Suppose $f\colon M\to\mathbb{R}$ is a smooth 0-form on our manifold $M$, and $X$ is a vector field with an induced flow $\varphi_t\colon M\to M$.

I am wanting to show that $\mathcal{L}_Xf = f$ implies that $\varphi_t^* f = e^t f$ but I am having trouble making any progress.

What I know is that we define the Lie derivative as $$\mathcal{L}_Xf := \lim_{t\to 0}\frac{\varphi_t^*f - f}{t},$$ but the problem is that I do not know exactly how to manipulate it into the desired form. If $$f(p)= \lim_{t\to 0}\frac{f(\varphi_t(p)) - f(p)}{t},$$ I don't exactly have any leads. Perhaps I am looking at it from the wrong angle and I should not use the limit definition?

I would be particularly happy with any hints, as I do not want to spoil the exercise too much. Thanks!

  • $\begingroup$ Can you do this around a point $p$ such that $X_p\neq0$? $\endgroup$ – Mariano Suárez-Álvarez Nov 29 '16 at 3:42
  • $\begingroup$ @MarianoSuárez-Álvarez My intuition was that this is kind of like the directional derivative of $f$ in the direction of $X$, so getting $f$ again would mean it is exponential of some sort. I don't know of any special conditions that occur when $X_p \neq 0$. Could you elabourate? $\endgroup$ – anakhro Nov 29 '16 at 3:54

Let $p \in M$ be fixed. Let $I \subset \mathbb{R}$ be the domain of $t \mapsto \phi_t(p)$, so that $I$ is an open interval containing $0$, and define a function $g_p : I \to \mathbb{R}$ by $g_p(t) := \phi_t^\ast f(p) = f(\phi_t(p))$. By the chain rule (or by the definition of smooth function on a manifold, depending on what reference you're using) the function $g_p$ is differentiable on all of $I$. What, then, is the relationship between the derivative $g_p^\prime : I \to \mathbb{R}$ of $g_p$ and $(\mathcal{L}_X f)(p)$, and what does it tell you about $g_p$? Note that this may require a tiny bit of care for $t \neq 0$.

  • $\begingroup$ Just making sure I understand correctly: would it be correct to say that---when evaluated at $t=0$---$g_p'$ and the Lie derivative are equal? $\endgroup$ – anakhro Nov 29 '16 at 4:06
  • 1
    $\begingroup$ Indeed. But then you can work a little harder to compute $g_p^\prime(t)$ for $t \neq 0$ by means of the group homomorphism property of the flow $\phi$, i.e., $\phi_{s + t}(p) = \phi_s(\phi_t(p))$ for $s$, $t \in I$ such that $s+t \in I$. $\endgroup$ – Branimir Ćaćić Nov 29 '16 at 4:07
  • $\begingroup$ Thank you, that helps me get a bit closer for now! $\endgroup$ – anakhro Nov 29 '16 at 4:08

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.