Compact complex submanifold $M \subset N$ which induces the zero map $H^2(N, \mathbb Q) \to H^2(M, \mathbb Q)$

Asked Apr 28 '25 at 07:54

Active Apr 28 '25 at 07:54

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Let $M \subset N$ be compact complex manifolds with $\dim M \geq 1$.

If $N$ is Kähler, one common argument is that the restriction map on second cohomology $$H^2(N, \mathbb R) \to H^2(M, \mathbb R)$$ is non-zero, because a Kähler class on $N$ will restrict to a Kähler class on $M$, which is necessarily non-zero.

I was wondering if the restriction map can be zero if $N$ is not a Kähler manifold? What is a good example? An example where the restriction is zero in cohomology with $\mathbb Z$-coefficients would be ideal.

asked Apr 28 '25 at 07:54

red_trumpet

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There are cases that $H^2(N)$ is zero, for example when $N$ is the hopf surface. – Arctic Char Apr 28 '25 at 08:02
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For an example with nonzero $H^2$(N)$, it seems that a blow up of the hopf surfaces at a point would work. – Arctic Char Apr 28 '25 at 08:05
@ArcticChar: This should be an answer. – Moishe Kohan Apr 28 '25 at 09:37
Related question. – Michael Albanese Apr 28 '25 at 10:18

Compact complex submanifold $M \subset N$ which induces the zero map $H^2(N, \mathbb Q) \to H^2(M, \mathbb Q)$

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