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

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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 dt α.dt β.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

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Introduction

The partition function ZN is the normalization constant for stationary probabilities (such that non-normalized probabilities of each state is polynomial in the parameters). ZN is a polynomial in

1 α, 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 ZN [Derrida & al] ◮ Generating function of lattice paths [Brak & al] ◮ Generating function of permutation tableaux, permutations

[Corteel Williams]

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Introduction

Theorem (Derrida & al)

Let D and E be matrices, W a row vector, V a column vector, such that: DE−qED = D+E, WE = 1 αW , DV = 1 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

WDEDV = q αβ2 + 1 β2 + 1 αβ .

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It follows that the sum of non-normalized probabilities, is ZN =

• m∈{D,E}N

WmV = W (D + E)NV It is interesting to consider an extra variable y and define ZN(y) = W (yD + E)NV This way, the coefficient of y k in ZN(y) corresponds to states with exactly k particles.

Example

Z2(y) = W

• y 2D2 + y(DE + ED) + EE
• V

= W

• y 2D2 + y
• (1 + q)ED + D + E
• + EE
• V

= y 2 β2 + y 1 + q αβ + 1 β + 1 α

• + 1

α2

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Let ˜ α = (1 − q) 1

α − 1, ˜

β = (1 − q) 1

β − 1.

It is possible to give explicit D, E, W , V satisfying DE − qED = D + E, WE = 1

αW , DV = 1 βV , WV = 1 [Derrida &

al]. D = 1 1 − q       1 + ˜ β 1 − q (0) 1 + ˜ βq 1 − q2 1 + ˜ βq2 1 − q3 1 + ˜ βq3 ... (0) ...       , E = 1 1 − q       1 + ˜ α (0) 1 − ˜ α ˜ β 1 + ˜ αq 1 − ˜ α˜ βq 1 + ˜ αq2 1 − ˜ α˜ βq2 1 + ˜ αq3 (0) ... ...       , W = (1, 0, 0, . . . ), V = (1, 0, 0, . . . )∗.

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Introduction

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

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Introduction

Motivated by the PASEP and the links with permutations, we want to evaluate ZN, characterized by (1 − q)NZN =

• Motzkin path P
• f length N

w(P) where the weight w(P) is the product of ✄ 1 − qh+1 for a step ր at height h to h + 1, ✄ (1 + y) + (˜ α + y ˜ β)qh for a step → at height h, ✄ y(1 − ˜ α ˜ βqh−1) for a step ց at height h to h − 1, where ˜ α = (1 − q) 1

α − 1, ˜

β = (1 − q) 1

β − 1.

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Introduction

Theorem (Blythe & al)

ZN(1) = 1 (1 − q)N

N

• n=0

RN,nBn where RN,n =

⌊ N−n

2

⌋

• i=0

(−1)iq(i+1

2 )n+i

i

• q
• 2N

N−n−2i

• −
• 2N

N−n−2i−2

• and

Bn =

n

• k=0

n k

• q

˜ αk ˜ βn−k.

Proof.

Diagonalize D + E in terms of q-Hermite polynomials, this gives an integral which can be simplified analytically.

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Introduction

Theorem

ZN(y) = 1 (1 − q)N

N

• n=0

RN,nBn where RN,n =

⌊ N−n

2

⌋

• i=0

(−y)iq(i+1

2 )n+i

i

• q

N−n−2i

• j=0

y j N

j

• N

n+2i+j

• −

N

j−1

• N

n+2i+j+1

• and

Bn =

n

• k=0

n k

• q

˜ αk(y ˜ β)n−k. This will be proved bijectively.

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Outline

To prove (1 − q)NZN =

N

• n=0

RN,nBn, the scheme is the following.

◮ Step 0. Define some sets ZN, RN,n, Bn of weighted Motzkin paths,

ZN having generating function (1 − q)NZN

◮ Step 1. Give a weight-preserving bijection ZN → N

• n=0

RN,n × Bn

◮ Step 2. Show that RN,n has generating function RN,n ◮ Step 3. Show that Bn has generating function Bn

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Outline

To prove (1 − q)NZN =

N

• n=0

RN,nBn, the scheme is the following.

◮ Step 0. Define some sets ZN, RN,n, Bn of weighted Motzkin paths,

ZN having generating function (1 − q)NZN

◮ Step 1. Give a weight-preserving bijection ZN → N

• n=0

RN,n × Bn

◮ Step 2. Show that RN,n has generating function RN,n ◮ Step 3. Show that Bn has generating function Bn

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Step 0

(1 − q)NZN is the g.f. of Motzkin paths of length N, with weights: ✄ 1 − qh+1 for a step ր at height h to h + 1, ✄ (1 + y) + (˜ α + y ˜ β)qh for a step → at height h, ✄ y(1 − ˜ α ˜ βqh−1) for a step ց at height h to h − 1. Instead of a weight 1 − qh+1 on steps ր, we can write 1 − qh+1 = (1 − q) + (q − q2) + · · · + (qh − qh+1) and consider that the weight is qi − qi+1 with 0 ≤ i ≤ h for a step ր at height h to h + 1.

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Step 0

Let ZN be the set of Motzkin paths of length N with weights ✄ qi − qi+1 with 0 ≤ i ≤ h for a step ր at height h to h + 1, ✄ 1 + y

• r

(˜ α + y ˜ β)qh for a step → at height h ✄ y or −y ˜ α˜ βqh−1 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)NZN.

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Step 0

Let RN,n be the set of Motzkin paths of length N with weights ✄ qi − qi+1 with 0 ≤ i ≤ h for a step ր at height h to h + 1, ✄ 1 + y or qh for a step → at height h n steps having a weight qh, ✄ y for a step ց Let Bn be the set of Motzkin paths of length n with weights ✄ qi − qi+1 with 0 ≤ i ≤ h for a step ր at height h to h + 1, ✄ (˜ α + y ˜ β)qh for a step → at height h, ✄ −y ˜ α˜ βqh−1 for a step ց at height h to h − 1.

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Outline

To prove (1 − q)NZN =

N

• n=0

RN,nBn, the scheme is the following.

◮ Step 0. Define some sets ZN, RN,n, Bn of weighted Motzkin paths,

ZN having generating function (1 − q)NZN

◮ Step 1. Give a weight-preserving bijection ZN → N

• n=0

RN,n × Bn

◮ Step 2. Show that RN,n has generating function RN,n ◮ Step 3. Show that Bn has generating function Bn

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Step 1

The bijection is best described in the direction

N

• n=0

RN,n × Bn − → ZN Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN. The definition is given via an example.

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α˜ βq −y ˜ α ˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α ˜ βq2 y −y ˜ α˜ β

Λ(H1, H2) =

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Step 1

Let (H1, H2) ∈ RN,n × Bn. We build Λ(H1, H2) ∈ ZN:

1 − q q q − q2 q2 q2 1 − q y y q y q0

H1 =

1 − q (˜ α + y ˜ β)q 1 − q −y ˜ α ˜ βq −y ˜ α˜ β

H2 =

1 − q q − q2 q − q2 (˜ α + y ˜ β)q3 q2 − q3 1 − q y y −y ˜ α˜ βq2 y −y ˜ α ˜ β

Λ(H1, H2) =

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Step 1

One can read Λ(H1, H2) from right to left and recover the paths H1, H2, this proves injectivity. If we forget the weights, there is a simple bijection between the two sets ZN and

N

• n=0

RN,n × Bn. (these are essentially bicolored involutions), this shows equality of cardinal. Thus Λ is a weight-preserving bijection Λ :

N

• n=0

RN,n × Bn → ZN

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Outline

To prove (1 − q)NZN =

N

• n=0

RN,nBn, the scheme is the following.

◮ Step 0. Define some sets ZN, RN,n, Bn of weighted Motzkin paths,

ZN having generating function (1 − q)NZN

◮ Step 1. Give a weight-preserving bijection ZN → N

• n=0

RN,n × Bn

◮ Step 2. Show that RN,n has generating function RN,n ◮ Step 3. Show that Bn has generating function Bn

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Step 2-3 (sketch)

Step 2: follows from the two different proofs for the particular case ZN(α = β = 1) [Corteel, JV, Rubey, Prellberg] Step 3: follows from known results on Al-Salam-Carlitz orthogonal polynomials.

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Outline

To prove (1 − q)NZN =

N

• n=0

RN,nBn, the scheme is the following.

◮ Step 0. Define some sets ZN, RN,n, Bn of weighted Motzkin paths,

ZN having generating function (1 − q)NZN

◮ Step 1. Give a weight-preserving bijection ZN → N

• n=0

RN,n × Bn

◮ Step 2. Show that RN,n has generating function RN,n ◮ Step 3. Show that Bn has generating function Bn

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Conclusion

We have a second proof of the formula for ZN(y) based on the

• perator point of view, using ˆ

D = (q − 1)D + I and ˆ E = (q − 1)E + I which satisfy ˆ D ˆ E − q ˆ E ˆ D = 1 − q, and combinatorics of rook placements in Young diagrams. Zn is essentially the moments of (rescaled) Al-Salam-Chihara

• rthogonal polynomials. There are analytical derivations of these

moments and some more general moments [Ismail Stanton].

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thanks for your attention

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