Monk algebras, completions, and atom structures Robin Hirsch and Ian - - PowerPoint PPT Presentation

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Monk algebras, completions, and atom structures Robin Hirsch and Ian - - PowerPoint PPT Presentation

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


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Monk algebras, completions, and atom structures

Robin Hirsch and Ian Hodkinson

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Aim To show that relation algebras can shed light on duality-related matters. To nail more nails in the coffin 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 define some relation algebras (like ‘Monk algebras’), based
  • n graphs.
  • 2. We use them to show that RRA is not closed under completions

(which we will define).

  • 3. If time, we use them to study the atom structures of atomic

relation algebras.

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  • 0. Relation algebra atom structures, complex algebras (revision)

Abstractly, a relation algebra atom structure is a structure S = (S, Id, ˘, C) satisfying certain first-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. If A is an atomic relation algebra, A ֒ → (At A)+.

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  • 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 v1, . . . , vk of distinct nodes of G, such that (v1, v2), . . . , (vk−1, vk), (vk, v1) 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.

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Relation algebras from graphs Fix a graph G. We define a relation algebra A(G). Its set of atoms is Q = {1,, rx, bx, gx : 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 define ˘ S = S, for all S ⊆ Q. We define 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,
  • (rx, ry, rz), (bx, by, bz), and (gx, gy, gz), whenever {x, y, z} is

independent in G. Fact: A(G) is a simple relation algebra (a kind of ‘Monk algebra’).

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Independent elements of A(G) For X ⊆ G, define RX = {rx : x ∈ X} ∈ A(G). Define BX, GX similarly. An element a ∈ A(G) is said to be independent if a = CX 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.

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Chromatic number and representability Let B ⊆ A(G) be a subalgebra. B is said to be balanced if

  • RG, BG, GG ∈ B,
  • for all X ⊆ G, RX ∈ B ⇐

⇒ BX ∈ B ⇐ ⇒ GX ∈ B. The chromatic number χ(B) of a balanced B ⊆ A(G) is the least k < ω such that RG 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) = ∞.

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Proof: B is representable ⇒ χ(B) = ∞ B is simple. Suppose that h : B → Re(U) is a representation. Because B is infinite, so is U. Pick distinct x0, x1, . . . ∈ U. Assume for contradiction that χ(B) < ∞. So −1, =

i≤n bi, for some independent elements b1, . . . , bn ∈ B.

h respects −1,. So for all i < j, we have (xi, xj) ∈ h(

i≤n bi).

h respects +. So there is bij ∈ {b1, . . . , bn} with (xi, xj) ∈ h(bij). By Ramsey’s theorem, we can assume bij is constant — say, bij = b7 for all i < j. Triangle consistency for (x0, x1, x2) gives (b01 ; b12) · b02 = 0. That is, (b7 ; b7) · b7 = 0. But b7 is independent, so by lemma 2, (b7 ; b7) · b7 = 0. Contradiction!

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Proof: χ(B) = ∞ ⇒ B is representable Assume χ(B) = ∞. Say b ∈ B is R-big if RG − b is the sum of finitely many independent elements of B. The set of R-big elements of B has the finite intersection property. So (using some form of AC) it extends to an ultrafilter Rµ of B. RG ∈ Rµ. If a ∈ B is independent, −a is R-big, so −a ∈ Rµ, so a / ∈ Rµ. So no element of Rµ is independent. Hence, by lemma 2, r ; r′ = 1 for any r, r′ ∈ Rµ. Choose ultrafilters Bµ, Gµ similarly. We now show that ∃ can use these ultrafilters to win the game Gu

ω(B)

  • f length ω played on ultrafilter networks over B. So B ∈ RRA.

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r ; s ∈ Nt(x, y). ∀ picks x, y ∈ Nt We assume no existing witness in Nt. and r, s ∈ B with ❄ ❄ Nt Nt(x, y) y x ③ ③ ✬ ✫ ✩ ✪ ③ ③

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∃ chooses ultrafilters Nt+1(x, z) ∋ r and Nt+1(z, y) ∋ s with Nt+1(x, z) ; Nt+1(z, y) ⊆ Nt(x, y). Nt+1(z, y) ∋ s Nt+1(x, z) ∋ r z PPPPPPP q PPPPPPP P q ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ❄ ❄ ③ Nt Nt(x, y) y x ③ ③ ✬ ✫ ✩ ✪ ③ ③

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RG + BG + GG ∈ Nt+1(x, z). Say, RG ∈ Nt+1(x, z). Similarly, say BG ∈ Nt+1(z, y). Nt+1(z, y) ∋ s Nt+1(x, z) ∋ r z PPPPPPP q PPPPPPP P q ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ❄ ❄ ③ Nt Nt(x, y) y x ③ ③ ✬ ✫ ✩ ✪ ③ ③

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Gµ Gµ ③ ③ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ Nt+1(z, y) ∋ s Nt+1(x, z) ∋ r z PPPPPPP q PPPPPPP P q ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✮ ❄ ❄ ③ ❄ Nt y x ③ ③ ✬ ✫ ✩ ✪

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Summing up We have proved that for infinite balanced B ⊆ A(G), B ∈ RRA ⇐ ⇒ χ(B) = ∞. In the rest of the talk, we will see some applications of this.

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  • 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.

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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.

Then χ(A(G)) = χ(G) = 2. ✲ t t t t t t t t t Let B be the subalgebra of A(G) whose elements are finite sums (unions) of {1,} and RX, BX, GX for all finite or cofinite X ⊆ Z. Can check this is a subalgebra. It is infinite and balanced. Then χ(B) = ∞. So B ∈ RRA. A(G) is a completion of B. But χ(A(G)) = 2. So A(G) / ∈ RRA.

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RRA is not Sahlqvist-axiomatisable Sahlqvist equations are defined 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.

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  • 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.)

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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˝
  • s graphs
  • ultraproducts

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Simple lemma Lemma 8 A relation algebra atom structure S is strongly representable iff S+ ∈ RRA.

  • Proof. TFAE:
  • 1. S is strongly representable
  • 2. every atomic relation algebra A with atom structure S is

representable

  • 3. S+ is representable.

We said yesterday that for any atomic relation algebra A, A ֒ → (At A)+ a → {x ∈ At A : x ≤ a}.

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Erd˝

  • s graphs

Theorem 9 (Erd˝

  • s, 1959) For any finite k, there is a finite graph Ek
  • f chromatic number > k and with no cycles of length < k.

For each n < ω, let Gn = ˙

  • k≥nEk (an infinite graph).

Then

  • χ(Gn) = ∞.

So by theorem 3, A(Gn) ∈ RRA.

  • Notation: for a graph G, write α(G) for At A(G). So

A(G) = α(G)+. By our simple lemma 8, α(Gn) ∈ SRAS.

  • Also, Gn has no cycles of length < n.

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SRAS not closed under ultraproducts Now take a non-principal ultrafilter D over ω. 1.

D Gn has no cycles (Ło´

s’s theorem). So χ(

D Gn) ≤ 2.

By theorem 3, A(

D Gn) /

∈ RRA.

  • 2. Let S =

D α(Gn).

S ∼ = α(

D Gn) (exercise)

By (1) and simple lemma 8, S / ∈ SRAS. Conclude: α(Gn) ∈ SRAS for all n, but an ultraproduct (S) of the α(Gn) is not in SRAS. Therefore, SRAS is not closed under ultraproducts, so (by Ło´ s’s theorem) is not elementary.

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Summary Venema: the class of weakly representable relation algebra atom structures is elementary. We proved that the class of strongly representable relation algebra atom structures is not elementary. This answers a question of Maddux (1982).

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Conclusion The A(G) are variants of Monk algebras. Simple and useful:

  • RRA not closed under Monk completions, not Sahlqvist-ax’ble
  • SRAS non-elementary (& similarly for strongly representable

atom structures of cylindric algebras)

  • IH–Venema, 2002: Any first-order (e.g., equational)

axiomatisation of RRA has infinitely many ‘non-canonical’ sentences (also true for McKinsey–Lemmon modal logic).

  • Goldblatt–IH–Venema, 2003: There are canonical modal logics

that are not the logic of any elementary class of frames. This answers a question of Fine (1975). Binary relations, and representations of RAs, are fundamental. The list suggests there is value in proving ‘negative’ results about them.

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References P . Erd˝

  • s, Graph theory and probability, Canad. J. Math. 11 (1959)

34–38.

  • R. Goldblatt, I. Hodkinson, and Y. Venema, Erd˝
  • s graphs resolve

Fine’s canonicity problem, Bull. Symb. Logic 10 (2004) 186–208.

  • R. Hirsch and I. Hodkinson, Strongly representable atom structures
  • f relation algebras, Proc. Amer. Math. Soc. 130 (2002), 1819–1831.
  • I. Hodkinson and Y. Venema, Canonical varieties with no canonical

axiomatisation, Trans. Amer. Math. Soc. 357 (2005), 4579–4605.

  • J. D. Monk, Completions of boolean algebras with operators,

Mathematische Nachrichten 46 (1970), 47–55.

  • Y. Venema, Atom structures and Sahlqvist equations, Algebra

Universalis 38 (1997) 185–199.

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