MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a Lie group and let $\xi.\eta$ be left invariant vector fields. We can now construct right invariant vector fields $X_\xi$ and $X_\eta$ by defining $X_\xi(e)=\xi(e)$ and $X_\eta(e)=\eta(e)$. For $GL_n$, it is true that $[X_\xi,X_\eta]=X_{[\eta,\xi]}$. Is it true for any Lie group?

share|cite|improve this question
Yes. This is often how the Lie bracket on a Lie algebra is defined. – Ben Webster Feb 17 '11 at 1:26
For example, this is what Frank Warner does in his Foundations of differentiable Manifolds and Lie Groups. – Mariano Suárez-Alvarez Feb 17 '11 at 1:32
Ben, I presume you mean that this is often how the Lie algebra of a Lie group is defined. I believe that the questioner knows this, and that the question is whether, when you switch from left invariant to right invariant, you simply get a sign change in the bracket. (The answer is yes, by using the map $g\to g^{-1}$ from the group to itself.) – Tom Goodwillie Feb 17 '11 at 1:41
Thanks Tom. Somehow I thought that that won't work. But it does. – Rex Feb 17 '11 at 2:09
If it's true for $GL(n)$, it's true for any subgroup of $GL(n)$. It's also true for any group that covers any of these groups. That takes care of "most" Lie groups and makes it probable that it's true for all Lie groups. – Deane Yang Feb 17 '11 at 2:22

For a left-invariant vector field $X$ on $G$, denote by $X^R$ the right-invariant vector field with the same value at the identity. From $\iota_*X=\iota_*L_{g*}X=R_{g^{-1}*}\iota_*X$, we see that $\iota_*X$ is right-invariant, where $\iota$ is the inversion map $g\mapsto g^{-1}$. Since $\iota_*X(1)=-X(1)$, we get $X^R=-\iota_*X$. Finally, $[X^R,Y^R]=[-\iota_*X,-\iota_*Y]=\iota_*[X,Y]=-[X,Y]^R$.

share|cite|improve this answer

The above statement ist true for GL(n). As Deane pointed out, this neceassrily implies that the statement must be true for any subgoup of GL(n) that is a Lie group as well, i.e., any subgriup of GL(n). It can be shown, however, that every Lie-Group can be considered as subgroups of some GL(n), where one uses representation theory. The above statement is really sloppy, but the essential point is that you can choose group representations that make the Lie group resemble some subgroup of GL(n). Consider for instanve the Spin Group which is either defined via the Clifford algebra or as the universal cover of SO(n).

EDIT: As I was fortunately pointed out, I mixed the terms Lie groups and Lie algebras. The theorem of Ado, which I was referring to, is valid for Lie algebras. Although it may be applied, the proof given above is more elementary and gives more "computational material". I am indebted to Ben whose comment raised my awareness to the mistake in my answer.

share|cite|improve this answer
Not quite. The universal covering group of $SL(2,\mathbb{C})$ is not a subgroup of any $GL(n)$. However, it is true that every Lie algebra has a faithful representation (Ado's theorem), and that every Lie group has a representation with discrete kernel, so that suffices to prove the result. However, much more elementary proofs suffice. – Ben McKay Nov 28 '11 at 19:21
@Ben: Sorry to nitpick, but I think you mean $SL(2,\mathbb{R})$ rather than $SL(2,\mathbb{C})$. $SL(2,\mathbb{C})$ is simply connected and hence is its own universal cover. – MTS Nov 28 '11 at 23:49
Yes, sorry, you are absolutely right, Ben, I was a bit confused and, apparently, I have messed the terms group and algebra. I will edit the answer to avoid misunderstandings. Thanks for the comment! – gggg gggg Dec 1 '11 at 16:32

Your Answer


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.