Intersection form on quotient manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:41:02Z http://mathoverflow.net/feeds/question/104253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104253/intersection-form-on-quotient-manifold Intersection form on quotient manifold Michel 2012-08-08T03:37:08Z 2012-08-08T19:38:21Z <p>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</p> <blockquote> <p>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 $|G|=2$.</p> <p>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 $|G|=2$.</p> </blockquote> <p>When I asked a <a href="http://math.stackexchange.com/questions/179118/intersection-form-on-quotient-manifold" rel="nofollow">similar question</a>, 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. </p> <p>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 $-id_{E_{2}}$? </p> http://mathoverflow.net/questions/104253/intersection-form-on-quotient-manifold/104257#104257 Answer by Will Sawin for Intersection form on quotient manifold Will Sawin 2012-08-08T04:56:06Z 2012-08-08T16:19:04Z <p>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.</p> <p>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!</p> <p>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.</p>