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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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    Is $\frak g$ the lie algebra of $G$?2012-07-31
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    @Andrew Yes, $\mathfrak{g}$ is the Lie algebra of $G$.2012-07-31
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    PS - not to be picky, but you can get a better looking dot for the action with \cdot2012-07-31
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    @Andrew Thanks! I'll keep that in mind for the future.2012-07-31
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    What is a geometric difference?2012-07-31
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    [This question](http://math.stackexchange.com/q/61387/5363) is somewhat related.2012-07-31
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    @t.b. Thank you! It definitely looks related. =)2012-07-31
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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.