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 .

  • 1
    $\begingroup$ Can you show us where you get stuck? The details are a bit of a mess, but the idea is straightforward. Start with the left side and compute at a point applied it to an arbitrary smooth function, i.e. write out $(L_X Y)_p f$ with the limit definition. You should be able to rearrange terms and rewrite things and cancel a little to get to the right hand side. $\endgroup$
    – Matt
    Apr 5, 2012 at 2:18

5 Answers 5


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

  • $\begingroup$ Why is $H$ differentiable? I deeply believe it is, but I'm having trouble justifying it $\endgroup$
    – rmdmc89
    Jul 18, 2018 at 19:36
  • 2
    $\begingroup$ It's a (if I recall correctly, fairly hard) theorem that the flows $\phi^{t}(x)$ generated by a vector field are smooth in both $x$ and $t$ (for small $t$). More generally, this should be the result of a theorem saying solutions to ordinary differential equations are smooth in the prescribed initial conditions. You can probably find it in Lang's manifolds book? $\endgroup$ Jul 18, 2018 at 19:59

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.

  • 2
    $\begingroup$ This answer is very helpful, but I am very confused about how $d\phi_t Y(f) = Y(f \circ \phi_t) \circ \phi_t^{-1}$. Where does the composition of $ \circ \phi_t^{-1}$ come from? As far as I know, the formula for pushforward is $(f^*Y)(g) = Y(g \circ f)$ Thank yuo! $\endgroup$ Jun 28, 2019 at 8:36

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 $\phi(p,t)$.


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 $\mathcal{L}_X\left(\frac{\partial}{\partial x_i}\right)=\sum_k-\frac{\partial a_k}{\partial x_i}\frac{\partial}{\partial x_k}=\sum_k\left[a_k\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i}\right]=\left[\sum_ka_k\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i}\right]=\left[X,\frac{\partial}{\partial x_i}\right]_\blacksquare$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.