Covering of ordinals - PDF Free Download

Covering of ordinals Laurent Braud IGM, Univ. Paris-Est Liafa, 8 January 2010 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 1 / 27

1 Covering graphs Ordinals MSO logic Fundamental sequence MSO-theory of covering graphs 2 Pushdown hierarchy Definition Iteration of exponentiation 3 Higher-order stacks presentation Definition Ordinal presentation Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 2 / 27

Ordinals An ordinal is a well-ordering, i.e. an order where each set has a smallest element each strictly decreasing sequence is finite During this talk, we confuse ordinal with graph of the order. 5 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 3 / 27

Ordinals An ordinal is a well-ordering, i.e. an order where each set has a smallest element each strictly decreasing sequence is finite During this talk, we confuse ordinal with graph of the order.... ω Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 3 / 27

Ordinals An ordinal is a well-ordering, i.e. an order where each set has a smallest element each strictly decreasing sequence is finite During this talk, we confuse ordinal with graph of the order.... ω + 1 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 3 / 27

Ordinals An ordinal is a well-ordering, i.e. an order where each set has a smallest element each strictly decreasing sequence is finite During this talk, we confuse ordinal with graph of the order. ω + 1 0 1 2 3 4... ω Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 3 / 27

Theorem (Cantor normal form, 1897) For α < ε 0, there is a unique decreasing sequence (γ i ) such that α = ω γ 0 + + ω γ n. It is enough to define ordinals with addition operation α ω α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 4 / 27

Addition α β Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 5 / 27

Addition α + β α β Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 5 / 27

Addition ω 2... Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 5 / 27

Addition ω + 2... Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 5 / 27

Addition ω + 2... 2 + ω = ω... Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 5 / 27

Exponentiation ω α ({decreasing finite sequences of ordinals < α}, < lex ) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 6 / 27

Exponentiation ω α ({decreasing finite sequences of ordinals < α}, < lex ) For instance, ω 2 = ω + ω + ω + ω... 2 = 0 1 decreasing sequences = (1,...,1, 0,...,0) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 6 / 27

Exponentiation ω α ({decreasing finite sequences of ordinals < α}, < lex ) For instance, ω 2 = ω + ω + ω + ω... 2 = 0 1 decreasing sequences = (1,...,1, 0,...,0) () (0) (0,0) (1) (1,0) (1,0,0)...... (1,1) (1,1,0) (1,1,0,0)...... We restrict to ordinals < ε 0 = ω ε 0. Notation : ω n = ω ω...ω n. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 6 / 27

Exponentiation ω α ({decreasing finite sequences of ordinals < α}, < lex ) For instance, ω 2 = ω + ω + ω + ω... 2 = 0 1 decreasing sequences = (1,...,1, 0,...,0) 0 1 2 ω ω + 1 ω + 2...... ω.2 ω.2 + 1 ω.2 + 2...... We restrict to ordinals < ε 0 = ω ε 0. Notation : ω n = ω ω...ω n. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 6 / 27

Monadic second-order logic first-order variables x,y... the structure : < set variables X,Y... and formulas x Y,,,, Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 7 / 27

Monadic second-order logic first-order variables x,y... the structure : < set variables X,Y... and formulas x Y,,,, { antisymmetry p, q( (p < q q < p)) strict order transitivity p, q, r((p < q (q < r) p < r) total order p, q(p < q q < p p = q) well order X, z X x(x X y(y X (x < y x = y))) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 7 / 27

MSO-logics and ordinals [Büchi, Shelah] MTh(S) = {ϕ S = ϕ}. Theorem For any countable α, MTh(α) is decidable. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 8 / 27

MSO-logics and ordinals [Büchi, Shelah] MTh(S) = {ϕ S = ϕ}. Theorem For any countable α, MTh(α) is decidable. α = ω γ 0 + + ω γ k }{{} β Theorem + ω γ k+1 + ω γ n }{{} where δ MTh(α) only depends on δ and whether β > 0. { γ0,...,γ k ω γ k+1,...,γ n < ω MTh(ω ω ) = MTh(ω ωω )... Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 8 / 27

Simplifying graphs 0 1 2 3... ω ω + 1 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 9 / 27

Simplifying graphs 0 1 2 3... ω ω + 1 Each countable limit ordinal is limit of an ω-sequence. How to define this sequence in fixed way? Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 9 / 27

Fundamental sequence [Cantor] Let α = ω γ 0 + + ω γ k 1 }{{} +ω γ k δ If γ n = 0, α a limit ordinal. There is an ω-sequence of limit α. { δ + ω γ.(n + 1) if γ α[n] = k = γ + 1 δ + ω γ k[n] otherwise. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 10 / 27

Fundamental sequence [Cantor] Let α = ω γ 0 + + ω γ k 1 }{{} +ω γ k δ If γ n = 0, α a limit ordinal. There is an ω-sequence of limit α. { δ + ω γ.(n + 1) if γ α[n] = k = γ + 1 δ + ω γ k[n] otherwise. Successor ordinals are a degenerate case : α β if { α = β[n] α + 1 = β. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 10 / 27

Covering graph of ω + 2 ω[n] = n + 1 G ω+2 0 1 2 3... ω ω + 1 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 11 / 27

Covering graph of ω 2 + 1 ω 2 [n] = ω.(n + 1) G ω2 +1 0 1 2 3... ω ω + 1... ω.2 ω.2 + 1... ω.3...... ω 2 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 12 / 27

Covering graph of ω ω Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 13 / 27

First result Proposition < is the transitive closure of. Theorem For α, β < ε 0, if α = β, then MTh(G α ) = MTh(G β ). Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 14 / 27

Proof sketch Proposition For α ω n, the out-degree of G α is at most n. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 15 / 27

Proof sketch Let σ be the sequence 0 σ, β σ if β is the largest s.t. β β, then β σ. Degree word : sequence of out-degrees of this sequence. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 16 / 27

Proof sketch Let σ be the sequence 0 σ, β σ if β is the largest s.t. β β, then β σ. Degree word : sequence of out-degrees of this sequence. Proposition The degree word is ultimately periodic, MSO-definable, injective. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 16 / 27

Pushdown hierarchy Many definitions : higher-order pushdown automata [Müller-Schupp, Carayol-Wöhrle], unfolding [Caucal] or treegraph [Carayol-Wöhrle] + MSO-interpretations or rational mappings prefix-recognizable relations [Caucal-Knapik,Carayol], term grammars [Dam, Knapik-Niwiński-Urzyczyn]... Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 17 / 27

Pushdown hierarchy Graph 0 (finite) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 18 / 27

Pushdown hierarchy Graph 0 (finite) I Treegraph Graph 1 (prefix-recognizable) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 18 / 27

Pushdown hierarchy Graph 0 (finite) I Treegraph Graph 1 (prefix-recognizable) I Treegraph Graph 2 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 18 / 27

Pushdown hierarchy Graph 0 (finite) I Treegraph Graph 1 (prefix-recognizable) I Treegraph Graph 2... Each graph in Graph n has a decidable MSO-theory. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 18 / 27

MSO-interpretations MSO-interpretation : I = {ϕ a (x, y)} a Γ where ϕ a is a formula over G. I(G) = {x a y G = ϕ a (x, y)} Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 19 / 27

MSO-interpretations MSO-interpretation : I = {ϕ a (x, y)} a Γ where ϕ a is a formula over G. I(G) = {x Exemple : transitive closure. a y G = ϕ a (x, y)} ϕ < (x, y) := X (x X closed (X) y X) x = y closed (X) := z X, z (z z z X) Proposition For α < ε 0, there is an interpretation I(G α ) = α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 19 / 27

MSO-interpretations MSO-interpretation : I = {ϕ a (x, y)} a Γ where ϕ a is a formula over G. I(G) = {x Exemple : transitive closure. a y G = ϕ a (x, y)} ϕ < (x, y) := X (x X closed (X) y X) x = y closed (X) := z X, z (z z z X) Proposition For α < ε 0, there is an interpretation I(G α ) = α. Proposition There is an interpretation from G α to G β with β α < ε 0. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 19 / 27

Treegraph Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 20 / 27

Main result Theorem (Bloom, Ésik) If α < ω 3 = ω ωω then α Graph 2. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 21 / 27

Main result Theorem (Bloom, Ésik) If α < ω 3 = ω ωω then α Graph 2. Proposition For α < ω n, G α Graph n 1. Theorem For α < ω n, α Graph n 1. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 21 / 27

Main result Theorem (Bloom, Ésik) If α < ω 3 = ω ωω then α Graph 2. Proposition For α < ω n, G α Graph n 1. Theorem For α < ω n, α Graph n 1. Proposition For n > 0 and α ω (3n + 1), G α / Graph n. Corollary G ε0 does not belong to the hierarchy. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 21 / 27

Treegraph on covering graphs... G ω Proposition There is a interpretation I such that I Treegraph(G α ) = G ω α for each α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 22 / 27

Treegraph on covering graphs............ Proposition There is a interpretation I such that I Treegraph(G α ) = G ω α for each α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 22 / 27

Treegraph on covering graphs ()... (0)... (1)... (2)... (0,0) (0,1) (1,0) (1,1) (2,0) (2,1) Proposition There is a interpretation I such that I Treegraph(G α ) = G ω α for each α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 22 / 27

Treegraph on covering graphs ()... (0)... (1)... (2)... (0,0) (1,0) (1,1) (2,0) (2,1) Proposition There is a interpretation I such that I Treegraph(G α ) = G ω α for each α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 22 / 27

Treegraph on covering graphs 0... 1... ω... ω 2... 2 ω + 1 ω.2 ω 2 + 1 ω 2 + ω Proposition There is a interpretation I such that I Treegraph(G α ) = G ω α for each α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 22 / 27

Treegraph on covering graphs 0 G ω ω 1 ω ω 2 2 ω + 1 ω.2 ω 2 + 1 ω 2 + ω Proposition There is a interpretation I such that I Treegraph(G α ) = G ω α for each α. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 22 / 27

Higher-order stacks [Carayol] a b a b a a b a b a push a pop 1 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 23 / 27

Higher-order stacks [Carayol] a b a b a a b a b a push a pop 1 c b a b a c b a b a a b a c b a b a c a c b a b a copy 2 pop 2 Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 23 / 27

Higher-order stacks [Carayol] Ops 1 = {push a, pop 1 } for n > 1, Ops n = {copy n, pop n } Ops n 1 Ops n is a monoïd for composition. Let L Ops n. A graph in Graph n can be represented with vertices in Stacks n. with s i, s i Stacks n 1. s 0... s k L s 0... s k Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 24 / 27

Presentation of ordinals Integers are represented by stacks over a unary alphabet. push(i) = i + 1 pop 1 (i + 1) = i For finite integers, α < β s α pop + 1 (s β) s β push + (s α ) Let dec 1 = pop + 1, inc 1 = push +. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 25 / 27

Presentation of ordinals Integers are represented by stacks over a unary alphabet. push(i) = i + 1 pop 1 (i + 1) = i For finite integers, α < β s α pop + 1 (s β) s β push + (s α ) Let dec 1 = pop + 1, inc 1 = push +. For infinite α, α = ω γ 0 + + ω γ k. Each γ i is representable by s γi Stacks n 1 s α = s γ0... s γk Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 25 / 27

For n > 1, if α < β, s α = s γ0... s γi... s γk Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 26 / 27

For n > 1, if α < β, s α = s γ0... s γi... s γk either s β = s γ0... s γi... s γk s γk+1... s γh s β (copy n.(id + dec n 1 )) + (s α ) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 26 / 27

For n > 1, if α < β, s α = s γ0... s γi... s γk either s β = s γ0... s γi... s γk s γk+1... s γh s β (copy n.(id + dec n 1 )) + (s α ) = tail + n (s α ) Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 26 / 27

For n > 1, if α < β, s α = s γ0... s γi... s γk either s β = s γ0... s γi... s γk s γk+1... s γh s β (copy n.(id + dec n 1 )) + (s α ) = tail + n (s α ) or s β = s γ0... s γi 1 s γ i >γ i... s γ k s β pop n.inc n 1.(copy n.(id + dec n 1 )) (s α ) inc n dec n = [pop n.inc n 1 + tail n ].tail n = pop n.[pop n + dec n 1 ].tail n Theorem If (α, <, >) is in Graph n, then (ω α, <, >) is in Graph n+1. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 26 / 27

Future work Obtain the stronger result α < ω (n + 1) α Graph n. Is there a definition of fundamental sequence so that the result remains true for further ordinals? α = β MTh(G α ) = MTh(G β ) Ordinals greater than ω ω are not selectable [Rabinovich, Shomrat]. Are covering graphs selectable? Extend the method to other linear orderings. Laurent Braud (IGM, Univ. Paris-Est) Covering of ordinals Liafa, 8 January 2010 27 / 27