I want to prove from the definition that the Lie bracket $[X,Y]$ of two left-invariant vector fields $X,Y: G \rightarrow TG$ where $G$ is a Lie group is again left-invariant.
Left-invariance basically means that for all $g \in G$ and $h \in G$ we have
$$DL_{g}(h)(X(h)) = X(gh).$$
But somehow I am stuck when I try to apply $$DL_{g}(h)([X,Y](g)).$$
I just don't see how I can end up with $$[X,Y](hg).$$
There must be a trick to end up with this result. Does anybody have an idea?
Best Answer
In general:
Let $\psi:M\to M_1$ a diffeomprohism. Then for $X,Y\in\mathfrak{X}(M)$ the relation $$[\psi_*X,\psi_*Y]=\psi_*[X,Y]$$ holds.
The proof are just calculations:
So let $f\in C^\infty(M)$. \begin{align} [\psi_*X,\psi_*Y](f)&=(\psi_*X)((\psi_*Y)(f))-(\psi_*Y)((\psi_*X)(f))\\ &=(\psi_*X)(Y(f\circ\psi)\circ\psi^{-1})-(\psi_*Y)(X(f\circ\psi)\circ\psi^{-1})\\ &=X(Y(f\circ\psi)\circ\psi^{-1}\circ\psi)\circ\psi^{-1}-Y(X(f\circ\psi)\circ \psi^{-1}\circ\psi)\circ\psi^{-1}\\ &=[X,Y](f\circ\psi)\psi^{-1}\\ &=\psi_*[X,Y](f) \end{align} Therefor if $X,Y$ are left-invariant, the commutator is left-invariant too.
Last we prove the following statement (as asked for in the comments): $$(\psi_*X)(f)=X(f\circ\psi)\circ\psi^{-1}, \text{ where } f\in C^\infty(M_1,\mathbb{R})$$ Let $m\in M_1$. Actually we use the transport of a vector field (more general: the transport of a section, because a vector field is just a section of the tangent bundle): \begin{align} (\psi_*X)(f)(m)&=(\psi'_{\psi^{-1}(m)}X_{\psi^{-1}(m)})_{m}(f)\\ &=X_{\psi^{-1}(m)}(f\circ\psi)\\ &=X(f\circ\psi)\circ\psi^{-1}(m) \end{align}