# Spectral sequence of filtered complex.

I am trying to understand the construction of a spectral sequence of a filtered complex. After reading through the entry in the nLab I came up with an example, that I don't understand: Consider the double complex $$C_{p,q}$$

$$\require{AMScd}$$ $$\begin{CD} \mathbb{Z} @<5<< \mathbb{Z} @<<<0@<<<0\\ @V V V @VV 3 V@V V V @VV V\\ 0 @<<< \mathbb{Z} @<3<<\mathbb{Z}@<<<0\\ @V V V @VV V@V 4 V V @VV V\\ 0 @<<< 0 @<<<\mathbb{Z}@<4<<\mathbb{Z}\\ \end{CD}$$ (with $$C_{0,0}$$ being in the lower left corner). Then the total complex $$Tot(C)$$ is as usually given by summing the antidiagonals and it is filtered by the value of $$p$$. Following the construction from the nLab (using r-almost zycles and boundaries) I get, that the $$E^1$$ page has the form

$$\require{AMScd}$$ $$\begin{CD} \mathbb{Z} @<<< 0 @<<<0@<<<0\\ @. @. @. @. \\ 0 @<<< \mathbb{Z} @<<<0@<<<0\\ @. @. @. @. \\ 0 @<<< 0 @<<<\mathbb{Z}/4@<<<\mathbb{Z}\\ \end{CD}$$

And the $$E^2$$-page has the form

$$\require{AMScd}$$ $$\begin{CD} \mathbb{Z}/5 @. 0 @.0@.0\\ @. @.@. @.\\ 0 @. \mathbb{Z}/3 @.0@.0\\ @. @. @. @. \\ 0 @. 0 @. \mathbb{Z}/4@.0\\ \end{CD}$$

And althought the $$E^2$$ page clearly is the homology of the double complex it is not the homology of the $$E^1$$ page.

What am I missing?

• I don't understand how you get your $E^1$ page (I believe it should be a $\mathbb{Z/3Z}$ in the second line). But more importantly, I don't understand how you get your $E^2$ page. The kernel of $\mathbb{Z\to Z/4Z}$ is not zero and should be isomorphic to $\mathbb{Z}$ (in fact it is $\mathbb{Z}$ since the differential is zero). There is no non zero differential on the $E^2$ page, however a last differential should appear on the $E^3$ page. – Roland Apr 12 at 10:01