# Convergence of right half-plane spectral sequence bounded on the right

This is a sequel to my previous question colimits of spectral sequences .

I think I've found the answer in S.A. Mitchell's paper "Hypercohomology spectra and Thomason's descent theorem". There the author states a "colimit lemma" (page 42) for ss of homotopy groups of spectra, which I think can be literally translated with no harm for cohomology groups of cochain complexes and it's exactly the result I was looking for.

However, my problem is more basic and shameful. Previously, but in the same page, Mitchell says that, having a right half-plane cohomology spectral sequence (coming, say, from a double complex, or a filtered complex), that is

$$E_2^{pq} = 0 \qquad \mbox{if} \quad p < 0 \ , \qquad \qquad \qquad [1]$$

then it converges if it is, for instance, bounded on the right, that is, if there exists $d$ such that

$$E_2^{pq} = 0 \qquad \mbox{if} \quad p > d . \qquad \qquad \qquad [2]$$

I assume the author uses the term "converge" in the sense of Cartan-Eilenberg (otherwise, I don't understand the proof of his "colimit lemma" at all), that is:

(a) We have an exhaustive filtration $F$ on the limit $G$, $\bigcup_p F^PG = G$, and also isomorphisms $E_\infty^p = F^pG / F^{p+1}G$, and

(b) The filtration on $G$ is Hausdorff, that is $\bigcap_p F^pG = 0$.

Now, I have no problem in assuming that in my particular ss the original filtration of my filtered complex is already exhaustive and hence so it is the induced one on the "limit" $G$ and the isomorphisms for $E_\infty$ (as it is the case for both filtrations of a double complex) and I think Mitchell is assuming this too implicitely, because these conditions seems "for free".

Also, the boundness conditions [1] and [2] will imply that the filtration on $G$ is in fact finite. So, if we had (b), the converge would be in the strong sense. Great! :-)

So the point is the Hausdorff condition of the filtration on $G$: why would [1] and [2] imply that the filtration on $G$ should also be Hausdorff?

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## 1 Answer

It doesn't -- just filter any chain complex trivially, the resulting spectral sequence has vanishing $E_2$ and obviously doesn't converge in any sense. So you have to look at how the spectral sequence is constructed. A very useful notion in this context is conditional convergence, which is probably satisfied in the example you are considering. In that case, if the derived functor of the $E_\infty$ term vanishes, the spectral sequence converges strongly -- in particular, this happens if the filtration is finite. I'm willing to bet that the spectral sequence you're looking at is conditionally convergent. For this notion, and other convergence questions about spectral sequences, I highly recommend Boardman's paper "Conditionally convergent spectral sequences."

Unfortunately, a lot of the literature is quite sloppy on convergence questions for spectral sequences...

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That paper by boardman is fantastic! Anytime i hear a question about convergence i think of that paper. –  Sean Tilson Jul 21 '10 at 17:14
@Tilman. Thank you very much. It seems likely that "my" particular ss will be conditionally convergent -whenever I understand the definition. :-) So, you should be right: together with the finiteness of the filtration, it will turn out to be strongly convergent, which is what I actually need. It's funny, because if this is true, it would be an example of a problem of ss that can't be tackled just with the "classical" -pre-Boardman- notions of convergence, but needs the "conditional" one. Amazing! (At least for me.) –  a.r. Jul 22 '10 at 7:44