0
$\begingroup$

this is exercise 5.1.1 in Weibel.

Suppose we have a double Complex $E_{\bullet,\bullet}$ with only the p and p-1 columns nonzero. Show that there is a SES:

0 $\rightarrow$ $E^2_{p-1,q+1}$ $\rightarrow$ $H_{p+q}(T=Tot_{\bullet}(E_{\bullet,\bullet}))$ $\rightarrow$ $E^2_{p,q}$ $\rightarrow$ 0

I don't understand what page $E_{\bullet,\bullet}$ is supposed to be on. I wrote out the chain for T but I don't see how its homology has to do with $E^2_{p-1,q+1}$, $E^2_{p,q}$.

Please Help!

1 Answers 1

1

$T$ is the total complex, that is $$T_n=\sum\limits_{p+q=n}E_{p,q}$$

Now $H_n(T)$ is filtered as $F_p(H_n(T))$. And the spectral sequence converges, in fact it stabilizes at the second page since all the maps are zero, cause of the assumption on the zero columns. This means that $$E^2_{p,q}=F_p(H_n(T))/F_{p-1}(H_n(T))$$ And the only nonzero ones are $E^2_{p-1,q+1}$ and $E_{p,q}^2$ so this implies that $F_{p-2}(H_n(T))=0$ and $F_{p}=H_n(T)$ thus your desired sequence is

$$0\rightarrow F_{p-1}(H_n(T))\rightarrow H_n(T)\rightarrow H_n(T)/F_{p-1}(H_n(T))\rightarrow 0$$

  • 0
    thanks!; can you quote the theorem that tells you a filtration that starts and ends at 0 and $H_n$ respectively exists? And by the way, this problem is given at the very beginning of the chapter before even spectral sequences are defined. Do you know of a way to find the SES using just the definition of $E^2_{p,q}$= $H_p(E^1_{\bullet,q})$ and $E^1_{p,q}$= $H_p(E^0_{p,\bullet})$ and the $T_{\bullet}$? – 2017-01-29
  • 0
    correct me if I'm wrong. We can construct the filtration explicitly using $F_p(T_n)$= $\oplus_{i+j=n,i>=p}E_{i,j}$ and Definition 12.21.5 from http://stacks.math.columbia.edu/tag/012K. $F_pH_n(T_{\bullet}) = Im(H_n(F_pT_{\bullet})→H_n(T_{\bullet}))$? – 2017-01-29