# When does a group of dilations/scalings exist in a metric space?

Notation: Let $(X,d)$ be a metric space. A similitude will be (by convention) a surjective (hence bijective) map $f: X \to X$ such that for all $x_1, x_2 \in X$, $d(f(x_1),f(x_2)) = r d(x_1, x_2)$ for some $r>0$. An isometry is a similitude for which $r = 1$. For each point $x \in X$, define $Sim(X,x)$ to be the group of all similitudes which fix $x$, i.e. $f(x) = x$, and define $Iso(X,x)$ to be the group of all isometries which fix $x$, then obviously $Iso(X,x) \subseteq Sim(X,x)$.

Question: (a) When, for a given $x \in X$, is $Iso(X,x)$ a normal subgroup of $Sim(X,x)$, so that the quotient $Sim(X,x)/Iso(X,x)$ is defined?

(b) When is the quotient $Sim(X,x)/Iso(X,x)$ isomorphic to a subgroup of $Sim(X,x)$, denoted by $Dil(X,x)$ (group of dilations centered at $x$) such that either of the following holds? $$Dil(X,x) \ltimes Iso(X,x) \cong Sim(X,x) \quad or \quad Dil(X,x) \rtimes Iso(X,x) \cong Sim(X,x)$$

I imagine that this question is probably too difficult to answer directly, so interesting examples and counterexamples will also suffice for answers.

Motivation: In a previous question I hypothesized that the existence of such a group might be equivalent to the existence of a notion of direction at that point $x$; see here. This question is also very analogous to another previous question of mine, the answer to which establishes that transitivity of the group action (in that case $Iso(X)$, in this case $Sim(X,x)$) is not sufficient.

Also the most obvious example is Euclidean space, as well as the motivating example — in this case $Iso(X,x) \cong O(n)$, $Sim(X,x) \cong ConformalGroup(n)$, and $Dil(X,x) \cong \mathbb{R}$. I would expect, but have not (yet?) proven, that $Dil(X,x) \cong \mathbb{R}$ for any metric vector space over the reals.

#### Solutions Collecting From Web of "When does a group of dilations/scalings exist in a metric space?"

About a., you can define a map from $Sim(X) \to \mathbb{R}_{> 0}$ by sending a similitude $g$ to that constant $r = r_g$. This is a homomorphism because $d(gf(x), gf(y)) = r_g d(f(x),f(y)) = r_g r_f d(x,y)$. The kernel is then the isometries, so $Isom(X)$ is always normal. Likewise if you insist on stabilizing a point.

About b., I don’t know a full answer. If $X$ has finite diameter (for example if $X$ is compact) then every similitude $g$ must have $g_r = 1$, i.e. must be an isometry: if $D$ is the diameter of $X$, then let $x,y$ be such that $d(x,y) = D$ (within $\epsilon$, so I’m really considering a sequence of $x$ and $y$ so that their distances converge to the diameter.) Then $d(g(x),g(y)) > D$ if $r_g > 1$. If $r_g < 1$, then $r_{g^{-1}} > 1$.

Let me know if it doesn’t make sense and why you disagree.