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Let $E_{1},E_{2}$ be elliptic curves over $\mathbb{C}$. We denote by $\iota_{i}$ the translation by a 2-torsion point on $E_{i}$. Then $G=\mathbb{Z}/2\mathbb{Z}$ acts freely on the the product $E_{1}\times E_{2}$ via the involution $\iota=(\iota_{1},\iota_{2})$. and the quotient $$ X=(E_{1}\times E_{2})/G $$ is a 4-dimensional manifold (complex surface). I would like to understand the intersection form on the middle cohomology $$ (-,-)_{X}:H^2(X,\mathbb{Z})\times H^2(X,\mathbb{Z}) \rightarrow H^4(X,\mathbb{Z})\cong \mathbb{Z} $$ via the cup product. I initially thought

There is a ono-to-one correspondence between $$ H^{2}(X,\mathbb{Z}) > \longleftrightarrow H^{2}(E_{1}\times > E_{2},\mathbb{Z})^{G}, $$ Since the 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 $|G|=2$.

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

When I asked a similar question, some people pointed out that the correspondence $$ H^{2}(X,\mathbb{Z}) \ \longleftrightarrow H^{2}(M\times N,\mathbb{Z})^{G}, $$ does not hold in general; there is the Hochschild-Serre spectral sequence $$ E^{p,q}=H^{p}(G,H^{q}(E_{1}\times E_{2},\mathbb{Z}))\Rightarrow H^{p+q}(X,\mathbb{Z}) $$ Here $E^{0,2}$ term corresponds to $H^{2}(M\times N,\mathbb{Z})^{G}$ above.

Having said that, I still don't quite understand the intersection form on $X$ (mainly due to my poor understanding of the Spectral sequence). I would appreciate it if anyone could describe the intersection form. What if $\iota_{2}$ is replaced by $-id_{E_{2}}$?

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If $\iota_2$ is replaced by $-id_{E_2}$ then $X$ is a hyperelliptic (also known as bielliptic) surface, en.wikipedia.org/wiki/Hyperelliptic_surface –  YangMills Aug 11 '12 at 22:50
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up vote 4 down vote accepted

Luckily, the $X$ thus described is a torus. This means that $H^2(X,\mathbb Z)$ has a nice explicit description: It is $\wedge^2 H^1(X,\mathbb Z)$. The intersection form is the the symmetric bilinear map to $\wedge^4 H^1(X,\mathbb Z)=\mathbb Z$. $H^1(X,\mathbb Z)$ is an index two sublattice of $H^1(E_1 \times E_2,\mathbb Z)$ , which I guess you can see from the exact sequence for homotopy groups of a fibration.

In fact, the action of $G$ on the cohomology groups is trivial, because it is homotopic to the identity, since the group of translations is connected!

So $H^2(X,\mathbb Z)$ lies in $H^2(E_1 \times E_2,\mathbb Z)$ and similarly $H^4(X,\mathbb Z)$ lies in $H^4(X,E_1 \times E_2)$, so you do indeed get a division by $|G|$ when you identify the $H^4$s with $\mathbb Z$. But this does not mean that all intersection forms in $H^2(E_1\times E_2,\mathbb Z)$ are even, just those for cocycles which are pullbacks of cocycles from $X$. The criterion for this is not $G$-invariance of the cohomology, as the $G$ action on the cohomology is trivial, it's $G$-invariance of the underlying cocycle.

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Even if the action of $G$ on the cohomology groups is trivial, the group cohomology may be non-trivial, isn't it? You are essentially saying that the correspondence in my question is correct in this case, and also we don't have to divide the form by 2. –  Michel Aug 8 '12 at 5:35
    
What I have in my mind is an Enriques surface $S=T/\langle \iota \rangle$, where $T$ is a K3 surface and $\iota$ is an involution without fixed point. The intersection form $U\oplus E_{8}$ on $S$ is obtained as "half" of the invariant lattice $U(2)\oplus E_{8}(2)$ of the K3 surface $T$. –  Michel Aug 8 '12 at 5:42
    
Michel -- to complement Will's answer: your $X$ is the quotient of the 4-torus $T=E_1\times E_2$ by a translation by some element $t$ of order 2. The topological automorphism group of $T$ acts transitively on 2-torsion, so we may assume that $t$ lives in, say, $E_1$. –  algori Aug 8 '12 at 16:00
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There is not a correspondence. There is actually an injection $H^2(X,\mathbb Z) \to H^2(E_1 \times E_2,\mathbb Z)^G=H^2(E_1\times E_2,\mathbb Z)$4. You do have to divide the form by $|G|$, in fact this is true in great generality. Let $f: Y \to X$ be a covering map of degree $n$ that is not necessarily the quotient by a group. If $A$ and $B$ are two cohomology cycles on $X$ whose degree adds to the real dimension of $X$, then $f^* A f^* B= n AB$. –  Will Sawin Aug 8 '12 at 18:03
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Well, it's not really about the class being $G$-invariant, it's about some cocycle in the appropriate cohomology theory being $g$-invariant. You can see why it has to be that way by looking at the $H^1$ of the circle under the degree $2$ map to itself. It's obvious from any number of arguments that the map on $H^1$ has to be multiplication by $2$, so only even classes lie in the image. I don't really know what "is sensible to $E_2$ part" means. –  Will Sawin Aug 9 '12 at 0:48
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