This question may be a simple problem for experts. Let $G$ be a connected compact Lie group and $T$ be its maximal torus. Let $\mathfrak{g}$ and $\mathfrak{t}$ be the corresponding Lie algebras. We know that $\mathfrak{t}\subset \mathfrak{g}$. The Lie group $G$ acts on $\mathfrak{g}$ by by the adjoint action.

For any $X\in \mathfrak{t}$, we denote by $G\cdot X$ the $G$-orbit of $X$ in $\mathfrak{g}$ and $$ F_X:=(G\cdot X) \cap \mathfrak{t}. $$

If $X$ is a regular element in $\mathfrak{t}$, i.e. $\exp(tX)$ is dense in $T$, then for $g\in G$, $Ad(g)X\in \mathfrak{t} $ implies $Ad(g)\mathfrak{t}\in \mathfrak{t}$, in other words $g$ is in the normalizer of $\mathfrak{t}$ and we can deduce that $F_X$ is exactly the Weyl group orbit of $X$ in $\mathfrak{t}$.

My question is: If $X$ is singular, is it still true that $F_X$ is the Weyl group orbit of $X$ in $\mathfrak{t}$?

In the singular case, $Ad(g)X\in \mathfrak{t}$ does not imply $g$ is in the normalizer of $\mathfrak{t}$. However maybe we can find another $\tilde{g}$ such that $Ad(g)X=Ad(\tilde{g})X$ and $\tilde{g}$ is in the normalizer of $\mathfrak{t}$.


Yes, it's still true that $G\cdot X\cap\mathfrak t=W\cdot X$.

The inclusion $\supset$ is clear. Conversely, suppose $g\cdot X\in\mathfrak t$. Since $\mathfrak t$ consists exactly of all $T$-fixed points in $\mathfrak g$, it follows that $t\cdot g\cdot X=g\cdot X$ for all $t\in T$. Hence $g^{-1}Tg$ is contained in the stabilizer $G_X$. So $T$ and $g^{-1}Tg$ are two maximal tori in $G_X$. Hence they are conjugate by some $h\in G_X$: $T = h^{-1}g^{-1}Tgh$. So now $gh$ normalizes $T$, and the Weyl group element it represents still sends $X$ to $g\cdot X$. QED.


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.