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.

  • $\begingroup$ Do you have an example of a homeomorphism that does not preserve rectifiability? I suspect they don't exist. Rectifiability amounts to all the coordinate functions having bounded variation, no? If you restrict a homeomorphism to a compact set you get a uniformly continuous function so I think that means it has to send rectifiable curves to rectifiable curves. $\endgroup$ – Ryan Budney Dec 22 '10 at 8:07
  • $\begingroup$ @Ryan So you're saying, for example, that no homeomorphism of the plane will take the unit interval to a segment of the Koch snowflake built on the interval. Can't one build such a homeomorphism iteratively? Move points up to form the central triangle; 4 smaller moves make 4 smaller triangles, etc. The map stabilizes after finitely many steps for points off the interval, so continuity and invertibility there. Some more work (both pinning down the construction and analyzing it) for continuity and injectivity on the interval itself. $\endgroup$ – David Feldman Dec 22 '10 at 10:22
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    $\begingroup$ @Ryan Or take your favorite continuous function $f$ on $[0,1]$ with unbounded variation. Now look at $(x,y)\mapsto (x,y+f(x))$. $\endgroup$ – David Feldman Dec 22 '10 at 10:24
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    $\begingroup$ Sullivan proved that every closed topological manifold of dimension other than $4$ has a unique Lipschitz structure in which rectifiability makes sense. $\endgroup$ – Igor Belegradek Dec 22 '10 at 13:03
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    $\begingroup$ Okay, so your example answers one of your questions David -- a square $[0,1]\times[0,1]$ admits at least two rectifiability structures, it's standard structure and the structure pulled-back via your map $f$. $\endgroup$ – Ryan Budney Dec 22 '10 at 15:59

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.

  • $\begingroup$ I disagree. As illustrated by the 1-dimensional example $x\mapsto x^3$, the Lipschitz pseudogroup is strictly smaller than the rectifiable pseudogroup (transformations that send rectifiable curves to rectifiable curves). However, one can rightfully argue that the Lipschitz pseudogroup is more interesting than the rectifiable pseudogroup. $\endgroup$ – André Henriques Dec 23 '10 at 12:42
  • $\begingroup$ Andre: I think the claim is that any rectifiable manifold admits a unique compatible Lipschitz structure. This is similar to to saying that any 2-manifold has a unique smooth structure, even though homeomorphisms need not be smooth. $\endgroup$ – Igor Belegradek Dec 23 '10 at 17:49
  • $\begingroup$ Andre: the map $x\mapsto x^3$ is locally Lipschitz. Can you clarify what exactly it shows? $\endgroup$ – Sergei Ivanov Dec 23 '10 at 18:42

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...)


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