It is known that torsion-free compact abelian groups are exactly the product of the maximal solenoid group $\Sigma_{(2,3,\cdots)}$ (which is the Pontryagin dual of the additive group $\mathbb{Q}$ of rational numbers equipped with the discrete topology) and the additive $p$-adic integers $\Delta_p$ (cf. 25.4 and 25.8 of Abstract Harmonic Analysis by Hewitt & Ross). This makes me wonder if a similar classification result exists for connected (instead of torsion free) compact abelian groups.

As in the torsion-free case, this reduces to the problem of classifying all discrete abelian groups for which the Pontryagin dual is connected, which seems a little intractable for me. Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$ (with products of $\mathbb{T}$ accounts for the torsion part and solenoids for the rest, if such an idea can be made precise)? In particular, is it true that any compact connected abelian group with a dense torsion subgroup a product of $\mathbb{T}$? More particularly, is it true that any compact connected abelian Lie group a finite product of $\mathbb{T}$? I think the last question has an affirmative answer (it is related to this question), but I don't know a rigorous proof.

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    $\begingroup$ For a discrete abelian group, having connected dual is equivalent to be torsion-free. The classification of torsion-free abelian groups is not easy but there's a large literature on it. $\endgroup$
    – YCor
    Nov 13, 2018 at 10:23
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    $\begingroup$ Yes any compact connected Lie group is a torus. By Pontryagin duality this is a restatement of the fact that any finitely generated torsion-free abelian group is free abelian. $\endgroup$
    – YCor
    Nov 13, 2018 at 10:24
  • $\begingroup$ @YCor Could you please point out a reference for the proof of the fact that a discrete abelian group has connected dual if and only if it is torsion-free? $\endgroup$ Nov 13, 2018 at 10:29
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    $\begingroup$ I'm fairly sure you'd find this in: Sidney A. Morris, Pontryagin Duality and the Structure of Locally Compact Abelian Groups, London Math. Soc. Lecture Notes 29, Cambridge U. Press, 1977. $\endgroup$
    – Todd Trimble
    Nov 13, 2018 at 12:05
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    $\begingroup$ @ToddTrimble Thanks, it is indeed the corollary 4 on page 99 of this nice little book. $\endgroup$ Nov 13, 2018 at 12:26

1 Answer 1


Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$

Following YCor's comments, the Pontryagin dual question is whether there exists a torsion-free abelian group which is not a direct sum of localizations of $\mathbb{Z}$. And the answer is yes: for example, you can consider the $p$-adic integers $\mathbb{Z}_p$ as a discrete group. Cf. the accepted answer at this MO question about simply presented abelian groups; YCor's answer is also educational and suggests simpler examples such as suitable subgroups of $\mathbb{Z}[1/p]^2$.

Edit: Let me see if I can flesh out YCor's claim about subgroups of $\mathbb{Z}[1/p]^2$. The interesting ones are the ones of rank $2$, and for simplicity I'll restrict my attention to subgroups containing $\mathbb{Z}^2$ (I thought I had an argument that every subgroup of rank $2$ is isomorphic to such a subgroup but now I'm not so sure). These correspond to subgroups of the quotient $\mathbb{Z}[1/p]^2/\mathbb{Z}^2 \cong \mu_{p^{\infty}}^2$, the product of two copies of the Prüfer $p$-group.

We can describe such subgroups using Goursat's lemma, which says in this case that if $H \subseteq \mu_{p^{\infty}}^2$ is a subgroup such that the two projections $\pi_1, \pi_2 : H \to \mu_{p^{\infty}}$ are surjective (this is the interesting case), then there exist subgroups $N_1, N_2 \subseteq \mu_{p^{\infty}}$ and an isomorphism $\varphi : \mu_{p^{\infty}}/N_1 \cong \mu_{p^{\infty}}/N_2$ such that

$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : \varphi(q_1) \equiv q_2 \bmod N_2 \}.$$

The proper subgroups of $\mu_{p^{\infty}}$ take the form $\mu_{p^n}$ for $n \in \mathbb{Z}_{\ge 0}$ (if we consider all of $\mu_{p^{\infty}}$ then $H$ is all of $\mu_{p^{\infty}}^2$). The quotient $\mu_{p^{\infty}}/\mu_{p^n}$ is isomorphic to $\mu_{p^{\infty}}$ again, and its automorphism group is the group of $p$-adic units $\mathbb{Z}_p^{\times}$, of which there are uncountably many.

So there are uncountably many choices for $H$, and hence uncountably many subgroups of $\mathbb{Z}[1/p]^2$ containing $\mathbb{Z}^2$. Of these, the subgroups isomorphic to $\mathbb{Z}^2, \mathbb{Z} \times \mathbb{Z}[1/p]$, or $\mathbb{Z}[1/p]^2$ are determined by the image of the copy of $\mathbb{Z}^2$ in each of them, and there are countably many choices for this image. So as promised, at most countably many subgroups can be isomorphic to a direct sum of localizations of $\mathbb{Z}$. Unfortunately I don't quite see how to explicitly exhibit a particular choice of $H$ which is not such a direct sum.

Edit #2: Let's take $N_1 = N_2 = 0$ to be trivial above, and $\varphi : \mu_{p^{\infty}} \cong \mu_{p^{\infty}}$ to be multiplication by a $p$-adic unit $u \in \mathbb{Z}_p^{\times}$. Then

$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : u q_1 = q_2 \}$$

is a subgroup of $\mu_{p^{\infty}}^2$ isomorphic to $\mu_{p^{\infty}}$, and it lifts to a subgroup

$$\widetilde{H} = \{ (q_1, q_2) \in \mathbb{Z}[1/p]^2 : u (q_1 \bmod 1) \equiv q_2 \bmod 1 \}.$$

Thinking of $H$ as a "line with slope $u$" suggests that $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ (in fact to $\mathbb{Z} \times \mathbb{Z}[1/p]$) iff $u$ is rational. Indeed, if $u = \frac{a}{b}$ is rational, so that $a, b$ are integers relatively prime to $p$ and each other, then the condition that $u (q_1 \bmod 1) \equiv q_2 \bmod 1$ is equivalent to the condition that $aq_1 - bq_2 \in \mathbb{Z}$, and consequently the map

$$\widetilde{H} \ni (q_1, q_2) \mapsto (q_1, aq_1 - bq_2) \in \mathbb{Z}[1/p] \times \mathbb{Z}$$

is an isomorphism.

Now suppose that $u$ is irrational. First let's show that $\widetilde{H}$ has no subgroup isomorphic to $\mathbb{Z}[1/p]$: equivalently, no nonzero element is $p$-divisible. If $(q_1, q_2) \in \widetilde{H}$ is a nonzero element, for it to be $p$-divisible would require that

$$u \left( \frac{q_1}{p^k} \bmod 1 \right) \equiv \frac{q_2}{p^k} \bmod 1$$

for all $k$, or equivalently that

$$u(q_1) - q_2 \in p^k \mathbb{Z}$$

for all $k$. Taking $k \to \infty$ gives $u = \frac{q_2}{q_1}$, but this contradicts $u$ irrational.

Hence in this case, if $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ then it can only be isomorphic to $\mathbb{Z}^2$. But the projection to either coordinate gives a surjection onto $\mathbb{Z}[1/p]$, which $\mathbb{Z}^2$ does not possess.

Some additional comments. Projection onto the first coordinate shows that $\widetilde{H}$ is an extension

$$0 \to \mathbb{Z} \to \widetilde{H} \to \mathbb{Z}[1/p] \to 0.$$

The argument above shows that this sequence splits iff $u$ is rational. In general, this extension is classified by a class in $\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z})$, which can be computed from $\text{Ext}^1(\mu_{p^{\infty}}, \mathbb{Z}) \cong \mathbb{Z}_p$ and the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Z}[1/p] \to \mu_{p^{\infty}} \to 0$ to give

$$\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \cong \mathbb{Z}_p/\mathbb{Z}.$$

The point of this computation is to show that for general homological reasons there are interesting nontrivial extensions of torsion-free abelian groups by torsion-free abelian groups; taking Pontryagin duals, there are interesting nontrivial extensions of compact connected abelian groups by compact connected abelian groups.

It should be possible to match up this $\mathbb{Z}_p$ with the $\mathbb{Z}_p$ that $u$ lives in above but I'm not entirely sure how.

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    $\begingroup$ Indeed, $\mathbf{Z}[1/p]^2$ has uncountably many subgroups, only countably of which are isomorphic to a product of rank-1 groups (which in this case means isomorphic to one of $\{0\}$, $\mathbf{Z}$, $\mathbf{Z}[1/p]$, $\mathbf{Z}^2$, $\mathbf{Z}\times\mathbf{Z}[1/p]$, $\mathbf{Z}[1/p]^2$). $\endgroup$
    – YCor
    Nov 13, 2018 at 22:30
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    $\begingroup$ To exhibit an example, one way is to use that the inclusion $\mathbf{Z}[1/p]\to\mathbf{Q}_p$ induces an isomorphism $\mathbf{Z}[1/p]/\mathbf{Z}\to\mathbf{Q}_p/\mathbf{Z}_p$. If $D$ is a line in $\mathbf{Q}_p$, it corresponds to the subgroup $u_D=(D+\mathbf{Z}_p^2)\cap\mathbf{Z}[1/p]^2$. Then $u_D$ is isomorphic to a product of rank 1 groups if and only if $D$ is rational. Hence, choose $D$ irrational to have an explicit example. $\endgroup$
    – YCor
    Nov 14, 2018 at 22:43
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    $\begingroup$ Also algebraic number theory provides examples. Let $x$ be an algebraic number of degree 2 over $\mathbf{Q}$, with conjugate $x'$, such that $x$ has negative $p$-valuation while $x'$ has zero $p$-valuation (so the minimal polynomial is split over $\mathbf{Q}_p$). Then the subring generated by $x$ has an additive structure with the required properties. $\endgroup$
    – YCor
    Nov 14, 2018 at 23:01
  • $\begingroup$ The proofs above have one or two mistakes in them; hopefully I'll get around to fixing them soon. $\endgroup$ Nov 24, 2018 at 4:19

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