I'm currently reading the article "A Generalization of a Poincaré–Bendixson Theorem to Closed Two-Dimensional Manifolds" by Arthur Shwartz. The paper first establishes a result for minimal sets, which are closed, non-empty subsets of $M$ which are invariant under the flow of some $C^2$ vector field, and contain no such proper subsets. In particular, the result can be summarized as follows:

Suppose that $M$ is a compact, connected, two-dimensional manifold and $\Sigma \subset M$ is a minimal set. Then $\Sigma$ is either:

  1. a fixed point,
  2. a periodic orbit (homeomorphic to $S^1$), or
  3. $M$ itself (which must be homeomorphic to $T^2$).

I was able to follow the proof of this theorem, but fail to understand the subsequent corollary, which may be stated as:

Suppose that $M$ is a compact, connected, orientable, two-dimensional manifold such that $M$ is not a minimal set and for some $x \in M$, the $\omega$-limit set of $x$$\omega(x)$—contains no fixed points. Then $\omega(x)$ is homeomorphic to $S^1$ and the flow of $x$$\varphi_{t}(x)$—winds towards $\omega(x)$ as $t \to \infty$.

The argument presented is that $\omega(x)$ must contain a minimal set (say $\Sigma$) which is homeomorphic to $S^1$ by the previous theorem. Now since $M$ is orientable, there exists a neighborhood $N$ of $\Sigma$ which is homeomorphic to $(-1, 1)\times S^1$ such that $\Sigma$ is mapped onto $\{0\}\times S^1$ and the line segments $(-1, 1)\times \{y\}$ are transversal arcs for $y \in S^1$. The setup is clear to me, but I don't follow the remaining steps of the proof: "It follows from the continuity of the flow, and the fact that $\Sigma \subset \omega(x)$, that there exist $0 < t_1 < t_2$, $q \in \omega(x)$, and a transversal $I_q$ through $q$ such that $\varphi_{t_2}(x)$ is between $\varphi_{t_1}(x)$ and $q$ on $I_q$. Thenceforth $\varphi_{t}(x)$ is "trapped" between the closed curve formed by $\varphi_{[t_1, t_2]}(x)$ and the portion of $I_q$ between $\varphi_{t_1}(x)$ and $\varphi_{t_2}(x)$ and $\Sigma$. Thus we find the next intersection of $\varphi_{t}(x)$ with $I_q$ between $\varphi_{t_2}(x)$ and $q$. Repetition of this argument yields the conclusion of the corollary."

I believe that we're aiming to show that $\omega(x) = \Sigma$ (in particular by a monotonicity-type argument on the surface of the cylinder analogous to that used in the proof of the Poincaré–Bendixson theorem in $\mathbb{R}^2$). But how can we be sure that $q \in N$, or that the flow remains within $N$ (and hence its image under the homeomorphism remains on the surface of the cylinder, which I understand to be what "traps" it)? Even if so, how would this ultimately show that $q \in \Sigma$?


1 Answer 1


To address your two questions:

  1. Why is $q \in N$?

If I understand correctly, they could have (and should have) just started with $q \in \Sigma$, since for any $q \in \Sigma$ the orbit $\phi_t(x)$ approaches $q$ arbitrarily closely.

To be precise, because $\Sigma \subset \omega(x)$, there's a $t_1 > 0$ such that $\phi_{t_1}(x) \in I_q \subset N$, and there's a $t_2 > t_1$ such that $\phi_{t_2}(x) \in I_q$ and is even closer to $q$ than $\phi_{t_1}(x)$. By orientability of $M$, $\phi_{t_2}(x)$ is between $q$ and $\phi_{t_1}(x)$.

  1. Why does the flow stay in $N$?

The proof doesn't need the flow to stay within $N$. Instead, it requires something weaker. Consider the subsegment $T$ of the transversal $I_q$ between $q$ and $\phi_{t_1}(x)$. By continuity, if we choose $t_1$ so that $\phi_{t_1}(x)$ is sufficiently close to $q$, then the orbit of any point $T$ will stay within $N$ up until it hits $I_q$ again.

That is enough to rule out any "weird manifold topology" and conclude there is a "trapped" region bounded by $\phi_{[t_1, t_2]}$, $\Sigma$, and a piece of $I_q$, as in the Poincare-Bendixson proof.

  • $\begingroup$ Thank you for your comment. I have a few issues remaining: $(1)$ If we assume that $q \in \Sigma$, I'm unsure how we can draw the conclusion that $\Sigma = \omega(x)$. That is, we know that there exists a sequence $(t_n)$ such that $\varphi_{t_n} \to q \in \Sigma$, but what's to stop some other sequence from approaching a point in $\omega(x) \setminus \Sigma$? $\endgroup$ Feb 24 at 22:15
  • $\begingroup$ $(2)$ The geometry is not totally clear to me in the final region. I understand that $\varphi_{[t_1, t_2]}(x)$ and $I_q$ combine to create a closed curve, but how does $\Sigma$ come into play with them? And how can we be sure that the region "traps" the flow if it is able to leave $N$ (for example, in the torus, a closed curve doesn't necessarily separate the space)? $\endgroup$ Feb 24 at 22:18
  • $\begingroup$ Think of the final region as bounded by the closed curve $\Sigma$, the piece of orbit $\varphi_{[t_1, t_2]}(x)$, and the segment of $I_q$ between $\varphi_{t_1}(x)$ and $\varphi_{t_2}(x)$. It's exactly the diagram you make in the planar version of Poincare-Bendixson. This is why they bother to construct $N$ homeomorphic to $S^1 \times (-1,1)$, by the way: not to be an invariant region, but to create a neighborhood where we can apply essentially planar reasoning. $\endgroup$ Feb 24 at 23:17
  • $\begingroup$ I understand that the aim is to use planar reasoning on the cylinder, but if the flow leaves $N$ between $t_1$ and $t_2$, then $\varphi_{[t_1, t_2]}(x)$ cannot be thought of as lying on the cylinder (rather, there would be two disjoint curve segments). In such a case, I don't see why the aforementioned curves necessarily enclose a region (for instance by utilizing some strange manifold topology outside of this effectively planar neighborhood). $\endgroup$ Feb 25 at 0:38
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    $\begingroup$ My apologies for taking so long to respond. It took some additional reasoning (and a few drawings), but I believe the argument makes sense now. Thank you for your answer! $\endgroup$ Mar 3 at 10:37

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