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Let me begin with some motivation. In calculating the Chern-Simons invariant of a $U(1)$ connection $A$ on a 3-manifold $M$, we can proceed by picking a bounding 4-manifold $X$ with $\partial X = M$ and an extension of $A$ to $X$ and write $$ CS(A)=\frac{k}{4 \pi^2}\int_X dA \wedge dA, $$ where the level $k$ must be an integer for this expression to be well-defined up to $2 \pi i$ when we choose a different bounding 4-manifold. However, if we have a spin structure everywhere, then on a closed 4-manifold, this expression is always an even multiple of $2\pi i$ since the intersection form is even. Thus, we can let $k$ be a half-integer and still get a well-defined invariant one we exponentiate.

Let me phrase this another way. If you have a spin 4-manifold $X$ and a map $X \to K(G,2)$, the pullback of a class in $H^4(K(G,2),\mathbb{R}/\mathbb{Z})$ is actually defined up to even integers.

I want to know why this happens for $H^4(K(G,2),\mathbb{R}/\mathbb{Z})$ and in which other cases it happens. I imagine the answer has to do with relations involving $w_2$ in the $p=2$ Steenrod algebra such as Wu's relation.

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  • $\begingroup$ I'm having trouble following this. For your question, what are you referring to by "what the spin structure gives you"? In relation to the motivation, the spin structure gives $w_2(M)=0$ which on a simply-connected 4-manifold is equivalent to having an even intersection form. I don't see why you switch gears to "characteristic classes of bundles other than the tangent bundle". $\endgroup$ Sep 24, 2013 at 21:28
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    $\begingroup$ I don't think I'm really switching gears. The Chern-Simons invariant is a sort of characteristic class of the $U(1)$ gauge bundle. Perhaps I should have chosen a more simple motivating example. By the way, does this definition of the Chern-Simons invariant only work for simply connected manifolds? $\endgroup$ Sep 25, 2013 at 0:21
  • $\begingroup$ Well first, CS is not a characteristic class, because $X$ has boundary (the Chern class arises from closed $X$). And I can choose my $X$ to be simply-connected, because the corresponding cobordism group is trivial. $\endgroup$ Sep 25, 2013 at 0:41
  • $\begingroup$ A post of mine from a while ago is related: mathoverflow.net/questions/100746/… $\endgroup$ Sep 25, 2013 at 5:13
  • $\begingroup$ Are you asking why the intersection form is even on $H^2(X,\mathbb{Z})$ when $X$ is a spin 4-manifold? This follows from the fact that the degree 2 Wu class, given in terms of the Stiefel-Whitney classes by $w_2 + w_1^2$, vanishes for spin manifolds. $\endgroup$ Sep 26, 2013 at 6:25

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I think your last sentence/question is not well posed. It is well known that there are flat manifolds admitting plenty of non-flat vector bundles. E.g. see J. Smillie, "Flat manifolds with non-zero Euler characteristics".

Sorry about the commentlike answer but I do not have enough points to properly comment.

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  • $\begingroup$ I'm not saying that this vanishing certainly happens. I'm interested in knowing when it happens and why. $\endgroup$ Sep 25, 2013 at 0:18

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