Vtyjkyuk

Let $X$ be a(n algebraic) K3 surface, i.e., $X$ is a smooth algebraic surface with trivial canonical bundle and $H^1(X,mathcalO)=0$. 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 $p:tildeXto X$, but $tildeX$ 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
beginequation
H^2(X,mathcalO_X)cong H^0(X,omega_X)^veecongmathbbC,
endequation
and by a dimensional argument $H^3(X,mathcalO_X)=0,,H^4(X,mathcalO_X)=0$.

Using the exponential sequence
beginequation
0tounderlinemathbbZtomathcalO_XtomathcalO_X^timesto0
endequation
and passing to long exact sequence in cohomology, one has:
begingather
0to H^0left(X,underlinemathbbZright)congmathbbZto H^0(X,mathcalO_X)congmathbbCxrightarrowexpH^0left(X,mathcalO_X^timesright)congmathbbC^timesto H^1left(X,underlinemathbbZright)to H^1(X,mathcalO_X)=0,\
0to H^1left(X,mathcalO_X^timesright)cong Pic(X)to H^2left(X,underlinemathbbZright)to H^2(X,mathcalO_X)congmathbbCto H^2left(X,mathcalO_X^timesright)to H^3left(X,underlinemathbbZright)to H^3(X,mathcalO_X)=0,\
0to H^3left(X,mathcalO_X^timesright)to H^4left(X,underlinemathbbZright)to H^4(X,mathcalO_X)=0,\
0to H^4left(X,mathcalO_X^timesright)to0;
endgather
so $H^3left(X,mathcalO_X^timesright)cong H^4left(X,underlinemathbbZright)$ and $H^4left(X,mathcalO_X^timesright)=0$.

By exactness of
beginequation
0tomathbbZtomathbbCxrightarrowexpmathbbC^timesto0
endequation
the previous sequence splits in
beginequation
0to H^1left(X,underlinemathbbZright)to H^1(X,mathcalO_X)=0
endequation
so $H^1left(X,underlinemathbbZright)cong H^1(X,mathbbZ)=0$.

By the Universal Coefficient Theorem for Cohomology, there exists a surjective morphism of Abelian groups
beginequation
0=H^1(X,mathbbZ)tohom(H_1(X,mathbbZ),mathbbZ)to0,
endequation
that is $H_1(X,mathbbZ)$ is either a non trivial torsion Abelian group or $0$.

Because $Pic(X)$ and $mathbbC$ are torsion free groups (see remark 1.2.5), then also $H^2left(X,underlinemathbbZright)cong H^2(X,mathbbZ)$ is a torsion free group; as consequence $H^dim_mathbbRX-2+1(X,mathbbZ)=H^3(X,mathbbZ)$ is a torsion free group; by Poincaré duality, $H_1(X,mathbbZ)$ is a torsion free group, and by previous reasoning $H_1(X,mathbbZ)=0$.

Remarks.

1. Let $X$ be a compact, without boundary, oriented (real) manifold of dimension $d$. The torsion subgroup of $H^p(X,mathbbZ)$ is isomorphic to torsion subgroup of $H^d-p+1(X,mathbbZ)$; where $pin1,dots,d$. (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, $H^3left(X,mathcalO_X^timesright)cong H^4left(X,underlinemathbbZright)cong H^4(X,mathbbZ)cong H_0(X,mathbbZ)congmathbbZ$ and $H^3left(X,underlinemathbbZright)cong H^3(X,mathbbZ)cong H_1(X,mathbbZ)=0$; so the non-trivial part of previous long exact sequence in cohomology is
   beginequation
   0to Pic(X)to H^2left(X,underlinemathbbZright)tomathbbCto H^2left(X,mathcalO_X^timesright)to0.
   endequation

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 $H_1(X,mathbbZ)$ 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

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 $H_1$.