If the pseudometrics inherited by two smooth curves are identical, must the curves be isometric?

Question

Let $\gamma_1,\gamma_2:\mathbb{R}\rightarrow\mathbb{R}^n$ ($n\geq 1$) be two smooth curves such that for every $\,t_2,t_1\in\mathbb{R}$ we have $|\gamma_1(t_2)-\gamma_1(t_1)|=|\gamma_2(t_2)-\gamma_2(t_1)|$. In otherwords, the pseudometrics on $\mathbb{R}$ given by: $(t_1,t_2)\mapsto |\gamma_1(t_2)-\gamma_1(t_1)|$ and $(t_1,t_2)\mapsto |\gamma_2(t_2)-\gamma_2(t_1)|$ are identical.

Question: Must there exist an isometry $f:\mathbb{R^n}\rightarrow \mathbb{R}^n$ such that $\gamma_2=f\circ\gamma_1$ ?

Thank you

Answer 1

Yes.

Sketch of proof:

Without loss of generality, we can assume that the affine hull of $\gamma_1(\mathbb{R})$ has dimension $n$, and that the affine hull of $\gamma_2(\mathbb{R})$ has dimension $\leq n$.

Choose $t_0, \ldots, t_n$ such that $$ \{\gamma_1(t_j) - \gamma_1(t_0)\}_{j = 1, \ldots, n} $$ are linearly independent. Let $A$ be the (unique) affine transformation such that $A(\gamma_1(t_j)) = \gamma_2(t_j)$ for every $j = 0, \ldots, n$. Since it is an affine transformation that fixes the pairwise distance between $n+1$ points, it is an isometry.

Now, observe that the system of equations $$ |x - \gamma_1(t_j)| = a_j , \qquad j\in 0, \ldots, n $$ has at most $1$ solution (this is the simultaneous intersection of $n+1$ spheres) this shows that $A$ must also in fact carry $\gamma_1(t)$ to $\gamma_2(t)$ for every $t$.