2
$\begingroup$

I am reading A User's Guide To Spectral Sequences and I don't understand the example in the informal introduction chapter:

We want to compute $H^*$, where $H^*$ is a graded $R$-module or a graded $k$-vector space [...] Suppose further that $H^*$ is filtered, that is, $H^*$ comes equipped with a sequence of sub objects $$ H^* \supset ... \supset F^n H^* \supset F^{n+1}H^* \supset ... \supset \{0\} $$ For the sake of clarity, let's assume for this chapter that $H^*$ is a graded vector space over a field $k$ and that $H^*=F^0 H^*$, that is, our filtration is bounded below by $H^*$ in the 0th filtration. For example [...] there is an obvious filtration, induced by the grading and given by $F^p H^*=\bigoplus_{n\geq p} H^n$.

[...]

A filtration of $H^*$, say $F^*$, can be collapsed into another graded vector space, called the associated graded vector space and defined by $E^p_0(H^*)=F^pH^*/F^{p+1}H^*$. In the case of a locally finite graded vector space (that is, $H^n$ if finite dimensional for each $n$), $H^*$ can be recovered up to isomorphism from the associated graded vector space by taking direct sums, that is $$ H\cong \bigoplus_{p=0}^\infty E^p_0(H^*) $$

I do not understand what is actually being computed here and from what. So, I see it, there is an unknown vector space $H^*$, what about $F^n H^*$ for $n>0$? Are these spaces given? Also, if we want to recover $H^*$ by taking direct sums as above, then don't we need to know what $H^*$ is, since the first summand is $\frac{F^0 H^*}{F^1 H^*}=\frac{H^*}{F^1 H^*}$, so we need to know that $H^*$ is?

$\endgroup$
1
$\begingroup$

In general, when using a spectral sequence, $H^*$ is the unknown piece. You then attempt to look at different parts of the spectral sequence. There will be groups $E_\infty^{pq}$, some of which may be easy to compute (the form of the spectral sequence gives you information on how to compute them). Each one is isomorphic to the quotient $F^pH^{p+q}/F^{p+1}H^{p+q}$. This is a piece of the graded vector space. You then look at different $E^{pq}_\infty$ groups to get different pieces of the graded vector space. For instance you could look at $E^{p+1,q-1}$. Each $E^{p+r,q-r}$ for $r\in\mathbb{Z}$ will give you a different piece of the same filtration quotient of $H^{p+q}$.

Hopefully, you can compute enough of the filtration via the $E_\infty$ groups to get some information about $H^*$, and sometimes you can completely recover $H^*$. This should become clear once you look at various examples.

$\endgroup$

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.