Does the equality $[u, v]=[X, Y](e)$ holds? 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?
 A: 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).
