# Do any of these concepts relating to similarity in general metric spaces have names?

Definition Let a metric space $(X,d)$ be given. A similitude is a function $f: X \to X$ such that, for all $x,y \in X$, $d(f(x),f(y)) = r \cdot d(x,y)$ for some positive real number $r>0$.

Questions:
1. What is the name of metric space in which every similitude is surjective?

For example $\mathbb{R}$ is such a space (I think) but $\mathbb{Z}$ is not.

2. The set of all surjective similitudes is always a group. However, unlike in $\mathbb{R}^n$, it is not always the case that, for each $r>0$, there exists a similitude $f:X \to X$ with scaling factor $r$.

What is the name of such a space, i.e. a space for which $SurjSim(X) \cong \mathbb{R}^+ \times Iso(X)$?

(Where $Iso(X)$ is the group of isometries of $X$.)

3. For any space as in 2., does $Iso(X)$ only determine $d$ up to a positive constant?

This is true for the Euclidean metric on $\mathbb{R}^n$.

4. Is 2. equivalent to the isometry group $Iso(X)$ being a homogeneous space with respect to $\mathbb{R}^+$? Or even to $X$ itself being a homogeneous space with respect to $\mathbb{R}^+$?

For example, any finite-dimensional vector space satisfies these properties -- is there a name for more general spaces which are "scaleable" like this? Homogeneous space can refer to any type of group action, including the group action of a group $G$ on its corresponding torsor which denotes a notion of translation, not of scaling.

Note: I think these may have something to do with the notions of (complete) convex metric spaces (for $r<1$) and (complete) externally convex metric spaces$^*$ (for $r>1$).

*See the third page, p. 231, here.

Also note that any geodesic space is a convex metric space as a result of Lemma 2.2. (ii) here.

So I think it is possible to show that an externally convex geodesic space satisfies 2., using the fact that any geodesic space is complete, Lemma 2.2(ii), and binary search.

• Are you sure that the two statements in your 2nd item are equivalent? Assuming that for $X$ there are similitudes for every $r \in \Bbb R$ (and moreover that $|X| > 1$) only gives us $SurjSim(X)/Iso(X) \cong \Bbb R^+$. To get a product, we would need a family $\lambda_r$ of surjective similitudes of factor $r\in \Bbb R^+$, each commuting with every isometry on $X$, satisfying $\lambda_{rs} = \lambda_r \circ \lambda_s$ for all $r, s>0$, and to me the existence of such a family is not obvious. – Marc Paul Apr 2 '17 at 21:32
• @MarcPaul No I am not certain at all. I thought I would make the bold claim hoping either (1) I was right or (2) someone would correct me and in the process figure out the answer. If you want, you can edit the post to split the claim into a separate question, since, as you correctly point out, it is extremely speculative. – Chill2Macht Apr 3 '17 at 5:49

Note: I said in the post I think that any externally convex geodesic space satisfies 2. -- however this is not necessarily true, even for an externally convex geodesic space which is also an M-space. At the very least, the simple method I had in mind (which works for Euclidean space) doesn't work, for example, for hyperbolic space. See the answer to this question: Is there a counterexample to this metric analog of the side-angle-side (SAS) postulate?

3. I believe the answer is yes, based on the following.

Claim: The condition in 2. is equivalent to the following:

For every $\lambda > 0$, the metric space $(X,d)$ is isometric to $(X, \lambda d)$.

Proof of claim: Assume 2. is true. Then for any $\lambda > 0$, there exists a surjective similitude $f: (X,d) \to (X,d)$ with scaling factor $r = \frac{1}{\lambda} > 0$. If we consider $f$ as a function $$\tilde{f}:(X,d) \to (X, \lambda d)$$ then $\tilde{f}$ is the same set map as $f$, and since $f$ is bijective, since it is surjective by assumption and any similitude is injective, $\tilde{f}$ is bijective. Moreover, $\tilde{f}$ is an isometry between $(X,d)$ and $(X, \lambda d)$ since: $$(\lambda d)(\tilde{f}(x_1), \tilde{f}(x_2)) = \lambda \cdot d(f(x_1),f(x_2))=\lambda \cdot \left(\frac{1}{\lambda} \cdot d(x_1, x_2) \right) = d(x_1, x_2)$$

Now assume that, for every $\lambda > 0$, the metric space $(X,d)$ is isometric to $(X, \lambda d)$.

Then for every $r > 0$, there exists a bijective function $\tilde{f}: (X,d) \to (X, \frac{1}{r}$d) such that $$\left(\frac{1}{r}d\right)(\tilde{f}(x_1), \tilde{f}(x_2)) = d(x_1,x_2)$$ If we define $f:(X,d) \to (X,d)$ such that it is the same set map as $\tilde{f}$, then we have that $$r \cdot d(x_1,x_2) = r \cdot \left(\frac{1}{r}d\right)(\tilde{f}(x_1), \tilde{f}(x_2)) = r \cdot \left(\frac{1}{r} \cdot d(f(x_1), f(x_2))\right) = d(f(x_1),f(x_2))$$ Thus $f$ is a similitude, and moreover $f$ is surjective since $\tilde{f}$ is bijective by assumption. $\square$

EDIT: I realize that the original post has a mistake -- in 2. it should say surjective similitude, not just any similitude. Although a space with the property as originally written would also be interesting -- it's just that the similitudes might be isometries into $(X, \lambda d)$, not isometric isomorphisms, i.e. not bijective -- any similitude is automatically injective, for the same reason any isometry is automatically injective, a fact I also should have mentioned in the original post.

Anyway, if we have 2., then by the argument above, this is equivalent to $(X,d)$ being isometric to $(X, \lambda d)$ for every $\lambda > 0$. Thus, any isometry $\tilde{f}$ in $Iso(X)$ is equivalent to a surjective similitude $f$ and vice versa. Thus it should follow that $Iso(X)$ only determines the metric $d$ up to a positive constant $\lambda > 0$, although I am not sure how to phrase this argument rigorously at the moment.

I think from 3. then it might follow at least that the group of surjective similitudes is at least the semidirect product of $\mathbb{R}^+$ and the isometry group, although I'm not sure. Anyway, then assuming that a space $X$ satisfying 2. is also a space where the group action of $Iso(X)$ is sufficiently nice, I think it follows that $X$ is homogeneous with respect to $\mathbb{R}^+$, although I am also not sure.