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 ${\partial }_x=⟨1,0⟩$ starting from a point $p=(a,b)$, which one easily confirms by differentiation is ${\mathrm{\Phi }}_t(a,b)=(a+t,b)$. Thus, as a diffeomorphism of the plane we have

$${\mathrm{\Phi }}_t:(x,y)\mapsto (x+t,y)$$

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

$$D{\mathrm{\Phi }}_t=\left(\begin{array}{ccc}\frac{\partial }{\partial x}(x+t)& \frac{\partial }{\partial y}(x+t)\frac{\partial }{\partial x}(y)& \frac{\partial }{\partial y}(y)\end{array}\right)=\left(\begin{array}{ccc}1& 0\text{0}& 1\end{array}\right)$$

Confirming our intuition that carrying a vector along with us via a coordinate vector field does not change its coordinate expression, since $({\mathrm{\Phi }}_t{)}_{\ast }W=(D{\mathrm{\Phi }}_t)W$. 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 ${\partial }_x,{\partial }_y$

$$W_{{\mathrm{\Phi }}_t}=W_{(x+t,y)}=\left(\begin{array}{c}a(x+t,y)b(x+t,y)\end{array}\right)$$

$$({\mathrm{\Phi }}_t{)}_{\ast }W=D{\mathrm{\Phi }}_{t}W=\left(\begin{array}{ccc}1& 0\text{0}& 1\end{array}\right)\left(\begin{array}{c}ab\end{array}\right)=\left(\begin{array}{c}a(x,y)b(x,y)\end{array}\right)$$

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:

$$ \mathcal{L}{\partial_x}W =\lim{t\to 0}$$\left(\begin{array}{c}\frac{a(x+t,y)-a(x,y)}{t}\\ fracb(x+t,y)-b(x,y)t\end{array}\right)$$=$$\left(\begin{array}{c}{a}_x{\text{b}}_x\end{array}\right)$$ $$

A More Interesting Example

Let $V={\partial }_x+y{\partial }_y=⟨1,y⟩$ 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 ${\mathrm{\Phi }}_t$ on the plane. By definition, the flow ${\mathrm{\Phi }}_t=(x(t),y(t))$ satisfies

$${\mathrm{\Phi }}_t^{\prime }={\left(\begin{array}{c}x(t)\text{y}(t)\end{array}\right)}^{\prime }=V(x(t),y(t))=\left(\begin{array}{c}1\text{y}(t)\end{array}\right)$$ so $x^{\prime }(t)=1$ and $y^{\prime }(t)=y(t)$. Solving these for the initial conditions $(x,y)$ yields the flow $${\mathrm{\Phi }}_t(x,y)=(x+t,y{e}^t)$$

To compute the pushforward we need the differential $D{\mathrm{\Phi }}_t$, which is diagonal $$D{\mathrm{\Phi }}_t=\left(\begin{array}{ccc}1& 0\text{0}& e^t\end{array}\right)$$

Thus for a vector field $W=⟨a,b⟩$, the terms in the numerator of the difference quotient become $$W_{{\mathrm{\Phi }}_t}=W_{(x+t,y{e}^t)}=\left(\begin{array}{c}a(x+t,y{e}^t)b(x+t,y{e}^t)\end{array}\right)$$

$$({\mathrm{\Phi }}_t{)}_{\ast }W=D{\mathrm{\Phi }}_{t}W=\left(\begin{array}{ccc}1& 0\text{0}& e^t\end{array}\right)\left(\begin{array}{c}a\text{b}\end{array}\right)=\left(\begin{array}{c}a(x,y)e^{t}b(x,y)\end{array}\right)$$

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

$$ \mathcal{L}{\partial_x}W =\lim{t\to 0}$$\left(\begin{array}{c}\frac{a(x+t,y{e}^t)-a(x,y)}{t}\\ fracb(x+t,y{e}^t)-e^{t}b(x,y)t\end{array}\right)$$ $$

We compute each component separately. For the ${\partial }_x$-component, consider the curve $\gamma (t)=(x+t,y{e}^t)$ passing through $(x,y)$ at $t=0$ (this is of course just the flow line ${\mathrm{\Phi }}_t(x,y)$). We may rewrite the difference quotient the derivative of a composition $\underset{t\to 0}{lim}\frac{a(\gamma (t))-a(\gamma (0))}{t}$ and then compute via the chain rule: $$(a\circ \gamma {)}^{\prime }(0)=D{a}_{\gamma (0)}{\gamma }^{\prime }(0)=(a_x,a_y)\left(\begin{array}{c}1y\end{array}\right)=a_x+y{a}_y$$

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,

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

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 $b_x+y{b}_y$. 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 $b_x+y{b}_y-b$ and

$${\mathcal{L}}_{V}W=\left(\begin{array}{c}{a}_x+y{a}_y{b}_x+y{b}_y-b\end{array}\right)$$

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 $U=V=⟨1,0⟩$, and in fact both have the same integral curve through this point: $t\mapsto (t,0)$. However,

$$({\mathcal{L}}_{U}W{)}_O=\left(\begin{array}{c}{a}_x{b}_x\end{array}\right)({\mathcal{L}}_{V}W{)}_O=\left(\begin{array}{c}{a}_x+0a_y{b}_x+0b_y-b\end{array}\right)=\left(\begin{array}{c}{a}_x{b}_y-b\end{array}\right)$$

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