$\newcommand{\al}{\alpha}$
$\newcommand{\ga}{\gamma}$
$\newcommand{\e}{\epsilon}$


Let $X,Y$ be Riemannian manifolds, such that $\dim(X) > \dim(Y)$.


I am trying to prove the following statement (mentioned by Gromov in his book on metric geometry):


There is no arcwise isometry (i.e length preserving map) from $X$ to $Y$.


However, the naive attempt to prove this hits an obstacle which I do not see how to pass:


Suppose by contradiction $f:X \to Y$ is an arcwise isometry. Then $f$ is $1$-Lipschitz, hence differeniable almost everywhere (by Rademacher's theorem).


**Question: Let $p \in X$ be a point where $f$ is differentiable. Is $df\_p:T\_pX \to T\_{f(p)}Y$ an isometry?**


(This will imply the claim of course).


*Here is what happens when trying to show this naively:*


Let $v \in T\_pX$, and let $\al:[0,1] \to X$ be a smooth path s.t $\al(0)=p,\dot \al(0)=v$.


Then $|\dot \alpha(s)| = |\dot \alpha(0)|+\Delta(s)=|v|+ \Delta(s)$ where $\lim\_{s \to 0} \Delta(s) =\Delta(0)= 0$, thus


$$ (1) \, \, L(\alpha|\_{[0,t]})=\int\_0^t |\dot \alpha(s)| ds=t|v|+\int\_0^t \Delta(s) ds.$$


$\al$ is Lipschitz and $f$ is $1$-Lipschitz, so $\ga:= f \circ \al$ is Lipschitz. By theorem 2.7.6 in the book ``A course in metric geometry'' ( Burago,Burago,Ivanov) it follows that:


$$ (2) \, \, L(\ga|\_{[0,t]})=\int\_0^t \nu\_{\ga}(s) ds, $$


where $\nu\_{\ga}(s):=\lim\_{\e \to 0} \frac{d\left( \ga(s),\ga(s+\e) \right)}{|\e|}$ is the *speed* of $\ga$.


Note that $\nu\_{\ga}(0)= |\dot \ga(0)|=|df\_p(v)|$.


Using the assumption $f$ preserves lengths, we would now like to compare $(1),(2)$ and take derivatives at $t=0$, to get $$|v|=\frac{d}{dt}\left. \right|\_{t=0} L(\al|\_{[0,t]})=\frac{d}{dt}\left. \right|\_{t=0} L(\ga|\_{[0,t]})=|df\_p(v)|.$$


However, it seems that the last equality is **false in general**; Even when the speed of a Lipschitz curve and the derivative of its length exist at a point, [they do not need to be equal](https://math.stackexchange.com/a/2057015/104576). 


It seems that the only thing we can say is that $ \frac{d}{dt}\left. \right|\_{t=0} L(\ga|\_{[0,t]}) \ge \nu\_{\ga}(0) =|df\_p(v)|$, so we are left with $|v| \ge |df\_p(v)|$ which doesn't help.


**Is there a way to "fix" the proof above?**



There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold
Your proof is correct, but you need to add words "amost everywhere" at ane more place.


We use [Rademacher's theorem](http://en.wikipedia.org/wiki/Rademacher%27s_theorem) and lemma about length of curve, which says that if a curve parametrized by length then its velocity is 1 almost everywhere, see 2.7.4 in Metric Geometry by Burgo, Burago and Ivanov.


Fix a chart $U\_{\subset\mathbb{R}^n}\to X$ and a vector $v\in \mathbb{R}^n$.
Let $g$ be the induced metric on $U$.


Note that from the lemma we get that $|d\_pv|=|v|\_g$ almost everywhere.
Repeat the same for $N=\tfrac12\cdot n\cdot(n+1)$ vectors $v\_1,\dots,v\_N$ in general position.
We get $|d\_pv\_i|=|v\_i|\_g$ for almost all $p$ and all $i$.
It follows that for all $w$,the identity $|d\_pw|=|w|\_g$ holds for almost all $p$.
Hence $\dim X \le \dim Y$.


You may want to check 


* Problem "Length-preserving map" [in may collection of problems](http://anton-petrunin.github.io/orthodox)
* [My paper](http://arxiv.org/abs/1003.5621) which discuss in particular length-preserving maps and dimension.


