0
$\begingroup$

I am wondering how the comparison theorem can be useful for resolving extension problem. Here is a quote from the book An Introduction to Homological Algebra by Weibel.

(Weibel) Comparison Theorem 5.2.12 Let {$E^r_{p,q}$} and {$E'^r_{p,q}$} converge to $H_*$ and $H_*'$, respectively. Suppose a map $h:H_*\rightarrow H'_*$ compatible with a morphism $f:E\rightarrow E'$ of spectral sequences. If $f^r$ is an isomorphism for all $p$ and $q$ and some $r$ (hence for $r=\infty$by the Mapping Lemma), then $h:H_*\rightarrow H'_*$ is an isomorphism.

And in the remark of the theorem, he said:

(Weibel) Remark: The same spectral sequence may converge to two different graded groups $H_*$, and it can be very difficult to reconstruct a picture of $H_*$ from this data. For example, knowing that a first quadrant spectral sequences has $E^\infty \simeq \mathbb{Z}/2$ for all $p$ and $q$ does not allow us to determine whether $H_3$ is $\mathbb{Z}/16$ or $\mathbb{Z}/2 \oplus \mathbb{Z}/8$. The Comparison Theorem 5.2.12 helps us reconstruct $H_*$ without the need for convergence.

I would like to know whether there is a good example whether we can use the comparison theorem to resolve extension problems. The only good example that I know is Borel's theorem. Is there any other good example? I have tried very hard to find it in the literature.

$\endgroup$
  • 1
    $\begingroup$ Whenever you have a map of filtered chain complexes that induces an isomorphism on (sufficiently high pages) of the spectral sequence, the chain map induces an isomorphism on homology. Then the hope is that one of your chain complexes is more computable than another either directly or by different methods (perhaps a different spectral sequence). But the actual resolving of extension problems happens by explicitly computing the homology some other way. $\endgroup$ – user98602 Oct 11 '18 at 22:41

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.