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.

• Could you please explain the notations $\hat X$ and $\hat Y$? Commented Nov 6, 2022 at 18:23
• 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 $\hat X$? Commented Nov 6, 2022 at 18:57
• If you indeed want to get an answer, please try to explain your notation.... Commented Nov 6, 2022 at 19:00
• I will type an answer tomorrow or on Tuesday. Commented Nov 6, 2022 at 20:12
• 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. Commented Nov 6, 2022 at 20:13

$$\DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\ad}{ad} \DeclareMathOperator{\Lie}{Lie} \newcommand{\g}{{\mathfrak g}} \newcommand{\z}{{\mathfrak z}} \newcommand{\s}{{\mathfrak s}} \newcommand{\O}{{\mathcal O}} \newcommand{\wh}{\widehat} \newcommand{\wt}{\widetilde}$$ 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 $$$$\label{e:*} T_\alpha(\O_r)\cong T_{\alpha_\s}(\O_{r_\s}).\tag{*}$$$$

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.