Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Corners in Tree–Like Tableaux Pawe� l Hitczenko & Amanda Lohss September 27, 2016 Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Tree–Like Tableaux Corners in Tree–Like Tableaux Occupied Corners in Tree–Like Tableaux Diagonal Boxes in Symmetric Tree–Like Tableaux Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Tree–Like Tableaux (Aval, Boussicault, Nadeau 2011) • • • • Definition • • A tree-like tableaux of size n is a Ferrers • diagrams of half-perimeter n + 1 such that, • • • 1 The box in the first column and first • row is pointed. • 2 Either all boxes to the left of a • pointed box is empty or all boxes above are empty. Figure: A tree–like tableaux of size 13. 3 Every row and every column contains There are n ! tableaux at least one point. of size n . Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Symmetric Tableaux Insertion Procedure Definition: The ”special point” is the right-most point among those that occur at the bottom of a column. Add a pointed column for each north step and a pointed row for each west step. If the step is below the special point, add a ribbon. • • • • • • • • • • • • • • • • • • • • • • • Pawe� l Hitczenko and Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Natural Tree Structure • • • • • • • • • • • • • • • • There is a bijection between tree–like tableaux and permutations which “preserves trees”. Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes The PASEP Tree-like tableaux provide a combinatorial formula for the partially asymmetric simple exclusion process (PASEP), an important particle model with applications in physics, biology and biochemistry. α q β u Figure: An example of the PASEP as defined by a Markov chain of size 8. Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Connection with the PASEP • • • • • • • • • • • • • ◦ • ◦ ◦ • • • ◦ ◦ • • ◦ Figure: A tree–like tableau and its associated state of the PASEP as represented by a Markov chain of size 12. Tree-like tableaux provide a combinatorial formula for the PASEP. Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Main Objective: Laborde Zubieta 2015 Conjecture: The number of corners in tree–like tableaux of size n is n ! × n +4 6 . Conjecture: The number of corners in symmetric tree–like tableaux of size 2 n + 1 is 2 n × n ! × 4 n +13 . 12 Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Permutation Tableaux Definition 0 1 0 0 1 1 A permutation tableaux of size n is a 0 0 1 1 Ferrers diagram of half-perimeter n such 0 1 1 1 that 0 There is at least one 1 in every 1 1 column. There is no 0 with a 1 above it and 2 a 1 to the left of it simultaneously. Figure: A permutation tableaux of size 12. Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes The Bijection: (Aval, Boussicault, Nadeau 2011) 1 1 1 • • • • 0 0 • ← → 0 1 • • Figure: An example of the bijection c ( T n ) = c ( P n ) + |{ P ∈ P n : M n ( P ) = S }| = c ( P n ) + ( n − 1)! Theorem (Hitczenko, L.) For permutation tableaux of size n, E n C n = n + 4 − 1 n . 6 Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Symmetric Tree–Like Tableaux • • • • • • • • • Figure: A symmetric tree–like tableaux of size 9. There are 2 n · n ! tableaux of size n . Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Type-B Permutation Tableaux 1 Definition 0 0 A type-B permutation tableaux of size n 0 1 1 is a shifted Ferrers diagram of half-perimeter n such that, 0 1 0 0 The rules of permutation tableaux 1 are satisfied. If there is a 0 on the diagonal, it is 2 Figure: A type-B a 0-row. permutation tableaux of size 6. Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes The Bijection: (Aval, Boussicault, Nadeau 2011) • • • • • • 1 • 0 0 ← → • • 0 1 • 1 • Figure: An example of the bijection between type–B permutation tableaux and symmetric tree–like tableaux. Add a column and a root point then point unrestricted rows. 1 Replace all 0 R ’s with points (except on 0-rows). 2 Replace all non-diagonal 1 T ’s with points. 3 Make symmetric. 4 Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Symmetric Tableaux The Bijection • • • • 1 • • • 0 1 • 0 ← → • 1 • • Figure: Transformation of the Shape Key Relationship: c ( T sym 2 n +1 ) = 2 c ( B n ) + 2 |{ B ∈ B n : M n ( B ) = S }| + |{ B ∈ B n : M 1 ( B ) = W }| c ( B n ) + 2 n ( n − 1)! + 2 n − 1 n ! , = Pawe� l Hitczenko and Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Technique for proof: Corteel and Hitczenko 2007 If U n − 1 is the number of unrestricted rows, then there are 2 U n − 1 +1 extensions. 2 1 P ( U n = U n − 1 + 1 | F n − 1 ) = 2 U n − 1 +1 = 2 U n − 1 . 1 �� U n − 1 � � U n − 1 �� P ( U n = k | F n − 1 ) = + 2 U n − 1 +1 k − 1 k − 1 1 � U n − 1 � = . 2 U n − 1 k − 1 Therefore, L ( U n |F n − 1 ) = 1 + Bin( U n − 1 , 1 / 2) . Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Relationship Between Measures For all B ∈ B n − 1 , P n ( B ) = 2 U n − 1 ( B )+1 = 2 U n − 1 ( B )+1 |B n − 1 | |B n | P n − 1 ( B ) |B n | Therefore, for any random variable X on B n − 1 , E n ( X ) = 2 |B n − 1 | E n − 1 (2 U n − 1 ( B n − 1 ) X ) |B n | Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Simple Illustration Proposition (known but new proof: Hitczenko, L.) For all n ≥ 0 , |B n | = 2 n n ! . Proof can be deduced using the following: L ( U n |F n − 1 ) = 1 + Bin( U n − 1 , 1 / 2) . E n ( X ) = 2 |B n − 1 | E n − 1 (2 U n − 1 ( B n − 1 ) X ) |B n | � a + 1 � � a Bin( m ) � = E 2 Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Corners in Type-B Permutation Tableaux Theorem (Hitczenko, L.) For type-B permutation tableaux of size n we have E n C n = 4 n + 7 − 1 2 n . 24 Proof. � n − 1 n − 1 � � � � � E n I M k = S , M k +1 = W = E n I M k = S , M k +1 = W k =1 k =1 Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Occupied Corners: Laborde Zubieta 2015 x oc ( T ) satisfies the The generating polynomial P n ( x ) := � T ∈T n following recurrence: � ′ ′ n ( x ) = nP n − 1 ( x ) + 2(1 − x ) P n − 1 ( x ) P P 0 ( x ) = 1 Theorem (Hitczenko, L.) As n → ∞ , the limiting distribution of the number of occupied corners in a random tree–like tableau of size n is Pois(1) . Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Proof. Theorem (Hitczenko, L.) k =0 a n , k x k satisfy, Let P n ( x ) = � m ′ ′ P n ( x ) = f n ( x ) P n − 1 ( x ) + g n ( x ) P n − 1 ( x ) where g n (1) = 0 , g ′ n ( x ) = g n = o ( f n ) , f ′ n ( x ) = 0 and f n P n − 1 (1) → c > 0 , as n → ∞ . P n (1) If a sequence of random variables X n is defined by a n , k a n , k P ( X n = k ) = P n (1) = j a n , j , � then d X n → Pois( c ) as n → ∞ . Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
Tree–Like Tableaux Corners Occupied Corners Diagonal Boxes Diagonal Boxes: Aval, Boussicault, Nadeau 2011 • • • • • • • • • Figure: A symmetric tree–like tableaux with three diagonal boxes. If B ( n , k ) is the number of symmetric tree–like tableaux of size 2 n + 1 with k k =1 B ( n , k ) x k satisfies the following recurrence, diagonal cells, then B n ( x ) = � n +1 � B n ( x ) = nx ( x + 1) B n − 1 ( x ) + x (1 − x 2 ) B ′ n − 1 ( x ) , B 0 ( x ) = x . Pawe� l Hitczenko & Amanda Lohss Corners in Tree–Like Tableaux
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