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Let $A$ and $B$ be topological groups. Sometimes there exist neighborhoods of the identity $U\subseteq A$ and $V\subseteq B$ such that any homomorphism $f\colon A\to B$ satisfying $f(U)\subseteq V$ is automatically continuous. Here are a couple examples of this:

  • Let $B=\mathbb{R}$. Let $U_0,U_1,\ldots\subseteq A$ be a neighborhood basis at the identity satisfying $U_{n+1}\cdot U_{n+1}\subseteq U_n$. Let $V_n=(-2^{-n},2^{-n})$. If $f\colon A\to B$ is a homomorphism satisfying $f(U_0)\subseteq V_0$, then $f(U_n)\subseteq V_n$, which implies continuity. I suspect that this sort of idea might work whenever $B$ is a Lie group. The case when $B$ is the circle group is important for proving local compactness of the Pontryagin dual (https://math.stackexchange.com/a/1510226/).

  • Let $A=B=\mathbb{Z_p}$. If $f\colon A\to B$ is a homomorphism, then $f(p^n\mathbb{Z}_p)\subseteq p^n\mathbb{Z}_p$, which implies continuity (without having to impose $f(U)\subseteq V$).

Do all locally compact Hausdorff topological groups have this continuity property? Or is there a nice condition that implies this equicontinuity property?

Thomas Browning
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1 Answers1

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For a counterexample, let $A=B=(\mathbb{Z}/2)^{\mathbb{N}}$ (or alternatively $G^\mathbb{N}$ for any compact $G$ such that there exists a discontinuous homomorphism $G\to G$). Then for any neighborhoods $U$ and $V$ of $1$ in $A$, there is a discontinuous homomorphism $f:A\to A$ such that $f(U)\subseteq V$. Indeed, any such $V$ contains a subgroup $C$ topologically isomorphic to $A$, and there is a discontinuous homomorphism $A\to C$.

Eric Wofsey
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