5
$\begingroup$

Since fedja's excellent comment on Joseph's question on probing a manifold with geodesics remained uncommented (especially by topologists), I'd like to make a question out of it:

Conjecture: Given an orientable 2-dimensional manifold and two closed curves on it which intersect transversally in exactly one point. Then the two curves cannot be homotopic.

(An immediate consequence of this would be that living on a surface with two such curves, one would know, that it is not homeomorphic to the sphere.)

How to proof this conjecture (if it's true)?

$\endgroup$
5
  • 6
    $\begingroup$ Counting signed intersections of curves gives a symplectic form on the first homology group of a closed surface. In particular, in such a surface two homologous curves never intersect transversally in exactly one point. $\endgroup$ Mar 14, 2013 at 17:42
  • 3
    $\begingroup$ Did you omit the word "orientable" from your conjecture? It's false as it stands. $\endgroup$
    – maproom
    Mar 14, 2013 at 17:43
  • 1
    $\begingroup$ (My comment refers to orientable surfaces, as well!, as otherwise there are no signed intersection numbers to get us started) $\endgroup$ Mar 14, 2013 at 17:50
  • $\begingroup$ @maproom: Indeed, I had only orientable surface in mind. So I added it. Thanks for the correction! $\endgroup$ Mar 14, 2013 at 17:51
  • 2
    $\begingroup$ If you really want to learn about curves on surfaces, I recommend "Travaux de Thurston" by Fathi, Laudenbach and Poeneru. If you are averse to reading French, Magalit and Kim have translated it into English ( Work of Thurston). Exposes 3 and 4 are the real stuff. $\endgroup$ Mar 15, 2013 at 2:38

3 Answers 3

7
$\begingroup$

I wanted to give a somewhat "low-tech" answer to this, here's the best I came up with:

Call the curves $C_1$ and $C_2$ and the surface that they lie in $\Sigma$. Take a small open neighborhood $N$ of $C_1 \cup C_2$. What does $N$ look like? Well, it's a neighborhood of $C_1$ stuck to a neighborhood of $C_2$ in an operation called a "plumbing" (which is the same operation as sticking together two strips of paper to make a cross).

The neighborhood of $C_i$ is a cylinder that has been plumbed to itself a few times (one plumbing for each self-intersection of $C_i$).

Now if $C_1$ can be homotoped to coincide with $C_2$ inside $\Sigma$ then it can be homotoped to $C_2$ inside the space $S$ that I make by crushing $\Sigma \setminus N$ to a point.

The idea is to convince yourself that $S$ is a torus with a finite number of points identified (one point for each boundary circle of $N$), and that $C_1$ wraps once around one direction of the torus while $C_2$ wraps once around the transverse direction.

$\endgroup$
1
  • $\begingroup$ Thanks for all the upvotes, but I've realized that my argument is not correct. The space space $S$ is not necessarily a torus with some points identified but just some genus $g \geq 1$ surface with a finite number of points identified. It's fixable though. Orient $C_i$ and choose an oriented loop $P_i$ on $C_i$ with no self-intersections such that $P_1$ and $P_2$ intersect once. Then crushing the complement of a neighborhood of $P_1 \cup P_2$ gives an honest torus such that $C_1$ wraps once in one direction and $C_2$ wraps once in the other direction. $\endgroup$ Mar 15, 2013 at 15:34
6
$\begingroup$

The conjecture is false. For example, in a Moebius band, the central circle self-intersects once with many transverse perturbations of itself.

If you want a conjecture like yours to be true, you'll have to assume that a tubular neighbourhood of one of the curves is trivial. That ensures a regular neighbourhood of the union is a punctured torus or Klein bottle.

$\endgroup$
11
  • $\begingroup$ Sorry, I just added "orientable" (thanks to maproom's hint). Is your additional assumption equivalent to require the manifold to be orientable? $\endgroup$ Mar 14, 2013 at 17:54
  • $\begingroup$ The additional assumption is weaker than assuming orientability -- the second curve can be the central core of a Moebius band provided the first is not. But in that situation there's a simpler proof that the two curves aren't homotopic, as one is an orientation-reversing path and the other is not. $\endgroup$ Mar 14, 2013 at 17:57
  • $\begingroup$ @Ryan: Can you tell - by the way - how to aquire a power of imagination like yours (which I envy you for). How can one come to see at once that "the central circle of a Moebius band self-intersects once with many transverse perturbations of itself"? (I assume this is something you see, but for ordinary people like me it requires a very lot of thinking about it.) The same question - by the way - goes to Mariano's first comment to my original question. So you don't have to take it too serious. $\endgroup$ Mar 14, 2013 at 18:13
  • 2
    $\begingroup$ It's a toolbox you build-up over time. The particular geometric tools you're referring to I think of as belonging to the "transversality/general position/tubular neighbourhoods/bundle mania" variety. If you haven't had the chance, the textbook by Guillemin and Pollack "Differential Topology" is one of the best self-learners for this kind of intuition. $\endgroup$ Mar 14, 2013 at 18:41
  • $\begingroup$ @Ryan: Great! (Didn't expect to get such a concise answer.) $\endgroup$ Mar 14, 2013 at 18:46
6
$\begingroup$

There are exactly two one plane bundles over the circle, the open annulus, and open Moebius band, where we take projection to be onto the central circle. We can now ask what the mod 2 self intersection of the zero section is. In the case of an annulus it is zero, in the case of the Moebius band it is one.

The tubular neighborhood theorem says that any smooth submanifold of a smooth manifold has a neighborhood homeomorphic to a k-plane bundle over the submanifold, where k is it's codimension.

You can't embed a Moebius band in an orientable surface, so the self intersection of any simple closed curve in an orientable surface is zero. The self intersection number is invariant under homotopy, so if gamma and gamma' are homotopic simple closed curves in a surface that intersect each other transversely then they intersect in an even number of points.

$\endgroup$
3
  • 1
    $\begingroup$ It took me a few seconds to realize that "one plane bundle" is the same as a "Line bundle" in your first line :-) $\endgroup$ Mar 14, 2013 at 19:03
  • $\begingroup$ I only begin to learn that and why orientability is crucial for my question. So excuse me for not having mentioned orientability in my original question.) $\endgroup$ Mar 14, 2013 at 19:05
  • $\begingroup$ @Mariano: I grinned over your "few seconds" - for others it takes a few years! $\endgroup$ Mar 14, 2013 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.