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Let $(X,dist)$ a compact metric space and $\phi$ a flow on $X$. We say that $x\in X$ is a spiral point if there is a continuous function $h:R\to R$ and $τ>0$ such that $h(t) − t > τ$ for all $t ≥ 0$ and $dist(\phi_t(x), \phi_{h(t)}(x)) → 0$ as $t → +∞$

I would appreciate it if you could help me prove the following statement: If $\phi$ is a continuous flow on the two-dimensional sphere then every point is spiral.

Adam
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helmonio
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1 Answers1

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Choose any $x\in S^2$. Poincaré-Bendixson tells us that if $\omega(x)$ (the $\omega$-limit set of $x$) is either (1) a fixed point, (2) a closed orbit (which in the case of the 2-sphere is necessarily periodic), or (3) a finite chain of fixed points connected by trajectories. In the first and second case, we let $\omega(x)$ be an orbit of period $T > 0$ (in the case that $\omega(x)$ consists of a single fixed point, this can be any positive number). Then, we note that $\phi(x, t)$ approaches $\omega(x)$ in the sense that $d(\phi(x, t), \omega(x))\to 0$, as otherwise, there is some $\{t_n\}_{n=1}^{\infty}\subset \{t > 0 : d(\phi(x, t), \omega(x)) > \epsilon\}$ for some $\epsilon > 0$ such that $t_n\to \infty$, and so we have some subsequence $\{t_{n_k}\}$ such that $\phi(x, t_{n_k})\to y'$ and $d(y', \omega(x)) > 0$. This implies $y'\notin \omega(x)$, which contradicts the definition of $\omega(x)$. Then, we let $h(t) = t+T$, so $h(t)-t = T > \tau$ for all $t\geq 0$ for any $\tau < T$, and we note that $$d(\phi(x, t), \phi(x, h(t))) = d(\phi(x, t), \phi(x, t+T)) = d(\phi(x, t), \phi(\phi(x, t), T))\to 0$$ as $d(\cdot, \cdot)$ and $\phi(\cdot, T)$ are continuous (and therefore uniformly continuous on $S^2\times S^2$ and $S^2$ respectively, as these are compact) and $d(y, \phi(y, T)) = 0$ for all $y\in \omega(x)$.

Now, we will handle the case where $\omega(x)$ is a chain of fixed points. If anybody has a nicer or less hand-wavy way of doing this, let me know (although, this logic theoretically also works in the case of the closed orbit). As before, $d(\phi(x, t), \omega(x))\to 0$. We choose a transversal $T$ to $\omega(x)$ and let $\{t_n\}_{n=1}^{\infty}\subset \mathbb{R}^+$ be the times at which $\phi(x, t)$ crosses $T$. Then, we let $h : \mathbb{R}\to \mathbb{R}$ be a continuous, increasing map such that $h(t_n) = t_{n+1}$ and for $S_{\epsilon}(t) := \{y\in \omega(x) : d(\phi(x, t), y) < \epsilon\}$ and some $\{\epsilon_n\}_{n=1}^{\infty}\subset \mathbb{R}$ decreasing to $0$ such that $d(\phi(x, t), \omega(x)) < \epsilon_n$ for all $t\in [t_n, t_{n+1}]$, we choose $h$ so that $S_{\epsilon_n}(t)\cap S_{\epsilon_n}(h(t))\neq \emptyset$ (basically, $h$ is chosen so that $\phi(x, h(t))$ "follows" $\phi(x, t)$ around $\omega(x)$, one pass ahead). Now, we note that $h(t)-t$ is bounded below, as otherwise, $\phi(x, t)$ travels around $\omega(x)$ arbitrarily fast, which implies that $\phi(\cdot, t)$ is not continuous near $\omega(x)$. Thus, we have some $\tau > 0$ such that $h(t)-t > \tau$. Furthermore, $d(\phi(x, t), \phi(x, h(t))) < 2\epsilon_n$ for $t\in [t_n, t_{n+1}]$, so $d(\phi(x, t), \phi(x, h(t)))\to 0$.

Michael L.
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  • Poincaré-Bendixson also tells that $\omega$-limit set could be a heteroclinic cycle or homoclinic loop. This is not a structurally stable case, but it still can be present. – Evgeny May 26 '17 at 09:03
  • On the 2-sphere, all positively recurrent orbits are periodic. Proof in Katok and Hasselblatt. – Michael L. May 26 '17 at 09:05
  • Actually, this is why the case of $S^2$ is so special. We don't have this nice property that every point on a compact subset of $\mathbb{R}^2$ is a spiral point under an arbitrary continuous flow. – Michael L. May 26 '17 at 09:15
  • Can you explain me, what is obstruction for taking this picture and making a flow on $\mathbb{S}^2$ out of it? – Evgeny May 26 '17 at 09:22
  • I think I partially understand what the catch is. Katok and Hasselblatt say "orbit", not set. Heteroclinic cycle or homoclinic loop are not orbits, they are union of orbits and equilibria. Still they can be attractors. – Evgeny May 26 '17 at 09:34
  • I think you're right, and it's a mismatch of terminology. Let me see if I can fill in the rest of this proof. – Michael L. May 26 '17 at 09:36
  • @Evgeny Does this proof make sense to you? – Michael L. May 26 '17 at 18:35
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    Sorry that I couldn't have answered earlier. This seems perfectly legit to me, if $T$ is a transversal to connecting orbit. By the way, $t_{n+1} - t_{n} \sim \gamma^{n+1}, ; \gamma > 1$: it can be proven from Grobman-Hartman. – Evgeny May 27 '17 at 05:32
  • Excellent, that's a nice result. Thanks. – Michael L. May 27 '17 at 05:39
  • I've checked my answer again: while it is true that passage time depends exponentially on distance from point on transversal to separatrix, the estimate for $t_n$ or $t_{n+1}-t_{n}$ seems to require more than this and the proof would be actually quite longer. So take this idea with caution. It can be done if we directly write Poincaré map for this transversal, it just needs more steps to prove this. – Evgeny May 27 '17 at 06:36
  • Fortunately, we don't really need $t_{n+1}-t_n$ to be exponential, and neither is that sufficient for this proof. We only it to be bounded below for any given transversal $T$. – Michael L. May 27 '17 at 06:52
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    Yeah, you are right. By the way, I think there is another way to prove that $t_{n+1}-t_{n}$ is greater than some constant $\tau$. This difference in times is spent on travelling near saddles and on travelling near connecting orbits. Time of travelling near connecting orbit is asymptotically constant (something like $\tau + \alpha_n, ; \alpha_n \rightarrow 0$) and gives lower bound for $t_{n+1}-t_n$. – Evgeny May 27 '17 at 09:08