# The Grothendieck Spectral Sequence

March 11, 2010 by | 1 Comment

Well, I meant to do lots more examples building up some more motivation for how powerful spectral sequences can be in some simple cases. But I’m just running out of steam on posting about them. Since we’ve done spectral sequences associated to a double complex, we may as well do the Grothendieck Spectral Sequence, then I might move on to another topic for a bit (I admit it is sort of sad to not prove the Kunneth formula using a SS).

I haven’t scoured the blogs to see whether these topics have been done yet, but I’m thinking about either basics on abelian varieties a la Mumford, or some curve theory, possibly building slowly to and culminating in Riemann-Roch.

In any case, we have the tools to do the Grothendieck Spectral Sequence (GSS) quite easily. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories with enough injectives. Let $\mathcal{A}\stackrel{G}{\to}\mathcal{B}\stackrel{F}{\to}\mathcal{C}$ be functors (and $FG:\mathcal{A}\to\mathcal{C}$ the composition). Suppose that $F$ and $G$ are left exact and for every injective $J\in\mathcal{A}$ we have $G(J)$ is acyclic. This just means that $R^iF(J)=0$ for all positive $i$.

Then there exists a spectral sequence (the GSS) with $E_2^{pq}\simeq (R^pF)(R^qG)(X)\Rightarrow R^{p+q}(FG)(X)$ with differential $d_{r}:E_r^{pq}\to E_r^{p+r, q-r+q}$.

The proof of this is just to resolve $X$ using the injectives that we know exist. This gives us a double complex. From a double complex, the way to get the $E_2$ term is to take vertical then horizontal homology, or horizontal and then vertical. Both of these will converge to the same thing. One way completed collapses to the “0 row” due to the fact that the exact sequence remained exact after applying the functor except at the 0 spot. Thus it stabilizes at this term and writing it out, you see that it is exactly $R^{p+q}(FG)(X)$. Taking homology in the other order gives us exactly $(R^pF)(R^qG)(X)$ by definition of a derived functor. This completes the proof.

I probably should write the diagram out for clarity, but really they are quite a pain to make and import into wordpress. The entire outline of the proof is here, so if you’re curious about the details, just carefully fill in what everything is from the previous posts.

This is quite a neat spectral sequence. It is saying that just by knowing the derived functors of $F$ and $G$ you can get to the derived functors of the composition of them. There are two important spectral sequence consequences of this one. They are the Leray SS and the Lyndon-Hochschild-Serre SS. The later computes group cohomolgy.

I promised early on to do the Leray SS for all the algebraic geometers out there. The Leray SS gives a way to compute sheaf cohomology. Let $\mathcal{A}=Sh(X)$ and $\mathcal{B}=Sh(Y)$ be the category of sheaves of abelian groups on X and Y. Let $\mathcal{C}=Ab$ the category of abelian groups. Let $f:X\to Y$ be a continuous map, then we have the functor $F=f_*$ and the two global section functors $\Gamma_X$ and $\Gamma_Y$.

Applying the GSS to these functors, we get that $H^p(Y, R^qf_*\mathcal{F})\Rightarrow H^{p+q}(X, \mathcal{F})$.

There are a few things to verify to make sure that the GSS applies, and we need the fact that $\Gamma_Y\circ f_*=\Gamma_X$. It would also be nice to have an example to see that this is useful. So maybe I’ll do those two things next time.

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Categories: algebra, algebraic geometry, algebraic topology | Tags: cohomology, double complex, grothendieck spectral sequence, leray spectral sequence, spectral sequence | Permalink.

### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

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