# Serre spectral sequence degeneration in homology vs cohomology

Let $$\pi\colon E \rightarrow B$$ be a fiber bundle with fiber $$F$$. I am not assuming that $$B$$ is simply-connected. We then have Serre spectral sequences in both rational homology and rational cohomology:

$$E^2_{pq} \cong H_p(B;H_q(F;\mathbb{Q})) \Longrightarrow H_{p+q}(E;\mathbb{Q})$$

$$E_2^{pq} \cong H^p(B;H^q(F;\mathbb{Q})) \Longrightarrow H^{p+q}(E;\mathbb{Q})$$

Since $$B$$ is not simply connected, the coefficient systems have to be regarded as local (twisted) coefficient systems on $$B$$. Assume that I know that one of these spectral sequences degenerates at the second page. Does it follow that the other one does as well?

• Don't you know this just from looking at the Betti numbers,? A spectral sequence degenerates if and only if the sum of all the Betti numbers is equal to the sum of all the $H^{p,q}$s. – Will Sawin Nov 16 '18 at 2:31
• The spectral sequences are dual to each other. – archipelago Nov 16 '18 at 8:56

By Will Sawin's comment, the claim will follow if we can show that $$\dim_\mathbb{Q} H^p(B;H^q(F;\mathbb{Q})) = \dim_\mathbb{Q} H_p(B;H_q(F;\mathbb{Q}))$$ for all $$p,q\ge0$$. The Universal Coefficient Theorem does not hold with twisted coefficients. However, note that the vector spaces on the right are by definition the homology groups of the chain complex $$C_*(\widetilde{B})\otimes_\pi H_q(F;\mathbb{Q}),$$ with differential induced by that of the chain complex $$C_*(\widetilde{B})$$ of the universal cover. Here $$\pi=\pi_1(B)$$ and we are taking tensor product of $$\mathbb{Z}\pi$$-modules.
Following archipelago's comment, let's see what happens if we dualize this chain complex by taking Hom into the rationals. By the tensor-hom adjunction there are natural isomorphisms $${\rm Hom}_\mathbb{Z}\big(C_*(\widetilde{B})\otimes_\pi H_q(F;\mathbb{Q}),\mathbb{Q}\big)\cong {\rm Hom}_\pi\big(C_*(\widetilde{B}),{\rm Hom}_\mathbb{Z}(H_q(F;\mathbb{Q}),\mathbb{Q})\big) \cong {\rm Hom}_\pi\big(C_*(\widetilde{B}), H^q(F;\mathbb{Q})\big),$$ and so we get precisely the cochain complex whose cohomology gives the vector spaces on the left. Since everything in sight is a rational vector space, the usual algebraic Universal Coefficient Theorem implies the claim.
• You're welcome. I just realized that my answer assumes that $E$ and $F$ are finite type. Maybe in the general case there's still scope for a counter-example. – Mark Grant Nov 18 '18 at 8:35