Isometry between vector spaces with different dimensions

Question

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.

Answer 1

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.

Answer 2

I wanted to mention two things. The first is that you do not need linearity for an isometry to be injective -- the link is misleading in that if $u,v \in \mathbb R^n$ and $T$ is the isometry, then $$ \|T(u)-T(v)\|=\|u-v\| $$ automatically, because $T$ preserves distance. (To state it more plainly, an isometry preserves distance and if two points are not the same, the distance between them is positive. Since their images are that same distance apart, they cannot be equal!)

The second point however is that, assuming that the distance function on both spaces $\mathbb R^n$ and $\mathbb R^m$ is the Euclidean one, if $T\colon \mathbb R^n \to \mathbb R^m$ is an isometry, then $T(x) =\alpha(x)+b$ where $\alpha$ is linear and $b\in \mathbb R^m$ -- that is, any isometry is a composition of a linear map and a translation, so that $T$ is "affine-linear". This of course implies that $m\geq n$, since $\alpha$ must be injective.

Since we can always compose an isometry $T$ with a translation so that the resulting isometry sends the zero vector $0_n \in \mathbb R^n$ to the zero vector $0_m\in \mathbb R^m$, the claim is equivalent to the following:

Proposition If $S\colon \mathbb R^n\to \mathbb R^m$ is an isometry such that $S(0_n)=0_m$, then $S$ is a linear map (which is then necessarily orthogonal).

Proof:

If $S$ is an isometry such that $S(0_n)=0_m$ then we have $\|S(v)\| = \|S(v)-S(0_n)\| = \|v-0_n\|=\|v\|$, that is, $S$ preserves the Euclidean norm. Moreover $$ \begin{split} \langle S(x),S(y)\rangle &= \frac{1}{2}\left(\|S(x)\|^2+\|S(y)\|^2-\|S(x)-S(y)\|^2\right)\\ &= \frac{1}{2}\left(\|x\|^2+\|y\|^2-\|x-y\|^2\right)\\ &= \langle x,y\rangle, \end{split} $$

Now this compatibility of $S$ with the inner products implies that if $v_1,v_2,v_3 \in \mathbb R^3$, and $\lambda \in \mathbb R$, then $$ \begin{split} \langle S(v_1+\lambda.v_2),S(v_3)\rangle &= \langle v_1+\lambda.v_2,v_3\rangle \\ &= \langle v_1,v_3\rangle + \lambda\langle v_2,v_3\rangle \\ &=\langle S(v_1),S(v_3)\rangle +\lambda \langle S(v_2),S(v_3)\rangle \\ &= \langle S(v_1)+\lambda.S(v_2),S(v_3)\rangle \end{split} $$

Thus the map $L_u \colon \mathbb R^n \to \mathbb R$ given by $L_u(v)= \langle S(v),S(u)\rangle$ is linear. But then if $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb R^n$, $\{S(e_1),\ldots,S(e_n)\}$ is an orthonormal set in $\mathbb R^m$ and $$ \tilde{S}(v)=\sum_{i=1}^n \langle S(v),S(e_i)\rangle S(e_i) = \sum_{i=1}^n L_{e_i}(v)S(e_i) $$ is the orthogonal projection of $S(v)$ onto $\text{span}\{S(e_1),\ldots,S(e_n)\}$ (by the first equality) and is a linear function of $v$ (by the second equality). But $$ \tilde{S}(v)\|^2 = \sum_{i=1}^n\langle S(v),S(e_i)\rangle^2 = \sum_{i=1}^n\langle v,e_i\rangle^2 = \|v\|^2 = \|S(v)\|^2, $$ hence $S(v) =\tilde{S}(v)$ and hence $S$ is linear as required.