# Cohomology of augmented double complex with exact rows using spectral sequences

I am learning about spectral sequences through Vakil's notes and wanted to try out a simple spectral sequence argument on a double complex. I recalled this little paragraph in Bott & Tu, p.97:

If all the rows of an augmented double complex are exact, then the total cohomology of the complex is isomorphic to the cohomology of the initial column.

(In their case this applies to the Čech-de Rham complex.) I think this is amenable to a proof via spectral sequences.

So the situation is that we have some double complex $C^{p,q}$ with $p \geq -1$ and $q \geq 0$. (The column $p = -1$ is the augmented one.) Since all the rows are exact, the first page using the horizontal spectral sequences already vanishes. Now we should compare this to the vertical spectral sequence, but this is where I get stuck. Since we have a first quadrant spectral sequence, all the terms will eventually stabilize to zero, but it seems that this might take some time very far into the double complex. Even for the very first terms, for instance, I find that $E_2^{-1,0} = 0$, but this only tells me that the $0$th cohomology of the augmented column injects into $H^{0,0}(C)$, not even that this is an isomorphism.

Can you help me finish this argument?

They could have made it clearer, but I think that Bott and Tu are making a distinction between "the double complex" and the "augmented double complex" with an extra initial column.

So in terms of spectral sequences, all they're saying is that if the rows of the augmented double complex are exact, then the cohomology of the rows of the unaugmented double complex just gives the initial column of the augmented complex. So the "rows first" spectral sequence for the unaugmented complex just degenerates at the $E_2$ page to the cohomology of the initial (augmenting) column.

The "columns first" spectral sequence isn't used here. All you need is that for a first quadrant double complex, either of the two spectral sequences converges to the cohomology of the total complex.

• Thanks, you are right. Actually, I just realized that the authors discuss this exact issue later in the book (page 166).
– anon
Mar 23, 2018 at 13:02

I'm currently studying that part of the book Differential Forms in Algebraic Topology and I think that the claim you are citing is not completely correct. To be more clear...Look at the the Lemma $4$ of these notes. You can easly prove the lemma by using spectral sequences. Indeed, take the s.s. ${}^IE^{pq}$ obtained by filtring the rows of the double complex, then you see that ${}^IE^{pq}$ collapses at the first sheet, i.e. the cohomology of the of the first non-exact row is isomorphic to the cohomology of the total complex. Bott and Tu want show that $H^\bullet_{dR}(M)\simeq H^\bullet(\mathcal{U},\mathbb{R})$, see the figure below.

By applying the Corollary $6$ of the notes what you get is exactly that Bott and Tu want check!

Notice that the topological assumption to apply the corollary is that $\mathcal{U}$ is a good cover, i.e. (algebraically) the columns eccept $\Omega^\bullet (M)$ are exact.

• Which part of the claim do you think is incorrect, precisely? Also, in Lemma 4 of the notes you linked, why is there non-vanishing horizontal cohomology at all? I was under the impression that the rows are exact everywhere (by Mayer-Vietoris), so that there is absolutely no horizontal cohomology.
– anon
Dec 5, 2017 at 21:50