In trying to get to grips with Lie derivatives I'm completely lost, just completely lost :( Is there anyone who could provide an example of calculating the Lie derivative of the most basic function you can, i.e. like in showing someone how to calculate the derivative you'd pick something like $f(x) \ = \ x^2$, by using all possible formuations of the lie derivative, simply as a means to illustrate the idea. Thus we could calculate it on $f(x) \ = \ x^2$ & $\vec{v}(x,y) \ = \ x^2i \ + \ 2y^2j$, I cannot find one simple calculation of this thing anywhere :(


1 Answer 1


The Lie derivative of a smooth function $f:M\to \mathbb{R}$ with respect to a tangent vector $X\in T_{p}(M)$ at a point $p$ is just the directional derivative of $f$ with respect to $X$ at $p$. So, in your example, $f:\mathbb{R}^2\to \mathbb{R}$ is a smooth function and $v$ defines a smooth vector field on $M$. If $(x,y)\in \mathbb{R}^2$, then what is the directional derivative of $f$ with respect to the vector $v_{(x,y)}=(x^2,2y^2)\in T_{(x,y)}(\mathbb{R}^2)$? You may recall from multivariable calculus that it is given by the dot product of this vector with the gradient of $f$ at this point; so,


$=x^2\frac{\partial f}{\partial x}+2y^2\frac{\partial f}{\partial y}$



Of course, $v$ is a smooth vector field on $\mathbb{R}^2$ and $f:\mathbb{R}^2\to \mathbb{R}$ is a smooth function. We've seen that for each $(x,y)\in \mathbb{R}^2$, $v_{(x,y)}(f)$ is a well-defined real number. Thus, $v(f):\mathbb{R}^2\to \mathbb{R}$ is a function defined by the rule $(x,y)\to v_{(x,y)}(f)$ and in this case it's smooth because it's given by the rule $v(f)(x,y)=2x^3$. In fact, this is a general rule in differential geometry: a smooth vector field can be thought of as an operator which swallows a smooth function and spits out another smooth function! The general principle is below:

Let the set of smooth functions $f:M\to \mathbb{R}$ be denoted by $C^{\infty}(M)$. If $X\in T_{p}(M)$ is a vector tangent to $M$ at $p$, then $X$ is a derivation $X:C^{\infty}(p)\to \mathbb{R}$. Note that $C^{\infty}(p)$ denotes the real vector space of germs at $p$, i.e., pairs $(f,U)$ where $U$ is an open neighborhood of $p$ and $f:U\to \mathbb{R}$ is a smooth function. Also, by a derivation $X:C^{\infty}(p)\to \mathbb{R}$, I mean a linear map satisfying the Leibniz rule: if $(f,U),(g,V)\in C^{\infty}(p)$, then $X(fg)=X(f)g+X(g)f$ on $U\cap V$.

I guess your question is: how in practice do we compute $X$? I've already claimed that $X$ can be thought of as a directional derivative operator. If $X\in T_{p}(M)$, then all we need is that we have a real-valued function $f$, smoothly defined in a neighborhood of $p$, and then we can talk about the directional derivative of $f$ at $p$ in the direction of $X$. In general, if $(U,\phi)$ is a coordinate neighborhood of $p\in M$, then we can write $X$ in these local coordinates: $X=\sum_{i=1}^{n} a_i \frac{\partial}{\partial x_i}$; here, $(x_1,\dots,x_n)$ are the local coordinates for $(U,\phi)$ and $\frac{\partial}{\partial x_i}$ for $1\leq i\leq n$ are the coordinate frames on $(U,\phi)$. The coordinate frames are just the directions on $(U,\phi)$ analogous to the case of $\mathbb{R}^n$ where we have coordinate axes. Note, however, that these "directions" are vectors in the tangent space to $M$ at various points in $U$!

Now, if $f:U\to \mathbb{R}$ is a smooth function, then $X(f)=\sum_{i=1}^{n} a_i\frac{\partial f}{\partial x_i}$. Technically, you need to prove that this formula is valid. All we know is that $X$ is linear and satisfies the Leibniz rule but it turns out that this is sufficient to restrict $X$ enough that it is given by the formula just stated.

In general, if $X$ is a smooth vector field on $M$, then a smooth function $f:M\to \mathbb{R}$ results in a smooth function $X(f):M\to \mathbb{R}$. In other words, $X:C^{\infty}(M)\to C^{\infty}(M)$ defines an operator:

Exercise 1: Prove that $X:C^{\infty}(M)\to C^{\infty}(M)$ is a derivation, i.e., it is real linear and satisfies the Leibniz rule.

Exercise 2: Prove that every derivation $X:C^{\infty}(M)\to C^{\infty}(M)$ is given by a smooth vector field on $M$ according to the recipe described above.

Exercise 3: What is the Lie derivative of the following functions with respect to the given smooth vector fields on $\mathbb{R}^3$:

(a) $f(x,y,z)=1$ and $v(x,y,z)$ is any smooth vector field on $\mathbb{R}^3$. Interpret the result geometrically.

(b) $f(x,y,z)=x^2+y^2+z^2$ and $v(x,y,z)=(x,y,z)$. Interpret the result geometrically.

(c) $f(x,y,z)=x^2+y^2+z^2$ and $v(x,y,z)$ is an smooth vector field on $\mathbb{R}^2$ tangent to the sphere of radius $r$ at any point of distance $r$ from the origin. Interpret this result geometrically.

Exercise 4: Let $f:S^1\to \mathbb{R}$ be the smooth function defined by assigning a point on the circle to its angle (measured in radians) from the x-axis in the counterclockwise direction; this is a number in $[0,2\pi)$ for each $x\in S^1$. Define local coordinates on $S^1$ by the rule $\theta\to (\cos \theta, \sin \theta)$ and let $X$ be the smooth vector field on $S^1$ given by $\frac{\partial}{\partial \theta}$ in these coordinates. Determine $X(f)$.

Exercise 5: If $X$ and $Y$ are smooth vector fields on $M$, then the Lie derivative of $Y$ with respect to $X$ is given by the derivation $Z:C^{\infty}(M)\to C^{\infty}(M)$ defined by the rule $Z(f)=X(Y(f))-Y(X(f))$ for all $f\in C^{\infty}(M)$. Of course, every such derivation corresponds to a smooth vector field on $M$ and in this case, $Z$ is commonly denoted by $[X,Y]$, the Lie bracket of $X$ and $Y$. In your example of $v(x,y)=(x^2,2y^2)$ for all $(x,y)\in\mathbb{R}^2$, define also $w(x,y)=(2y^2,x^2)$ for all $(x,y)\in \mathbb{R}^2$. Determine the Lie bracket of $v$ and $w$, that is, explicitly write down this smooth vector field on $\mathbb{R}^2$ in the form $(g(x,y),h(x,y))$ for all $(x,y)\in \mathbb{R}^2$.

I hope this helps! Let me know if you have further questions.

  • $\begingroup$ Thanks, this stuff helped a lot. My only problem now is with calculating the lie derivative using the difference quotient definition. If I try to calculate the lie derivative of $$G(x,y) \ = \ (x^2 \ + \ y,2y)$$ along $$F(x,y) \ = \ (2x \ + \ 3y,5x)$$ using the definition $$\mathcal{L}_F(G(\vec{x}_0)) \ = \ \ \lim_{t \rightarrow 0} \frac{(\theta_-t)_*(G(\theta_t(\vec{x}_0))) \ - \ G(\vec{x}_0)}{t}$$, where I (hopefully correctly) Calculated the flow of $F$ to be $$\theta(x,y,t) \ = \ (xe^{-3t},ye^{5t})$$ I get stuck in the difference quotient, just not working & I don't know why. $\endgroup$
    – bolbteppa
    Jul 8, 2013 at 15:57
  • $\begingroup$ I just can't even get an answer, let alone one that equals my calculations of $[F,G]$, if you wouldn't mind helping me out a bit with this I'd really appreciate it! $\endgroup$
    – bolbteppa
    Jul 8, 2013 at 15:59
  • $\begingroup$ @bolbteppa I don't think your calculation of the flow $\theta:\mathbb{R}^2\times \mathbb{R}\to \mathbb{R}^2$ of $F$ is correct. The initial conditions are clearly satisfied but your flow satisfies the system of ODE's $x'=-3x$ and $y'=5y$ which isn't the system of ODE's under consideration. Can you calculate the flow correctly? I'm more than happy to give pointers if you're stuck. If you do that, then you just have to compute the Jacobian matrix of the diffeomorphism $\theta_t:\mathbb{R}^2\to \mathbb{R}^2$ for each fixed $t$. That's the first step in computing the numerator of the quotient. $\endgroup$ Jul 8, 2013 at 16:54
  • $\begingroup$ @bolbteppa I'm going to sleep now. If you post again, then I'll respond to any questions you might have in about 8 - 10 hours. If you're still stuck, then I'll try to write out fairly explicit details of the computation so it won't be difficult for you to finish it off. $\endgroup$ Jul 8, 2013 at 16:56
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    $\begingroup$ @AmiteshDatta You're a beautiful person - this is a great example of how a simple function like $f(x)=x^{2}$ can provide so much content. Wonderful. If I could +2 I would. I can only +1. $\endgroup$
    – user284001
    Nov 11, 2015 at 14:54

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