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Let $G$ be an algebraic reductive group over an algebraically closed field $k$, $T$ a maximal torus and $B = TU$ a Borel subgroup containing it. I'm interested in computing $H^*(G/U,\mathcal O_{G/U})$ [corrected typo; I had written $B/U$] (coherent cohomology) (in terms of the representation theory of $G$?). I suppose this is well known, but I can't find it anywhere....

Any suggestions?


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Why do you write $B/U$ for $T$? Are you sure you want $B/U$? As $T$ is affine there is no higher coherent cohomology. – Wilberd van der Kallen Nov 14 '11 at 8:25
It might be a typo, as the title refers to $G/U$. – Dror Speiser Nov 14 '11 at 8:53
All of the higher cohomology groups are zero, and the natural pullback map from k to H^0 is an isomorphism. – Jason Starr Nov 14 '11 at 13:08
Oops. I though it was G/B not G/U. Since G/U --> G/B is affine, one can compute by pushing forward to G/B. Then one gets, essentially, the direct sum of all invertible sheaves on G/B. The problem of computing the cohomology of each of these is Borel-Weil-Bott. – Jason Starr Nov 14 '11 at 15:16
up vote 4 down vote accepted

Assuming that unknown means $G/U$, $H^{*}(G/U,\mathcal{O}_{G/U})$ has been computed as a $D_{G/U}$-module by Levasseur: "Differential operators on the base affine space". The global sections are identified with a certain quotient of global differential operators and the $i$th cohomology is given as a sum of simple $D_{G/U}(G/U)$-modules indexed by elements of length $i$ in the Weyl group.

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Thanks for the useful reference! (I assume you mean Levasseur-Stafford, ) – David Ben-Zvi Nov 14 '11 at 15:21

Chris Brav's answer gives a nice description of the cohomology in the $D$-module context. Just to expand a bit on that, I'd like to give a direct description, and also say a word about positive characteristic, since the paper of Levasseur-Stafford only discusses the characteristic-0 case.

As Jason Starr mentions above, we can decompose this cohomology in terms of line bundles on $G/B$. Fix an isomorphism $B/U \cong T$; in particular, this gives $T$ and $k[T]$ structures of $B$-modules. For any $B$-module $M$ let $ \mathcal L(M) $ denote the $G$-equivariant bundle on $G/B$ with fiber $M$. Then we have a $G$-equivariant isomorphism $$ H^*(G/U, \mathcal O_{G/U}) \cong H^* \big(G/B, \mathcal L ( k[T] ) \big) . $$ So, we basically want to understand the structure of $k[T]$ as a $B$-module.

Let $X(T)$ denote the character group of $T$. Then $k[T] \cong k( X(T) )$, the group algebra of $X(T)$ over $k$. As Jason pointed out, we now get a direct sum of line bundles on $G/B$ corresponding to the elements of $X(T)$. However, note that we may not get all of the line bundles on $G/B$, since $G$ might not be simply-connected; $X(T)$ may be a proper subset of the full weight lattice of $G$. Here isogeny will play a role. (In Levasseur-Stafford, for example, they assume $G$ to be simply connected). In any event, the characteristic 0 story will now follow from Borel-Weil. The postive-characteristic answer, on the other hand, is still an open question, since the full cohomology of line bundles on $G/B$ isn't completely known there (although a lot is known, cf Jantzen's book "Representations of Algebraic Groups").

Regardless of characteristic, though, we have a nice description of the global sections. Let $X^+(T)$ denote the set of dominant weights in $X(T)$; then we get $$ H^0 \big(G/B, \mathcal L ( k[T] ) \big) \cong \bigoplus_{\mu \in X^+(T)} H^0( G/B, \mathcal L(\mu) ) , $$ a direct sum of standard modules for $G$. In characteristic 0 these modules are all simple, but they are not all simple in positive characteristic.

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Thanks for the answer, Chuck. My original intention was to somehow invert the order and try to deduce the cohomology of $H^*(G/B,L)$ from the knowledge of the cohomology of $H^*(G/U,O)$. That is, I wanted to compute the latter cohomology group independently of Borel-Weil-Bott (I don't have any reason to think this should work). – Nicolás Nov 14 '11 at 19:55
This can be done in one special case: you can compute $ H^0( G/U, \mathcal O_{G/U} ) $ directly without using Borel-Weil-Bott, by using Frobenius reciprocity. My gut feeling is that this approach becomes a lot harder for computing the higher cohomology groups. If you look at Levasseur-Stafford, they ultimately push down to the flag variety and invoke Borel-Weil. (continued . . .) – Chuck Hague Nov 14 '11 at 20:40
(continued . . .) In general, that's pretty much always what one does when computing cohomology of bundles on homogeneous spaces: at some point you push down to a flag variety or partial flag variety because you can get traction there. – Chuck Hague Nov 14 '11 at 20:40

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