Computing Lie Derivatives: Worked Examples

Step-by-step Lie derivative computations in the plane.

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 $X$.