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The Mazur-Ulam theorem doesn't hold for complex Hilbert spaces because antiunitary operators are origin preserving surjective isometries, but they aren't linear. Is it true, that every $f:H\to H$ surjective isometry of a complex Hilbert space $H$ has the form $$f(x)=f(0)+U(x)$$ where $U$ is unitary (hence linear) or antiunitary (hence antilinear)?

mma
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  • Is not this an instance of Wigner's theorem? – მამუკა ჯიბლაძე Dec 20 '19 at 08:31
  • @მამუკაჯიბლაძე Not, or at least not obviously. Wigner's theorem is about the transition probability keeping transformations of the projectiv space $\mathcal P(H)$ – mma Dec 20 '19 at 08:43
  • But is not any isometry a symmetry transformation? – მამუკა ჯიბლაძე Dec 20 '19 at 09:04
  • No.Symmetry transformations aren't even transformations of the Hilbert space $H$. They act on the "unit ray space", i.e, on the projective space $\mathcal P(H)$. And they preserve the transition probabilities, what isn't a metric. Wigner's theorem is just about these distinctions. – mma Dec 21 '19 at 07:24
  • Maybe one could still use it: an isometry $f$ defines a symmetry transformation $T$ sending a ray to its image under $f-f(0)$; by Wigner's theorem there is either unitary or antiunitary $U$ with $f(x)-f(0)=\lambda(x)U(x)$ for some scalar function $\lambda$. Cannot one determine $\lambda$ using that $f$ is an isometry? Sorry if you already tried something like this and found that it does not work – მამუკა ჯიბლაძე Dec 21 '19 at 11:37
  • Yes, by Wigner, symmetries on $\mathcal P(H)$ are induced by unitary or antiunitary isometries. But perhaps there are more isometries which induce the same symmetry as a given unitary or antiuitary one. – mma Dec 21 '19 at 21:11
  • If they induce the same symmetry transformation they should differ by a scalar function. And since these are isometries, this scalar fuction must, I believe, take values in the circle group. Probably some further restrictions can be imposed on it? – მამუკა ჯიბლაძე Dec 22 '19 at 06:38
  • If they induce the same symmetry transformation they should differ by a scalar function. - Why? – mma Dec 22 '19 at 07:08
  • If two isometries, say $F$ and $F'$, induce the same symmetry transformation $T$ this means that the lines through $F(x)$ and through $F'(x)$ are equal, both being the $T$-image of the line through $x$, no? – მამუკა ჯიბლაძე Dec 22 '19 at 15:10
  • Yes, it seems good. But this doesn't imply that these two isometries differ by a scalar function, because we cannot assume linearity for the isometries. Am I right? – mma Dec 22 '19 at 20:56
  • That $F(x)$ and $F'(x)$ belong to the same line, means $F'(x)=\lambda F(x)$ for some scalar $\lambda$. Maybe we cannot conclude that $\lambda$ depends linearly on $x$, but still it is a scalar function. Moreover if we look at underlying real spaces, I think by Mazur-Ulam it will be $\mathbb R$-linear? – მამუკა ჯიბლაძე Dec 23 '19 at 04:11
  • Oh, yes, sorry, I was tired a bit yesterday evening. So, you've proved that for every origin keeping isometry $F:H\to H$ there is a function $\lambda:H\to\mathbb T$ (where $\mathbb T$ is the circle group) that $F(x)=\lambda(x) U(x)$ for some unitary or antiunitary $U$. And This means that $\langle F(x),F(x)\rangle =\langle\lambda(x)U(x),\lambda(x)U(x)\rangle=\langle U(x),U(x)\rangle$. And $\langle F(x),F(y)\rangle = \langle U(x),U(y)\rangle$ follows from the Parallelogram law! Are we ready? – mma Dec 23 '19 at 07:15
  • Hmm I do not really see the last part. I thought one still needs Mazur-Ulam for the underlying real normed space to complete the argument. – მამუკა ჯიბლაძე Dec 23 '19 at 19:32
  • You are right, my thought about the proof by the parallelogram law was wrong. – mma Dec 25 '19 at 02:18
  • Note that Mazur-Ulam does say that $x\mapsto f(x) - f(0)$ is $\mathbb R$-linear. – Atom Mar 09 '23 at 09:29

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The map $(z, w)\in \mathbf C^2 \mapsto (z,\overline{w})\in \mathbf C^2$ is norm-preserving, but not linear nor antilinear.

Giafazio
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