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I feel like the above must be true but embarrassingly cannot seem to prove it. Take linearly independent, commuting vector fields $X$ and $Y$ on a manifold and corresponding flows $\Phi^t_X$, $\Phi^t_Y$. Suppose that the diffeo associated flow $t\rightarrow \Phi^t_X \circ \Phi^t_Y$ is the identity for some $t_1>0$. Does it follow that $\Phi^{t_1}_X= \Phi^{t_1}_Y=\text{Id}$? What if $X$ or $Y$ vanish at some isolated points? Does the "result" (which is not currently a result) still hold?

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    $\begingroup$ If we are on the flat torus $\mathbb R^2/\mathbb Z^2$ and your vector fields are constant, aren't you just asking whether the fact that the sum of 2 linearly independent vectors has integer coordinates implies that each of them has integer coordinates? I presume it does not, but, perhaps, I misunderstand the setup? $\endgroup$
    – fedja
    Jun 3, 2018 at 19:22
  • $\begingroup$ I think in the specific case that the vector fields are on torus and constant (by constant I assume you mean invariant wrt to the Lie group structure on the torus), then maybe the question reduces to what you said, though I am not quite sure what you mean by "integer coordinates". Of course the v.f. could have whatever coordinates as long as the ratio is rational. $\endgroup$
    – R Mary
    Jun 3, 2018 at 19:54
  • $\begingroup$ I mean take two translation flows $\Phi_a(x)=x+at$ and $\Phi_b(x)=x+bt$ on $\mathbb R^2/\mathbb Z^2$ with the usual Euclidean metric. At time $1$ the composition shift is $a+b$. If it is in $\mathbb Z^2$, it is the identity on the torus. $\endgroup$
    – fedja
    Jun 3, 2018 at 21:51
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    $\begingroup$ If $X=(1,\epsilon)$ and $Y=(1,-\epsilon)$ then the sum will have a periodic orbit on the standard torus $\mathbb R^2/\mathbb Z^2$ for $t=\frac12$, but the fields $X$ and $Y$ will not. $\endgroup$ Jun 4, 2018 at 11:09
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    $\begingroup$ @MikhailKatz, ah yes of course how silly of me. I only just understood the point of fedja . If someone wants to write this up as an answer I will accept, otherwise I will do so myself :) $\endgroup$
    – R Mary
    Jun 4, 2018 at 11:20

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