# Do isometries send lines to lines?

My question is if any isometry $f:V\to W$ between real normed spaces sends lines to lines. I've seen several questions/answers about this but only in euclidean spaces.

So I thought it was false on general (real) normed spaces. However I found this theorem of Mazur-Ulam: any surjective isometry $f:V\to W$ is an affine map, hence it maps lines to lines.

But if my isometry is not surjective, would this still apply? I think that considering the image space $f(V)$ it would be the same, because $f:V\to f(V)$ is affine and any line $L$ would be sent to a line $f(L)$ in $f(V)$ which is also a line in $W$.

Is this correct?

Thank you.

## 1 Answer

No, this is not correct, because $f(V)$ doesn't have to be a vector space.

Consider, for instance, the norm $\bigl\|(x,y)\bigr\|=\max\{|x|,|y|\}$ in $\mathbb{R}^2$ and the usual norm in $\mathbb R$. Now, consider the map$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb{R}^2\\&x&\mapsto&\bigl(x,\sin(x)\bigr).\end{array}$$Then $f$ is an isometry, but it is not affine and it does not send lines into lines.

Of course, the only important property of the sine function here is that $(\forall x\in\mathbb{R}):|\sin x|\leqslant|x|$. I could have even used a function which is discontinuous everywhere, such has $\chi_{\mathbb Q}$.

• Does this send it into lines of some non-Euclidean surface or is it truly not sending lines into any other form of lines? – The Great Duck Jul 22 '17 at 23:06
• The codomain is $\mathbb{R}^2$. How do you want to find a surface there? Besides, the only line in $\mathbb R$ is $\mathbb R$ itself and its image is the graph of the sine function. – José Carlos Santos Jul 23 '17 at 7:26
• Oh yeah, good point. I was more-or-less addressing the other potential interpretation of their question which would be that isometries map lines to geodesics of other surfaces. It wasn't clear whether or not they were referring to Euclidean lines or not. Thank you for the edit though. That definitely shows that the result isn't required to be a curve. – The Great Duck Jul 24 '17 at 1:29