# Arbitrary torsion in cohomology of Kähler manifolds

Fix an integer $$i\geq 3$$ and a finite abelian group $$G$$.

Is there a connected closed Kähler manifold $$M$$ such that $$H^i(M, \mathbb{Z})\approx \mathbb{Z}^n\oplus G$$ for some integer $$n\geq 0$$?

• For context, there is a positive answer for $i = 2$, but not for $i = 1$ as $H^1$ is always free abelian. Oct 8 '20 at 12:24

The answer is positive and can be deduced from Proposition 15 of "Sur la topologie des varietes algebriques en characteristique p" by Serre. According to this proposition for any finite group $$G$$ there exists a complete intersection $$X$$ on which $$G$$ is acting freely. Set $$Y=X/G$$. Then $$\pi_1(Y)=G$$. Let now $$G$$ be your abelian group. Then $$\pi_1(Y)\cong G$$ and $$H_1(Y,\mathbb Z)\cong G$$. It follows that $$H^2(Y,\mathbb Z)\cong \mathbb Z\oplus G$$. To get torsion $$G$$ is higher cohomologies take the product $$Y\times \mathbb CP^n$$.
• what if $i$ is odd?
• If $i$ is odd, take the product with a complex torus.