# What conditions are sufficient for the Leray-Hirsch theorem to be a Künneth formula?

This was originally posted on MSE, and since it didn't receive much attention, I'll try here. Let me know if this is not the appropriate place.

Given a fiber bundle $$F \to E \to B$$ over a paracompact base $$B$$, assume its cohomology satisfies all the required properties for the Leray-Hirsch theorem to hold. This tells us that, as groups, $$H^n (E,R) \cong \bigoplus_{p+q=n} H^p (F,R) \otimes H^q (B, R).$$

But this does NOT tell us that, as rings, we have a ring isomorphism $$H^* (E,R) \cong H^* (F,R) \otimes H^* (B,R).$$

I'm wondering when this stronger isomorphism actually does hold. What extra structure on the fiber bundle is sufficient, without it being a trivial bundle?

I know of the classic counterexample with the bundle $$\mathbb{C}P^3$$ over $$S^4$$ with fiber $$S^2$$. I know that principle bundles can have the module isomorphism of Leray-Hirsch upgraded to the ring isomorphism. What else do we know? Any references on this subject would be highly appreciated.

EDIT: How about the following rather strong condition on our fiber bundle. Say both the integral cohomology of $$F$$ and $$B$$ lie in even degrees only, and that the bundle is torsion-free (one of the two cohomology rings are free and finitely generated, I believe is the criteria). Then, when applying the Serre spectral sequence, we would immediately get that it collapses for degree reasons. I only have a basic understanding of the SSS, but would this be enough to show that $$H^* (E, \mathbb{Z})$$ splits as a tensor products of the two other cohomology rings?

(Not really an answer, but too long for a comment, bear with me.) I'd say that those conditions will be pretty subtle and I doubt anything both universal and meaninful can be said. For example, let's restrict attention to sphere bundles which come as unit sphere bundles of vector ones of rank $$r$$, so we have fibration $$S^{r-1} \to E \to B$$.
If cohomology coefficients are 2-divisible, then top Stiefel-Whitney class determines $$H^*(E)$$. It's more or less folklore, usually attributet to R. Thom afair. But, if you look at integral cohomology, then you need top three Stiefel-Whitney classes to vanish for cohomology to split as a product. (I'm not going to write a proof here, because the one I'm able to produce is lengthy and is based on lengthy and totally unilluminating diagram chasing in spectral sequence associated to the universal bundle over $$BSO(r)$$ on cochain level)
• I see, so part of the subtlety lies in the chosen ring $R$. What about if $R$ is as well-behaved as we like, say a nice field of characteristic $0$? Jan 26 at 2:58
• Definitely not much better if your spaces are not simply connected. If you have a surface bundle over a circle $X \to E \to S^1$, $H^*(E) = H^*(X) \otimes H^*(S^1)$ additively if monodromy acts trivially on $H^*(X)$, i. e. lands into Torelli subgroup of mapping class group of the fiber. But as an algebra (even rationally!) $H^*(E)$ is a product only when monodromy lies in kernel of Johnson homomorpism, which is much smaller than Torelli subgroup. Jan 26 at 3:17