Easy proof of the fact that isotropic spaces are Euclidean - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:39:15Z http://mathoverflow.net/feeds/question/41211 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean Easy proof of the fact that isotropic spaces are Euclidean Sergei Ivanov 2010-10-05T21:52:12Z 2012-01-19T09:47:30Z <p>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).</p> <p>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.</p> <p>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).</p> <p><strong>Added.</strong> 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?</p> <p>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.</p> <p>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:</p> <ul> <li>Does the fact hold true in infinite dimensions (say, for separable Banach spaces)?</li> </ul> http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean/41213#41213 Answer by Greg Kuperberg for Easy proof of the fact that isotropic spaces are Euclidean Greg Kuperberg 2010-10-05T21:59:41Z 2010-10-05T21:59:41Z <p>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.</p> <p>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.</p> <hr> <p>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.</p> http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean/41215#41215 Answer by rpotrie for Easy proof of the fact that isotropic spaces are Euclidean rpotrie 2010-10-05T22:28:14Z 2010-10-05T22:28:14Z <p>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). </p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean/41224#41224 Answer by Dick Palais for Easy proof of the fact that isotropic spaces are Euclidean Dick Palais 2010-10-06T00:44:57Z 2010-10-06T15:06:38Z <p>There is a classic paper by Jordan and von Neumann where they prove results that allows this question is settled in an elementary way.</p> <p>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.</p> <p>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:</p> <p>4.The condition that every $&lt;= 2$-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 $&lt;= 2$ 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.</p> <p>[SEE BELOW: The following sentence does NOT complete the proof !]<br> And as rpotrie has pointed out in another answer, the two dimensional case follows from the assumed transitivity condition. </p> <p>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.) </p> <p>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.)</p> http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean/41302#41302 Answer by Bill Johnson for Easy proof of the fact that isotropic spaces are Euclidean Bill Johnson 2010-10-06T17:40:44Z 2010-10-06T18:25:32Z <p>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 $2$. </p> <p>For a recent paper related to the Banach-Mazur rotation problem, which contains some other references related to the problem, see</p> <p><a href="http://arxiv.org/abs/math/0110202" rel="nofollow">http://arxiv.org/abs/math/0110202</a>.</p> http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean/86085#86085 Answer by Jose Navarro for Easy proof of the fact that isotropic spaces are Euclidean Jose Navarro 2012-01-19T09:47:30Z 2012-01-19T09:47:30Z <p>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 $X^*$:</p> <p>If $B$ is the unit ball of $X$, for any linear functions $\omega , \omega' \in X^*$, define:</p> <p>$$\langle \omega , \omega' \rangle := \int_B \omega \ \omega' dm$$</p> <p>where $m$ is the Lebesgue measure on $X$, normalized so that $m(B)=1$.</p> <p>(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)</p>