Let $\Phi _t$ the flow of a vector field $X$, and $\mu$ an $m$ form on a manifold $M$. I want to prove the following relation:
$${d\over dt} \Phi _t^* \mu= \Phi_t^* L_X\mu$$
I was looking for an elementary proof of the fact. Thank for your help.
Show that ${d\over dt} \Phi _t^* \mu= \Phi_t^* L_X\mu$, $\;\Phi _t$ the flow of $X$, $\mu$ a form.
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$\begingroup$
manifolds
differential-forms
vector-fields
lie-derivative
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0This should just amount to writing down the limit definitions, changing $\lim\limits_{t\to t_0}$ to $\lim\limits_{t\to 0}$. – 2017-01-20
1 Answers
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In case you don't get any other answers, this is proved in Proposition 12.36 in my Introduction to Smooth Manifolds (2nd ed.). The proof is only a few lines long.
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0Thanks for the reference. It seem a very complete and useful book. On the proof, I'm having trouble understanding this step: ${d\over ds} |_{s = 0}\; d(\theta_{t_0})^{*}_{p}\, d(\theta_s)^{*}_{\theta_{t_0}(p)} (A_{\theta_s (\theta_{t_0} (p))}) = d(\theta_{t_0})^{*}_{p}\; {d\over ds} |_{s = 0}\; d(\theta_s)^{*}_{\theta_{t_0}(p)} (A_{\theta_s (\theta_{t_0}(p))})$. Can you please explain this step. – 2017-01-21
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0@user95747: It's just the fact that $d(\theta_{t_0})_p^*$ is a fixed linear map between two fixed vector spaces, and does not depend on $t$. If $A:V\to W$ is any linear map and $f:\mathbb R\to V$ is a differentiable function, then $\frac d {ds}(A\circ f)(s) = A\big(\frac d {ds} f(s)\big)$. – 2017-01-22