5
$\begingroup$

In Lecture Notes in Algebraic Topology by Davis & Kirk, on page 241, there is written:

enter image description here

What do $E^\infty$ and $\lim_{r\to\infty}E^r_{p,q}$ mean? If a spectral sequence is not first-quadrant and is not bounded (each diagonal having only finitely nonzero modules), such as the Leray-Serre-Atiyah-Hirzebruch spectral sequence, how is convergence defined? Shouldn't condition 1 be written so that we get an isomorphism $E^r_{p,q}\cong E^{r+1}_{p,q}$ instead of only a surjection?

$\endgroup$
4
$\begingroup$

$E^{\infty}$ is grid of modules (without any arrows between them). The only condition it needs to satisfy is that the elements along the diagonal $E_{p,n-p}$ (of of total sum /weight $n$) must be graded pieces of some filtration of an $A$-module.

The expression $E^{\infty}_{p,q}=\lim_{r \rightarrow \infty} E^{r}_{p,q}$ means that by condition 1, for every $(p,q)$ the module $E_{p,q}^{r}$ stabilizes (becomes identical) after finitely many $r$'s. The value of $E^{\infty}_{p,q}$ (module sitting at position $(p,q)$ in that grid) is that stable limiting module.

In spectral sequence language: $$E^{r+1}_{p,q} = \ker(d^{r}_{p,q}: E^{r}_{p,q} \rightarrow E^{r}_{p+r, q-r+1})/ \operatorname{im}(d^{r}_{p-r,q+r-1}: E^{r}_{p-r,q+r-1} \rightarrow E^{r}_{p,q}).$$ Now vanishing of $d^{r}_{p,q}$ depends only on $(p,q)$ so you can only claim surjection and not isomorphism. But that is not a problem because increasing $r_{0}$ (which you are allowed to do ) you will get the isomorphism, but that $r_{0}$ wouldn't have a clean description as the condition (1) above.

I am not sure what you mean by Leray spectral sequence not being bounded, however a spectral sequence needn't converge (and then it is usually useless).

$\endgroup$
  • 2
    $\begingroup$ Just a suggestion: you may want to look at the section on spectral sequences in Gelfand Manin's Methods in Homological Algebra for general discussion. $\endgroup$ – DBS Jul 9 '13 at 0:39
  • $\begingroup$ What does condition 1 have to do with $E^r_{p,q}$ becoming identical after finitely many steps? If $E^r_{p,q}$ is an arbitrary module (e.g. a $\mathbb{Z}$-module $\bigoplus_{k\in\mathbb{N}}\mathbb{Z}_{2^k}$), then taking its quotients might never stabilize, so $E^\infty_{p,q}$ might not exist. For example, the LSAH s.s. lies in the first and fourth quadrant, so its diagonals potentially have infinitely many nonzero modules. $\endgroup$ – Leon Jul 9 '13 at 11:55
  • $\begingroup$ If the differentials starting and terminating at point $(p,q)$ respectively vanish beyond some $r_{0}$ (as claimed) by the proposition then the homology at position $(p,q)$ will be the same for all higher $r$. The so called diagonals in a spectral sequence are lattice (integer) points on the line $x+ y =n$ (and not $x-y = n$, so if you have terms only in the first ans fourth quadrants, the diagonals are still finite. $\endgroup$ – DBS Jul 9 '13 at 18:01
  • $\begingroup$ Of course I am assuming a noetherian situation with all my modules finitely generated, so the situation you mention wouldn't arise. I do not think there is a good theory in the case of non-noetherian modules. Also the Leray spectral sequence in topology is retricted to first quadrant (we take one sided projective or injective resolutions) but if you could write the example that is troubling you... I can probably say something more useful then. $\endgroup$ – DBS Jul 9 '13 at 19:37
  • $\begingroup$ How can diagonals be finite? The $n$-th diagonal $E^2_{p,n-p}$ for $p\in\mathbb{N}$ is infinite when we have a 1st&4th quadrant s.s. What proposition are you talking about? The differential going out of $E^2_{p,q}$ will eventually be zero, but the differential goint into $E^2_{p,n-p}$ can be always nonzero, so we get a sequence of quotients $E^r_{p,q}$ for $r\!=\!2,3,\ldots$, which may never stabilize. $\endgroup$ – Leon Jul 9 '13 at 22:15
1
$\begingroup$

I was told by my professor that $\lim_{n\to\infty}E_{p,n-p}^r$ means the direct limit of the direct system of modules $(E_{p,n-p}^r)_{r=r_0}^\infty$ and quotient projection homomorphisms $E_{p,n-p}^r\rightarrow E_{p,n-p}^{r+1}$, which exists by condition 1.

Furthermore, for a 1st & 4th quadrant spectral sequence (such as the Leray-Serre-Atiyah-Hirzebruch spectral sequence), condition 1 is always satisfied, since for fixed $p$ and $q$, the homomorphism $d^r:E_{p,q}^r \rightarrow E_{p-r,q+r-1}^r$ will be $0$ for large enough $r$, because if $r > p$, then $p-r<0$ and $E_{p-r,q+r-1}^r=0$.

$\endgroup$

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.