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For any compact Kahler manifold $X$, we have a decomposition $$H^n(X,\mathbb{C})\cong \bigoplus_{i+j=n}H^{i,j}(X).$$ The cohomology of spaces is a homotopy invariant. Then it seems natural to ask if the decomposition of cohomology is also a homotopy invariant. That is to say, if we denote by $X,Y$ two homotopy equivalent compact Kahler manifolds, do we have $$H^{i,j}(X)\cong H^{i,j}(Y)?$$ The reason I have hope that this might be true is that any homotopy between smooth manifolds can in fact be assumed to be smooth.

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    Nope, it's not even diffeomorphism invariant: https://math.stackexchange.com/questions/279837/dolbeault-cohomology-is-invariant-under-homeomorphisms – Qiaochu Yuan Aug 07 '20 at 22:10

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