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.

  • $\begingroup$ By the Kummer variety do you mean the singular quotient or its resolution once the 16 singular points are blown up? $\endgroup$ – José Figueroa-O'Farrill Jun 18 '10 at 14:13
  • $\begingroup$ I mean the singular quotient. I suppose I really should have said singular Kummer surface in the title. $\endgroup$ – AJ Stewart Jun 18 '10 at 15:12
  • $\begingroup$ In that case, why is not just the invariant part of the cohomology? (Perhaps I'm missing something obvious, though.) $\endgroup$ – José Figueroa-O'Farrill Jun 18 '10 at 16:22
  • $\begingroup$ I edited the title to better reflect the question, by the way. $\endgroup$ – José Figueroa-O'Farrill Jun 18 '10 at 16:38
  • 2
    $\begingroup$ 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 $3>1$. $\endgroup$ – Tom Goodwillie Jun 18 '10 at 17:52

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.