7
$\begingroup$

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!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Thank you for your answer @LSpice! But could you explain please what do you mean by " since regular elements in this case are strongly regular" ? $\endgroup$
    – Maria
    Commented Nov 13, 2020 at 13:39
  • 1
    $\begingroup$ It is possible for an element to be regular, in the sense that no root vanishes on it, but not strongly regular, in the sense that it can still be fixed by some Weyl-group element. I think, though I'm not positive, that, for Lie-algebra elements this does not happen in characteristic $0$, but it can happen: for example, in adjoint $\mathsf B_2$ over $\mathbb F_4$, with $\alpha$ long simple and $\beta$ short simple, $\varpi_\alpha^\vee(1) + \varpi_\beta^\vee(\theta)$ (with $\theta \notin \mathbb F_2$) is regular but fixed by $s_\beta$. $\endgroup$
    – LSpice
    Commented Nov 13, 2020 at 16:48
  • 1
    $\begingroup$ (I don't know why I went with such a complicated example, which, moreover, works on the wrong side of the duality; it can happen in $\mathfrak{su}(3)^*$, over $\mathbb F_3$, as evidenced by $\varpi_\alpha + \varpi_\beta$, where $\alpha$ and $\beta$ are simple roots swapped by the Galois involution, which is regular but fixed by the long element $s_\alpha s_\beta$.) $\endgroup$
    – LSpice
    Commented Nov 13, 2020 at 18:52
  • 1
    $\begingroup$ This is good to know ! Thanks a lot ! $\endgroup$
    – Maria
    Commented Nov 13, 2020 at 21:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .