The first object we'll need is the

Examples

*Clifford algebra*. Fix a field k (in this series it will always be either the set of real numbers R or the set of complex numbers C).Algebras

I'll start from a quick reminder of what an algebra is. Suppose A is a vector space over k. Suppose further that a mapping m: A x A -> A is given. For convenience sake, given a, b in A we denote m(a, b) by ab and call m

*multiplication*. A is called a*unital associative**k-algebra*(just*k-algebra*in the sequel) when the following conditions hold:- m is
*bilinear*(i.e. linear in each of the arguments separately). In details, it means that

Given a, b, c in A: (a + b)c = ac + bc*additivity in the 1st argument / left distributivity*

Given x in k and a, b in A: (xa)b = x(ab)*homogenuity in the 1st argument*

Given a, b, c in A: c(a + b) = ca + cb*additivity in the 2nd argument / right distributivity*

Given x in k and a, b in A: a(xb) = x(ab)*homogenuity*in the 2nd argument - Given a, b, c in A: (ab)c = a(bc)
*associativity* - There exists an element "1" in A such that for any a in A: a1 = 1a = a
*unit*

An algebra A is called commutative when given a, b in A we have ab = ba.

**Examples**

- Consider V a k-vector space. Consider End(V) the set of all endomorphisms of V, i.e., linear operators V -> V. End(V) is a k-algebra where multiplication corresponds to composition of linear operators. For dim V <>
- Consider V a k-vector space, W a subspace. Consider End(V, W) the set of endomorphisms of V leaving W invariant (non-standard notation). That is, given a in End(V, W), w in W we have aw also in W. End(V, W) is an algebra. It is a subalgebra of End(V), that is, a linear subspace closed under multiplication. For dim V <>
- Fix n a natural number. The set of n x n matrices with coefficients in k forms an algebra: Mat(n, k). It is isomorphic to End(V) for dim V = n. Obviously, dim Mat(n) = n^2.
- Fix n a natural number. The set of upper-triangular n x n matrices with coefficients in k forms an algebra: UT(n, k) (non-standard notation). We have dim UT(n, k) = n (n + 1) / 2.
- Consider k[x] the set of polynomials with coefficients in k in the variable x. k[x] is a

k-algebra. It is infinite-dimensional. It is commutative. - Fix n a natural number. We have k^n the set of column vectors of size n with coefficients in k. We can define multiplication in k^n by multiplying each vector entry separately. It makes k^n into an algebra. Obviously dim k^n = n. k^n is commutative.

Ideals

*A subset I of A is called a right ideal*when the following conditions hold:

- I is a linear subspace of A
- Given a in A, b in I, ba is also in I

s1 a1 + s2 a2 + ... + sn an

where:

s1, s2 ... sn are elements of S

s1, s2 ... sn are elements of S

a1, a2 ... an are elements of A

**Claim:**SA is a right ideal.

SA is called

*the right ideal of A generated by S.*

*A subset I of A is called a left ideal*when the following conditions hold:

- I is a linear subspace of A
- Given a in A, b in I, ab is also in I

Consider S an arbitrary subset of A. Denote AS to be the collection of all elements of A of the form

a1 s1 + a2 s2 + ... + an sn

where:

s1, s2 ... sn are elements of S

s1, s2 ... sn are elements of S

a1, a2 ... an are elements of A

**Claim:**AS is a left ideal.

AS is called

*the left ideal of A generated by S.*

A subset I of A is called a

*two-sided ideal*when it is simultaneously a left ideal and a right ideal.Consider S an arbitrary subset of A. Denote ASA to be the collection of all elements of A of the form

a1 s1 b1 + a2 s2 b1 + ... + an sn bn

where:

s1, s2 ... sn are elements of S

s1, s2 ... sn are elements of S

a1, a2 ... an are elements of A

b1, b2 ... bn are elements of A

**Claim:**ASA is a two-sided ideal.

ASA is called

*the two-sided ideal of A generated by S.*

**Claim:**Suppose A is a commutative algebra. Then a subset of A is a left ideal if and only if it is a right ideal if and only if it is a two-sided ideal.

Thus for a commutative algebra all three notions coincide hence we speak simply of

*ideals*.

**Examples**

- Consider V a k-vector space. Consider the algebra End(V). Consider W a subspace of V. Define I = {a in End(V) | Im a lies in W}. I is a right ideal of End(V). Define

J = {a in End(V) | Ker a contains W}. J is a left ideal in End(V). - Fix n a natural number. Consider the algebra Mat(n, k). Fix m <= n another natural number. Define I = {a in Mat(n, k) | the first m rows are zero}. I is a right ideal of Mat(n, k). Define J = {a in Mat(n, k) | the first m columns are zero}. J is a left ideal of Mat(n, k).
- Fix n a natural number. Consider the algebra UT(n, k). Fix m <= n another natural number. Define J = {a in UT(n, k) | the first m columns are zero}. I is a two-sided ideal of UT(n, k).
- Consider the algebra k[x]. Consider S a finite subset of k[x].

Define I = {p in k[x] | for any a in S: p(a) = 0}. I is an ideal of k[x]. It is generated by the polynomials {x - a} where a traverses elements of S. Now fix n a natural number. Define

J = {p in k[x] | for any m natural with m <= n: p^(m)(a) = 0}. J is an ideal of k[x]. It is generated by the single polynomial x^(m + 1). - Fix n a natural number. Consider the algebra k^n. Consider m <= n another natural number. Define I = {v in k^n | the first m entries of v are zero}. I is an ideal.

Quotient Algebra

Consider A an algebra and I a two-sided ideal. Then we may take the vector space quotient A/I. That is, we consider the set of equivalence classes of A under the following equivalence relation: Given a, b in A they are equivalent when a - b is in I.

It is easy to see the operation of multplication in A defines an operation of multiplication in A/I as well, that is, makes A/I into an algebra on its own right. For this to work, it is crucial that I is a two-sided ideal. A/I is called

*the quotient algebra of A by I.***Examples**

- Fix n a natural number. Consider the algebra UT(n, k). Fix m <= n another natural number. Define J = {a in UT(n, k) | the first m columns are zero}. Then UT(n, k) / J is naturally isomorphic to UT(m, k).
- Consider the algebra k[x]. Consider S a finite subset of k[x].

Define I = {p in k[x] | for any a in S: p(a) = 0}. Then k[x] / I is naturally isomorphic to k^n where n is the number of elements of S. - Fix n a natural number. Consider the algebra k^n. Consider m <= n another natural number. Define I = {v in k^n | the first m entries of v are zero}. Then k^n / I is naturally isomorphic to k^m.

Generators and Relations

One of the simplest ways to construct an algebra is using generators and relations. This is done as follows. Suppose G is an abritrary set (possibly infinite). Consider F = k the algebra of

*non-commutative polynomials*with coefficients in k and variables G. For G non-empty this algebra is infinite-dimensional. It is also called the

*free algebra*

*over G*.

Now take R an arbitrary subset of F. We have I = FRF a two-sided ideal. We obtain the algebra A = F / I. A is called

*the algebra generated by G with relations R*. In this context, elements of G are called*generators*and elements of R*relations.*It is often convenient to define relation using equations. For example, suppose f, g, h are elements of G and x, y, z are elements of k. Then the relationxf^2 = ygh + zhg

means that the element xf^2 - ygh - zhg of F is in R.

*Quadratic Spaces*

Consider V a vector space over k. V is called a

*quadratic space*when it is equipped with a*symmetric bilinear non-degenerate form*Q. A quick reminder of what that means:- Q is mapping V x V -> k
- Q is
*bilinear (i.e. linear in each of the arguments separately):*

Given u, v, w in A: Q(u + v, w) = Q(u, w) + Q(v, w)*additivity in the 1st argument*

Given x in k and u, v in A: Q(xu, v) = xQ(u, v)*homogenuity in the 1st argument*

Given u, v, w in A: Q(w, u + v) = Q(w, u) + Q(w, v)*additivity in the 2nd argument*

Given x in k and u, v in A: Q(u, xv) = xQ(u, v)*homogenuity in the 2nd argument* - Q is
*symmetric*, that is, given u, v in V: Q(u, v) = Q(v, u) - Q is
*non-degenerate*: Suppose u in V is such that for any v in V we have Q(u, v) = 0. Then u = 0.

*isomorphic*when there exists a linear mapping i: V -> W such that

- i is
*injective*: Given u, v in V, i(u) = i(v) implies u = v. Equivalently, Given u in V, i(u) = 0 implies u = 0. - i is
*surjective*: Given w in W, there exists v in V such that i(v) = w. - i preserves the quadratic stucture, that is, given u, v in V: Q(u, v) = R(i(u), i(v))

*isomorphism*between V and W. Two isomorphic quadratic spaces are "essentially the same".

In the sequel, we'll only care about finite-dimensional quadratic spaces.

**Proposition:**

- Suppose V, W are quadratic spaces over k = C. Then V is isomorphic to W if and only if

dim V = dim W. - Suppose V is a quadratic space over k = C of dimension n. Then there exists a basis

e1, e2 ... en of V such that:

Q(ei, ei) = 1

Q(ei, ej) = 0 for i =/= j

**Proposition:**

- Suppose V is a quadratic space over k = R of dimension n. Then there exists a natural number s and a basis of V e1, e2 ... en such that:

For i <= s: Q(ei, ei) = 1 For i > s: Q(ei, ei) = -1

Q(ei, ej) = 0 for i =/= j

We call s the "s-number" of V and denote it sn V (this is not standard terminology). - (Trivial) Suppose V, W are quadratic spaces over k = R. Then V is isomorphic to W if and only if dim V = dim W, sn V = sn W.

Fix V a vector space. The tensor algebra T(V) is the algebra generated by V with the following relations:

- Given u, v, w in V with u + v = w, we take u + v = w to be a relation.
- Given u, v in V, x in k with xu = v we take xu = v to be a relation.

Suppose V is a quadratic space. The Clifford algebra C(V) is the algebra generated by V with the following relations:

- The relations we used for T(V).
- Given u, v in V: uv + vu = -2Q(u, v)

**Claim:**dim C(V) = 2^dim V

We use k = R in these examples.

- Suppose dim V = 0. Then C(V) is isomorphic to R.
- Suppose dim V = 1, sn V = 1. Then C(V) is isomorphic to C.
- Suppose dim V = 2, sn V = 2. Then C(V) is isomorphic to the
*quaternion algebra*H. - For dim V > 2, C(V) is no longer a
*division algebra*, that is, it doesn't have an*inverse*for each non-zero element.