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This is probably a very elementary question. Nevertheless I decided to post it on MO. Consider a smooth manifold $M$ and a smooth complete vector field $v:M\rightarrow TM$. Consider an autonomous differential equation defined by $v$: $\frac{d}{dt}x(t)=v(x(t)$. Let $\Phi_t(\cdot):M\rightarrow M $ be the evolution map corresponding to time $t$. That is $\Phi_t(x)$ is the solution of differential equation with the initial condition $x\in M$ after time $t$.

Question

Assume that we know that $\Phi_t(\cdot)$ has at least one fixed point in $M$ for each $t\in\mathbb{R}_+$ (or even $t\in\mathbb{R}$). Under what conditions imposed on the manifold $M$ and a vector field $v$ one can be sure that there is a stationary point of the vector field $v$ in $M$?

Counterexample

Let $M= \{(x,y) \in\mathbb{R}^2 |1< \sqrt{x^2+y^2} <2 \}$ and let the vector field $v$ has in cylindrical coordinates $(r,\phi)$ the following form: $v=\frac{r-1}{2-r}\frac{\partial}{\partial \phi}$. It is clear that for each $T>0$ we have a circle of periodic orbits with period T. Therefore $\Phi_T(\cdot)$ has fixed points that do not correspond to stationary points of the vector field $v$.

Modified counterexample

I above example it is sufficient to consider $v=(r-1)\frac{\partial}{\partial \phi}$

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1 Answer

The implication holds e.g. on compact manifolds. To see this, you just have to convince yourself that in this case the flow of a vector field without zeros cannot have fixed points for very small times.

EDIT (more details): Assume $v$ is a vector field without zeros on a closed manifold $M$ and let $\phi_t$ be its flow. For small $t$ you can Taylor-expand (let's say in local charts) $\phi_t(x)=x+tv(x)+\ldots$. So, using compactness to control the error term uniformly, you can find $\varepsilon>0$ such that $\phi_t(x)\neq x$ for all $x\in M$ and all $0 < t < \varepsilon$.

Non-compact case: Modifying your counterexample somewhat, you can construct one on a complete Riemannian manifold where the vector field has bounded norm (think of circles around a hyperbolic cusp), and with a bit more effort one should also get an example on a manifold with bounded geometry, etc. So, I doubt that there are useful/simple conditions in the non-compact case.

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Thanks. I am not sure I understand your answer. You wrote: "The implication holds e.g. on compact manifolds. To see this, you just have to convince yourself that in this case the flow of a vector field without zeros cannot have fixed points for very small times." I would rather say in this context that that: "for a vector field without zeros associated $\Phi_t$ cannot have fixed points for very large $t$" –  Michal Oszmaniec Mar 10 '12 at 12:45
    
I have added more details, is it clear for you now? Also, $\phi_t$ obviously can have fixed points for $t$ large even if $v$ has no zeros (just think of closed orbits). –  Robert Haslhofer Mar 10 '12 at 13:28
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