# Computing Lie derivative

Can anyone help me with computing Lie derivative ${L}_{X}Y$ using its definition for these two vector fields: $X=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}, Y=x^2\frac{\partial}{\partial x}$ ? I know how to find it using definition of Lie bracket but I don't quit understand definition of Lie derivative and I can't find anywhere simple example showing how it should work.

• Can you point to which definition of the Lie derivative you are interested in, if not the Lie bracket? Commented Nov 11, 2015 at 17:27
• I am interested in using this definition ${L}_{X}Y=\lim \limits_{h \to 0} \frac{1}{h} ({Y}_{p}-({{\phi}_{h*}Y}_{p}))$ Commented Nov 11, 2015 at 18:01

The Lie derivative is usually defined either to be equal to the Lie bracket $L_XY=\left[ X,Y\right]$, or by using the flow of $Y$ along $X$. You have said you already are familiar with the Lie bracket, so here is the second definition. $$(L_XY)_p=\lim_{t\to 0}\frac{\mathrm d\Phi_X^{-t}Y_{\Phi^t_X(p)}-Y_p}{t}$$ Here $p$ is a point in the manifold, $\Phi^s_X$ is the flow transformation that carries a point $r$ a certain amount $s$ along the integral curve of $X$ to another point $q$ in the manifold, and $\mathrm d\Phi^s_X$ is the tangent map of the flow transformation which takes a vector from the tangent space at $r$ to the tangent space at $q$. The $-t$ in the tangent map compared to the $t$ in the flow transformation means that the tangent map operator is acting on the inverse of the flow transformation: just the flow transformation going the same amount in the opposite direction along $X$.

So, intuitively, the Lie derivative goes an infinitesimal amount along $X$ from $p$, finds the vector there, pushes it back to $p$, and compares it to the original vector at $p$ to find the rate of change.

To compute all this we need the integral curves of $X$ in order to find the flow transformation, then the tangent map (also called the pushforward) of the flow transformation, then you can compute the limit. Unfortunately I am not familiar enough with the pushforward to compute it (I attempted to do it for a while but was not getting the correct answer). But perhaps you can see why the Lie bracket is usually used to define $L_XY$. :)

EDIT: Found my mistake. Somehow I had calculated that the integral curves of your given $X$ were hyperbolas when they are in fact circles (going counterclockwise around the origin at a rate of 1 radian per unit time). This means that the flow transformation is just clockwise rotation of the point, and the pushforward happens to act the same way on the tangent vectors (probably since our base manifold is already a vector space and the flow transformation is linear on it).

So, if $p=(x,y)$ and $Y_p=x^2\partial_x$, we have: $$\Phi^t_X(p)=(x\cos t+y\sin t,y\cos t-x\sin t)$$ $$\mathrm d\Phi_X^{-t}Y_{\Phi^t_X(p)}=(x\cos t+y\sin t)^2\cos t\partial_x-(x\cos t+y\sin t)^2\sin(-t)\partial_y$$ $$(L_XY)_p=\lim_{t\to 0}\frac{(x\cos t+y\sin t)^2\cos t\partial_x+(x\cos t+y\sin t)^2\sin t\partial_y-x^2\partial_x}{t}$$ Using the limits $\lim_{t\to 0}\cos t=1$, $\lim_{t\to 0}\sin t=0$, $\lim_{t\to 0}(\sin t)/t=1$, and $\lim_{t\to 0}(\cos^3t-1)/t=0$ this becomes: $$L_XY=2xy\partial_x+x^2\partial_y$$ The same answer obtained with the Lie bracket.

• Ok. I understand, more or less, the idea of this formula but actually i still have a problem with putting it into practice.. But, thank you anyway :) Commented Nov 11, 2015 at 19:32
• I added an example calculation after finding my error. Commented Nov 11, 2015 at 19:44
• Ok, great. I think I see now how it works. Thank you very much! Commented Nov 11, 2015 at 19:45

\begin{eqnarray*} L_{y\partial_x-x\partial_y}(x^2\partial_x) &=&[y\partial_x-x\partial_y\ ,\ x^2\partial_x]\\ &=&(y\partial_x-x\partial_y)(x^2\partial_x) -x^2\partial_x(y\partial_x-x\partial_y)\\ &=&y\partial_x(x^2\partial_x)-x\partial_y(x^2\partial_x)-x^2\partial_x(y\partial_x)+x^2\partial_x(x\partial_y)\\ &=&y(2x\partial_x+x^2\partial_x\partial_x)-x^3\partial_y\partial_x-x^2y\partial_x\partial_x+x^2(\partial_y+x\partial_x\partial_y)\\ &=&2xy\partial_x+x^2\partial_y \end{eqnarray*}

• Thank you for the answer but I'd like to compute it using this formula ${L}_{X}Y=\lim \limits_{h \to 0} \frac{1}{h} ({Y}_{p}-({{\phi}_{h*}Y}_{p}))$. I know that results should be the same but I'd like to know how to use other formula than Lie bracket. Commented Nov 11, 2015 at 18:06
• @janmarqz You forgot to distribute a negative sign going from lines 2 to 3 so your final answer should be $2xy\partial_x+x^2\partial_y$. :) Commented Nov 11, 2015 at 19:46
• i was checking for one who is awoke ; ) Commented Nov 11, 2015 at 23:14