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The usual proof of the Kunneth formula (say for either the homology or cohomology of manifolds) is essentially pure homological algebra. I was wondering if there was a more geometric proof, i.e., one that does not go through Kunneth for the tensor product of complexes.

This is a bit vague, so here are two possible versions:

  • Given smooth manifold $X$ and $Y$ and a de Rham cohomology class $\alpha$ on $X \times Y$, is there some analytic way to find a closed form on $X \times Y$ representing $\alpha$ that is a sum of products of closed forms on $X$ and $Y$?

I would be interested even in the case of Kahler manifolds.

  • Given manifolds $X$ and $Y$ and an (integral) homology class $\beta \in H_*(X \times Y)$, is there a multiple of $\beta$ that is represented by a submanifold $M \subseteq X \times Y$, and a geometric procedure for degenerating $M$ to a sum of submanifolds that are products.

Taking a multiple of $\beta$ is necessary because general homology classes need not be represented by submanifolds, and it also deals with the issue that the Kunneth formula is more complicated with integer coefficients.

Such degenerations sometimes show up, e.g. when $X = Y$ is a toric variety and $M \subseteq \Delta(X)$, then one can degenerate $M$ by acting with a generic $1$-parameter subgroup.

My motivation is the Kunneth usually fails in algebraic geometry (say for Chow groups), and when it holds, the proof is very non-formal. I realize that the proof of Kunneth for topological spaces is not very hard, but it would be nice to have a more geometric proof.

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    $\begingroup$ The Künneth formula is formal for homology: since it holds for all kind of spaces, there is no reason to expect that we should need any impressive geometric argument for it. What is at stake for the Künneth formula in cohomology, is that it involves a finiteness property: it is deduced from the Künneth formula in homology by duality, and the duality operator is stronlgy monoidal only under finiteness hypothesis. The proof of finiteness theorems in cohomology usually requires non-formal arguments from geometry (e.g. constructing a finite CW-structure, or using resolution of singularities). $\endgroup$
    – D.-C. Cisinski
    Apr 5, 2021 at 21:07
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    $\begingroup$ There is at the least a Morse theory proof; "counting flowlines" is geometric to me. $\endgroup$
    – Chris Gerig
    Apr 5, 2021 at 21:59
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    $\begingroup$ Take a look at the proof of the Kunneth formula for compactly supported de Rham cohomology in Bott-Tu "Differential forms in algebraic topology" (around page 50). See if you find it sufficiently geometric. $\endgroup$
    – Moishe Kohan
    Apr 5, 2021 at 22:31
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    $\begingroup$ The proof in Bott-Tu is the most geometric argument that I've seen, but I still think that one could hope for something more explicit (especially in special cases, like for Kahler manifolds) $\endgroup$
    – Matt Larson
    Apr 6, 2021 at 17:43
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    $\begingroup$ One could presumably use Hodge theory, if $X$ and $Y$ are both compact. Fix Riemmannian metrics on $X$ and $Y$, and take the product metric on $X \times Y$. Then the Laplacians are related as $\nabla^2_{X \times Y} = \nabla^2_X + \nabla^2_Y$, so a form is harmonic on $X \times Y$ if and only if it is separately harmonic for $\nabla^2_X$ and $\nabla^2_Y$. Maybe it wouldn't be too hard to show that this implies that a harmonic form on $X \times Y$ is a linear combination of wedges of harmonic forms pulled back from $X$ and from $Y$? $\endgroup$
    – David E Speyer
    Apr 7, 2021 at 19:00

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