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A Künneth formula by Grothendieck/Schwartz states the following:

Let $A, B$ be chain complexes of nuclear Fréchet spaces. If the differentials $d_A, d_B$ are topological homomorphisms (meaning in this setting: if they have closed ranges), then we have the Künneth formula $$H(A \, \hat \otimes \, B ) \cong H(A) \, \hat \otimes \, H(B). \quad \quad\quad (*)$$

I would like to apply this statement to more than two product factors, and to do so, it would suffice that the range of the differential on the product space be closed again. Is there an easy way to see why/if this is true? I've tried to deduce it from the proof given in their paper, but somehow I don't see it, because the range of the product differential does not seem to come up explicitly.

On the one hand, the closed range property seems quite fickle and is in general not closed under linear combinations, and I fear that an expression like $d_A \otimes \text{id} + \text{id} \otimes d_B$ might be too general to hope for the closed range property again.

On the other hand, the RHS of $(*)$ is canonically a Fréchet space again, the LHS only if the range of the product differential is closed. If this isomorphism held, but the product differential range was not closed, I feel that would be quite strange...

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  • $\begingroup$ So are these chain complexes in the category of Frechet spaces with closed ranges (bounded linear operators)? $\endgroup$
    – Ben MacAdam
    Commented Mar 1, 2021 at 19:02
  • $\begingroup$ Yes, all graded components of $A$ and $B$ are nuclear Fréchet spaces, and the differentials between these Fréchet spaces are bounded linear operators with closed range. The homologies are simply the usual ones, so $\ker d_A / \text{im } d_A$, in particular we do not take the closure of the image. $\endgroup$
    – user126256
    Commented Mar 1, 2021 at 19:08
  • $\begingroup$ So if the RHS is canonically a Frechet space, and the LHS is a Frechet spaces iff the product differential is closed, and $LHS \cong RHS$, wouldn't that put a Frechet space structure on $LHS$ proving your result? $\endgroup$
    – Ben MacAdam
    Commented Mar 1, 2021 at 19:44
  • $\begingroup$ First of all (this relates to Ben MacAdam's comment), could you clarify whether the Kuenneth formula that you cite is asserting the at LHS and RHS are isomorphic as not-necessarily-Hausdorff TVS, or just isomorphic as abstract vector spaces? $\endgroup$
    – Yemon Choi
    Commented Mar 1, 2021 at 22:28
  • $\begingroup$ I suspect that if the result you want is true, it is going to require digging into the proof of the usual "Kuenneth isomorphism". There is a version of this in the Banach-space setting under some strong conditions, spelled out in a paper of Gourdeau--Lykova--White, who were precisely concerned to ensure that one could "iterate the Kunneth formula"; it may be the case that in the nuclear Frechet case enough things go right that their results still hold. (Some of their results are stated for Frechet spaces but I can't remember exactly which ones) $\endgroup$
    – Yemon Choi
    Commented Mar 1, 2021 at 22:32

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The discussion in the comments catapulted me onto the right track! It seems the solution is exactly to note that the isomorphism $H(A \, \hat \otimes \, B ) \cong H(A) \, \hat \otimes \, H(B)$ is not only an isomorphism of abstract vector spaces, but indeed an isomorphism of topological vector spaces, where all homology spaces $\ker d/\text{im }d$ are equipped with the respective quotient topologies. If this is done, then the right-hand-side is a Fréchet space, so the left-hand-side is, too, and the quotient topology of a Fréchet space by a subspace if Fréchet if and only if the subspace was closed.

Proving this takes a bit more effort than what is done in the Schwartz/Grothendieck paper, but the paper "A Künneth formula in topological homology and its applications to the simplicial cohomology of $ \ell^1 (\mathbb{Z}_+^{k}) $." by Gourdeau, Lykova and White, mentioned by Yemon Choi in the comments, deals with this. They shows the precise statement in Corollary 5.3, under the assumption that $A$ and $B$ are bounded from below as complexes.

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    $\begingroup$ Glad this reference was helpful! I apologize for only making comments rather than writing a proper answer $\endgroup$
    – Yemon Choi
    Commented Mar 2, 2021 at 16:20

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