I'm trying to understand the Lie bracket operation $[X, Y]$ as the rate of change of $Y$ as seen by an observer moving along the flow of $X$.

Example 1

Suppose $X=\{1, x\}^T$ and $Y=\{1, 0\}^T$, then $[X, Y]=\{0, -1\}$ and the flow of $X$ is $\{t+x_0,\frac{1}{2}t^2+ t\ x_0+ y_0\}$.

But I'm failing to see how along any of the flows below, the rate of change of vector $\{1, 0\}$ is $\{0,-1\}$.

enter image description here

Example 2

Suppose $X=\{x, x\}^T$ and $Y=\{1, 0\}^T$, then $[X, Y]=\{-1, -1\}$ and the flow of $X$ is $\{e^t\ x_0,e^t\ x_0-x_0+y_0\}$.

Again from plot below I'm not able to see how the directional derivative makes sense. enter image description here


Adding except from the source. enter image description here


1 Answer 1


Too long for a comment.

$Y=\{1,0\}^\top$ which I prefer to write as $Y=\partial_y$ is a constant vector field. Its directional derivative w.r.t. any other vector field $X$ is zero. The Lie dervative ${\cal L}_XY$ of a vector field $Y$ w.r.t. $X$ is the difference of the directional derivatives: $$ {\cal L}_XY=[X,Y]=\partial_XY-\partial_YX\,. $$

  • Note that the commutator $[X,Y]$ is by definition the difference of two second order differential operators, while the directional derivatives are first order differential operators.

In your first case, \begin{align} X&=\partial_x+x\partial_y\,,&Y&=\partial_x\,,&XY&=\partial_x^2+x\partial_y\partial_x\,,&YX&=\partial_x^2+\partial_y+x\partial_x\partial_y \end{align} so that the commutator becomes \begin{align}\require{cancel} [X,Y]&=XY-YX=\cancel{\partial_x^2}+\bcancel{x\partial_y\partial_x}-\cancel{\partial_x^2}-\partial_y-\bcancel{x\partial_x\partial_y}=-\partial_y\,. \end{align} On the other hand, the directional derivatives are calculated by applying $\partial_x,\partial_y$ to the components of $X$ and $Y$ and taking the dot product with the components of $Y$, resp. $X\,:$ $$ \partial_XY^\nu=X^x\partial_xY^\nu+X^y\partial_yY^\nu\,. $$ This is the $\nu$-th component of $\partial_XY\,.$

In your first case $X^x=1,X^y=x,Y^x=1,Y^y=0$ so that \begin{align} \partial_XY&=0\,,&\partial_YX=\partial_y\,. \end{align}

  • 1
    $\begingroup$ You seem to have a misunderstanding about commutator and directional derivative. I tried to address this in an edit. $\endgroup$
    – Kurt G.
    Nov 6, 2023 at 4:06
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    $\begingroup$ Thanks. This shows that -with the flow definition- ${\cal L}_XY=[X,Y]$ holds as well. Conclusion: unlike what I wrote in a now deleted comment: The Lie derivative is in general not a directional derivative $\partial_XY\,.$ It is the difference of two such directional derivatives. Finally, ${\cal L}_XY=[X,Y]$ it is the rate of change of $Y$ as seen by an observer that moves with the flow of $X\,.$ $\endgroup$
    – Kurt G.
    Nov 6, 2023 at 20:20
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    $\begingroup$ The screenshot from the book shows a slightly more readable version of something that is often presented in ponderous ways like this. Some DGs seem to believe that the more intimidating a notation is the better. Trying to visualise this you have my full sympathy. I think that the point is that in the $t$-derivative we have to find ways to compare vectors that live in different tangent spaces $T_xM$ and $T_{\Psi(x,t)}M\,.$ ... $\endgroup$
    – Kurt G.
    Nov 7, 2023 at 5:29
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    $\begingroup$ Getting closer. $\endgroup$
    – Kurt G.
    Nov 7, 2023 at 11:17
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    $\begingroup$ This looks pretty good, too. $\endgroup$
    – Kurt G.
    Nov 7, 2023 at 11:30

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