T-bundles and the Borel-Weil-Bott theorem

Hi,

Let $G$ be a reductive, connected group, $T$ a maximal torus, and $B$ a Borel subgroup containing $T$ with unipotent radical $U$. Then it turns out that the functions on the algebraic variety $G/U$ give a representation of $G$ where each irreducible representation appears exactly once. Geometrically, $G/U$ is a $B/U = T$-bundle over the flag manifold $G/B$, and I think one can deduce Borel-Weil-Bott by studying this $T$-bundle.

This much was explained to me some time ago, and now I would like to understand this circle of ideas better, but I can't find it anywhere... Any explanations/details/references/etc. would be appreciated!

Thanks!

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 It's useful to clarify that you are working with a Lie group (or linear algebraic group) over a field like $\mathbb{C}$, while the original setting for work of Borel-Weil and Bott was mostly the associated compact Lie groups. There are by now many viewpoints on B-W-B, ranging from the setting of several complex variables in J.L. Taylor's 2002 AMS graduate text to the algebraic geometry techniques used in Demazure's influential short proofs in Invent. Math. There's also an interesting paper by Bernstein-Gelfand-Gelfand on the "base affine space" $G/U$, etc. – Jim Humphreys Oct 28 2011 at 14:43

I'm very skeptical about the possibility of getting the full Borel–Weil–Bott theorem just by studying $G/U \to G/B$. Probably the closest thing I can think of is Bott's original proof of his theorem, which involves studying certain $\mathbb P^1$-bundles $G/B \to G/P$. On the other hand, you can prove the Borel–Weil theorem by studying the function space $\mathcal{O}(G/U)$, but even here you need to know a little more than just that this space contains every irrep of $G$ exactly once. More specifically, you want to know how each irrep shows up. Let me sketch the argument. To be safe, I assume we're working over $\mathbb C$, but what follows probably works over any algebraically closed field of characteristic zero.
To start off, note that $G$ acts on $\mathcal{O}(G)$ by left and right translation. Viewing $\mathcal{O}(G)$ under the latter action, we can think of $$\mathcal{O}(G/U) = \{ f \in \mathcal{O}(G) \colon f(gu) = f(g) \text{ for all } g \in G, u \in U \}$$ as the space $\mathcal{O}(G)^U$ of $U$-invariants. Now recall that there's a $G\times G$-equivariant decomposition $$\mathcal{O}(G) = \bigoplus V \otimes V^\ast \qquad \text{[an algebraic Peter–Weyl theorem]}$$ where the sum runs over the irreps of $G$, and $G$ acts on $V$ by left translation and on $V^\ast$ by right translation. Therefore we find that $$\mathcal{O}(G/U) = \mathcal{O}(G)^U = \bigoplus V \otimes (V^\ast)^U.$$ Let's assume that $U$ is built up using negative roots, so that $(V^\ast)^U$ is the lowest weight space of $V^\ast$, and in particular is one-dimensional. This shows that every irrep of $G$ appears in $\mathcal{O}(G/U)$ exactly once. But that's not all: using the right $G$-action, we can "capture" the irrep of highest weight $\lambda$. Indeed, as a $T$-module, $(V^\ast)^U = \mathbb C_\mu$, where $\mu$ is the lowest weight of $V^\ast$, or said differently, $-\mu$ is the highest weight of $V$. So, using the fact that $\text{Hom}_T(\mathbb C_\lambda, \mathbb C_\mu) = \delta_{\lambda\mu} \mathbb C_\lambda$, we see that the irrep of $G$ of highest weight $\lambda$ can be gotten as $$\text{Hom}_T(\mathbb C_{-\lambda}, \mathcal{O}(G/U)) = \bigoplus V \otimes \text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U).$$
We can re-write the left side of the above as \begin{align} (\mathbb C_\lambda \otimes \mathcal{O}(G/U))^T &= \{ f \in \mathcal{O}(G) \colon f(gtu) = \lambda(t)^{-1} f(g) \text{ for all } g \in G, t \in T, u \in U \} \\ &= \{ f \in \mathcal{O}(G) \colon f(gb) = \lambda(b)^{-1} f(g) \text{ for all } g \in G, b \in B \}, \end{align} which of course we can think of as the space of global sections of the line bundle $L_\lambda = G \times_\lambda \mathbb C$ over $G/B$. This proves the first part of the Borel–Weil theorem, namely that if $\lambda$ is dominant then $H^0(G/B,L_\lambda)$ is the irrep of highest weight $\lambda$. The other part, that $H^0(G/B,L_\lambda)=0$ if $\lambda$ is not dominant also follows easily. Indeed, all of the above works just as well for such $\lambda$, except in this case we have $\text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U)=0$ for all irreps $V$.
 Thanks for the helpful answer, Faisal. – unknown Oct 28 2011 at 13:28 Besides being a nice approach to the ideas of B-W-B, this set-up carries over in many ways to analogous reductive algebraic groups in prime characteristic even though the finite dimensional highest weight representations involved are usually no longer irreducible (Weyl modules or dual Weyl modules). – Jim Humphreys Oct 28 2011 at 14:47 P.S. It's not clear to me how close one gets here to Bott's Theorem, since that involves the Weyl group more heavily. In any case, the ongoing complications in prime characteristic are addressed in a recent MO question 78153. – Jim Humphreys Oct 28 2011 at 17:13 Dear Jim- I agree: Bott's theorem seems out of reach. I think a fair assessment of the situation is that even though studying the quotient map $G/U \to G/B$ can naturally lead you to the line bundles $L_\lambda$, you still have to do some work to compute their cohomology. – Faisal Oct 28 2011 at 17:58