Hints of showing an identity of Lie derivative

Question

Let $X$, $Y$ be vector fields on manifold $M$ and $f$ is a real smooth function on $M$, then may I get some hints of showing $(L_X L_Y-L_Y L_X)(f)=L_{[X,Y]}f$, where $[X,Y]$ means the Lie bracket?

Answer 1

Here's the solution. The Lie derivative of $f \in C^\infty(M)$ with respect to $V$, which we denoted by $\mathscr{L}_V f$, is defined

\begin{align} (\mathscr{L}_V f)_p =\frac{d}{dt}\bigg|_{t=0}f(\sigma_t(p)) =\lim_{t\rightarrow 0} \frac{f(\sigma_t(p))-f(p)}{t} \end{align}

where $\sigma_t(p)$ is flow associated to $V$ such that $\sigma_0(p) = p$ and \begin{align} \frac{d}{dt}\sigma_t = V(\sigma_t). \end{align}

From the above definition, one can show that

\begin{align} \mathscr{L}_V f = V f \end{align}

since \begin{align} (\mathscr{L}_V f)(p) = \frac{d}{dt}\bigg|_{t=0} f(\sigma_t(p)) = df(p)[V(\sigma_0)] = df(p)[V(p)] = Vf(p). \end{align}

Using the above fact, we have that

\begin{align} \mathscr{L}_Y\mathscr{L}_X f = \mathscr{L}_Y[Xf]= YXf \end{align}

since $Xf \in C^\infty(M)$. Hence it follows \begin{align} [\mathscr{L}_X\mathscr{L}_Y -\mathscr{L}_Y\mathscr{L}_X]\ f = XYf-YXf =[X, Y]\ f= \mathscr{L}_{[X, Y]}\ f. \end{align}