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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} $ ?

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  • $\begingroup$ Does 'generic' in "generic coadjoint orbit" have a technical meaning here? $\endgroup$
    – LSpice
    Commented Sep 21, 2022 at 14:24
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    $\begingroup$ Ah, OK. I am used to "strongly regular semisimple" in that context. $\endgroup$
    – LSpice
    Commented Sep 21, 2022 at 14:31
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    $\begingroup$ 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 $\mathfrak g^*$.) $\endgroup$
    – LSpice
    Commented Sep 21, 2022 at 15:29
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    $\begingroup$ @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. $\endgroup$
    – Mira
    Commented Sep 21, 2022 at 15:43
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    $\begingroup$ It seems to be on Google Books. $\endgroup$
    – LSpice
    Commented Sep 26, 2022 at 14:27

2 Answers 2

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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 $X_\alpha \in (\mathfrak g_{\mathbb C})_\alpha$.

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    $\begingroup$ 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 $\mathfrak{g}$. $\endgroup$
    – Mira
    Commented Sep 22, 2022 at 19:28
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    $\begingroup$ 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! $\endgroup$
    – Mira
    Commented Sep 22, 2022 at 19:30
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Sep 22, 2022 at 19:59
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Sep 22, 2022 at 20:08
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    $\begingroup$ 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.) $\endgroup$
    – LSpice
    Commented Sep 23, 2022 at 0:58
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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

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