I am studying through Introduction to mechanics and symmetry by Marsden and Ratiu, specifically the chapter on Momentum maps, and wanted some confirmation as to whether my argument for the following problem is correct. I have added quite a bit of context to the original problem to make it more readily understandable.
Let $G$ be a finite dimensional Lie Group. We take $P=T^*G$ a Poisson manifold with bracket induced by the canonical symplectic form, that is, $\{ F,G\}=\omega(X_F,X_G)=X_F[G]$, where $\omega=-d\Theta$, for $\Theta$ the canonical one form on $P$.
We take the left action on $T^*G$ to be the cotangent lift of conjugation on $G$, namely,$\Phi_g: G\to G$, by $\Phi_g h:=ghg^{-1}$ and $\Psi_g: T^*G\to T^*G$, $\Psi_g(\alpha_h)=T^*\Phi_g(\alpha_h)$. This means that for $v\in T_{g^{-1}hg}G$, $$\Psi_g(\alpha_{h})(v)=T^*\Phi_g(\alpha_{h})(v)=\langle \alpha_{h}, T\Phi_g\cdot v\rangle $$
We then Let $\mathbf{J}: T^*G\to \mathfrak{g}^* $ be the moment map: $\langle \mathbf{J}(z),\xi\rangle = \Theta(\xi_P)$, where $\xi_P(h):= \frac{d}{dt}\vert_{t=0}\Psi_{\exp(t\xi)}(\alpha_h)$. I would like to show that $\mathbf{J}$ is equivariant under our group action, i.e. $\langle \mathbf{J}(\alpha), \mathrm{Ad}_{g^{-1}}\xi\rangle=\langle \mathbf{J}(\Psi_g\alpha), \xi\rangle $.
I am a bit shaky on some of my details, so I wanted to make sure that my proof is correct.
We have: $$\langle \mathbf{J}(\Psi_g\alpha_h), \xi\rangle=\mathbf{i}_{\xi_P(\Psi_g\alpha_h)}\Theta=\mathbf{i}_{\Psi^*_{g}(\xi_P(\alpha_h))}\Theta=\mathbf{i}_{(\mathrm{Ad}_{g^{-1}}\xi)_P(\alpha_h)}\Theta=\langle \mathbf{J}(\alpha_h), \mathrm{Ad}_{g^{-1}}\xi\rangle$$
Does this seem correct? It seems as though I did not use the fact that $\Theta$ is the canonical one form on $T^*G$, though I may have implicitly done this when applying the pushforward. Any comments would be greatly appreciated.
EDIT:
I've been looking into this some more, particularly in the expression I have that seems to imply $\xi_p(\Psi_g\alpha_h)=\Psi^*_g(\xi_P(\alpha_h))$.
By definition: $\xi_P(\Psi_g(\alpha_h))=\frac{d}{dt}\vert_{t=0}\Psi_{\exp(t\xi)}\Psi_{g}(\alpha_h)$. I'll investigate this by considering a smooth local vector field $X$ defined on a neighborhood $U$ of $ghg^{-1}\in G$ for for which $\exp(t\xi)ghg^{-1}\exp(-t\xi)\in U$ for $t$ small. We have \begin{align}\langle \Psi_{\exp(t\xi)}\Psi_{g}(\alpha_h),X(t)\rangle&=\langle T^*\Phi_{\exp(t\xi)} T^*\Phi_g(\alpha_h),X(t)\rangle\\ &=\langle T^*(\Phi_{g}\circ \Phi_{\exp(t\xi)})(\alpha_h), X(t)\rangle\\ &=\langle \alpha_h, T\Phi_{g}\cdot T\Phi_{\exp(t\xi)}(X(t))\rangle \end{align}
This seems to imply that my statement was correct, i.e. $\xi_P(\Psi_g(\alpha_h))=\Psi^*_g(\xi_P(\alpha_h))$ but I am unsure how to come to this conclusion formally.
