Can we always parametrize simple closed curve with a rectifiable curve?

Question

Let $\gamma:[0,1]\rightarrow \mathbb{C}$ be a simple closed curve. ( i.e. $\gamma$ is injective on $[0,1)$ and $\gamma(0)=\gamma(1)$).

Then, does there exist a closed rectifiable curve $\alpha:[0,1]\rightarrow \mathbb{C}$ such that $\alpha([0,1])=\gamma([0,1])$?

Answer 1

Nice question. But if $\gamma$ is itself nonrectifiable then $\alpha$ must also be nonrectifiable. I like the question because the proof uses lots of different bits and pieces of first year topology.

First let me reduce the general case to a special case:

• We may assume that $\alpha(0)=\alpha(1)=\gamma(0)=\gamma(1)$, that $\alpha(t) \ne \alpha(0)$ for $0<t<1$, and that the range restriction $\alpha : [0,1] \to \text{Image}(\gamma) = \gamma([0,1])$ is path homotopic to $\gamma$, meaning very roughly that $\alpha$ goes around the circle $\text{Image}(\gamma)$ exactly one time and in the same direction as $\gamma$ itself.

The proof of this reduction involves using a universal covering map $p : \mathbb{R} \to \text{Image}(\gamma)$ of the circle $\text{Image}(\gamma)$, with deck transformation group $\mathbb{Z}$ acting on $\mathbb{R}$ by translations. By covering space theory, the map $\alpha$ lifts to a map $$\tilde\alpha : [0,1] \to \mathbb{R} $$ such that $|\tilde\alpha(1)-\tilde\alpha(0)|$ is an integer. Since $\alpha$ surjects onto $\text{Image}(\gamma)$, it follows that the compact, connected set $\text{Image}(\tilde\alpha)$ must contain an interval of length $1$, having the form $T,T+1$ for some $T$. After possibly truncating the domain of $\alpha$ and reversing its orientation, we may assume that $\tilde\alpha(0)=T$, $\tilde\alpha(1)=T+1$, and $\tilde\alpha(t) \in (T,T+1)$ for $0<t<1$. This truncation/reversal is allowed, because if the result is nonrectifiable then the original $\alpha$ is nonrectifiable. After cutting and pasting $\gamma$, we may assume that $T=0$, completing the reduction.

Now let's assume $\gamma$ is non rectifiable. Choose a real number $L>0$. Next, choose a partition $0=s_0 < s_1 < … < s_K = 1$ such that $$\sum_{k=1}^K \left| \gamma(s_k) - \gamma(s_{k-1}) \right| \ge L $$ Define a partition $0=t_0 < t_1 < … < t_K = 1$ as follows: $t_k$ is the least number in the interval $(t_{k-1},1]$ such that $\alpha(t_k) = \gamma(s_k)$ (the existence of this partition is where we use the special case). It follows that $$\sum_{k=1}^K \left| \alpha(t_k) - \alpha(s_{k-1}) \right| \ge L $$ Therefore, $\alpha$ is not rectifiable.