Lambda Calculus between Algebra and Topology. Antonino Salibra



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Lambda Calculus between Algebra and Topology Antonino Salibra

Thanks Special thanks to Laboratoire PPS (Univ. Paris Diderot) for supporting my research with financial support and very interesting collaborations The Organizers Antonio Bucciarelli Alberto Carraro Thomas Ehrhard Giulio Manzonetto The invited Speakers The sponsors ANR, CNRS, Univ. Paris Diderot and Univ. Paris 13

Lambda Calculus between Algebra and Topology From Lambda calculus to Universal Algebra: (i) Lambda abstraction algebras (ii) Church algebras (iii) Boolean-like-algebras (iv) Semi-Boolean-like algebras (v) n-subtractive algebras From Universal Algebra to Lambda calculus: (i) The structure of the lattice of the λ-theories (ii) Boolean algebras, Stone representation theorem and the indecomposable semantics (iii) The order incompleteness problem and other problems

Lambda Calculus between Algebra and Topology Topology refines partial orderings through the separation axioms: A space (X, τ) is T 0 iff the preorder τ, defined by x τ y iff x y, is a partial order. Every partial order is the specialization order of a space. From Lambda calculus to Topology: (i) new axioms of connectedness (ii) new axioms of separation (iii) topological algebras (iv) Visser spaces and Priestley spaces From Topology to Lambda calculus: (i) Topological incompleteness/completeness theorems (ii) The order-incompleteness problem (iii) Topological models

Models of lambda calculus Church (around 1930): Lambda calculus λ-theory = congruence w.r.t. application and λ-abstractions containing αβ-conversion Scott: First model and Continuous Semantics (1969) A continuous model D is a reflexive object in the category CPO. Stable semantics (Berry), Strongly stable semantics (Bucciarelli- Ehrhard). All known models of λ-calculus admits a compatible (w.r.t. application) partial order and are topological algebras (w.r.t. Scott topology). Each model M defines a λ-theory T h = (M) = {(P, Q) : P and Q have the same interpretation} and an order theory T h (M) = {(P, Q) : interpretation of P interpretation of Q}

Theory (In)Completeness Problem A class C of models of λ-calculus is theory complete if ( consistent λ-theory T )( M C) T h = (M) = T. Theory incomplete, otherwise. Theorem 1 (Theory incompleteness) All known semantics are theory incomplete. Honsell-Ronchi: Scott semantics; Bastonero-Gouy: stable semantics; Salibra: strongly stable semantics and all pointed po-models. Selinger (1996) asked: Are partial orderings intrinsic to computations? A po-model is a pair (M, ), where M is a model and is a nontrivial partial ordering on M making the application monotone. The order-completeness problem by Selinger asks whether the class PO of po-models is theory complete or not. ANSWER: Unknown. The best we know about order-incompleteness: Theorem 2 (Carraro-S. 2012) There exists a λ-theory T such that, for every po-model (M, ), T h = (M) T (M, ) has infinite connected components and the connected component of the looping term Ω is a singleton set.

Two Other Open Problems Equational Completeness Problem asks whether any two terms equal in all Scott continuous models are β-convertible. T h = (D) = λβ? ANSWER: Unknown D Scott Equational Consistency Problem (Honsell-Plotkin 2006) asks whether, for every finite set E of equations between λ-terms consistent with the λ-calculus, there exists a continuous model contemporaneously satisfying all equations of E. ( E finite set of identities)[(e λβ consistent) ( D Scott) D = E]? ANSWER: No (Carraro-S. 2012/2013). These problems and the order incompleteness problem are interconnected

Part I Algebras: Stone, Boole and Church

Stone and Boole Theorem 3 (Stone Representation Theorem) Every Boolean algebra is isomorphic to a field of sets. Every Boolean algebra can be embedded into a Boolean product of indecomposable Boolean algebras (2 is the unique indecomposable Boolean algebra!). Then every Boolean algebra is isomorphic to a subalgebra of 2 I = P(I) for a suitable set I. Generalisations to other classes of algebras by Pierce (rings with unit) Comer and Vaggione. Combinatory algebras (CA) and λ-abstraction algebras (LAA) satisfy an analogous theorem...

Church algebras The untyped λ-calculus has truth values 0, 1 and if-then-else construct q(x, y, z) of programming: λ-calculus (LAA): 1 λxy.x; 0 λxy.y; q(e, x, y) = (ex)y Combinatory logic (CA): 1 k; 0 sk; q(e, x, y) = (ex)y Boolean algebras: q(e, x, y) = (e x) ( e y) Rings with unit 1: q(e, x, y) ex + (1 e)y. Definition 1 (Manzonetto-Salibra 2008) An algebra A is a Church algebra if it admits two constants 0, 1 and a ternary term q(x, y, z) satisfying: q(1, x, y) = x; q(0, x, y) = y. There are equations which are contemporaneously satisfied by 0 and 1: for example, q(1, x, x) = x; q(0, x, x) = x.

Central elements An element e of a Church algebra A is central if A = A/Cong(e = 1) A/Cong(e = 0). Lemma 1 Let A be a Church algebra et e A. The following conditions are equivalent: e is central; e satisfies the following identities: (i) q(e, x, x) = x. (ii) q(e, q(e, x, y), z) = q(e, x, z) = q(e, x, q(e, y, z)). (iii) q(e, f(x), f(y)) = f(q(e, x 1, y 1 ),..., q(e, x n, y n )), operation f (iv) e = q(e, 1, 0). Central elements are the unique way to decompose the algebra as Cartesian product. A is indecomposable if the unique central elements are 0, 1.

Stone, Boole and Church Theorem 4 The central elements of a Church algebra A constitute a Boolean algebra: e d = q(e, 1, d); e d = q(e, d, 0); e = q(e, 0, 1) Let V be a variety of Church algebras, A V and F be the Boolean space of maximal ideals of the Boolean algebra of central elements of A. Then the map defined by f : A Π I F (A/θ I ), f(x) = (x/θ I : I F), gives a weak Boolean product representation of A. The quotient algebras A/θ I are directly indecomposable if the indecomposable members of V constitute a universal class. (True for CA and LAA!)

Central elements at work in lambda calculus! The indecomposable CAs (models of λ-calculus) are the building blocks of CA. The indecomposable semantics is the class of models which are indecomposable as combinatory algebras. Theorem 5 Scott is always simple! Proof: Scott continuous semantics (and the other known semantics of λ- calculus) are included within the indecomposable semantics, because every Scott model is simple (i.e., it admits only trivial congruences) as a combinatory algebra.

Central elements at work in lambda calculus! Theorem 6 The algebraic incompleteness theorem: There exists a continuum of λ-theories which are not equational theories of indecomposable models. Proof: 1.Decomposable CAs are closed under expansion. 2. Ω (λx.xx)(λx.xx) is a non-trivial central element in the term algebra of a suitable λ-theory φ, because - the λ-theory ψ 1 generated by Ω = λxy.x is consistent; - the λ-theory ψ 2 generated by Ω = λxy.y is consistent; - Ω is central in the term algebra of φ = ψ 1 ψ 2. 3. All models of φ are decomposables! The algebraic incompleteness theorem encompasses all known theory incompleteness theorems: (Honsell-Ronchi 1992) Scott continuous semantics; (Bastonero-Guy 1999) Stable semantics; (Salibra 2001) Strongly stable semantics.

Central elements at work in universal algebra! 1. Boolean-like algebras: Church algebras (of any algebraic type), where all elements are central. 2. Semi-Boolean-like algebras: Church algebras (of any algebraic type), where all elements are semi-central. Theorem 7 A double pointed variety is discriminator iff it is idempotent semi-boolean-like and 0-regular. 3. Lattices of equational theories

Part II Topology: Selinger, Scott, Visser, Priestley

Selinger A model D is orderable if it admits a non-trivial compatible (w.r.t. application) partial ordering. Theorem 8 (Selinger 1996) The term algebra of λβ (or λβη) is unorderable (but it can be a subalgebra of a big orderable algebra). Corollary 1 T h = (D) = λβ and D orderable 1. The term denotations are an antichain. 2. T h = (D) = T h (D).

The Equational Completeness Problem Partial answers to the Equational Completeness Problem: technique Given a class C of models, we construct an effective model E (maybe not in the class C!) such that T h (E) T h (M), for all M C. Since the model E is effective, then {N : E = N λx.x} is co-r.e. Consequence: T h (M) is not r.e. and T h = (M) λβ, λβη. Assume, by contraposition, T h (M) to be r.e. Then {N : E = N λx.x} co-r.e. {N : M = N λx.x} r.e. Visser open Visser closed {N : M = N λx.x} = Λ Why? VISSER topology is hyperconnected on Λ!

The Equational Completeness Problem RESULTS: (Berline-Manzonetto-S. 2010) λβ (and λβη) is not the theory of an effective model of λ-calculus. The order theory of an effective model is not r.e. (Carraro-S. 2009) λβη is not the theory of a model living in the category of Scott domains (Carraro-S. 2012) λβ (and λβη) cannot be the theory of a filter model living in CPO.

Leaving Lambda Calculus Towards Computability Theory (Work in Progress) An enumerated set is a pair X = ( X, φ X ), where φ X : N X is an onto map (Mal cev 1964) Example: φ CF : N CF mapping the program n into the function φ n computed by the program n. Example: φ Λ : Λ Λ/λβ mapping a λ-term M into its equivalence class [M] β. Given an enumerated set X, we define Y X r.e. (co-r.e., decidable) if φ 1 X (Y ) is r.e. (co-r.e., decidable).

The Visser topology The r.e. sets of X are a ring R E X of sets generating a topology τ E on X. The co-r.e. sets of X are a ring R V X of sets generating a topology τ V. ( X, τ E, τ V ) is a bitopological space τ E is the Ershov topology and τ V is the Visser topology Example: τ E on CF is the Scott topology; while τ V iff graph(f) is decidable. is T 0 with f τ V -open Example: τ E on Λ/λβ is the discrete topology, while τ V is nontrivial, hyperconnected and T 1.

The Visser topology Lemma 2 τ E is T 0 iff τ V is T 0. Proposition 1 Let τ E be T 0. 1. a E b iff b V a. 2. τ V contains the lower topology. 3. τ E contains the upper topology Proof: x is Visser-closed and x is Ershov-closed. Proposition 2 If the enumerated set X is pre-complete then 1. (Visser) τ V is hyperconnected; 2. τ E is compact iff τe has a bottom element. CF and Λ/T are precomplete.

The join of Ershov and Visser Hereafter we always assume that τ E is T 0. Proposition 3 τ E τ V is zero-dimensional (i.e., it has a base of clopens), Hausdorff and satisfies the Priestley separation axiom (w.r.t. E ). Proof: Let a, b X. Since τ E is T 0, either a E b or b E a. In the first case there is an r.e. open U such that a U but b / U. X \ U is a co-r-e open such that b X \ U and a / X \ U. Let x E y. Then there is an r.e. set U such that x U but y / U. U is E-upper. The complement is E-down which contains y but not x. Lemma 3 If τ E = Scott, then τ E τ V Proof: By Prop. 1. contains the Lawson topology. Example: τ E τ V on CF is not compact. Example: τ E τ V on Λ/T? Unknown, in general (it is the discrete topology if the theory T is r.e., but it is zero-dimensional and Hausdorff on the theory of Bohm trees)

The Priestley space of computability Let X be an enumerated set. If ( X, τ E ) a T 0 -space, then the compactfication (Y, τ c ) of ( X, τ E τ V ) is a Priestly space and ( X, τ E τ V ) is a dense subspace of (Y, τ c ): Proof: Consider 2 = {0, 1} with three topologies: The discrete top τ d ; The top τ 0 with 0 < 1; The top τ 1 with 1 < 0. We have τ d = τ 0 τ 1. We consider the ring R E X of r.e. subsets of X and consider the product topology on 2 RE X. We have: The topology τ 0 on 2 RE X is the Scott topology w.r.t. ; The topology τ 1 on 2 RE X is the Scott topology w.r.t. ; The topology τ d = τ 0 τ 1 on 2 RE X is a Priestley space. Consider the closed subspace of lattice homomorphisms HOM(R E X, 2). It is Priestley (because closed), and ( X, τ E τ V ) embeddes into HOM(R E X, 2) as a dense subspace.