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Consider a normed vector space $V$. Suppose that for every pair of unit vectors $v,w$ there exists a linear isometry which sends $v$ to $w$ (and leaves the subspace spanned by $v$ and $w$ invariant).

Does it follow that $V$ is an inner product space?

By the formulation and the parallelogram law the problem immediately reduces to the two-dimensional case. I couldn't find a nice argument for that case. It would be nice to have a relatively elementary argument for this.

It seems to me that the answer for $\dim{V} = 2$ should be yes: the group $G$ of linear isometries of $V$ is a closed subgroup of $GL(2,\Bbb R)$ and since the unit sphere is compact, $G$ is compact. However, there aren't that many infinite compact subgroups of $GL(2,\Bbb R)$: up to conjugation, they are $O(2)$ and $SO(2)$. I would like this to tell me that I really have a circle as a unit sphere but I don't really know how to make this precise and how to conclude.

Could you please help me finishing this up or give me an alternative argument (or a counterexample)?

Bonus question: What happens if I drop the condition on leaving the subspace spanned by $v$ and $w$ invariant?

Thanks!

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You're almost there! If $v$ is a unit vector, the unit sphere must be the orbit of $v$ under the action of $\text{O}(2)$ or $\text{SO}(2)$ which in either case is just the usual unit circle (up to scale). –  Qiaochu Yuan Aug 6 '12 at 19:20
    
Thank you for confirming that I didn't hallucinate! –  geezer Aug 6 '12 at 19:28
    
@geezer: The two accounts have been merged, and cookies from the most recent should give you access. Welcome to the forum, again! –  mixedmath Aug 8 '12 at 9:38
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@mixedmath: thank you very much for your help and the welcome! Having registered my account, I expect no further trouble with lost cookies :-) Sorry for the inconvenience. –  geezer Aug 8 '12 at 11:17
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1 Answer

up vote 3 down vote accepted

The answer to the bonus question in full generality is no. A Banach space is called transitive if for every unit vectors $u,v$ there is a surjective linear isometry mapping $u$ to $v$. There are nonseparable transitive Banach spaces which are not Hilbert spaces. The Banach-Mazur rotation problem asks whether every separable transitive Banach space is a Hilbert space. This remains unsolved. You can find a bunch of references by googling "transitive Banach space" or "Banach-Mazur problem". There is an MO thread with a proof for finite dimensions and a nonseparable counterexample.

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Thank you very much for this very interesting and helpful information, I wasn't aware that this was a well-known problem! –  geezer Aug 7 '12 at 8:40
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