# Geometric structure of flag manifolds, Borel -Weil-Bott theorem

I want to know if there is proof of Borel Weil Bott theorem, that is as geometric as it can be.

Let $G$ be a semisimple compact Lie group and $T$ be a maximal torus. We know that $G/T$ is a projective manifold. A way to show this is to consider the complexification of $G$ and identify $G/T$ as $G_{\mathbb C}/B$.

It seems to me that the following are true:

(1) The root space of $G$ corresponds to $H^{1, 1}\big( G/T\big)\cap H^2(G/T, \mathbb R)$: as $G/T$ is simply connected, $H^2(G/T, \mathbb Z)\cong H_2(G/T, \mathbb Z) \cong \pi_2(G/T)$ and $\pi_2(G/T)$ are generated by those "$\mathbb P^1$" corresponds to each $\alpha \in \Delta^+$ (the $su(2)$ subalgebra $S_\alpha$, to be precise).

(2) The positive Weyl Chamber $C$ corresponds to the Kahler cone of $G/T$, for similar reason as in (1).

So my question is, is (1) and (2) actually holds?

For the following two, I am not so sure. It is related to the proof of Borel Weil Bott theorem.

(3) Is $\rho = \frac{1}{2}\sum_{\alpha\in \Delta^+} \alpha$ corresponds to the First Chern class of $G/T$? Because we know that if a integral weight $\lambda$ satisfies $\lambda + \rho \in C$, then all higher cohomology $H^p(G/T, L_\lambda)$ ($p>0$) vanishes. If $C$ is the Kahler cone, then it seems that the statement is a consequence of Kodaira vanishing theorem, if $\rho$ actually corresponds to the first Chern class.

(4) When $\lambda + \rho$ is not in $C$, there is some correspondance between $H^p(G/T, L_\lambda)$ and $H^0(G/T, L_\mu)$ for some $\mu$ such that $\mu + \rho \in C$. Is there an intuitive geometric reason behind this?

To sum up a bit, can anyone suggest a reference related to these "geometrical aspect" of $G/T$? Thanks in advance.

1. Correct. You can be fairly explicit here. For each root $\alpha$, let $\omega_\alpha \in \mathfrak g^\ast$ be a left-invariant form on $G$ that is dual to $\mathfrak g_\alpha$. Then for $\lambda \in \mathfrak t^\ast$, we have $d\lambda = \sum_{\alpha\in\Phi^+} \langle \lambda,\alpha\rangle \omega_\alpha \wedge \omega_{-\alpha}$ (up to some scalar). This defines a form on $G/T$ that is of type (1,1) and is positive iff $\lambda$ is in the interior of $C$. Another way of thinking about this whole thing is to note that we have, via transgression, an isomorphism $H^2(G/T;\mathbb R) \cong H^1(T;\mathbb R)$ and this latter space may be identified with $\mathfrak t^\ast$ in the usual manner.
3. Not quite. It's $2\rho$ that corresponds to $c_1(G/T)$. I'm not sure how you're applying Kodaira, but note that the first Chern class of the canonical bundle of $G/T$ is $-2\rho$.
4. You have to be a bit careful about what you mean here. In any case, what you're noticing is essentially a manifestation of Serre duality (keep in mind that $L_{-2\rho}$ is the canonical bundle of $G/T$).
• The easy way to do calculation #3 is to compute the $T$-equivariant first Chern class, which one can do after restriction to fixed points. By the $G$-invariance of the tangent bundle, this $c_1$ will be $W$-invariant, so it's enough to calculate it at the basepoint of $G_{\mathbb C}/B$. The tangent space there is ${\mathfrak g}_{\mathbb C}/{\mathfrak b} \cong {\mathfrak n}_-$, with weights the negative roots. Taking the Euler class adds those up. I'm not going to fight over the sign. – Allen Knutson Nov 2 '13 at 4:44