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 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
- My paper which discuss in particular length-preserving maps and dimension.
I am completenig some details based on Anton's answer:
We prove the following theorem:
Let $X,Y$ be Riemannian manifolds, and let $f:X \to Y$ be a length preserving map. Then $df$ is an isometry almost everywhere.
The proof is composed of 3 steps:
Step I:
It is enough to prove the claim where $X=(\mathbb{R}^n,g)$ where $g$ is an arbitrary metric.
Step II:
Assume step I, i.e we consider $f:(\mathbb{R}^n,g) \to Y$. Let $v \in \mathbb{R}^n$ fixed . Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$.
Step III:
Conclude that $df$ is an isometry almost everywhere.
We begin with proofs of steps I,III, since these are easy.
Proof of step I:
$f$ is $1$-Lipschitz, hence by Rademacher's Thm it is differentiable a.e. Since any manifold admits a countable cover of charts, we finished.
Proof of step III:
$f:(\mathbb{R}^n,g) \to Y$.
Let $v_1,...,v_n$ be a basis for $\mathbb{R}^n$. By applying step II for every $v \in Q=\{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, we get that $|df_x(v)|_{f(x)}=|v|_x$ for all $v \in Q$ and for almost every $x \in \mathbb{R}^n$. Now it is a simple linear algebra fact that at all these points $x$, $df_x$ is an isometry. Indeed,
$$ \langle df_x(v_i),df_x(v_j) \rangle = \frac{1}{2}(|df_x(v_i)+df_x(v_j)|^2 - |df_x(v_i)|^2 - |df_x(v_j)|^2) = \frac{1}{2}(|df_x(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2) $$ $$ = \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle.$$
Comment: Here we see why restricting to a chart was so helpful! We can use the "same" basis vectors $v_i$ for different points $x$ (on a general non-parallelizable manifold, this is meaningless).
Proof of step II:
Fix $v \in \mathbb{R}^n$. Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$.
Proof:
Fix $x \in \mathbb{R}^n$, and define $\alpha_x(t)=x+tv$. Putting $\gamma_x=f \circ \alpha_x$ (this is a Lipschitz path), we get by theorem 2.7.6 that $$ (1): \, \, \nu_{\alpha_x}(t) =\frac{d}{dt}\left. \right|_{t=0} L(\alpha_x|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga_x|_{[0,t]})=\nu_{\ga_x}(t) \, \, \text{ for almost every $t$}$$
For any path $\beta$ in a Riemannian manifold, at every point of differentiability, it holds that $\nu_{\beta}(t)=|\dot \beta(t)|_{\beta(t)}$.
Specializing this for $\beta = \alpha_x$ (and combining $(1)$) we obtain
$$ (1)': \, \, |v|_{\alpha_x(t)}=\nu_{\ga_x}(t) \, \, \text{ for almost every $t$}$$
Specializing this for $\beta = \gamma_x$ leads to
observation I:
$\nu_{\gamma_x}(t)=|df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}$ for those $t$ where $f$ is differentiable at $\alpha_x(t)$.
We define the sets
$ B:=\{(x,t) \in \mathbb{R}^n \times \mathbb{R} | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \, \text{ is differentiable at } \, \alpha_x(t) \} \subseteq \mathbb{R}^{n+1}, $
$B_x=\{ t \in \mathbb{R} | \, \, (x,t) \in B \} \subseteq \mathbb{R}$ and
$B^t= \{x\in\Bbb R^n: (x,t)\in B\} \subseteq \mathbb{R}^n$.
We also define $h:\mathbb{R}^{n+1} \to \mathbb{R}^n$ by $h(x,t)=\alpha_x(t)$.
$h(B)=\{\alpha_x(t) | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \, \text{ is differentiable at } \, \alpha_x(t) \}=\{\alpha_x(t) | \, \, |df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}=|v|_{\alpha_x(t)} \}$
where the last equality follows from observation I.
Hence, it is enough to show $h(B)^c$ has measure zero in $\mathbb{R}^n$.
This essentially follows from Tonelli's theorem:
Since $h$ is surjective, $h(B)^c \subseteq h(B^c)$. Thus, it is enough to show $\mu(B^c)=0$ in $\mathbb{R}^{n+1}$.
$$ B^c=\{ (x,t) | \, \, \neg ( \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} ) \} \cup \{ (x,t) | \, \, f \, \text{ is not differentiable at } \, \alpha_x(t) \}:=A \cup D$$
$(1)'$ implies that $\lambda_1(A_x)=0$ for every $x \in \mathbb{R}^n$, hence (by Tonelli) $\mu(A)=0$.
$D^t = \{ x \in \mathbb{R}^n | f \, \text{ is not differentiable at } \, x+tv \} = W -tv$ (where $W \subseteq \mathbb{R}^n$ is the set of points of non-differentiability of $f$), so $\lambda_n(D^t)=\lambda_n(W)=0$
Again, Tonelli implies $\mu(D)=0$.