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.

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    $\begingroup$ 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. $\endgroup$ – HenrikRüping Sep 3 '12 at 20:52
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    $\begingroup$ 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. $\endgroup$ – Lucas Sep 3 '12 at 21:08
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    $\begingroup$ Perhaps one of those who voted to close would care to elaborate? $\endgroup$ – Gerry Myerson Sep 3 '12 at 22:37
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    $\begingroup$ 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. $\endgroup$ – Kevin Walker Sep 3 '12 at 22:57
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    $\begingroup$ A long collection of characterizations of Hilbert spaces also is in this old thread: mathoverflow.net/questions/11192 A characterization of Hilbert spaces along the lines of item 3 also came up on math.SE a while ago: 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. $\endgroup$ – Theo Buehler Sep 4 '12 at 4:32

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.


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