Are non-bijective isometries necessarily affine functions?

Question

Any isometry on $\mathbb{E}^n$, in the sense of a distance-preserving bijective function, is an affine function (see here, here and here). An affine function is defined in the following way:

If $X$ and $Y$ are affine spaces, then every affine transformation $f\colon X\to Y$ is of the form $ x\mapsto Mx+b$, where $M$ is a linear transformation on $X$ and $b$ is a vector in $Y$.

Now, let's consider functions which preserve distances between $\mathbb{E}^n$ and $\mathbb{E}^m$ (according to Wikipedia, these are still called isometries). My question is: are these non-bijective isometries still affine functions? If they're not, then can you show me a counter-example?

Answer 1

Yes, this is true. For ease, let's write $U=\mathbb{E}^n$ and $V=\mathbb{E}^m$. First let's suppose that $f:U\to V$ is inner product preserving, namely for any $u,v\in U$, $\langle f(u),f(v)\rangle=\langle u,v\rangle$. I claim that $f$ must be linear. Indeed, let $u_1,\dots,u_n$ be an orthonormal basis for $U$. Let's first note that $\langle u_i,u_j\rangle=\langle f(u_i),f(u_j)\rangle$, so certainly $f(u_1),\dots,f(u_n)$ are orthonormal. Notice that for any $c_1,\dots,c_n$, we have $$c_i=\bigg\langle \sum_j c_j u_j,u_i\bigg\rangle=\bigg\langle f\biggl(\sum_j c_j u_j\biggr),f(u_i)\bigg\rangle,$$ so we have $f(\sum_j c_ju_j)=\sum_j c_jf(u_j)+x$ where $x\in V$ is orthogonal to $\text{span}\{f(u_1),\dots,f(u_n)\}$. We now see that $$\sum_j c_j^2=\bigg\langle \sum_j c_j u_j,\sum_j c_ju_j\bigg\rangle=\bigg\langle \sum_j c_jf(u_j)+x,\sum_j c_jf(u_j)+x\bigg\rangle=\sum_j c_j^2+\Vert x\Vert^2,$$ so $x=0$. Hence, we have $f(\sum_j c_j u_j)=\sum_j c_j f(u_j)$. Since $u_1,\dots,u_n$ was a basis for $U$, this tells us that $f$ is a linear map from $U$ to $V$, i.e. $f(x)=Ax$ for some $m\times n$ matrix $A$.

Now, suppose $f:U\to V$ is distance preserving, i.e. $\Vert u-v\Vert=\Vert f(u)-f(v)\Vert$. If $b=f(0)$, I claim that $g(x)=f(x)-b$ is inner product preserving. Indeed, we have $$\Vert f(u)-f(v)\Vert^2=\Vert g(u)-g(v)\Vert^2=\Vert g(u)\Vert^2+\Vert g(v)\Vert^2-2\langle g(u),g(v)\rangle=\Vert f(u)-f(0)\Vert^2+\Vert f(v)-f(0)\Vert^2-2\langle g(u),g(v)\rangle=\Vert u-0\Vert^2+\Vert v-0\Vert^2-2\langle g(u),g(v)\rangle.$$ On the other hand, $$\Vert f(u)-f(v)\Vert^2=\Vert u-v\Vert^2=\Vert u\Vert^2+\Vert v\Vert^2-2\langle u,v\rangle,$$ so we must have $\langle u,v\rangle=\langle g(u),g(v)\rangle$, so $g$ is inner product preserving. Thus, we know that $g(x)=Ax$ where $A$ is an $m\times n$ matrix. We conclude that $f(x)=Ax+b$, which is affine.

Answer 2

We show that a distance preserving $f$ such that $f(0) = 0$ is linear (in the general case consider the function $f'$ given by $f'(a) = f(a) - f(0)$ which satisfies this assumption).

Let $X_1 = \mathbb{E}^n \times \ell^2$, $X_2 = \mathbb{E}^m \times \ell^2$ (products taken as Hilbert spaces). There are isomorphisms $\varphi_i : X_i \to \ell^2$. Define $F : X_1 \to X_2, F(a,u)= (f(a),u)$. Then $F$ is distance preserving and $F(0) = 0$. Use the answer of Distance preserving function on Hilbert space to see that $T = \varphi_2 F \varphi_1^{-1}$ (and hence also F) is linear. This is possible only for a linear $f$.