I am having some difficulties understanding this proof. Let $G$ be a closed matrixsubgroup of the general linear group.
We have a right translation $Y(g):=dR_g(e) Y(e)$ on the Lie algebra $Y \in \mathfrak{g}$ that is just given by right multiplication with the matrix $g \in G,$ i.e. $Y(g):=Y(e)g.$ Similarly, $X$ is also a right-inv. vector field.
Now, I want to show that $[X(g),Y(g)] = [X(e),Y(e)]g.$
We actually showed it in class and went like that ($\textbf{first expression of Lie bracket(!)}$)
$$[X(g),Y(g)]_{i,j} = \sum_{k,l=1}^{n} Y(g)_{k,l} \frac{\partial X(g)_{i,j}}{\partial g_{k,l}}- X(g)_{k,l} \frac{\partial Y(g)_{i,j}}{\partial g_{k,l}}$$
$$= \sum_{k,l,m=1}^{n} Y(g)_{k,l}X(e)_{i,m} \frac{\partial g_{m,j}}{\partial g_{k,l}}- X(g)_{k,l}Y(e)_{i,m} \frac{\partial g_{m,j}}{\partial g_{k,l}}$$ using linear independence of the coordinates we get $$= \sum_{k=1}^{n} Y(g)_{k,l}X(e)_{i,k} - X(g)_{k,l}Y(e)_{i,k} $$ which is nothing but ($\textbf{second expression of Lie bracket(!)}$) $$(X(e)Y(e)-Y(e)X(e))g)_{i,j} = ([X(e),Y(e)]g)_{i,j}$$
which was to be shown.
Now, the first and last equality are the ones that bother me.
Afaik from wikipedia the Lie bracket is $[X,Y]= X(Y)-Y(X),$ but the first line suggests that we are using the different convention$[X,Y]= Y(X)-X(Y)$ in this proof. Now, I would understand that the proof still works, if we would have sticked to this convention, but in the last line we use that $[X(e),Y(e)]=X(e) \circ Y(e)-Y(e) \circ X(e).$ This is nothing but the more canonical convention. Thus, we used two different definitions of Lie brackets.
So somehow the conventions don't add up here. Could anybody explain this fact to me?
Edit: As my question caused apparently some confusion, I want to clarify the problem. The thing is that I suspect that in this proof, there were two different definitions of the Lie bracket used. This would mean that the proof is wrong. On the other hand, the proof only consists of working out the calculation and thus, it should be possible to identify whether one can still get the correct result, by modifying the two positions appropriately?
$\textbf{Edit2:}$ First, I want to thank Ted Shifrin for shedding some light on what is going on. Despite, there is still the problem that the "proof" given here somehow contradicts the right-invariance statement of the Lie-bracket. Cause if we start with $[X(g),Y(g)]=Y(g)X(g)-X(g)Y(g)$ (which is in the proof according to the notation proposed by Ted Shifrin and which notation I will assume in this paragraph) and show that this is equal to $(X(e)Y(e)-Y(e)X(e))g$ then we would have actually shown that the Lie bracket is not(!) right-invariant. As the latter is $-[X(e),Y(e)]g$ where the minus sign is crucial, instead of $[X(e),Y(e)]g$ which we would have needed in order to show right-invariance. So something is still very wrong here.
I would moreover want to ask how to interpret Lie brackets appearing in representation theory, as for instance: $Ad_g(X)=gXg^{-1}$(at least in the setting of matrix groups) and then $ad_{\xi}(\eta)= [\xi , \eta]= \xi \eta - \eta \xi.$
Thus, I feel that in Lie theory setting the canonical Lie bracket $[A,B] = AB-BA$ is more common in some way. Maybe you Ted Shifrin or anybody else could also comment on this. Even if this does not immediately help with the problem.
If anything is unclear, please let me know.
As I explained in the answer to this question, the Lie bracket of left-invariant vector fields is $[X,Y]=XY-YX$, whereas the Lie bracket of right-invariant vector fields has the opposite sign. (It's not a matter of convention. It's a computation of the Lie derivative in either case.) I don't know where your lecture got the second expression for the Lie bracket, but there is ostensibly a sign error there.
However, if you're trying to show that the Lie bracket of right-invariant vector fields is again right-invariant, this follows directly from the standard fact proved in an introductory differentiable manifolds course that if $f\colon M\to N$ is smooth, then $[f_*X,f_*Y] = f_*[X,Y]$ for any vector fields $X,Y$ on $M$.