# M721: Index Theory

## Clifford algebras and Spin groups

Still lot of editing needed. the pdf file can be obtained here

As a provisional definition, clifford algebraover a field ${k}$ can be defined as

$\displaystyle Cl_{n}(k) = k[x_{1}, \ldots, x_{n}] /R$

where,

$\displaystyle R = < x_{i}^{2} = -1, x_{i} x_{j} = - x_{j}x_{i} \forall i \neq j >$

Easy to see that ${Clifford}$ ${Algebra}$ has dimension ${2^{n}}$. ${Cl_{n}(k)}$ can also be thought of as

$\displaystyle Cl_{n}(k) = TV / R$

where ${V = k^{n}}$ as vector space, ${TV}$ is the tensor algebra, ${R}$ is same as above, except that ${x_{i}}$‘s are standard basis for ${V}$. This observation leads to a more general definition of clifford algebra, where ${V}$ is a vector space equipped with a symmetric bilinear form

Definition 1 Let ${V}$ be a vector space with symmetric biliear form ${\beta}$ and quadratic form ${q = \beta(x,x)}$. Then the clifford algebraover ${V}$ can be defined as$\displaystyle Cl(V, q) = TV / < x \otimes x + q(x)>$

Remark 1 These are some of the properties that ${CL(V, q)}$ enjoys

1. There is a natural inclusion of ${i: V \hookrightarrow Cl(V,q)}$.
2. If ${q(x) = 0 \forall x \in V}$ then ${Cl(V, q) = \bigwedge(V)}$ the exterior algebra
3. Let ${.}$ denote the clifford multiplication( induced by the tensor product of ${TV}$), then$\displaystyle x . y + y . x = -2 \beta(x,y)$
4. Universal Property : Let ${f: V \longrightarrow A}$, where ${A}$ is a ${k}$-algebra, such that ${f(x)^{2} = -q(x)}$, then there exists an unique map ${F}$ such that ${f = F \circ i}$, that is the following diagram commutes$\displaystyle \xymatrix@C-1pc{ V\ar[d]^{i} \ar[r]^{f}&A\\ Cl(V,q)\ar@{.>}[ur]_{F} }$
5. A map ${\phi: ( V,q) \longrightarrow (V',q')}$, such that ${q'(\phi (x)) = q(x)}$, extends to a ${k}$-algebra homomorphism$\displaystyle \Phi: Cl(V,q) \longrightarrow Cl(V', q')$Thus the orthogonal group ${\mathbb{O}(n)}$ has an action on ${Cl_{n}(\mathbb{R})}$

Proposition 2 ${Cl(v,q)}$ is a filtered algebra whose associate graded is ${\bigwedge(V)}$

Before proving the theorem, recall the following definition

Definition 3 If ${A}$ is a ${k}$-algebra then a filtration ${(A,\mathfrak{F} )}$ of ${A}$ is sequence of subspaces$\displaystyle F_{0}A \subset F_{1}A \subset \dots \subset A = \bigcup_{r}F_{r}A$

The associate gradedof ${(A, \mathfrak{F})}$ is defined as$\displaystyle Gr(A, \mathfrak{F}) = \bigoplus_{r} F_{r}A/F_{r-1}A$

Proof: Let ${\pi}$ be the quotient map

$\displaystyle \pi : TV \longrightarrow Cl(V,q)$

Define, ${G_{r} = V \otimes \ldots \otimes V}$ (${r}$ fold tensor product). Define filtration on ${TV}$ by setting

$\displaystyle F_{r}TV = \bigoplus_{i \leq r} G_{i}$

Define filtration ${\pi_{*}F_{r} TV}$ be the filtration on the clifford algebra. Note ${x \otimes x = q(x) \in \pi_{*}F_{0}Cl(V, q)}$. Hence, in the associated graded ${x \otimes x = 0}$. On the other hand the relation ${ x_{i}x_{j} = -x_{j} x_{i} }$ prevails in the associated graded. Hence the associate graded is isomorphic to ${\bigwedge V}$. $\Box$

Remark 2 ${Cl(V,q)}$ is a ${ \mathbb{Z}/2}$-graded algebra.$\displaystyle Cl^{0}(V,q) = \pi_{*} TV^{even}$

$\displaystyle Cl^{1}(V,q) = \pi_{*} TV^{odd}$

Definition 4 Recall,${S^{n-1} \subset V \hookrightarrow Cl(V,q)}$. Define,$\displaystyle Pin(n) = S^{n-1} \subset Cl^{\times}(V,q)(units)$

and$\displaystyle Spin(n) = S^{n-1} \cap Cl^{0}(V,q)$

On ${Cl(V,q)}$ we have an involution map, which is induced by the involution on ${TV}$ given by,

$\displaystyle \overline{x_{1} \otimes \ldots \otimes x_{r}} = (-1)^{r} x_{1} \otimes \ldots \otimes x_{r}$

If ${x \in S^{n-1} \subset Cl(V.q)}$ then

$\displaystyle x .\overline{x} = - \beta (x,x )= q(x) = 1$

Let ${v \in V \subset Cl(V,q)}$ and ${x \in Pin(n)}$, then observe

$\displaystyle q(-x.v.\overline{x}) = \beta(-x.v .\overline{x},-x.v.\overline{x}) = x.(-q(v)).x =- q(v).(-q(x)) = q(v)$

Lemma 5 There exist short exact sequences$\displaystyle 1 \longrightarrow \lbrace-1, +1 \rbrace \longrightarrow Pin(n) \xrightarrow{p}\mathbb{O}(n) \longrightarrow 1$

and$\displaystyle 1 \longrightarrow \lbrace-1, +1 \rbrace \longrightarrow Spin(n) \xrightarrow{p} \mathbb{SO}(n) \longrightarrow 1$

where ${p}$ is the map which sends$\displaystyle x \mapsto ( v\mapsto-x.v.\overline{x)}$

Let ${k}$ be a field. Recall, tensor product of ${k}$-algebras ${A}$ and ${B}$ is a ${k}$-algebra, denoted by ${A\otimes_{k}B }$ and multiplication is given by

$\displaystyle (a \otimes b).(a' \otimes b')= aa' \otimes bb'$

moreover if ${M_{j}(A)}$ denotes the set of all ${j \times j}$ matrices. Then we have the following isomorphism

$\displaystyle M_{j}(A) \otimes M_{k}(B) \cong M_{jk}(A \otimes B)$

Define

$\displaystyle Cl_{p,q}(k) = Cl(k^{p+q},x_{1}^{2}+ \ldots+ x_{p}^{2} - x_{p+1}^{2} - \dots x_{p+q}^{2} )$

Remark 3 If ${k = \mathbb{C}}$, then$\displaystyle Cl_{p,q}(\mathbb{C}) \cong Cl_{p+q,0}(\mathbb{C})$

This follows from the fact that the quadratic forms ${q_{p,q}(x)=x_{1}^{2}+ \ldots+ x_{p}^{2} - x_{p+1}^{2} - \dots x_{p+q}^{2}}$ and ${q_{p+q,0}(x)=x_{1}^{2}+ \dots+ x_{p+q}^{2}}$ induces isomorphic innerproduct structure on ${\mathbb{C}^{p+q}}$ where the isomorphism sends$\displaystyle e_{t} \mapsto e_{t}: 0 \leq t \leq p$

$\displaystyle e_{t} \mapsto ie_{t}: p < t \leq p+q$

Theorem 6 If ${k= \mathbb{C}}$, then we have the following isomorphisms

1. ${Cl_{0}(\mathbb{C}) = \mathbb{C}}$
2. ${Cl_{1}(\mathbb{C}) = \mathbb{C} \times \mathbb{C}}$
3. ${Cl_{2}(\mathbb{C}) = M_{2}(\mathbb{C})}$
4. ${Cl_{n+2}(\mathbb{C}) = M_{2}(Cl_{n}(\mathbb{C}))}$

Corollary 7 (Bott Periodicity) As a consequence of (iv) we have$\displaystyle Cl_{n}(\mathbb{C}) = M_{2^{n}}(\mathbb{C})$

if ${n}$ even, and$\displaystyle Cl_{n}(\mathbb{C}) = M_{2^{n}}(\mathbb{C}) \times M_{2^{n}}(\mathbb{C})$

if ${n}$ is odd.

To work out the case when the underlying field is ${\mathbb{R}}$. For any field ${k}$ we have the following isomorphisms.

Lemma 8 For any field ${k}$

1. ${Cl_{n+2,0}(k) \cong Cl_{0,n}(k) \otimes Cl_{2,0}(k) .}$
2. ${ Cl_{o,n+2}(k) \cong Cl_{n,0}(k) \otimes Cl_{0,2}(k).}$

Proof: Let ${e_{i}}$ denote the standard basis of ${k^{n+2}}$ and cannonical generatoring set of the Clifford algebra ${ Cl_{n+2,0}(k)}$.

1. To get the first isomorphism we simply produce a map given by sending$\displaystyle e_{i} \mapsto e_{i} \otimes e_{i}e_{2} \forall 1\leq i \leq n$and$\displaystyle e_{n+1} \mapsto 1 \otimes e_{1}$$\displaystyle e_{n+2} \mapsto 1 \otimes e_{2}$It is easy to check that the above map is an isomorphism.
2. is similar to ${(i)}$.

$\Box$ One can explicitly check some of the lower dimension cases( ${ n = 0, 1, 2}$). Then one can repeatedly use the isomorphisms in previous lemma. One has to work upto dimension ${8}$ when ${k= \mathbb{R}}$, before one sees the patern, which is called the Bott periodicity. TSome of the calculations are as follows calculations are as follows

1. ${Cl_{0,0}(\mathbb{R}) \cong \mathbb{R}}$
2. ${Cl_{1,0}(\mathbb{R})\cong \mathbb{C}}$
3. ${Cl_{0,1}(\mathbb{R}) \cong \mathbb{R} \times \mathbb{R}}$
4. ${Cl_{2,0}(\mathbb{R}) \cong \mathbb{H}}$
5. ${Cl_{0,2}(\mathbb{R}) \cong M_{2}(\mathbb{R})}$
6. ${Cl_{3,0}(\mathbb{R}) \cong Cl_{0,1}(\mathbb{R}) \otimes Cl_{2,0}(\mathbb{R})\cong (\mathbb{R} \times \mathbb{R})\otimes \mathbb{H}\cong \mathbb{H}\times \mathbb{H} }$
7. ${Cl_{4,0}(\mathbb{R}) \cong Cl_{0,2}(\mathbb{R}) \otimes Cl_{2,0}(\mathbb{R})\cong M_{2}(\mathbb{R})\otimes \mathbb{H}\cong M_{2}(\mathbb{H}) }$
8. In general one gets,
${Cl_{n+4,0}(\mathbb{R}) \cong Cl_{0,n+2}(\mathbb{R}) \otimes Cl_{2,0} \cong Cl_{n,0}(\mathbb{R}) \otimes Cl_{0,2}(\mathbb{R}) \otimes Cl_{2,0}(\mathbb{R})}$

Putting all these observations together we get

Theorem 9 The Bott periodicity in case of real number looks like$\displaystyle Cl_{8k+r}(\mathbb{R}) = \left\lbrace \begin{array}{cccccc} M_{2^{4k}}(\mathbb{R})& r=0\\ M_{2^{4k}}(\mathbb{C})& r=1\\ M_{2^{4k}}(\mathbb{H})& r=2\\ M_{2^{4k}}(\mathbb{H}) \times M_{2^{4k}}(\mathbb{H})& r=3\\ M_{2^{4k+1}}(\mathbb{H})& r=4\\ M_{2^{4k+2}}(\mathbb{C})& r=5\\ M_{2^{4k+3}}(\mathbb{R})& r=6\\ M_{2^{4k+3}}(\mathbb{R}) \times M_{2^{4k+3}}(\mathbb{R})& r=7\\ \end{array} \right.$

Lemma 10 As ${k}$-algebras ${Cl_{n}^{0}(k) \cong Cl_{n-1}(k)}$ Proof:The isomorphism is given explicitly by the map induced by sending

$\displaystyle e_{i} \mapsto e_{i}.e_{n}$

Easy to check that this is an isomorphism of algebras. $\Box$