Manifolds with rectifiable curves

Question

To begin with, observe that the notion of a rectifiable curve makes sense in, say, a smooth or a PL-manifold but not in merely a topological manifold. Indeed if $f:[0,1]\rightarrow U\subset{\Bbb R}^n$, $U$ open, represents a rectifiable curve in $U$, and $g:U\rightarrow V$ a mere homeomorphism, we would generally have no reason to expect a rectifiable curve from the composition $g\circ f$.

Certain homomorphisms (e.g. smooth, PL) do preserve rectifiable curves; the totality of these forms a pseudogroup and we may use this pseudogroup to equip manifolds with just enough geometric structure to distinguish a class of curves as rectifiable.

Do manifolds with just this much structure (I'll call it a rectifiability structure till I know better) occur in the literature? And the usual questions: do all manifolds support a rectifiability structure? if one exists, is it unique up to homeomorphism?

I'd be interested in similar questions relative to higher dimensional objects - manifolds where embedded $n$-disk "have" an $n$-volume ("have" in scare quotes because I only mean relative to each patch of any atlas).

Just a final remark: biLipschitz homeomorphisms preserve rectifiability (right?), but not all rectifiability preserving homeomorphisms are biLipschitz (e.g. $x \mapsto x^{1/3}$), so if I've reinvented the wheel, I haven't reinvented that wheel.

Answer 1

This is just Lipschitz structure: only locally Lipschitz maps preserve rectifiability of all curves.

What is wrong with the $x\mapsto x^{1/3}$ map is explained in Tapio Rajala's answer. (A more explicit example is the path $t\mapsto t^2\sin(1/t)$, $t\in[0,1]$, and you can make a non self-overlapping example if you go to dimension 2). Here are the missing details of Tapio's answer in the general case.

If a homeomorphism $f:U\to U$ is not Lipschitz on some compact set, then there exist sequences $x_n$ and $y_n$ converging to some $p\in U$ rapidly (e.g. such that $|x_n-p|<2^{-n}$ and $|y_n-p|<2^{-n}$) and such that $|f(x_n)-f(y_n)|>n^2|x_n-y_n|$. Consider a curve that oscillates between $x_n$ and $y_n$ approximately $(n^2|x_n-y_n|)^{-1}$ times, then goes to $x_{n+1}$ and oscillates between $x_{n+1}$ and $a_{n+1}$, and so on. It is rectifiable (its length is bounded above by something like $2\sum_n (n^{-2}+2^{-n})$) but its $f$-image is not: each oscillating part adds at least 1 to the length.

Answer 2

Because this is my first post here, I could not post parts of this as comments which I would have preferred. These are mainly (trivial) observations rather than an answer.

As David Feldman noted, the notion of a rectifiable curve needs a metric. To see this, consider a manifold $M$ with a metric $d$ that has a rectifiability structure. Then the same topological manifold with a snowflaked metric $d^t$, $0 < t < 1$, does not have a rectifiability structure.

The example $$ f \colon \mathbb{R} \to \mathbb{R} \colon x \mapsto x^{1/3} $$ is a bit misleading. Suppose you are not using the Hausdorff measure $\mathcal{H}^1$ to measure the length of your curve but instead measure the overlapping parts multiple times. If you define a curve $\gamma \colon [0,1] \to \mathbb{R}$ with $\gamma(x) = k^{-2}d(2^{k+2}x,3)$, when $2^{-(k+1)} < x \le 2^{-k}$ and $k \in \mathbb{N}$, then $$l(\gamma) = \sum_{k=0}^\infty \frac{2}{k^{2}} < \infty$$ and $$l(f(\gamma)) = \sum_{k=0}^\infty \frac{2}{k^{2/3}} = \infty.$$ (If you instead use $\mathcal{H}^1$ to measure the length of a curve, the $f$ indeed preserves the rectifiability structure.)

With the above construction of $\gamma$ in mind consider a non-Lipschitz homeomorphism $f \colon \Omega \to \Omega'$, where $\Omega, \Omega' \subset \mathbb{R}^n$, $n \ge 2$, are bounded open sets. Then for all $L \in \mathbb{N}$ you can find points $x_L,y_L \in \Omega$ so that $d(f(x_L),f(y_l)) \ge Ld(x,y)$. Now around any accumulation point of $\{x_L\}$ you should be able to construct a curve $\gamma$ in a similar fashion as above so that $\mathcal{H}^1(\gamma)< \infty$ and $\mathcal{H}^1(f(\gamma))= \infty$. (I did not check the details of this...)