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Consider a double complex $C^{*,*}$, where each $C^{*,*}$ are bigraded modules over $R$ (in fact they are vector spaces). Let the horizontal and vertical differentials be $d_1,d_2$, so that $d_1,d_2$ have bidegrees $(1,0),(0,1)$ respectively and that $\{d_1,d_2\}=0$. I am interested in the cohomology of the total complex $H(C^{*,*},d_1+d_2)$.

This double complex that I got is unbounded, and that $H^i(C^{*,*},d_1)$ is non-zero only at degree (0,0). On the other hand $H^i(C^{*,*},d_2)$ is zero everywhere.

If I understand correctly what this is saying is that the spectral sequence associated to the bicomplex $(C^{*,*},d_1,d_2)$ stabilizes on page 1 and is non-zero at (0,0). But the spectral sequence associated to $(C^{*,*},d_2,d_1)$ vanishes on page 1. Is this a contradiction?

Could it be that only one of these spectral sequence converges to the total complex $H(C^{*,*},d_1+d_2)$?

Any references on this would be highly appreciated. Thanks!

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With unbounded double complexes, convergence (in the usual sense, as compared to conditional convergence) is hard to determine. So it is possible that neither spectral sequence converges to the cohomology of the total complex. One standard reference for spectral sequences is Boardman's paper "Conditionally convergent spectral sequences". See section 10 for double complexes.

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What you are saying is that the horizontal spectral sequence stabilizes on page 1 and the vertical one vanishes on page 1. (Therefore both spectrals are at infinity on page 1.) Although your double complex is unbounded is it bounded below? If yes, you have convergence on one of these spectrals (both are exhaustive) so you may compare the desired cohomology with the page 1 of that spectral. There is a more general condition to convergence; you may check this on Weibel's book "An Introduction to Homological Algebra".

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