Lie derivative of a vector field equals the lie bracket

Here is a simple proof which I found in the book "Differentiable Manifolds: A Theoretical Phisics Approach" of G. F. T. del Castillo. Precisely it is proposition 2.20.

We denote $(\mathcal{L}_XY)_x=\frac{d}{dt}(\phi_t^*Y)_x|_{t=0},$ where $(\phi^*_tY)_x=(\phi_{t}^{-1})_{*\phi_t(x)}Y_{\phi_t(x)}.$

Recall also that $(Xf)_x=\frac{d}{dt}(\phi_t^*f)_x|_{t=0}$ where $\phi_t^*f=f\circ\phi_t$ and use that $(\phi^*_tYf)_x=(\phi^*_tY)_x(\phi^*_tf).$

We claim that $(\mathcal{L}_XY)_x=[X,Y]_x.$

Proof:$$(X(Yf))_x=\lim_{t\to 0}\frac{(\phi_t^*Yf)_x-(Yf)_x}{t}=\lim_{t\to 0}\frac{(\phi^*_tY)_x(\phi^*_tf)-(Yf)_x}{t}=\star$$

Now we add and subtract $(\phi^*_tY)_xf.$ Hence $$\star=\lim_{t\to 0}\frac{(\phi^*_tY)_x(\phi^*_tf)-(\phi^*_tY)_xf+(\phi^*_tY)_xf-Y_xf}{t}=$$ $$=\lim_{t\to 0}(\phi^*_tY)_x\frac{(\phi^*_tf)-f}{t}+\lim_{t\to 0}\frac{(\phi^*_tY)_x-Y_x}{t}f=Y_xXf+(\mathcal{L}_XY)_xf.$$ So we get that $XY=YX+\mathcal{L}_XY.$

Here is a proof that is more-or-less equivalent to the one given by Fallen Apart, but with details further explicated/clarified.

Let $M$ be a smooth manifold and $X$, $Y$ smooth vector fields. The Lie derivative is defined as $$(\mathcal{L}_{X} Y) (p) = \lim_{t\to 0} \frac{\phi^{-t}_{\star} Y (\phi^{t} (p)) - Y(p)}{t}$$ where $\phi^t$ denotes the flow of $X$. Taking a test function $f \in C^{\infty} (M)$, we apply the Lie derivative to $f$ to obtain $$(\mathcal{L}_{X} Y)(p)f = \lim_{t\to 0} \frac{\phi^{-t}_{\star} Y (\phi^{t} (p))f - Y(p)f}{t} = \lim_{t\to 0} \frac{Y(\phi^t (p))(f\circ \phi^{-t}) - Y(p)}{t}$$ Thus, setting $$H(x,y) = Y(\phi^x (p))(f\circ \phi^{-y})$$ we have that $$(\mathcal{L}_{X} Y)(p) f = \frac{d}{dt} \Big\vert_{t = 0} H(t,t) = \frac{\partial H}{\partial x} (0,0) + \frac{\partial H}{\partial y} (0,0)$$ where the second equality is due the chain rule. So, to complete the calculation we just have to compute the partial derivatives of $H$. We find $$\frac{\partial H}{\partial x} (0,0) = \frac{\partial}{\partial x} \Big\vert_{x= 0} (Yf)(\phi^x (p)) = X(p)(Yf)$$ For the other partial derivative of $H$, we introduce a curve $\alpha: (-\epsilon, \epsilon) \to M$ such that $\alpha(0) = p$ and $\alpha'(0) = Y(p)$. Then we have $$\frac{\partial H}{\partial y} (0,0) = \frac{\partial}{\partial y} \Big\vert_{y = 0} Y(p) (f\circ \phi^{-y}) = \frac{\partial}{\partial y} \Big\vert_{y = 0} \frac{d}{ds} \Big\vert_{s= 0} (f\circ \phi^{-y} \circ \alpha)(s)$$ $$= \frac{d}{ds} \Big\vert_{s= 0} \frac{\partial}{\partial y}\Big\vert_{y = 0} (f\circ \phi^{-y} \circ \alpha)(s) = \frac{d}{ds} \Big\vert_{s= 0} -(Xf)(\alpha(s)) = -Y(p)(Xf)$$ We conclude $$(\mathcal{L}_{X} Y)(p)f = X(p) (Yf) - Y(p) (Xf) \implies \mathcal{L}_{X} Y = XY - YX =: [X, Y]$$

Let $X$ and $Y$ two vector fields then the Lie derivative $L_{X}Y$ is the commutator $[X, Y]$.

the proof:

we have

$$L_{X}Y=\lim_{t\to 0}\frac{d\phi_{-t}Y-Y}{t}(f)$$ $$=\lim_{t\to 0}d\phi_{-1}\frac{Y-d\phi_{t}Y}{t}(f)$$ $$=\lim_{t\to 0}\frac{Y(f)-d\phi_{t}Y(f)}{t}$$ $$=\lim_{t\to 0}\frac{Y(f)-Y(f\circ\phi_{t})\circ\phi_{t}^{-1}}{t}$$

we put $\phi_{t}(x)=\phi(t,x)$ and we apply the Taylor formula with integral remains, then there exists $h(t,x)$ such that:

$$f(\phi(t,x))=f(x)+th(t,x)$$ where $h(0,x)=\frac{\partial}{\partial t}f(\phi(t,x))(0,x)$

by defintion of tangent vector: $X(f)=\frac{\partial}{\partial t}f\circ\phi_{t}(x)(0,x)$

then we have $h(o,x)=X(f)(x)$ so:

$$L_{X}Y(f)=\lim_{t\to 0}\left(\frac{Y(f)-Y(f)\circ \phi_{t}^{-1}}{t}-Y(h(t,x))\circ \phi_{t}^{-1}\right)$$ $$=\lim_{t\to 0}\left(\frac{(Y(f)\circ\phi_{t}-Y(f))\circ\phi_{t}^{-1}}{t}-Y(h(t,x))\circ\phi_{t}^{-1}\right)$$

we have $\lim_{t\to 0}\phi_{t}^{-1}=\phi_{0}^{-1}=id.$

then we conclude that

$$L_{X}Y(f)=\lim_{t\to 0}\left(\frac{Y(f)\circ\phi_{t}-Y(f)}{t}-Y(h(0,x))\right)$$ $$= \frac{\partial}{\partial t}Y(f)\circ\phi_{t}(x)-Y(h(0,x))$$ $$= X(Y(f)) -Y(X(f))= [X,Y]$$

This completes the proof.