I was reading through Ravi Vakil's book/lecture notes on spectral sequences, but I came to an impasse. He leaves as an exercise the construction of the $d_2$ differentials of the spectral sequence (associated with a double complex). I know how to construct it using the big bad machinery of filtered complexes, but in the notes, he gives a hint that there is a more elementary proof that is pretty similar to the proof of the Snake lemma.

It's not clear to me how a diagram chase could really help though, since I don't see any short exact or even long exact sequences, which are used extensively in the proof of the Snake lemma.

Am I missing something? Could someone fill in some of the details?


1 Answer 1


Write $d = d^v + d^h$ for the vertical and the horizontal differential in the double cochain complex $C^{\bullet,\bullet}$. Take $x \in E_2^{pq}$ and lift it to an element $x' \in E_1^{pq}$. Then $d_1(x')=0$, which means precisely that if we lift $x'$ further to an element $x'' \in C^{pq}$ then $d^h(x'')$ is in the image of $d^v$, say $d^v(y) = d^h(x'')$. The element $d^h(y)$ vanishes under the vertical differential, so it gives an element of $E_1^{p+2,q-1}$, and then it also vanishes under the $E_1$-differential, so it gives an element of $E_2^{p+2,q-1}$. One can show that this element is independent of choices by a messy diagram chase.

  • $\begingroup$ Great! Just what I was looking for. Glad that I was on the right track at least. $\endgroup$ Aug 17, 2018 at 10:08
  • $\begingroup$ The conventions in this comment are opposite to Vakil's: Dan's $d_0$ is vertical, and the indices are the other way around. $\endgroup$
    – Lukas
    May 20, 2021 at 22:04

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