map which is

1. Symmetric
2. Non-degenerate
3. Positive-definite
##### Forms

Let . We say is a bilinear form if

for , and .

A symmetric bilinear form on is a bilinear map such that .

Given two 1-forms at define a symmtric bilinear form on by

where .

Note that and we denote .

REDEFINE WITHOUT THE ADDED TANGENT-SPACE STRUCTURE, BUT ONLY USING VECTOR SPACE STRUCTURE.

A symmetric tensor on is a map which assigns to each a symmetric bilinear form on ; it can be written as

where are smooth functions on .

Remember, we're using Einstein notation.

##### Skew-linear and sesquilinear form

We say the mapping between complex vector spaces is skew-linear if

Let be a vector space over equipped with the inner product

Since this mapping is skew-linear in the second argument, i.e.

we say this is a sesquilinear form.

##### Hermitian

Let be a vector space over . If the sequilinear form is symmetric, i.e.

we say is a Hermitian.

#### Theorems

Let be a vectors in an inner product space. Then

with equailty if and only if and are linearly dependent.

Let be an inner product space. then two endomorphism are called adjoint to if the following holds:

Let and be the adjoint of . We say is self-adjoint if and only if

Let be a finite-dimensional inner product space and let be a self-adjoint linear mapping.

Then has an orthonormal basis consisting of eigenvectors of .

We'll prove this using induction on . Let denote the an inner product space with with the self-adjoint linear mapping .

Base case: . Let denote an eigenvector of , i.e.

where is the underlying field of the vector space . Then clearly spans .

General case: Assume the hypothesis holds for , then

where

Further, let and with be an eigenvector of . Then

since is self-adjoint. This implies that

Therefore, restricting to we have the mapping

Where is also self-adjoint, since is self-adjoint on the entirety of . Thus, by assumption, we know that there exists a basis of consisting of eigenvectors of , which are therefore orthogonal to the eigenvectors of in . Hence, existence of basis of eigenvectors for implies existence of basis of eigenvectors of .

Thus, by the induction hypothesis, if is a self-adjoint operator on the inner product space with , , then there exists a basis of eigenvectors of for , as claimed.

### Jordan Normal Form

#### Notation

• is an endomorphism of the finite dimensional F-vector space .
• Characteristic equation of :

• Polynomials

• Generalized eigenspace of wrt. eigenvalue

• denotes the dimension of
• Basis

• Restriction of

• defined

is well-defined and injective.

#### Definitions

An endomorphism of an F-vector space is called nilpotent if and only if such that

#### Motivation

• be a finite dimensional vector space
• an endomorphism
• a choice of ordered basis for determines a matrix representing wrt. basis
• Another choice of basis leads to a different representation; would like to find the simplest possible matrix that is conjugate to a given matrix

#### Theorem

Given an integer , we define the matrix

or equivalently,

which we call the nilpotent Jordan block of size .

Given an integer and a scalar define an matrix as

which we is called the Jordan block of size and eigenvalue .

Let be an algebraically closed field, and be a finite dimensional vector space and be an endomorphism of with characteristic polynomial

where and , for distinct .

Then there exists an ordered basis of s.t.

with such that

with .

That is, in the basis is block diagonal with Jordan blocks on the diagonal!

##### Proof of Jordan Normal Form
• Outline

We will prove the Jordan Normal form in three main steps:

1. Decompose the vector space into a direct sum

according to the factorization of the characteristic polynomial as a product of linear factors:

for distinct scalars , where for each :

2. Focus attention on each of the to obtain the nilpotent Jordan blocks.
3. Combine Step 2 and 3
• Step 1: Decompose

Rewriting as

where are the eigenvalues of .

For define

There exists polynomials such that

For each , let

be a basis of , where is the algebraic multiplicity of with eigenvalue .

1. Each is stable under , i.e.
2. For each , such that

In other words, there is a direct sum decomposition

3. Then

is a basis of , so in particular . The matrix of the endomorphism wrt. to basis is given by the block diagonal matrix

with .

1. Let is that

Then

Hence , i.e. is stable under .

2. By Lemma 6.3.1 we have and so evaluating this at , we get

Therefore, , we have

Observe that

where we've used Cailey-Hamilton Theorem for the second equality. Let

then

hence all can be written as a sum of , or equivalently,

as claimed.

3. Since is a basis of for each , and since

we have

form a basis of . Consider the ordered basis , then the can be expressed as a linear combination of the vectors with . Therefore the matrix is block diagonal with i-th block having size .

From this, one can prove that any matrix can be written as a Jordan decomposition:

Let , then there exists a diagonalisable (NOT diagonal) matrix and nilpotent matrix such that

In fact, this decomposition is unique and is called the Jordan decomposition.

• Step 2: Nilpotent endomorphisms

Let be a finite dimensional vector space and such that

for some , i.e. is nilpotent. Further, let be minimal, i.e. but .

For define

If then

i.e. , hence

Moreover, since and , we have and . Therefore we get the chain of subspaces

We can now develop an algorithm for constructing a basis!

• Constructing a basis
1. Choose arbitrary basis for
2. Choose basis of of by mapping using and choosing vectors linearly independent of .
3. Repeat!

Or more accurately:

1. Choose arbitrary basis for :

2. is linearly independent.

3. Choose vectors

such that

is a basis of .

4. Repeat!

Now, the interesting part is this:

Let be a finite dimensional vector space and such that

for some , i.e. is nilpotent.

Let be the ordered basis of constructed as above

Then

where denotes the nilpotent Jordan block of size .

It follows from the explicit construction of the basis that

Since is defined by how it maps the basis vectors in , and in the basis becomes a vector of all zeros except the entry corresponding to the j-th basis-vector chosen for , where it is a 1.

Hence is a nilpotent Jordan block as claimed.

Concluding step 2; for all nilpotent endomorphisms there exists a basis such that the representing matrix can be written as a block diagonal matrix with nilpotent Jordan blocks along the diagonal.

• Step 3: Bringing it together

Again considering the endomorphisms restricted to , we can apply Proposition 6.3.9 to see that this endomorphism can be written as a block diagon matrix of the form stated for a suitable choice of basis.

The endomorphism restricted to is of course , thus the matrix wrt. the chosen basis is just . Therefore the matrix for (when restricted to ) is just plus , i.e. .

Thus, each appearing in this theorem is exactly of the form we stated in Jordan Normal form.

#### Algorithm

1. Calculate the eigenvalues, of

2. For each and , let

1. Compute and

2. Let

3. Set

i.e. the difference in dimension between each of the nullspaces.

1. Let be the largest integer such that .
2. If does not exist, stop. Otherwise, goto step 3.
3. Let and .
4. Let
1. Change to for .
3. Let the full basis be the union of all the , i.e. .

## Tensor spaces

The tensor product between two vector spaces is a vector space with the properties:

1. if and , there is a "product"
2. This product is bilinear:

for , and .

Let be a vector space.

If is a bilinear map, then there exists a unique linear map such that

I find the following instructive to consider.

In the case of a Cartesian product, the vector spaces are still "independent" in the sense that any element can be expressed in the basis

which means

Now, in the case of a tensor product, we in some sense "intertwine" the spaces, making it so that we cannot express elements as "one part from and one part from ". And so we need the basis

Therefore

By universal property of tensor products we can instead define the tensor product between two vector spaces as the dual vector space of bilinear forms on .

If , , then is defined as the map

for every bilinear form .

That is,

and

by

Observe that this satisfies the universal property of tensor product since if we are given some , for every , i.e. , we have , i.e. is a bilinear form on . Furthermore, this dual of bilinear forms on is then (since we are working with finite-dimensional vector spaces).

From universal property of tensor product, for any , there exists a unique such that

Letting the map

be defined by

where is defined by . By the uniqueness of , and linearity of the maps under consideration, this defines an isomorphism between the spaces.

Suppose

• is a basis for
• is a basis for

Then a bilinear form is fully defined by (upper-indices since these are the coefficients of the co-vectors). Therefore,

Since we are in finite dimensional vector spaces, the dual space then has dimension

as well.

Furthermore, form a basis for . Therefore we can write the elements in as

First observe that if , then is fully defined by how it maps the basis elements , and

Then observe that if , then

Hence, we have a natural homomorphism which simply takes , with coefficients to the corresponding element with the same coefficients! That is, the isomorphism is given by

### Tensor algebra

Now we'll consider letting . We define tensor powers as

with , , etc. We can think of as the dual vector space of k-multilinear forms on .

Combining all tensor powers using the direct sum we have define the tensor algebra

whose elements are finite sums

of tensor products of vectors .

The "multiplication" in is defined by extending linearly the basic product

The resulting algebra is associative, but not commutative.

### Two viewpoints

There are mainly due two useful ways of looking at tensors:

1. Using Lemma lemma:homomorphisms-isomorphic-to-1-1-tensor-product, we may view tensors as linear maps . In other words,

• Furthermore, we can view as the dual of , i.e. consider a tensor as a "multilinear machine" which takes in vectors and co-vectors, and spits out a real number!
2. By explicitly considering bases of and of , the tensors are fully defined by how they map each combination of the basis elements. Therefore we can view tensors as multi-dimensional arrays!

#### Tensors (mainly as multidimensional arrays)

Let be a vector space where is some field, is addition in the vector space and is scalar-multiplication.

Then a tensor is simply a linear map from some q-th Cartesian product of the dual space and some p-th Cartesian product of the vector space to the reals . In short:

where denotes the (p, q) tensor-space on the vector space , i.e. linear maps from Cartesian products of the vector space and it's dual space to a real number.

Tensors are geometric objects which describe linear relations between geometric vectors, scalars, and other tensors.

A tensor of type is an assignment of a multidimensional array

to each basis of an n-dimensional vector space such that, if we apply the change of basis

Then the multi-dimensional array obeys the transformation law

We say the order of a tensor is if we require an n-dimensional array to describe the relation the tensor defines between the vector spaces.

Tensors are classified according to the number of contra-variant and co-variant indices, using the notation , where

• is # of contra-variant indices
• is # of co-variant indices

Examples:

The tensor product takes two tensors, and , and produces a new tensor, , whose order is the sum of the orders of the original tensors.

When described as multilinear maps, the tensor product simply multiplies the two tensors, i.e.

which again produces a map that is linear in its arguments.

On the components, this corresponds to multply the components of the two input tensors pairwise, i.e.

where

• is of type
• is of type

Then the tensor product is of type .

Let be a basis of the vector space with , and be the basis of .

Then is a basis for the vector space of over , and so

Given , we define

Components:

A contraction is basis independent:

Generalized to by summing over combinations of the indices, alternating sign if wanting anti-symmetric. See the general definition of a wedge product for more for example.

##### Examples
• Linear maps - matrices

A linear map is represented as a matrix, and we say this is a tensor of order 2, since it requires 2-dimensional array to describe the relation.

## Clifford algebra

First we need the following definition:

Given a commutative ring and modules M\$ and , an quadratic function s.t.

• (cube relation): For any we have

• (homegenous of degree 2): For any and any , we have

A quadratic R-module is an module equipped with a quadratic form: an R-quadratic function on with values in .

The Clifford algebra of a quadratic R-module can be defined as the quotient of the tensor algebra by the ideal generated by the relations for all ; that is

where is the ideal generated by .

In the case we're working with a vector space (instead of a module), we have

Since the tensor algebra is naturally graded, the Clifford algebra is naturally graded.

### Examples

#### Exterior / Grassman algebra

• Consider a Clifford algebra generated by the quadratic form

i.e. identically zero.

• This apparently gives you the Exterior algebra over
• A bit confused as to why