$H^2(X,\mathbb{Z})$,$H_2(X,\mathbb{Z})$of smooth complex projective variety$X$

Question

is there some example that: $X$ is a smooth complex projective variety, second singular cohomology $H^2(X,\mathbb{Z})$ has nontrivial torsion subgroup?

Answer 1

The universal coefficient theorem says that $H^2(X;\Bbb Z) \cong \text{Hom}(H_2(X;\Bbb Z),\Bbb Z) \oplus \text{Ext}(H_1(X;\Bbb Z),\Bbb Z)$. Because complex projective varieties are compact and triangulable, $H_2(X)$ and $H_1(X)$ are finitely generated, so $\text{Hom}(H_2(X);\Bbb Z),\Bbb Z)$ is torsion-free, and $\text{Ext}(H_1(X;\Bbb Z),\Bbb Z)$ is the torsion subgroup of $H_1(X;\Bbb Z)$.

So, equivalently, your question asks "Are there smooth projective complex varieties with torsion in $H_1(X;\Bbb Z)$?" And the answer is yes; take an Enriques surface, for instance, which has $\pi_1(E) = \Bbb Z/2$ because it has the simply-connected $K3$ surface as a double cover.

Answer 2

Feng, try this:

Consider a surface $\Sigma$ in $\mathbb{P}^3$. There is an action of $\mu_p$ (roots of unity) on $\mathbb{P}^3$ where the coordinate $z_i$ is sent to $\zeta^i z_i$ under the action of $\zeta \in \mu_p$. Assume $\Sigma$ is fixed by this action. Then $H^2(\Sigma/\mu_p,\mathbb{Z})_{tors} = \mathbb{Z}/p\mathbb{Z}$. (Recall that by Lefschetz theorem $\Sigma$ is simply connected.) Why? Consider a nontrivial character of $\mu_p$. This gives you descent data for descending the trivial line bundle on $\Sigma$ to $\mathcal{L}$ on $\Sigma/\mu_p$. This will have $p$-torsion Chern class $c_1(\mathcal{L})$ in $H^2$.