# Number of spin structures

I am probably missing something obvious, but still...

Consider an oriented Riemannian $n$-dimensional vector bundle $\pi: E\rightarrow X$ over compact manifold $X$ with $\omega_2(E)=0$ so it has spin structures. Let $P_{SO}(E)$ denote the associated principal $SO_n$ frame bundle. We would like to count the number of different spin structures over $E$.

$1)$ From the fibration $SO_n\xrightarrow{i}P_{SO}(E)\xrightarrow{\pi}X$ we have an exact sequence $0\rightarrow H^1(X, \mathbb{Z}_2)\xrightarrow{{\pi}^*}H^1(P_{SO}(E), \mathbb{Z}_2)\xrightarrow{i^*}H^1(SO_n , \mathbb{Z}_2)$ from which we see that the number of different spin structures over $E$ is $H^1(X, \mathbb{Z}_2)$.

$2)$ On the other hand the short exact sequence $0\rightarrow\mathbb{Z}_2\rightarrow Spin_n\rightarrow SO_n\rightarrow 1$ gives an exact sequence $H^0(X, SO_n)\xrightarrow{\delta} H^1(X, \mathbb{Z}_2)\rightarrow H^1(X, Spin_n)\rightarrow H^1(X, SO_n)$ from which the number of different spin-structures over $E$ is $H^1(X , \mathbb{Z}_2)/\delta(H^0(X, SO_n))$.

So why do we have different answers for the same object?

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Do you know an example where the image of $\delta$ is nonzero? –  S. Carnahan Sep 28 '12 at 6:54
yes, there are a lot of examples. It can even be surjective. –  Axel Sep 28 '12 at 7:25

Your first answer is correct, and the second one is almost correct, but the problem is that they count different things:

In the first case, you count all twofold covers of $P = P_{SO} E$ that restrict to a nontrivial cover of each fibre. In other words, you count $Spin (n)$-bundles $Q$, together with an isomorphism $\phi:Q \times_{Spin(n)} SO(n) \cong P$ (up to isomorphism of $(P,\phi)$). Here $Q \times_{Spin(n)} SO(n)$ is just a different way to write $Q/\mathbb{Z}_2$. This is the usual notion of a spin structure on $E$.

Your second answer (correctly) counts the number of $Spin (n)$-principal bundles $Q \to X$ such that $Q /\mathbb{Z}_2$ \emph{admits} an isomorphism with $P$.

The point is that an abstract $Spin (n)$-bundle $Q \to X$ can yield many different spin structures on $P = Q/\mathbb{Z}_2$. Say if $P$ is trivial, then the set of homotopy classes of isomorphisms $Q /\mathbb{Z}_2 \to P$ is in bijection with $[X;SO(n)]$. This accounts for the division by the image $\delta$ in your second answer.

For nontrivial $P$, your second answer is not quite correct, as your exact sequence is only a sequence of sets.

To see this is an example, let $X=S^1$ and $n \geq 3$. As $SO(n)$ $Spin(n)$ are connected, all principal bundles on $S^1$ are trivial. In this case $H^0 (X;SO(n))=[S^1;SO(n)]=Z/2$; $\delta$ is an isomorpism and the two terms to the left are null. The generator of $[S^1;SO(n)]$ gives an isomorphism of the trivial $SO(n)$-bundle that transforms the two spin structures into each other.

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Johannes, thank you very much for such a complete answer. To make things more clear, may I slightly rephrase your main point and you tell me if I got you correctly? So, having a fixed bundle $P$ -the $SO_n$ bundle over $X$ we: $1)$ Find all abstract $Spin_n$ bundles $Q$ over $X$. $2)$ Take all $Q$ such that the associated $SO_n$ bundles $Q\cross SO_n$ are $\textbf{somehow}$ (as $SO_n$-bundle over $X$) isomorphic to $P$ $3)$ For the first answer it is important $\textbf{how}$, for the second answer $\textbf{somehow}$ is enough. Is it the issue? –  Axel Sep 29 '12 at 21:44
@Axel: Yes, that is precisely the issue: A spin structure on an $SO(n)$ bundle $P$ is a $Spin(n)$-bundle $Q$, \emph{together with} an isomorphism $Q \times_{Spin(n)} SO(n) \cong P$. –  Johannes Ebert Sep 30 '12 at 16:15
Great, thank you! –  Axel Oct 1 '12 at 7:03
Look a little further in the long exact sequence: $Ker(\delta)=Im(\phi)$ where $\phi:H^0(X,Spin_n)\to H^0(X,SO_n)$, and $Im(\phi)=H^0(X,SO_n)$ so that $\delta=0$. Indeed, $Spin_n$ double-covers $SO_n$, and $H^0(X;G)$ is |$\pi_0X$| copies of $G$.
No; $\phi$ is not surjective; you can't lift any map $X \to SO(n)$. –  Johannes Ebert Sep 28 '12 at 10:22
Sorry you're right, this doesn't work because our coefficients ($G$) here are not abelian groups. –  Chris Gerig Sep 28 '12 at 18:54
''because our coefficients (G) here are not abelian groups'': that's not the point; it is because $H^0 (X;G)$ is the set of homotopy classes of maps $X \to G$ and $G$ is not discrete. –  Johannes Ebert Sep 30 '12 at 16:16