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I have a simple algebraic topology question. Let $M$ and $N$ be 2-dimensional oriented manifolds (say $H^{2}(M,\mathbb{Z})\cong \mathbb{Z}\alpha_{M}$ and $H^{2}(M,\mathbb{Z})\cong \mathbb{Z}\alpha_{N}$). Assume that a finite group $G$ acts on $M$ and $N$ in such a way that $G$ preserves the orientation of $M$ and $N$ and the induced action of $G$ on $M\times N$ has no fixed point. Then $$ X=(M\times N)/G $$ is a 4-dimensional oriented manifold.

I would like to understand the intersection form on the middle cohomology $H^{2}(X,\mathbb{Z})$. There is a ono-to-one correspondence between $$ H^{2}(X,\mathbb{Z}) \ \longleftrightarrow H^{2}(M\times N,\mathbb{Z})^{G}, $$ i.e. any $G$-invariant element of $H^{2}(M\times N,\mathbb{Z})$ descends to $H^{2}(X,\mathbb{Z})$ and any element of $H^{2}(X,\mathbb{Z})$ can be pulled back to $H^{2}(M\times N,\mathbb{Z})^{G}$ by the quotient map. Since the $G$-action is free, the intersection form on $H^{2}(X,\mathbb{Z})$ is given by the intersection form on $H^{2}(M\times N,\mathbb{Z})^{G}$ divided by $|G|$. So, any intersection number on $H^{2}(M\times N,\mathbb{Z})^{G}$ must be a multiple of $|G|$.

On the other hand we have $$ p_{1}^{*}(\alpha_{M}), \ p_{2}^{*}(\alpha_{N})\in H^{2}(M\times N,\mathbb{Z})^{G} $$ (because $G$ preserves both $M$ and $N$) and $$ p_{1}^{*}(\alpha_{M})\cup \ p_{2}^{*}(\alpha_{N})=\alpha_{M\times N} $$ where $p_{i}$ is the $i$-th projection of $M\times N$ and $H^{2}(M\times N,\mathbb{Z})\cong \mathbb{Z}\alpha_{M\times N}$. This means that the intersection number $p_{1}^{*}(\alpha_{M})\cup \ p_{2}^{*}(\alpha_{N})$ is 1, not divisible by $|G|$ (unless $G$ is trivial).

Could anyone point out what is wrong with my argument?

Thank you in advance.

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Dear Michel, I didn't look carefully at your argument, but your claim about the cohomology of $X$ coinciding with the invariant cohomology is incorrect in general; rather, there is the Hochschild--Serre spectral sequence $E_2^{p,q} := H^p\bigl(G,H^q(M\times N,\mathbb Z) \bigr) \implies H^{p+q}(X,\mathbb Z).$ (Note that if we used $\mathbb Q$ coefficients rather than $\mathbb Z$-coefficients, then, since the higher group cohomology of the finite group $G$ with coefficients in a $\mathbb Q$-vector space vanishes, we would just get that the $\mathbb Q$-cohom. of $X$ equals the invariant ... –  Matt E Aug 6 '12 at 2:51
    
... $\mathbb Q$-cohom. of $M\times N$. But this certainly needn't be true for $\mathbb Z$-cohomology. (As a toy example, consider the case of quoienting out $S^2$ by the free $\mathbb Z/2\mathbb Z$-action to get $\mathbb R P^2$, as discussed in this answer.) Regards, –  Matt E Aug 6 '12 at 3:43
    
Thank you for the detailed answer. I was not aware of the Hochschild--Serre spectral sequence. I still think that my argument about the intersection form works, mod torsion part. –  Michel Aug 6 '12 at 5:28
2  
Dear Michel, Still without having thought through the details of your situation, note that the HS spectral sequence won't just come into the torsion part, but can also cause some map that you might naively guess to be an isomorphism to instead by an injection whose image has finite, but non-trivial, index in the target. This index would be caused by the non-vanishing of some higher group cohomology, and so would involve primes dividing $|G|$. Since your problem involves exactly the problem of whether a certain number is divisible by $|G|$, it seems that the phenomenon I'm discussing ... –  Matt E Aug 6 '12 at 6:02
    
... might play a role. Regards, –  Matt E Aug 6 '12 at 6:04
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1 Answer

Why not try out both arguments on a specific example?

Take $M = N = S^1\times S^1$ and $G = \mathbb{Z}/2\mathbb{Z}$ acting by $-1 *(\theta,\phi) = (\theta +\pi,\phi +\pi)$. Then this action is free on each of $M$ and $N$, so the induced action on $M\times N$ is certainly free.

In this case, $(M\times N)/ G \cong (S^1)^4$, since, for example, $(M\times N)/ G$ is a compact abelian Lie group. (I'm sure there's an easy direct way to see it).

So, what do both computations give you? Which one must be wrong?

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More generally, there is a free action of $G=\mathbb{Z}/n\mathbb{Z}$ on $M$ (whose quotient is still $M$) , so $(M\times N)/G \cong (S^1)^4$ for any of these $G$. That should really help narrow things down. –  Jason DeVito Aug 5 '12 at 19:00
    
Your action has a fixed point, namely the origin. $G=\mathbb{Z}/2\mathbb{Z}$ must act on $S^{1}\times S^{1}$ as translation by a 2-torsion point. I still don't understand what's wrong in my argument. –  Michel Aug 5 '12 at 19:09
    
@Michael: You're right about the action, I'll correct it now. With out you supplying details to your first argument, I'm not sure why it's wrong. I'm not sure I could figure it out any way, as I'm only vaguely familiar with invariant (co)homology. –  Jason DeVito Aug 5 '12 at 21:43
    
I added some argument. I have seen the correspondence between invariant cohomology and cohomology downstairs many times, so this must be OK. Then the latter half seems trivial. –  Michel Aug 5 '12 at 23:46
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