# Cohomology of derived tensor product of complexes and Künneth spectral sequence

Let $$R$$ be any commutative ring, let $$V^\bullet$$ and $$W^\bullet$$ be (co)chain complexes of $$R$$-modules, indexed cohomologically. We can also assume that they have both cohomology in nonpositive degrees. Using K-flat resolutions we can define the derived tensor product:

$$V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet.$$ I'm looking for assumptions on $$V^\bullet$$ or $$W^\bullet$$ that ensure that the following "Künneth formula" holds: $$H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet),$$ where the right hand side is the tensor product of graded $$R$$-modules. Searching the literature, it seems that there should be some spectral sequence involving Tor of the cohomologies, as mentioned for instance in the nLab entry.

• Is there a more precise reference for such a result?
• Looking at these Künneth formulas and the spectral sequence, I suspect that the following is true: $$H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet)$$ holds if $$V^\bullet$$ or $$W^\bullet$$ has flat cohomologies, namely $$H^k(V^\bullet)$$ (say) is a flat $$R$$-module for all $$k$$. Is it correct?

A little more wildly, could one expect that without assumptions one has $$H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \overset{\mathbb L}{\otimes}_R H^*(W^\bullet),$$ so taking K-flat resolutions of the cohomology graded $$R$$-module if necessary?

This isn't a complete answer, but here are some thoughts. I'm sceptical that it is often true.

If $$R$$ is semisimple then the result holds, because every module over a semisimple ring is projective.

We may as well take $$V$$ and $$W$$ to be complexes of projectives, and then ask when is $$H(V\otimes W)\cong H(V)\otimes H(W)$$? If $$R$$ is a field this is the usual Kuenneth theorem.

When $$V$$ has flat cohomology the spectral sequence is not hard to obtain: let $$V' \to V$$ and $$W' \to W$$ be flat resolutions; because both $$V$$ and $$W$$ are cohomologically bounded above we can take $$V'$$ and $$W'$$ to be bounded above. The direct sum total complex of the double complex $$V' \otimes W'$$ computes the Tor groups. Taking cohomology in the vertical direction and using flatness of $$W'$$ tells us that the $$E_1$$ page of the associated spectral sequence is $$H(V)\otimes W'$$. Now taking cohomology in the horizontal direction and using flatness of $$H(V)$$ tells us that the $$E_2$$ page is $$H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$$, as desired.

The same argument when $$V$$ does not necessarily have flat cohomology gives a spectral sequence with $$E_2$$ page $$\mathrm{Tor}(HV,W)\Rightarrow \mathrm{Tor}(V,W)$$. In general, Tor spectral sequences tend to look like this (as in e.g. https://stacks.math.columbia.edu/tag/061Y) - a spectral sequence of the form $$H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$$ will not exist without this flatness assumption on cohomology (just think about when $$V$$ and $$W$$ are genuine $$R$$-modules!). Of course, when $$V$$ has flat cohomology then the tensor product $$H(V)\otimes H(W)$$ is already the derived tensor product.

However, this doesn't answer your main question in the flat cohomology case: you have a spectral sequence, and you are interested in knowing when it collapses (or more generally when the $$E_2$$ page equals the $$E_\infty$$ page).

As for your wild question, when $$V$$ and $$W$$ are genuine modules, you are asking that $$V\otimes^\mathbb{L}W$$ be a formal complex. This is true in certain situations: for example the HKR theorem is true on the level of cochains and gives a quasi-isomorphism between the Hochschild complex and the graded module of polyvector fields. Or if $$R$$ is a ring, $$r_1,\ldots, r_n$$ is a (finite) regular sequence, $$K$$ the Koszul complex, and $$M$$ an $$R$$-module annihilated by all of the $$r_i$$ then you can easily check that $$R/(r_1,\ldots, r_n)\otimes^\mathbb{L}_R M \simeq K\otimes_R M$$ is formal. In general it won't be true, but I can't think of an example off the top of my head.