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Is there a simple example of an isometry between normed vector spaces that is not an affine map?

Name
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2 Answers2

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The note by Jussi Väisälä linked to by the Wikipedia article about the Mazur–Ulam theorem contains the following example:

An isometry need not be affine. To see this, let $E$ be the real line $\mathbf{R}$, let $F$ be the plane with the norm $\lVert x \rVert = \max(|x_1|, |x_2|)$, and let $\phi: R \to R$ be any function such that $|\phi(s)-\phi(t)| \le |s-t|$ for all $s, t \in\mathbf{R}$, for example, $\phi(t) = |t|$ or $\phi(t) = \sin t$. Setting $f(s) = (s, \phi(s))$ we get an isometry $f : E \to F$, which is usually not affine.

(But of course this is not a bijection.)

Hans Lundmark
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    +1 for linking to Väisälä's article. For those with the relevant university subscription or near a library may want to look at the published version in The American Mathematical Monthly Vol. 110, No. 7 (Aug. - Sep., 2003), pp. 633-635. – t.b. Aug 24 '11 at 09:31
  • @Theo: And thank you too, for linking to the real thing. :) – Hans Lundmark Aug 24 '11 at 09:42
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Yes.


Let $\mathbb{C}$ be a vector space over itself with absolute value as its norm.

Define $ \; \; f : \mathbb{C} \to \mathbb{C} \; \; $ by $ \; \; f(z) = \overline{z} \; \; $ .

$f$ is a non-linear (bijective) isometry that satisfies $\; f(0) = 0 \;$ .


See the Mazur–Ulam theorem.

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    This feels a little like cheating, since the corresponding transformation on $\mathbb{R}^2$ is linear. ;-) – Hans Lundmark Aug 24 '11 at 05:49
  • @Hans: I agree. The question was posed on normed vector spaces, and as a map on normed vector spaces, $z\mapsto\bar{z}$ is linear. – robjohn Aug 24 '11 at 07:05
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    No, it's: as a map on real normed vector spaces, $z\mapsto \overline{z}$ is linear. –  Aug 24 '11 at 07:10
  • @RickyDemer Could you explain how it is non-linear? –  Feb 12 '18 at 18:06
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    @VivianL. ​ ​ ​ $f(i\hspace{-0.02 in}\cdot \hspace{-0.04 in}1) : = : f(i) : = : \overline{i} : = : -i : \neq : i : = : i\cdot 1 : = : i\cdot \overline{1} : = : i\cdot f(1) \hspace{1.54 in}$ –  Feb 12 '18 at 18:57
  • @RickyDemer When we choose a scalar, we have to make sure it only scales rather than rotate a vector. How can you choose $i$ as a scalar? It probably will rotate a vector. Sorry, I am new to linear algebra. –  Feb 12 '18 at 19:27
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    @VivianL. ​ I can choose $i$ as a scalar because I specified $\mathbb{C}$ as a vector space $\hspace{1.43 in}$ over itself (rather than over $\mathbb{R}$). ​ ​ ​ ​ –  Feb 12 '18 at 20:16