I have some question about the answer of Lie algebra of $GL_n(\mathbb{C})$. In the picture below:
First,the definition of left action. What is the mean of $xX(x^{-1}y)$, I understand $X(x^{-1}y)$ as the value of vector field $X$ at $x^{-1}y$. Then ,how $x$ act on $X(x^{-1}y)$ ? Is it $x_*(X(x^{-1}y))$ (I use the denote in Riemannian geometry)?
Second, What is $\widetilde{[a,b]}$ ? I don't know the mean of ~ over $[a,b]$.
$a\in G={\rm Gl}_n (\mathbb{C}),\ x\in M_n( \mathbb{C})$ Define a vector field $X(a):=a\cdot x\ \ast$ where multiplication is matrix multiplication So $$ L_b\ X(a)=bax =X(ba) $$ Hence $X$ is a left invariant vector field
If $e^{tx}$ is an integral curve at $I$, then $ae^{tx}$ is integral curve at $a$ : $$ \frac{d}{dt} ae^{tx}= L_a X(t)=X(ae^{tx}) $$
Recall the definition of Lie bracket in Lie group : $$ Ad_a : T_IG\rightarrow T_IG,\ Ad_a (x)= \frac{d}{dt}\bigg|_{t=0} ae^{tx}a^{-1} $$
$$ [y,x](e):=\frac{d}{dt} Ad_{e^{ty}} (x) $$
Note that $$[y,x](e)=yx-xy = \frac{\partial }{\partial t} \frac{\partial }{\partial s} e^{ty}e^{sx}e^{-ty}$$
Here $$ df\ L_a\ [y,x](e):=\frac{\partial }{\partial t} \frac{\partial }{\partial s} f(ae^{ty}e^{sx}e^{-ty} )$$
In further recall the definition of $[Y,X](a)$ in Riemannian manifold : If $\phi$ is flow of $Y$ \begin{align*} [Y,X](a)&= \frac{d}{dt} d\phi_{-t} X_{\phi_t(a)} \\&= \frac{\partial}{\partial t}\frac{\partial}{\partial s} \phi_t(a)e^{sx} e^{-ty} \\&= \frac{\partial}{\partial t}\frac{\partial}{\partial s} ae^{ty} e^{sx} e^{-ty} \end{align*}
So we complete the proof