# Lie bracket as a directional derivative

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 $$\{0,-1\}$$.

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.

Edit

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}

• You seem to have a misunderstanding about commutator and directional derivative. I tried to address this in an edit. Nov 6, 2023 at 4:06
• 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 $X\,.$ Nov 6, 2023 at 20:20
• 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 $T_{\Psi(x,t)}M\,.$ ... Nov 7, 2023 at 5:29
• Nov 7, 2023 at 11:17
• This looks pretty good, too. Nov 7, 2023 at 11:30