# Appearances of $\mathbb{Q}/\mathbb{Z}$ in Pontryagin duality for profinite groups

(This is a somewhat lazy question which came up as I'm reading about Pontryagin duality for the first time)

For a locally compact abelian topological group $$G$$, its Pontryagin dual is the group of continuous homomorphisms $$G^* := \text{Hom}(G,\mathbb{R}/\mathbb{Z})$$ with the compact-open topology ($$\mathbb{R}/\mathbb{Z}$$ is given the usual topology).

If $$G$$ is profinite abelian, then any such homomorphism cannot be surjective, and hence its image is a proper closed subgroup, hence finite, and hence there is a natural isomorphism $$\text{Hom}(G,\mathbb{R}/\mathbb{Z}) \cong \text{Hom}(G,\mathbb{Q}/\mathbb{Z})$$ where $$\mathbb{Q}/\mathbb{Z}$$ is given the topology induced from the inclusion $$\mathbb{Q}/\mathbb{Z}\subset\mathbb{R}/\mathbb{Z}$$. The same isomorphism holds if $$G$$ is discrete torsion abelian.

This topology on $$\mathbb{Q}/\mathbb{Z}$$ is clearly totally disconnected, but neither discrete nor compact, hence not profinite.

On the other hand, there is another appearance of $$\mathbb{Q}/\mathbb{Z}$$, now viewed as an ind-finite abelian topological group, hence a discrete abelian group, namely under the identification $$\widehat{\mathbb{Z}}^*\cong \mathbb{Q}/\mathbb{Z}$$.

Can some wise people say a word about these two topologies on $$\mathbb{Q}/\mathbb{Z}$$ and how they're related?

Certainly if $$G$$ is discrete abelian, then $$\text{Hom}(G,\mathbb{Q}/\mathbb{Z})$$ does not depend on the topology on $$\mathbb{Q}/\mathbb{Z}$$, and if $$G = \varprojlim G_i$$ is profinite, then we have $$\text{Hom}(\varprojlim G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\text{Hom}(G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\text{Hom}(G_i,\mathbb{Z}/|G_i|\mathbb{Z})$$ and hence the result seems to also be independent of the choice of topology on $$\mathbb{Q}/\mathbb{Z}$$.

Is there any situation where one must be careful which topology to consider on $$\mathbb{Q}/\mathbb{Z}$$?

• If $G$ is a locally compact group, every closed subgroup of countable index is open. It follows that every continuous homomorphism into $(Q/Z)_i$ (endowed with the topology of inclusion in $R/Z$) is continuous into $(Q/Z)_d$ (discrete topology). However, the topology (uniform convergence on compact subsets) of $Hom(G,(Q/Z)_i)$ and $Hom(G,(Q/Z)_d)$ can differ, e.g., when $G$ is discrete and infinite cyclic. When $G$ is profinite, or more generally locally elliptic, however, all homomorphisms $Hom(G,(Q/Z)_d)\to Hom(G,(Q/Z)_i)\to Hom(G,R/Z)$ are topological isomorphisms. – YCor Oct 3 '18 at 22:16
• Thanks for this, @YCor. I was disquieted by the idea that differing topologies on $\Bbb Q/\Bbb Z$ could matter, but l lacked the clarity to see why. – Lubin Oct 3 '18 at 22:33