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

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.

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?**

metricisometries, take for instance $\alpha(t)=(\cos t,\sin t)$. Also, even when they are injective isometries, they are not always surjective ($(x \to (x,0)$). $\endgroup$ – Asaf Shachar Dec 14 '16 at 16:12