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Let $G=GL_n(\mathbb{C})$.

Scenerio 1: Let $G$ act on $T^*(G)=G\times \mathfrak{g}^*$ by $ g.(x,y)=(gx,y). $

Scenerio 2: Let $G$ act on $T^*(G)=G\times \mathfrak{g}^*$ by $ g.(x,y)=(xg^{-1},gyg^{-1}). $

Is there a geometric difference between the two scenerios?

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    @t.b. What is puzzling about Scenerio 1 is that the group action $G$ on $G$ by left multiplication doesn't induce any action on on its Lie algebra, whereas in Scenerio 2, the group action $G$ on $G$ by right inverse does induce an action on $\mathfrak{g}^*$, but as Andrew has shown below, the two actions are supposed to be equivalent...2012-07-31

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The actions are equivalent. Consider the map $\phi:G\times\frak{g}^*$ $\to G\times\frak{g}^*$ defined by $(x,y)\mapsto (x^{-1},xyx^{-1}),$ where we consider the domain with the first action, and the codomain with the second. Then this map is equivariant, since $\phi(g\cdot(x,y)) = \phi((gx,y))=(x^{-1}g^{-1},(gx)y(x^{-1}g^{-1}))= g\cdot(x^{-1},xyx^{-1})=g\cdot\phi(x,y).$

To see that $\phi$ is a bijection is straightforward, thus the $G$-actions are equivalent.