Monk algebras, completions, and atom structures Robin Hirsch and Ian Hodkinson
Aim To show that relation algebras can shed light on duality-related matters. To nail more nails in the cof fi n of the hope that RRA is docile. But also perhaps, to show there is value in proving more ‘negative’ results about RRA . Outline 1. We de fi ne some relation algebras (like ‘Monk algebras’), based on graphs. 2. We use them to show that RRA is not closed under completions (which we will de fi ne). 3. If time, we use them to study the atom structures of atomic relation algebras. 1
0. Relation algebra atom structures, complex algebras (revision) Abstractly, a relation algebra atom structure is a structure S = ( S, Id, ˘ , C ) satisfying certain fi rst-order conditions (the ‘correspondents’ of Tarski’s RA axioms). If A is an atomic relation algebra then At A is such a structure. Conversely, given such an S , we can form its complex algebra: S + = ( ℘ ( S ) , ∪ , , ∅ , S, 1 , , ˘ , ; ) , where • 1 , = { x ∈ S : S | = Id ( x ) } • ˘ a = { ˘ x : x ∈ a } (for a ⊆ S ) • a ; b = { z ∈ S : ∃ x ∈ a ∃ y ∈ b ( S | = C ( x, y, z )) } (for a, b ⊆ S ). S + is an atomic relation algebra, and At S + ∼ = S . → (At A ) + . If A is an atomic relation algebra, A ֒ 2
1. Graphs and relation algebras Here, graphs are undirected: G = ( V, E ) where E ⊆ V × V is symmetric. They can have loops ( ( x, x ) ∈ E ), but they won’t in this talk. A subset X ⊆ V is independent if E ∩ ( X × X ) = ∅ . For k < ω , a k -colouring of G is a partition of V into ≤ k independent sets. The chromatic number χ ( G ) of G is the least k < ω such that G has a k -colouring, and ∞ if there is no such k . A cycle of length k in a graph G is a sequence v 1 , . . . , v k of distinct nodes of G , such that ( v 1 , v 2 ) , . . . , ( v k − 1 , v k ) , ( v k , v 1 ) are edges of G . Fact 1 χ ( G ) ≤ 2 iff G has no cycles of odd length. From now on, we write x ∈ G , X ⊆ G , instead of x ∈ V , X ⊆ V , etc. 3
Relation algebras from graphs Fix a graph G . We de fi ne a relation algebra A ( G ) . Its set of atoms is Q = { 1 , , r x , b x , g x : x ∈ G } — red, blue and green copies of nodes of G . A ( G ) is the full complex algebra over Q : A ( G ) = ( ℘ ( Q ) , ∪ , − , ∅ , Q, { 1 , } , ˘ , ; ) . We de fi ne ˘ S = S , for all S ⊆ Q . We de fi ne composition by listing the inconsistent triples of atoms — those ( a, b, c ) such that ( a ; b ) · c = 0 . They are: • (1 , , a, b ) , ( a, 1 , , b ) , and ( a, b, 1 , ) , whenever a � = b , • ( r x , r y , r z ) , ( b x , b y , b z ) , and ( g x , g y , g z ) , whenever { x, y, z } is independent in G . Fact: A ( G ) is a simple relation algebra (a kind of ‘Monk algebra’). 4
Independent elements of A ( G ) For X ⊆ G , de fi ne R X = { r x : x ∈ X } ∈ A ( G ) . De fi ne B X , G X similarly. An element a ∈ A ( G ) is said to be independent if a = C X for some C ∈ { R , B , G } and some independent X ⊆ G . Lemma 2 Let a ∈ A ( G ) . 1. If a is independent, then ( a ; a ) · a = 0 . 2. If a ≤ − 1 , is not independent, then a ; a = 1 . 5
Chromatic number and representability Let B ⊆ A ( G ) be a subalgebra. B is said to be balanced if • R G , B G , G G ∈ B , • for all X ⊆ G , R X ∈ B ⇐ ⇒ B X ∈ B ⇐ ⇒ G X ∈ B . The chromatic number χ ( B ) of a balanced B ⊆ A ( G ) is the least k < ω such that R G is the sum of k independent elements of B , and ∞ if there is no such k . Remark: χ ( G ) = χ ( A ( G )) ≤ χ ( B ) . Theorem 3 For infinite balanced B ⊆ A ( G ) , we have B ∈ RRA ⇐ ⇒ χ ( B ) = ∞ . 6
Proof: B is representable ⇒ χ ( B ) = ∞ B is simple. Suppose that h : B → Re ( U ) is a representation. Because B is in fi nite, so is U . Pick distinct x 0 , x 1 , . . . ∈ U . Assume for contradiction that χ ( B ) < ∞ . So − 1 , = � i ≤ n b i , for some independent elements b 1 , . . . , b n ∈ B . h respects − 1 ,. So for all i < j , we have ( x i , x j ) ∈ h ( � i ≤ n b i ) . h respects + . So there is b ij ∈ { b 1 , . . . , b n } with ( x i , x j ) ∈ h ( b ij ) . By Ramsey’s theorem, we can assume b ij is constant — say, b ij = b 7 for all i < j . Triangle consistency for ( x 0 , x 1 , x 2 ) gives ( b 01 ; b 12 ) · b 02 � = 0 . That is, ( b 7 ; b 7 ) · b 7 � = 0 . But b 7 is independent, so by lemma 2, ( b 7 ; b 7 ) · b 7 = 0 . Contradiction! 7
Proof: χ ( B ) = ∞ ⇒ B is representable Assume χ ( B ) = ∞ . Say b ∈ B is R -big if R G − b is the sum of fi nitely many independent elements of B . The set of R -big elements of B has the fi nite intersection property. So (using some form of AC) it extends to an ultra fi lter R µ of B . R G ∈ R µ . If a ∈ B is independent, − a is R -big, so − a ∈ R µ , so a / ∈ R µ . So no element of R µ is independent. Hence, by lemma 2, r ; r ′ = 1 for any r, r ′ ∈ R µ . Choose ultra fi lters B µ , G µ similarly. We now show that ∃ can use these ultra fi lters to win the game G u ω ( B ) of length ω played on ultra fi lter networks over B . So B ∈ RRA . 8
✬ ✩ ∀ picks x, y ∈ N t and r, s ∈ B with ③ r ; s ∈ N t ( x, y ) . x N t ③ We assume no existing witness in N t . N t ( x, y ) ❄ ③ ③ ❄ y ✫ ✪ 9
∃ chooses ultra fi lters N t +1 ( x, z ) ∋ r and N t +1 ( z, y ) ∋ s ✬ ✩ with N t +1 ( x, z ) ; N t +1 ( z, y ) ⊆ N t ( x, y ) . ③ x N t ③ PPPPPPP N t +1 ( x, z ) ∋ r q P PPPPPPP ③ N t ( x, y ) ❄ z q ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ③ ✏ ✏ ③ ✏ N t +1 ( z, y ) ∋ s ✏ ✏ ✏ ✏ ✏ ✮ ❄ y ✫ ✪ 10
R G + B G + G G ∈ N t +1 ( x, z ) . Say, R G ∈ N t +1 ( x, z ) . ✬ ✩ Similarly, say B G ∈ N t +1 ( z, y ) . ③ x N t ③ PPPPPPP N t +1 ( x, z ) ∋ r q P PPPPPPP ③ N t ( x, y ) ❄ z q ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ③ ✏ ✏ ③ ✏ N t +1 ( z, y ) ∋ s ✏ ✏ ✏ ✏ ✏ ✮ ❄ y ✫ ✪ 11
✬ ✩ ③ x N t ③ PPPPPPP ❤ ❤ ❤ ❤ N t +1 ( x, z ) ∋ r ❤ ❤ ❤ ❤ ❤ ❤ G µ ❤ q P ❤ ❤ PPPPPPP ❤ ❤ ③ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❄ ❤ z q ✭ ✭ ✏ ✭ ✭ ✏ ✭ ✭ ✏ ✭ ✭ ✏ ✭ ✭ ✏ ✭ ✭ ✏ ✭ G µ ✭ ✏ ✮ ✏ ✭ ③ ✏ ✭ ✭ ✏ ✭ ✭ ③ ✏ ✭ N t +1 ( z, y ) ∋ s ✭ ✏ ✭ ✭ ✏ ✭ ✭ ✏ ✏ ✮ ✏ ❄ ❄ y ✫ ✪ 12
Summing up We have proved that for in fi nite balanced B ⊆ A ( G ) , B ∈ RRA ⇐ ⇒ χ ( B ) = ∞ . In the rest of the talk, we will see some applications of this. 13
2. Completions A completion of a relation algebra A is a relation algebra A such that 1. A ⊆ A , 2. A is complete as a boolean algebra: � S exists for all S ⊆ A , 3. A is dense in A : ∀ c ∈ A \ { 0 } ∃ a ∈ A \ { 0 } ( a ≤ c ) . Monk (1970): every relation algebra has a completion, which is unique up to isomorphism. Easy fact: If B is an atomic relation algebra, then its completion is (At B ) + — the complex algebra over its atom structure. 14
Theorem (IH, 1997; this proof by R Hirsch–IH 2002) Theorem 4 RRA is not closed under completions. This answers an implicit question of Monk (1970). Proof. Let G = ( Z , E ) where ( x, y ) ∈ E iff | x − y | = 1 . t t t t t t t t t Then χ ( A ( G )) = χ ( G ) = 2 . ✲ Let B be the subalgebra of A ( G ) whose elements are fi nite sums (unions) of { 1 , } and R X , B X , G X for all fi nite or co fi nite X ⊆ Z . Can check this is a subalgebra. It is in fi nite and balanced. Then χ ( B ) = ∞ . So B ∈ RRA . A ( G ) is a completion of B . But χ ( A ( G )) = 2 . So A ( G ) / ∈ RRA . � 15
RRA is not Sahlqvist-axiomatisable Sahlqvist equations are de fi ned syntactically. E,g., all positive equations are Sahlqvist. Corollary 5 (Venema, 1997) RRA is not axiomatisable by Sahlqvist equations. Proof. By [Givant–Venema 1999], Sahlqvist equations are preserved under completions of ‘conjugated BAOs’ (e.g., relation algebras). By theorem 4, RRA is not closed under completions. � 16
3. Weakly and strongly representable atom structures Representability of an atomic relation algebra is not determined by its atom structure. There are two atomic relation algebras ( B and A ( G ) ) with the same atom structure, one representable, the other not. So let’s distinguish two kinds of relation algebra atom structure S : • weakly representable: some atomic relation algebra with atom structure S is representable, • strongly representable: every atomic relation algebra with atom structure S is representable. Theorem 6 (Venema, 1998) The class of weakly representable relation algebra atom structures is elementary. (He proved a general result for varieties of completely additive BAOs.) 17
What about the strongly representable atom structures? Write SRAS for the class of strongly representable atom structures. Theorem 7 (Hirsch–IH, 2002) SRAS is not elementary. The proof uses • a simple lemma on strongly representable atom structures • the ‘Monk algebras’ A ( G ) • Erd ˝ os graphs • ultraproducts 18
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