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By a kummer surface, I mean a quotient of two-dimensional complex torus by multiplication by $-1$. It is a K3 surface and known to be simply connected. But it is not clear to me why it is simply-connected from its construction, which is a quotient of a surface with infinite fundamental group.

Is there a direct explanation or any references?

Thanks.

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    $\begingroup$ You need to resolve the singularities of the quotient to get the K3 surface. $\endgroup$ Commented Oct 28, 2016 at 15:35
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    $\begingroup$ Yes, the quotient is sinulgar. I am wondering why the quotient (or its crepant resolution) is simply-connected. $\endgroup$ Commented Oct 28, 2016 at 15:55
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    $\begingroup$ All complex K3 surfaces are diffeomorphic. So the resolution of the Kummer is diffeomorphic to a smooth quartic surface in $\mathbb{P}^3$, which is well-known to be simply-connected. $\endgroup$ Commented Oct 28, 2016 at 16:22
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    $\begingroup$ That's right. But I would like to see more direct proof. $\endgroup$ Commented Oct 28, 2016 at 17:21

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In the paper "The homology of Kummer manifolds" E. Spanier proves among other things that the fundental group of the resolution $K$ of singularities of the quotient $$(S^1\times S^1)^{\times n}/(\mathbb{Z}/2)$$ is simply connected (theorem 1). The proof is elementary and can be easily adapted to the singular quotient itself.

In your case $n=2$, $K$ is a K3-surface and the quotient is the Kummer surface.

Paper available here: http://www.ams.org/journals/proc/1956-007-01/S0002-9939-1956-0087188-3/S0002-9939-1956-0087188-3.pdf

Edit : let me give some more details, let us consider $$\mathcal{K}=(S^1\times S^1 \times S^1\times S^1)/(\mathbb{Z}/2)$$ it has 16 singular points $\{s_1,\dots,s_{16}\}$, the link of each singular point is homeomorphic to $\mathbb{R}P^3=S^3/(\mathbb{Z}/2)$. Let $$\mathcal{K}_{r}=\mathcal{K}-\bigcup_{i=1}^{16} U_i$$ where $U_i$ is a "nice" neighbourhood of the singular point $s_i$ i.e. $$U_i\cong c \mathbb{R}P^3$$ it is homeomorphic to the open cone of the link $\mathbb{R}P^3$.

Now you notice that $\mathcal{K}_r$ is a manifold with boundary, it has $16$ boundary components and the main point is to prove that the morphism $$\mathcal{I}:\coprod_{i=1}^{16}\pi_1(\mathbb{R}P^3)\rightarrow \pi_1(\mathcal{K}_r)$$ induced by the inclusions of each boundary components into $\mathcal{K}_r$ is surjective.

Using the fact that each neighbourhood $U_i$ is contractible you get that $$\pi_1(\mathcal{K})\cong\pi_1 (\mathcal{K}_r)/Im(\mathcal{I})=\{e\}.$$

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I haven't checked details but the following might work. Coverings of the Kummer surface will lift to coverings of the abelian surface (whose fundamental group is clearly $\mathbb{Z}^4$), so the fundamental group of the Kummer surface is abelian. Now it's enough to show that the first homology group is trivial. The latter I expect you will find in textbooks, e.g. as a computation of the Hodge diamond of a K3 surface.

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  • $\begingroup$ But is it clear that the resulting cover of the abelian surface will be étale (possibly after blowing down some exceptional divisors)? Since there is no morphism $A \to K$, but only a morphism $\widetilde{A} \to K$ where $\widetilde{A}$ is $A$ blown up in $A[2]$, base-changing an étale cover of $K$ gives an étale cover of $\widetilde{A}$, but (as far as I can see) not necessarily one of $A$. $\endgroup$ Commented Oct 29, 2016 at 4:50
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    $\begingroup$ @René: but $\widetilde{A}$ has the same fundamental group as $A$. $\endgroup$ Commented Oct 29, 2016 at 6:20
  • $\begingroup$ @potentiallydense: ah, good point. $\endgroup$ Commented Oct 29, 2016 at 8:54
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    $\begingroup$ Although one can show that $h^{1,0}=0$, the fundamental group may still have a torsion part. Am I correct? $\endgroup$ Commented Oct 30, 2016 at 2:38
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    $\begingroup$ @Lee Yes, in principle, but the argument shows that the fundamental group is a subgroup of a free abelian group, so is also free abelian. $\endgroup$ Commented Oct 30, 2016 at 3:46

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