6
$\begingroup$

Let $G$ be a semisimple, simply-connected, complex algebraic group. Fix a Borel subgroup $B$ and let $P$ be a parabolic subgroup properly containing $B$. If $M$ is a $B$-module, then we have the Leray-Serre spectral sequence corresponding to the fibration $P/B\to G/B\to G/P$

$E^{p,q}_2=H^p(G/P, H^q(P/B,M))\implies H^{p+q}(G/B,M)$

The notation $H^•(G/B, M)$ means the sheaf cohomology of the sections of the bundle $G\times_B M\to G/B$ and similarly with $G,B,M$ replaced by another group $G$ with closed subgroup $B$ and $B$-module $M$.

My question is, how can the module on the right be seen as a graded version of the module on the left? What is the filtration?

$\endgroup$
8
  • $\begingroup$ I'm unclear about the formulation if $M$ isn't acted on by $P$. (Also, how does the question depend on working in characteristic 0?) $\endgroup$ Commented Jul 12, 2017 at 22:20
  • $\begingroup$ $H^q(P/B,M)$ means the $q$th sheaf cohomology of the sheaf of sections of $P\times_B M$. I think this has a natural left $P$ action, but I'm not an expert at sheaf cohomology. Characteristic may not be an issue in this generality, which would be interesting. $\endgroup$ Commented Jul 12, 2017 at 23:52
  • $\begingroup$ (Of course "minimal parabolic" means "minimal parabolic among non-Borels". Slightly smaller than regular is 'subregular'; I wonder if slightly bigger than minimal is 'superminimal'?) $\endgroup$
    – LSpice
    Commented Jul 13, 2017 at 0:09
  • $\begingroup$ Yes, that is what I mean. However, I would also be interested in any parabolic subgroup containing $B$. $\endgroup$ Commented Jul 13, 2017 at 1:06
  • 1
    $\begingroup$ I'm unclear as to what you're asking. Even if the spectral sequence collapses at the $E_2$ page, the left and right sides are not isomorphic. The right is the associated graded object of a filtration on the left. $\endgroup$
    – Mark Grant
    Commented Jul 13, 2017 at 9:48

1 Answer 1

2
$\begingroup$

In the vein of @MarkGrant's comment: Since this is a first quadrant spectral sequence, for a given pair $(p,q)$ the individual terms $E_2^{p,q},E_3^{p,q},E_4^{p,q},\ldots$ occurring on each subsequent page of the spectral sequence will eventually reach a stable value $E_\infty^{p,q}$. (The larger $p+q$ is, the more pages of the spectral sequence it may take for $E_r^{p,q}$ to reach its stable value.) In general $E_\infty^{p,q}$ will only be a subquotient of $E_2^{p,q}$. Then the statement that the spectral sequence converges to $H^*(G/B,M)$ means that $H^{p+q}(G/B,M)$ admits a decreasing filtration $F^* H^{p+q}(G/B,M)$ with $$F^p H^{p+q}(G/B,M) / F^{p+1} H^{p+q}(G/B,M) = E_\infty^{p,q}.$$ If you happen to know that the spectral sequence collapses at the $E_2$-page, and hence for all $p,q$ that $E_2^{p,q} = E_\infty^{p,q}$, then this tells you how to think of the $H^p(G/P,H^q(P/B,M))$ as the filtration layers of $H^{p+q}(G/B,M)$.

$\endgroup$
1
  • $\begingroup$ Thanks! That helps clear up some confusion. I'd be really interested to know what this filtration looks like in the case I'm describing. $\endgroup$ Commented Jul 13, 2017 at 22:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .