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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Oct 9, 2012 at 6:39
  • $\begingroup$ 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 $\mathbb R$ $\endgroup$ Commented Oct 9, 2012 at 7:12
  • $\begingroup$ 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. $\endgroup$ Commented Oct 9, 2012 at 7:14
  • $\begingroup$ @Aaron (I know I am super late...) I just posted a proof using just the definition of a Lie group. $\endgroup$ Commented May 5, 2021 at 6:20

3 Answers 3

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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 $i_*X = -X$.

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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.

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Take $X\in T_eG$ and let $\alpha\colon I\longrightarrow G$ be a $C^\infty$ curve (where $I$ is an open interval of $\Bbb R$ such that $0\in I$) such that $\alpha(0)=e$ and that $\alpha'(0)=X$. For each $t\in I$, let $\alpha^{-1}(t)=(\alpha(t))^{-1}$. Then $\alpha^{-1}$ is also a $C^\infty$ curve such that $\alpha^{-1}(0)=e$. Besides, $\alpha^{-1}=i\circ\alpha$, and therefore$$\left(\alpha^{-1}\right)'(0)=Di_e(\alpha'(0))=Di_e(X).$$On the other hand, $\alpha\cdot\alpha^{-1}$ is constant, and therefore $\left(\alpha\cdot\alpha^{-1}\right)'(0)=0$. But$$\left(\alpha\cdot\alpha^{-1}\right)'(0)=\alpha'(0)+\left(\alpha^{-1}\right)'(0)=X+Di_e(X).$$Therefore, $Di_e(X)=-X$.

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  • $\begingroup$ Well, your proof is identical to mine posted below. $\endgroup$ Commented May 29 at 14:04

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