Epilogue: counting the spin structures

A spin structure on $(M, \eps)$ is an equivalence class of pairs $(\sS,C)$, but ¿what can be said about the equivalence relation?

First, $\sS$ has a class $[\sS] \in \Mrt(B, A)$: these are classified by $\rH^2(M;\bZ)$. If $(\sS_1, C_1)$ is another spin structure, then $C_1\: \sS_1^\3 \to \sS_1$ comes from a $B$-$A$-bimodule isomorphism $T_1\: \sS_1 \to \sS_1$. But now $\sS_1 \isom \sS \ox \sL$ for some $\sL$, where $[\sL] \in \rH^2(M;\bZ)$ is well defined. Thus we get

and therefore $(\sS_1, C_1) \sim (\sS, C)$ if this diagram commutes. Now
$$\begin{align*} \sS_1 &\isom \sS_1^\3 \ox_A \Hom_B(\sS_1^\3, \sS_1) \isom \sS_1... ...\Hom_B(\sS^\3, \sS) \ox_A \sL \isom \sS_1^\3 \ox_A \sL \ox_A \sL, \end{align*}$$
since $\Hom_B(\sS^\3, \sS)$ is trivial: the existence of $T$ shows that $[\sS^\3] = [\sS]$ in $\Mrt(B,A)$. The conclusion is that $\sS_1 \isom \sS_1^\3 \ox_A \sL \ox_A \sL$. Thus $\sS_1$ is also selfdual if and only if $\sL \ox_A \sL$ is trivial: $(\x2)_*[\sL] = 0$ in $\rH^2(M;\bZ)$. But, using the long exact sequence (), we find that $\ker(\x2)_* = \im\{\del\: \rH^1(M;\bZ_2) \to \rH^2(M;\bZ)\}$.

Conclusion: Those $[\sS \ox_A \sL] \in \Mrt(B,A)$ for which $\sL \ox_A \sL$ is trivial, but $\sL$ is not, i.e., the distinct spin structures on $(M, \eps)$, are classified by $\rH^1(M,\bZ_2)$.