The dual group of the profinite integers

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?

• – Watson Nov 26 '18 at 10:36

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 $\mathbb{Q}/\mathbb{Z}$.

• Thank you very much! Indeed, the result is that "$\operatorname{Hom}_\mathbb{Z} (-,\mathbb{Q}/\mathbb{Z})$ gives an equivalence between the category of discrete torsion groups and the category of profinite groups", but I forgot that the morphisms in the latter category are assumed to be continuous. – Al F Jan 26 '17 at 20:37

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.

• There is one thing that bothers me: the contravariant $\operatorname{Hom}_\mathbb{Z} (-,\mathbb{Q}/\mathbb{Z})$ should convert colimits to limits, but $\hat{\mathbb{Z}}$ is a limit of $\mathbb{Z}/n\mathbb{Z}$, not a colimit. How do you pass from a limit to a colimit? – Al F Jan 26 '17 at 19:30
• I am probably missing something, but in general we have only $\operatorname{Hom} (\varinjlim_i X_i,Y) = \varprojlim_i \operatorname{Hom} (X_i,Y)$ and $\operatorname{Hom} (X,\varprojlim_i Y_i) = \varinjlim_i \operatorname{Hom} (X,Y_i)$. – Al F Jan 26 '17 at 19:34
• @AlF no, that's right, I had things backwards from so many years since taking the basics. You actually go through the proof you list, not the dual proof, then use the double dual fact I alluded to to prove the other fact. So in short, the proof you already have is basically what you want. My answer has been edited to reflect this. – Adam Hughes Jan 26 '17 at 19:41
• Thank you. It's indeed the same thing that I wrote above, and I was trying to figure out if the dual to $\hat{\mathbb{Z}}$ was possible to compute directly, without using that the double dual of a group is isomorphic to the same group. (I edited the question to clarify what I want.) – Al F Jan 26 '17 at 20:02
• @AlF if there is it will be much more complicated, as the standard proof of double dualization is stripped down to only (1) the basic fact of how hom works as a contravariant functor (when the second argument is fixed), (2) finite groups are self dual, and (3) the definition of the profinite completion of $\Bbb Z$. So you have a definition, a classical fact, and a basic theory result. Getting any simpler while proving something substantial is probably not possible based on experience. – Adam Hughes Jan 26 '17 at 20:07