# Classification of compact connected abelian groups

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.

• 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. – YCor Nov 13 '18 at 10:23
• 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. – YCor Nov 13 '18 at 10:24
• @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? – Rick Sternbach Nov 13 '18 at 10:29
• 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. – Todd Trimble Nov 13 '18 at 12:05
• @ToddTrimble Thanks, it is indeed the corollary 4 on page 99 of this nice little book. – Rick Sternbach Nov 13 '18 at 12:26

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.

• 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 $\{0\}$, $\mathbf{Z}$, $\mathbf{Z}[1/p]$, $\mathbf{Z}^2$, $\mathbf{Z}\times\mathbf{Z}[1/p]$, $\mathbf{Z}[1/p]^2$). – YCor Nov 13 '18 at 22:30
• 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. – YCor Nov 14 '18 at 22:43
• 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. – YCor Nov 14 '18 at 23:01
• The proofs above have one or two mistakes in them; hopefully I'll get around to fixing them soon. – Qiaochu Yuan Nov 24 '18 at 4:19