Is it possible to have an isometry between two spaces of different dimension? Suppose we have $f : \mathbb{R}^d \rightarrow \mathbb{R}$ with $d \neq 1$ such that $|f(x) – f(y)| = \|x -y\|_2$ for all $x,y \in \mathbb{R}^d$. Does a mapping like this exist and if yes, what are the restrictions (if any) induced by such definition?
My initial guess was probably not as isometries are injective and injectivity between two finite-dimensional vector spaces $V,W$ is granted iff $\dim(V)=\dim(W)$. But this only holds if we assume $f$ to be linear.
Best Answer
Consider the basis vectors $u_1, \dots, u_d$ of $\mathbb R^d$. Then, $d(0, u_j) = d(0, -u_j) = 1$. Since $d \neq 1$, there are at least two distinct $j$, thus $4$ distinct vectors $v_i$ with $d(0,v_i) = 1$. Let $f$ be an isometry from $\mathbb R^d$ to $\mathbb R$, then $d(f(0),f(v_i)) = 1$. But $f$ is injective, thus there have to be 4 distinct points $x_i \in \mathbb R$ with $d(f(0),x_i)=|f(0) - x_i|=1$, which is clearly not the case.
This does not easily generalize to the nonexistence of isometries between $\mathbb R^d$ and $\mathbb R^n$, $d \neq n$ though, since it depends on the fact that there are only finitely many real numbers of a certain norm.