# Lie bracket of Invariant Vector fields

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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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, $[X^R,Y^R]=[-\iota_*X,-\iota_*Y]=\iota_*[X,Y]=-[X,Y]^R$.
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