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Let $\mathcal L$ be the set of all group topologies on $\Bbb Z$. It is known that $(\mathcal L,\subseteq)$ is a modular complete lattice [1].

Is $(\mathcal L,\subseteq)$ distributive?

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[1] Lamper, Milan. Complements in the lattice of all topologies of topological groups. Arch. Math. (Brno) 10 (1974), no. 4, 221--230 (1975).

Minimus Heximus
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  • What is a group topology on Z? Can it be isomorphic to a group topology on Q? On Z_p? – The Masked Avenger Sep 12 '14 at 15:42
  • A group topology on a group $G$ is a topology $\mathcal T$ on the set $G$ with which $(G,\mathcal T)$ is a topological group, that is, the function $(x,y)\mapsto xy^{-1}$ is continuous. – Minimus Heximus Sep 12 '14 at 15:45
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    Note that $(\mathcal{L}, \subseteq)$ is distributive if and only if the lattice $M_3$ ( http://en.wikipedia.org/wiki/Distributive_lattice#mediaviewer/File:M3_1xyz0.svg ) cannot be embedded in $(\mathcal{L}, \subseteq)$. So one idea might be to find three group topologies $\tau_1,\tau_2,\tau_3$ with identical infimum and identical supremum, or show that this cannot be done. – Dominic van der Zypen Sep 17 '14 at 07:41
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    There is also this result of Lukacs-Palfy: a group $G$ is abelian iff the subgroups lattice of $G \times G$ is modular (see a proof here). – Sebastien Palcoux Sep 25 '14 at 15:37
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    "something in the structure of $\mathbb{Z}$ other than abelianness, must be used to prove distributivity": Yes, perhaps the distributivity of its subgroups lattice. More generally (Ore's theorem) a group is locally cyclic iff its subgroups lattice is distributive (see a proof here) – Sebastien Palcoux Sep 26 '14 at 16:23
  • Where does the question come from? – Andreas Thom Nov 08 '14 at 20:48
  • @AndreasThom: I am working on some generalization of group topologies which I call f-subgroups. the lattice of f-subgoups is not distributive on any non-locally-cyclic group. This is a counterexample for locally cyclic case. I can investigate more deeply which groups allow distributive lattices. – Minimus Heximus Nov 08 '14 at 21:01

1 Answers1

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For each $\alpha \in S^1$, the map $\varphi_{\alpha} \colon \mathbb Z \to S^1$, given by $n \mapsto \alpha^n$, induces a topology $\tau_{\alpha}$ on $\mathbb Z$.

A basis of neighborhoods of $0$ for $\tau_{\alpha}$ is given by the sets $$U_{n,\alpha} := \left\{k \in \mathbb Z \mid |\alpha^k-1| < \frac1n \right\}, \quad n \in \mathbb N.$$

I denote by $\tau_{\alpha} \wedge \tau_{\beta}$ the largest group topology that is contained in $\tau_{\alpha} \cap \tau_{\beta}$ - which is in general different from $\tau_{\alpha} \cap \tau_{\beta}$.

Claim 1: If $\alpha,\beta \in S^1$ are irrational and such that $\alpha/\beta \in S^1$ is also an irrational angle, then $\tau_{\alpha} \wedge \tau_{\beta}$ is trivial.

Proof: Let $U \in \tau_{\alpha} \wedge \tau_{\beta} \subset \tau_{\alpha} \cap \tau_{\beta}$ and $0 \in U$. There exists some $V \in \tau_{\alpha} \wedge \tau_{\beta}$ such that $V-V \subset U$ and $0 \in V$. Then, there exists $n$, such that $U_{n,\alpha} - U_{n,\beta} \subset V - V \subset U$. However, $\mathbb Z=U_{n,\alpha} - U_{n,\beta}$. Indeed, for any $k \in \mathbb Z$, there exists some $m$, such that $|\alpha^m-1|<\frac1n$ and $|\beta^{k+m}-1|=|\beta^{m} - \beta^{-k}|<\frac1n$, because $(\alpha,\beta)$ generates a dense subgroup of $S^1 \times S^1$ and thus can approximate the point $(1,\beta^{-k})$ arbitrarily well. q.e.d.

Claim 2: $\tau_{\alpha} \vee \tau_{\beta}$ is the topology induced from the map $\mathbb Z \to S^1 \times S^1$, $n \mapsto (\alpha^n,\beta^n)$.

In particular, applying various automorphisms of $S^1 \times S^1$, we see that $$\tau_{\alpha} \vee \tau_{\beta} = \tau_{\alpha} \vee \tau_{\alpha\beta} = \tau_{\beta} \vee \tau_{\alpha\beta}.$$

Let $\alpha,\beta \in S^1$ be irrational angles, such that $\alpha/\beta$ is also irrational and consider the topologies $\tau_{\alpha},\tau_{\beta},\tau_{\alpha \beta}$. Any pairwise meet (in the lattice of group topologies) is the trivial topology (by Claim 1) and any pairwise join yields the same topology (by Claim 2). Hence,

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embeds as a sub-lattice in the lattice of group topologies. Hence, the lattice of group topologies cannot be distributive.

Andreas Thom
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  • By $V+V \subset U$ do you mean $V-V \subset U$? By an irrational angle $\alpha$ do you mean the angle between positive reals and the line passing through point $\alpha$ and 0 is irrational? – Minimus Heximus Nov 08 '14 at 20:08
  • I corrected $V+V$ to $V-V$. An angle $\alpha \in S^1$ is irrational, if $\alpha = \exp(2\pi i t)$, for some irrational real number $t$. – Andreas Thom Nov 08 '14 at 20:10
  • Thank you. I'll examine the answer more carefully and I'll award the bounty after it disappeared from featured questions. – Minimus Heximus Nov 08 '14 at 20:13
  • It seems I can prove your last double equations directly but I do not understand your proof using Claim2. Will you explain more? – Minimus Heximus Nov 10 '14 at 09:53
  • Yes, there are many proofs and it is elementary to see it directly. Using the automorphisms (for example $(z,w) \mapsto (z,zw)$) of $S^1 \times S^1$ just seemed more natural to me. Claim 2 is just the observation that the join of topologies is related to the diagonal embedding into the the product of topological spaces - again almost a triviality. – Andreas Thom Nov 10 '14 at 14:12
  • Let $\phi:\Bbb T^2\to \Bbb T^2$ be an automorphism and $\theta(k)=(\alpha^k,\beta^k)$. Do you say $\theta \circ \phi$ and $\theta$ induce equal topologies on $\Bbb Z$? – Minimus Heximus Nov 10 '14 at 14:24
  • Yes, isn't this obvious if $\phi$ is continuous. – Andreas Thom Nov 10 '14 at 18:38
  • All I can deduce, at first glance, is that the topologies are homeomorphic but I'm not sure about equality! – Minimus Heximus Nov 10 '14 at 18:47
  • Well, $U \subset \mathbb Z$ is open in the topology induced by $\theta$, if and only if there exists an open subset $V \subset \mathbb T^2$, such that $V \cap \theta(\mathbb Z)=\theta(U)$. But $V$ is open if and only of $\phi(V)$ is open and $\phi(\theta(U)) = \phi(V) \cap \phi(\theta(\mathbb Z))$. Thus, $U$ is also open in the topology induced by $\phi \circ \theta$. Am I missing something? – Andreas Thom Nov 10 '14 at 19:27