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?

  • 5
    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Nov 16, 2018 at 2:31
  • 2
    $\begingroup$ The spectral sequences are dual to each other. $\endgroup$ Nov 16, 2018 at 8:56

1 Answer 1


The answer is yes, and this follows from the comments, but it's worth spelling out the (non-obvious) details.

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.

  • $\begingroup$ 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. $\endgroup$
    – Mark Grant
    Nov 18, 2018 at 8:35
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    $\begingroup$ There is a version of the universal coefficient theorem for homology with twisted coefficients. You more or less gave a proof for it. $\endgroup$ Nov 18, 2018 at 15:44
  • $\begingroup$ @archipelago: You're right, I wasn't aware of this but now I see it is an exercise in Spanier's book (on page 283). Do you know another reference for this? $\endgroup$
    – Mark Grant
    Nov 18, 2018 at 17:08
  • $\begingroup$ @MarkGrant: I hadn't noticed the finite-type assumption, but thankfully that condition is satisfied in all the examples I care about! $\endgroup$
    – Laura
    Nov 19, 2018 at 16:37
  • $\begingroup$ Just a comment about the twisted coefficient UCT: see Maxime Ramzi's A note on the universal coefficient theorem with twisted coefficients sites.google.com/view/maxime-ramzi-en/notes/twisted-uct $\endgroup$
    – David Roberts
    Sep 29, 2023 at 3:48

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