Pushforward of inverse map at the identity?

Question

Let $G$ be a Lie group and $i:G \rightarrow G$ denote the inversion map $i(x) = x^{-1}$.

(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.

Answer 1

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.

Answer 2

This is a proof using just the definition of a Lie group. Let $m: G\times G \to G$ be the multiplication, then $$ dm_e : T_{(e,e)} (G\times G) \cong T_eG \oplus T_eG \to T_e G$$ is given by $(X, Y) \mapsto X+Y$. This is true as clearly $dm_e (X, 0) =X$, $dm_e (0, Y) = Y$ and that $dm_e$ is linear.

The composition

$$ G \overset{(\operatorname{id}, i)}{\to} G\times G \overset{m}{\to}G$$ is constant (everything maps to $e$). Using the chain rule we have (for all $X\in T_eG$)

$$ 0= dm_e ( \operatorname{id}_* X, i_* X) = X + i_*X,$$

thus $i_*X = -X$.