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Let us develop the question:

  1. Let us focus on finite real vector space, equiped with a norm. A priori, one does not make the hypothesis that the norm is derived from a scalar product.

  2. Which structural condition could be asked on such space that is equivalent to say this space is euclidean?

  3. Example: in such normed space, given two vectors u and v of same length, there is a global isometry T, such that u=T(v). By global isometry T, it is meant a bijection that conserves the norm/length.

Regarding 3 (the direction I am currenty exploring):

  • It would be interesting to get a complete demonstration for that or a counter example.

  • The Mazur–Ulam theorem, applied to such T that maps zero vector to itself, implies that T is linear. This is a start.

  • Even such set of T transformations is a group (easy to check). How to move forward?

  • Some hint: since the pair of equal length (u,v) can be as closed as possible ( norm(u-v) < epsilon ), such group of T transformations should be continous. But how to prove it?
    Then, how to use such constraint to move forward?

  • Another hint: target the conclusive step that the polarization identity is necessarly implied.

Lucas
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    Maybe it is a bit tautological but if a norm comes from a scalar product you can reconstruct the scalar product via $s(v,w)=(||v+w||^2-||v||^2-||w||^2)/2$. So one could require that this expression is bilinear.

    Now I wonder whether 3. is sufficient. I don't think so but couldn't find a counterexample.

     – HenrikRüping Sep 03 '12 at 20:52
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    A norm does not necessarly derive from a scalar product. If the norm satisfies the polarization identity then a scalar product can be defined as your mentionned. But such "polarization identity" is not what is meant by structural condition. Item 3 is the kind of condition that is looked for. – Lucas Sep 03 '12 at 21:08
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    Perhaps one of those who voted to close would care to elaborate? – Gerry Myerson Sep 03 '12 at 22:37
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    I voted to close. The question has been edited several times since it was closed. Originally, it was not clear what, exactly, the question was. If there are experiences MO users who think the question should now be reopened I will be happy to add my vote to theirs. – Kevin Walker Sep 03 '12 at 22:57
  • oops -- "experiences" should be "experienced" above. – Kevin Walker Sep 03 '12 at 22:57
  • the question now seems (just about) reasonable in its present form, but I suspect it may already be answered by http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean – Yemon Choi Sep 03 '12 at 23:16
  • A "(just about) reasonable" endorsement from Yemon is good enough for me, so I will cast a vote to reopen. I don't feel strongly one way or the other. – Kevin Walker Sep 03 '12 at 23:23
  • Have cast second vote to re-open, since one may as well post the link to the older question as a (possible) answer. – Yemon Choi Sep 04 '12 at 01:18
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    A long collection of characterizations of Hilbert spaces also is in this old thread: http://mathoverflow.net/questions/11192/ A characterization of Hilbert spaces along the lines of item 3 also came up on math.SE a while ago: http://math.stackexchange.com/q/179606 You might also wish to consult Day's book Normed linear spaces chapter VII, §2, p. 151f in the third edition for many characterizations of Hilbert spaces. – Theo Buehler Sep 04 '12 at 04:32
  • For reference: http://tea.mathoverflow.net/discussion/1436/reopening-question-on-characterizing-euclidean-norm-via-isometry-group/ – JRN Sep 04 '12 at 06:27
  • Dan Amir has written a whole book on this topic (Characterizations of inner-product spaces). – jbc Sep 04 '12 at 14:57
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    Lucas (who seems to have lost control of an unregistered account) says, "Thanks for this material. This is a big step for me. Now I have to digest all that. Lucas." – S. Carnahan Sep 10 '12 at 06:12

1 Answers1

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Question 3 seems to be answered by the result mentioned in Sergei Ivanov's earlier question on MO: if a finite-dimensional (real) Banach space has an isometry group acting transitively on its sphere, then the norm is Euclidean. The question itself outlines a proof of this result, and the answers suggest some other variations.

Community
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Yemon Choi
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