Understanding proof that $[X_e,Y_e]\in T_e G$ generates the right-invariant vector field $[X,Y]$ on $G$

Question

Let $X_e,Y_e \in T_eG$ be vectors and $G = GL(n).$ Then the right translation is given by $Y_g = Y_e g$ and $X_g = X_e g.$

Now, I have a proof showing that $[X_e,Y_e] \in T_eG$ is the element generating the right- invariant vector field $[X,Y]$ on $G.$

Now, the proof starts with

$$([X,Y]_g)_{i,j}) = \sum_{k,l=1}^{N} \left((Y_g)_{k,l} \frac{\partial (X_g)_{i,j}}{\partial g_{k,l}}- (X_g)_{k,l} \frac{\partial (Y_g)_{i,j}}{\partial g_{k,l}} \right),$$

but I don't understand where the derivatives come from in this expression. Can anybody explain this equality to me?

Answer 1

Two ingredients here:

(1) The Lie bracket is actually the Lie derivative $\mathscr L_X Y$ of the vector fields. By definition, this is $$\mathscr L_X Y(p) = \frac d{dt}\Big|_{t=0} \phi_{-t*}Y_{\phi_t(p)},$$ where $\phi_t$ is the flow of $X$.

(2) The flow of the right-invariant vector field $X$ is given by left multiplication by $\exp(tX_e)$. (Solve $\dfrac{dg}{dt}=X_eg$.)

Putting these together, we have \begin{align*} \mathscr L_X Y(p) &=\frac d{dt}\Big|_{t=0} \phi_{-t*}Y_{\phi_t(p)} \\ &= \frac d{dt}\Big|_{t=0} \exp(-tX_e)Y_e\exp(tX_e) = -X_eY_e+Y_eX_e, \end{align*} as desired.

Answer 2

This is really the definition of Lie bracket (defined on $GL(n)$ as a manifold). The definition is

$$\begin{split} [X, Y]_g &= XY - YX\\ & = \sum_{k,l} X_{k,l}\partial_{g_{k,l}} \ Y - Y_{k,l} \partial_{g_{k,l}} \ X \end{split}$$

That's why you have to differentiate the vector fields. (For some reason I don't know why, it's of by a sign)