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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.

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  • $\begingroup$ Do you want me to try an answer for that one, as well? Although I really think that you just need to follow your nose and try some simple examples. Start out with $G=S^1$, and after it $SO(3;\mathbb{R})$. $\endgroup$
    – CvZ
    Sep 9, 2015 at 2:07
  • $\begingroup$ @CvZ could you maybe briefly sketch in a comment how you want to show that this is a manifold? (maybe mention some theorems that you want to use)? $\endgroup$
    – user167575
    Sep 9, 2015 at 2:11
  • $\begingroup$ You wanted to see if this is an application of the regular value theorem, is it not? So try it for $G=S^1$, check if the momentum mapping satisfies the hypothesis of the theorem. You do understand what happens for the momentum mapping of an integrable system, do you not? $\endgroup$
    – CvZ
    Sep 9, 2015 at 2:16
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    $\begingroup$ ah, then $H_{\xi}(q,p) = \xi(q_1p_2-q_2p_1),$we can differentiate w.r.t to any coordinate and get $dJ(q,p) = (p_2,-p_1,-q_2,q_1).$ So not any $x$ is a regular value, as $x=0$ does not do it, as then $dJ(0,0)=(0,0,0,0) $ in particular. $\endgroup$
    – user167575
    Sep 9, 2015 at 2:45
  • $\begingroup$ I recommend you to change your title. Every smooth action of Lie group on any manifold generates smooth manifolds: the orbits. It is a misleading title and not that helpful for other users. Try something with the words 'smooth manifolds and preimage of momentum mappings'. $\endgroup$
    – CvZ
    Sep 9, 2015 at 16:46

1 Answer 1

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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 $G$.


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.

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  • $\begingroup$ I am really tired right now, so I might be wrong about this. Morrow is another day and I shall revise this answer. $\endgroup$
    – CvZ
    Sep 9, 2015 at 4:42
  • $\begingroup$ so you say: Yes it is a manifold, because it is diffeomorphic to $G$? I think you can also see it from my explicit representation of the preimage in the question, right? $\endgroup$
    – user167575
    Sep 9, 2015 at 11:23
  • $\begingroup$ Sorry for not see a direct answer. I was really tired and had to rewrite things in a way that I was able to work it out. However, It is really useful to work with left and right trivialisations of $TG$ and $T^*G$. $\endgroup$
    – CvZ
    Sep 9, 2015 at 16:43

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