The variety generated by completions of representable relation algebras

Maddux recently defined the variety V generated by the completions of representable relation algebras. In this note, we observe that V is canonical, answering Maddux’s problem 1.1(3), and show that the variety of representable relation algebras is not finitely axiomatisable over V.


Introduction
Last year, Roger Maddux defined a completely new variety of relation algebras! It is generated by the completions of representable relation algebras.
It has attracted some attention.
In this talk, we show that the representable relation algebras are not finitely axiomatisable within this new variety.

Plan
• Unary relations -boolean algebras BA is a nice class: • elementary, defined by equations -i.e., a variety • finitely axiomatisable by a few explicit equations.
So we feel we understand algebras of unary relations fairly well.

Binary relations -representable relation algebras
A binary relation on a set U is a subset of U × U . Let where for a, b ⊆ U × U ,
So the nice simple picture (BA) for unary relations is not replicated.

Completions of relation algebras
This is a partial order on B (corresponds to ⊆).
We say that B is complete if meets and joins (infs and sups) wrt. ≤ of all subsets of B exist in B.

Monk (1970):
Every relation algebra A has a completion: an algebra A c such that: • the boolean reduct of A c is a complete boolean algebra.
A c is unique up to isomorphism over A.
Monk showed that A c is also a relation algebra.

Maddux (2018):
Let V be the variety generated by RRA c . That is, V = HSP RRA c . Clearly, Maddux gave new examples of algebras separating V from RRA. So the gap between RRA and V is substantial.
Maybe V is quite close to RRA after all?
No: we'll sketch a proof that RRA is not finitely axiomatisable over V .
Idea of proof: find a sequence of algebras in V \ RRA having an ultraproduct in RRA.
We do it using relation algebras built from graphs.

Relation algebras from graphs
• one of x, y, z is 1 , and the other two are equal, or • {x, y, z} ⊆ V × 3, and {x, y, z} is not independent in G × 3.
We define an algebra where for a, b ⊆ A: a = a, a ; b = {z ∈ A : (x, y, z) is consistent for some x ∈ a, y ∈ b}.

Chromatic number and representability of A(G)
Recall: the chromatic number χ(G) of a graph G = (V, E) is the least n such that V is the union of n independent sets, and ∞ if no such n.
Well known: χ(G) ≤ 2 iff G has no cycles of odd length.

A(G)
is always a relation algebra (the '3' in G × 3 is used here).

Proof sketch that
Then by algebra, we can suppose A(G) ⊆ Re(U ) for some set U . As G is infinite, U must be infinite.
Then in Re(U ) we have a 1 ∪ · · · ∪ a n = (U × U ) \ Id U . So for each i < j < ω, (x i , x j ) lies in some a k (1 ≤ k ≤ n).
By Ramsey's theorem, we can assume that k is constant.
But a k is independent, so (a k ; a k ) ∩ a k = ∅. Contradiction.
Key point: since χ(G) = ∞, for each i < 3 there is an ultrafilter µ i of A(G) such that • µ i contains no independent sets in G × 3.
Theorem. RRA is not finitely axiomatisable over V .
Proof. Below, k, n are integers with k ≥ 0 and n ≥ 1.
• Let K n be a complete graph with n nodes. So χ(K n ) = n.
• Let E n be a graph with χ(E n ) ≥ n and no cycles of length ≤ n. Erdős (1959) constructed finite examples.
Write +, for disjoint union of graphs. Let

Remark. These are infinite graphs with chromatic number ∞. So
A(G k n ) ∈ RRA for all k, n.
Now for each k ≥ 0 define non-principal ultraproducts (for some non-principal We saw A(G k n ) ∈ RRA for all k, n. RRA is elementary, so A k ∈ RRA. Hence, its completion A(G k ) is in Maddux's variety V = HSP RRA c .
(As promised, this shows RRA c ⊆ RRA.) We have A(G k ) ∈ V \ RRA for each k ≥ 0.
Now one final lot of ultraproducts: As before, B c ∼ = A(G). In particular, B ⊆ A(G) up to isomorphism.
Thus, an ultraproduct B of algebras A(G k ) in V \ RRA lies in RRA. By Łoś's theorem, RRA is not finitely axiomatisable over V . I have deleted the short 'proof' of this theorem given in the talk, since it had a gap -I claimed without justification that V = HSPUp{S c : S ∈ RRA simple}. This is in fact true, but the proof is no simpler than [3].
See [3] for a proof of the theorem.
Andréka-Givant-Németi [2] extended: constructed infinitely many C. Again, they are apparently very far from RRA -not even in RA 5 . They are non-integral.