6
$\begingroup$

Betti numbers are a complete invariant of chain complexes of vector spaces modulo chain homotopy equivalence.

Can we similarly find complete invariants for (say, finite dimensional) filtered chain complexes? In particular, do the dimensions of the spaces appearing in the spectral sequence give a complete invariant?

$\endgroup$
  • 2
    $\begingroup$ Every chain complex of vector spaces is quasi-isomorphic to its homology, so homology is a complete invariant (every chain complex here is bi-fibrant so no issues worrying about chain homotopy equivalent versus quasi-isomorphic). $\endgroup$ – Connor Malin Jan 14 at 21:46
  • 2
    $\begingroup$ About the second question: You might want to add the assumptions that each chain complex is complete with respect to the filtrations. Otherwise you get the following counterexample. Let $C_*$ be $\bigoplus_\mathbb{N} k$ concentrated in degree $0$ and let $D_*$ similarly be the direct product, both equipped with the canonical decreasing filtrations given by the submodules where the first $k$-coordinates are zero. You get the same spectral sequences, but these two chain complexes cannot be homotopy equivalent, since $H_0$ is different. $\endgroup$ – HenrikRüping Jan 14 at 21:49
  • $\begingroup$ @ConnorMalin thanks, edited the question accordingly. $\endgroup$ – alesia Jan 14 at 21:55
  • $\begingroup$ @HenrikRüping thanks, added finiteness requirement to make things simple $\endgroup$ – alesia Jan 14 at 21:55

Your Answer

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

Browse other questions tagged or ask your own question.