Non-zero index implies index $1$?

Question

Suppose $G\subset \Bbb C$ is open, $0\notin G$, and some closed curve in $G$ has non-zero index about the origin. Does it follow that some closed curve has index $1$ about the origin?

(To avoid an XY problem: All I really need to know is that if the index is always even then it is always $0$.)

Seems clear, but as sometimes happens in topology I have no idea how to prove it.

My work so far: Oh gimme a break.

Context: complex analysis.

If the curve is not simple, for example $e^{it}$ with $t\in [0,4\pi]$ the answer is not — Martín Vacas Vignolo, Jan 28 '18 at 19:06
@MartínVacasVignolo If that curve is contained in $U$ then so is $e^{it}$, $t\in[0,2\pi]$. — Angina Seng, Jan 28 '18 at 19:08
@LordSharktheUnknown oh yes, i read wrong (that the same curve has index 1) — Martín Vacas Vignolo, Jan 28 '18 at 19:10
@MartínVacasVignolo I don't understand your comment at all. Possibly you misunderstood the question - the two occurrences of the phrase "some closed curve" do not refer to the same curve. — David C. Ullrich, Jan 28 '18 at 19:11
This question was asked in various forms many times, see for instance https://math.stackexchange.com/questions/2393771/alternative-definition-of-simple-connectedness?noredirect=1&lq=1 (disclaimer: my solution is way too complicated for the problem). — Moishe Kohan, Jan 29 '18 at 17:53
@MoisheCohen I don't see how that's the same question at all... — David C. Ullrich, Jan 29 '18 at 19:30

score 4 · Accepted Answer · answered Jan 28 '18 at 19:48

4

This is more like a sketch of how a solution might go, rather than an actual solution. There may be some insurmountable error within.

You can replace your original path with a polygonal closed path with the same winding number about $0$. You can also assume that all the edges of this path have different slopes. There will probably be some nontrivial self-intersections. If so break the path into a "sum" of simple closed paths. One of these must have nonzero winding number. Now appeal to the (polygonal) Jordan curve theorem to see that the winding number is this curve is $\pm1$.

answered Jan 28 '18 at 19:48

Angina Seng

161,540

A little informal but seems convincing to me. The fact that any closed polygonal path (with distinct slopes so self-intersections are just points) is a sum of simply polygonal closed paths seems like a trivial bit of graph theory: Say the verticies of the polygonal path and the self-intersections are vertices of the graph, and define the edges to be the edges. Then evey vertex has even index so the graph is a union of disjoint circuits. – David C. Ullrich Jan 28 '18 at 20:01

score 0 · Answer 2 · edited Jun 19 '18 at 14:52

0

Argue by induction on the number of self intersections, $n$, of the polygonal path. If $n=0$ then apply the Jordan curve theorem. If $n=k+1$ write the path as the sum of $2$ paths with self intersection numbers $0$ and $m<k+1$.

I have been trying for a long time to find a proof that does not use some version of the Jordan curve theorem.

edited Jun 19 '18 at 14:52

user1729

32,369

answered Jun 19 '18 at 14:32

William Cohn

1

In the above, m<k+1. I have been trying for a long time to find a proof which does not use some version of the Jordan curve theorem. – William Cohn Jun 19 '18 at 14:34
Welcome to MSE. It is in your best interest that you use MathJax. – José Carlos Santos Jun 19 '18 at 14:36

Non-zero index implies index $1$?

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