T-bundles and the Borel-Weil-Bott theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:40:24Z http://mathoverflow.net/feeds/question/79350 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79350/t-bundles-and-the-borel-weil-bott-theorem T-bundles and the Borel-Weil-Bott theorem unknown 2011-10-28T05:13:50Z 2011-10-29T19:59:02Z <p>Hi,</p> <p>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.</p> <p>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!</p> <p>Thanks!</p> http://mathoverflow.net/questions/79350/t-bundles-and-the-borel-weil-bott-theorem/79364#79364 Answer by Faisal for T-bundles and the Borel-Weil-Bott theorem Faisal 2011-10-28T08:13:05Z 2011-10-29T19:59:02Z <p>I'm very skeptical about the possibility of getting the full Borel–Weil–Bott theorem just by studying <code>$G/U \to G/B$</code>. Probably the closest thing I can think of is Bott's original proof of his theorem, which involves studying certain <code>$\mathbb P^1$</code>-bundles <code>$G/B \to G/P$</code>. On the other hand, you can prove the Borel–Weil theorem by studying the function space <code>$\mathcal{O}(G/U)$</code>, but even here you need to know a little more than just that this space contains every irrep of <code>$G$</code> 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 <code>$\mathbb C$</code>, but what follows probably works over any algebraically closed field of characteristic zero.</p> <p>To start off, note that <code>$G$</code> acts on <code>$\mathcal{O}(G)$</code> by left and right translation. Viewing <code>$\mathcal{O}(G)$</code> under the latter action, we can think of <code>$$\mathcal{O}(G/U) = \{ f \in \mathcal{O}(G) \colon f(gu) = f(g) \text{ for all } g \in G, u \in U \}$$</code> as the space <code>$\mathcal{O}(G)^U$</code> of <code>$U$</code>-invariants. Now recall that there's a <code>$G\times G$</code>-equivariant decomposition <code>$$\mathcal{O}(G) = \bigoplus V \otimes V^\ast \qquad \text{[an algebraic Peter–Weyl theorem]}$$</code> where the sum runs over the irreps of <code>$G$</code>, and <code>$G$</code> acts on <code>$V$</code> by left translation and on <code>$V^\ast$</code> by right translation. Therefore we find that <code>$$\mathcal{O}(G/U) = \mathcal{O}(G)^U = \bigoplus V \otimes (V^\ast)^U.$$</code> Let's assume that <code>$U$</code> is built up using negative roots, so that <code>$(V^\ast)^U$</code> is the lowest weight space of <code>$V^\ast$</code>, and in particular is one-dimensional. This shows that every irrep of <code>$G$</code> appears in <code>$\mathcal{O}(G/U)$</code> exactly once. <em>But that's not all:</em> using the right <code>$G$</code>-action, we can "capture" the irrep of highest weight <code>$\lambda$</code>. Indeed, as a <code>$T$</code>-module, <code>$(V^\ast)^U = \mathbb C_\mu$</code>, where <code>$\mu$</code> is the lowest weight of <code>$V^\ast$</code>, or said differently, <code>$-\mu$</code> is the highest weight of <code>$V$</code>. So, using the fact that <code>$\text{Hom}_T(\mathbb C_\lambda, \mathbb C_\mu) = \delta_{\lambda\mu} \mathbb C_\lambda$</code>, we see that the irrep of <code>$G$</code> of highest weight <code>$\lambda$</code> can be gotten as <code>$$\text{Hom}_T(\mathbb C_{-\lambda}, \mathcal{O}(G/U)) = \bigoplus V \otimes \text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U).$$</code></p> <p>We can re-write the left side of the above as <code>\begin{align} (\mathbb C_\lambda \otimes \mathcal{O}(G/U))^T &amp;= \{ 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 \} \\ &amp;= \{ f \in \mathcal{O}(G) \colon f(gb) = \lambda(b)^{-1} f(g) \text{ for all } g \in G, b \in B \}, \end{align}</code> which of course we can think of as the space of global sections of the line bundle <code>$L_\lambda = G \times_\lambda \mathbb C$</code> over <code>$G/B$</code>. This proves the first part of the Borel–Weil theorem, namely that if <code>$\lambda$</code> is dominant then <code>$H^0(G/B,L_\lambda)$</code> is the irrep of highest weight <code>$\lambda$</code>. The other part, that <code>$H^0(G/B,L_\lambda)=0$</code> if <code>$\lambda$</code> is not dominant also follows easily. Indeed, all of the above works just as well for such <code>$\lambda$</code>, except in this case we have <code>$\text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U)=0$</code> for all irreps <code>$V$</code>.</p>