For a (bounded) double complex (of abelian groups or vector spaces) one can consider two spectral sequences that converge to the cohomology of the totalization: one can first compute either the cohomology of rows, or the cohomology of columns. Suppose that one of these spectral sequences degenerates at $E_1$ (i.e. the cohomology of rows yields the factors of the induced filtration of the limit). Do any 'nice' properties of the second spectral sequence follow?

1$\begingroup$ Why the ktheory tag? $\endgroup$– Dylan WilsonJan 29, 2012 at 9:52

2$\begingroup$ And what is meant by nice? If you're asking whether the second spectral sequence has to degenerate quickly, I'm pretty sure one can cook up, for each $r$, fairly simple complexes where one spectral sequence degenerates immediately, and the other lasts til $E_r$. Take two exact sequences and place them far apart, maybe? $\endgroup$– Dylan WilsonJan 29, 2012 at 9:57

2$\begingroup$ Is there an arxiv tag 'homology' without ktheory?:) $\endgroup$– Mikhail BondarkoJan 29, 2012 at 18:45
2 Answers
There is a basic way to see whether things like this should be true. Any bounded double complex of vector spaces over a field $k$ is (noncanonically) the direct sum of complexes of the following two sorts:
Squares: $$\begin{matrix} k & \rightarrow & k \\ \uparrow & & \uparrow \\ k & \rightarrow & k \end{matrix}$$
Staircases:$$\begin{matrix} k & \rightarrow & k & & & & \\ & & \uparrow & & & & \\ & & k & \rightarrow & k & & \\ & & & & \uparrow & & \\ & & & & k & \rightarrow & k\\ \end{matrix}$$ We'll say that the "length" of a staircase is the number of nonzero entries in it, so the above stair case has length $6$. Staircases may have even odd or even length, and may start and end either with vertical or horizontal maps.
The operation of "forming the spectral sequence" commutes with direct sum, so it is enough to know what the spectral sequences of each of these look like.
The spectral sequence of a square is zero on every page except for the square itself; in particular, it converges in one step.
The spectral sequence of an odd length staircase is one dimensional on every page except for the stair case itself. The one nonzero term is at one end of the staircase for the horizontal spectral sequence and the other for the vertical spectral sequence. So it also converges in one page.
The spectral sequence of an even staircase is the one which violates your claim. In one direction, it is zero on every page after the staircase itself. In the other direction, it there are $m$ pages with two nonzero terms, where the length of the staircase was $2m$. These terms are at the two ends of the stair case, and they annihilate each other on the $(m+1)$st page.
In particular, a double complex which consists simply of a length $2m$ staircase will die on the first page in one direction, but will survive for $m$ pages in the other.

$\begingroup$ Thank you very much for such a simple and clear explanation!! $\endgroup$ Jan 29, 2012 at 18:47
Let $E,F$ be the two spectral sequences of the double complex and for simplicity assume they are in the first quadrant. If $E_1$ degenerates, say, $E_1^{i,j}=0$ for $j>0$ then you know that $F$ converges to $E_2$, i.e. $F_2^{\;i,j} \Rightarrow E_2^{i+j,0}$.
In general, I don't think much can be said about the properties of the second spectral sequence. As an example consider the LHS spectral sequence of a group extension $1 \to H \to G \to Q \to 1$. The double complex is $$Hom_{kQ}(X,Hom_{kH}(Y,M))$$ where $k$ is a ring, $M$ is a $kG$module and $X,Y$ are projective resolutions. Now we have $$E_1^{i,j} = H^iHom_{kQ}^\ast(X,Hom_{kH}^j(Y,M))=H^i(Q;Hom_{kH}^j(Y,M))=0 \text{ if } j>0.$$ Thus $E_1$ degenerates, but the second spectral sequence $$F_2^{\; i,j} = H^i(Q;H^j(H;M)) \Rightarrow H^{i+j}(G;M)$$ can be arbitrarily complicated.