Let $X$ be a finite-dimensional Banach space whose isometry group acts transitively on the set of lines (or, equivalently, on the unit sphere: for every two unit-norm vectors $x,y\in X$ there exist a linear isometry from $X$ to itself that sends $x$ to $y$). Then $X$ is a Euclidean space (i.e., the norm comes from a scalar product).


I can prove this along the following lines: the linear isometry group is compact, hence it admits an invariant probability measure, hence (by an averaging argument) there exists a Euclidean structure preserved by this group, and then the transitivity implies that the Banach norm is proportional to that Euclidean norm.


But this looks too complicated for such a natural and seemingly simple fact. Is there a more elementary proof? I mean something reasonably short and accessible to undegraduates (so that I could use it in a course that I am teaching).


**Added.** As Greg Kuperberg pointed out, there are many other ways to associate a canonical Euclidean structure to a norm, e.g. using the John ellipsoid or the inertia ellipsoid. This is much better, but is there something more "direct", avoiding any auxiliary ellipsoid/scalar product construction?


For example, here is a proof that I consider "more elementary", under the stronger assumption that the isometry group is transitive on two-dimensional flags (that is, pairs of the form (line,plane containing this line)): prove this in dimension 2 by any means, this implies that the norm is Euclidean on every 2-dimensional subspace, then it satisfies the parallelogram identity, hence it is Euclidean.


Looking at this, I realize that perhaps my internal criterion for being "elementary" is independence of the dimension. So, let me try to transform the question into a real mathematical one:


* Does the fact hold true in infinite dimensions (say, for separable Banach spaces)?



Easy proof of the fact that isotropic spaces are Euclidean
It is a famous problem (due to Banach and Mazur) whether a separable infinite dimensional Banach space which has a transitive isometry group must be isometrically isomorphic to a Hilbert space. Of course, if every two dimensional subspace has a transitive isometry group, then the space is a Hilbert space since then the norm satisfies the parallelogram identity. For counterexamples in the non separable setting, consider the $\ell\_p$ sum of uncountably many copies of $L\_p(0,1)$ with $p$ not $2$. 


For a recent paper related to the Banach-Mazur rotation problem, which contains some other references related to the problem, see


<http://arxiv.org/abs/math/0110202>.


