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On a manifold $M$, let there be two integrable vector fields $X$ and $Y$. Let $\phi_t^{X}:M\to M$ (respectively $\phi_t^{Y}:M\to M$) be the diffeomorphism obtained by following the flows of $X$ (respectively $Y$) for 'time' $t$.

I'm trying to construct an integrable vector field $Z$ such that $\phi^Z_1=\phi_1^X\circ \phi^Y_1$. Am I right in saying that that vector field $Z$ is obtained by taking the derivative of $\phi_t^X\circ \phi^Y_t$ w.r.t. $t$?

dennis
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    Interesting question. What attempts, related facts, and ideas do you have so far? – aschepler Mar 09 '23 at 13:39
  • On $S^2$ every vector field vanishes somewhere (https://en.wikipedia.org/wiki/Hairy_ball_theorem), so the corresponding flows will have fixed points. But you can cook up examples (such as rotation along different axes) whose composition doesn't have fixed points. – red whisker Mar 09 '23 at 14:30
  • @redwhisker The composition of two rotations is another rotation and has two fixed points. Are you sure there are examples? – aschepler Mar 09 '23 at 14:33
  • @aschepler that's right (I didn't have a real example when I made the comment). I imagined this would lead to a counter example – red whisker Mar 09 '23 at 14:51
  • @aschepler I've edited the question proposing a solution – dennis Mar 09 '23 at 14:51
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    The answer is negative, see the link. – Moishe Kohan Mar 09 '23 at 15:17
  • @MoisheKohan Hi Moishe. If a diffeomorphism $M\to M$ is generated by some vector field (by following its flow lines for 1 unit of time), do you know how one can extract the vector field from the diffeo? – dennis Mar 09 '23 at 17:03
  • It is in general nonunique, so you cannot. – Moishe Kohan Mar 09 '23 at 17:06
  • @MoisheKohan It is said that the Lie algebra of the diffeomorphism group is the set of vector fields equipped with the Lie bracket of vector fields. Wouldn't that mean that an element of the group is the exponentiation of a Lie algebra element which is basically a diffeomorphism generated by the flow of some vector field? – dennis Mar 11 '23 at 17:00
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    @dennis: No, it does not mean surjectivity of the exponential map (even in the case of ordinary Lie groups). In fact, even local surjectivity will fail (unlike the case of ordinary Lie groups). See here for references. – Moishe Kohan Mar 11 '23 at 21:33

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