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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.

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By the Kummer variety do you mean the singular quotient or its resolution once the 16 singular points are blown up? –  José Figueroa-O'Farrill Jun 18 '10 at 14:13
    
I mean the singular quotient. I suppose I really should have said singular Kummer surface in the title. –  AJ Stewart Jun 18 '10 at 15:12
    
In that case, why is not just the invariant part of the cohomology? (Perhaps I'm missing something obvious, though.) –  José Figueroa-O'Farrill Jun 18 '10 at 16:22
    
I edited the title to better reflect the question, by the way. –  José Figueroa-O'Farrill Jun 18 '10 at 16:38
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There is $2$-torsion; the third mod $2$ homology group is nontrivial. Quick proof: Topologically this is what you get from the product of four circle groups by identifying each element with its inverse. As such, it contains as a retract the analogous quotient space of a product of three circle groups. The latter has nontrivial third mod $2$ homology, because it is a $3$-manifold except for singularities of codimension $3>1$. –  Tom Goodwillie Jun 18 '10 at 17:52

1 Answer 1

up vote 5 down vote accepted

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).

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