To make the connection to the Lie derivative, let $t \mapsto \Phi^V_t$ be the 1-parameter group of diffeomorphisms (or flow) generated by the vector field $ V $. The differential $ d\Phi^V_t $ of each diffeomorphism maps the vector field Y to a new vector field $ \mathrm{d}\Phi^V_{t}(Y) $. To pull-back the vector field one applies the differential of the inverse, $ d ((\Phi^V_{t})^{-1})= d \Phi_{-t}^V $.

So what is $(\Phi^V_{t})^{-1}$ really saying? According to my knowledge, $\Phi^V_{t}$ is basically a point on flow curve (induced by vector field $V$ of some manifold $M$) given by input $x$ on manifold $M$ and some "time" $t$. So what would inverse map to?


$\Phi_t^V:M\to M$ is invertible and $(\Phi_t^V)^{-1}=\Phi_{-t}^V$. Indeed, if you flow $t$ units of time and then you flow $-t$ units of time, i.e. you flow back $t$ units of time, you end at the same point.

Formally, this is a direct consequence of the "group law" for flows: $\Phi_{s+t}=\Phi_s \circ \Phi_t$, and the fact that $\Phi_0=\operatorname{id}$.

So, applying $(\Phi_t^V)^{-1}=\Phi_{-t}^V$ to a point of $M$ you flow back $t$ units of time.


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.