Let $$G$$ be a compact connected Lie group. We denote by $$\mathfrak{g}$$ the Lie algebra of $$G$$ and by $$\mathfrak{g}^*$$ the dual space of $$\mathfrak{g}$$. Let $$\mathcal{O}_r: = G\cdot r$$ be a generic coadjoint orbit of $$G$$.

The coadjoint orbit $$\mathcal{O}_r$$ endowed with the Kirillov–Kostant–Souriau $$\omega$$ is a symplectic manifold. I've read that it is also a Kähler manifold; meaning that there exists a unique almost complex structure $$J$$ on $$\mathcal{O}_r$$ which is compatible with $$\omega$$ and such that the form $$g(\cdot,\cdot):= \omega(\cdot,J\cdot )$$ is a Riemannian metric on $$\mathcal{O}_r$$.

Given an element $$\beta \in \mathcal{O}_r$$, then the tangent space of $$\mathcal{O}_r$$ at $$\beta$$ is $$T_\beta \mathcal{O}_r = \lbrace \xi_{\mathcal{O}_r}(\beta), \xi \in \mathfrak{g}\rbrace$$ , where $$\xi_{\mathcal{O}_r}(\beta) = \frac{d}{dt}\rvert_ {t=0} e^{-t \xi}\cdot\beta$$. What is $$J(\xi_{\mathcal{O}_r}(\beta) )$$, $$\xi \in \mathfrak{g}$$ ?

• Does 'generic' in "generic coadjoint orbit" have a technical meaning here? Sep 21 at 14:24
• Ah, OK. I am used to "strongly regular semisimple" in that context. Sep 21 at 14:31
• I'm also a little confused by your description of $T_\beta\mathcal O_r$, which seems to make no reference to $r$. Are you sure it's correct? (At least $\beta$ should be an element of $G\cdot r$, not just of $\mathfrak g^*$.) Sep 21 at 15:29
• @LSpice, you are right $\beta$ should be an element of $G.r$, I'll fix that! However the description of $T_\beta \mathcal{O}_r$ is correct.
– asma
Sep 21 at 15:43
• It seems to be on Google Books. Sep 26 at 14:27

Put $$T = G_r$$. We may, and do, assume that $$\beta = r$$, and simply describe a $$T$$-invariant complex structure on $$\operatorname T_{\mathcal O_r}(r)$$.

Instead of having one 1-dimensional subspace of $$\mathfrak g$$ for every root $$\alpha$$, we get a $$2$$-dimensional subspace $$\{X_\alpha + \overline{X_\alpha} \mathrel: X_\alpha \in (\mathfrak g_{\mathbb C})_\alpha\}$$ for every pair of roots $$\{\alpha, \overline\alpha = -\alpha\}$$.

Our Kähler structure treats this $$2$$-dimensional space, which I will provocatively call $$\mathfrak g_{\pm\alpha}$$ since its complexification is $$(\mathfrak g_{\mathbb C})_\alpha \oplus (\mathfrak g_{\mathbb C})_{-\alpha}$$, as a $$\mathbb C$$-vector space via the (isomorphic) projection to $$(\mathfrak g_{\mathbb C})_\alpha$$, and then rotates by $$i$$—but we must choose $$i$$ appropriately to get a negative definite metric. After our discussion in the comments, I think I have finally cleaned up the relevant signs.

Fix a root $$\alpha$$ of $$T$$ in $$\mathfrak g_{\mathbb C}$$. Put $$i_\alpha = -\lambda\lvert\lambda\rvert^{-1}$$, where $$\lambda = r(\mathrm d\alpha^\vee(1))$$ ($$H_\alpha \mathrel{:=} \mathrm d\alpha^\vee(1)$$ is sometimes called the coroot, but I prefer to reserve that terminology for $$\alpha^\vee$$ itself), so that $$i_\alpha$$ is a square root of $$-1$$. Then $$J$$ carries $$\xi_{\mathcal O_r}(r)$$, where $$\xi = X_\alpha + \overline{X_\alpha}$$, to $$\xi'_{\mathcal O_r}(r)$$, where $$\xi' = i_\alpha(X_\alpha - \overline{X_\alpha})$$, for every $$X_\alpha \in (\mathfrak g_{\mathbb C})_\alpha$$.

• I realize now that I don't know why is the metric $g$ definite positive. Here is my attempt: suppose that we are in the simplest situation where $\mathfrak{g}_\mathbb{C}= \mathfrak{t}_\mathbb{C} \oplus {(\mathfrak{g}_\mathbb{C})}_\alpha \oplus {(\mathfrak{g}_\mathbb{C})}_{-\alpha}$. Let $\xi= X_\alpha + \overline{X_\alpha}$. Let $<.,.>$ an Ad-invariant inner product on $\mathfrak{g}$.
– asma
Sep 22 at 19:28
• Then, $g_r(\xi_{\mathcal{O}_r}, \xi_{\mathcal{O}_r})= \omega(\xi_{\mathcal{O}_r}, J(\xi_{\mathcal{O}_r}))=<H_r, [\xi,J(\xi)]> = -2i<[H_r, X_\alpha],\overline{X_\alpha}>= -2i\alpha(H_r)<X_\alpha,\overline{X_\alpha}>.$ From here I don't know how to continue , any help please!
– asma
Sep 22 at 19:30
• Re$\newcommand\l{\overline}\newcommand\p[2]{\langle#1,#2\rangle}$, $\alpha(H_r)$ is non-$0$ pure imaginary, and $\p{X_\alpha}{\l{X_\alpha}}=\p{s X_\alpha}{s\overline{X_\alpha}}=\lvert c\rvert^2\p{\l{X_\alpha}}{X_\alpha}=\lvert c\rvert^2\l{\p{X_\alpha}{\l{X_\alpha}}}$, where $s$ is a representative in $G$ of the reflexion in $\alpha$ and $c$ is the non-$0$ complex number such that $s X_\alpha$ equals $c\overline{X_\alpha}$. So the pairing is either positive definite or negative definite. If it's negative definite, then take $-J$ instead. Sep 22 at 19:59
• This issue (of maybe having to change the complex structure after the fact) can be fixed by choosing the square root $i$ of $-1$ carefully. I have updated my answer accordingly. Sep 22 at 20:08
• You are right. My argument shows that $\langle X_\alpha, \overline{X_\alpha}\rangle$ is real, but not that it is positive. I will think about it. (TeX note: please use $\langle\rangle$ \langle\rangle instead of $<>$ <>; compare the spacing in $\langle X, Y\rangle = 0$ \langle X, Y\rangle = 0 to the spacing in $<X, Y> = 0$ <X, Y> = 0.) Sep 23 at 0:58

For the affine case for coadjoint orbits of Souriau, see Jean-Louis Koszul book "Introduction to symplectic geometry" in chapter 4 and 5 where it is explained the Souriau cocycle in cas of non null cohomology: https://link.springer.com/book/10.1007/978-981-13-3987-5