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The d-dimensional simplex $\mathbb{S}^d$ defined as follows is isometric to $\mathbb{R}^n$ (see Egozcue et al, 2003). This is because it is a Euclidean vector space with the same number of dimension as $\mathbb{R}^d$ \begin{align} \label{def_simplex} \mathbb{S}^d = \Bigg\{ \mathbf{x} = [x_1,..,x_D] : x_i > 0, \sum^{D}_{1} x_i = 1 \Bigg\} \end{align} A necessary condition for two metric spaces to be isometric is that an isomorphism should exist between the two spaces, hence the isometric log ratio transformation defined by Egozcue et al, 2003.

One thing which I don't understand is that when looking at spaces which are not homeomorphic to each other, for example a sphere and a plane, an argument evoked to prove that they are not homeomorphic is that the sphere is compact while the plane isn't. Yet, the metric space of the d-dimensional simplex is also compact and however it is apparently isometric to the Euclidean space $\mathbb{R}^d$.

Is compactness necessary for two spaces to be isometric ?

To provide more context, the authors of the paper justify the isometry between $\mathbb{S}^d$ and $\mathbb{R}^d$ with the following :

There are also mathematical reasons for introducing this new family of transformations. In fact, all Hilbert spaces are isometric whenever they have the same dimension (Berberian, 1961). Thus, a natural undertaking is to find an isometry between the simplex, which is a Hilbert space—and hence an Euclidean space— and a real space of the same dimension (Billheimer, Guttorp, and Fagan, 2001; Pawlowsky-Glahn and Egozcue, 2001, 2002).

EDIT : added justification of Egozcue et al for the isometry between $\mathbb{S}^d$ and $\mathbb{R}^d$.

outofthegreen
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    Two spaces that are isometric must be homeomorphic. Compactness (or not) is a property that is kept under homeomorphisms. Note that $\Bbb S^d$ is defined with the condition that all $x_i>0$, thereby it is not compact (in other words, the boudary faces are removed). – Hagen von Eitzen Mar 24 '22 at 16:44
  • The paper doesn't seem to talk about standard isometries, but something called "isometric logratio (ilr) transformation associated to an Aitchison-orthonormal basis". I don't know what that means, but definitely not standard isometry. In fact this ilr transformation is not even continuous (because it is a bijection $S^d\to\mathbb{R}^{d-1}$). – freakish Mar 24 '22 at 16:44
  • I edited my original post to add the justification of the authors. By isometric, the authors mean that the norm of a vector in $\mathbb{S}^d$, this norm being measured w.r.t Aitchison's geometry in the simplex : Let $\mathbf{x} \in \mathbb{S}^d$, its norm $||\mathbf{x}||$ is defined as : \begin{align} ||\mathbf{x}||a = \Delta(x,e) = \sqrt{\sum^D{i=1} \Bigg(log \frac{x_i}{g(x)} \Bigg)}
    \end{align} Hence $$ \forall \mathbf{x} \in \mathbb{S}^d : ||\mathbf{x}||_a = ||ilr(\mathbf{x})||$$     – outofthegreen Mar 24 '22 at 16:57
  • @HagenvonEitzen Thank you for your comment. Indeed the simplex is not compact but it is still bounded, whereas $\mathbb{R}^d$ isn't so how could these two spaces could be isometric ? – outofthegreen Mar 24 '22 at 17:01

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An isometry is a homeomorphism, and so isometric spaces share all topological properties, including being compact. Moreover they share metric properties like being bounded.

But the author of the paper defines a custom norm on the $S^d$ simplex, and it is not the Euclidean one. The author defines it as

$$\lVert x\rVert_a=\sqrt{\sum_{i=1}^d \bigg(\ln\frac{x_i}{g(x)}\bigg)^2}$$

where $g(x)$ is the geometric mean. The author claims that $S^d$ is a normed space with this norm and some custom vector space structure (which I didn't check), and this alone implies that $S^d$ cannot be bounded.

All in all, the isometry $S^d\to\mathbb{R}^{d-1}$ is with the custom norm on the left side and the Euclidean on the right. Otherwise this indeed doesn't make much sense.

freakish
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