22
$\begingroup$

Suppose $F^m \to E^{m + n} \to B^n$ is a fiber bundle of closed oriented manifolds. I'm interested in understanding how the Serre spectral sequences for homology and cohomology of $E$ interact with each other via Poincaré duality. Specifically:

  • Does the Poincaré duality isomorphism for $E$ (as realized by the cap product with the fundamental class) respect the filtrations on $H_*(E)$ and $H^*(E)$ induced by the filtration on $E$ by the sub-bundles over the $p$-skeleta in $B$? Were this to be the case, I would expect that $F_*^p$ (in the homological filtration) would be dual to $F^{m+n-*}_{n-p}$ (in the cohomological filtration). Does this happen?
  • Amalgamating Poincaré dualities for $B$ and for $F$ will yield isomorphisms between the $E_2$-page and the $E^2$-page: $E_2^{p,q} \cong E^2_{n-p, m-q}$. Is this isomorphism induced from Poincaré duality on $E$ in some way?
  • Does the above isomorphism on $E_2$-pages commute with the differentials? Is there a Poincaré duality $E_\infty \cong E^\infty$ induced from the Poincaré duality on the $E_2$-page?

In a related question, Tyler Lawson explains how the homology and cohomology versions of the Serre spectral sequence (over a field) are dual to each other under the universal coefficient theorem. The key observation is that dualizing sends the exact couple giving rise to the homology SSS to the exact couple giving rise to the cohomology SSS. I'd like to be able to apply this principle in this setting, except that the pieces of the filtration on $E$ are not in general submanifolds of $E$ so I don't even know how to define a Poincaré duality map between these exact couples.

$\endgroup$
  • 3
    $\begingroup$ Two suggestions: (1) instead of using the skeletal filtration and exact couples to describe the Serre spectral sequence, use the fact that it's isomorphic to the Leray spectral sequence, or (2) if you insist on using the skeletal filtration, use Verdier duality on the skeleta. Of course either way you'll have to check or use all kinds of non-obvious functoriality and compatibilities, but I suspect you might have an easier time finding these in the literature. $\endgroup$ – Daniel Litt Nov 4 '15 at 22:29
6
$\begingroup$

After thinking about this for a little while, it seems like there is a satisfactory way of carrying this out in group cohomology using the Hochschild-Serre spectral sequence of a group extension $$1 \to H \to G \to Q \to 1$$ (henceforth we will call this the HSSS), and since my original motivating situation happened to be a fibration of aspherical manifolds, this will suffice for my purposes. I will refrain from selecting this as the answer, though, since I would still be curious to know how to carry this out in the topological setting.

For simplicity, I'll restrict attention to the trivial coefficient module $\mathbb Z$. Suppose that $H$ is a $PD_{m}$-group, $Q$ is a $PD_{n}$-group, and $G$ is a $PD_{m+n}$-group. Chapter VII of Brown's Cohomology of Groups contains a construction of the HSSS. In the homological setting, the idea is to show that there is a particular chain complex $C$ for which the homology with coefficients in the chain complex $H_*(Q, C)$ is isomorphic to $H_*(G)$; the HSSS is then derived from one of the filtrations on the bigraded chain complex underlying the definition of $H_*(Q, C)$.

In fact, letting $F$ be a projective resolution of $\mathbb Z$ over $\mathbb Z G$ (which can be assumed to be of length $m+n$), Brown explains why $C = F_H$ suffices. If we let $F'$ be a projective resolution of $\mathbb Z$ over $\mathbb Z Q$, the upshot is that the tensor product of chain complexes $F' \otimes_Q F_H$ computes $H_*(G)$.

Turning to cohomology, similar arguments show that the homology of the cochain complex $$ \operatorname{Hom}_Q(F', \operatorname{Hom}_H(F, \mathbb Z)) $$ is given by $H^*(G)$.

The crucial observation is that for a $PD_k$-group $\Gamma$, if $F$ is a projective resolution of finite length $k$, then $\overline F : = \operatorname{Hom}_\Gamma(F, \mathbb Z \Gamma)$ is also a projective resolution of $\mathbb Z$ over $\mathbb Z \Gamma$. (A technical point which explains how the shift in degrees arises is that one must monkey with the grading on $\overline F$ in the proper way).

We can apply tensor-Hom adjunction to see that $$ (*)\quad \operatorname{Hom}_Q(F', \operatorname{Hom}_H(F,\mathbb Z)) \cong \overline{F'} \otimes_Q \operatorname{Hom}_H(F,\mathbb Z) \cong \overline{F'} \otimes_Q (\operatorname{Hom}_H(F,\mathbb Z H) \otimes_H \mathbb Z). $$ As $Q$ is a $PD$-group by assumption, we may use $\overline{F'}$ as our projective resolution in the computation of $H_*(Q,F_H)$. I next claim that after an adjustment to the grading, (a portion of) the chain complex $\operatorname{Hom}_H(F,\mathbb Z)$ is weakly equivalent to $F_H$. This follows from the fact that $H$ is also a $PD$-group. More specifically, the assumption that $H$ is a $PD_m$ group is equivalent to the condition that $H^i(H, \mathbb Z H)$ vanish unless $i = m$, in which case it should be infinite cyclic. This says that the portion $$ \operatorname{Hom}_H(F_0, \mathbb Z H) \to \dots \to \operatorname{Hom}_H(F_{m-1}, \mathbb Z H) \to \operatorname{Hom}_H(F_{m}, \mathbb Z H) $$ of the chain complex $\operatorname{Hom}_H(F, \mathbb Z H)$ forms a projective resolution of $\mathbb Z$ over $\mathbb Z H$, from which the claim follows.

Paying attention to the bigradings, the $(p,q)$-component of the left-hand side in $(*)$ corresponds to the $(n - p, m-q)$-component on the right.

Now the properties I was looking for above will follow automatically from the spectral sequence machinery.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.