Classification of spinor modules

In this section, $A = C(M)$ and as before, $B = \Ga(M,\bCl(T^*M))$ or $B = \Ga(M,\bCl^0(T^*M))$, according as the dimension of $M$ is even or odd.

Consider now the set $\Mrt(B,A)$ of isomorphism classes of $B$-$A$-bimodules: we have seen that $\delta(B) = 0$ if and only if $\Mrt(B,A)$ is nonempty. We shall assume from now on that indeed $\delta(B) = 0$, so that there exists at least one $B$-$A$-bimodule $\sS = \Ga(M,S)$ --continuous sections, for the moment-- such that at each $x \in M$, $S_x$ is an irreducible representation of the simple algebra $B_x$. Therefore, any such $\sS$ has a partner $\sS^\3 = \Hom_A(\sS, A)$ such that $\sS \ox_A \sS^\3 \isom B$ and $\sS^\3 \ox_B \sS \isom A$: in other words, $\sS$ is an equivalence $B$-$A$-bimodule, and its isomorphism class $[\sS]$ is an element of $\Mrt(B,A)$.

Since $\sS^\3 \isom \Ga(M,S^*)$ where $S^* \to M$ is the dual vector bundle to $S \to M$, we can write this equivalence fibrewise: $\sS_x \ox_\bC \sS_x^* = \End_\bC(\sS_x) \isom B_x$ and then $\sS_x^* \ox_{B_x} \sS_x \isom \bC$, for $x \in M$.

To proceed, we explain how $B$ acts on $\sS^\3 = \Hom_A(\sS, A)$. The spinor module $\sS$ carries an $A$-valued hermitian pairing () given by the local scalar products defined in the construction of $\sS$, that may be written

$$\pairing{\psi}{\phi} : x \mapsto \<\psi_x\vert\phi_x>, \word{for} x \in M.$$ (8)

We can identify elements of $\sS^\3$ with ``bra-vectors'' $\bra{\psi}$ using this pairing, namely, we define $\bra{\psi}$ to be the map $\phi \mapsto \pairing{\psi}{\phi} \in A$. Since $A$ is unital, there is a ``Riesz theorem'' for $A$-modules showing that all elements of $\sS^\3$ are of this form. Now the left $B$-action is defined by

$$b \bra{\psi} := \bra{\psi} \circ \chi(b^!).$$

Recall that $b \mapsto \chi(b!)$ is a linear antiautomorphism of $B$.

Thus, the ``mod 2 reduction'' $j_*[\sL_{\sS}] \in \rH^2(M;\bZ_2)$, coming from the short exact sequence of abelian groups $0 \to \bZ \xrightarrow{\x2} \bZ \xrightarrow{j} \bZ_2 \to 0$, is independent of $[\sS]$. Indeed, it defines an invariant $\ka[B] \in \rH^2(M;\bZ)$. This is clear, when one takes into account the corresponding long exact sequence in Cech cohomology and the governing assumption that $\delta(B) = 0$:

$$\cdots \to \rH^1(M;\bZ_2) \xrightarrow{\del} \rH^2(M;\bZ) ... ... \rH^2(M;\bZ_2) \xrightarrow{\del} \rH^3(M;\bZ) \to \cdots$$ (9)

¿What is the meaning of the condition $\ka(B) = 0$? It means that, by replacing any original choice of $\sS$ by a suitably twisted $\sS \ox_A\sL$, we can arrange that $\sL_{\sS}$ is trivial, i.e. $\sL_{\sS} \isom A$, or better yet, that

$$\sS^\3 \isom \sS \quad\text{as $B$-$A$-bimodules}.$$

We now reformulate this condition in terms of a certain antilinear operator $C$; later on, in the context of spectral triples, we shall rename it to $J$.

The antilinear operator $C\: \sS \to \sS$, which becomes an antiunitary operator on a suitable Hilbert-space completion of $\sS$, is called the charge conjugation. It exists if and only if $\ka(B) = 0$.

¿What, then, are and spin structures on $M$? We choose on $M$ a metric (without losing generality), and also an orientation $\eps$, which organizes the action of $B$, in that a change $\eps \mapsto -\eps$ induces $c(\ga) \mapsto -c(\ga)$, which either

(i)

reverses the $\bZ_2$-grading of $\sS = \sS^+ \oplus \sS^-$, in the even case; or

(ii)

changes the action on $\sS$ of each $c(\al)$ to $-c(\al)$, for $\al \in \sA^1(M)$, in the odd case --recall that $c(\al) := c(\al\ga)$ in the odd case.

In the long cohomology exact sequence there is a boundary homomorphism

$$\rH^2(M;\bZ_2) \xrightarrow{\del} \rH^3(M;\bZ).$$

By examining the definitions of the various Cech cocyles that we have obtained so far, one can show that $\delta(B) = \del(\ka(B))$.