Note: Lie Derivative Example

This note has some quick explicit computations of the Lie derivatives of vector fields from its definition

To keep things simple, we work in $\mathbb{R}^2$ equipped with the coordinates $x,y$. We denote the coordinate vector fields $\partial_x$, $\partial_y$, and the dual 1-forms $dx$ and $dy$.

A Warmup Example

First, let’s think about the Lie derivative along the coordinate vector field $\partial_x$ of an arbitrary field $W=a(x,y)\partial_x + b(x,y)\partial_y$.

The first step is to find the flow associated to $\partial_x=\langle 1,0\rangle$ starting from a point $p=(a,b)$, which one easily confirms by differentiation is $\Phi_t(a,b)=(a+t,b)$. Thus, as a diffeomorphism of the plane we have

$$\Phi_t\colon (x,y)\mapsto (x+t,y)$$

The differential of this flow at time $t$ is the identity

$$D\Phi_t = \begin{pmatrix}\frac{\partial}{\partial x}(x+t) & \frac{\partial}{\partial y}(x+t)\
\frac{\partial}{\partial x}(y) & \frac{\partial}{\partial y}(y) \end{pmatrix}=\begin{pmatrix}1&0\0&1\end{pmatrix}$$

Confirming our intuition that carrying a vector along with us via a coordinate vector field does not change its coordinate expression, since $(\Phi_t)_\ast W = (D\Phi_t)W$. Thus at any fixed time $t$ we can explicitly evaluate the numerator of our difference quotient as a vector written in the coordinate vector fields $\partial_x, \partial_y$

$$W_{\Phi_t}=W_{(x+t,y)}=\begin{pmatrix}a(x+t,y)\ b(x+t,y)\end{pmatrix}$$

$$(\Phi_t)_\ast W = D\Phi_t W = \begin{pmatrix}1&0\0&1\end{pmatrix} \begin{pmatrix}a\ b\end{pmatrix}=\begin{pmatrix}a(x,y)\ b(x,y)\end{pmatrix}$$

Forming the difference quotient and taking the limit as $t\to 0$ confirms our guess, each coordinate is precisely the $x$-partial derivative of the original:

$$ \mathcal{L}{\partial_x}W =\lim{t\to 0}\begin{pmatrix}\frac{a(x+t,y)-a(x,y)}{t}\\frac{b(x+t,y)-b(x,y)}{t}\end{pmatrix}=\begin{pmatrix}a_x\b_x\end{pmatrix} $$

A More Interesting Example

Let $V= \partial_x + y\partial_y=\langle 1,y\rangle$ be the vector field along which we flow. This vector field is exponentially spreading out from the $x$ axis as it flows, so what happens to $W=a\partial_x+b\partial_y$ as we flow along? Again, we start by solving the differential equation posed by $V$ for a flow $\Phi_t$ on the plane. By definition, the flow $\Phi_t=(x(t),y(t))$ satisfies

$$\Phi_t^\prime = \begin{pmatrix}x(t)\y(t)\end{pmatrix}^\prime=V(x(t),y(t))=\begin{pmatrix}1\y(t)\end{pmatrix}$$ so $x^\prime(t)=1$ and $y^\prime(t)=y(t)$. Solving these for the initial conditions $(x,y)$ yields the flow $$\Phi_t(x,y) = (x+t, ye^t)$$

To compute the pushforward we need the differential $D\Phi_t$, which is diagonal $$D\Phi_t = \begin{pmatrix}1&0\0&e^t\end{pmatrix}$$

Thus for a vector field $W=\langle a,b\rangle$, the terms in the numerator of the difference quotient become $$W_{\Phi_t}=W_{(x+t,ye^t)}=\begin{pmatrix}a(x+t,ye^t)\ b(x+t,ye^t)\end{pmatrix}$$

$$(\Phi_t)_\ast W = D\Phi_t W = \begin{pmatrix}1&0\0&e^t\end{pmatrix} \begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}a(x,y)\ e^t b(x,y)\end{pmatrix}$$

Forming the difference quotient and taking the limit as $t\to 0$ gives the Lie derivative here:

$$ \mathcal{L}{\partial_x}W =\lim{t\to 0}\begin{pmatrix}\frac{a(x+t,ye^t)-a(x,y)}{t}\\frac{b(x+t,ye^t)-e^tb(x,y)}{t}\end{pmatrix} $$

We compute each component separately. For the $\partial_x$-component, consider the curve $\gamma(t)=(x+t,ye^t)$ passing through $(x,y)$ at $t=0$ (this is of course just the flow line $\Phi_t(x,y)$). We may rewrite the difference quotient the derivative of a composition $\lim_{t\to 0}\frac{a(\gamma(t))-a(\gamma(0))}{t}$ and then compute via the chain rule: $$(a\circ\gamma)^\prime(0)=Da_{\gamma(0)}\gamma^\prime(0)=\left(a_x,a_y\right)\begin{pmatrix}1\ y\end{pmatrix}=a_x+ya_y$$

For the $\partial_y$ component, we have this additional factor of $e^t$ to deal with, so we perform a little trick. Adding and subtracting $b(x,y)$ from the numerator allows us to separate it into two limits,

$$\begin{align}\frac{b(x+t,ye^t)-e^tb(x,y)}{t}&=\frac{b(x+t,ye^t)-e^tb(x,y)+b(x,y)-b(x,y)}{t}\
&=\frac{b(x+t,ye^t)-b(x,y)}{t}-\frac{e^tb(x,y)-b(x,y)}{t}\
\end{align}$$

The first of these is identical to the previous term (with $a$ swapped for $b$) so in the limit as $t\to 0$ we know it approaches $b_x+yb_y$. The second term has a factor of $b$ in common which pulls out of the numerator, yielding the limit of $\frac{e^t-1}{t}$. This is the derivative of $e^t$ at $0$ which is 1, so the second term contributes a factor of $b$. Overall then we have $b_x+yb_y-b$ and

$$\mathcal{L}_V W = \begin{pmatrix}a_x+ya_y\ b_x+yb_y-b \end{pmatrix}$$

Dependence on the Flow

These two examples above illustrate an important property of the Lie derivative: it depends very much on the entire vector field you flow along, not just the value of that field at a point (or even, an integral curve of that field). Notice that for our vector fields $U=\partial_x$ (the first example) and $V= \partial_x + y\partial_y$ (the second example), at the origin $O$ both have the same value $U=V=\langle 1,0\rangle$, and in fact both have the same integral curve through this point: $t\mapsto (t,0)$. However,

$$(\mathcal{L}_U W)_O=\begin{pmatrix}a_x\ b_x\end{pmatrix}\hspace{1cm}(\mathcal{L}_V W)_O=\begin{pmatrix}a_x+0a_y\ b_x+0b_y-b\end{pmatrix}=\begin{pmatrix}a_x\ b_y-b\end{pmatrix}$$

Precisely, we have explicit vector fields $U$ and $V$ and point $p=(0,0)$ where $U_p=V_p$ but $(\mathcal{L}_UW)_p\neq (\mathcal{L}_VW)_p$. This is in stark contrast to the covariant derivative $\nabla$, where if $V$ and $W$ are any two vector fields which agree at $p$, then $(\nabla_V X)_p = (\nabla_W X)_p$ for every vector field $X$.