# Pushforward of Inverse Map around the identity?

Let $G$ be a Lie group and $i:G \rightarrow G$ denote the inversion map.

(Notation: $f_*$ is the pushforward map $F_*:T_pG \rightarrow T_{i(p)}G$ which takes $(F_{*}X)(f)=X(f\circ F)$ and $X$ is a tangent vector, $X\in T_pG$.)

I wish to show that $i_{*}:T_{e}G\rightarrow T_{e}G$ is given by $i_{*}(X)=-X$

As a first step, it is trivial to prove that $i_*$ is an involution as $\mbox{Id}_{*}=(i\circ i)_{*}=i_{*}\circ i_{*}$ but I can't seem to make any further progress. Any help would be appreciated.

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Here's a possible suggestion (I haven't worked through the details, though). It is easy to prove the proposition for $G=\mathbb R$. Now, $\mathbb R$ embeds inside $G$ in the form of one parameter subgroups, with one subgroup for every tangent vector at $e$. Thus you can compute $i_*(X)$ be restricting it to the one parameter subgroup through $X$, I think. –  Aaron Oct 9 '12 at 6:39
You don't need to use extrinsic coordinates. You just need that inverses commute with homomorphisms, I think. Let me attempt to write up a full solution. I don't know that you can avoid one parameter subgroups, because the problem doesn't seem to have a place to get started except in the simple case of $\mathbb R$ –  Aaron Oct 9 '12 at 7:12
Thanks for the help Aaron! In fact I just realized that extrinsic coordinates were not needed and deleted my earlier comment in embarrasment. Assuming the existence of one parameter subgroups, (which by definition satisfy $\gamma(-x)=i(\gamma(x))=i\circ\gamma$) this is indeed simple as for each $\gamma$ corrosponding to X, $i_{*}X=(i\circ\gamma_{i})'(0)=-X$ ... I think. –  Sam Oct 9 '12 at 7:14

When $G=\mathbb R$, $i(x)=-x$, and so $i_*(X)=-X$.
Suppose that $\varphi:H \to G$ is a homomorphism of (Lie)-groups, and $i_H, i_G$ are the inversion maps. We can write the fact that homomorphisms preserve inverses as $i_G \circ \varphi = \varphi i_H$. Therefore $(i_G)_* \circ \varphi_* = \varphi_* (i_H)_*$.
Consider a one parameter subgroup $\varphi:\mathbb R\to G$. Then combining the two above observations, we have
$$(i_G)_*(\varphi_*(X)) = \varphi_* (-X)=-\varphi_* (X)$$
for $X\in T_e \mathbb R$. Thus, $i_*(Y)=-Y$ for every $Y\in T_e G$ that is in the image of (the derivative of) a one parameter subgroup. Since we can find a one parameter subgroup through each vector at the identity, the proposition is proved.