How can one show $G/T$ is a coadjoint orbit for a compact Lie group $G$ and $T$ its maximal torus?

Question

$\newcommand{\g}{\mathfrak{g}}$Let $G$ be a compact Lie group and $\g$ its Lie algebra. I came across the the very important result that $G/T$ ($T$ a maximal torus of $G$) can be identified to a coadjoint orbit. However it is not at all clear to me how one can show this result. I guess we must somehow prove that there always exists an $F\in \g^*$ such that $\operatorname{Stab}(F)\cong T$. But this is not clear at all from the definition of $\operatorname{Stab}(F)$ so I guess the proof for that fact must be using more elegant ideas than that. So can anyone explain what ideas are used to show this or reference somewhere?

Answer 1

To be very explicit let's take a look at the case $G = U(n), T = U(1)^n$. As Allen says, by finding a suitable invariant form we can look at adjoint orbits rather than coadjoint orbits. Here $\mathfrak{g} = \mathfrak{u}(n)$ consists of the $n \times n$ Hermitian matrices. The generic element of $X \in \mathfrak{g}$ is semisimple with distinct eigenvalues, and for such an element it's a straightforward exercise to see that the stabilizer with respect to conjugation by $G$ is a conjugate of $T$ (namely the conjugate which is diagonal in a basis of eigenvectors for $X$).

Answer 2

Use the Haar measure on $G$ (compact!) to average a metric, obtaining a $G\times G$-invariant metric, and thus an identification $\mathfrak g \cong \mathfrak g^*$. Also, the geodesic spray $\mathfrak g \to G$ defined using this metric is $G$-equivariant. Hence a $G$-orbit in $\mathfrak g^*$ near enough $0$ is $G$-isomorphic to a conjugacy class in $G$.

Now take a topological generator $t$ of $T$, i.e. $\overline{\langle t\rangle} = T$. Then the conjugacy class $G\cdot t$ of $t$ is $G/Z_G(t) = G/Z_G(T)$. Now you need to know the standard fact that for $G$ connected, tori are self-centralizing. (Personally, I prove that by using the Bruhat decomposition of $G\cdot t$ to show it's simply-connected, and therefore the covering map $G/T \to G/Z_G(T)$ must be trivial.)

Answer 3

Fix a regular element $\lambda$ in $Lie(T)\subset Lie(G)$, then the coadjoint orbit $Ad(G) \lambda$ is isomorphic to $G/T.$ Best,

Answer 4

Although there are three good answers already, perhaps you would like the explicit description $F = \sum \mathrm d\alpha$, where $\alpha$ runs over some system of simple roots.