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!
 A: 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 $V$.
