# "Halving" neighborhoods in topological groups

Let $$S=B\left(0,\frac{\epsilon}{2}\right)$$ be a ball in $$\mathbb{R}^n$$.  Given a translate $$S+t$$, there is the nice property that, for all $$x,y\in S+t$$, we have $$x-y\in S-S\subseteq B(0,\epsilon)$$.  I would like to generalize this to arbitrary topological groups.

### Fix a topological group $$G$$ with identity $$e$$.  For each neighborhood $$U$$ of $$e$$, does there exist another neighborhood $$V\subseteq U$$ of $$e$$ such that $$\{vw^{-1}:v,w\in V\}\subseteq U$$?

In the above, neighborhoods are assumed a fortiori open.

I have not assumed $$G$$ abelian, but feel free to do so if it changes the answer.

Some observations on counterexamples
The discrete topology is not a counterexample: take $$V=\{e\}$$. Similarly, for the indiscrete topology, $$V=U=G$$.

One possible route is to look for an arithmetic obstruction.  $$C_3$$ is not a counterexample because it is discrete, but perhaps if $$\{g:\nexists x(x^2=g)\}$$ is dense in $$U$$, then there exists no $$V$$? (One line of inquiry suggested by this nonexample is to considere $$C_3^{\omega}$$ with the product topology.)

Obviously, the argument for $$\mathbb{R}^n$$ goes through immediately if $$G$$ is locally metrizable. By Nagata-Smirnov, that means that a (hypothetical) counterexample violates regularity, Hausdorffness, or $$\sigma$$-local finitude in $$U$$.

Checking Steen & Seebach for metrization counterexamples doesn't really help: most don't have a natural group structure.  The remainder are the Long Line (45), the converging rational topology (65), $$\beta\mathbb{Z}$$ (111), and the radial interval topology (141).  All of these violate metrizability by having neighborhoods of the origin that are "too large" (i.e. not $$\sigma$$-locally finite), but that's the wrong place to look. If $$V$$ is large, then so is $$U$$. Instead, counterexamples should arise from groups that are not Hausdorff.

Yes, this is almost immediate from the definition of a topological group. The map $$f:G\times G\to G$$ defined by $$f(g,h)=gh^{-1}$$ is continuous (since the group product and inverse maps are continuous). Since $$U$$ is a neighborhood of $$e$$ and $$f(e,e)=e$$, this means $$f^{-1}(U)$$ is a neighborhood of $$(e,e)$$. That is, there exist neighborhoods $$V_1$$ and $$V_2$$ of $$e$$ such that $$V_1\times V_2\subseteq f^{-1}(U)$$. The set $$V=V_1\cap V_2\cap U$$ then satisfies your requirements.