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?

  • $\begingroup$ 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. $\endgroup$ – Roland Apr 12 at 10:01

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