Let $G$ be a topological group with unit element $e$.
We say that $D\subseteq G$ is discrete if for all $x\in D$ there is a unit-neighborhood $U\subseteq G$ such that $x^{-1}D\cap U=\{e\}$. We say that $D\subseteq G$ is uniformly discrete if there is a unit-neighborhood $V\subseteq G$ such that $|x^{-1}D\cap V| \le 1$ for all $x\in G$, equivalently $D^{-1}D\cap V^{-1}V=\{e\}$. (Note that, we do not consider Abelian groups, so this is technically left uniform discreteness, however for simplicity, we stick to left translates while noting that similar considerations can be done for right translates.)
Note that if $D^{-1}D$ is discrete, $D$ is even uniformly discrete, as for $e\in D^{-1}D$ there is a unit-neighborhood $U\subseteq G$ with $D^{-1}D\cap U=\{e\}$. The converse question is much harder. There are uniformly discrete sets $D$ with discrete difference sets (which is what I call $D^{-1}D$) but even slight modifications lead to losing discreteness. For example, $D=\{n+\tfrac{1}{n}:n\in\mathbb{N}\}$ is uniformly discrete (in $(\mathbb{R},+)$) and $D-D=\bigcup_{k\in\mathbb{N}_0}\left\{\pm k \left(1\mp \tfrac{1}{n(n+k)}\right):n\in\mathbb{N}\right\}$ is still discrete. However, even just adding the single point $\{1\}$ to $D$ results in the difference set containing $1=2-1$ and $1-\tfrac{1}{n(n+1)}$ for all $n\in\mathbb{N}$ which loses discreteness.
I found a condition under which the converse does hold. I could not find literature, so I just made up a term, namely I call a set $A\subseteq G$ with $e\in A$ locally translation invariant (name is subject to change) if for every $a\in A$ there is a unit-neighborhood $U\subseteq G$ such that $a^{-1}A\cap U\subseteq A$ (or $A\cap aU\subseteq aA$). Note that every discrete set containing $e$ is locally translation invariant.
It turns out that this is exactly what we need. Namely, $D^{-1}D$ is discrete if and only if $D$ is uniformly discrete and $D^{-1}D$ is locally translation invariant.
Proof. Let $D^{-1}D$ be discrete, then $D$ is uniformly discrete and for $x,y\in D$ there is unit-neighborhood $U\subseteq G$ such that $D^{-1}D\cap x^{-1}y U=\{x^{-1}y\}\subseteq x^{-1}yD^{-1}D$, so $D^{-1}D$ is locally translation invariant. Conversely, find unit-neighborhood $U_0\subseteq G$ such that $D^{-1}D\cap U_0=\{e\}$ by uniform discreteness. Then for $x,y\in D$, we obtain $x^{-1}yD^{-1}D \cap x^{-1}yU_0=\{x^{-1}y\}$. By local translation invariance find unit-neighborhood $U$ with (w.l.o.g.) $U\subseteq U_0$ such that $D^{-1}D\cap x^{-1}y U\subseteq x^{-1}y D^{-1}D$, then it follows that $D^{-1}D\cap x^{-1}yU\subseteq x^{-1}yD^{-1}\cap x^{-1}y U_0=\{x^{-1}y\}$, so $D^{-1}D$ is discrete. $$\tag*{$\blacksquare$}$$ Finally to my question. I would like to find a geometric interpretation of what I called local translation invariance or even find out if there is some literature on this problem. Thank you for your help.
