How does one view the De Rham spectral sequence as a Grothendieck spectral sequence? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:40:19Z http://mathoverflow.net/feeds/question/68899 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68899/how-does-one-view-the-de-rham-spectral-sequence-as-a-grothendieck-spectral-sequen How does one view the De Rham spectral sequence as a Grothendieck spectral sequence? James D. Taylor 2011-06-27T02:36:54Z 2011-06-27T05:49:25Z <p>I was rereading basic results on de Rham cohomology, and this led me inevitably to the fact that $H^q(X,\Omega^p)$ converges to $H^*(X)$ for any smooth proper variety (over any field). How does one view this spectral sequence "maturely" as a Grothendieck spectral sequence?</p> http://mathoverflow.net/questions/68899/how-does-one-view-the-de-rham-spectral-sequence-as-a-grothendieck-spectral-sequen/68903#68903 Answer by Torsten Ekedahl for How does one view the De Rham spectral sequence as a Grothendieck spectral sequence? Torsten Ekedahl 2011-06-27T05:42:13Z 2011-06-27T05:42:13Z <p>If by "Grothendieck spectral sequence" you mean the spectral sequence associated to the composite of functors (fulfilling the Grothendieck condition) then I am skeptical as to whether this is possible. Also I do not see that there would be any particular point in being able to view it in that light (unless the functors involved would be interesting in themselves). As you seem to be looking for some general principle that would give the dRss I would like to a mention one such which seem to give most spectral sequences in use very quickly. Usually one constructs the dRss by taking a Cartan-Eilenberg resolution of the de Rham complex and then consider the spectral sequence associated to a double complex. However, one can instead use the Massey exact couple construction; once one has an exact couple one automatically gets a spectral sequence. One systematic way of constructing such exact couples is to start with a triangulated category $T$, a sequence of morphisms $\cdots\to X_{i-1}\to X_i\to X_{i+1}\to\cdots$ and a homological functor $H$ on $T$. This gives a spectral sequence starting with $H^i(Y_j)$ and going towards $\varinjlim_iH^\ast(Y_i)$. (Convergence is not assured but is OK for instance of $Y_i$ is $0$ for small $i$ and equal if $i$ is large.) Starting with the naive truncations of de Rham complex gives the dRss (and starting with canonical truncations in the algebraic case gives the conjugate spectral sequence).</p> <p>This setup works very generally, for instance it is how Adams first constructed the Adams spectral sequence.</p> http://mathoverflow.net/questions/68899/how-does-one-view-the-de-rham-spectral-sequence-as-a-grothendieck-spectral-sequen/68904#68904 Answer by Mariano Suárez-Alvarez for How does one view the De Rham spectral sequence as a Grothendieck spectral sequence? Mariano Suárez-Alvarez 2011-06-27T05:49:25Z 2011-06-27T05:49:25Z <p>You probably want one (the second one...) of the two hypercohomology spectral sequences which compute $\mathbb H^\bullet(X,\Omega^\bullet)$, the hypercohomology of the de Rham complex. A reference for this is Weibel's book.</p> <p>I doubt you can view this as a Grothendieck spectral sequence, but it has sufficiently much <em>hyper</em> in it to be considered mature, I guess.</p>