# Understanding quotients $\mu^{-1}(0)/G$

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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The reason that $g$ doesn't move $y$ is that your definition of $g.(x,y)$ says so; this has nothing to do with trasitivity of a $G$ action. Also the answer to "is this the right map?" is certainly negative: exactly because the $G$ action doesn't touch $y$, the dual action doesn't "see" the second component of $T^*G$. –  Marc van Leeuwen Aug 1 '12 at 8:46
@MarcvanLeeuwen Thank you! So for Scenerio 1, a map $\mu_1: T^*G\rightarrow \mathfrak{g}^*$ doesn't exist or is this the zero map? –  math-visitor Aug 1 '12 at 21:19
It would help if you said exactly what dualizing means; this is not clear to me. The you can either come to the conclusion that it cannot be properly defined, or that it yields the zero map (I can imagine some operation that yields a zero map, but I don't think it makes much sense). –  Marc van Leeuwen Aug 3 '12 at 19:48
@MarcvanLeeuwen Thank you, your comments are certainly helpful. I meant when $G$ is acting on $G$ by left multiplication, then taking the derivative of this action induces a map of Lie algebras: $\mathfrak{g}\rightarrow T_e(G)=Vect(G)$. Compose this map with the tangent bundle of $G$ to obtain $\mathfrak{g}\rightarrow T_e(G)=Vect(G)\rightarrow TG=G\times \mathfrak{g}$ (I am not certain yet what the maps should be concretely). Then we dualize the composition to obtain $T^*G =G\times \mathfrak{g}^* \rightarrow \mathfrak{g}^*$ where the map must be $(x,y)\mapsto 0$ or maybe the map is undefined. –  math-visitor Aug 3 '12 at 19:57
@MarcvanLeeuwen When you say you can imagine some operation that yields a zero map, can you give me an example or explain such cases? It is possible that I need to re-evaluate and think through geometrically regarding Scenerio 1. But I do know that in Scenerio 1, as $G$ moves points in the domain, we can think of the tangent vectors over $x$ not being affected by the action of $G$. As for Scenerio 2, as $G$ is acting on $G$, it also moves tangent vectors over the point $x$ (for Scenerio 2, recall that I defined $g.(x,y)=(xg^{-1},gyg^{-1})$). –  math-visitor Aug 3 '12 at 20:03