# Let $X$ be a K3 surface, show that $H_1(X,\mathbb Z)=0$

Let $X$ be a(n algebraic) K3 surface, i.e., $X$ is a smooth algebraic surface with trivial canonical bundle and $H^1(X,\mathcal{O})=0$. This assumption directly implies that $H^1(X,\mathbb C)=0$, so $H_1(X,\mathbb C)=0$ by Poincaré duality. In particular $H_1(X,\mathbb Z)$ is a torsion group.

Next, to show that $H_1(X,\mathbb Z)$ is actually $0$, the book says otherwise there is a nontrivial torsion element in fundamental group which allows us to pass to a finite covering $p:\tilde{X}\to X$, but $\tilde{X}$ is still a K3 surface, and considering any K3 surface has Euler characteristic 24, so $p$ has to be an identity map, which is a contradiction.

Here is my question: First homology group is the Abelianzation of fundamental group $H_1=\pi_1/[\pi_1,\pi_1]$, but why a nontrivial torsion element in $H_1$ has a nontrivial torsion representative in $\pi_1$?

First proof.

By GAGA principle, $$X$$ is a (connected compact) complex surface; using the Serre duality, one has $$$$H^2(X,\mathcal{O}_X)\cong H^0(X,\omega_X)^{\vee}\cong\mathbb{C},$$$$ and by a dimensional argument $$H^3(X,\mathcal{O}_X)=0,\,H^4(X,\mathcal{O}_X)=0$$.

Using the exponential sequence $$$$0\to\underline{\mathbb{Z}}\to\mathcal{O}_X\to\mathcal{O}_X^{\times}\to0$$$$ and passing to long exact sequence in cohomology, one has: $$\begin{gather} 0\to H^0\left(X,\underline{\mathbb{Z}}\right)\cong\mathbb{Z}\to H^0(X,\mathcal{O}_X)\cong\mathbb{C}\xrightarrow{\exp}H^0\left(X,\mathcal{O}_X^{\times}\right)\cong\mathbb{C}^{\times}\to H^1\left(X,\underline{\mathbb{Z}}\right)\to H^1(X,\mathcal{O}_X)=0,\\ 0\to H^1\left(X,\mathcal{O}_X^{\times}\right)\cong Pic(X)\to H^2\left(X,\underline{\mathbb{Z}}\right)\to H^2(X,\mathcal{O}_X)\cong\mathbb{C}\to H^2\left(X,\mathcal{O}_X^{\times}\right)\to H^3\left(X,\underline{\mathbb{Z}}\right)\to H^3(X,\mathcal{O}_X)=0,\\ 0\to H^3\left(X,\mathcal{O}_X^{\times}\right)\to H^4\left(X,\underline{\mathbb{Z}}\right)\to H^4(X,\mathcal{O}_X)=0,\\ 0\to H^4\left(X,\mathcal{O}_X^{\times}\right)\to0; \end{gather}$$ so $$H^3\left(X,\mathcal{O}_X^{\times}\right)\cong H^4\left(X,\underline{\mathbb{Z}}\right)$$ and $$H^4\left(X,\mathcal{O}_X^{\times}\right)=0$$.

By exactness of $$$$0\to\mathbb{Z}\to\mathbb{C}\xrightarrow{\exp}\mathbb{C}^{\times}\to0$$$$ the previous sequence splits in $$$$0\to H^1\left(X,\underline{\mathbb{Z}}\right)\to H^1(X,\mathcal{O}_X)=0$$$$ so $$H^1\left(X,\underline{\mathbb{Z}}\right)\cong H^1(X,\mathbb{Z})=0$$.

By the Universal Coefficient Theorem for Cohomology, there exists a surjective morphism of Abelian groups $$$$0=H^1(X,\mathbb{Z})\to\hom(H_1(X,\mathbb{Z}),\mathbb{Z})\to0,$$$$ that is $$H_1(X,\mathbb{Z})$$ is either a non trivial torsion Abelian group or $$0$$.

Because $$Pic(X)$$ and $$\mathbb{C}$$ are torsion free groups (see remark 1.2.5), then also $$H^2\left(X,\underline{\mathbb{Z}}\right)\cong H^2(X,\mathbb{Z})$$ is a torsion free group; as consequence $$H^{\dim_{\mathbb{R}}X-2+1}(X,\mathbb{Z})=H^3(X,\mathbb{Z})$$ is a torsion free group; by Poincaré duality, $$H_1(X,\mathbb{Z})$$ is a torsion free group, and by previous reasoning $$H_1(X,\mathbb{Z})=0$$.

Remarks.

1. Let $$X$$ be a compact, without boundary, oriented (real) manifold of dimension $$d$$. The torsion subgroup of $$H^p(X,\mathbb{Z})$$ is isomorphic to torsion subgroup of $$H^{d-p+1}(X,\mathbb{Z})$$; where $$p\in\{1,\dots,d\}$$. (cfr. exercise 54)
2. By this proof, one has the following (partial) result: $$\pi_1(X)$$ is a perfect group. For exact, $$\pi_1(X)=0$$; see the second proof.
3. By Poincaré duality, $$H^3\left(X,\mathcal{O}_X^{\times}\right)\cong H^4\left(X,\underline{\mathbb{Z}}\right)\cong H^4(X,\mathbb{Z})\cong H_0(X,\mathbb{Z})\cong\mathbb{Z}$$ and $$H^3\left(X,\underline{\mathbb{Z}}\right)\cong H^3(X,\mathbb{Z})\cong H_1(X,\mathbb{Z})=0$$; so the non-trivial part of previous long exact sequence in cohomology is $$$$0\to Pic(X)\to H^2\left(X,\underline{\mathbb{Z}}\right)\to\mathbb{C}\to H^2\left(X,\mathcal{O}_X^{\times}\right)\to0.$$$$

Second proof.

Any K3 surface is diffeomorphic to a Fermat quartic surface (see example 1.1.3.i, remark 1.3.6.i and theorem 7.1.1), in particular they are simply connected (corollary 7.1.4); that is $$\pi_1(X)=0$$ and the relevant Abelianization $$H_1(X,\mathbb{Z})$$ is trivial.

References

• Davis J. F., Kirk P. (2001) Lectures Notes in Algebraic Topology, Graduate Studies in Mathematics 35 American Mathematical Society
• Huybrechts D. (2016) Lectures on K3-surfaces, Cambridge University Press
• It is not correct that $H^0(X,O_X^\times)=0$: that group is $\mathbf C^\times$. But the map from $H^0(X,O_X)=\mathbf C$ is a surjection, which is enough for this argument. Aug 14, 2018 at 11:38
• How do you conclude $H_1(X,\Bbb Z) = 0$ from $\hom(H_1(X,\Bbb Z), \Bbb Z) = 0$? Aug 14, 2018 at 12:58
• When I must work in algebraic topology kingdom, my brain goes catch on a loop! Aug 14, 2018 at 14:19
• How do you prove that K3 surfaces are all diffeomorphic to each other? Aug 14, 2018 at 16:29
• Dear @Armandoj18eos you write “as consequence $H^{\dim_{\mathbb{R}}X-2+1}(X,\mathbb{Z})=H^3(X,\mathbb{Z})$ is a torsion free group”. Where does this follow? Aug 15, 2018 at 3:29

What book is this? The argument is incorrect. The existence of a nontrivial finite cover corresponds to the existence of a nontrivial subgroup of finite index, not the existence of a nontrivial torsion element. And this is guaranteed by the existence of a nontrivial finite quotient of $\pi_1$, which is in turn guaranteed by the existence of a nontrivial finite quotient of $H_1$.

• Dear Qiaochu, thanks for your explanation. I wrote "there is a nontrivial torsion element in fundamental group" off top of my head, it was not the precise statement of the book. Aug 14, 2018 at 23:50