Prove that the Lie derivative of a vector field equals the Lie bracket: $\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y]$

Question

Let $X$ and $Y$ be vector fields on a smooth manifold $M$, and let $\phi_t$ be the flow of $X$, i.e. $\frac{d}{dt} \phi_t(p) = X_p$. I am trying to prove the following formula:

$\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y],$

where $[X,Y]$ is the commutator, defined by $[X,Y] = X\circ Y - Y\circ X$.

This is a question from these online notes: http://www.math.ist.utl.pt/~jnatar/geometria_sem_exercicios.pdf .

Answer 1

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 $XY=YX+\mathcal{L}_XY.$

Answer 2

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 $$(\mathcal{L}_{X} Y)(p)f = X(p) (Yf) - Y(p) (Xf) \implies \mathcal{L}_{X} Y = XY - YX =: [X, Y]$$

Answer 3

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.

Answer 4

Consider two vector fields $X$ and $Y$. Note that starting from the point $p$, the vector field $X$ generates the curve $\phi(p,t)$ via the following equation \begin{align*} \frac{d}{dt}(f(\phi(p,t))|_t = Xf|_t \end{align*} with $\phi(p,0)=p$. Also, note that for small $t$, $f(\phi(p,t))=f(p)+Xf|_pt$. We can define the Lie derivative of $Y$ with respect to $X$ as \begin{align*} (\mathcal{L}_XY)_pf = &\lim_{t \to 0} \frac{(Yf)(\phi(p,t))-(\phi_{*t}Y_p)f}{t} \\ = &\lim_{t \to 0} \frac{(Yf)(p)+X(Yf)|_pt-Y_p(f+X_pft)}{t}\\ =&\lim_{t \to 0} \frac{(Yf)(p)+(XY)_pft-(Yf)_p-(YX)_pft}{t}\\ =&(XY-YX)_pf. \end{align*} In the first line of the above equation $\phi_{*t}Y_p$ denotes the push forward of $Y_p$ along the curve $\phi(p,t)$.

Answer 5

Here is another approach in 4 steps (just one of them is hard):

1. Verify that $\mathcal{L}_X(Y+Z)=\mathcal{L}_X(Y)+\mathcal{L}_X(Z)$ for any fields $Y,Z$;

2. Verify that $\mathcal{L}_X(fY)=f\mathcal{L}_X(Y)+X(f)Y$ for any smooth function $f$;

3. Show that $\mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)=\left[X,\frac{\partial}{\partial x_i}\right]$;

4. Conclude $\mathcal{L}_X(Y)=[X,Y]$.

1. Obvious by $\mathbb{R}$-linearity of $d\phi_{-t}$ and $\frac{d}{dt}$. $_\blacksquare$

2. Just notice that $(d\phi_{-t})(f\,Y)=f(d\phi_{-t})(Y)$ and use Leibnitz's rule. $_\blacksquare$

3. This is the delicate part. Using coordinates, write $X=\sum_ia_i\frac{\partial}{\partial x_i}$ for functions $a_i$ and $\phi(t,x)=(\phi_1(t,x),...,\phi_n(t,x))$ where $x=(x_1,...,x_n)$. Because $\phi(0,x)=x$ we have $\frac{\partial \phi_k}{\partial x_j}(0,x)=\delta_{jk}$ and $\frac{\partial^2 \phi_k}{\partial x_\ell\partial x_j}(0,x)=0$. So: \begin{align*} \mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)_p&=\left.\frac{d}{dt}\right|_{t=0}(d\phi_{-t})_{\varphi_t(p)}\left(\left.\frac{\partial}{\partial x_i}\right|_{\phi_t(p)}\right)\\ &=\left.\frac{d}{dt}\right|_{t=0}\sum_k\frac{\partial \phi_k}{\partial x_i}(-t,\phi_t(p))\left.\frac{\partial}{\partial x_i}\right|_p\\ &=\sum_k\left(\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(-t,\phi_t(p))\right)\left.\frac{\partial}{\partial x_i}\right|_p \end{align*}

We will apply the chain rule to calculate the derivative inside the sum. Since, $\frac{\partial^2\phi_k}{\partial x_j\partial x_i}=0$, we only need to worry about the derivative of $\frac{\partial \phi_k}{\partial x_i}$ with respect to the time coordinate. With that in mind, we see that $\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(-t,\phi_t(p))=\left(\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(t,p)\right)\cdot\left(\left.\frac{d}{dt}\right|_{t=0}-t\right)=-\left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(t,p)$. Now:

\begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{\partial \phi_k}{\partial x_i}(t,p)&=\left.\frac{d}{dt}\right|_{t=0}\left.\frac{\partial}{\partial x_i}\right|_p\phi_k\\ &=\left.\frac{\partial}{\partial x_i}\right|_p\underbrace{\left.\frac{d}{dt}\right|_{t=0}\phi_k}_{=a_k}\\ &=\frac{\partial a_k}{\partial x_i}(p)\\ \end{align*} Therefore $\mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)=\sum_k-\frac{\partial a_k}{\partial x_i}\frac{\partial}{\partial x_k}=\sum_k\left[a_k\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i}\right]=\left[\sum_ka_k\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i}\right]=\left[X,\frac{\partial}{\partial x_i}\right]_\blacksquare$

4. For $Y=\sum_kb_k\frac{\partial}{\partial x_k}$ use 1), 2), 3) and the fact that $\left[X,b_k\frac{\partial}{\partial x_k}\right]=b_k\left[X,\frac{\partial}{\partial x_k}\right]+X(b_k)\frac{\partial}{\partial x_k}$. $_\blacksquare$