Spectral sequence with two non zero rows

Question

I'm trying to solve exercise 5.2.2 of Weibel's Introduction to homological algebra : if a spectral converging to $H$ has $E_{p,q}^2=0$ expect for $q=0,1$ then there is a long exact sequence $$\cdots \to H_{p+1} \to E_{p+1,0}^2 \overset{d}{\to} E_{p-1,1}^2 \to H_p \to E_{p,0}^2 \overset{d}{\to} E_{p-2,1}^2 \to H_{p-1} \to \cdots$$ Let $p\in\mathbb{N}$, since $\{E_{p,q}^r\}$ converges to $H$ there is a finite filtration $0=F_{-1}H_p \subseteq \ldots \subseteq F_pH_p=H_p$ and we have the following short exact sequences $$0 \to F_{p-1}H_p \to H_p \to E_{p,0}^2 \to 0$$ $$0 \to F_{p-2}H_p \to F_{p-1}H_p \to E_{p-1,1}^2 \to 0$$ $$0 \to \ker(d_{p,0}^2) \to E_{p,0}^2 \overset{d}{\to} E_{p-2,1}^2 \to 0$$ My problem is that I don't know if the fact that $F_{p-1}H_p$ surjects into $E_{p-1,1}^2$ is enough to conclude and if not, I can't find any short exact sequence of the form $$0 \to K_1 \to E_{p-1,1}^2 \to K_2 \to 0$$ Can anyone help me please?

Answer 1

Ultimately, the filtration comes from the $E^\infty$ page. So let's attempt to calculate explicitly what this page is. Well, from the $E^2$ page we can only immediately infer the $E^3$ page - so let's focus on that first.

Obviously if $q\neq 0,1$ we have that $E^3_{p,q}=0$; the only subquotient of $0$ is $0$.

If $q=0$, $E^3_{p,0}$ is the quotient of $\ker(E^2_{p,0}\to E^2_{p-2,1})$ by $\mathrm{im}(E^2_{p+2,-1}=0\to E^2_{p,0})=0$ i.e. it is just $\ker(E^2_{p,0}\to E^2_{p-2,1})$.

If $q=1$, $E^3_{p,1}$ is the quotient of $\ker(E^2_{p,1}\to0=E^2_{p-2,2})=E^2_{p,1}$ by $\mathrm{im}(E^2_{p+2,0}\to E^2_{p,1})$ i.e. just $E^2_{p,1}/\mathrm{im}(E^2_{p+2,0}\to E^2_{p,1})$.

Putting it all together, this tells us there are exact sequences: $$0\to E^3_{p+2,0}\to E^2_{p+2,0}\to E^2_{p,1}\to E^3_{p,1}\to0$$

Can we calculate the $E^4$ page? Yes, and it's actually very easy. The $E^3$-differentials have vertical grading $2$, and we know $E^3$ vanishes outside of the $0,1$ rows so all differentials are necessarily zero. Think about it; if I start with some $E^3_{p,0}$, I'm mapping to $E^3_{p-3,2}=0$ and if I start with $E^3_{p,1}$, I'm mapping to zero and I'm also being mapped to by $E^3_{p+3,-1}=0$... everything is zero. Thus $E^4=E^3$ and indeed $E^3=E^4=E^5=E^6=\cdots=E^\infty$ by the same arguments.

The filtration quotients $F_pH_{p+q}/F_{p-1}H_{p+q}\cong E^3_{p,q}$ vanish for $q\neq 0,1$ so the only nontrivial ones are $F_pH_p/F_{p-1}H_p$ and $F_pH_{p+1}/F_{p-1}H_{p+1}$; if we fix some $n$ then that means $F_nH_n/F_{n-1}H_n\cong E^3_{n,0}$ and $F_{n-1}H_n/F_{n-2}H_n\cong E^3_{n-1,1}$ and all other quotients in total degree $n$ vanish. If you have a bounded below filtration, as you seem to do in the question, then $F_{n-m}=0$ for large enough $m$ and we inductively infer from $F_r/F_{r-1}=0$ for $r\le n-2$ that $F_{n-2}=0$ as well. If you have a bounded above filtration you likewise get to infer $F_nH_n=H_n$. I should note that this is true, and the long exact sequence we find at the end is correct, for arbitrary bounded (above and below) filtrations and spectral sequences concentrated in rows $q=0,1$ but not necessarily in the first quadrant. We don't need a - what Weibel calls - "canonically bounded" sequence and filtration, just a bounded one.

Thus there are short exact sequences: $$F_{n-2}H_n=0\to E^3_{n-1,1}\cong F_{n-1}H_n/F_{n-2}H_n\cong F_{n-1}H_n\to F_nH_n=H_n\to F_nH_n/F_{n-1}H_n\cong E^3_{n,0}\to0$$Less messily but without explanations, there is an exact sequence: $$0\to E^3_{n-1,1}\to H_n\to E^3_{n,0}\to0$$

Now we can splice that with the sequence relating $E^3$ with $E^2$: $H_n$ surjects onto $E^3_{n,0}$ which injects into $E^2_{n,0}$ as a kernel, $E^2_{n-1,1}$ surjects onto $E^3_{n-1,1}$ (a quotient) which injects into $H_n$ and we can eliminate the $E^3$ terms without affecting exactness. Overall we can obtain a long exact sequence: $$\cdots\to E^2_{n-1,1}\twoheadrightarrow E^3_{n-1,1}\hookrightarrow H_n\twoheadrightarrow E^3_{n,0}\hookrightarrow E^2_{n,0}\to E^2_{n-2,1}\twoheadrightarrow E^3_{n-2,1}\hookrightarrow\cdots$$But even better we can throw away the $E^3$, leaving them implicitly as the images and kernels in question and get some: $$\cdots\to H_{n+1}\to E^2_{n+1,0}\to E^2_{n-1,1}\to H_n\to E^2_{n,0}\to E^2_{n-2,1}\to H_{n-1}\to E^2_{n-1,0}\to E^2_{n-3,1}\to H_{n-2}\to\cdots$$Make sure you believe why exactness of this last sequence follows from exactness of the previous sequence.