Easy proof of the fact that isotropic spaces are Euclidean - MathOverflow

Easy proof of the fact that isotropic spaces are Euclidean

2010-10-05T21:52:12Z

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)?

Answer by Greg Kuperberg for Easy proof of the fact that isotropic spaces are Euclidean

2010-10-05T21:59:41Z

The heart of the matter is to define a canonical inner product for any norm in finite dimensions. Since it is canonical, an $X$-isometry is also an isometry of the inner product. If the group is transitive on lines, you thus immediately get that norm is Euclidean.

There are two popular ways to do this. One is John's theorem: The ellipsoid in the unit ball $K$ of $X$ (which is any convex, centrally symmetric body) with the largest volume is unique. Or of course you could use John's theorem dually, taking the smallest ellipsoid that contains $K$. The other popular, canonical ellipsoid is the Legendre ellipsoid of $K$, by definition the ellipsoid $L$ that has the same moment of inertia matrix as $K$, assuming that both $L$ and $K$ have uniformly distributed mass.

On the other hand, the averaging argument is also "slick" in my opinion, and I don't really see anything wrong with it even for undergraduates. Arguably the problem with any slick argument is that it is too adroit for some students.

Answer by rpotrie for Easy proof of the fact that isotropic spaces are Euclidean

2010-10-05T22:28:14Z

Maybe this is not good enough, but in dimension two, you can fix a unit vector $v$ and since you must have that $\langle v, w \rangle = cos \alpha$ where $\alpha$ is the angle between $v$ and $w$ (where $w$ is another unit vector).

Now, you consider $A_w$ the (unique oritentation preserving) isometry that sends $v$ to $w$ and you get that $det(A_w-Id)$ should be $2-2cos(\alpha)$ so you can have a well defined inner product between unit vectors which you can extend linearly.

It seems that extending this argument to higher dimensions may involve averaging (between the isometries that send $v$ to $w$) and it may be the same argument you had in mind.

Answer by Dick Palais for Easy proof of the fact that isotropic spaces are Euclidean

2010-10-06T00:44:57Z

There is a classic paper by Jordan and von Neumann where they prove results that allows this question is settled in an elementary way.

On Inner Products in Linear, Metric Spaces Author(s): P. Jordan and J. V. Neumann, The Annals of Mathematics, Second Series, Vol. 36, No. 3 (Jul., 1935), pp. 719-723.

They first prove by elementary arguments (their Theorem I) the so-called Jordan v. Neumann criterion, that a Banach space is Hilbert iff for all $x$ and $y$, $(*) ||x + y||^2 - ||x - y||^2 = 2||x||^2 + 2 ||y||^2$. They then show from this that a Banach space is Hilbert iff every 1 and 2 dimensional subspace is Euclidean. Here is their argument:

4.The condition that every $<= 2$-dimensional subspace $L'$ of $L$ be isometric to a Euclidean space, is obviously necessary for the existence of an inner-product in the generalized linear,metric space L. It is sufficient,too, because if it is fulfilled, one can argue as follows:If $f_o,g_0\in L$ the space $L'$ of all $\alpha f_0 + \beta g_0$ ($\alpha,\beta$ arbitrary complex numbers) is $<= 2$ dimensional,thus (*) holds in $L'$ (as in every Euclidean space). Therefore it holds in particular for $f = f_0, g = g_0$,and as $f_0,g_0$ are arbitrary, Theorem I proves the existence of an inner product.

[SEE BELOW: The following sentence does NOT complete the proof !]
And as rpotrie has pointed out in another answer, the two dimensional case follows from the assumed transitivity condition.

ERROR NOTICE: I noticed a serious error in the above reasoning! If the isometry group $G$ of a Banach space $V$ is transitive on the unit sphere of $V$, it does NOT follow in any obvious way that the isometry group of a subspace $V'$ of $V$ is transitive on the unit sphere of $V'$. (If $e_1,e_2$ are unit vectors in $V'$, then an element $g$ of $G$ that carries $e_1$ to $e_2$ need not leave $V'$ invariant.)

I did not at first realize how remarkable the conclusion is that transitivity on the unit sphere implies Euclidean. It can be rephrased as saying that transitivity on $S$ implies $2$-transitivity, which to me at least seems even more remarkable. (It was realizing this fact that let me see my silly error.)

Answer by Bill Johnson for Easy proof of the fact that isotropic spaces are Euclidean

2010-10-06T17:40:44Z

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.

Answer by Jose Navarro for Easy proof of the fact that isotropic spaces are Euclidean

2012-01-19T09:47:30Z

As Greg says, the heart of the matter is to define a canonical inner product for any norm in finite dimensions; and this can easily be achieved on the dual $X^*$:

If $B$ is the unit ball of $X$, for any linear functions $\omega , \omega' \in X^*$, define:

$$ \langle \omega , \omega' \rangle := \int_B \omega \ \omega' dm $$

where $m$ is the Lebesgue measure on $X$, normalized so that $m(B)=1$.

(Observe this inner product is just the one in $L_2(B)$ restricted to $X^* \subset L_2(B)$, so this construction applies to any borelian set $B \subset X$, not necessarily a convex, symmetric body)