TITLE: Question about an example in symplectic geometry
QUESTION [6 upvotes]: Let M be a coadjoint orbit of dimension 6 of $SU(3)$, and let T be the maximal torus in $SU(3)$. If we denote $\mu : M \longrightarrow \mathbb{R}^2$ the moment map associated to the action of T on M, then the image of the moment map is a hexagon with vertices are image of $M^T$ by $\mu $.
My questions are:
$1.$ What is  $M^T$? (My attempt was to choose a regular element $ X \in \mathfrak{t} \simeq  \mathfrak{t}^*$, and consider M to be the orbit of X, and then I get  $M^T=\lbrace y=gxg^{-1} \in M, ty=yt ,\forall t\in T \rbrace=   \mathfrak{t} ?).$
$2.$ why is the image of the moment map a hexagon? Well, I know from convexity theorem that the  image of the moment map, $\mu(M)$, is the convex hull of $\lbrace \mu(F)$, F connected component of $M^T\rbrace$, and that $\mu$ is constant on each connected component of $M^T$ and this implies that the set $\mu(M^T)$ is finite, but how can we find the components of $M^T$ and the cardinal number of the set $\mu (M^T)$ without having an explicit formula of $\mu$?.
Any feedback would be greatly appreciated!

REPLY [4 votes]: I'll preserve your notation:  $M$ is the coadjoint orbit of a regular semisimple element $X \in \mathfrak t^*$ (which you seem to also call $x$).  I also assume we're working in characteristic $0$, or at least not $3$.
The orbit $M$ is neither contained in, nor contains, $\mathfrak t^*$.  Rather, a conjugate of $X$ lies in $\mathfrak t^*$ if and only if it is a conjugate by the Weyl group $W = \operatorname N_{\operatorname{SU}(3)}(T)/T$.  Thus, since regular elements in this case are strongly regular, $M^T = M \cap \mathfrak t^*$ has order $6$.  These are the vertices of your hexagon.