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 $n\geq 0$?
1 Answer
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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 $Y\times \mathbb CP^n$.
answered Oct 8, 2020 at 13:21
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$\begingroup$ what if $i$ is odd? $\endgroup$– user164740Oct 8, 2020 at 13:32
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6$\begingroup$ If $i$ is odd, take the product with a complex torus. $\endgroup$– abxOct 8, 2020 at 14:32
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3$\begingroup$ Then I guess it would be more efficient to take product with a complex torus in all cases. $\endgroup$– user164740Oct 8, 2020 at 17:13