Combinatorics of the PASEP partition function Matthieu Josuat-Verg` - PowerPoint PPT Presentation

Combinatorics of the PASEP partition function Matthieu Josuat-Verg` es Universit´ e Paris-sud 11 FPSAC 2010 August 2nd 1 / 35

Introduction The Partially Asymmetric Self-Exclusion Process (PASEP) is a probabilistic model describing the evolution of particles in a finite number N of sites. q . dt dt q . dt β. dt α. dt dt In the time interval dt , possible events are: ◮ if ◦ , the leftmost site becomes • with probability α. dt ◮ if • , the rightmost site becomes ◦ with probability β. dt ◮ •◦ becomes ◦• with probability dt ◮ ◦• becomes •◦ with probability q . dt 2 / 35

Introduction The partition function Z N is the normalization constant for stationary probabilities (such that non-normalized probabilities of each state is polynomial in the parameters). Z N is a polynomial in α , 1 1 β , q . Many physical quantities can be obtained from the partition function (phase diagram, correlation functions, currents, density profile...) There are various ways to obtain the partition function: ◮ Matrix-product form of Z N [Derrida & al] ◮ Generating function of lattice paths [Brak & al] ◮ Generating function of permutation tableaux, permutations [Corteel Williams] 3 / 35

Introduction Theorem (Derrida & al) Let D and E be matrices, W a row vector, V a column vector, such that: WE = 1 DV = 1 DE − qED = D + E , α W , β V , WV = 1 , then for any m word in D and E of length N, ◮ WmV defines a unique polynomial in 1 α , 1 β , q, ◮ and it is the non-normalized probability of the state m (under D ↔ • , E ↔ ◦ ) Example DED = qEDD + DD + ED , so the non-normalized probability of • ◦ • is αβ 2 + 1 q β 2 + 1 WDEDV = αβ . 4 / 35

It follows that the sum of non-normalized probabilities, is � WmV = W ( D + E ) N V Z N = m ∈{ D , E } N It is interesting to consider an extra variable y and define Z N ( y ) = W ( yD + E ) N V This way, the coefficient of y k in Z N ( y ) corresponds to states with exactly k particles. Example � y 2 D 2 + y ( DE + ED ) + EE � Z 2 ( y ) = W V � y 2 D 2 + y � � � = W (1 + q ) ED + D + E + EE V = y 2 � 1 + q + 1 β + 1 + 1 � β 2 + y α 2 αβ α 5 / 35

α − 1, ˜ α = (1 − q ) 1 β = (1 − q ) 1 Let ˜ β − 1. It is possible to give explicit D , E , W , V satisfying DE − qED = D + E , WE = 1 α W , DV = 1 β V , WV = 1 [Derrida & al]. 1 + ˜ 1 − q (0)  β  1 + ˜ 1 − q 2 β q   1  1 + ˜  β q 2 1 − q 3 D = ,   1 − q ...   1 + ˜ β q 3   ... (0)  1 + ˜ α (0)  α ˜ 1 − ˜ β 1 + ˜ α q 1   α ˜  α q 2  E = 1 − ˜ β q 1 + ˜ ,   1 − q   α ˜ β q 2 α q 3 1 − ˜ 1 + ˜   ... ... (0) V = (1 , 0 , 0 , . . . ) ∗ . W = (1 , 0 , 0 , . . . ) , 6 / 35

Introduction yD + E is tridiagonal and can be thought as a transfer matrix, then   1 0   Z N ( y ) = (1 , 0 , 0 , . . . )( yD + E ) N   0    .  . . shows that Z N ( y ) is a generating function of weighted Motzkin paths. 7 / 35

Introduction Motivated by the PASEP and the links with permutations, we want to evaluate Z N , characterized by (1 − q ) N Z N = � w ( P ) Motzkin path P of length N where the weight w ( P ) is the product of 1 − q h +1 for a step ր at height h to h + 1, ✄ α + y ˜ β ) q h (1 + y ) + (˜ for a step → at height h , ✄ α ˜ β q h − 1 ) y (1 − ˜ for a step ց at height h to h − 1, ✄ α = (1 − q ) 1 α − 1, ˜ β = (1 − q ) 1 where ˜ β − 1. 8 / 35

Introduction Theorem (Blythe & al) N 1 � Z N (1) = R N , n B n (1 − q ) N n =0 where ⌊ N − n ⌋ 2 ( − 1) i q ( i +1 �� �� 2 ) � n + i � 2 N 2 N � � � R N , n = − i N − n − 2 i N − n − 2 i − 2 q i =0 and n � n � α k ˜ � β n − k . B n = ˜ k q k =0 Proof. Diagonalize D + E in terms of q -Hermite polynomials, this gives an integral which can be simplified analytically. 9 / 35

Introduction Theorem N 1 � Z N ( y ) = R N , n B n (1 − q ) N n =0 where ⌊ N − n ⌋ N − n − 2 i 2 � N ( − y ) i q ( i +1 2 ) � n + i y j �� N �� � � N N � �� � �� R N , n = − n +2 i + j j − 1 n +2 i + j +1 i q j i =0 j =0 and n � n � � α k ( y ˜ β ) n − k . B n = ˜ k q k =0 This will be proved bijectively. 10 / 35

Outline N To prove (1 − q ) N Z N = � R N , n B n , the scheme is the following. n =0 ◮ Step 0. Define some sets Z N , R N , n , B n of weighted Motzkin paths, Z N having generating function (1 − q ) N Z N N ◮ Step 1. Give a weight-preserving bijection Z N → � R N , n × B n n =0 ◮ Step 2. Show that R N , n has generating function R N , n ◮ Step 3. Show that B n has generating function B n 11 / 35

Outline N To prove (1 − q ) N Z N = � R N , n B n , the scheme is the following. n =0 ◮ Step 0. Define some sets Z N , R N , n , B n of weighted Motzkin paths, Z N having generating function (1 − q ) N Z N N ◮ Step 1. Give a weight-preserving bijection Z N → � R N , n × B n n =0 ◮ Step 2. Show that R N , n has generating function R N , n ◮ Step 3. Show that B n has generating function B n 12 / 35

Step 0 (1 − q ) N Z N is the g.f. of Motzkin paths of length N , with weights: 1 − q h +1 for a step ր at height h to h + 1, ✄ α + y ˜ β ) q h (1 + y ) + (˜ for a step → at height h , ✄ α ˜ β q h − 1 ) y (1 − ˜ for a step ց at height h to h − 1. ✄ Instead of a weight 1 − q h +1 on steps ր , we can write 1 − q h +1 = (1 − q ) + ( q − q 2 ) + · · · + ( q h − q h +1 ) and consider that the weight is q i − q i +1 with 0 ≤ i ≤ h for a step ր at height h to h + 1. 13 / 35

Step 0 Let Z N be the set of Motzkin paths of length N with weights q i − q i +1 with 0 ≤ i ≤ h for a step ր at height h to h + 1, ✄ α + y ˜ β ) q h 1 + y or (˜ for a step → at height h ✄ α ˜ β q h − 1 y or − y ˜ for a step ց at height h to h − 1. ✄ It is bigger than the original set of Motzkin paths since there are several possible choices on each step but the generating function is also (1 − q ) N Z N . 14 / 35

Step 0 Let R N , n be the set of Motzkin paths of length N with weights q i − q i +1 with 0 ≤ i ≤ h for a step ր at height h to h + 1, ✄ 1 + y or q h for a step → at height h ✄ n steps having a weight q h , y for a step ց ✄ Let B n be the set of Motzkin paths of length n with weights q i − q i +1 with 0 ≤ i ≤ h for a step ր at height h to h + 1, ✄ α + y ˜ β ) q h (˜ for a step → at height h , ✄ α ˜ β q h − 1 − y ˜ for a step ց at height h to h − 1. ✄ 15 / 35

Outline N To prove (1 − q ) N Z N = � R N , n B n , the scheme is the following. n =0 ◮ Step 0. Define some sets Z N , R N , n , B n of weighted Motzkin paths, Z N having generating function (1 − q ) N Z N N ◮ Step 1. Give a weight-preserving bijection Z N → � R N , n × B n n =0 ◮ Step 2. Show that R N , n has generating function R N , n ◮ Step 3. Show that B n has generating function B n 16 / 35

Step 1 The bijection is best described in the direction N � R N , n × B n − → Z N n =0 Let ( H 1 , H 2 ) ∈ R N , n × B n . We build Λ( H 1 , H 2 ) ∈ Z N . The definition is given via an example. 17 / 35

Step 1 H 1 = Let ( H 1 , H 2 ) ∈ R N , n × B n . We build Λ( H 1 , H 2 ) ∈ Z N : 1 − q Λ( H 1 , H 2 ) = q q − q 2 q 2 q 2 1 − q 1 − q q − q 2 y q − q 2 y α + y ˜ β ) q 3 (˜ q q 2 − q 3 y 1 − q q 0 y y α ˜ H 2 = β q 2 − y ˜ y α ˜ 1 − q − y ˜ β α + y ˜ (˜ β ) q 1 − q α ˜ − y ˜ β q α ˜ − y ˜ β 18 / 35

Step 1 H 1 = Let ( H 1 , H 2 ) ∈ R N , n × B n . We build Λ( H 1 , H 2 ) ∈ Z N : 1 − q Λ( H 1 , H 2 ) = q q − q 2 q 2 q 2 1 − q 1 − q q − q 2 y q − q 2 y α + y ˜ β ) q 3 (˜ q q 2 − q 3 y 1 − q q 0 y y α ˜ H 2 = β q 2 − y ˜ y α ˜ 1 − q − y ˜ β α + y ˜ (˜ β ) q 1 − q α ˜ − y ˜ β q α ˜ − y ˜ β 19 / 35

Step 1 H 1 = Let ( H 1 , H 2 ) ∈ R N , n × B n . We build Λ( H 1 , H 2 ) ∈ Z N : 1 − q Λ( H 1 , H 2 ) = q q − q 2 q 2 q 2 1 − q 1 − q q − q 2 y q − q 2 y α + y ˜ β ) q 3 (˜ q q 2 − q 3 y 1 − q q 0 y y α ˜ H 2 = β q 2 − y ˜ y α ˜ 1 − q − y ˜ β α + y ˜ (˜ β ) q 1 − q α ˜ − y ˜ β q α ˜ − y ˜ β 20 / 35

Step 1 H 1 = Let ( H 1 , H 2 ) ∈ R N , n × B n . We build Λ( H 1 , H 2 ) ∈ Z N : 1 − q Λ( H 1 , H 2 ) = q q − q 2 q 2 q 2 1 − q 1 − q q − q 2 y q − q 2 y α + y ˜ β ) q 3 (˜ q q 2 − q 3 y 1 − q q 0 y y α ˜ H 2 = β q 2 − y ˜ y α ˜ 1 − q − y ˜ β α + y ˜ (˜ β ) q 1 − q α ˜ − y ˜ β q α ˜ − y ˜ β 21 / 35

Step 1 H 1 = Let ( H 1 , H 2 ) ∈ R N , n × B n . We build Λ( H 1 , H 2 ) ∈ Z N : 1 − q Λ( H 1 , H 2 ) = q q − q 2 q 2 q 2 1 − q 1 − q q − q 2 y q − q 2 y α + y ˜ β ) q 3 (˜ q q 2 − q 3 y 1 − q q 0 y y α ˜ H 2 = β q 2 − y ˜ y α ˜ 1 − q − y ˜ β α + y ˜ (˜ β ) q 1 − q α ˜ − y ˜ β q α ˜ − y ˜ β 22 / 35