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.

  2. Yes, this is correct as well.

  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 $G/T$).

As for references, a good place to start is the pair of articles by Borel and Hirzebruch from the 1950s. It is in these articles that the Borel–Weil–Bott theorem was first conjectured. Bott's original paper (Annals, 1957) is definitely worth a read too, even though it is a bit "heavy-handed" (to quote Bott himself). The proof he gives is very geometric indeed—all the Lie algebra cohomology business can more or less be ignored, if you want. Some papers of Griffiths and Schmid from that era are also fairly geometric and contain a wealth of information. Schmid also has more recent survey articles on geometric representation theory that are worth a look. Finally, let me recommend Dennis Snow'swell-written survey article on homogeneous vector bundles. (Caveat: All these references are a bit short on symplectic geometry and the coadjoint picture. I don't know what to recommend for that.)

  • 2
    $\begingroup$ 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. $\endgroup$ – Allen Knutson Nov 2 '13 at 4:44
  • $\begingroup$ Isn’t the pair of articles by Borel–Hirzebruch actually a trio? $\endgroup$ – LSpice May 24 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.