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?
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$\begingroup$ Yes. This is often how the Lie bracket on a Lie algebra is defined. $\endgroup$– Ben Webster ♦Commented Feb 17, 2011 at 1:26
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$\begingroup$ For example, this is what Frank Warner does in his Foundations of differentiable Manifolds and Lie Groups. $\endgroup$– Mariano Suárez-ÁlvarezCommented Feb 17, 2011 at 1:32
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8$\begingroup$ 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.) $\endgroup$– Tom GoodwillieCommented Feb 17, 2011 at 1:41
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$\begingroup$ Thanks Tom. Somehow I thought that that won't work. But it does. $\endgroup$– RexCommented Feb 17, 2011 at 2:09
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1$\begingroup$ 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. $\endgroup$– Deane YangCommented Feb 17, 2011 at 2:22
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2 Answers
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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, $[X^R,Y^R]=[-\iota_*X,-\iota_*Y]=\iota_*[X,Y]=-[X,Y]^R$.
answered Nov 28, 2011 at 18:43
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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.
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2$\begingroup$ 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. $\endgroup$ Commented Nov 28, 2011 at 19:21
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1$\begingroup$ @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. $\endgroup$– MTSCommented Nov 28, 2011 at 23:49
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$\begingroup$ 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! $\endgroup$ Commented Dec 1, 2011 at 16:32