Momentum map for coadjoint orbit

Question

Let $G\times \mathfrak{g}^*\to \mathfrak{g}^*$ be the coadjoint action defined by $f\cdot q=Ad^*_fq$ where $Ad:G\to GL(\mathfrak{g})$ is the adjoint representation of $G$ with dual $Ad^*$. For $p\in \mathfrak{g}^*$, let $W_p$ be the coadjoint orbit of $p$, i.e., $W_p=\{Ad^*_fp:f\in G\}$. Let's define the map $\mathcal{J}:G\to W_p$ by the inclsuion $q\mapsto q$. My problem is t show this map is a momentum map for the coadjoint action.

Recall that a momentum map for a symplectic action $G\times M\to M$ is a smooth map $\mathcal{J}:M\to \mathfrak{g}^*$ such that for any $X\in \mathfrak{g}$, $$i_{\xi_X}\omega=d\langle \mathcal{J}(\cdot ),X\rangle$$ where $(M,\omega)$ is a symplectic manifold and $\xi_X$ is the infinitesimal generator of the action, i.e., the vector field on $M$ defined by $(\xi_X)^q\equiv \frac{d}{dt}\Big|_{t=0}e^{tX}\cdot q$.

My try: One can easily check that the infinitesimal generator of the coadjoint action is $\xi_X=ad^*_X$ where $ad^*_X=\frac{d}{dt}\Big|_{t=0}Ad^*_{e^{tX}}$ and symplectic form on $W_p$ is defined by $\omega_q (ad^*_Xq,ad^*_Yq)\equiv \langle q,[X,Y]\rangle$ because $T_qW_p=\{ad^*_Xq:X\in \mathfrak{g}\}$. I want to show that $i_{ad^*_X}\omega =d\langle \mathcal{J}(\cdot ),X\rangle$ where $\mathcal{J}$ is the inclusion. In doing so, let $\gamma (t)$ be a path in $W_p$ such that $\gamma (0)=q$ and $\dot{\gamma}(0)=ad^*_Y q\in T_qW_p$. On the one hand, we have $$(i_{ad^*_X}\omega)_{q}\dot{\gamma}(0)=\omega_{q}((ad^*_X)_{q},\dot{\gamma}(0))=\omega_{q}(ad^*_X q,ad^*_Yq)=\langle q,[X,Y]\rangle=\langle q,ad_X (Y)\rangle=\langle ad^*_Y q,X\rangle .$$ On the other hand, since $ad^*_X=\xi_X$ is the infinitesimal generator of the coadjoint action of $G$ on $W_p$, we have $$(i_{ad^*_X}\omega)_{q}\dot{\gamma}(0))=(i_{\xi_X}\omega)_{q}\dot{\gamma}(0).$$ Now I don't know how to compute $(d\langle \mathcal{J}(\cdot ),X\rangle)_q \dot{\gamma}(0)$ to show that it equals $\langle ad^*_Y q,X\rangle$. Actually, I don't know how to work with 1-form $d\langle \mathcal{J}(\cdot ),X\rangle$.

I would be really appreciate it if somepne could explain it to me. Thanks in advance.

Answer 1

I think you meant to write $\mathcal{J}:W_p\mapsto \mathfrak{g}^*$ (given by the inclusion map). In other words $$ \mathcal{J} = \operatorname{id}_{\mathfrak{g}^*}\vert_{W_p}, $$ the restriction of the identity map $\operatorname{id}_{\mathfrak{g}^*}$ to the coadjoint orbit. The goal is to show this is a momentum map for the coadjoint action on the coadjoint orbit.

Then taking an arbitrary vector $\operatorname{ad}_Y^*q$ in the tangent space to $W_p$ at $q$ $$ (d\langle\mathcal{J}(\cdot),X\rangle)_q(\operatorname{ad}_Y^*q) = \frac{d}{dt}\Bigg\vert_{t=0}\langle\operatorname{id}_{\mathfrak{g}^*} (q+t\operatorname{ad}_Y^*q),X\rangle = \langle\operatorname{ad}_Y^*q,X\rangle $$ which is the same as your calculated value of $(i_{\operatorname{ad}_X^*}\omega)_q(\operatorname{ad}_Y^*q)$. Hence $$ d\langle\mathcal{J}(\cdot),X\rangle = i_{\operatorname{ad}_X^*}\omega, $$ as required.