## Introduction

In a recent paper [9, 1.10], Maddux defined the variety $$V=\mathbf{H}\mathbf{S}\mathbf{P}\,\mathsf{RRA}^c$$, where $$\mathsf{RRA}$$ is the variety of representable relation algebras and $$\mathsf{RRA}^c=\{\mathfrak {A}^c:\mathfrak {A}\in \mathsf{RRA}\}$$. Here, $$\mathfrak {A}^c$$ denotes the completion of the relation algebra $$\mathfrak {A}$$. For details of these notions and further ones used below, see the very attractive introduction to [9].

In [9, problem 1.1(3)], Maddux asked whether V is closed under canonical extensions.

### Theorem 1.1

V and $$\mathbf{S}\,\mathsf{RRA}^c$$ are closed under canonical extensions.

### Proof

By [4, theorem 3.8], if $$\mathsf K$$ is a class of relation algebras that is closed under ultraproducts, and $$\mathsf{K}^c=\{\mathfrak {A}^c:\mathfrak {A}\in \mathsf{K}\}$$, then $$\mathbf{S}\,\mathsf{K}^c$$ and $$\mathbf{H}\mathbf{S}\mathbf{P}\,\mathsf{K}^c$$ are both closed under canonical extensions. Theorem 1.1 follows by taking $$\mathsf{K}=\mathsf{RRA}$$, which is a variety and so closed under ultraproducts. $$\square$$

RA denotes the class of all relation algebras. Using the known facts that RA is closed under completions [10] but RRA is not [7], Maddux noted that

\begin{aligned} \mathsf{RRA}\subsetneq V\subseteq \mathsf{RA}, \end{aligned}

and he showed that V contains a number of non-representable ‘Monk algebras’, so that the gap between $$\mathsf{RRA}$$ and V is substantial. In [9, problem 1.1(1)], he asked whether $$V=\mathsf{RA}$$. This was answered negatively by Andréka and Németi [1], where it is shown that in fact there are continuum-many varieties lying between V and $$\mathsf{RA}$$. That might suggest that V is ‘nearer’ to $$\mathsf{RRA}$$ than to $$\mathsf{RA}$$, but as ‘evidence’ in the other direction, we show below that $$\mathsf{RRA}$$ is not finitely axiomatisable over V.

## $$\mathsf{RRA}$$ is not finitely axiomatisable over V

It suffices to show that RRA contains an ultraproduct of algebras in $$V{\setminus }\mathsf{RRA}$$. To this end, we use a construction from [6] of relation algebras from graphs.

### Graphs

Graphs here are undirected and loop-free. Let G be a graph whose set of nodes is N, say. Recall that a cycle of length$$l\ge 3$$inG is a subset $$\{v_0,\dots {},v_{l-1}\}\subseteq N$$ of size l such that $$(v_i,v_{(i+1)\bmod l})$$ is an edge of G for each $$i<l$$. A subset $$X\subseteq N$$ is said to be independent if no pair of nodes in X is an edge of G. The chromatic number$$\chi (G)$$ of G is the least natural number n such that N is the union of n (possibly empty) independent sets, and $$\infty$$ if there is no such n. It is well known (see, e.g., [2, 1.6.1]) that $$\chi (G)\le 2$$ iff G has no cycles of odd length. We let $$+$$ and $$\sum$$ denote disjoint union of graphs. Then if $$G_i$$$$(i\in I)$$ are graphs, $$\chi (\sum _{i\in I}G_i)$$ is the least upper bound (possibly $$\infty$$) of $$\{\chi (G_i):i\in I\}$$.

Let n be a positive integer. Let $$K_n$$ be a complete graph with precisely n nodes. Clearly, $$\chi (K_n)=n$$. Let $$E_n$$ be a graph with $$\chi (E_n)\ge n$$ and with no cycles of length at most n—finite examples were constructed by Erdős [3].

We use these graphs to construct some infinite graphs ($$G^k_n$$, $$G^k$$, $$G^\omega$$), and compute their chromatic numbers. First define

\begin{aligned} G^0_n&= \sum _{m\ge n}E_m, \end{aligned}
(2.1)
\begin{aligned} G^k_n&= G^0_n+K_k \quad \text{ if } 0<k<\omega . \end{aligned}
(2.2)

Since $$\{\chi (E_m):m\ge n\}$$ is unbounded, each $$G^k_n$$ has chromatic number $$\infty$$.

Next, fix a non-principal ultrafilter D over $$\omega {\setminus }1$$. Define the ultraproduct

\begin{aligned} G^k=\prod _D \{G^k_n:0<n<\omega \}, \quad \text{ for } \text{ each } k<\omega . \end{aligned}
(2.3)

First consider the case when $$k=0$$. Observe that $$G^0_n$$ in (2.1) has no cycles of length$${}\le n$$. Now for each $$l\ge 3$$, the property of having no cycles of length l can be expressed by a first-order sentence, and is true for all but finitely many $$G^0_n$$. So by Łoś’s theorem, $$G^0$$ in (2.3) has no cycles of length l. This holds for each l, so in fact $$G^0$$ has no cycles at all, and hence $$\chi (G^0)\le 2$$.

What about $$\chi (G^k)$$ for $$k>0$$? By standard ultraproduct considerations,

\begin{aligned} G^k\cong G^0+K_k, \end{aligned}

so we further obtain $$\chi (G^k)=\max (\chi (G^0),\chi (K_k))\le \max (2,k)<\infty$$.

Finally let

\begin{aligned} G^\omega =\prod _D\{G^k:0<k<\omega \}. \end{aligned}
(2.4)

For each $$m>0$$, $$K_m$$ embeds into $$G^k$$ for every $$k\ge m$$. It follows by Łoś’s theorem that $$K_m$$ embeds into $$G^\omega$$, so plainly, $$\chi (G^\omega )\ge m$$. This holds for every m, so $$\chi (G^\omega )=\infty$$.

### Relation algebras from graphs

Let G be an infinite graph with set of nodes N. We write $$N\times 3$$ for the set $$N\times \{0,1,2\}$$, and $$G\times 3$$ for the graph whose set of nodes is $$N\times 3$$ and where ((vi), (wj)) is an edge of $$G\times 3$$ iff $$i\ne j$$ or (vw) is an edge of G. In simple words, $$G\times 3$$ consists of three disjoint copies of G, with all possible edges added between the copies.

We now define a relation algebra atom structure isomorphic to one in [6, section 4] and [5, chapter 14]. We stipulate that $$A=(N\times 3)\cup \{1^{\mathrm{'}}\}$$, $$I=\{1^{\mathrm{'}}\}$$, $${\breve{x}}=x$$ for every $$x\in A$$, and for each $$x,y,z\in A$$, C(xyz) holds iff

1. (1)

one of xyz is $$1^{\mathrm{'}}$$ and the other two are equal, or

2. (2)

$$\{x,y,z\}\subseteq N\times 3$$ and $$\{x,y,z\}$$ is not independent (in the graph $$G\times 3$$).

One can check that for any graphs $$G_n$$$$(0<n<\omega )$$,

\begin{aligned} \alpha \left( \prod _DG_n \right) \cong \prod _D\alpha (G_n). \end{aligned}
(2.5)

We write $$\mathfrak {A}(G)$$ for the complex algebra of $$\alpha (G)$$ (see [9, 1.13]). By [8, lemma 6.2], $$\mathfrak {A}(G)$$ is a relation algebra; of course it is atomic and its atom structure is $$\alpha (G)$$. We will need the following fact about it, from [6, theorems 10–11] or [5, theorems 14.12–13] or (for $$\Leftarrow$$) [5, exercise 14.2(7)]. It uses our assumption that G is infinite.

### Fact 2.1

$$\mathfrak {A}(G)\in \mathsf{RRA}$$ if and only if $$\chi (G)=\infty$$.

### Why $$\mathsf{RRA}$$ is not finitely axiomatisable over V

Let $$k<\omega$$. For $$0<n<\omega$$, we saw that $$\chi (G^k_n)=\infty$$, so $$\mathfrak {A}(G^k_n)\in \mathsf{RRA}$$ by fact 2.1. Define the ultraproduct

\begin{aligned} \mathfrak {A}^k=\prod _D\{\mathfrak {A}(G^k_n):0<n<\omega \}. \end{aligned}
(2.6)

Then $$\mathfrak {A}^k\in \mathsf{RRA}$$, since $$\mathsf{RRA}$$ is a variety and closed under ultraproducts. So by definition of V,

\begin{aligned} \mathfrak {C}^k\,=(\mathfrak {A}^k)^c\in V. \end{aligned}

By Łoś’s theorem, $$\mathfrak {A}^k$$ is atomic. So (e.g., [5, remark 2.67]) its completion $$\mathfrak {C}^k$$ is isomorphic to the complex algebra of the atom structure of $$\mathfrak {A}^k$$. This atom structure is $$\prod _D\alpha (G^k_n)\cong \alpha (\prod _DG^k_n)=\alpha (G^k)$$ by (2.62.52.3). Hence, $$\mathfrak {C}^k\cong \mathfrak {A}(G^k)$$. We saw that $$\chi (G^k)<\infty$$, so by fact 2.1, $$\mathfrak {C}^k\notin \mathsf{RRA}$$.

Finally let $$\mathfrak {C}$$ be the ultraproduct $$\prod _D\{\mathfrak {C}^k:0<k<\omega \}$$. As before, this is an atomic relation algebra with atom structure isomorphic to $$\prod _D\alpha (G^k)\cong \alpha (\prod _DG^k)=\alpha (G^\omega )$$, so its completion $$\mathfrak {C}^c$$ is isomorphic to $$\mathfrak {A}(G^\omega )$$. But $$\chi (G^\omega )=\infty$$, so $$\mathfrak {A}(G^\omega )\in \mathsf{RRA}$$ by fact 2.1. Then $$\mathfrak {C}\subseteq \mathfrak {C}^c\in \mathsf{RRA}$$, and as RRA is closed under subalgebras, we obtain $$\mathfrak {C}\in \mathsf{RRA}$$.

We can now prove our main theorem.

### Theorem 2.2

$$\mathsf{RRA}$$ is not finitely axiomatisable over V.

### Proof

We have shown that $$\mathfrak {C}^k\in V{\setminus }\mathsf{RRA}$$ for $$k>0$$, and $$\mathfrak {C}=\prod _D\mathfrak {C}^k\in \mathsf{RRA}$$. It follows by Łoś’s theorem that $$\mathsf{RRA}$$ is not finitely axiomatisable over V. $$\square$$