MHB What Is the Pushforward of the Inversion Mapping at the Identity of a Lie Group?

[Euge]

Here is this week's POTW:

-----
Let $G$ be a Lie group, and let $i : G \to G$, $i(g) = g^{-1}$, be the inversion mapping. Compute the pushforward of $i$ at the identity of $G$.
-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

Let $e$ be the identity of $G$; let $m : G \times G \to G$ be the multiplication mapping given by $m(g,h) = gh$ for all $g,h\in G$. Let $X\in T_eG$. By linearity of the pushforward $m_{*,(e,e)}$,

$$m_{*,(e,e)}(X,i_{*,e}(X)) = m_{*,(e,e)}(X,0) + m_{*,(e,e)}(0,i_{*,e}(X)).$$

If $c(t)$ is an integral curve of $X$ starting at $e$, then $i(c(t))$ is an integral curve of $i_{*,e}(X)$ starting at $e$. Thus $(c(t),e)$ and $(e,i(c(t))$ are integral curves of $(X,0)$ and $(0,i_{*,e}(X))$, respectively, starting at $(e,e)$. Hence

$$m_{*,(e,e)}(X,0) = \frac{d}{dt}\bigg|_{t = 0} m(c(t),e) = \frac{d}{dt}\bigg|_{t = 0} c(t) = X$$

and

$$m_{*,(e,e)}(0,i_{*,e}(X)) = \frac{d}{dt}\bigg|_{t = 0} m(e,i(c(t)) = \frac{d}{dt}\bigg|_{t = 0} i(c(t)) = i_{*,e}(X).$$

This establishes the equation

$$m_{*,(e,e)}(X, i_{*,e}(X)) = X + i_{*,e}(X).$$

On the other hand, since $(c(t),i(c(t))$ is an integral curve of $(X,i_{*,e}(X))$ starting at $(e,e)$,

$$m_{*,(e,e)}(X,i_{*,e}(X)) = \frac{d}{dt}\bigg|_{t = 0} m(c(t),i(c(t)) = \frac{d}{dt}\bigg|_{t = 0} e = 0.$$

Therefore

$$0 = X + i_{*,e}(X),$$

yielding $i_{*,e}(X) = -X$. This proves that $i_{*,e}$ is given by the negative.