What is the argument for the fact that each arc in the interior of a 2-manifold can be "thickend" to obtain a 2-cell containing the arc in its interior and being disjoint from any preassigned compact set in the complement of the arc? Thanks in advance!
$\begingroup$
$\endgroup$
4
-
$\begingroup$ What about, say, taking $V_{\delta/2}$, the set of points lying at most a distance $\delta/2$ away from the arc? Here $\delta > 0$ would be the distance between said arc and the preassigned compact set. $\endgroup$– Leo MoosCommented Aug 29, 2022 at 21:46
-
$\begingroup$ Are we assuming that the surface comes with a metric? $\endgroup$– Sam NeadCommented Aug 29, 2022 at 21:47
-
$\begingroup$ Also, taking a $\delta/2$--neighbourhood need not give a two-cell - you need to take the constant small enough to avoid the neighbourhood overlapping itself. $\endgroup$– Sam NeadCommented Aug 29, 2022 at 21:50
-
3$\begingroup$ This is a delicate question if you want a real proof, and not "just" some handwaving about "oh, this follows from the Schoenflies theorem". $\endgroup$– Sam NeadCommented Aug 29, 2022 at 23:53
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Suppose that $S$ is the surface, $\alpha$ is the arc, and $K$ is the compact set.
Proof 1: Find an arc $\alpha'$ so that $\gamma = \alpha \cup \alpha'$ is a Jordan curve (+). The Jordan/Schoenflies [J/S] theorems provide an annulus (or Möbius) neighbourhood $A$ of $\gamma$ and a homeomorphism on $A$ making $\gamma$ "straight". Shrink $A$ as needed to avoid $K$ and then use $A$ to find the desired two-cell.
Proof 2: Let $T$ be the double branched cover of $S$, branched over the two points of $\partial \alpha$. Let $\beta$ and $\beta'$ be the two preimages of $\alpha$ in $T$. So the union $\gamma = \beta \cup \beta'$ is a Jordan curve. Apply [J/S] to find an annulus neighbourhood $A$ of $\gamma$. Shrinking $A$ as needed, we may assume that it is disjoint from the preimage of $K$. We further arrange matters so that $A$ is invariant under the deck transformation, as are two of its cutting arcs (meeting $\gamma$ in exactly the points $\partial \beta$). So the image of $A$ is the desired two-cell.
Of course, in Proof 1, sentence (+) is saying that endpoints of $\alpha$ are "accessible". Showing such things is part of various proofs of [J/S]. The branched cover trick in Proof 2 is designed to avoid (+). However, I'll guess that no proof can avoid using [J/S], or at least some of its "working parts".
-
2$\begingroup$ Belevant: mathoverflow.net/questions/57766/… $\endgroup$ Commented Aug 30, 2022 at 0:35
-
1$\begingroup$ Yes, very relevant! Also, I notice that I, sometime in the past voted, up that post and some of the answers... We should probably mark this question (and my answer) as a duplicate. $\endgroup$– Sam NeadCommented Aug 30, 2022 at 1:25
-
$\begingroup$ Unfortunately I don't have enough reputation to write a comment, so I have to use the answer field to ask for the clarification for the first proof idea here. @SamNead, could you please explain how one can apply the Jordan-Schoenflies to a surface? The statement of the theorem holds for now for a plane only. Then, it's also not so clear, why, for exmaple, a spiral arc $[0, 1] \to \mathbb{C}$, $t \mapsto (1 - t)e^{i \frac{1}{1 - t}}$, completed to a Jordan curve, has necesseraly an annulus neighbourhood. How exactly does it follow from the Jordan-Schoenflies? Finally, I've read the thread https $\endgroup$ Commented Sep 9, 2022 at 9:55
-
$\begingroup$ [Text continued from Palina Severyna's answer which was converted to a comment:] Finally, I've read the thread mathoverflow.net/questions/57766/… (you marked my question as duplicate to it) and couldn't understand in the argument of Bill Thurston, why the original arc can be identified with one of its two pre-images under the branched cover $z \to z^2$ of a sphere. What is a homeomorphism between them? I also would like to remark that I may not assume that the surface permits a triangulation. $\endgroup$ Commented Sep 9, 2022 at 11:29