5
$\begingroup$

$\newcommand{\<}{\langle} \newcommand{\>}{\rangle} $

It is a fact that for every norm $\| \|$ on a finite dimensional (real) vector space, its isometry group $\text{ISO}(|| \cdot ||)$ is contained in some isometry group of a suitable inner product. (see this question).

Now assume we have a norm $\| \|$ such that $\text{ISO}(|| \cdot ||)=\text{ISO}(\<,\>)$ for some inner product.

Is $\| \|$ necessarily induced by an inner product?

Update:

The answer is yes. The key fact is the transitivity of the isometry group. Actually, as pointed out in this question in MO, the following statement is true:

Let $X$ be finite-dimensional normed space whose isometry group acts transitively on the unit sphere (i.e, for every two unit-norm vectors $x,y∈X$ 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).

The proof is decomposed of two steps:

(1) showing there exists an inner product $\< ,\>$ whose isometry group $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$. The basic idea is this:

$\text{ISO}(|| \cdot ||)$ is compact, hence it admits an invariant probability measure, hence (by an averaging argument) there exists a Euclidean structure preserved by it. (The details can be found here).

(2) The transitivity of $\text{ISO}(|| \cdot ||)$ together with (1) impliy that the original norm is proportional to that Euclidean norm.

$\endgroup$
3
$\begingroup$

I think the answer is yes and the corresponding inner product is proportionnal to $\langle \cdot , \cdot \rangle$. Let's denote $E$ the vector space and $S$ the sphere associated to the inner product $\langle \cdot , \cdot \rangle$, i.e. the set $S=\lbrace x \in E \; | \; \langle x , x \rangle =1 \rbrace$.

The key for the demonstation is the fact that $ISO(\langle \cdot , \cdot \rangle)$ acts transitively on $S$. For the following, we will fix an element $x_0$ of $S$.

Let $x\in E$ be any element of $E$. Defining $y=\dfrac{x}{\sqrt{\langle x , x \rangle}}$, we have $y \in S$. Thus, by transitivity, there exists $u \in ISO(\langle \cdot , \cdot \rangle)$ such that $y=u(x_0)$. Then, we have : $$\begin{array}{rcl} \|x\|^2 & = & \| \sqrt{\langle x , x \rangle} y \|^2 \\ & = & \langle x , x \rangle \|y\|^2 \\ & = & \langle x , x \rangle \|u(x_0)\|^2 \\ & = & \langle x , x \rangle \|x_0\|^2 \; \; \; \; \text{as} \; u\in ISO(\langle \cdot , \cdot \rangle)=ISO(\| \cdot \|)\\ \end{array}$$

Thus, defining $\lambda=\|x_0\|^2$, we have, for all $x \in E$ : $$ \boxed{\| x \|^2 = \lambda \langle x , x \rangle}$$ which ends the demonstration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.