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 $\pi_1$?


2 Answers 2


First proof.

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

Using the exponential sequence \begin{equation} 0\to\underline{\mathbb{Z}}\to\mathcal{O}_X\to\mathcal{O}_X^{\times}\to0 \end{equation} 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 \begin{equation} 0\to\mathbb{Z}\to\mathbb{C}\xrightarrow{\exp}\mathbb{C}^{\times}\to0 \end{equation} the previous sequence splits in \begin{equation} 0\to H^1\left(X,\underline{\mathbb{Z}}\right)\to H^1(X,\mathcal{O}_X)=0 \end{equation} 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 \begin{equation} 0=H^1(X,\mathbb{Z})\to\hom(H_1(X,\mathbb{Z}),\mathbb{Z})\to0, \end{equation} 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 $H_1(X,\mathbb{Z})=0$.


  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 an Abelian 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 \begin{equation} 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. \end{equation}

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.


  • 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
  • 2
    $\begingroup$ 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. $\endgroup$ Aug 14, 2018 at 11:38
  • 1
    $\begingroup$ How do you conclude $H_1(X,\Bbb Z) = 0$ from $\hom(H_1(X,\Bbb Z), \Bbb Z) = 0$? $\endgroup$
    – Claudius
    Aug 14, 2018 at 12:58
  • $\begingroup$ When I must work in algebraic topology kingdom, my brain goes catch on a loop! $\endgroup$ Aug 14, 2018 at 14:19
  • $\begingroup$ How do you prove that K3 surfaces are all diffeomorphic to each other? $\endgroup$ Aug 14, 2018 at 16:29
  • 1
    $\begingroup$ 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? $\endgroup$
    – AG learner
    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 $H_1$.

  • $\begingroup$ 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. $\endgroup$
    – AG learner
    Aug 14, 2018 at 23:50

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