Calculate the Lie Derivative

Solution 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,

$v_{(x,y)}(f)$

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

$=x^2(2x)+2y^2(0)$

$=2x^3$.

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.