# The orbit $(G\cdot X) \cap \mathfrak{t}$ for $X\in \mathfrak{t}$ singular

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 $\mathfrak{t}$.

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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.