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

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How about translation? –  Qiaochu Yuan Aug 24 '11 at 2:26
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@Q Translations are affine by definition. –  whuber Aug 24 '11 at 4:15
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@whuber : They are (generally) not linear, which is what the question was asking when QY commented. –  Ricky Demer Aug 24 '11 at 5:53
    
@Ricky Thanks. The translation comment makes perfect sense in light of the original version. –  whuber Aug 24 '11 at 6:38
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@Name: This string of comments shows why it's a good idea to mark your edits when you change the content of the question. –  joriki Aug 24 '11 at 9:27
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2 Answers

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.)

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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 9:31
    
@Theo: And thank you too, for linking to the real thing. :) –  Hans Lundmark Aug 24 '11 at 9:42
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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 5:49
    
@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 7:05
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No, it's: as a map on real normed vector spaces, $z\mapsto \overline{z}$ is linear. –  Ricky Demer Aug 24 '11 at 7:10
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