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Spin representations, Modeling divisions

Clifford action

Andrius: The Clifford action is the key to the nature of spin representations. So I am trying to understand that. I need to describe it properly.

We want to describe the spin representations of the spin group {$\textrm{Spin}(n)$}. We can define the spin group {$\textrm{Spin}(n)$} as a subgroup of the complex Clifford algebra {$Cl_n$}. In particular, we can define the spin group's Lie algebra as the bivectors of {$Cl_n$}. Furthermore, we can specify an action of the Clifford algebra that will allow us to identify the Clifford algebra with endomorphisms of a vector space, which is to say, with a matrix algebra.

Given the complex vector space {$V\cong\mathbb{C}^n$} we may have {$n=2m$} or {$n=2m+1$}. We specify an orthonormal basis and pair up its elements with possibly one element left over, yielding {$\{x_1, y_1,\cdots,x_m,y_m,u_1\}$}. We define the quadratic form {$Q(\sum_j q_jx_j+r_jy_j+s_1u_1)=\sum_j q_j^2+r_j^2$}. We write {$a_j=\frac{1}{2}(x_j+iy_j), \bar{a}_j=\alpha_j=\frac{1}{2}(x_j-iy_j)$} and note that {$Q(a_j)=\frac{1}{4}Q(x_j)-\frac{1}{4}Q(y_j)=\frac{1}{4}-\frac{1}{4}=0,Q(\bar{a}_j)=\frac{1}{4}Q(x_j)-\frac{1}{4}Q(y_j)=\frac{1}{4}-\frac{1}{4}=0$}, which is to say, they are isotropic. We define subspaces {$W=\langle a_1,\cdots, a_m \rangle$} and {$W^*=\langle \bar{a}_1,\cdots, \bar{a}_m \rangle$} which are both totally isotropic with respect to {$Q$}. We have {$V=W\oplus W^*\oplus U$} where {$U$} is zero-dimensional {$\langle 0 \rangle$} or one-dimensional {$\langle u_1 \rangle$} and, in either case, {$U$} is totally isotropic. Note that the notation {$a_j$} and {$\bar{a}_j$} conceives of conjugation as with regard to the explicit {$i$} and not with regard to any {$i$} implicit in the choice of {$x_j$} or {$y_j$}.

We define the Clifford algebra {$Cl_n(V,Q)$}. We describe the Clifford action of {$Cl_n(V,Q)$} on the exterior algebras {$S=\wedge^\bullet W$} and {$S'=\wedge^\bullet W^*$}.

Pierre Deligne describes the Clifford action on {$S'$} in Proposition 2.2 in his Notes on Spinors.

Clifford action for {$n=2m$}

If {$n=2m$}, then {$W$} and {$W^*$} are dual vector spaces. (Deligne writes {$W^\wedge$} for {$W^*$}.) The quadratic form on {$v=w+w*$} defines the inner product {$Q(w+w^*)=\langle w, w^*\rangle = w^*(w)$}. In particular,

• {$Q(a_j)=\langle a_j, 0\rangle = 0$}
• {$Q(\bar{a_j})=\langle 0, \bar{a}_j\rangle =0$}
• {$Q(x)=Q(a_j+\bar{a}_j)=\langle a_j,\bar{a}_j\rangle = \bar{a}_j(a_j) = 1$}
• {$Q(y)=Q(-ia_j + i\bar{a}_j) = \langle -ia_j, i\bar{a}_j \rangle = i\bar{a}_j(-ia_j)=i(-i)\bar{a}_j(a_j)=1$}

Note that the brackets indicate the nondegenerate bilinear mapping, called the natural pairing of a vector space with its dual space, and not the complex inner product.

Clifford action on {$S'$}

One defines the super algebra isomorphism {$\Phi:Cl_n(W\oplus W^*)\rightarrow \textrm{End}(S')$} as follows.

Map {$\bar{a}_j$} to the endomorphism {$\bar{a}_j\wedge: s\rightarrow \bar{a}_j\wedge s$}. Note that if {$s$} is a monomial that contains {$\bar{a}_j$}, then {$\bar{a}_j\wedge s = 0$}. If {$s$} is a monomial that does not contain {$\bar{a}_j$}, then we add it, but we will need to change the sign if we use an odd number of swaps to position it.

Map {$a_j$} to the interior product {$\iota_{a_j}(s): s\rightarrow \langle a_j, s\rangle$}. Note that if {$s$} is a monomial that does not contain {$\bar{a}_j$}, then {$\iota_{a_j}(s)=0$}. If {$s$} is a monomial that contains {$\bar{a}_j$}, then we remove it upon positioning it at the front of the monomial, which will change the sign if that involves an odd number of swaps.

We decompose {$W=\oplus_j W_j$} where {$W_j=\langle a_j \rangle$}, {$W^*=\oplus_jW^*_j$} where {$W^*_j=\langle \bar{a}_j\rangle$} and {$V=\oplus_jV_j$} where {$V_j=W_j\oplus W^*_j$}.

The superalgebra {$Cl(V)=\otimes_j Cl(V_j)$} is a tensor product of superalgebras. The graded vector space {$S'=\wedge^\bullet W^*=\otimes_j S'_j$} is the tensor product of graded vector spaces {$S'_j=\wedge^\bullet W^*_j$}. The isomorphism {$\Phi$} is a tensor product of the morphisms {$\Phi_j:C(V_j)\rightarrow \textrm{End}(S'_j)$}.

Clifford action on {$S$}

We want to define the super algebra isomorphism {$\Phi:Cl_n(W\oplus W^*)\rightarrow \textrm{End}(S)$}. I think we can do this as follows.

Map {$a_j$} to the endomorphism {$a_j\wedge: s\rightarrow a_j\wedge s$}. Note that if {$s$} is a monomial that contains {$a_j$}, then {$a_j\wedge s = 0$}. If {$s$} is a monomial that does not contain {$a_j$}, then we add it, but we will need to change the sign if we use an odd number of swaps to position it.

Map {$\bar{a}_j$} to the interior product {$\iota_{\bar{a}_j}(s): s\rightarrow \langle s, \bar{a}_j \rangle$}. Note that if {$s$} is a monomial that does not contain {$a_j$}, then {$\iota_{\bar{a}_j}(s)=0$}. If {$s$} is a monomial that contains {$a_j$}, then we remove it upon positioning it at the front of the monomial, which will change the sign if that involves an odd number of swaps.

Then we define and tensor morphisms as above.