I've got a question about the definition of plane curves as equivalence classes under the relation of reparametrization.
It is said that whenever two plane curves have different trace, i.e. their graphical representation on the plane differs, the curves belong to distinct equivalence classes.
The converse is not true, however. The two curves and defined by are different in the above sense, in spite of the fact that they have the same trace. This is odd, since I can reparametrize one to the other by means of the map .
Is it just because the cirve is closed and periodic? How do we actually unambiguously distinguish closed curves/periodic curves?
Nov 14th 2011, 05:49 AM
I suspect what is meant by "plane curves as equivalence classes under the relation of reparameterization" is that all the curves are assumed to have domain [0,1], and reparameterizing is defined as precomposing with a homeomorphism of [0,1].
Nov 14th 2011, 06:15 AM
Aha...I mean we have defined a reparametrization map to be a diffeomorphism of intervals, but without the requirement that the intervals be the same. Althought it is certainly reasonable to require that for periodic curves, since as in the example above the curves have different lenghts, which is not what we actually wish if we consider two similar curves.
OK, so this is done in order to incorporate the case of periodic curves, isn't it?
Nov 14th 2011, 09:53 AM
Once you have decided to "mod out by" diffeomorphisms of the domain, it's no longer meaningful to ask whether the intervals are the same, since any closed interval is diffeomorphic to any other. For convenience you may as well take them all to be [0,1]. The reason why the two you mentioned are not equivalent is that one goes once around a circle, while the other goes twice around. One is injective (okay, except at the endpoints) and the other is not. No diffeomorphism of the domain will change one into the other.
Nov 14th 2011, 10:18 AM
one way to distinguish between two closed paths is to calculate their winding numbers.
provided there is a point on the region of the plane that is bounded by the curve that the curve does not pass through, we can remove that point from the plane.
is homotopy-equivalent to , so we can calculate the homotopy class of the deformation of c(t) to a path on ,
which will give us an integer (since ).
in your example, the winding number of is 1, while the winding number of is 2, so they are not homotopic.
Nov 14th 2011, 11:55 AM
I don't know if this is quite what you're looking for, but it would answer the question for me if I were you. Suppose you have some trace of a curve in the plane and you wish to find out which of two periodic curves is the least "redundant". Of course, this amounts to asking what the smallest interval you can really take is. An interesting fact that lets you know this is always answerable (in the sense that there really is a smallest period) is that the set of periods of some smooth curve is, not surprisingly, a subgroup of . In fact, it's not hard to prove that is a discrete subgroup of (i.e. a topological group whose inherited topology from is discrete). Then, the basic theory of discrete subgroups allows us to conclude that there exists some such that . Clearly then the curve you really want is the one whose domain is on an interval of length --the fundamental period. Thus, in your example your curve has fundamental period and moreover any other curve will have period for some . This allows you to differentiate between closed curves based on what multiplicity of the fundamental period they have.
More information (including a self-contained proof of some of the discrete subgroup facts) can be found here on my blog.