Intersection of Three Closed Curves

An exploration of topological equivalence in curve triples


Given three closed curves intersecting transversely with no triple points, we say the triple of curves is equivalent to another such triple if it is possible to transform one into the other without creating any triple points during the transformation. The question is, when are two curve triples equivalent?

First, define an interior intersection (II) of a triple as follows: An intersection of two of the curves in the triple is an II if it is inside the third curve. Note that the number of II's in a triple is the sum of the II's for each pair of curves.

1) Two triples are equivalent only if the number of II's in the two triples is equivalent mod 2.

This statement is fairly simple to prove. Suppose we have 3 curves-- A, B, and C-- with a total of x II's. We start to transform the cuves by pulling and stretching them, gaining or losing intersections in the process. We have a single legal move (and its reverse):

Curve C can never pass between the intersections of A and B, or we would create a triple point (shown in yellow) in the process of the move:

So suppose we are pulling A and B apart (or pushing them together) using a legal move. If the move takes place entirely inside C, we have lost (gained) 2 II's. If it takes place entirely outside C, we have no cnage in the number of II's. Now suppose that one of the intersections of A and B is inside C and the other outside, so that we would lose (gain) exactly one II by completing the move. This means that curve C must pass between the two intersections an odd number of times, because one must pass through a curve an odd number of times to move from inside to outside or vice versa. In particular, C must pass between the intersections of A and B at least once, and therefore, the move is illegal.

Therefore when performing a legal move, both intersections are of the same type (II or not II), and in either case, the number of II's gained (lost) is equivalent to 0 mod 2. So it is impossible to transform a triple with and even number of II's to one with and odd number, or vice versa.

2) Two triples may not be equivalent even if they have the same number of II's mod 2.

As shown above, we can't move A relative to B if C crosses between the intersections of A and B. In particular, we cannot perform the following move, because it creates two triple points:

Of course, we could pull the above configuration apart by moving C first:

But consider the following triple, in which only the II's are shown in black:

We see that each pair of curves has two intersections interior to the third curve, and that the third curve passes twice between those intersections regardless of which way around the two curves we look. In other words, we have a situation similar to the diagram, except that it is impossible to choose a C which we can pull apart first. So there is no way to reduce the number of II's of this triple. Therefore it is not possible to transform it into, for example, the triple below, even though they both have an even number of II's.

In fact, there is an entire family of inequivalent triples like the one above:

It is impossible to reduce the number of II's in any of these triples. Each triple has 3 more II's than the preceding triple, one for each pair of colors.

So, are all triples equivalent to one of the triples in this set?

No. Here is another irreducible triple, but this one has eight interior intersections, whereas the others all had multiples of three:

Work in progress. To be continued...


This page was inspired by a remark about curve triples made by Martin Gardner on p. 52 of The Universe in a Handkerchief, New York: Copernicus, 1996.

If you found this page interesting, you might also enjoy my exploration of knots and surfaces.


Sharon Goldwater

Last modified: Tue Mar 18 1997