# Algebra

## Table of Contents

## Notation

- denotes a group
**CANI**stands for:**C**ommutative**A**ssociative**N**eutral element (e.g. 0 for addition)**I**nverse

## 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.

### 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**:

**A**ssociativity:**D**istributivity over field-addition:**D**istributivity over field-multiplication:**U**int:

### Ring

### 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

- be a group homomorphism

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

In particular, if is *surjective*, then

Le be a group, and .

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

since . So so that ,

And now we check that the

Let , . Wan to show that .

- Let , . Then since , so , for all .
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 .

- The stabilizer of each left coset is the conjugate
- 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 .

### 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 .

### 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

### Vector Spaces

#### Notation

- 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 :

- is a
**basis**, i.e. linearly independent generating set - is
**minimal among all generating sets**, i.e. does not generate for any - 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:

- is an abelian group
- is a monoid
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.

#### 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:

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

- I.e. we stay in even when multiplied by elements from

**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 subset of a ring . Then is a *subring* if and only if

- has a multiplicative identity
- is closed under subtraction:
- 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.

- If then or
- and then

#### Factor Rings

Let be a ring and and ideal of .

The mapping

Has the following properties:

- is
*surjective* 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.

#### 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 .

Equipped with addition and s-multiplication

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 .

The mapping defined by

is

*surjective*.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 .

Let be a *ring*.

The **determinant** is a mapping given by

Let be a *commutative* ring, then

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.

#### 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

- Symmetric
- Non-degenerate
- 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

#### Theorems

Let be a vectors in an inner product space. Then

with equailty if and only if and are linearly dependent.

#### Adjoints and Self-adjoints

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:

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 :- Focus attention on each of the to obtain the nilpotent Jordan blocks.
- 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 .

- Each is
*stable*under , i.e For each , such that

In other words, there is a

**direct sum decomposition**Then

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

*block diagonal matrix*with .

Let is that

Then

Hence , i.e. is

*stable*under .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.

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 thatIn fact, this decomposition is

*unique*and is called the**Jordan decomposition**. - Each is
- 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 subspacesWe can now develop an algorithm for constructing a basis!

- Constructing a basis

- Choose arbitrary basis for
- Choose basis of of by mapping using and choosing vectors linearly independent of .
- Repeat!

Or more accurately:

Choose arbitrary basis for :

Since is

*injective*, by the fact that the image of a set of linear independent vectors is a linearly independent set if the map is*injective*, thenis linearly independent.

Choose vectors

such that

is a

*basis*of .- 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.

- Constructing a basis
- 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

Calculate the eigenvalues, of

For each and , let

Compute and

Let

Set

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

- Let be the largest integer such that .
- If does not exist, stop. Otherwise, goto step 3.
- Let and .
- Let

- Change to for .

- 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:

- if and , there is a "product"
This product is

*bilinear*:for , and .

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:

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!

- Furthermore, we can view as the
- 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:

**Scalar**:**Vector**:**Matrix**:- 1-form:
- Symmetric bilinear form:

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.

## 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**