Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.