Local convergence for random permutations The case of uniform - PowerPoint PPT Presentation

Infinite rooted permutations Definition random permutations when the size tends to infinity. and study limits of GOAL: Define a notion of local convergence in namely, the set of (possibly infinite) rooted permutations. We set rooted at 0. We underline that infinite rooted permutations can be thought as 4 We call infinite rooted permutation a pair ( A , � ) where A is an infinite interval of integers containing 0 and � is a total order on A . We denote the set of infinite rooted permutations by S ∞ • . S • := S • ∪ S ∞ ˜ • ,

Infinite rooted permutations rooted at 0. random permutations when the size tends to infinity. namely, the set of (possibly infinite) rooted permutations. Definition We set 4 We underline that infinite rooted permutations can be thought as We call infinite rooted permutation a pair ( A , � ) where A is an infinite interval of integers containing 0 and � is a total order on A . We denote the set of infinite rooted permutations by S ∞ • . S • := S • ∪ S ∞ ˜ • , GOAL: Define a notion of local convergence in ˜ S • and study limits of

Restriction function around the root Definition by 5 σ = 4 2 5 8 3 6 1 7 r 2 i = 5 2 ≤ σ,i − 3 ≤ σ,i 0 ≤ σ,i − 4 ≤ σ,i − 2 ≤ σ,i 1 ≤ σ,i 3 ≤ σ,i − 1 2 ≤ σ,i 0 ≤ σ,i − 2 ≤ σ,i 1 ≤ σ,i − 1 The restriction function around the root is defined, for every h ∈ N , ˜ r h : S • − → S • ( ) ( A , � ) �→ A ∩ [ − h , h ] , � .

Restriction function around the root Definition by 5 σ = 4 2 5 8 3 6 1 7 r 2 i = 5 2 ≤ σ,i − 3 ≤ σ,i 0 ≤ σ,i − 4 ≤ σ,i − 2 ≤ σ,i 1 ≤ σ,i 3 ≤ σ,i − 1 2 ≤ σ,i 0 ≤ σ,i − 2 ≤ σ,i 1 ≤ σ,i − 1 The restriction function around the root is defined, for every h ∈ N , ˜ r h : S • − → S • ( ) ( A , � ) �→ A ∩ [ − h , h ] , � .

Restriction function around the root Definition by 5 σ = 4 2 5 8 3 6 1 7 r 2 i = 5 2 ≤ σ,i − 3 ≤ σ,i 0 ≤ σ,i − 4 ≤ σ,i − 2 ≤ σ,i 1 ≤ σ,i 3 ≤ σ,i − 1 2 ≤ σ,i 0 ≤ σ,i − 2 ≤ σ,i 1 ≤ σ,i − 1 The restriction function around the root is defined, for every h ∈ N , ˜ r h : S • − → S • ( ) ( A , � ) �→ A ∩ [ − h , h ] , � .

This topology is metrizable by a local distance d . 6 Definition rooted permutation as a dense subset. of finite separable and complete. Moreover it contains the space d is a compact Polish space, i.e. , compact, The metric space Theorem Local distance for ˜ S • We say that a sequence ( A n , � n ) n ∈ N of rooted permutations in ˜ S • is locally convergent to an element ( A , � ) ∈ ˜ S • , if for all H > 0 there exists N ∈ N such that for all n ≥ N , r H ( A n , � n ) = r H ( A , � ) .

6 Theorem Definition rooted permutation as a dense subset. of finite separable and complete. Moreover it contains the space d is a compact Polish space, i.e. , compact, The metric space Local distance for ˜ S • We say that a sequence ( A n , � n ) n ∈ N of rooted permutations in ˜ S • is locally convergent to an element ( A , � ) ∈ ˜ S • , if for all H > 0 there exists N ∈ N such that for all n ≥ N , r H ( A n , � n ) = r H ( A , � ) . This topology is metrizable by a local distance d .

Theorem Definition rooted permutation as a dense subset. 6 Local distance for ˜ S • We say that a sequence ( A n , � n ) n ∈ N of rooted permutations in ˜ S • is locally convergent to an element ( A , � ) ∈ ˜ S • , if for all H > 0 there exists N ∈ N such that for all n ≥ N , r H ( A n , � n ) = r H ( A , � ) . This topology is metrizable by a local distance d . The metric space ( ˜ S • , d ) is a compact Polish space, i.e. , compact, separable and complete. Moreover it contains the space S • of finite

Our goal Study limits of random permutations when the size tends to infinity Limiting objects : Permutons separable permutations, substitution-closed classes, Mallows permutations, ... Corresponding statistic : ? Concrete examples : ? 7 1. Scaling limits: Corresponding statistic : � occ ( π, σ ) , for all π ∈ S Concrete examples : ρ -avoiding permutations for | π | = 3 , 2. Local limits: Limiting objects : ?

Our goal Study limits of random permutations when the size tends to infinity Limiting objects : Permutons separable permutations, substitution-closed classes, Mallows permutations, ... Corresponding statistic : ? Concrete examples : ? 7 1. Scaling limits: Corresponding statistic : � occ ( π, σ ) , for all π ∈ S Concrete examples : ρ -avoiding permutations for | π | = 3 , 2. Local limits: Limiting objects : Rooted permutations i.e., total orders

Local convergence: the consecutive occurrences characterization

Local convergence We want to study limits of unrooted permutations w.r.t. the local distance, therefore we need to choose a root. QUESTION: How do we make this choice? ANSWER: Uniformly at random among the indices of the permutation. Observation In this way, a fixed permutation naturally identifies a random variable i with values in the set 8

Local convergence We want to study limits of unrooted permutations w.r.t. the local distance, therefore we need to choose a root. QUESTION: How do we make this choice? ANSWER: Uniformly at random among the indices of the permutation. Observation In this way, a fixed permutation naturally identifies a random variable i with values in the set 8

Local convergence We want to study limits of unrooted permutations w.r.t. the local distance, therefore we need to choose a root. QUESTION: How do we make this choice? ANSWER: Uniformly at random among the indices of the permutation. Observation In this way, a fixed permutation naturally identifies a random variable i with values in the set 8

Local convergence We want to study limits of unrooted permutations w.r.t. the local distance, therefore we need to choose a root. QUESTION: How do we make this choice? ANSWER: Uniformly at random among the indices of the permutation. Observation 8 In this way, a fixed permutation σ naturally identifies a random variable ( σ, i ) with values in the set S • .

n be a permutation of size n TFAE: Link: BS for some random rooted infinite permutation (b) There exists an infinite vector of non-negative real numbers such that c - occ n for all patterns r h (a) h 1 for all h all 2 h 1 n Weak-local convergence: the deterministic case Definition BS Benjamini–Schramm converges to a random rooted permutation law let 9 Theorem [B.] law For any n We say that a sequence ( σ n ) n ∈ N of elements in S σ ∞ , if → σ ∞ , ( σ n , i n ) − w.r.t. the local distance d . → σ ∞ instead of ( σ n , i n ) → σ ∞ . − − We write σ n

Weak-local convergence: the deterministic case Theorem [B.] 1 2 h all h for all 1 h r h Link: (b) There exists an infinite vector of non-negative real numbers BS Definition 9 BS law law Benjamini–Schramm converges to a random rooted permutation We say that a sequence ( σ n ) n ∈ N of elements in S σ ∞ , if → σ ∞ , ( σ n , i n ) − w.r.t. the local distance d . → σ ∞ instead of ( σ n , i n ) → σ ∞ . − − We write σ n For any n ∈ N , let σ n be a permutation of size n . TFAE: → σ ∞ , for some random rooted infinite permutation σ ∞ . (a) σ n − (∆ π ) π ∈S such that � c - occ ( π, σ n ) → ∆ π , for all patterns π ∈ S .

Weak-local convergence: the deterministic case BS for all (b) There exists an infinite vector of non-negative real numbers BS Definition Theorem [B.] law 9 Benjamini–Schramm converges to a random rooted permutation law We say that a sequence ( σ n ) n ∈ N of elements in S σ ∞ , if → σ ∞ , ( σ n , i n ) − w.r.t. the local distance d . → σ ∞ instead of ( σ n , i n ) → σ ∞ . − − We write σ n For any n ∈ N , let σ n be a permutation of size n . TFAE: → σ ∞ , for some random rooted infinite permutation σ ∞ . (a) σ n − (∆ π ) π ∈S such that � c - occ ( π, σ n ) → ∆ π , for all patterns π ∈ S . ( ) r h ( σ ∞ ) = ( π, h + 1 ) h ∈ N , all π ∈ S 2 h + 1 . Link: P = ∆ π ,

n n is a sequence of random permutations: n aBS n qBS w.r.t. the product topology law n c - occ qBS: n c - occ aBS: Weak-local convergence: deterministic & random case Theorem [B.] If BS BS: 10 If ( σ n ) n ∈ N is a sequence of deterministic permutations: � → σ ∞ − ⇐ ⇒ c - occ ( π, σ n ) → ∆ π , ∀ π ∈ S σ n

n qBS Weak-local convergence: deterministic & random case aBS: w.r.t. the product topology law n c - occ qBS: Theorem [B.] 10 BS BS: If ( σ n ) n ∈ N is a sequence of deterministic permutations: � → σ ∞ − ⇐ ⇒ c - occ ( π, σ n ) → ∆ π , ∀ π ∈ S σ n If ( σ n ) n ∈ N is a sequence of random permutations: → σ ∞ E [ � − ⇐ ⇒ c - occ ( π, σ n )] → ∆ π , ∀ π ∈ S σ n aBS

Weak-local convergence: deterministic & random case aBS: w.r.t. the product topology law qBS: Theorem [B.] 10 BS BS: If ( σ n ) n ∈ N is a sequence of deterministic permutations: � → σ ∞ − ⇐ ⇒ c - occ ( π, σ n ) → ∆ π , ∀ π ∈ S σ n If ( σ n ) n ∈ N is a sequence of random permutations: → σ ∞ E [ � − ⇐ ⇒ c - occ ( π, σ n )] → ∆ π , ∀ π ∈ S σ n aBS ( � ) → µ ∞ − ⇐ ⇒ c - occ ( π, σ n ) − → ( Λ π ) π ∈S σ n qBS π ∈S

Our goal • Study limits of random permutations when the size tends to infinity 1. Scaling limits: Limiting objects : Permutons Concrete examples : separable permutations, substitution-closed classes, ... Limiting objects : Rooted permutations Corresponding statistic : ? Concrete examples : ? 11 Corresponding statistic : � occ ( π, σ ) , for all π ∈ S 2. Local limits:

Our goal • Study limits of random permutations when the size tends to infinity 1. Scaling limits: Limiting objects : Permutons Concrete examples : separable permutations, substitution-closed classes, ... Limiting objects : Rooted permutations Concrete examples : ? 11 Corresponding statistic : � occ ( π, σ ) , for all π ∈ S 2. Local limits: Corresponding statistic : � c - occ ( π, σ ) , for all π ∈ S

Local limit for uniform 231-avoiding permutations

231-avoiding permutations Definition for all 12 For all n > 0 we define the following probability distribution on Av n ( 231 ) , P 231 ( π ) := 2 | LRMax ( π ) | + | RLMax ( π ) | , π ∈ Av n ( 231 ) . 2 2 | π |

231 such that n qBS n aBS 231-avoiding permutations 231 231 and 231 and 1 2 h for all P 231 1 h r h Theorem [B.] for all h There exists a random infinite rooted permutation Corollary for all Prob then 13 Let σ n be a uniform random permutation in Av n ( 231 ) for all n ∈ N , � c - occ ( π, σ n ) − → P 231 ( π ) , π ∈ Av ( 231 ) .

231-avoiding permutations for all and and for all Theorem [B.] Corollary 13 Prob then Let σ n be a uniform random permutation in Av n ( 231 ) for all n ∈ N , � c - occ ( π, σ n ) − → P 231 ( π ) , π ∈ Av ( 231 ) . There exists a random infinite rooted permutation σ ∞ 231 such that for all h ∈ N , ( ) r h ( σ ∞ π ∈ S 2 h + 1 , P 231 ) = ( π, h + 1 ) = P 231 ( π ) , → L ( σ ∞ → σ ∞ σ n qBS − 231 ) σ n aBS − 231 .

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

A bijection between 231 -avoiding permutations & binary trees 14 σ = 4 1 3 2 10 5 7 6 9 8 ↔

consider a sequence of uniform binary trees T n with n nodes; Steps of the proof A binary Galton-Watson tree is a random rooted tree defined as of 2. We give to each node children according to an independent copy 1. We start with the root; • We build the random tree T recursively: with distribution equivalently, a random variable on 0 L R 2 or, • We consider a probability distribution follow: Remark 0 1 offspring distribution • We also consider a family of binary Galton-Watson trees T with sequence of uniform 231-avoiding permutations of size n , we can • Thanks to the previous bijection, instead of considering a 15 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E

Steps of the proof A binary Galton-Watson tree is a random rooted tree defined as of 2. We give to each node children according to an independent copy 1. We start with the root; • We build the random tree T recursively: with distribution equivalently, a random variable on 0 L R 2 or, • We consider a probability distribution follow: Remark 0 1 offspring distribution • We also consider a family of binary Galton-Watson trees T with sequence of uniform 231-avoiding permutations of size n , we can • Thanks to the previous bijection, instead of considering a 15 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E consider a sequence of uniform binary trees T n with n nodes;

Steps of the proof A binary Galton-Watson tree is a random rooted tree defined as of 2. We give to each node children according to an independent copy 1. We start with the root; • We build the random tree T recursively: with distribution equivalently, a random variable on 0 L R 2 or, • We consider a probability distribution follow: Remark sequence of uniform 231-avoiding permutations of size n , we can • Thanks to the previous bijection, instead of considering a 15 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E consider a sequence of uniform binary trees T n with n nodes; • We also consider a family of binary Galton-Watson trees T δ with offspring distribution η ( δ ) , δ ∈ ( 0 , 1 ) .

Steps of the proof A binary Galton-Watson tree is a random rooted tree defined as of 2. We give to each node children according to an independent copy 1. We start with the root; • We build the random tree T recursively: with distribution equivalently, a random variable on 0 L R 2 or, • We consider a probability distribution follow: Remark sequence of uniform 231-avoiding permutations of size n , we can • Thanks to the previous bijection, instead of considering a 15 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E consider a sequence of uniform binary trees T n with n nodes; • We also consider a family of binary Galton-Watson trees T δ with offspring distribution η ( δ ) , δ ∈ ( 0 , 1 ) .

Steps of the proof Remark of 2. We give to each node children according to an independent copy 1. We start with the root; • We build the random tree T recursively: follow: A binary Galton-Watson tree is a random rooted tree defined as 15 sequence of uniform 231-avoiding permutations of size n , we can • Thanks to the previous bijection, instead of considering a [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E consider a sequence of uniform binary trees T n with n nodes; • We also consider a family of binary Galton-Watson trees T δ with offspring distribution η ( δ ) , δ ∈ ( 0 , 1 ) . • We consider a probability distribution η on { 0 , L , R , 2 } or, equivalently, a random variable ξ with distribution η.

Steps of the proof Remark 2. We give to each node children according to an independent copy 1. We start with the root; • We build the random tree T recursively: follow: A binary Galton-Watson tree is a random rooted tree defined as 15 sequence of uniform 231-avoiding permutations of size n , we can • Thanks to the previous bijection, instead of considering a [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E consider a sequence of uniform binary trees T n with n nodes; • We also consider a family of binary Galton-Watson trees T δ with offspring distribution η ( δ ) , δ ∈ ( 0 , 1 ) . • We consider a probability distribution η on { 0 , L , R , 2 } or, equivalently, a random variable ξ with distribution η. of ξ .

1 P 231 • Applying singularity analysis and reusing the bijection: Steps of the proof E Av 231 for all P 231 n c - occ O 1 T c - occ • With a long recursion we prove that 4 16 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E • We relate T δ and the sequence ( T n ) n ∈ N by + ∞ ∑ [ ] [ ] F ( T δ ) = F ( T n ) · P ( | T δ | = n ) E n = 1 ( 1 − δ 2 ) n + ∞ ∑ [ ] = 1 + δ F ( T n ) · C n · ; E 1 − δ n = 1

• Applying singularity analysis and reusing the bijection: Steps of the proof E Av 231 for all P 231 n c - occ • With a long recursion we prove that 4 16 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E • We relate T δ and the sequence ( T n ) n ∈ N by + ∞ ∑ [ ] [ ] F ( T δ ) = F ( T n ) · P ( | T δ | = n ) E n = 1 ( 1 − δ 2 ) n + ∞ ∑ [ ] = 1 + δ F ( T n ) · C n · ; E 1 − δ n = 1 [ ] = δ − 1 · P 231 ( π ) + O ( 1 ); c - occ ( π, T δ ) E

Steps of the proof E for all • With a long recursion we prove that 4 16 [ � ] c - occ ( π, σ n ) → P 231 ( π ) , for all π ∈ Av ( 231 ) FIRST STEP: Prove that E • We relate T δ and the sequence ( T n ) n ∈ N by + ∞ ∑ [ ] [ ] F ( T δ ) = F ( T n ) · P ( | T δ | = n ) E n = 1 ( 1 − δ 2 ) n + ∞ ∑ [ ] = 1 + δ F ( T n ) · C n · ; E 1 − δ n = 1 [ ] = δ − 1 · P 231 ( π ) + O ( 1 ); c - occ ( π, T δ ) E • Applying singularity analysis and reusing the bijection: [ � ] c - occ ( π, σ n ) → P 231 ( π ) , π ∈ Av ( 231 ) . E

n 2 using similar We finally apply the Second moment method. Steps of the proof 2 Av 231 for all 0 n Var c - occ • Therefore Av 231 for all P 231 n 2 c - occ techniques and we obtain, c - occ • We study the second moment Prob 17 SECOND STEP: Prove that � c - occ ( π, σ n ) − → P 231 ( π ) , for all π ∈ Av ( 231 )

Steps of the proof for all We finally apply the Second moment method. Av 231 for all 0 n Var c - occ • Therefore 17 techniques and we obtain, using similar Prob SECOND STEP: Prove that � c - occ ( π, σ n ) − → P 231 ( π ) , for all π ∈ Av ( 231 ) [ � c - occ ( π, σ n ) 2 ] • We study the second moment E [ � c - occ ( π, σ n ) 2 ] → P 231 ( π ) 2 , π ∈ Av ( 231 ); E

Steps of the proof using similar for all Var • Therefore for all techniques and we obtain, 17 Prob SECOND STEP: Prove that � c - occ ( π, σ n ) − → P 231 ( π ) , for all π ∈ Av ( 231 ) [ � c - occ ( π, σ n ) 2 ] • We study the second moment E [ � c - occ ( π, σ n ) 2 ] → P 231 ( π ) 2 , π ∈ Av ( 231 ); E ( � ) → 0 , π ∈ Av ( 231 ) . c - occ ( π, σ n ) We finally apply the Second moment method.

Thanks for your attention Article and slides available at: http://www.jacopoborga.com (from midnight also on arXiv) Questions? 17

Back-up slides

• With probability 1 2 we do one of the two following construction: for all • We sample a second (possibly empty) permutation; • We root it at its maximum; 2 • We sample a first non-empty permutation; 4 The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 .

• With probability 1 2 we do one of the two following construction: for all • We sample a second (possibly empty) permutation; • We root it at its maximum; 2 • We sample a first non-empty permutation; 4 The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 .

• With probability 1 2 we do one of the two following construction: for all • We sample a second (possibly empty) permutation; • We root it at its maximum; 2 • We sample a first non-empty permutation; 4 The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 .

• With probability 1 2 we do one of the two following construction: for all • We sample a second (possibly empty) permutation; • We root it at its maximum; 2 • We sample a first non-empty permutation; 4 The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 .

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

• We sample a first non-empty permutation; 2 4 for all • We sample a second (possibly empty) permutation; • We root it at its maximum; The construction of the random order σ ∞ 231 . • We consider the following Boltzmann distribution on Av ( 231 ) : ( 1 ) | π | P ( π ) = 1 , π ∈ Av ( 231 ) , P ( ∅ ) = 1 2 . • With probability 1 / 2 we do one of the two following construction:

Local limit for uniform 321-avoiding permutations

321-avoiding permutations Definition Example otherwise. 0 1 1 For all n > 0 , we define the following probability distribution on Av n ( 321 ) ,  | π | + 1  if π = 12 ... | π | ,   2 | π | P 321 ( π ) := if c - occ ( 21 , π − 1 ) = 1 ,  2 | π |  

321-avoiding permutations Example 0 1 Definition otherwise. For all n > 0 , we define the following probability distribution on Av n ( 321 ) ,  | π | + 1  if π = 12 ... | π | ,   2 | π | P 321 ( π ) := if c - occ ( 21 , π − 1 ) = 1 ,  2 | π |   π − 1 = π =

Av 231 are deterministic: 321 such that n qBS n aBS r h and 321 and 1 2 h for all P 321 1 h 321 321-avoiding permutations for all h Theorem [B.] There exists a random infinite rooted permutation Corollary Since the limiting objects P 321 for all Prob then 321 Let σ n be a uniform random permutation in Av n ( 321 ) for all n ∈ N , � c - occ ( π, σ n ) − → P 321 ( π ) , π ∈ Av ( 321 ) .

321-avoiding permutations for all and Theorem [B.] Corollary for all Since the limiting objects then Prob and Let σ n be a uniform random permutation in Av n ( 321 ) for all n ∈ N , � c - occ ( π, σ n ) − → P 321 ( π ) , π ∈ Av ( 321 ) . ( ) P 321 ( π ) π ∈ Av ( 231 ) are deterministic: There exists a random infinite rooted permutation σ ∞ 321 such that for all h ∈ N , ( ) r h ( σ ∞ π ∈ S 2 h + 1 , 321 ) = ( π, h + 1 ) = P 321 ( π ) , P → L ( σ ∞ → σ ∞ − 321 ) − 321 . σ n qBS σ n aBS

A bijection between 321 -avoiding permutations & trees It is well known that 321-avoiding permutations can be broken into two increasing subsequences, the first above the diagonal and the second below the diagonal:

A bijection between 321 -avoiding permutations & trees 0 11 Pre-order (from 0) Post-order (from 1) ↔ 1 5 6 6 7 10 3 2 1 2 4 3 4 8 7 9 9 5 10 8