# What is the scope of validity of Kunneth formula for de Rham?

In books like Bott-Tu or all pdf texts I have found on internet, the Kunneth formula for manifolds $M$ and $N$ and their de Rham cohomology $$H^{\bullet}_{dR}(M \times N) \simeq H^{\bullet}_{dR}(M) \otimes H^{\bullet}_{dR}(N)$$ is proved under various finiteness hypothesis : one of the two manifolds is compact, or with finite dimensional de Rham spaces, or admitting a finite good cover.

On another side, for singular cohomology of topological spaces $X$ and $Y$ and a PID $A$ (let's say $Z$) there is the following more general Kunneth formula : $$H^n(X \times Y;A) \simeq (\sum_{i+j=n} H^i(X;A)\otimes H^j(Y;A))\oplus(\sum_{p+q=n+1} Tor(H^p(X;A),H^q(Y;A))$$

If $X$ and $Y$ are manifolds and we take $A \equiv \mathbb{R}$, the $H^p(X;\mathbb{R})$ ans $H^q(Y;\mathbb{R})$ are vector spaces, so the $Tor$ part is null, and using the de Rham theorem we end up with a Kunneth formula for de Rham without any finitenes hypothesis.

But Bott-Tu p 108 give an explicit counterexample to the Kunneth formula when both manifolds have infinite dimensional cohomology, and write "that some sort of finiteness hypothesis is necessary for Kunneth and Leray-Hirsch to hold".

So what is wrong with the Kunneth formula for de Rham "deduced" from kunneth for singular above ?

And what is the real scope of Kunneth formula for de Rham i.e. the minimal hypothesis for the formula to hold ?

• I think the assumptions you mention are simply there to provide a simpler proof. If it's true for singular cohomology, then it's true for de Rham cohomology, by de Rham's theorem. I'm pretty sure the most general statement you can get for de Rham's cohomology is that you just want $H^k(Y)$ to be finite-dimensional for all $k$. CW is not even necessary (thanks to CW approximations), and "free" is automatic over a field (well, if you assume the axiom of choice I guess). – Najib Idrissi Apr 4 '18 at 15:25

I think the assumptions (good cover / compact) you mention are simply there to provide a simpler proof. If it's true for singular cohomology, then it's true for de Rham cohomology, by de Rham's theorem. I'm pretty sure the most general statement you can get for de Rham's cohomology is that you just want $H^k(Y)$ to be finite-dimensional for all $k$. CW is not even necessary (thanks to CW approximations), and "free" is automatic over a field (well, if you assume the axiom of choice I guess).
There is a Künneth formula for homology, which has no assumptions about finiteness, but if you pass to cohomology without finiteness assumptions, you run into problem. The actual statement for cohomology is $H^n(X \times Y; F) \cong \bigoplus_{i+j=n} H^i(X; H^j(Y;F))$ for a field $F$. But if $H^j(Y;F)$ is not finitely generated, you don't necessarily have $H^i(X; H^j(Y;F)) \cong H^i(X;F) \otimes_F H^j(Y;F)$. So there is no contradiction here.
• @ychemama Yes, that formula is wrong without finiteness hypotheses. The counterexample in the book of Bott and Tu shows it. Probably the best way to go is to try and prove the formula from the (correct) formula for homology: you will quickly see what goes wrong. It's also in Hatcher's book, Chapter 3.B. As I said, the key issue is the difference between $H^i(X; H^j(Y))$ and $H^i(X) \otimes H^j(Y)$, which in turns come from the difference between $\hom(U, V \otimes W)$ and $\hom(U, V) \otimes W$ when you don't have finite dimensional vector spaces. – Najib Idrissi Apr 6 '18 at 14:48
• Is your formula correct even if $F$ is only assumed to be a PID? Years ago, I saw this problem on Bott & Tu, but now I still cannot decide whether it is correct. Generally if we have a ring spectrum $R$ and a right $R$-module $M$ and a left $R$-module $N$, we have a spectral sequence $\pi_*M\otimes_{\pi_*R}\pi_*N\implies\pi_*(M\otimes_RN)$, but I don't know when it splits. Also a version for mapping spectra, but I don't know how to make it precise. – Yai0Phah Nov 27 '18 at 13:32