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!



Question about an example in symplectic geometry
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.


