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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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    I think this is precisel$y$ the cause! As for the reference: Ken Brown's *Cohomology of Groups*.2012-08-06

1 Answers 1

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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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    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.2012-08-05