Theorem.

Let G be a separable abelian group with a bounded invariant metric d. Then d extends to a (bounded by the same constant) metric D on G⊕C, where C is a cyclic group which is dense in G⊕C. In particular, G embeds into a monothetic metric group.

Proof.

Instead of the metric, we shall work with the corresponding norm, i.e. the distance of an element from the group zero, which we shall still denote d, or D. That is, d⁢(g) denotes d⁢(g,0). Without loss of generality, suppose that d is bounded by 1. Let H={hn:n∈ℕ} be a countable dense subgroup of G and let π:ℕ→ℕ2 be some bijection; we denote by π⁢(n)⁢(1), resp. π⁢(n)⁢(2), the respective coordinates of π⁢(n). Suppose the cyclic group C is generated by some c. By induction, we shall construct a partial norm D′ on H⊕C which extends d, i.e. a partial function satisfying D′⁢(g)=0 if and only if g=0, D′⁢(g)=D′⁢(-g) and D′⁢(g1+…+gn)≤D′⁢(g1)+…+D′⁢(gn), whenever the corresponding elements are in the domain of D′. Moreover, we shall produce a strictly increasing sequence (kn)n such that D′⁢(ckn-hπ⁢(n)⁢(1))≤1/π⁢(n)⁢(2). At the end, we may define D by D⁢(x)=min⁡{1,inf⁡{∑i=1mD′⁢(xi):x=∑i=1mxi,(xi)i⊆dom⁢(D′)}}, for all x∈H⊕C. Since D′ was a partial norm, D extends D′, thus it extends d, and by the induction we will have guaranteed that C is dense in H⊕C.

At the first step of the induction, we set D′ to be equal to d on H, then we set k1=1 and define D′⁢(c-hπ⁢(1)⁢(1))=D′⁢(hπ⁢(1)⁢(1)-c)=1/π⁢(1)⁢(2). It follows that D′ is clearly a partial norm. Suppose we have done the first n-1 steps, found k1,…,kn-1 and defined D′ appropriately so that it is a partial norm. Let δ=min⁡{1/π⁢(i)⁢(2):i<n}; let kn>kn-1 be arbitrary satisfying kn>kn-1/δ. Then we set D′⁢(ckn-hπ⁢(n)⁢(1))=D′⁢(hπ⁢(n)⁢(1)-ckn)=1/π⁢(n)⁢(2). We claim that D′ is still a partial norm. Suppose on the contrary that the triangle inequality is broken, i.e. there are x,x1,…,xn∈dom⁢(D′) such that x=x1+…+xn and D′⁢(x)>D′⁢(x1)+…+D′⁢(xn). We shall suppose that x∈H; that is the most important case, and the other case is treated analogously. For any z∈H⊕C, denote by k⁢(z) the unique integer such that z can be written as h⊕ck⁢(z). Since k⁢(x)=0, we must have ∑i=1nk⁢(xi)=0. For at least one i≤n we must have that k⁢(xi) is kn or -kn, since before the extension at the n-th step, D′ was a partial norm. Also, since H is abelian, we may suppose that for no i,j≤n we have k⁢(xi)=-k⁢(xj)≠0, since in that case xi+xj=0 and we may remove xi and xj from the decomposition of x. Indeed, note that by the inductive construction for each i≤n there is a unique element u in the domain of D′ such that k⁢(u)=ki, and a unique element v such that k⁢(v)=-kn; and obviously v=-u. Let I={i≤n:0≠k⁢(xi)≠kn}. It follows that |∑i∈Ik(xi)|≥kn, so by definition of kn we have |I|>1/δ, so ∑i∈ID′⁢(xi)>1, a contradiction. ∎

Corollary (Morris, Pestov).

Every separable abelian topological group G embeds into a monothetic group.

Proof.

Let H be a countable dense subgroup of G and let (ρα)α<κ be a collection of pseudonorms bounded by 1 which give the topology of G, and H. By the previous proof we may extend each ρα on H⊕C, where C is cyclic, using the same numbers (kn)n and π:ℕ→ℕ2. By either completing H⊕C with respect to these pseudonorms or extending the pseudonorms on G⊕C by amalgamation, we obtain a monothetic group to which G embeds. ∎

Problem.

Prove a metric version of the Higman–Neumann–Neumann theorem. Either for separable groups with general left-invariant metrics, or for groups with bi-invariant metric.