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.

  • 1
    $\begingroup$ 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

$GL(n,R)$ 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 $Y(g)=gx$, $-DX(Y(g))+DY(X(g))=-gyx+gxy=[xy-yx](g)$.


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


| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.