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This question is related to this link: Geometric difference between two actions of $GL_n(\mathbb{C})$ on $G\times \mathfrak{g}^*$

Further analyzing Scenerio 1: Let $G=GL_n(\mathbb{C})$ act on $G$ by left multiplication ($g.x=gx$). Taking the derivative of this action $\mathfrak{g}\rightarrow Vect(G)$, we see that $G$ acts on $TG=G\times \mathfrak{g}$ by $ g.(x,y)=(gx,y), $ where $\mathfrak{g}$ is the tangent space of $G$ at the identity.

Why doesn't $g$ move the vector $y$? Is it because the $G$-action on $G$ is transitive?

However from Scenerio 2 at the above link, we see that $G$ acting on $G$ by right inverse ($g.x=xg^{-1}$) does induce an action on the tangent space at the identity (the action is by conjugation).

Back to Scenerio 1: Now dualize the infinitesimal $G$-action on $TG$ to obtain $\mu_1:T^*G\rightarrow \mathfrak{g}^*$ where the map is $(x,y)\mapsto y$ (is this the right map?). Then do we have $\mu_1^{-1}(0)/G=T^*(G/G)=\{\mathrm{I}\}\times \{\mathrm{I}\}\cong pt$, where $\mathrm{I}$ is the identity matrix?

Recalling Scenerio 2 from above link: Similarly do we have $\mu_2:T^*G\rightarrow \mathfrak{g}^*$ where $(x,y)\mapsto xyx^{-1}$ with $\mu_2^{-1}(0)/G=T^*(G/G)=pt$?

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    I think your construction is not correct. Probably you mean $\mathfrak g$ instead of $Vect(G)$ (which as it is makes little sense). But "compose with the tangent bundle of $G$" is not really meaningful (a bundle is not a function), and there is no reasonable map $\mathfrak g\to TG\simeq G\times\mathfrak g$. The way you want to dualize is unclear to me as well.2012-08-05

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