0
$\begingroup$

Let $G$ be a Lie group and denote by $T_e G$ the tangent space of $G$ at the identity element $e$.

How do I show that the identification of $T_e G$ with the space of right- invariant vector fields induces the opposite Lie algebra structure ?

I am familiar with the fact that $T_e G$ is isomorphic to the space of left- invariant vector fields on $G$ and I guess that the same holds for right-invariant vector fields.

There is a hint saying that one should consider the Lie group $\bar{G}$ equal to $G$ as a smooth manifold with the multiplication $\bar{gh} = hg$. Moreover, one should use the following lemma :

Let $G$ and $H$ be two Lie groups and let $\varphi : G \rightarrow H$ be a Lie group morphism. Then the differential $d_e \varphi : T_e G \rightarrow T_e H$ is a Lie algebra morphism.

So, if I apply this lemma to $\varphi : G \rightarrow \bar{G}, g \mapsto g^{-1}$, then $d_e \varphi : T_e G \rightarrow T_e \bar{G}$ is a Lie algebra morphism.

Unfortunately, I am stuck and would be pleased if someone can help me to solve this problem.

Thanks.

$\endgroup$
1
$\begingroup$

The sketch you give is almost complete, one has only to conclude from this the result. Let me explain: You start with the Lie group $G$ and for obvious reasons also the opposite group (what you called $\overline{G}$) is a Lie group (with multiplication in the opposite order, whence opposite group). We now have the following facts mentioned already by you:

  1. $\iota \colon G \rightarrow \overline{G} , x \mapsto x^{-1}$ is a morphism of Lie groups. Computing $T_e \iota$ (what you denote $d_e\iota$) one obtains the well known (i.e. contained in many introductory Lie theory texts) formula $T_e \iota(v) = - v$
  2. As mentioned, $T_e G$ is isomorphic (as Lie algebra!) to the Lie algebra of left invariant vector fields on $G$ and $T_e \overline{G}$ is isomorphic (as Lie algebra) to the left invariant vector fields on $\overline{G}$.
  3. Using that $\iota \colon G \rightarrow \overline{G}$ is a Lie group morphism, $T_e \iota$ is a Lie algebra morphism and we have by bilinearity and antisymmetry of the Lie bracket. $$-[x,y]_G = T_e \iota ([x,y]_G) = [T_e \iota (x),T_e \iota (y)]_{\overline{G}} = [x,y]_{\overline{G}} = [y,x]_G$$ (here subscripts denote Lie brackets with respect to the group structure of $G$ or $\overline{G}$ respectively).

Observe now that due to the opposite composition in $\overline{G}$ left invariant vector fields on $\overline{G}$ correspond exactly to right invariant vector fields on $G$. Thus the Lie bracket on $T_e \overline{G} = T_e G$ induced by the right invariant vector fields is the Lie bracket induced by the opposite group $\overline{G}$, which is the negative of the Lie bracket induced by $G$ (due to 3.). We conclude that the right-invariant vector fields induce the opposite Lie algebra structure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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