# The Unapologetic Mathematician

## Calculating the Lie Derivative

It’s all well and good to define the Lie derivative, but it’s not exactly straightforward to calculate it from the definition. For one thing, it requires knowing the flow $\Phi$ of the vector field $X$, which requires solving a differential equation that might be difficult in practice. Luckily, there’s an easier way.

But first, a lemma: if $I$ is an interval containing $0\in\mathbb{R}$, $U\subseteq M$ is an open set, and $f:I\times U\to\mathbb{R}$ is a differentiable function with $f(0,p)=0$ for all $p\in U$, then there is another differentiable function $g:I\times U\to\mathbb{R}$ such that

\displaystyle\begin{aligned}f(t,p)&=tg(t,p)\\\frac{\partial f}{\partial t}(t,p)&=g(0,p)\end{aligned}

This is basically just like a lemma we proved for functions on star-shaped neighborhoods. Indeed, it suffices to set

$\displaystyle g(t_0,p)=\int\limits_0^1\frac{\partial f}{\partial t}(st_0,p)\,ds$

Now let $p\in M$, $f:M\to\mathbb{R}$, and $\Phi:I\times U\to M$ is the local flow of a vector field $X$ on a region $U$ containing $p$. Consider the function $(t,q)\mapsto f\left(\Phi(t,q)\right)-f(q)$, which satisfies the condition of the above lemma. We can thus write

$\displaystyle\left[f\circ\Phi_t\right](q)-f(q)=tg(t,q)$

for some $g:I\times U\to\mathbb{R}$. Or if we write $g_t(q)=g(t,q)$ we can write

$\displaystyle f\circ\Phi_t=f+tg_t$

We also can see that $g_0=Xf$. And so we can write

\displaystyle\begin{aligned}\Phi_{-t*}Y_{\Phi_t(p)}(f)&=Y_{\Phi_t(p)}\left(f\circ\Phi_{-t}\right)\\&=Y_{\Phi_t(p)}\left(f-tg_{-t}\right)\\&=\left[(Yf)\circ\Phi_t\right](p)-t\left[(Yg_{-t})\circ\Phi_t\right](p)\end{aligned}

So now we calculate

\displaystyle\begin{aligned}(L_XY)_pf&=\lim\limits_{t\to0}\frac{\Phi_{-t*}Y_{\Phi_{t}(p)}(f)-Y_pf}{t}\\&=\lim\limits_{t\to0}\frac{\left[(Yf)\circ\Phi_t\right](p)-t\left[(Yg_{-t})\circ\Phi_t\right](p)-Yf(p)}{t}\\&=\lim\limits_{t\to0}\frac{\left[(Yf)\circ\Phi_t\right](p)-Yf(p)}{t}-\lim\limits_{t\to0}\left[(Yg_{-t})\circ\Phi_t\right](p)\end{aligned}

Last time we wrote the action of $X$ on a smooth function $\phi$ as

$\displaystyle X_p(\phi)=\lim\limits_{t\to0}\frac{\left[\phi\circ\Phi_t\right](p)-\phi(p)}{t}$

which pattern we can recognize in our formula. We thus continue

\displaystyle\begin{aligned}(L_XY)_pf&=\lim\limits_{t\to0}\frac{\left[(Yf)\circ\Phi_t\right](p)-Yf(p)}{t}-\lim\limits_{t\to0}\left[(Yg_{-t})\circ\Phi_t\right](p)\\&=X_pYf-Yg_0(p)\\&=X_pYf-Y_pXf\\&=[X,Y]_pf\end{aligned}

That is, for any two vector fields the Lie derivative $L_XY$ is actually the same as the bracket $[X,Y]$.