Let $G$ be a Lie group, $\mathfrak{g}$ its Lie algebra and $\mathfrak{h}\subseteq \mathfrak{g}$ a vector subspace. I defined two smooth vector fields $X, Y:G\rightarrow TG$ setting $X(g)=DR_g(e)u$ and $Y(g)=DR_g(e)v$ where $R_g:G\rightarrow G$ is the right translation and $u, v\in \mathfrak{h}$ are fixed elements. Is it true that $[u, v]=[X, Y](e)$? I need that in a proof but I can't justify it, can anyone help me?
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Does taking $\mathfrak h=\mathfrak g$ make any difference? If your equation holds for the whole space, then it holds for a subspace. – Marc van Leeuwen May 05 '13 at 19:38
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but I don't know if the equality holds $[u, v]=X, Y$ for the whole space, does it? – PtF May 05 '13 at 19:52
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I just meant that is the answer to you question "Is it true..." is yes, then it is in particular true for $\mathfrak h=\mathfrak g$ (since you said nothing that excludes this case), and conversely if the answer for the particular case $\mathfrak h=\mathfrak g$ is "yes", then it must be "yes" for all $\mathfrak h$, by restriction. So the question really boils down to whether it holds for $\mathfrak h=\mathfrak g$. – Marc van Leeuwen May 06 '13 at 04:20
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Your statement does not hold, but requires only throwing in a minus sign on one side to be repaired. Citing from Wikipedia,
For a Lie group $G$, the corresponding Lie algebra is the tangent space at the identity, which can be identified with the left invariant vector fields on $G$. The Lie bracket of the Lie algebra is then the Lie bracket of the left invariant vector fields, which is also left invariant.
This means that your statement would be true if you had defined $X,Y$ to be left-invariant vector fields with values $u,v$ respectively at the origin, as $X'(g)=DL_g(e)u$ and $Y'(g)=DL_g(e)v$; then $Z'(g)=DL_g(e)[u,v]$ is a left-invariant vector field with value $[u,v]$ et $e$, and according to the cited statement one has $Z'=[X',Y']$, and in particular $[u,v]=Z'(e)=[X',Y'](e)$.
To see that for right-invariant vector fields one has to throw in a minus sign, one may reason that right-invariant vector fields on $G$ are left-invariant vector fields on the opposite group $G^{op}$, which is just $G$ but with multiplication in the opposite order. The Lie bracket $[X,Y]$ of vector fields does not depend on the group structure, but the Lie bracket $[u,v]$ on the Lie group changes sign if we replace $G$ by $G^{op}$ (this is easy to check when the the Lie bracket is given by a commutator).
Marc van Leeuwen
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Perfect, I adapted this question from a proof I was doing and I had already shown $X$ and $Y$ were both right invariant vector fields.. Where can I find the proof of the fact that the Lie algebra bracket and the Lie bracket of left invariant vector fields coincide? Thanks for the help =D – PtF May 06 '13 at 10:50
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I think the Lie bracket on the tangent space at identity is often motivated (maybe even defined) through the Lie bracket of the corresponding left-invariant vector fields. Although this seems to be part of the fundamental theory of Lie groups, I have no reference available. – Marc van Leeuwen May 06 '13 at 12:01
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