Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

up vote 11 down vote accepted

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.

share|improve this answer
    
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 $G$.

share|improve this answer
4  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.