A map which preserve lengths of paths has distortion $1$?

Question

Let $X,Y$ be connected Riemannian manifolds, and suppose $f:X \to Y$ preserves lengths of paths ($f$ is an arcwise isometry). I don't assume anything about the regularity of $f$.

Is it true that $$ \lim_{y \to x} \frac{d^Y(f(x),f(y))}{d^X(x,y)}=1 \text{ for all } x \in X ?\tag{1}$$

Edit: I am not sure that the limit $(1)$ needs to exist at all even when $f$ is an arcwise isometry. I give an example in an answer below where $$ \liminf_{y \to x} \frac{d^Y(f(x),f(y))}{d^X(x,y)}<1.$$

However, I am not sure what happens if the limit does exist. Is it possible that in that case the limit always equals $1$?

(In high dimensional settings the existence of a limit is "rare"- if $f$ is differentiable at $x$, than probably all the singular values of $df_x$ should be $1$).

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What happens for the special case where $\dim X=\dim Y$?

(I suspect that when $\dim X < \dim Y$ there might be much shorter paths in $Y$, that won't be realized as images of paths in $X$ (via $f$), since in $Y$ there is "more room" than in $X$, but I am not sure how to construct an example).

Answer 1

(1) ${\rm dim}\ X< {\rm dim}\ Y$ : Consider a unit speed curve $$ f: [0,l]\rightarrow \mathbb{E}^2,\ f(0)=p_0,\ f(l)=O,\ p_n=(0,\frac{1}{2^{n} } ),\ O:=(0,0)$$

Consider a unit speed curve $f_0$ in $B(O,|p_0|)$ and out $B(O,|p_1|)$ s.t. it is shortest curve satisfying the following $$ f_0(0)=p_0,\ f_0(l_0)=p_1,\ f_0(t)=(0,-|p_1|)$$ for some $0<t<l_0$ where $l_0$ is length of $f_0$.

Then $$ \pi < {\rm length}\ f_0 <2\pi $$

Define $g_0$ to be line segment $[p_0p_1]$

Also define scalings $$f_n:= \frac{1}{4^n} f_0,\ g_n=\frac{1}{4^n} g_0$$

Finally $$ f=\bigcup_{i=0}^\infty\ f_n\cup g_n $$ where $l$ is sum of lengths of $f_n,\ g_n$.

Hence if $X=f([0,l]),\ Y=\mathbb{E}^2$, then $$ \frac{|O-p_{2n}|}{{\rm length}\ f_n} < \frac{ \frac{1}{4^n} }{ \pi \frac{1}{4^n} } < 1 $$

(2) ${\rm dim}\ X= {\rm dim}\ Y$ : $B$ is closed unit ball in $\mathbb{R}^m$ Assume that $f: B\rightarrow \mathbb{R}^m$ is a path isometry. That is $f$ is short.

If $N_i$ is $\epsilon_i$-net for $B$, assume that $f_i=f$ on $N_i$. Then there is piecewise distance preserving $F_i$ which is extension of $f_i$ (cf. Brehm's theorem).

Hence it is claimed that $f$ is uniform limit of $F_i$ :

For $x$ there is $y\in N_i $ s.t. $|x-y|<\epsilon_i$ So $$|f(x)-F_i(x)| \leq |f(x)-f(y)|+|f(y)-F_i(y)| \leq \epsilon_i $$

Note that $F_i$ satisfies the condition in OP except measure $0$ set so that so does $f$.

Answer 2

First we note that since $f:X \to Y$ is arcwise isometry, it is short ($1$-Lipschitz) hence its distortion (if it exists) is not greater than $1$:

$$ \lim_{y \to x} \frac{d^Y(f(x),f(y))}{d^X(x,y)} \le \lim_{y \to x} \frac{d^X(x,y)}{d^X(x,y)}=1.$$

So, our only chance is to construct an arcwise isometry with distortion smaller than $1$. The idea is to build an "inefficient" map which goes "back and forth":

Let us build a map $f:I \to \mathbb{R}$:

$f$ is a unit speed curve defined as follows:

$f$ starts at $-\frac{1}{2}$, then goes to $\frac{1}{2}$, then to $-\frac{1}{4}$, then to $\frac{1}{4}$ etc... i.e

$$-\frac{1}{2} \to \frac{1}{2} \to -\frac{1}{4} \to \frac{1}{4} \to -\frac{1}{8} \to \frac{1}{8} \to ...$$

So:

$$f(0)=-\frac{1}{2},f(1)=\frac{1}{2},f(t_2)=\frac{1}{2^2},f(t_3)=\frac{1}{2^3},f(t_n)=\frac{1}{2^n},$$

and we get that:

$$t_{n+1}=t_n+(\frac{1}{2^n}+\frac{1}{2^{n+1}})+2\frac{1}{2^{n+1}}=t_n + \frac{5}{2^{n+1}}.$$

Finally, define $t=\lim_{n \to \infty} t_n < \infty$, and $f(t)=0$.

Then

$$ \frac{d^{\mathbb{R}}(f(t_n),f(t))}{d^{\mathbb{R}}(t_n,t)}=\frac{\frac{1}{2^n}}{t-t_n} \le \frac{\frac{1}{2^n}}{t_{n+1}-t_n}=2^{n+1}\frac{\frac{1}{2^n}}{5}=\frac{2}{5}<1.$$