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Let X be a nowhere vanishing complete vector field on a manifold M, $\gamma: \mathbb{R} \to M$ be its flow with $\gamma(0)=p \in M$ and suppose it is not periodic. If $\gamma(\mathbb{R})$ is closed, is $\gamma$ a proper map?

The only counter-example I can think of is a curve looping back to itself such that $\lim_{t\to + \infty} \gamma(t) = \gamma(\hat{t})$ for some $\hat{t} \in \mathbb{R}$ but I don't know if this can arise from a nowhere vanishing complete vector field.

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    $\begingroup$ What is meant by proper map? $\endgroup$ – Sergei Akbarov Jul 7 '17 at 23:33
  • $\begingroup$ I think it is the usual definition, that is $\gamma$ is a homeomorphisme on its image. Equivalently, $\gamma$ is one-to-one and the pre-image of a compact set is a compact set. $\endgroup$ – M. Dus Jul 14 '17 at 9:02
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I assume that your vector field is at least continuous. I will answer your second question. Such a situation cannot arise. Assume that $\gamma(t)$ converges to some point $p\in M$. Then, $X(\gamma(t))$ converges to $X(p)$, so $\gamma'(t)$ converges to some vector $a$. Using local coordinates, you can assume that $M=\mathbb{R}^n$, for simplicity. Since $\gamma(t)$ converges, $\gamma'(t)$ is integrable, so that its limit (since it does exist) is zero. Thus, $X(p)=0$.

In general, non-properness means that there is a sequence $t_n$ tending to infinity such that $\gamma(t_n)$ converges to some point $p$.

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Assume the vector field $X$ to be of class $C^1$. As hinted by M. Dus, to answer the first question it suffices to exclude the case that there is $t_n \to \infty$ (say) such that $\gamma(t_n) \to \gamma(\tau) (=: p) $. Take a closed flow box $U$ of $p$, with transversal $T$. Clearly, $p$ is not isolated in $T \cap \gamma(\mathbb{R})$. Since the flow maps are homeomorphisms, the closed set $T \cap \gamma(\mathbb{R})$ contains no isolated points, hence is perfect. Therefore $T \cap \gamma(\mathbb{R})$ has cardinality $\mathfrak{c}$, which contradicts the second countability of the real line.

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  • $\begingroup$ I don't really understand the last argument. I mean, the Cantor set is perfect but embeds as a subset of the real line. Maybe I missed the argument. $\endgroup$ – M. Dus Mar 19 '18 at 11:44
  • $\begingroup$ A flow box is a closed nbhd of the form $\{ \Phi(t, \xi): -\tau \le t \le \tau, \xi \in T \}$ for some $\tau > 0$ and some $(n-1)$-dimensional disk $T$ (transversal), transverse to the vector field. Consequently, for any two $\gamma(t_1), \gamma(t_2) \in T$ one has $\lvert t_1 - t_2 \rvert > 2 \tau$. If $\gamma(\mathbb{R}) \cap T$ were uncountable then there would be an uncountable subset of $\mathbb{R}$ whose mebers are $> 2 \tau$ apart. $\endgroup$ – user539887 Mar 19 '18 at 12:59
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    $\begingroup$ Incidentally, using the results contained in: C. Calcaterra, A. Boldt, Lipschitz flow-box theorem, J. Math. Anal. Appl. 338 (2008), 1108-1115, one can relax the assumption on the vector field to being only (locally) Lipschitz. $\endgroup$ – user539887 Mar 19 '18 at 13:13
  • $\begingroup$ Oh yeah, of course they are $2\tau$ apart. Thanks for answering, this is nice! $\endgroup$ – M. Dus Mar 19 '18 at 17:30

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