smooth manifolds and preimage of momentum mappings

Question

Let $G$ be a Lie group acting on itself as $\phi(h)(g)= L_h(g)$ as a left translation. Then we can consider the cotangent lift of this action, namely $\Phi: G \times T^*G \rightarrow T^*G$ as $\Phi(h)(g,p) = (hg,(dL_{h^{-1}}(hg))^*p).$

It can now be shown that such a map induces a canonical Hamilton function with moment map on the cotangent bundle

$H_{\xi}(g,p) = J(q,p)(\xi):=(dR_g)^*(e)(p)(\xi)$ for some $\xi \in \mathfrak{g}.$

This is now my motivation for the question:

If we consider $J^{-1}(x)$ for $x \in \mathfrak{g}^*$ then this set is given by $$J^{-1}(x) = \{ (g , (dR_{g^{-1}})^*(g)(x));g \in G \}.$$

My question is: Why is it a manifold? (I admit that it looks very much like an application of the regular value theorem, but I don't see why it applies)

If there is anything unclear about my question, please let me know.

Answer 1

Using the left multiplication of $G$ on itself one can show that $T^*G\cong \mathfrak{g}^*\times G$, and under that identification the lifted action is simply $h\cdot (\xi,g)=(\xi,h\cdot g)$, and the momentum mapping is $\mathfrak{g}^*\times G\ni (\xi,g)\mapsto(\mathrm{Ad}_{g^{-1}})^*(\xi)\in\mathfrak{g}^*$.

The preimage by the momentum mapping of a point $\zeta\in\mathfrak{g}^*$ is the set of points $(\xi,g)\in\mathfrak{g}^*\times G$ satisfying $(\mathrm{Ad}_{g^{-1}})^*(\xi)=\zeta$; thus, $\{((\mathrm{Ad}_g)^*(\zeta),g)\}_{g\in G}\subset\mathfrak{g}^*\times G$.

The mapping $G\ni g\mapsto ((\mathrm{Ad}_g)^*(\zeta),g)\in\mathfrak{g}^*\times G$ provides a diffeomorphism between the The preimage by the momentum mapping of a point $\zeta\in\mathfrak{g}^*$ and $G$.

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Edit:

In the Original Poster notation, $J^{-1}(x)=\{(g,(dR_{g^{-1}})^*(g)(x)); \ g\in G\}$ and the mapping $G\ni g\mapsto (g,(dR_{g^{-1}})^*(g)(x))\in T^*G$ provides a diffeomorphism between $G$ and $J^{-1}(x)$; therefore the preimage $J^{-1}(x)$ is a smooth manifold.