An example in symplectic geometry

Question

$\DeclareMathOperator\SU{SU}$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 $A$, $B$, $C$, $D$, $E$, $F$ the images of the elements $M^T$ by $\mu $.

For $P \subset \mathfrak{t}^*$ an affine space with vectorial direction $\overrightarrow{P}$, let $P^\perp \mathrel{:=} \lbrace \xi \in \mathfrak{t} \mathrel| \langle y, \xi \rangle =0, \forall y \in {(\overrightarrow{P})}^\perp\rbrace $, and let $T_P$ be the sub-torus generated by $\operatorname{Exp}(P^\perp)$.

If $\Sigma \mathrel{:=} \lbrace \text{$P$ convex polytope in $\mathfrak{t}^*$} \mathrel| \exists \text{$Z$ connected component of $M^{T_P}$ s.t $ \mu (Z)= P$}\rbrace$, how can I prove that $\Sigma = \lbrace\text{faces of $ \mu (M)$}\rbrace \cup\lbrace [AD],[BE], [FC]\rbrace$?

Answer 1

$\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\Cent{C}\DeclareMathOperator\Norm{N}\newcommand\fg{\mathfrak g}\newcommand\fl{\mathfrak l}\newcommand\ft{\mathfrak t}\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}$Based on your previous question Question about an example in symplectic geometry and this one, it looks like you are working through some notes on symplectic geometry and moment maps. It might be a good idea to gather your questions and see if you can unify them into one big one, rather than asking several different but closely related ones.

Put $G = \operatorname{SU}(3)$. Let $\{\alpha, \beta\}$ be a system of simple roots of $T$ in $G$, and let $X^*$ be an element of $M$ that lies in the subset of $\fg^*$ that vanishes on all root spaces of $T$ in $\fg_\C$, which we identify with $\ft^*$.

For future reference, suppose that $g \in G$ is such that $\Ad^*(g)X^*$ is trivial on every root space in $\fg_\C$ other than the $\pm\alpha$-root spaces. Let $L$ be the subgroup of $G$ whose complexified Lie algebra is the sum of the Lie algebra of $T$ and the $\pm\alpha$-root spaces in $G_\C$ (so $L$ is, as it were, $\operatorname S(\operatorname U(2) \times \operatorname U(1))$). Then we may identify $\fl^*$ with the set of elements of $\fg^*$ trivial on every root space in $\fg_\C$ other than the $\pm\alpha$-root spaces. Clearly, $\Ad^*(L\cdot\Norm_G(T))X^*$ is contained in $\fl^*$. On the other hand, suppose that $g \in G$ is such that $\Ad^*(g)X^*$ lies in $\fl^*$. Then $T = \Cent_L(X^*)$ and $g T g^{-1} = \Cent_L(\Ad^*(g)X^*)$ are both maximal in $L$, hence are conjugate by an element of $L$; so $L g$ intersects $\Norm_G(T)$.

Let $T'$ be a subtorus of $T$. We have that the fixed points of $T'$ in $\fg^*$ are those $Y^* \in \fg^*$ that vanish on every root subspace in $\fg_\C$ associated to a root that is non-trivial on $T'$. In particular, this fixed-point space depends only on the collection of roots trivial on $T'$. If this set of roots is non-empty (equivalently, if $T' \ne T$) and does not contain all roots (equivalently, $T'$ is non-trivial), then it is a singleton, hence a Weyl conjugate of $\{\alpha\}$. Then we have shown that, up to Weyl conjugacy, $M^{T'}$ equals $\Ad^*(L\cdot\Norm_G(T))X^*$, whose components, indexed by $\Norm_L(T)\backslash\Norm_G(T)$, are $\Ad^*(L)X^*$, $\Ad^*(L s_\beta)X^*$, $\Ad^*(L s_\beta s_\alpha)X^*$. The corresponding $P$s are, respectively, the edge between $\mu(X^*)$ and $\mu(s_\alpha X^*)$; the edge between $\mu(X^*)$ and $\mu(s_\beta X^*)$; and the diagonal between $\mu(X^*)$ and $\mu(s_\beta s_\alpha X^*)$. In your labelling, these might be $[AB]$, $[AF]$, and $[AD]$. Taking Weyl conjugates gives the other faces and diagonals.