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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 ...2012-08-06
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    ... $\mathbb Q$-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](http://math.stackexchange.com/a/38728/221).) Regards,2012-08-06
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    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.2012-08-06
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    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 ...2012-08-06
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    ... might play a role. Regards,2012-08-06
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    I see. Now that your answer seems very much relevant to my problem. Could you introduce me a good reference? I cannot find a kind of HS spectral sequence you mentioned above.2012-08-06
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    I think this is precisely the cause! As for the reference: Ken Brown's *Cohomology of Groups*.2012-08-06

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