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](https://math.stackexchange.com/questions/179118/intersection-form-on-quotient-manifold), 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 $-id\_{E\_{2}}$? 



Intersection form on quotient manifold
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.


