Image of a coadjoint orbit of SU(3) under a moment map

Question

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!

Answer 1

The elements in $su(3)$ satisfy $X+X^T=0$ and are unitarily diagonalizable, with purely imaginary eigenvalues. This means that, identifying $su(3)$ with its dual and adjoint and coadjoint actions, the orbits are labeled by the triples of real numbers, the imaginary parts of the eigenvalues. The highest dimensional orbits are the ones corresponding to distinct eigenvalues. A maximal torus consists of those matrices in $SU(3)$ that are diagonal in some basis. We can take the standard basis, and then "the" maximal torus consists of unitary diagonal matrices, $diag\{e^{\theta_1}, e^{i\theta_2}, e^{i\theta_3}\}$ with $\theta_1+\theta_2+\theta_3=0$. Fix an orbit corresponding to $\lambda_1, \lambda_2, \lambda_3$. The matrices in this orbit invariant under the torus are diagonal matrices, of which, having fixed the triple $\lambda_1, \lambda_2, \lambda_3$, we get 6 -- all the diagonal matrices with all the permutations of the eigenvalues $i\lambda_1, i\lambda_2, i\lambda_3$ on the diagonal. These give the 6 vertices of the hexagon you are after.

Remark 1: The $t^*$ can be thought of as the subspace $v_1+v_2+v_3=0$ inside a 3D space "dual to tangential directions to $\theta$s". The images of the 6 fixed point under momentum map are permuted when we permute the coordinates in this 3D space, so they are either one point, vertices of a triangle, or of a hexagon. These should correspond to the cases of all equal eigenvalues, two equal eigenvalues, and all eigenvalues distinct, respectively.

Remark 2: The permutation group taking the 6 fixed points to each other is the Weyl group of $su(3)$. In general, there should be a description of everything in appropriate representation-theoretic language; but I'm actually not an expert on this, so I'll leave it to someone else to supply it.

Answer 2

I won't answer the question directly, but I'll give you a vivid view of the underlying algebra, which shows how the hexagon arises.

The linear Poisson bracket associated with $U(3)$ is determined from the following: $$\{q^i_j, q^k_l\} = δ^k_j q^i_l - δ^i_l q^k_j, \hspace 1em (i,j,k,l ∈ \{1,2,3\}),$$ where $δ^k_j$ is the Kroenecker delta ($δ^k_j = 1$ if $k = j$, $δ^k_j = 0$ if $k ≠ j$). The coadjoint orbits of $SU(3)$ comprise the symplectic leaves within this Poisson manifold for which $T = q^1_1 + q^2_2 + q^3_3 = 0$. On all of the symplectic leaves of this manifold, $T$ is an invariant: the "trace" invariant, itself. So, the constraint $T = 0$ yields the symplectic leaves associated with the trace-zero subalgebra $su(3)$ of $u(3)$.

The remaining $q$ coordinates can be arrayed as: $$(R,O,Y,G,B,P) = \left(q^1_2, q^1_3, q^2_3, q^2_1, q^3_1, q^3_2\right),$$ reminiscent of the colors {"red", "orange", "yellow", "green", "blue", "purple"} on a color wheel. Defining $$H(Θ) = \frac{q^1_1 - q^2_2}{2} \cos Θ + \frac{q^1_1 + q^2_2 - 2 q^3_3}{2 \sqrt{3}} \sin Θ,$$ one can write: $$\begin{align} \{R, G\} &= 2 H(0˚) &= q^1_1 - q^2_2, \\ \{Y, P\} &= 2 H(120˚) &= q^2_2 - q^3_3, \\ \{B, O\} &= 2 H(240˚) &= q^3_3 - q^1_1. \end{align}$$ Those are three of the six cardinal points of a hexagon, the other three being their respective negatives $$\begin{align} \{G, R\} &= 2 H(180˚) &= q^2_2 - q^1_1, \\ \{P, Y\} &= 2 H(300˚) &= q^3_3 - q^2_2, \\ \{O, B\} &= 2 H(60˚) &= q^1_1 - q^3_3. \end{align}$$

The remaining Poisson brackets for the "colored" coordinates are: $$ \{R, Y\} = O, \hspace 1em \{Y, B\} = G, \hspace 1em \{B, R\} = P,\\ \{G, O\} = Y, \hspace 1em \{O, P\} = R, \hspace 1em \{P, G\} = B, $$ with the remaining brackets, for "adjacent" elements all being zero: $$ \{R, O\} = \{O, Y\} = \{Y, G\} = \{G, B\} = \{B, P\} = \{P, R\} = 0.$$

The toroidal sub-algebra is associated with the coordinates $H(Θ)$, which satisfy the identities: $$H(Θ) \cos Θ' + H(Θ + 90˚) \sin Θ' = H(Θ + Θ'), \hspace 1em \{H(Θ), H(Θ')\} = 0.$$

The remaining fundamental brackets are given by: $$\begin{align} \{H(Θ), R\} &= \cos Θ R, &\hspace 1em \{H(Θ), O\} &= \cos(Θ - 60˚) O, \\ \{H(Θ), Y\} &= \cos(Θ - 120˚) Y, &\hspace 1em \{H(Θ), G\} &= \cos (Θ - 180˚) G, \\ \{H(Θ), B\} &= \cos(Θ - 240˚) B, &\hspace 1em \{H(Θ), P\} &= \cos(Θ - 300˚) P. \end{align}$$

So, when they use the term "color" for the charge associated with $SU(3)$, in quantum chromodynamics, they really do mean color. The analogy runs deep. The coordinates $H(0˚)$ and $H(90˚)$ play the analogous role of "chromaticity coordinates", while $T$ - which is factored out when going from $u(3)$ to $su(3)$ - plays the role of "brightness". The hexagon is the color wheel, itself.