formal languages and pervin spaces - irifmgehrke/jep.pdf · liafa, cnrsanduniversityparisdiderot...

LIAFA, CNRS and University Paris Diderot

Formal languages and Pervin spaces

Jean-Eric Pin1

(joint work with Mai Gehrke and Serge Grigorieff)

1LIAFA, CNRS and University Paris Diderot

March 2013, Paris


Outline

(1) The algebraic approach to regular languages

(2) Duality

(3) Pervin spaces

(4) Perspectives and conclusion


Part I

The algebraic approach

Three levels:

• Local approach: attach to each regularlanguage a finite (ordered) monoid.

• Global approach: attach to each variety ofregular languages a variety of finite monoids.

• Profinite approach: Replace finite monoids byprofinite monoids.


Syntactic ordered monoid of a subset L of A∗

Syntactic preorder [Schutzenberger 1956]:u 6L v iff, for every x, y ∈ A∗,

xuy ∈ L⇒ xvy ∈ L

Syntactic (or context, observational) congruence

u ∼L v iff u 6L v and v 6L u

iff xuy ∈ L⇔ xvy ∈ L

Syntactic monoid: A∗/∼L

Syntactic ordered monoid: (A∗/∼L,6L /∼L)


The syntactic ordered monoid of (ab)∗

1 2

a

bElements

1 2

1 1 2

a 2 0

b 0 1

aa 0 0

ab 1 0

ba 0 2

Relations

bb = aaaba = abab = b

ab a b ba

aa

1

Syntactic order


Nonregular languages

Let Equality = {u ∈ {a, b}∗ | |u|a = |u|b}.Its ordered syntactic monoid is (Z,=).

Let Majority = {u ∈ {a, b}∗ | |u|a > |u|b}.Its syntactic monoid is (Z,>).


The local algebraic approach

Theorem (Kleene-Nerode)

Let L be a subset of A∗. Are equivalent:

(1) L is a regular language,

(2) The congruence ∼L has finite index,

(3) The syntactic monoid of L is finite.

R Study regular languages through properties oftheir syntactic [ordered] monoids.


Star-free languages

Star-free languages = smallest class of languagescontaining the finite languages and closed underBoolean operations and product.

Theorem (Schutzenberger 1965)

A language is star-free iff its syntactic monoid isfinite and aperiodic.

A finite monoid M is aperiodic if there is an n > 0such that for all x ∈M , xn = xn+1.


Piecewise testable languages

Piecewise testable languages = Booleancombination of languages of the formA∗a1A

∗a2 · · ·A∗akA

∗, where a1, . . . , ak are letters:

Theorem (Simon 1972)

A language is piecewise testable iff its syntacticmonoid is finite and division-free.

A monoid is division-free (or J -trivial) if any twoelements that divide each other are equal.(x divides y if y = sxt for some s, t ∈M).


The global approach: Varieties of languages

A variety of languages is a class of regular languagesV such that:

(1) for each alphabet A, V(A∗) is closed underBoolean operations and quotients.

(2) if L ∈ V(B∗), then for each monoid morphismα : A∗ → B∗, then α−1(L) ∈ V(A∗).

Quotients: x−1Ly−1 = {u ∈ A∗ | xuy ∈ L}

Regular languages, star-free languages, piecewisetestable languages form varieties of languages.


The global approach: Varieties of finite monoids

Variety of finite monoids = class of finite monoidsclosed under taking submonoids, quotient monoidsand finite direct products.

Examples:

• All finite monoids

• Commutative monoids

• Idempotent monoids

• Aperiodic monoids

• Division-free monoids

• Groups


Global approach: Eilenberg’s variety theorem

Theorem (Eilenberg 1976)

There is bijection between varieties of monoids andvarieties of languages.

Examples

Finite monoids Regular languages

Aperiodic monoids Star-free languages

Division-free monoids Piecewise testable languages


Positive varieties

A positive variety of languages is a class of regularlanguages closed under union, intersection,quotients and inverses of morphisms.

A variety of finite ordered monoids is a class of finiteordered monoids closed under taking finite products,ordered submonoids and quotient monoids.

Theorem (Pin 1995)

There is a bijection between positive varieties oflanguages and varieties of finite ordered monoids.


Profinite metrics (1): Separating words

A monoid morphism ϕ : A∗ →M separates twowords u and v of A∗ if ϕ(u) 6= ϕ(v).

The morphism u→ |u| mod 2 (from A∗ intoZ/2Z) separates abaabaaba and abaabaabab.

Proposition

One can always separate two distinct words by afinite monoid [group, division-free monoid ].

However, commutative monoids cannot separate abfrom ba.


Profinite metrics: Step 1

Let V be a variety of finite monoids. The relation

u ∼V v iff no monoid of V can separate u from v

is a congruence. The quotient A∗/∼V is a monoid.

Exemple. If V is the variety of all finitecommutative monoids, u ∼V v iff for all a ∈ A,|u|a = |v|a and A∗/∼V = N

A.

Up to changing A∗ to A∗/∼V, we will assume thatV separates the elements of A∗.


The profinite metric dV

Let V be a variety of finite monoids separating theelements of A∗. Let u and v be two words. Put

rV(u, v) = min{|M | M ∈ V and

separates u from v}

dV(u, v) = 2−rV(u,v)

Then dV is an ultrametric, that is

(1) d(x, y) = 0 ⇐⇒ x = y,

(2) d(x, y) = d(y, x),

(3) d(x, z) 6 max{d(x, y), d(y, z)}


The free pro-V monoid

The completion of the metric space (A∗, dV) is

denoted FV(A). It is a compact space [nontrivialresult].

Further, the product on A∗ is uniformly continuousand has a unique uniformly continuous extension tothe completion of A∗, making FV(A) a compactmonoid, called the free pro-V monoid.

For V = {all finite monoids}, the completion is also

denoted A∗ and its elements are called profinitewords.


Properties of the free pro-V monoid

Theorem

(1) A finite A-generated monoid belongs to V iff

it is a continuous quotient of FV(A).

(2) If V ⊆W then FV(A) is a continuous

quotient of FW(A). In particular, FV(A) is a

continuous quotient of A∗.


Reiterman’s theorem

Let u and v be profinite words of A∗. A finitemonoid satisfies the identity u = v if, for all

continuous morphisms ϕ = A∗ →M , one hasϕ(u) = ϕ(v).

Theorem (Reiterman 1982)

A class of finite monoids is a variety of finitemonoids iff can be defined by a set of profiniteidentities.


The operator ω

In a finite semigroup, every element x has a uniqueidempotent power, denoted xω.

In a compact semigroup, the closed subsemigroupgenerated by an element x contains a uniqueidempotent, denoted xω.

Equivalent definition in A∗:For each profinite word u, the sequence un!

converges to an idempotent denoted uω.


Examples of identities

A finite monoid is aperiodic iff it satisfies theidentity xω+1 = xω.

A finite monoid is division-free iff it satisfies theidentities

xω+1 = xω and (xy)ωx = (xy)ω = y(xy)ω

iff it satisfies the identitiesxω+1 = xω and (xy)ω = (yx)ω.

A finite monoid is a group iff it satisfies the identityxω = 1.


The algebraic approach to regular languages

Monadic second

order logic MSO[<]

Regular

languages

Finite

monoids

Kleene 56

Buchi 69


The algebraic approach to regular languages (2)

First order

logic FO[<]

Star-free

languages

xω = xω+1Aperiodic

monoids

DecidabilitySchutzenberger 65

Mc Naughton 71



BΣ1[<]

formulas

Piecewise testable

languages

xω = xω+1

(xy)ω = (yx)ωJ -trivial

monoids

DecidabilityI. Simon 75

Thomas 87



BΣ1[<,+1]

formulas

Locally

testable

xωyxωyxω = xωyxω

xωyxωzxω = xωzxωyxω

local

semilattices

DecidabilityBrzozowski-Simon 72

Thomas 87



FO2[<]

Σ2 ∩ Π2

Unambiguous

star-free

xω = xω+1

(xy)ω(yx)ω(xy)ω = (xy)ωDA

DecidabilitySchutzenberger 76

SchThV 01, ThW 98

EVW 97, PW 97


Summary of the first part

Local approach: study regular languages throughproperties of their syntactic monoid.

Global approach: characterize varieties of regularlanguages through profinite identities. Study thepro-V monoids.

Is it possible to use the local/global approach formore general classes of regular languages thanvarieties? And beyond regular languages?


Summary of the first part

Local approach: study regular languages throughproperties of their syntactic monoid.

Global approach: characterize varieties of regularlanguages through profinite identities. Study thepro-V monoids.

Is it possible to use the local/global approach formore general classes of regular languages thanvarieties? And beyond regular languages?

Yes we can...


Part II

Duality

Lattice? You mean

bounded distributive

lattice? Then it has

a dual !

That’s the way its all started. . .


Duality in a nutshell

Definition

The dual space of a bounded distributive lattice isthe set of its prime filters.

Elements ←→ Prime filters

Boolean algebras ←→ Stone spaces

Distributive lattices ←→ Ordered Stone spaces

Sublattices ←→ Quotient spaces

n-ary operations ←→ (n+ 1)-ary relations

Key idea: A lattice of subsets of A∗ is completelycharacterized by its dual space.


Duality for regular languages

[Almeida ↑89] Duality between varieties of regularlanguages and clopen sets of profinite monoids.

[Pippenger 97] Slightly more general results: Stoneduality explicitly mentioned.

[GGP 08] Extended Stone duality for any lattice ofregular languages. The product is the dual of theresiduation operations.

XY ⊆ Z ⇐⇒ Y ⊆ X\Z ⇐⇒ X ⊆ Z/Y

Equational theory for lattices of regular languages.


Lattices, filters and ideals

A lattice of subsets of X is a set of subsets of Xcontaining ∅ and X and closed under finiteintersection and finite union. A Boolean algebra is alattice closed under complement.

Let L be a lattice of subsets of X. A filter is anonempty subset F of L such that:

(1) if K ∈ F and K ⊆ L, then L ∈ F ,

(2) if K,L ∈ F , then K ∩ L ∈ F .

An ideal is a nonempty subset I of L such that:

(1) if K ∈ I and L ⊆ K, then L ∈ I,

(2) if K,L ∈ I, then K ∪ L ∈ I.


Prime filters

A prime filter on L is a subset F of L such that

(1) X ∈ F , ∅ /∈ F ,

(2) if K ∈ F and K ⊆ L, then L ∈ F ,

(3) if K,L ∈ F , then K ∩ L ∈ F ,

(4) if K ∪ L ∈ F , then either K ∈ F or L ∈ F .

In other words. . .

(1) A prime filter is nontrivial,

(2) closed under extension,

(3) closed under intersection,

(4) and has to choose. . .


Stone-Priestley duality

Let S be the dual space of L. The map e : L → 2S

e(L) = { prime filters containing L }

is an injective lattice morphism.

Topology on S:

Take as a basis of open sets the sets of the forme(L). Then S is a compact, totally disconnectedspace (not necessarily Hausdorff). It is ordered by

the specialization order: x 6 y ⇐⇒ {x} ⊆ {y}.


Some examples of dual spaces

• Let L be a regular language and let L be thelattice generated by the languages u−1Lv−1

(u, v ∈ A∗). Its dual space is the orderedsyntactic monoid of L.

• Regular languages→ the free profinite monoidon A. [Almeida]

• Star-free languages→ the free pro-aperiodicmonoid on A. [Almeida]


More examples of dual spaces

• Finite languages ∪ {A∗} → A∗ ∪ {∞},one-point compactification of A∗.

• All languages→ βA∗, the Stone-Cechcompactification of A∗.

One can define βA∗ as the closure of the range ofA∗ in

∏ϕ ϕ(A

∗), where ϕ is any function from A∗

into a compact space.


Reconciling duality and metric spaces

Back to the Boolean algebra of star-free languages.The profinite approach says that its dual space isthe completion of a metric space.

Is it possible to define the dual space of any latticeas the completion of some suitable space?

Metric spaces do not suffice: βA∗ is not metrizable.

Uniform spaces work well for Boolean algebras butquasi-uniform spaces are needed for lattices.


Reconciling duality and metric spaces (2)

Potential problem: completion is well-defined foruniform spaces but is quite messy for quasi-uniformspaces (three competing definitions).

What kind of (quasi)-uniform spaces do we get?

• transitive (→ ultrametric)• totally bounded (→ The completion of thespace is compact.)

This leads to consider transitive, totally boundedquasi-uniform spaces.


Reconciling duality and metric spaces (2)

Potential problem: completion is well-defined foruniform spaces but is quite messy for quasi-uniformspaces (three competing definitions).

What kind of (quasi)-uniform spaces do we get?

• transitive (→ ultrametric)• totally bounded (→ The completion of thespace is compact.)

This leads to consider transitive, totally boundedquasi-uniform spaces.

And then a miracle occurs...


The miracle

Theorem

A quasi-uniform space is a Pervin space iff it istransitive and totally bounded.

The miracle is that the definition of Pervin spaces isreally very simple !


Part III

Pervin spaces

W. J. Pervin, Quasi-uniformization oftopological spaces, Math. Ann. 147 (1962),316–317.


Pervin spaces

Definition

A Pervin space is a pair (X,L) where L is a latticeof subsets of X.

The elements of L form a basis of a topology inwhich each open set is a (possibly infinite) union ofelements of L.


Examples of Pervin spaces

Examples on X = {0, 1}

• The Boolean space: L ={∅, {0}, {1}, {0, 1}

}

• The Sierpinski space: L ={∅, {1}, {0, 1}

}.

Examples on X = N

• L = {∅} ∪ {cofinite subsets of N}.

• L = {N} ∪ {finite subsets of N}.

• L = {finite/cofinite subsets of N}.


Examples on X = A∗

• L ={Regular languages

}

• L ={Star-free languages

}

• L ={Group languages

}

• L ={All languages

}

• L ={Commutative languages

}

• L = {A∗} ∪{Finite languages

}


The preorder of a Pervin space (X,L)

Preorder on X: x 6L y iff, for all L ∈ L,

x ∈ L⇒ y ∈ L

Equivalence relation: x ∼L y iff, for all L ∈ L,

x ∈ L⇔ y ∈ L

We often write 6 for 6L and ∼ for ∼L.

This preorder coincides with the specialisationpreorder.


Examples

Let L be a language and let L be the latticegenerated by the quotients of L. Then ∼ is thesyntactic congruence of L and 6L is its syntacticorder.

Let X = A∗ and L = {commutative languages}.Then ∼ is the commutative equivalence:

u ∼ v iff |u|a = |v|a for all a ∈ A

and A∗/∼ = NA.


The quotient space (X/∼,L/∼)

Let L ∈ L. If x ∈ L and x ∼ y, then y ∈ L.

Thus the quotient space X/∼, the quotient lattice

L/∼ = {L/∼ | L ∈ L}

and the Pervin space (X/∼,L/∼) are well-definednotions.

Proposition

If L is the lattice generated by the quotients of alanguage L of A∗, then A∗/∼ is the syntacticmonoid of L.


The Kolmogorov quotient of a Pervin space (X,L)

Proposition (trivial)

The following conditions are equivalent:

(1) 6 is an order,

(2) ∼ is the equality relation,

(3) X is a Kolmogorov (T0) space.

Kolmogorov: open sets distinguish points.

In particular, the quotient space (X/∼,L/∼) isKolmogorov.


Uniform continuity

Let (X,K) and (Y,L) be Pervin spaces.

Definition

A function ϕ : X → Y is uniformly continuous if,for each L ∈ L, ϕ−1(L) ∈ K.

Remark. Any uniformly continuous map iscontinuous and order preserving, but the converse isnot true.


Completion of a Pervin space

Step one: Take the quotient X/∼.

Step two: All the completions available forquasi-uniform spaces are equivalent for Pervinspaces. Take the one you prefer!

A. Csaszar, D-completions of Pervin-typequasi-uniformities., Acta Sci. Math. 57,1-4(1993), 329–335.


Valuations on a lattice of subsets L

Definition

A valuation on L is a lattice morphism from L intothe Boolean algebra {0, 1}.

Thus it is a map v : L → {0, 1} such that, for allL1, L2 ∈ L,

(1) v(∅) = 0, v(X) = 1,

(2) v(L1 ∩ L2) = v(L1)v(L2),

(3) v(L1 ∪ L2) = v(L1) + v(L2).

where the sum and the product are the Booleanoperations.


Valuations and prime filters are the same

If v : L → {0, 1} is a valuation, then the set v−1(1)is a prime filter.

If p is a prime filter, the map

v(L) =

{1 if L ∈ p

0 otherwise

is a valuation.


Completion of a Pervin space (X,L)

From now on, we assume that 6 is an order on X.

For each L ∈ L, let

L = {v is a valuation such that v(L) = 1}

In particular, X is the set of all valuations on L.

Definition

The completion of a Pervin space (X,L) is the

Pervin space (X, L), where L is the lattice of

subsets of X defined by L = {L | L ∈ L}.


Embedding (X,L) into (X, L)

For each x ∈ X, let vx be the valuation defined by

vx(L) =

{1 if x ∈ L

0 if x /∈ L

Proposition

The map x→ vx defines an injective and uniformlycontinuous embedding from (X,L) into (X, L).

Further, X is dense in X.


Properties of the completion of a Pervin space

L = {v is a valuation such that v(L) = 1}

X = {all valuations} L = {L | L ∈ L}

Theorem

The lattice L is the set of all compact open subsetsof X. In particular, X is compact.


Pervin spaces and duality

Theorem (Duality theorem)

The maps L 7→ L and K 7→ K ∩X are mutuallyinverse lattice isomorphisms between L and L.

Corollary

The completion of the Pervin space (X,L) is equalto the dual of L.

For instance, the completion of (X,P(X)) is theStone-Cech compactification of X.


The prime filter theorem

Theorem

Let I be an ideal and let F be a filter disjoint fromI. Then there is a valuation v on L such thatv(L) = 1 for all L ∈ F and v(L) = 0 for all L ∈ I.

1

0

I

F


Example 1

Let X = N and L = {∅} ∪ {cofinite subsets of N}.

This space is compact, but not complete. Indeed,the valuation v given by v(L) = 1 for each cofiniteset L defines a new element, denoted ∞.

Thus X = X ∪ {∞} and

L = {∅} ∪ {cofinite subsets of N containing ∞}.

The order on X is the equality relation, but in X ,x 6∞ for all x ∈ X .


Example 2

Let X = N and L = {N} ∪ {finite subsets of N}.

This space is neither compact nor complete. Indeed,the valuation v given by v(X) = 1 and v(L) = 0 foreach finite set L defines a new element −∞.

Thus X = X ∪ {−∞} and

L = {N} ∪ {finite subsets of N}.

The order on X is the equality relation, but in X ,−∞ 6 x for all x ∈ X .


Example 3

Let X = N and L = {finite/cofinite subsets of N}.

This space is neither compact nor complete. Indeed,the valuation v given by

v(L) =

{1 if L is cofinite

0 if L is finite

defines a new element ∞. Thus X = X ∪ {∞} and

L = {finite subsets of N} ∪

{cofinite subsets of N containing ∞}.


Example 4

Let Ln = {1k| 0 < k 6 n} and

X =

{1

n| n is a positive integer

}

L = {X} ∪ {Ln | n > 0}

This space is compact, but not complete. Indeed,the valuation v given by v(X) = 1 and v(Ln) = 0

defines a new element 0. Thus X = X ∪ {0} and

L = {∅} ∪ {finite subsets of X containing 0}.

The order on X is 0 6 · · · 6 1n6 · · · 6 1

2 6 1.


Syntactic monoid and syntactic space

Let L be a lattice closed under quotients.

Definition

The syntactic monoid of L is the monoid (A∗/∼).The syntactic space of L is the completion of thePervin space (A∗/∼,L/∼).

By construction, the syntactic space of L is the dualof L. It is compact but it is not always a monoidbecause the product in A∗/∼ might not beuniformly continuous.


Some examples

The syntactic space of ...

• the lattice of all regular languages of A∗ is thefree profinite monoid on A.

• the set of all star-free languages of A∗ is thefree pro-aperiodic monoid on A.

• the lattice of finite or full languages is theone-point compactification of A∗.

• the set of all languages of A∗ is the Stone-Cechcompactification of A∗.


The case of a single language

Let L be a language. The syntactic space of L isobtained as follows:

(1) Compute the syntactic monoid M of L andthe image P of L in M the usual way).

(2) Let L be the lattice generated by thequotients of P in M .

(3) The syntactic space of L is the completion of(M,L).

If L is a regular language, its syntactic monoid isfinite and equal to its completion. Thus, for regularlanguages, only the algebraic properties of thesyntactic monoid are important.


Syntactic space of Majority

The completion is Z = Z ∪ {−∞,+∞}. Theclosure of the addition (considered as a subset of

Z3) is the relation + given in the following table:

+ i −∞ +∞

j {i+ j} {−∞} {+∞}

−∞ {−∞} {−∞} Z

+∞ {+∞} Z {+∞}


A nonregular example

Let M = (Z,+) and let L be the Boolean algebraof finite or cofinite subsets of Z. The syntacticspace of L is Z = Z ∪ {∞}.

The closure of the addition on Z is the relation +

+ i ∞

j {i+ j} {∞}

∞ {∞} Z


The Boolean algebra Rec(Z)

Let Rec(Z) be the set of recognizable subsets of Z,that is, the finite unions of subsets of the form{a+ nZ | n > 1, 0 6 a < n}.

Its syntactic monoid is Z and its syntactic space Z

is the one-generated profinite free group. Theclosure of the addition is here the addition of Z.

We denote by i 7→ i the natural embedding of Zinto Z and by + the addition on Z.


Recognizable + finite subsets of Z

Let L be the Boolean algebra generated by the finitesubsets and the recognizable subsets of Z. Its syntac-tic monoid is Z and its syntactic space is the disjointunion Z ∪ Z: Z corresponds to the principal ultrafil-ters of L and the profinite group Z corresponds tothe nonprincipal ultrafilters.

The closure + of the addition on Z is commutativebut nonfonctional. It extends + on Z and, for i ∈ Z

and u, v ∈ Z, one has i + u = i+ u and

u + v =

{{k, k} if u + v = k with k ∈ Z

{u+ v} otherwise.


Duality versus Pervin spaces

A subtle difference. Duality deals with abstractlattices (pointless topology). Pervin spaces onlydeals with lattices of subsets.

Pros of duality.

• Duality for lattices with operations.

• Existing results in duality ready for use.

Pros of Pervin spaces.

• Completion in two steps.

• Definitions are simpler than for spectral spaces.

• Intuition from (ultra)metric spaces available,but most results need to be rewritten.


Uniformly continuous extensions

Theorem

Every uniformly continuous mapϕ : (X,LX)→ (Y,LY ) admits a unique uniformly

continuous extension ϕ : (X, LX)→ (Y , LY ).

Corollary

Let ϕ1 and ϕ2 be two uniformly continuous mapsfrom (X,LX) to (Y,LY ) and let ϕ1 and ϕ2 be their

uniformly continuous extensions from (X, LX) to

(Y , LY ). If ϕ1 6 ϕ2, then ϕ1 6 ϕ2.


Principe du prolongement des identites

Proposition

Let ϕ1 and ϕ2 be two uniformly continuous mapsfrom (X, LX) to (Y , LY ). If, for all x ∈ X,ϕ1(x) 6 ϕ2(x), then ϕ1 6 ϕ2. In particular, if ϕ1

and ϕ2 coincide on X, then they are equal.


Equations: the profinite case

Let L be a lattice of regular languages closed underquotients. Let (ML,6) = (A∗/∼L,6L) be thesyntactic ordered monoid of L and let η : A∗ →MLbe the quotient map.

Then η is a uniformly continuous function from(A∗,Reg(A∗)) to (ML,L/∼). It extends to a

uniformly continuous function from A∗ to ML.

Let (u, v) be a pair of profinite words. We say thatL satisfies the equation u 6 v if η(u) 6 η(v).


Equations: the general case

Let L be a lattice of languages closed underquotients. Let (ML,6) = (A∗/∼L,6L) be thesyntactic ordered monoid of L and let η : A∗ →MLbe the quotient map.

Then η is a uniformly continuous function from(A∗,P(A∗)) to (ML,L/∼). It extends to a

uniformly continuous function from βA∗ to ML.

Let (u, v) be a pair of elements of βA∗. We saythat L satisfies the equation u 6 v if η(u) 6 η(v),where η : A∗ → A∗/∼ is the syntactic map.


Equations

Theorem

A set of regular languages of A∗ is a lattice closedunder quotients iff it can be defined by a set ofequations of the form u 6 v, where u, v areprofinite words.

Theorem

A set of languages of A∗ is a lattice closed underquotients iff it can be defined by a set of equationsof the form u 6 v, where u, v ∈ βA∗.


Open maps

Theorem

Let ϕ : (X,LX)→ (Y,LY ) be a uniformlycontinuous map and let L ∈ LX . Are equivalent:

(1) There is a smallest K ∈ LY such thatϕ(L) ⊆ K,

(2) The upper set generated by ϕ(L) is open.


Best approximation property

Best approximation property: for all L1, L2 ∈ L,there is a smallest L such that L1L2 ⊆ L.

Proposition

Let L be a Boolean algebra of regular languages ofA∗ closed under quotients. Are equivalent:

(1) L has the best approximation property,

(2) L is closed under product,

(3) The product on the completion of (A∗,L) isan open map.


Closure under product

Proposition

Suppose that L contains the finite languages. ThenL has the best approximation property iff it isclosed under product.

A group language is a language whose syntacticmonoid is a finite group. The group languages ofA∗ form a Boolean algebra closed under quotient,which has the best approximation property but isnot closed under product.


Metrizability

Proposition

Let L be a Boolean algebra of subsets of A∗. Areequivalent:

(1) The associated Pervin space is metrizable,

(2) The uniformity has a countable basis,

(3) L is countable.

Similar results for lattices/quasi-metrizability.


Translations and quotients

Let L be a lattice of subsets of A∗.

Proposition

The translations u 7→ xu and u 7→ uy are alluniformly continuous iff L is closed under quotients.

A monoid in which the translations are uniformlycontinuous is called a semiuniform monoid.


Uniform continuity of the product

Let (M,L) be a Pervin space (with 6 an order).

Proposition

Suppose that M is a monoid and that the productis uniformly continuous. Then the product admits a

unique uniformly continuous extension to M , which

defines a structure of monoid on M .


When is the product uniformly continuous?

Let L be a Boolean algebra of subsets of A∗ closedunder quotients and let M be its syntactic monoid.

Theorem (GGP 2010)

Are equivalent:

(1) the product on M uniformly continuous,

(2) the completion of M is a compact monoid,

(3) the closure of the product of M is functional,

(4) the elements of L are all regular languages.


Part IV

Perspectives and conclusion


Related notions

Syntactic monoids are very successful for regular lan-guages, but are not doing so well beyond regular lan-guages (noticeable exception: the theory of context-free groups [Muller Schupp 83]).

Sakarovitch [1976] proposed to use the pair formedby a syntactic monoid and the image of the languagein its syntactic monoid (pointed monoid).

Our new definition is an extension of this idea. Theimage of the language is used to define a lattice onthe syntactic monoid, and then consider the comple-tion of the resulting Pervin space.


Dreams

Any lattice closed under quotients can be defined bya set of equations of the form u 6 v, whereu, v ∈ βA∗.

Further, one has L1 ⊆ L2 iff L1 satisfies theequations defining L2 iff the syntactic space of L1 isa quotient of the syntactic space of L2.

Thus in principle, one could separate two classes oflanguages L1 and L2 by finding an equationsatisfied by one of the classes and not by the other.

Home work. Find a set of equations defining yourown favorite complexity class or logical fragment. . .


Logic and circuit complexity

Let N be the class of all numerical predicates.Then the FO[N ]-definable languages of A∗ form aBoolean algebra, whose syntactic space is βN.

It is known that FO[N , a] defines AC0, the class oflanguages computed by unbounded fanin,polynomial size, constant-depth Boolean circuits.

What is the syntactic space of the Boolean algebraof all FO[N , a]-definable languages?


Beyond recognizable languages

[Barrington, Compton, Straubing, Therien 1992]proved that

FO[N , a] ∩ Reg(A∗) = FO[<,MOD, a]

Is it possible to prove this result by using syntacticspaces?

This would permit to attack difficult conjectures incircuit complexity.


Why to be pessimistic

• Eilenberg’s variety theorem doesn’t help muchfor proving Schutzenberger’s theorem.

• βN is a very complex object (its nickname isthe three-headed monster).

• βN is not a compact monoid. Chasingidempotents does not work as in the finite orcompact case.

• Duality gives a nice encoding, but intrinsicdifficulties may just remain.

• Known results on βN indicate that settheoretic problems may occur on the way.


Why to be optimistic

• This approach is very successful for lattices ofregular languages.

• It is a global approach and syntactic spacescontain a lot of information.

• It gives access to the power of topology and tohigher mathematics.

• Stone duality has been successful in other areasof mathematics, notably in algebraic geometry.

• For circuit complexity classes, known partialresults may guide intuition.

formal languages and pervin spaces - irifmgehrke/jep.pdf · liafa, cnrsanduniversityparisdiderot...

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