Intersection form on quotient manifold

2012-08-08T03:37:08Z

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 $-id_{E_{2}}$?

Answer by Will Sawin for Intersection form on quotient manifold

2012-08-08T04:56:06Z

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.

Intersection form on quotient manifold - MathOverflow

Intersection form on quotient manifold

Answer by Will Sawin for Intersection form on quotient manifold