Betti Cohomology of singular Kummer Surface

2010-06-18T02:35:32Z

Let $A$ be a complex torus of (complex) dimension 2 and $X$ the associated Kummer variety $A/\sigma$, where $\sigma(x)=-x$. I would like to compute the cohomology of $X$ with $\mathbb{Z}$ coefficients. My initial instinct was to use Mayer-Vietoris, but the exact sequence involves the cohomology of the quadratic cone minus a point which is also proving to be difficult for me. My hope is that as in the case with $\mathbb{Q}$ coefficients $$ H^1(X,\mathbb{Z})=H^3(X,\mathbb{Z})=0,\qquad H^0(X,\mathbb{Z})=H^4(X,\mathbb{Z})=\mathbb{Z}\quad\text{ and }\quad H^2(X,\mathbb{Z})=\wedge^2H^1(A)\ $$ Any tips as to how to compute $H^i(X,\mathbb{Z})$ or, equivalently, places to find tips in the literature would be very helpful. Thank you.

Answer by Torsten Ekedahl for Betti Cohomology of singular Kummer Surface

2010-06-19T05:36:40Z

I missed that the question concerned the singular Kummer surface (which I think historically was what was what was called the Kummer surface but our current fixation on non-singularity has changed that) so one needs a few more steps than Barth, Peters, van de Ven: Compact complex surfaces (which will be my reference below).

Let $\pi\colon\tilde X\rightarrow X$ be the minimal resolution of singularities and consider the Leray spectral sequence for $\pi$. We have $\pi_\ast\mathbb Z=\mathbb Z$ and $R^2\pi_\ast\mathbb Z$ the skyscraper sheaf with one $\mathbb Z$ at each of the 16 singular points. The Leray s.s. thus gives that $H^i(X,\mathbb Z)=H^i(\tilde X,\mathbb Z)$ for $i\neq2,3$ and hence $H^i(X,\mathbb Z)=\mathbb Z$ for $i=0,4$ and $H^1(X,\mathbb Z)=0$ as well as a short exact sequence $$ 0\rightarrow H^2(X,\mathbb Z)\rightarrow H^2(\tilde X,\mathbb Z)\rightarrow \bigoplus_{v\in V}\mathbb Zv\rightarrow H^3(X,\mathbb Z)\rightarrow0, $$ where $V$ is the set of singular points. Now, it is easy to see that $H^2(\tilde X,\mathbb Z)\rightarrow \mathbb Zv$ is given by $f\mapsto \deg(f_{E_v})$, where $E_v:=\pi^{-1}(v)$. We have $\deg(f_{E_v})=\langle e_v,f\rangle$, where $e_v\in H^2(\tilde X,\mathbb Z)$ is the fundamental class of $E_v$. Hence, we get to begin with that $H^2(X,\mathbb Z)$ is the orthogonal complement in $H^2(\tilde X,\mathbb Z)$ of the $e_v$. By Cor. 5.6 (of BPV) this can be identified with $H^2(A,\mathbb Z)$. On the other hand, the image of $H^2(\tilde X,\mathbb Z)$ in $\bigoplus_{v\in V}\mathbb Zv$ contains the linear functions given by the $e_v$ and $e_v(v')=-2\delta_{v,v'}$ so that we may consider the image of $H^2(\tilde X,\mathbb Z)$ in $\bigoplus_{v\in V}\mathbb Z/2v$. By the fact that the cup product pairing on $H^2(\tilde X,\mathbb Z)$ is perfect (by Poincaré duality) and by Prop. 5.5 we get that this image is dual to the subspace of affine functions of $\bigoplus_{v\in V}\mathbb Z/2v$ (where $V$ is identified by the kernel of multiplication by $2$ in $A$) and hence we get an identification of $H^3(X,\mathbb Z)$ with the dual of the $\mathbb Z/2$-space of affine functions of $V$, in particular it has dimension $5$.

Remark: It is interesting to note that while the quotient $A/\sigma$ as a topological space does not use the complex structure of $A$ it still seems easier to use it (in a very weak form, the blowing up only uses that a conical neighbourhood has a certain form) as we consider the complex blow up of the singular points. Indeed, the use of Mayer-Vietoris tried by the poser does look more difficult (of course that would also use the local form of the singularity but somehow in a less complex fashion).

Betti Cohomology of singular Kummer Surface - MathOverflow

Betti Cohomology of singular Kummer Surface

Answer by Torsten Ekedahl for Betti Cohomology of singular Kummer Surface