# Can a normed vector space be reconstructed from its metric?

Okay, me previous question Can a vector space be reconstructed form its norm? was solved by a nice (but in hindsight embarassingly simple) counter-example. Perhaps a better question emerges from that asnwer as well as from the discussions in the comments there:

Let $$(V,+,\cdot,\|\cdot\|)$$ be a normed $$\Bbb R$$ vector space. Assume we are only given the underlying set $$V$$, the zero vector $$0$$, and the induced metric $$d\colon V\times V\to\Bbb R$$, $$(v,w)\mapsto\|v-w\|$$. Can we reconstruct $$+$$, $$\cdot$$, and $$\|\cdot\|$$ from this?

The last part is of course trivial because $$\|v\|=d(v,0)$$. On the other hand, if we were not given $$0$$ as a base point, we'd be doomed because $$d$$ is translation invariant. (Or we might instead be satisfied with reconstructing the affine structure)

At least in case of the Euclidean norm on a finite-dimensional space, the answer is Yes: First, we can determine $$\dim V$$ as one less than the maximal number of vertices in a regular simplex. If we pick such a maximal simplex with side length $$1$$, say, and with one vertex $$0$$, then we can "triangulate" all other points of $$V$$ by their distances to the vertices, and this allows us to reconstruct $$+$$ and $$\cdot$$.

Remains the question, what the situation looks like in all other cases. Actually, this single question automatically splits into two parts:

Q1: What is the situation with other norms on finite-dimensional spaces? "Triangulation" becomes a non-trivial task and even finding a "nice" simplex may prove difficult.

Q2: What is the situation with infinite dimensional spaces? Those can be even weirder ...

When pressed to pick one, I'd be more interested in answers to Q1 ...

EDIT: To clarify, added $$\Bbb R$$ as scalar field.

• How is one expected to know the scalar field (with respect to which $V$ is a vector space)? May 4 '20 at 21:29
• Operation $+$ is given a priory, without it there's nothing to do, how do you take $v-w$? May 4 '20 at 21:33
• @janmarqz I do $v-w$ to compute $d$ from $\|\cdot\|$. But if I forget the other structure and remember only $d$, I cannot immediatly subtracr May 4 '20 at 22:00

I'll assume the scalar field is $$\mathbb R$$. The Mazur-Ulam theorem asserts that any surjective isometry of normed linear spaces is affine. So (up to choice of origin) the metric does uniquely determine the vector space structure.