Is a normed vector space with an isometry group acting transitively on its unit ball an inner product space?

Question

In section 2.3 of The Octonions, Baez discusses a proof for the Hurwitz theorem on normed division algebras via Clifford algebras that goes through the following argument. Supposing that $\mathbb K$ is a normed division algebra, and $a \in \mathbb K$ with $\Vert a\Vert = 1$, the operators

$$ L_a : \mathbb K \rightarrow \mathbb K\\\qquad\, x\mapsto ax$$

are norm-preserving and map any point of the unit sphere to any other point (because $\mathbb K$ is assumed to be a division algebra). Baez then writes that “the only way the unit sphere in $\mathbb K$ can have this much symmetry is if the norm on $\mathbb K$ comes from an inner product”.

Is it true that a normed vector space must in fact be an inner product space if there is a group of isometries acting transitively on its unit sphere? If so, how so? If not, did Baez mean something else?

Answer 1

To convert my comments to an answer:

I will assume that your normed division algebra $K$ is real and finite-dimensional. I am nearly sure that Baez assumes both.

Lemma. Let $V$ be a finite-dimensional real normed vector space. Suppose that the subgroup $G$ of linear isometries of $K$ acts transitively on the unit sphere of the norm on $V$. Then the norm on $V$ comes from an inner product.

Proof. Let $B$ denote the (closed) unit ball of the norm of $V$. Then $B$, as any compact convex solid in $V$, admits the Jones ellipsoid which is the unique least volume ellipsoid containing $B$. (A formal definition of a closed ellipsoid is that it is the unit ball of a norm given by an inner product.) By the uniqueness of $E$, the group $G$ preserves $E$. But then $G$ also preserves the (nonempty) set of points where the boundaries of $B$ and $E$ meet. By the transitivity assumption, it then follows that $\partial B=\partial E$, i.e. $B=E$. Hence, $B$ comes from an inner product. qed

Edit. According to the survey by Pelczynski and Bessaga,

Some aspects of the present theory of Banach spaces, page 255,

the above lemma is a result due to Auerbach and von Neumann (independently). At the same time, according to the same source, if $\mu$ is a non-sigma-finite non-atomic measure, then for each $p\in [1,\infty)$, the Banach space $V=L^p(\mu)$ has the property that the group of linear isometries acts transitively on the unit sphere. (I did not check this.) In this example, the Banach space $V$ is not a Hilbert space. It is an open problem, due to Banach and Mazur if every separable Banach space whose group of linear isometries acts transitively on the unit sphere is necessarily a Hilbert space.

Answer 2

Here is a quick argument that came to my mind, though I am a little bit out of my comform zone with some statements (I highlighted them with a $\color{red}{??}$, and maybe someone more knowledgeable can correct or affirm me).

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If $V$ is your (finite-dimensional, real) normed vector space with norm $\|\cdot\|$ and transitive symmetry group $G\subseteq\mathrm{GL}(V)$ on $S:=\{v\in V\mid\|v\|=1\}$, then $G$ is a compact linear group ($\color{red}{??}$).

I believe that it is a well-known fact ($\color{red}{??}$) from the representation theory of compact groups that there is a $G$-invariant inner product on $V$, that is, $\langle gv,gw\rangle_* = \langle v,w\rangle_*$ for all $v,w\in V$ and $g\in G$. Then, also the norm $\|v\|_*^2:=\langle v,v\rangle_*$ is $G$-invariant.

Set $\alpha:=\|v\|^2/\|v\|_*^2$ (which is independent of $v$ because both norms are $G$-invariant). Then $\langle v,w\rangle:=\alpha \langle v,w\rangle_*$ is an inner product to the norm $\|\cdot\|$ of $V$.