Equality involving Lie Brackets I have a question concerning Lie brackets: Consider the Lie bracket $$[, ]:\mathfrak{g}\times \mathfrak{g}\rightarrow \mathfrak{g},$$ where $\mathfrak{g}=T_eG$ is the Lie algebra of a Lie group $G$. Also, consider the application $$R_g:G\rightarrow G, $$ given by $R_g(a)=ag$. Suppose $\mathfrak{h}$ is a subspace of $\mathfrak{g}$. If $u, v\in \mathfrak{h}$ then does the equality holds $$[DR_p(e)u, DR_p(e)v]=[u, v](p)?$$
 A: A priori, it depends on the way you extend your vector $u,v\in \mathfrak{g}$ to vector fields in $TG$ and of the subspace $\mathfrak h$ in general.
Usually, the vector field associated to $u$ in $TG$ is the unique left-invariant vector field $X$ such that $X_e=u$.
And as you might notice, $[DR_p(e)u,DR_p(e)v]=DR_p(e)[u,v]$ so you are asking if for every $u,v\in \mathfrak{h}$, $DR_p(e)[u,v]=[u,v](p)$.
If you want it to be true for every point $p\in G$, then this is equivalent to say that $[X,Y]$ is a right-invariant vector field where $X,Y\in TG$ are the unique left-invariant vector fields associated to $u,v\in \mathfrak g$. This is in some sense a very strong statement.
For example, if you look at the Lie group $GL_2(\mathbb R)$ and the two vectors $u=\begin{pmatrix}1&0\\0&0\end{pmatrix}, v=\begin{pmatrix}0&0\\1&0\end{pmatrix}\in \mathfrak{gl}_2(\mathbb R)$, then you can check that $DR_p(I_2)[u,v]\neq [u,v](p)$ for any matrix of the form $p=\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$ with $\alpha\neq 0$.
