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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)?
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When I was an undergraduate I read substantial portions of your book (with Burago & Burago) on metric geometry, and I remember encountering this as an exercise and getting hung up on it for about two weeks. I finally asked Ralph Spatzier who suggested more or less the exact argument that you outlined. I could barely understand the argument and went on for years with the problem of simplifying it in the back of my mind. I don't mind the more sophisticated argument anymore, but it would nevertheless be very gratifying for me personally to see this issue resolved. So thanks for the question! – Paul Siegel Oct 6 '10 at 1:10
Not sure I understand your last question, Sergei. It is a famous problem 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 $\ell_p(A)$ with $p$ not $2$ and $A$ uncountable. – Bill Johnson Oct 6 '10 at 1:42
@Bill Johnson: thanks, you answered it. I was not aware about this being an open problem. – Sergei Ivanov Oct 6 '10 at 7:28
@Sergei Ivanov: The example I mentioned in the non separable setting is wrong. An example is the $\ell_p$ sum of uncountably many copies of $L_p(0,1)$, not the $\ell_p$ sum of uncountably many copies of the real line. – Bill Johnson Oct 6 '10 at 8:04
@Bill: You should post your remarks as an answer to the revised question. It's an open problem for separable Banach spaces and there are inseparable counterexamples. Sweet! – Greg Kuperberg Oct 6 '10 at 10:45
up vote 10 down vote accepted

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

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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.

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Yes, thanks, but this is not quite what I am looking for. This is intended for students that probably never heard about Euclidean vs. non-Euclidean norm thing before. I'll try to edit question to make it more clear. – Sergei Ivanov Oct 5 '10 at 22:15
I'll have to trust Bill Johnson's answer that the new question is an open problem. As for the original question, I learned what John's ellipsoid is before I really knew what a norm is. If you state everything in terms of convex bodies, then I don't see how an undergraduate could object to the John ellipsoid argument. It's a wonderful and elementary fact that there is only one largest ellipsoid, and even only one local maximum of the volume function for ellipsoids. – Greg Kuperberg Oct 6 '10 at 10:47
I learned about John ellipsoid much later, and did so the hard way (discovering the John representation theorem myself). So it is not a common knowledge to me. But I agree with you - using John ellipsoid is much better from the pedagogical perspective, even if I claim its uniqueness without proof (or leave it as an exercise). After all, I'm not going to use the result in the course, its sole purpose its to impress the students. – Sergei Ivanov Oct 6 '10 at 16:56
@Sergei I think, uniqueness of the John ellipsoid is quite easy to prove (we are searching for a non-negative quadratic form $Q$ such that $(Qx,x)\leq \|x\|^2$ and $\det Q$ is maximal possible. If it is not unique, say $Q_1$ and some $Q_2\ne Q_1$ satisfy, then $(Q_1+Q_2)/2$ has strictly larger determinant), and your students, if they are not freshmen, should know all the necessary linear algebra. – Fedor Petrov Oct 7 '10 at 22:53

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.)

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Right, Dick. You could solve the Banach-Mazur problem if you could prove that transitivity implies transitivity of every two dimensional subspace. – Bill Johnson Oct 6 '10 at 18:28
Owing to fat fingers I accidentally flagged Bill's comment when I meant to +1 it - apologies – Yemon Choi Sep 3 '12 at 23:18
Perhaps someone may fix the typo in (*). – Włodzimierz Holsztyński May 16 '13 at 0:02

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.

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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)

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