# When representation of two different coadjoint orbits are equivalent?

Let $G$ be a compact connected Lie group and $\mu:T\to S^1$ be a representation of a maximal torus $T \subset G$ and $\lambda=d\mu$ be a weight for some $\lambda\in\mathfrak{t}^*$ (where $\mathfrak{t}$ is the Lie algebra of the maximal torus). Let $O_{\lambda}$ be a generic coadjoint orbit and $O_{\lambda'}$ be another coadjoint orbit ($\lambda'\in \mathfrak{g}^*$). When are the two representations $\rho \colon G \to GL(\Gamma(O_{\lambda},G\times_{\mu} \mathbb{C}))$ (where $\Gamma(O_{\lambda},G\times_{\mu} \mathbb{C})$ is the set of holomorphic sections) and $\rho \colon G \to GL(\Gamma(O_{\lambda'},G\times_{\xi} \mathbb{C}))$ (where $\xi:O_{\lambda'}\to S^1$ is a character) equivalent?

I think the representations are equivalent if and only if $\lambda$ and $\lambda'$ are conjugate under the Weyl group. To see this, note that the irreducible representations of $G$ can be parametrized by dominant weights, or equivalently, Weyl-orbits of weights.
• It's not clear what it would mean for $\lambda$ and $\lambda'$ to be conjugate under the Weyl group, since $\lambda'$ isn't assumed to be $\mathfrak{t}^*$ (I think; the notation in the question is a mess, so it's hard to be sure). If $\lambda'$ is in $\mathfrak{t}^*$ and conjugate to $\lambda$ then they live in the same coadjoint orbit. – Ben Webster Feb 5 '14 at 19:08
• I admit that I hastily assumed $\lambda'\in\mathfrak{t}^*$. Of course, one can write any coadjoint orbit as the coadjoint orbit containing some $\lambda'\in\mathfrak{t}^*$. Since $\mathcal{O}_{\lambda}$ and $\mathcal{O}_{\lambda'}$ coincide if and only if $\lambda$ and $\lambda'$ are Weyl-conjugate, we have arrived at exactly the same answer. – Peter Crooks Feb 5 '14 at 20:50