The dual group of the profinite integers

Question

I am trying to figure out the following easy example of Pontryagin duality (see Pontrjagin duality for profinite and torsion abelian groups).

For an abelian group $A$ (torsion or profinite), let its dual $A^D$ be the group of homomorphisms $\operatorname{Hom}_\mathbb{Z} (A, \mathbb{Q}/\mathbb{Z})$.

For example, the dual of $\mathbb{Q}/\mathbb{Z}$ are the profinite integers $\hat{\mathbb{Z}}$: $$\operatorname{Hom}_\mathbb{Z} (\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \operatorname{Hom}_\mathbb{Z} (\varinjlim_m \mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \varprojlim_m \operatorname{Hom}_\mathbb{Z} (\mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \varprojlim_m \mathbb{Z}/m\mathbb{Z} \cong \hat{\mathbb{Z}},$$ as the $m$-torsion part of $\mathbb{Q}/\mathbb{Z}$ is given by $\left[\frac{i}{m}\right]$ for $i = 0, 1, \ldots, m-1$. So we have $$\mathbb{Q}/\mathbb{Z}^D \cong \hat{\mathbb{Z}}.$$

Then we should have $\operatorname{Hom}_\mathbb{Z} (\hat{\mathbb{Z}}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}$, but what is the right and easy proof? We just saw that $\hat{\mathbb{Z}}^D \cong (\mathbb{Q}/\mathbb{Z}^D)^D$, but how do I see directly that $(\mathbb{Q}/\mathbb{Z}^D)^D \cong \mathbb{Q}/\mathbb{Z}$ without invoking the Pontryagin duality?

Answer 1

It is not true that $\operatorname{Hom}_\mathbb{Z} (\hat{\mathbb{Z}}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}$. Indeed, $\hat{\mathbb{Z}}$ is an uncountable torsion-free abelian group, so tensoring with $\mathbb{Q}$ gives an uncountable-dimensional $\mathbb{Q}$-vector space $\hat{\mathbb{Z}}\otimes\mathbb{Q}$. You can then use this to construct uncountably many different homomorphisms $\hat{\mathbb{Z}}\to\mathbb{Q}/\mathbb{Z}$ (for instance, pick a basis for $\hat{\mathbb{Z}}\otimes\mathbb{Q}$ consisting of elements of $\hat{\mathbb{Z}}$ and consider the maps to $\mathbb{Q}$ given by each basis element composed with the map $\mathbb{Q}\stackrel{1/2}{\to}\mathbb{Q}/\mathbb{Z}$; for each basis element this gives a homomorphism $\hat{\mathbb{Z}}\to\mathbb{Q}/\mathbb{Z}$ sending that basis element to $1/2$ but every other basis element to $0$).

What is true is that the group of continuous homomorphisms $\hat{\mathbb{Z}}\to\mathbb{Q}/\mathbb{Z}$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$, where you give $\hat{\mathbb{Z}}$ the profinite topology and $\mathbb{Q}/\mathbb{Z}$ the quotient topology from $\mathbb{Q}$ (or the discrete topology, if you prefer; it turns out to make no difference in this case). To prove this, note that since $\hat{\mathbb{Z}}$ is compact, the image of any homomorphism $f:\hat{\mathbb{Z}}\to\mathbb{Q}/\mathbb{Z}$ is compact. But the only compact subgroups of $\mathbb{Q}/\mathbb{Z}$ are finite subgroups, so this means the image of $f$ is finite. There is then some $n$ such that $nf(x)=0$ for all $x$, so $f$ factors through the quotient $\hat{\mathbb{Z}}\to\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\cong\mathbb{Z}/n\mathbb{Z}$.

Thus every continuous homomorphism $\hat{\mathbb{Z}}\to\mathbb{Q}/\mathbb{Z}$ factors through one of the finite quotients $\mathbb{Z}/n\mathbb{Z}$ of $\hat{\mathbb{Z}}$. This means the functor $\operatorname{Hom}_c (-, \mathbb{Q}/\mathbb{Z})$ (consisting of only continuous homomorphisms) actually does preserve the limit in question. So the group of continuous homomorphisms $\hat{\mathbb{Z}}\to\mathbb{Q}/\mathbb{Z}$ is the colimit of the dual groups $(\mathbb{Z}/n\mathbb{Z})^D\cong\mathbb{Z}/n\mathbb{Z}$, and this colimit is $\mathbb{Q}/\mathbb{Z}$.

Answer 2

That's basically all there is to the "right and easy" proof for this.

$$\hom\left(\Bbb Q/\Bbb Z,S^1\right)=\hom\left(\varinjlim_n \Bbb Z/n, S^1\right).$$

By definition of categorical duals you get this to be

$$\varprojlim_n\hom\left(\Bbb Z/n, S^1\right)=\varprojlim_n\Bbb Z/n=\widehat{\Bbb Z}$$

since finite groups are self-dual. You can bootstrap from this and the fact that the double dual is canonically isomorphic to the original group and you can show the Pontryjagin dual of $\widehat{\Bbb Z}$ is $\Bbb Q/\Bbb Z$. I always find this particularly pleasing since torsion free dualizes to connected and $\Bbb Q/\Bbb Z$ is a completely torsion group just as $\widehat{\Bbb Z}$ is totally disconnected.

This is basically what you wrote in fewer words, but it's definitely the "right" proof in some sense of the word: it appeals only to canonical categorical facts and is based on simple, straightforward classical facts from the basic theory.