# Are all isometries of subsets affine?

I've found at least two questions that deal with whether isometries are affine, Are isometries always linear? and Should isometries be linear?

However, both of these questions assume we are dealing with entire vector spaces. What if, instead, we have two normed vector spaces $$V$$ and $$W$$ over $$\mathbb{R}$$ with $$X\subset V$$ and $$f:X\rightarrow W$$ an isometry (i.e., $$\vert\vert f(x_1)-f(x_2)\vert\vert_W=\vert\vert x_1-x_2\vert\vert_V$$ for all $$x_1,x_2$$ in $$X$$)? Since a general subset $$X$$ isn't a vector space, the question must be made more precise: is $$f$$ the restriction of some affine transformation $$g:V\rightarrow W$$ to $$X$$? By the Mazur-Ulam theorem, this is equivalent to asking whether $$f$$ can always be extended to a surjective isometry on all of $$V$$. If not, what additional conditions have to be met for $$f$$ to be affine in this sense?

My motivation for asking this is that often we say rigid motions must be given by orthogonal transformations. For example, a rigid cube flying through the air can only rotate. But since the time evolution of the cube is only a map from the cube's initial configuration to its present, saying that such motion must be affine because isometries of $$\mathbb{R}^3$$ as a whole are affine doesn't seem to follow - I don't care about $$\mathbb{R}^3$$, I only care about the cube specifically.

• Your motivation is very different from the question you asked. For instance, mapping $(x,y,0)$ to $(x,y)$ is an isometry from a subset of $\mathbb{R}^3$ to $\mathbb{R}^2$, but it can't be extended to an isometry from all of $\mathbb{R}^3$ since it's already subjective. A better question would be to restrict to the same domain and codomain and probably to really just restrict to just $\mathbb{R}^n$ with the standard norm, in which case the result is true. – Charlie Cunningham May 24 at 4:30
• @CharlieCunningham Silly that I missed that example. Would it prehaps suffice to also require that there exist an isomorphism between $V$ and some subspace of $W$ (so that $W$ is "at least as big")? – J_P May 24 at 8:29
• That's not enough. You really need that V is finite dimensional. For instance, consider the map from $\mathbb{\ell}^2$ to $\mathbb{\ell}^2$ given by $(0, a_1, a_2, .... )$ gets sent to $(a_1, a_2, .... )$. Again, this is a surjective isometry, and so can't be extended to a full isometry on the domain, but the domain and codomain are isomorphic. – Charlie Cunningham May 24 at 15:14
• math.stackexchange.com/questions/1687436/… – Charlie Cunningham May 24 at 16:27
• Finite dimensional normed vector spaces over $\mathbb{R}$ are all equivalent to $\mathbb{R}^n$ with the standard norm. – Charlie Cunningham May 24 at 16:29