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Let $G$ be a smooth manifold and recall that smooth vector fields on $G$ correspond $1-1$ to derivations of $C^\infty(G)$. If I want to compute the Lie bracket of two vector fields, I usually need to use the Lie bracket of the associated derivations:

$[X,Y](g)(f) = L_{[X,Y]}(f)(g)=[L_X,L_Y](f)(g)=L_X(L_Y(f))(g)-L_Y(L_X(f))(g)$.

At this point I could notice that

$L_X(f)(g)=X(g)(f):=D_gf(X(g))$, so that $L_X(f)=D_\bullet f(X(\bullet)) \in C^\infty(G)$.

So now I would get, for example, that $L_X(L_Y(f))(g)=X(g)(L_Y(f)) = D_g(L_Y(f))(X(g))=D_g(D_\bullet f(Y(\bullet)))(X(g)).$

Is there a way to evaluate such a thing in general, or even in a specific case, and to understand what the map $L_X(f)=D_\bullet f(X(\bullet)) \in C^\infty(G)$ does? It sounds very weird to me that I should take the differential of something given as a differential, in the last equation.

In the case $G=GL(n,\mathbb{R})$, for instance, I wanted to use such a computation to find the value of $[X,Y](1)$, where I know that $X(g)=gx$ and $Y(g)=gy$ for $x,y,g \in M_{nn}(\mathbb{R}) \cong T_1G$. The value should be $[X,Y](1)=xy-yx$ (this shows that the Lie bracket on $G$ is the usual one, by the way), but I cannot compute it, or rather: I can do it in local coordinates, but I am looking for a more general approach.

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    $\begingroup$ It shouldn't be weird. If you think of a Lie derivative as a partial derivative in the direction of $X$, then the Lie bracket of $X$ and $Y$ measures precisely how much the mixed partials $X \circ Y$ and $Y \circ X$ fail to commute. (You never used the structure of a group anywhere in this, so no, this isn't about Lie groups - it's just about smooth manifolds.) $\endgroup$ – user98602 Jul 30 '16 at 20:57
  • $\begingroup$ I agree that it is not about Lie groups (although the example came from a chapter on Lie groups), I edited that. And I appreciate the interpretation you gave me. Do you have any hint on how to use it in the example? $\endgroup$ – 57Jimmy Jul 30 '16 at 21:12
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$GL(n,R)$ is an open subset of the vector space $M_{n\times n}(R)$. The classical formula for the Lie bracket of the vectors fields $X,Y$ defined on an open subset of a vector space is $[X,Y]=-DX(Y)+DY(X)$.

The vector field defined by $X(g)=gx$ is a linear map on $M_{nn}(R)$, thus equal to its differential, we deduce that if $Y(g)=gx$, $-DX(Y(g))+DY(X(g))=-gyx+gxy=[xy-yx](g)$.

N.B

The formula $[X,Y]=-DX.Y+DY.X$ comes from $L_X(L_Y(f)-L_Y(L_X(f))$.

$L_X(f)=df(x).X(x)$ implies that $L_Y(L_X(f))=d^2f.X(x).Y(x)+df(x).DX(Y(x))$. Since $d^2f.X(x).Y(x)=d^2f.Y(x).X(x)$, we deduce that $L_X(L_Y(f))-L_Y(L_X(f))=-df(x).DX(Y(x))+df(x).DY(X(x))=df(x).(-DX.Y+DY.X)(x)$. This implies that

$[X,Y]=-DX.Y+DY.X$.

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