Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}.$ Denote by $\mathfrak{g}^*$ the dual space of $\mathfrak{g}$. Let $r$ be an element of $\mathfrak{g}^*$ such that $G_r$ the stabilizer of $r$ under the coadjoint action is a maximal torus of $G$. Denote by $\mathcal{O}_r$ the coadjoint orbit of $G$ which pass through $r$.

$\mathcal{O}_r$ is endowed with a 2-form which is a symplectic form $$\omega_\alpha(\hat{X},\hat{Y})= -\alpha([X,Y]), \alpha \in \mathfrak{g}^*, \quad X,Y, \in \mathfrak{g}. $$ Often it is more convenient to choose an inner product $\langle . , . \rangle$ on $\mathfrak{g}$ to identify $\mathfrak{g}^*$ with $\mathfrak{g}$. Once such an inner product has been chosen, we can write the 2-form as $$\omega_\lambda(\hat{X},\hat{Y}) = −\langle \lambda, [X,Y] \rangle, \lambda, X, Y \in \mathfrak{g}.$$

If $G$ is semisimple, then we choose the killing form denoted $k$ to define $\omega$: $$\omega_\lambda(\hat{X},\hat{Y}) = −k(\lambda, [X,Y]).$$

$\textbf{Question}$: In the case where $G$ is a compact connected Lie group (not necessarily semisimple), why does the 2-form $$\omega_\lambda(\hat{X},\hat{Y}) = −k(\lambda, [X,Y]), \lambda, X, Y \in \mathfrak{g}.$$ defined using the killing form defines a symplectic form on thecoadjoint orbit $ \mathcal{O_r}$ of $G$ ?

Ps: This question appeared under bounty for 100 points here https://math.stackexchange.com/questions/4563148/symplectic-form-of-a-coadjoint-orbit-of-a-compact-connected-lie-group but didn't receive any answer so far.

  • $\begingroup$ Could you please explain the notations $\hat X$ and $\hat Y$? $\endgroup$ Commented Nov 6, 2022 at 18:23
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    $\begingroup$ The first formula $$\omega_\alpha(\hat{X},\hat{Y})= -\alpha([X,Y]), \alpha \in \mathfrak{g}^*, \quad X,Y, \in \mathfrak{g}.$$ is not clear. What is the relation between $X$ and $\hat X$? $\endgroup$ Commented Nov 6, 2022 at 18:57
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    $\begingroup$ If you indeed want to get an answer, please try to explain your notation.... $\endgroup$ Commented Nov 6, 2022 at 19:00
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    $\begingroup$ I will type an answer tomorrow or on Tuesday. $\endgroup$ Commented Nov 6, 2022 at 20:12
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    $\begingroup$ For today: Write $\frak g=\frak z \oplus \frak s$, $\frak g^*=\frak z^* \oplus \frak s^*$, where $\frak z$ is the center of $\frak g$, and $\frak s=[\frak g,\frak g]$ is the derived Lie algebra. Then the natural projection $\frak g^* \to \frak s^*$ induces an isomorphism of the symplectic varieties $G\cdot r$ and $G^{\rm ad}\cdot r_{\frak s}$ and preserves the Killing form. Here $G^{\rm ad}=G/Z(G)$ , and $r_{\frak s}$ denotes the projection of $r$ to $\frak s^*$. Note that ${\frak s}={\rm Lie}\,G^{\rm ad}$ is a semisimple Lie algebra. $\endgroup$ Commented Nov 6, 2022 at 20:13

1 Answer 1


$\DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\ad}{ad} \DeclareMathOperator{\Lie}{Lie} \newcommand{\g}{{\mathfrak g}} \newcommand{\z}{{\mathfrak z}} \newcommand{\s}{{\mathfrak s}} \newcommand{\O}{{\mathcal O}} \newcommand{\wh}{\widehat} \newcommand{\wt}{\widetilde}$ Let $G$ be a connected compact Lie group, and let $\g$ denote its Lie algebra. Write $\z$ for the center of $\g$ and set $\s=[\g,\g]$, which is a semisimple Lie algebra. Then $$ \g=\z\oplus\s.$$

Write $Z=Z(G)^0$ (the identity component of the center of $G$), $\ \wt S=[G,G]$, $S=G/Z(G)$. Then $Z$ is a torus, whereas $\wt S$ and $S$ are semisimple Lie groups. We can identify $$ \z=\Lie Z,\quad \Lie S=\s=\Lie\wt S.$$

The group $G$ acts on $\g$ by adjoint representation. Moreover, $G$ acts on $\g$ via the canonical surjective homomorphism $\pi\colon G\to S$. We write $$ g\cdot X=s\cdot X,\quad\text{where} \ g\in G,\ s=\pi(g)\in S,\ X\in\g.$$

We write an element $X\in \g$ as $$ X=X_\z+X_\s\ \quad \text{with}\ X_\z\in\z,\ X_\s\in \s\,.$$ Then $$g\cdot X=X_\z+g\cdot X_\s=X_\z+s\cdot X_\s\,.$$

Let $\g^*$ denote the dual space for $\g$. Then $$\g^*=\z^*\oplus \s^*.$$ For $r\in \g^*$ we may write $$r=r_\z+r_\s\quad\text{with}\ r_\z\in\z^*, \ r_\s\in \s^*.$$ Then for $g\in G$ we have $$g\cdot r=r_\z+g\cdot r_\s=r_\z+s\cdot r_\s\,.$$

Let $X\in\g$, $\ X=X_\z+X_\s\,$. Then $[X_\z\,,X_\s]=0$. It follows that $$\exp -tX=(\exp -tX_\z)\cdot( \exp -tX_\s) \quad\text{with}\ \exp -tX_\z\in Z,\ \exp -tX_\s\in\wt S,$$ whence $$(\exp -tX)\cdot \alpha=\alpha_\z+(\exp-t X_\s)\cdot \alpha_\s\quad \text{for}\ \alpha=\alpha_\z+\alpha_\s\in \g^*.$$ Write $$ \wh X=\frac d{dt}\Big|_{t=0}(\exp -tX)\cdot \alpha.$$ Then $\wh X=\wh{X_\s}$, where $$ \wh {X_\s}=\frac d{dt}\Big|_{t=0}(\exp -tX_\s)\cdot \alpha_\s.$$

For $r=r_\z+r_\z \in\g^*$ write $\O_r=G\cdot r$, $\ \O_{r_\s}=S\cdot r_\s\,$. Then $$ \O_r=r_\z+S\cdot r_\s=r_\z+\O_{r_\s}\,.$$ Thus for $\alpha=g\cdot r\in \O_r$ we have \begin{equation}\label{e:*} T_\alpha(\O_r)\cong T_{\alpha_\s}(\O_{r_\s}).\tag{$*$} \end{equation}

Consider the adjoint representation $$\ad\colon \g\to\mathfrak{gl}(\g),\quad (\ad X)\cdot Y=[X,Y].$$ For $X=X_\z+X_\s\in\g$ we have $\ad X=\ad X_\s\,$.

Consider the Killing form $$ k\colon \g\times\g\to{\mathbb R}, \quad (X,Y)\mapsto \Tr\!\big ((\ad X)\cdot (\ad Y)\big).$$ Then $$ k(X,Y)=\Tr\!\big ((\ad X_\s)\cdot (\ad Y_\s)\big)=k(X_\s\,,Y_\s).$$

For $\lambda \in\g$, we define a skew-symmetric form on $T_\alpha(\O_r)$ by $$\omega_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y]).$$ Since $[X,Y]=[X_\s\,,Y_\s]$, for $\lambda=\lambda_\z+\lambda_\s$ we have $$\omega_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y]) =-k(\lambda_\s,[X_\s\,,Y_\s])=\omega_{\lambda_\s}(\wh{X_\s}\,, \wh{ Y_\s\, }\,).$$

We can identify the tangent spaces $T_\alpha(\O_r)$ and $T_{\alpha_\s}(\O_{r_\s})$ by \eqref{e:*}. Then the formula above is probably what you need.

  • $\begingroup$ Many thanks for your detailed answer @Mikhail Borovoi. $\endgroup$
    – Mira
    Commented Nov 9, 2022 at 22:45

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