# Poincaré–Bendixson Theorem on a compact, connected, orientable, two-dimensional manifold

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. 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.

• 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 $\omega(x) \setminus \Sigma$? Feb 24 at 22:15
• $(2)$ 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)? Feb 24 at 22:18
• 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. Feb 24 at 23:17
• 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). Feb 25 at 0:38
• 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! Mar 3 at 10:37