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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
Main result
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
1 Pad ,bd ?
2 hbk [ak xx] ?3 Frob ?
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 2 / 29
Main result
Main result
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
1 Pad ,bd ?
2 hbk [ak xx] ?3 Frob ?
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 2 / 29
Main result
Main result
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
1 Pad ,bd ?
2 hbk [ak xx] ?3 Frob ?
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 2 / 29
Main result
Main result
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
1 Pad ,bd ?2 hbk [ak xx] ?
3 Frob ?
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 2 / 29
Main result
Main result
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
1 Pad ,bd ?2 hbk [ak xx] ?3 Frob ?
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 2 / 29
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 3 / 29
Rectangular parking functions
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 4 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
Dyck paths
Dyck paths(square n × n)
Catalan numbers1
n+1
(2nn
)
Dyck paths(rational a× b)(a ∧ b = 1)
RationalCatalan numbers
1a+b
(a+ba
)Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 5 / 29
Rectangular parking functions
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.
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 6 / 29
Rectangular parking functions
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.
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 6 / 29
Rectangular parking functions
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.
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 6 / 29
Rectangular parking functions
Rectangular Dyck paths
Dyck paths(in a rectangle m × n)
(m, n) = (ad , bd), a ∧ b = 1
Dm,n set of (rectangular) Dyck paths m × n.
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 6 / 29
Rectangular parking functions
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.
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 6 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Rectangular parking functions
Rectangular parking functions
Rectangular parking functions(m × n)
1
2
3
4
6
5
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 7 / 29
Symmetric functions
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 8 / 29
Symmetric functions
Symmetric functions
xx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
Symmetric functions
Symmetric functionsxx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
Symmetric functions
Symmetric functionsxx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
Symmetric functions
Symmetric functionsxx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
Symmetric functions
Symmetric functionsxx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
Symmetric functions
Symmetric functionsxx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
Symmetric functions
Symmetric functionsxx = x1, x2, . . .
pk(xx) =∑i≥1
xki
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 9 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
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
2
)h11(xx)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 10 / 29
Symmetric functions
Complete 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
n
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Symmetric functions
Complete 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
n
)Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 11 / 29
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 12 / 29
Frobenius characteristic
Symmetric group action
Sn symmetric group
|Sn| = n!group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group
|Sn| = n!group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group
|Sn| = n!
group of permutations over n letters
σ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group
|Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n
3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n
3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =
( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =( nρ(γ)
)
= n!r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group actionSn symmetric group |Sn| = n!
group of permutations over n lettersσ = σ1 σ2 · · · σn
γ ∈ Bm,n3
6
1
4
5
2
(6, 8, 4, 6, 6, 4)
Lγ increasing labellings of γ
|Lγ | =( nρ(γ)
)= n!
r1! r2! ··· rl ! for ρ(γ) = (r1, r2, . . . , rl)
∑γ∈Bm,n
|Lγ | = mn
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 13 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
3
6
1
4
5
2
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 13
6
1
4
5
2
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 15
1
4
6
3
2
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Symmetric group action
Sn acts on Lγ by permuting the labels
σ = 4 2 5 6 3 1
5
1
4
6
3
2
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
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 14 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 15 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Frobenius characteristic
Frobenius characteristic
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]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 16 / 29
Proof of the 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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 17 / 29
Proof of the main result
The formula
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 18 / 29
Proof of the main result
The formula
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 18 / 29
Proof of the main result
γ ∈ Bm,n 2 low points
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n
2 low points
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n
2low points
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n
2 lowpoints
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n
2 lowpoints
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n
2 low points
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
γ ∈ Bm,n 2 low points (1 ≤ t ≤ d)
Rotation
9
87
6
54
32
1
87
6
54
32
19
7
6
54
32
19
8
6
54
32
19
87
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 19 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
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]]
←→
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 20 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx)
=1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx)
=1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx)
=1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx)
=1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx) =
1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx) =
1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx) =
1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑α∈Dt
m,n
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
1t
Φtd(xx) =
1ad
∑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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 21 / 29
Proof of the main result
∑t>0
1t
Φtd(xx) =
1ad
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
1ad
hbd [ad xx] =(− log(1− P(xx; z))
)∣∣zd
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 22 / 29
Proof of the main result
∑t>0
1t
Φtd(xx) =
1ad
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
1ad
hbd [ad xx] =(− log(1− P(xx; z))
)∣∣zd
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 22 / 29
Proof of the main result
∑t>0
1t
Φtd(xx) =
1ad
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
1ad
hbd [ad xx] =(− log(1− P(xx; z))
)∣∣zd
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 22 / 29
Proof of the main result
∑t>0
1t
Φtd(xx) =
1ad
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
1ad
hbd [ad xx] =(− log(1− P(xx; z))
)∣∣zd
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 22 / 29
Proof of the main result
P(xx; z) = 1− exp(−∑k>0
1ak
hbk [ak xx] zk)
Φtd(xx) =
(P(xx; z)
)t∣∣zd
=⇒ Φd(xx) =∑t>0
Φtd(xx) =
(1
1− P(xx; z)
)∣∣zd
∑d≥0
Φd(xx) zd =1
1− P(xx; z)= exp
(∑k>0
1ak
hbk [ak xx] zk)
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 23 / 29
Proof of the main result
P(xx; z) = 1− exp(−∑k>0
1ak
hbk [ak xx] zk)
Φtd(xx) =
(P(xx; z)
)t∣∣zd
=⇒ Φd(xx) =∑t>0
Φtd(xx) =
(1
1− P(xx; z)
)∣∣zd
∑d≥0
Φd(xx) zd =1
1− P(xx; z)= exp
(∑k>0
1ak
hbk [ak xx] zk)
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 23 / 29
Proof of the main result
P(xx; z) = 1− exp(−∑k>0
1ak
hbk [ak xx] zk)
Φtd(xx) =
(P(xx; z)
)t∣∣zd
=⇒ Φd(xx) =∑t>0
Φtd(xx) =
(1
1− P(xx; z)
)∣∣zd
∑d≥0
Φd(xx) zd =1
1− P(xx; z)= exp
(∑k>0
1ak
hbk [ak xx] zk)
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 23 / 29
Proof of the main result
P(xx; z) = 1− exp(−∑k>0
1ak
hbk [ak xx] zk)
Φtd(xx) =
(P(xx; z)
)t∣∣zd
=⇒ Φd(xx) =∑t>0
Φtd(xx) =
(1
1− P(xx; z)
)∣∣zd
∑d≥0
Φd(xx) zd =1
1− P(xx; z)= exp
(∑k>0
1ak
hbk [ak xx] zk)
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 23 / 29
Proof of the main result
P(xx; z) = 1− exp(−∑k>0
1ak
hbk [ak xx] zk)
Φtd(xx) =
(P(xx; z)
)t∣∣zd
=⇒ Φd(xx) =∑t>0
Φtd(xx) =
(1
1− P(xx; z)
)∣∣zd
∑d≥0
Φd(xx) zd =1
1− P(xx; z)= exp
(∑k>0
1ak
hbk [ak xx] zk)
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 23 / 29
Consequences
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 24 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
∣∣Pad ,bd ∣∣ zd = exp
(∑k≥1
1ak
(ak)bk zk
)
= exp
(∑k≥1
(ak)bk−1 zk
)
∑d≥0
∣∣Dad ,bd
∣∣ zd = exp
(∑k≥1
1ak
(ak + bk − 1
bk
)zk
)
= exp
(∑k≥1
1ak + bk
(ak + bk
bk
)zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 25 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
∣∣Pad ,bd ∣∣ zd = exp
(∑k≥1
1ak
(ak)bk zk
)
= exp
(∑k≥1
(ak)bk−1 zk
)
∑d≥0
∣∣Dad ,bd
∣∣ zd = exp
(∑k≥1
1ak
(ak + bk − 1
bk
)zk
)
= exp
(∑k≥1
1ak + bk
(ak + bk
bk
)zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 25 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
∣∣Pad ,bd ∣∣ zd = exp
(∑k≥1
1ak
(ak)bk zk
)
= exp
(∑k≥1
(ak)bk−1 zk
)
∑d≥0
∣∣Dad ,bd
∣∣ zd = exp
(∑k≥1
1ak
(ak + bk − 1
bk
)zk
)
= exp
(∑k≥1
1ak + bk
(ak + bk
bk
)zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 25 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
∣∣Pad ,bd ∣∣ zd = exp
(∑k≥1
1ak
(ak)bk zk
)
= exp
(∑k≥1
(ak)bk−1 zk
)
∑d≥0
∣∣Dad ,bd
∣∣ zd = exp
(∑k≥1
1ak
(ak + bk − 1
bk
)zk
)
= exp
(∑k≥1
1ak + bk
(ak + bk
bk
)zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 25 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
∣∣Pad ,bd ∣∣ zd = exp
(∑k≥1
1ak
(ak)bk zk
)
= exp
(∑k≥1
(ak)bk−1 zk
)
∑d≥0
∣∣Dad ,bd
∣∣ zd = exp
(∑k≥1
1ak
(ak + bk − 1
bk
)zk
)
= exp
(∑k≥1
1ak + bk
(ak + bk
bk
)zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 25 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
∑d≥0
∣∣Pad ,bd ∣∣ zd = exp
(∑k≥1
1ak
(ak)bk zk
)
= exp
(∑k≥1
(ak)bk−1 zk
)
∑d≥0
∣∣Dad ,bd
∣∣ zd = exp
(∑k≥1
1ak
(ak + bk − 1
bk
)zk
)
= exp
(∑k≥1
1ak + bk
(ak + bk
bk
)zk
)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 25 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
Frob(P2a,2b
)=
12a
h2b[2a xx] +12
(1ahb[a xx]
)2
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
Frob(P2a,2b
)=
12a
h2b[2a xx] +12
(1ahb[a xx]
)2
Frob(P3a,3b
)=
13a
h3b[3a xx]+
(12a
h2b[2a xx]
)(1ahb[a xx]
)+16
(1ahb[a xx]
)3
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Consequences
∑d≥0
Frob(Pad ,bd
)zd = exp
(∑k≥1
1ak
hbk [ak xx] zk
)
Frob(Pad ,bd
)=∑µ`d
1mµ!
Ha/bµ (xx)
µ = (µ1, µ2, . . . , µ`) = (1m1 , 2m2 , . . . )mµ! =
∏j mj !
Ha/bµ (xx) =
∏`i=1
1aµi
haµi [bµi xx]
|D2a,2b| =
(1
2a + 2b
(2a + 2b
2a
))+
12
(1
a + b
(a + b
a
))2
|D3a,3b| =
(1
3a+3b
(3a+3b3a
))+
(1
a+b
(a+b
a
))(1
2a+2b
(2a+2b2a
))+
16
(1
a + b
(a + b
a
))3
[Bizley 1954]
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 26 / 29
Generalization (Schröder)
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 27 / 29
Generalization (Schröder)
Generalization - Schröder parking functions
Rectangular Schröder parking functions(m × n)
3
2
4
1
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 28 / 29
Generalization (Schröder)
Generalization - Schröder parking functions
Rectangular Schröder parking functions(m × n)
3
2
4
1
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 28 / 29
Generalization (Schröder)
Generalization - Schröder parking functions
Rectangular Schröder parking functions(m × n)
3
2
4
1
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 28 / 29
Generalization (Schröder)
Generalization - Schröder parking functions
Rectangular Schröder parking functions(m × n)
3
2
4
1
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 28 / 29
Generalization (Schröder)
Generalization - Schröder parking functions
Rectangular Schröder parking functions(m × n)
3
2
4
1
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 28 / 29
Generalization (Schröder)
Generalization - Schröder parking functions
Rectangular Schröder parking functions(m × n)
3
2
4
1
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)
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 28 / 29
Références
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.
Jean-Christophe Aval Rectangular Parking Functions IHP - 01/06/2017 29 / 29