Expanding Lattice Ordered Abelian Groups to Riesz Spaces

An abelian l-group G is the reduct of a Riesz space if and only if G is Riesz extendable.

Proof. All conditions are clearly necessary since G(b) is the set of all real multiples of b.

There is an l-group G with at least two different expanding families.

Proof. By the examples given first (to our knowledge) in [5], there is an l-group G with at least two Riesz space structures ρ ≠ ρ′. But if G(b),G′(b) are the corresponding expanding families, then G(b) ≠ G′(b) for some b. In fact, suppose G(b) = G′(b) for every b ∈ G⁺. Then ρ(q, b) = ρ′(q, b) for every q ∈ Q, and for every real r, ρ(r, b) is the unique element of G(b) such that ρ(q, b) < ρ(r, b) < ρ(q′, b) for every pair of rationals q < r < q′, and ρ′(r, b) is the unique element of G′(b) such that ρ(q, b) < ρ(r, b) < ρ(q′, b) for every pair of rationals q < r < q′. So ρ(r, b) = ρ′(r, b) for every real r and b ∈ G⁺, so ρ = ρ′. □

The categories of Riesz spaces and expanded l-groups are equivalent.

Proof. Given a Riesz space G we associate the expanded l-group (G, G(b)) where G(b) is the group of the real multiples of G. This association is functorial and is an equivalence by Theorem 3.1. □

Let G be an l-group such that G(b) is convex for every b. Then G is Archimedean.

Proof. Suppose for a contradiction that G is not Archimedean. Then there are b, ∊ ∈ G⁺ such that n∊ ≤ b for every n ∈ N. Now b ≤ b + ∊ ≤ 2b, so b + ∊ ∈ G(b) and ∊ ∈ G(b). But every nonzero element of G(b) dominates b, whereas ∊ dominates b. This contradiction concludes the proof. □

Let G be a non-Archimedean l-group embedded in a product of totally ordered l-groups and closed under finite variant. Then G admits either zero or more than one Riesz space structure.

Proof. Let G ⊆ Π_i∈IG_i, where each G_i is a totally ordered abelian group. Since G is closed under finite variant, for every i ∈I, G contains the vector u_i consisting of 1 in position i and 0 elsewhere. Let ρ be a Riesz space structure on G. Let r ∈ R⁺ and q, q′ ∈ Q such that 0 < q < r < q′. Then ρ(r, u_i) is between ρ(q, u_i) and ρ(q′, u_i), hence ρ(r, u_i) must have zero in all components different from i. Moreover, suppose any v ∈ G has v_i =0 for some i ∈ I. Then v is orthogonal to u_i (i.e. v Λ u_i = 0), and by definition of Riesz structure, ρ(v) Λ ρ(u_i) = 0. That is, ρ(v) has i-th coordinate equal to zero. By additivity, if v and w have the same i-th component, then ρ(r, v) and ρ(r, w) have the same i-th component. So, for every i ∈ I, there is a map ρ_i: G_i × R → G_i such that ρ_i(r, v_i) = ρ(r, v)_i, and ρ_i is a Riesz space structure on G_i. In other words, all the Riesz space structures on G are products of Riesz space structures on G_i.

Now suppose G has a Riesz space structure ρ. By the argument above, every G_i has a Riesz space structure ρ_i. Since G is non-Archimedean, some G_i must be non-Archimedean. Any such G_i must have another Riesz space structure ρi′. Now, let ρ′: R × G → G be the map such that ρ′(r, v) = w if and only if w_i = ρi′(r, v_i) and w_j = ρ_i(r, v_j) for every j ≠ i. ρ′ is well defined because ρ′(r, v) is a finite variant of ρ(r, v), and is a Riesz space structure on G. Since ρ_i ≠ ρi′, we conclude ρ ≠ ρ′. □

There is a totally ordered abelian group G with a strong unit u, such that G has two non-isomorphic Riesz space structures ρ₁and ρ₂.

Proof. Like in [9], the idea is to build a group with two “asymmetric” Riesz structures.

Let R^a be the field of the real algebraic numbers. R and R^a are real closed fields, so they are elementarily equivalent. By Frayne’s Theorem there is an embedding j₁: R → *R^a, where *R^a = (R^a)^I/U is an ultrapower of R^a (so I is a set and U is an ultrafilter over I).

Let j₂ : R → R^I /U be the diagonal embedding. So the field K = R^I/U has two natural Riesz space structures ρ₁, ρ₂, where ρ₁(r, x) = j₁(r)x and ρ₂(r, x) = j₂(r)x.

Let K₀ be the set of finite sums Σ_ij₁(r_i)j₂(s_i) where r_i, s_i ∈ R. K₀ is a Riesz subspace of K in both Riesz structures ρ₁ and ρ₂, and has a strong unit j₁(1) (note that j₁(1) = j₂(1)).

Note that K₀ has the cardinality of the continuum, so K₀ is included in at most 2^ℵ₀ Archimedean classes (to our knowledge the exact number of Archimedean classes of K₀ is not known, note that K₀ is defined in an indirect way by an ultraproduct construction).

The idea is to consider certain sequences of elements of K₀ indexed by a regular cardinal Λ sufficiently large. More precisely, we fix two regular cardinals η, Λ such that 2^ℵ₀ < η < Λ.

Note that any two elements of K₀ have Archimedean distance less than η.

Let us equip the group K0Λ with the lexicographic ordering. That is, we let g < h if and only if the first nonzero component of h – g is positive. In this way K0Λ is a totally ordered abelian group.

Let G ⊆ K0Λ be the set of all sequences g ∈ K0Λ such that for every α < Λ, G(α) can be written, in the vector space (K₀, ρ₁), as a real linear combination of some finite set F ⊆ K₀ independent of α. In other words, the range of g has finite dimension in (K₀, ρ₁). In symbols,

where r_i,α ∈ R, F is finite, and k_i ∈ K₀.

Note that G inherits from K0Λ (and from K) the two vector space structures above, which we will still call ρ₁ and ρ₂.

An example of strong unit of G is simply u = (j₁(1), 0, 0,...). Note that in order to have a strong unit, we do not need the condition (present in [9]) that the components of the elements of G are bounded.

Since G is totally ordered, the absolute value of an element g of G, written |g|, is simply g if g ≥ 0, and –g if g < 0; and the Archimedean classes (id est equidominance classes) of G are also totally ordered in the natural way.

Let us call Archimedean distance between g, h ∈ G the number, possiby infinite, of Archimedean classes between g and h.

More simply than [9], we let Γ = j₁(R)^Λ. Similarly to [9] we have:

Let γ_ij the Λ-sequence such that γ_ij(α) = j₁(r_i,α)j₁(r_ij). Then γ_ij ∈ Γ. Moreover

and, letting α range over Λ, we have

that is, g is a linear combination of Γ in the vector space (G, ρ₂).

The following corollary, instead, is new.

Let g_n be a sequence of elements of G. A weak sum of g_n, if it exists, is an element s ∈ G such that, for every n, s – g₁ – g₂ – ··· g_n is dominated by g_n at a distance at least η. Note that a weak sum is not necessarily unique, because the components α < Λ of s beyond the first nonzero components of all g_n are not specified (and such components exist because Λ is an uncountable regular cardinal).

We have the following key lemma.

Like in [9] we say that an enriched Riesz space is a triple (G, ρ, B), where (G, ρ) is a Riesz space and B is a subset of G. An isomorphism of enriched Riesz spaces (G, ρ, B) and (G′,ρ′, B′) is an isomorphism between the Riesz spaces (G, ρ) and (G′, ρ′) which sends B bijectively onto B′.

Now suppose by contradiction that (G, ρ₁) is isomorphic to (G, ρ₂). Then for some subset △ of G, the enriched Ries space (G, ρ₂, Γ) is isomorphic to (G, ρ₁, △). So △ must satisfy Corollary 6.1 and Lemma 6.2 (up to replacing ρ₂ with ρ₁). Let us choose a sequence (t_n) of real transcendental numbers linearly independent over the subfield R^a of R. Then already in [9] it was observed:

The idea of the following lemma (which gives the main construction) is to use the sequence j₂(t_n) and define a sequence δ_n of positive elements of △ such that Lemma 6.2 may be applied to δ_n.

We denote by nη the ordinal η + η + ··· + η (a sum with n occurrences of η).

The inductive step n + 1 is as follows. We have △_i, δ_i, k_i for 1 ≤ i ≤ n. By definition of G, the components of every element of △₁ ∪ ··· ∪ △_n have finite dimension in (K₀, ρ₁), whereas the sequence j₂(t_n) has infinite dimension. So we can find a number k_n+1 ∈ N so high that j₂(t_kn+1) is not generated by △₁ ∪ △₂ ∪ ··· ∪ △_n.

Let f_n+1 be the quasiconstant element of G associated to k_n+1. By Corollary 6.1, f_n+1 is generated in (G, ρ₁) by a finite set △_n+1 of positive elements of △ with distance less than η from f_n+1. In particular, f_n+1((n + 1)η) = j₂(t_kn+1) is generated in (K₀, ρ₁) by the elements δ((n + 1)η) with δ ∈ △_n+1. So, by definition of k_n+1, there must be δ_n+1 ∈ △_n+1 such that δ_n+1((n + 1)η) is also linearly independent in (K₀, ρ₁) from the components of △₁ ∪ △₂ ∪ ··· ∪ △_n.

The inductive construction is thus completed. □

Now by Lemma 6.2 the positive sequence δ_n ∈ △ constructed in the previous lemma admits a weak sum s ∈ G. By definition of weak sum, for every n ∈ N, we have

By construction, for every n ∈ N, δ_n(nη) is linearly independent in (K₀, ρ₁) from all components of the sequences δ₁,...,δ_n–1. So also s(nη) is linearly independent in (K₀, ρ₁) from s(η),..., s(n – 1η). Summing up, we conclude that the range of s has infinite dimension in (K₀, ρ₁), so s ∉ G, a contradiction. □

An MV-algebra A is the reduct of a Riesz MV-algebra if and only if the abelian l-group image Ξ(A) of A in the Mundici equivalence (Γ, Ξ) is the reduct of a Riesz space.

Proof. Suppose A is the reduct of a Riesz MV-algebra R. Then by [7], Ξ(A) is the l-group reduct of the Riesz space Ξ′(R).

Conversely, if the l-group Ξ(A) is the reduct of a Riesz space S = Ξ′(R), then A is the reduct of the Riesz MV-algebra R. □

An MV-algebra A is the reduct of a Riesz MV-algebra if and only if A is Riesz-extendable.

Proof. The conditions are necessary since R(a) is the set of ra for r ∈ [0, 1].

Conversely, suppose all conditions are met. Let r ∈ [0, 1] and a ∈ A. If a = 0 then we let ra = 0. If a > 0 then we let ra be the image of r in the unique difference isomorphism from [0, 1] to R(a). This is a Riesz MV-algebra structure on A. □

There is a totally ordered MV-algebra with two non-isomorphic Riesz MV-algebra structures.

Proof. Let (G, u) be a totally ordered abelian group with a strong unit u such that G has two non-isomorphic Riesz space structures (such a group exists by Theorem 6.1). Consider the MV-algebra A = Γ_M(G, u), where Γ_M is the Mundici functor of [11]. So the universe of A is the set {x ∈ G|0 ≤ x ≤ u} and the MV-algebra operations are x ⊕ y = min(x + y, u) and ¬x = u – x.

We note (see [7: Theorem 3]) that every Riesz space structure on G gives a Riesz MV-algebra structure on A. Actually, in [7] there is an equivalence Γ_DL between the category of Riesz spaces with strong unit and Riesz MV-algebras, which coincides with Γ_M when restricted to the MV-algebra reducts of Riesz MV-algebras and the abelian l-group reducts of Riesz spaces.

So, the structures ρ₁ and ρ₂ on A cannot be isomorphic, otherwise by the functor Γ_DL we should have an isomorphism between (G, ρ₁) and (G, ρ₂), contrary to Theorem 6.1. □