Take the tour ×

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?

share|improve this question
Is $\frak g$ the lie algebra of $G$? –  Andrew Jul 31 '12 at 0:25
@Andrew Yes, $\mathfrak{g}$ is the Lie algebra of $G$. –  math-visitor Jul 31 '12 at 0:26
PS - not to be picky, but you can get a better looking dot for the action with \cdot –  Andrew Jul 31 '12 at 0:27
@Andrew Thanks! I'll keep that in mind for the future. –  math-visitor Jul 31 '12 at 0:27
This question is somewhat related. –  t.b. Jul 31 '12 at 6:30
show 3 more comments

1 Answer

up vote 5 down vote accepted

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.

share|improve this answer
add comment

Your Answer


