Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to follow a proof that the Lie derivative of a vector field $Y$ in the direction of a vector field $X$ is the commutator $[X,Y]$. We write $\Phi_t$ for the flow of $X$.

How do I get from $$\frac{d}{dt}\vert _{t=0}\Phi^*_t(Y(\Phi^*_{-t}(f)))$$ to

$$\frac{d}{dt}\vert _{t=0} (\Phi^*_t(Y(f))-Y(\Phi^*_t(f))$$ ? I'm sure it's simple but I'm just not seeing it.

share|cite|improve this question
up vote 2 down vote accepted

Formally, this is because of the product rule (e.g. $\frac{d}{dt}\vert_{t=0} (A(t)BC(t)) = A'(0) B C(0) + A(0) B C'(0)$): $$ \frac{d}{dt}\vert_{t=0} \Phi_t^*(Y(\Phi_{-t}^* (f))) = (\frac{d}{dt}\vert_{t=0} \Phi_t^*) (Y(\Phi_0^* f)) + \Phi_0^* Y(\frac{d}{dt}\vert_{t=0} \Phi_{-t}^* f). $$ Now, use the fact that since $\Phi_t$ is a flow, $\Phi_0$ is the identity map and for the second term make the substitution $t \mapsto -t$. The $\frac{d}{dt}\vert_{t=0}$ comes out in the second term since $Y$ is a linear operator and independent of $t$.

To convince yourself that you can apply the product rule to these pullbacks of diffeomorphisms you can either prove it using the same proof of the product rule for functions or you can observe that locally this is all just matrix multiplication (with functions as entries) and that the product rule holds for matrix multiplication of function valued matrices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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