semflajolet.math.cnrs.frsemflajolet.math.cnrs.fr/uploads/Main/Aval-slides-IHP.pdf · Références...

Rectangular Parking Functions

Jean-Christophe Aval

Joint work with F. Bergeron (UQAM, Montréal)

IHP - 01/06/2017

Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 1 / 29

Main result

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

1 Pad ,bd ?

2 hbk [ak xx] ?3 Frob ?

Main result

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

1 Pad ,bd ?

Main result

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

1 Pad ,bd ?

Main result

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

1 Pad ,bd ?2 hbk [ak xx] ?

3 Frob ?

Main result

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

1 Pad ,bd ?2 hbk [ak xx] ?3 Frob ?

Main result

Outline

1 Main result

2 Rectangular parking functions

3 Symmetric functions

4 Frobenius characteristic

5 Proof of the main result

6 Consequences

7 Generalization (Schröder)

Rectangular parking functions

Outline

1 Main result

6 Consequences

Dyck paths

Dyck paths(square n × n)

Catalan numbers1

Dyck paths(rational a× b)(a ∧ b = 1)

RationalCatalan numbers

Dyck paths

Catalan numbers1

Dyck paths

Catalan numbers1

Dyck paths

Catalan numbers1

Dyck paths

Catalan numbers1

Dyck paths

Catalan numbers1

Dyck paths

Catalan numbers1

)Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29

Rectangular Dyck paths

Dyck paths(in a rectangle m × n)

(m, n) = (ad , bd), a ∧ b = 1Dm,n set of (rectangular) Dyck paths m × n.

(m, n) = (ad , bd), a ∧ b = 1

Dm,n set of (rectangular) Dyck paths m × n.

Rectangular parking functions(m × n)

Pm,n set of (rectangular) parking functions m × n

Lα set of parking functions with (Dyck) path α

|Pa,b| = ab−1 (a ∧ b = 1)

|Pn+1,n| = (n + 1)n−1 [Konheim-Weiss 1966]

|Pa,b| = ab−1 (a ∧ b = 1)

|Pn+1,n| = (n + 1)n−1

[Konheim-Weiss 1966]

|Pa,b| = ab−1 (a ∧ b = 1)

Symmetric functions

Outline

1 Main result

6 Consequences

Symmetric functions

xx = x1, x2, . . .

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk(xx + yy) = pk(xx) + pk(yy)

pk [2 xx] = pk(xx + xx) = 2 pk(xx)

pk [m xx] = mpk(xx)

Symmetric functions

Symmetric functionsxx = x1, x2, . . .

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk [m xx] = mpk(xx)

Symmetric functions

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk [m xx] = mpk(xx)

Symmetric functions

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk [m xx] = mpk(xx)

Symmetric functions

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk [m xx] = mpk(xx)

Symmetric functions

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk [m xx] = mpk(xx)

Symmetric functions

pk(xx) =∑i≥1

λ = (λ1, λ2, . . . , λ`)pλ = pλ1 pλ2 · · · pλ`

pk [m xx] = mpk(xx)

Symmetric functions

Complete symmetric functions

hk complete symmetric function of degree k∑k≥0

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2(xx + yy) = h2(xx) + h2(yy) + h1(xx) h1(yy)

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

h2[m xx] = mh2(xx) +(m

)h11(xx)

Symmetric functions

hk complete symmetric function of degree k

∑k≥0

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2(xx + yy) =

h2(xx) + h2(yy) + h1(xx) h1(yy)

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] =

h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hk tk =

∏i≥1

1(1− xi t)

hλ = hλ1 hλ2 · · · hλ`

h2(xx) = x21 + x2

2 + · · ·+ x1x2 + x1x3 + x2x3 + · · ·

h2[2 xx] = h2(xx + xx) = 2 h2(xx) + h11(xx)

)h11(xx)

Symmetric functions

hn[m xx] =

∑u1+u2+···+um=n

hu1(xx) hu2(xx) · · · hum(xx)

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

Bm,n set of rectangular paths m × n (with last step = East)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =

∑u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =∑

u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =∑

u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =∑

u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =∑

u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =∑

u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Symmetric functions

hn[m xx] =∑

u1+u2+···+um=n

γ∈Bm,n

hρ(γ)(xx) (h0 = 1)

(0, 0, 0, 2, 3, 0, 1, 0, 0)

γ ∈ Bm,n ρ(γ) = (3, 2, 1)

|Bm,n| =

(m + n − 1

Frobenius characteristic

Outline

1 Main result

6 Consequences

Symmetric group action

Sn symmetric group

|Sn| = n!group of permutations over n lettersσ = σ1 σ2 · · · σn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

Lγ increasing labellings of γ

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

Symmetric group actionSn symmetric group

|Sn| = n!group of permutations over n lettersσ = σ1 σ2 · · · σn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

|Sn| = n!

group of permutations over n letters

σ = σ1 σ2 · · · σn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

|Sn| = n!

group of permutations over n lettersσ = σ1 σ2 · · · σn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

Symmetric group actionSn symmetric group |Sn| = n!

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =

( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =( nρ(γ)

= n!r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

γ ∈ Bm,n3

(6, 8, 4, 6, 6, 4)

|Lγ | =( nρ(γ)

r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)

∑γ∈Bm,n

|Lγ | = mn

Sn acts on Lγ by permuting the labels

σ = 4 2 5 6 3 1

Transitive action on orbit = Lγ

Stabilizer of π ∈ Lγ = S2 × S3 × S1= Sr1 × Sr2 × · · · × Srl for ρ(γ) = (r1, r2, . . . , rl)

−→ Trivial action of Sr1 × Sr2 × · · · × Srl induced on Sn

σ = 4 2 5 6 3 1

σ = 4 2 5 6 3 13

σ = 4 2 5 6 3 15

σ = 4 2 5 6 3 1

Stabilizer of π ∈ Lγ

= S2 × S3 × S1= Sr1 × Sr2 × · · · × Srl for ρ(γ) = (r1, r2, . . . , rl)

σ = 4 2 5 6 3 1

Stabilizer of π ∈ Lγ = S2 × S3 × S1

= Sr1 × Sr2 × · · · × Srl for ρ(γ) = (r1, r2, . . . , rl)

σ = 4 2 5 6 3 1

Action of Sn with character χ

Cyclic type σ = 4 2 5 6 3 1 = (1, 4, 6) (2) (3, 5) λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Cyclic type σ = 4 2 5 6 3 1 = (1, 4, 6) (2) (3, 5) λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Cyclic type

σ = 4 2 5 6 3 1 = (1, 4, 6) (2) (3, 5) λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Cyclic type σ = 4 2 5 6 3 1

= (1, 4, 6) (2) (3, 5) λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Cyclic type σ = 4 2 5 6 3 1 = (1, 4, 6) (2) (3, 5)

λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Cyclic type σ = 4 2 5 6 3 1 = (1, 4, 6) (2) (3, 5) λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Cyclic type σ = 4 2 5 6 3 1 = (1, 4, 6) (2) (3, 5) λ(σ) = (3, 2, 1)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

• Frob is an isomorphism

• Frob(χλ) = sλ (Schur)

• Frob(ηn) = hn ηλ = IndSnSλ1×···×Sλ`ηλ1 × · · · × ηλ` Frob(ηλ) = hλ

Thus for γ ∈ Bm,nFrob(Lγ) = hρ(γ)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

• Frob(ηn) = hn

ηλ = IndSnSλ1×···×Sλ`ηλ1 × · · · × ηλ` Frob(ηλ) = hλ

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

• Frob(ηn) = hn ηλ = IndSnSλ1×···×Sλ`ηλ1 × · · · × ηλ`

Frob(ηλ) = hλ

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Frob(χ) =1n!

∑σ∈Sn

χ(σ)pλ(σ)(xx)

∑γ∈Bm,n

Frob(Lγ) =∑

γ∈Bm,n

hρ(γ) = hn[m xx]

Proof of the main result

Outline

1 Main result

6 Consequences

The formula

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

The formula

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

γ ∈ Bm,n 2 low points

Rotation

γ ∈ Bm,n

2 low points

Rotation

γ ∈ Bm,n

2low points

Rotation

γ ∈ Bm,n

2 lowpoints

Rotation

γ ∈ Bm,n

2 lowpoints

Rotation

γ ∈ Bm,n

2 low points

Rotation

γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)

Rotation

Properties of the rotation

• for γ′ image of γ by rotation :

ρ(γ′) = ρ(γ) =⇒ Frob(Lγ′) = Frob(Lγ)

• the number t of low points is preserved

• bijection : Dtm,n × [[1,m]] −→ Btm,n × [[1, t]]

←→

ρ(γ′) = ρ(γ)

=⇒ Frob(Lγ′) = Frob(Lγ)

←→

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

ad,bdFrob(Lα) a and b fixed

Φ1d(xx) =

∑α∈Dad,bd

primitiveFrob(Lα) (m = ad)

∑t>0

Φtd(xx)

∑t>0

∑γ∈Btm,n

hρ(γ) =1ad

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

∑t>0

Φtd(xx)

∑t>0

∑γ∈Btm,n

hρ(γ) =1ad

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

primitiveFrob(Lα)

(m = ad)

∑t>0

Φtd(xx)

∑t>0

∑γ∈Btm,n

hρ(γ) =1ad

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

primitiveFrob(Lα)

(m = ad)

∑t>0

Φtd(xx)

∑t>0

∑γ∈Btm,n

hρ(γ) =1ad

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

primitiveFrob(Lα)

(m = ad)

∑t>0

Φtd(xx) =

∑t>0

∑γ∈Btm,n

hρ(γ)

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

∑t>0

Φtd(xx) =

∑t>0

∑γ∈Btm,n

hρ(γ)

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

∑t>0

Φtd(xx) =

∑t>0

∑γ∈Btm,n

hρ(γ) =1ad

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑α∈Dt

Frob(Lα)×m =∑

γ∈Btm,n

Frob(Lγ)× t

Φtd(xx) :=

∑α∈Dt

Φ1d(xx) =

∑α∈Dad,bd

∑t>0

Φtd(xx) =

∑t>0

∑γ∈Btm,n

hρ(γ) =1ad

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

∑t>0

Φtd(xx) =

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

P(xx; z) :=∑

j>0 Φ1j (xx) z j

Φtd(xx) =

(P(xx; z)

)t∣∣zd

hbd [ad xx] =(− log(1− P(xx; z))

)∣∣zd

∑t>0

Φtd(xx) =

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

P(xx; z) :=∑

j>0 Φ1j (xx) z j

Φtd(xx) =

(P(xx; z)

)t∣∣zd

)∣∣zd

∑t>0

Φtd(xx) =

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

P(xx; z) :=∑

j>0 Φ1j (xx) z j

Φtd(xx) =

(P(xx; z)

)t∣∣zd

)∣∣zd

∑t>0

Φtd(xx) =

hbd [ad xx]

Φtd(xx) =

∑c1,c2,...,ct>0

c1+c2+···+ct=d

Φ1c1(xx) Φ1

c2(xx) · · · Φ1ct (xx)

P(xx; z) :=∑

j>0 Φ1j (xx) z j

Φtd(xx) =

(P(xx; z)

)t∣∣zd

)∣∣zd

P(xx; z) = 1− exp(−∑k>0

hbk [ak xx] zk)

Φtd(xx) =

(P(xx; z)

)t∣∣zd

=⇒ Φd(xx) =∑t>0

Φtd(xx) =

1− P(xx; z)

)∣∣zd

∑d≥0

Φd(xx) zd =1

1− P(xx; z)= exp

(∑k>0

hbk [ak xx] zk)

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

P(xx; z) = 1− exp(−∑k>0

hbk [ak xx] zk)

Φtd(xx) =

(P(xx; z)

)t∣∣zd

=⇒ Φd(xx) =∑t>0

Φtd(xx) =

1− P(xx; z)

)∣∣zd

∑d≥0

Φd(xx) zd =1

1− P(xx; z)= exp

(∑k>0

hbk [ak xx] zk)

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

P(xx; z) = 1− exp(−∑k>0

hbk [ak xx] zk)

Φtd(xx) =

(P(xx; z)

)t∣∣zd

=⇒ Φd(xx) =∑t>0

Φtd(xx) =

1− P(xx; z)

)∣∣zd

∑d≥0

Φd(xx) zd =1

1− P(xx; z)= exp

(∑k>0

hbk [ak xx] zk)

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

P(xx; z) = 1− exp(−∑k>0

hbk [ak xx] zk)

Φtd(xx) =

(P(xx; z)

)t∣∣zd

=⇒ Φd(xx) =∑t>0

Φtd(xx) =

1− P(xx; z)

)∣∣zd

∑d≥0

Φd(xx) zd =1

1− P(xx; z)= exp

(∑k>0

hbk [ak xx] zk)

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

P(xx; z) = 1− exp(−∑k>0

hbk [ak xx] zk)

Φtd(xx) =

(P(xx; z)

)t∣∣zd

=⇒ Φd(xx) =∑t>0

Φtd(xx) =

1− P(xx; z)

)∣∣zd

∑d≥0

Φd(xx) zd =1

1− P(xx; z)= exp

(∑k>0

hbk [ak xx] zk)

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Consequences

Outline

1 Main result

6 Consequences

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

∣∣Pad ,bd ∣∣ zd = exp

(∑k≥1

(ak)bk zk

(∑k≥1

(ak)bk−1 zk

∑d≥0

∣∣Dad ,bd

∣∣ zd = exp

(∑k≥1

(ak + bk − 1

(∑k≥1

1ak + bk

(ak + bk

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

(∑k≥1

(ak)bk zk

(∑k≥1

(ak)bk−1 zk

∑d≥0

∣∣Dad ,bd

∣∣ zd = exp

(∑k≥1

(ak + bk − 1

(∑k≥1

1ak + bk

(ak + bk

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

(∑k≥1

(ak)bk zk

(∑k≥1

(ak)bk−1 zk

∑d≥0

∣∣Dad ,bd

∣∣ zd = exp

(∑k≥1

(ak + bk − 1

(∑k≥1

1ak + bk

(ak + bk

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

(∑k≥1

(ak)bk zk

(∑k≥1

(ak)bk−1 zk

∑d≥0

∣∣Dad ,bd

∣∣ zd = exp

(∑k≥1

(ak + bk − 1

(∑k≥1

1ak + bk

(ak + bk

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

(∑k≥1

(ak)bk zk

(∑k≥1

(ak)bk−1 zk

∑d≥0

∣∣Dad ,bd

∣∣ zd = exp

(∑k≥1

(ak + bk − 1

(∑k≥1

1ak + bk

(ak + bk

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

∑d≥0

(∑k≥1

(ak)bk zk

(∑k≥1

(ak)bk−1 zk

∑d≥0

∣∣Dad ,bd

∣∣ zd = exp

(∑k≥1

(ak + bk − 1

(∑k≥1

1ak + bk

(ak + bk

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

Frob(P2a,2b

h2b[2a xx] +12

(1ahb[a xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

Frob(P2a,2b

h2b[2a xx] +12

(1ahb[a xx]

Frob(P3a,3b

h3b[3a xx]+

h2b[2a xx]

)(1ahb[a xx]

(1ahb[a xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Consequences

∑d≥0

Frob(Pad ,bd

)zd = exp

(∑k≥1

hbk [ak xx] zk

Frob(Pad ,bd

)=∑µ`d

Ha/bµ (xx)

µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =

∏j mj !

Ha/bµ (xx) =

∏`i=1

haµi [bµi xx]

|D2a,2b| =

2a + 2b

(2a + 2b

(a + b

|D3a,3b| =

(3a+3b3a

(2a+2b2a

(a + b

[Bizley 1954]

Generalization (Schröder)

Outline

1 Main result

6 Consequences

Generalization - Schröder parking functions

Rectangular Schröder parking functions(m × n)

Sm,n(k) set of (rectangular) Schröder parking functions m × nwith k diagonal steps

∑k≥0

FrobS(k)m,n(xx) yk

= FrobPm,n(xx + y)

∑k≥0

FrobS(k)m,n(xx) yk

= FrobPm,n(xx + y)

∑k≥0

FrobS(k)m,n(xx) yk

= FrobPm,n(xx + y)

∑k≥0

FrobS(k)m,n(xx) yk

= FrobPm,n(xx + y)

∑k≥0

FrobS(k)m,n(xx) yk

= FrobPm,n(xx + y)

∑k≥0

FrobS(k)m,n(xx) yk = FrobPm,n(xx + y)

Références

J.-C. Aval, F. Bergeron, Interlaced Rectangular Parking Functions,preprint (2015) arXiv:1503.03991.

J.-C. Aval, F. Bergeron, Rectangular Schröder Parking FunctionsCombinatorics, preprint (2016) arXiv:1603.09487.

semflajolet.math.cnrs.frsemflajolet.math.cnrs.fr/uploads/Main/Aval-slides-IHP.pdf · Références...

Documents

Transcript of semflajolet.math.cnrs.frsemflajolet.math.cnrs.fr/uploads/Main/Aval-slides-IHP.pdf · Références...