1
$\begingroup$

Let $\mathfrak{g}$ be a real semisimple Lie algebra. We know that a coadjoint orbit $\mathcal{O} \hookrightarrow \mathfrak{g}^*$ carries a natural symplectic form $\omega$, namely the Kirillov-Kostant-Souriau form. Is there something tying $\omega$ to the Killing form $\kappa$ of $\mathfrak{g}$? It seems to me like there should be something like that, since both objects are intrinsic to $\mathfrak{g}$. I'm thinking something along the lines of, "if $\mathcal{O}$ is a semi-Riemannian submanifold of $\mathfrak{g}^*$, then the Levi-Civita connection of $\kappa$ (defined on the dual $\mathfrak{g}^*$ in the obvious way) induces a symplectic connection on $\mathcal{O}$", or something.

$\endgroup$
  • $\begingroup$ Your body deals with an arbitrary semi-Riemannian submanifold, whereas your title deals with a coadjoint orbit. Surely this is too much generality, since, for example, a submanifold could have odd dimension. $\endgroup$ – LSpice Feb 11 at 23:43
  • $\begingroup$ I'm not sure I follow. My body deals with a coadjoint orbit that happens to be a semi-Riemannian submanifold of the dual of the Lie algebra, i.e. the pullback of the Killing form on it is non-degenerate. $\endgroup$ – user18063 Feb 12 at 0:05
  • $\begingroup$ Oh, I see. So "if $\mathcal O$ is a semi-Riemannian submanifold of $\mathfrak g^*$" means "if, in addition to being a coadjoint orbit, $\mathcal O$ is …"? I thought you were introducing a new meaning of $\mathcal O$. $\endgroup$ – LSpice Feb 12 at 0:17
  • $\begingroup$ A obvious observation: Since the map $\mathfrak{g} \to \mathfrak{g}^*$ induced by the Killing form is an Ad-equivariant isomorphism, you can use it transport the symplectic structure from the coadjoint orbits to a symplectic structure on the adjoint orbits. That's the only construction involving the KKS symplectic structure and the Killing form that I'm aware of. $\endgroup$ – Tobias Diez Feb 12 at 20:35

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.