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Does there exist a notion of Jordan curve homotopy?

In particular, suppose we have two Jordan curves $C_0 : S^1 \rightarrow \mathbb{R}^2$ and $C_1 : S^1 \rightarrow \mathbb{R}^2$. When does there exist a continuous function $f: S^1 \times [0,1] \rightarrow \mathbb{R}^2$ such that:

$f(x,0) = C_0(x)$, $f(x,1) = C_1(x)$, and for all $t \in [0,1]$, the function $C_t: S^1 \rightarrow \mathbb{R}^2$ defined by $C_t(x) = f(x,t)$ is a Jordan curve.

My intuition tells me that such a function always exists, but I'm unsure about how to go about proving this. Also, if this is a known result, are there similar results for manifolds other than $\mathbb{R}^2$?

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2  
The orientation of $C_0$ and of $C_1$ should match for such an homotopy to exist, at the very least. –  Mariano Suárez-Alvarez Jul 21 '10 at 2:33
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It always exists if the orientations of C_0 and C_1 match up. First, homotope C_0 so C_1 lies in the precompact region of R^2-C_0 ("blow it up" -- you can achieve this using affine transformations). Using the Schönflies theorem, you can then deduce that C_0 and C_1 bound an embedded annulus in the plane, which you can then use as a guide to homotope C_0 to C_1. –  Andy Putman Jul 21 '10 at 2:41
    
Louis H. Kauffman gives a cute proof in his Knots and physics for smooth curves. –  Mariano Suárez-Alvarez Jul 21 '10 at 2:45

3 Answers 3

up vote 8 down vote accepted

Homotopies through embeddings are usually called isotopies.

There is a subtlety called local flatness that comes up in higher dimensions. Let $E$ be any embedding of $\mathbb R$ in $\mathbb R^3$ such that $E(s)=(s,0,0)$ when $s<-1$ or $s>1$. Define a homotopy $H_t$ with $E_0(s)=(s,0,0)$ for all $s\in \mathbb R$ and $E_1=E$, as follows: $E_t(s)=tE(s/t)$ if $-t\le s\le t$ and otherwise $E_t(s)=(s,0,0)$. This is a homotopy through embeddings, but it (un)ties the knot. This is easily adapted to apply to examples of embeddings of $S^1$ in $\mathbb R^3$, for example.

The way to fix this problem is to only consider embeddings that are locally flat and isotopies that are locally flat. An embedding $E:M\to N$ is (topologically) locally flat if for every point $p\in M$ there exist charts around $p$ and $E(p)$ such that $E$ looks like $(x_1,\dots,x_m)\mapsto (x_1,\dots,x_m,0,\dots,0)$. An isotopy $E_t$ is locally flat if for each point $p\in M$ and time $\tau\in I$ there are charts around $(p,\tau)\in M\times I$ and around $(E_\tau(p),\tau)\in N\times I$, both of them using projection to $I$ as last coordinate, such that locally $(x,t)\mapsto (E_t(x),t)$ looks like $(x_1,\dots,x_m,t)\mapsto (x_1,\dots,x_m,0,\dots,0,t)$. Local flatness is automatic when $m=1$ and $n=2$. The example I gave (with $m=1$ and $n=3$) was such that the isotopy was not locally flat although if the original embedding $E_1$ was locally flat then for every $t$ the embedding $E_t$ was, too.

The Alexander horned sphere ($m=2$, $n=3$, not locally flat) can be smoothed out by such a procedure, too.

Another way of limiting oneself to the right kind of isotopies is to use ambient isotopies: to require $E_t$ to be $H_t\circ E_0$ where $H_t$ is a homeomorphism $N\to N$ depending on $t$. (Local flatness in the case $m=n$ follows from invariance of domain.)

Another way is to limit oneself to smooth embeddings (meaning, as usual, smooth maps that are topological embeddings and that are one to one on the tangent-space level, or equivalently locally flat in the smooth category) and smooth isotopies.

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Thanks Tom! This really helps. –  Kerry Jul 21 '10 at 4:39

At least in dimension 2, results like this are true in great generality. A Jordan curve in the plane is simply an embedded curve. In other surfaces, we have that two simple closed curves $\gamma_1$ and $\gamma_2$ are homotopic if and only if they are regularly homotopic (that is, homotopic through embeddings). This is due to Baer and can be found as Prop 1.7 in Farb and Margalit's book "A primer on mapping class groups" (available here)

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I am not sure if this is right.

Since both curves are Jordan curves, you can "enlarge" $C_{1}$ by affine transformations such that it has no intersection with $C_{2}$ since the diameter of $C_{i}$ is bounded(since $S_{1}$ is compact, its image is also compact, therefore closed and bounded).Then you just need some sort of radial homotopy $C_{rt}=rC_{1}(t)+(1-r)C_{2}(t)$ with $r\in [0,1]$. This will be the homotopy from $C_{1}$ to $C_{2}$ you need.

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But that will in general not result in a Jordan curve for all $r$. –  Mariano Suárez-Alvarez Jul 21 '10 at 3:46
    
I sketched a proof that is somewhat similar to this in the comments above. A radial homotopy will not work, however. –  Andy Putman Jul 21 '10 at 3:52
    
I don't know how to prove it be homeomorphic to $S_{1}$, but it should not have self-intersections. Yes I can't prove it must be a Jordan curve. –  Kerry Jul 21 '10 at 3:53
    
The intermediate steps of a straight-line homotopy will often have self-intersections. Draw some mildly complicated examples and you will see... –  Andy Putman Jul 21 '10 at 3:56
    
Thanks! I looked at the picture in wikipedia, the mistake is obvious. –  Kerry Jul 21 '10 at 3:58

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