Collapsing spectral sequence

Question

In Weibel's definition of a homology spectral sequence collapsing at $E^r$ he says that if it converges to $H_*$ then $H_n$ is the unique non zero $E_{pq}^r$ such that $p+q=n$. I am trying to prove it but so far, I haven't been able to go very far, I am trying to use the fact that $E_{pq}^\infty\cong F_pH_n/F_{p-1}H_n$ but I haven't been really successful and I don't have any idea on how to see that. Does anyone have any help for me?

Answer 1

Okay I think I've managed to find the answer: Since $E_{p,q}^r\neq 0$ for only one index $p_0$ then consider a filtration $$0=F_sH_n \subseteq \ldots \subseteq F_{p_0}H_n \subseteq \ldots \subseteq F_tH_n=H_n.$$ By assumption for all $(p,q)\in\mathbb{Z}^2$ and $n=p+q$, we have $$E_{p,q}^\infty\cong F_pH_n/F_{p-1}H_n$$ Now this implies that $F_{s+1}H_n/F_sH_n\simeq E_{s,q}^\infty=0$ hence $F_{s+1}H_n=0$. By induction, we show that for $k\in[|s;p_0-1|]$, $F_kH_n=0$. Now we have $E_{p_0,q}^\infty\cong F_{p_0}H_n$ and $E_{p_0+1,q}^\infty\cong F_{p_0+1}H_n/F_{p_0}H_n$ hence $F_{p_0+1}H_n\simeq E_{p_0,q}^\infty$ and we show that it is the case for all $k\in[|p_0;t|]$. This shows that the term $F_tH_n=H_n\cong E_{p_0,q}^\infty$ which is said non zero term.

Answer 2

It seems that some authors use the verb "degenerate" to mean that the spectral sequence has no further nonzero differentials at that page or later, reserving "collapse" as given here; with these senses, "collapse" is stronger than "degenerate". Others use "collapse" to mean just the first, weaker, notion. To give one illustration, I think of nLab is pretty reliable, but see Examples 3.4 and 3.5 on the page https://ncatlab.org/nlab/show/spectral+sequence+of+a+filtered+complex. Since the term "collapse" is ambiguous, maybe we should all avoid it and/or spell out exactly what we mean when we use it. (In contrast, saying that a spectral sequence "degenerates at the $E_r$-page" is not, I think ambiguous.)