M721: Index Theory

Ruminations of a Graduate Class

Clifford algebras and Spin groups

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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


\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)


\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 nand\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


Written by prasit0605

October 27, 2012 at 6:41 pm

Posted in Uncategorized

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