Michelle Wachs University of Miami Based on joint … › files › wam › mwachs.pdfMichelle Wachs University of Miami Based on joint work with John Shareshian Binomial Coe cients

Eulerian Polynomials and Beyond

Michelle WachsUniversity of Miami

Based on joint work with John Shareshian

Binomial Coefficients and Eulerian numbers

11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

) ⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩

• coeff’s of polynomial(t + 1)n =

∑nj=0

(nj

)t j

• coeff’s of Eulerian polynomial

An(t) =∑n−1

j=0

⟨nj

⟩t j

• rows add to 2n • rows add to n!

• subsets of {1, 2, . . . , n} ofsize j

• permutations in Sn withj descents

• Rows are palindromic and unimodal.


11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

)

⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩


∑nj=0

(nj

)t j


An(t) =∑n−1

j=0

⟨nj

⟩t j






11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

) ⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩


∑nj=0

(nj

)t j


An(t) =∑n−1

j=0

⟨nj

⟩t j






11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

) ⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩


∑nj=0

(nj

)t j


An(t) =∑n−1

j=0

⟨nj

⟩t j






11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

) ⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩


∑nj=0

(nj

)t j


An(t) =∑n−1

j=0

⟨nj

⟩t j






11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

) ⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩


∑nj=0

(nj

)t j


An(t) =∑n−1

j=0

⟨nj

⟩t j






11 1

1 2 11 3 3 1

1 4 6 4 1

11 1

1 4 11 11 11 1

1 26 66 26 1

(nj

)=(n − 1

j − 1

)+(n − 1

j

) ⟨nj

⟩= (n − j)

⟨n − 1j − 1

⟩+(j + 1)

⟨n − 1

j

⟩


∑nj=0

(nj

)t j


An(t) =∑n−1

j=0

⟨nj

⟩t j





Eulerian polynomials - Euler’s definition

∑i≥1

t i =t

1− t

∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1

Euler’s definition∑i≥1

int i =t An(t)

(1− t)n+1

Leonhard Euler(1707-1783)

Euler’s exponential generating function formula∑n≥0

An(t)zn

n!=

1− t

e(t−1)z − t=

(1− t)ez

etz − tez


∑i≥1

t i =t

1− t∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1


int i =t An(t)

(1− t)n+1



An(t)zn

n!=

1− t

e(t−1)z − t=

(1− t)ez

etz − tez


∑i≥1

t i =t

1− t∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1


int i =t An(t)

(1− t)n+1



An(t)zn

n!=

1− t

e(t−1)z − t=

(1− t)ez

etz − tez


∑i≥1

t i =t

1− t∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1


int i =t An(t)

(1− t)n+1



An(t)zn

n!=

1− t

e(t−1)z − t=

(1− t)ez

etz − tez


∑i≥1

t i =t

1− t∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1


int i =t An(t)

(1− t)n+1



An(t)zn

n!=

1− t

e(t−1)z − t=

(1− t)ez

etz − tez


∑i≥1

t i =t

1− t∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1


int i =t An(t)

(1− t)n+1



An(t)zn

n!=

1− t

e(t−1)z − t

=(1− t)ez

etz − tez


∑i≥1

t i =t

1− t∑i≥1

it i =t

(1− t)2

∑i≥1

i2t i =t(1 + t)

(1− t)3

∑i≥1

i3t i =t(1 + 4t + t2)

(1− t)4

Euler’s triangle

11 1

1 4 11 11 11 1

1 26 66 26 1


int i =t An(t)

(1− t)n+1



An(t)zn

n!=

1− t

e(t−1)z − t=

(1− t)ez

etz − tez

Eulerian numbers - combinatorial interpretation

For σ ∈ Sn,

Descent set: DES(σ) := {i ∈ [n − 1] : σ(i) > σ(i + 1)}

σ = 3.25.4.1 DES(σ) = {1, 3, 4}

Define des(σ) := |DES(σ)|. So

des(32541) = 3

Excedance set: EXC(σ) := {i ∈ [n − 1] : σ(i) > i}

σ = 32541 EXC(σ) = {1, 3}

Define exc(σ) := |EXC(σ)|. So

exc(32541) = 2


For σ ∈ Sn,

Descent set: DES(σ) := {i ∈ [n − 1] : σ(i) > σ(i + 1)}

σ = 3.25.4.1 DES(σ) = {1, 3, 4}

Define des(σ) := |DES(σ)|. So

des(32541) = 3

Excedance set: EXC(σ) := {i ∈ [n − 1] : σ(i) > i}

σ = 32541 EXC(σ) = {1, 3}

Define exc(σ) := |EXC(σ)|. So

exc(32541) = 2


S3 des exc

123 0 0

132 1 1

213 1 1

231 1 2

312 1 1

321 2 1

∑σ∈S3

tdes(σ) = 1+4t+t2

∑σ∈S3

texc(σ) = 1+4t+t2

11 1

1 4 11 11 11 1

1 26 66 26 1

Eulerian polynomial

An(t) =n−1∑j=0

⟨nj

⟩t j =

∑σ∈Sn

tdes(σ) =∑σ∈Sn

texc(σ)

MacMahon (1905) showed equidistribution of des and exc.Carlitz and Riordin (1955) showed equals An(t).


S3 des exc

123 0 0

132 1 1

213 1 1

231 1 2

312 1 1

321 2 1

∑σ∈S3

tdes(σ) = 1+4t+t2

∑σ∈S3

texc(σ) = 1+4t+t2

11 1

1 4 11 11 11 1

1 26 66 26 1

Eulerian polynomial

An(t) =n−1∑j=0

⟨nj

⟩t j =

∑σ∈Sn


texc(σ)



S3 des exc

123 0 0

132 1 1

213 1 1

231 1 2

312 1 1

321 2 1

∑σ∈S3

tdes(σ) = 1+4t+t2

∑σ∈S3

texc(σ) = 1+4t+t2

11 1

1 4 11 11 11 1

1 26 66 26 1

Eulerian polynomial

An(t) =n−1∑j=0

⟨nj

⟩t j =

∑σ∈Sn


texc(σ)


Unimodality of Eulerian Polynomials

Probably not a new formula

An(t) =

b n+12c∑

m=1

∑k1,...,km≥2

(n

k1 − 1, k2, . . . , km

)tm−1

m∏i=1

[ki − 1]t

where[k]t := 1 + t + · · ·+ tk−1.

Sum & Product Lemma: Let A(t) and B(t) be Positive, Unimodal,Palindromic with respective centers of symmetry cA and cB . Then

A(t)B(t) is PUP with center of symmetry cA + cB .

If cA = cB then A(t) + B(t) is PUP with center of symmetrycA.

Center of symmetry:

(m − 1) +m∑i=1

ki − 2

2=

1

2(n − 1).



An(t) =

b n+12c∑

m=1

∑k1,...,km≥2

(n

k1 − 1, k2, . . . , km

)tm−1

m∏i=1

[ki − 1]t

where[k]t := 1 + t + · · ·+ tk−1.




Center of symmetry:

(m − 1) +m∑i=1

ki − 2

2=

1

2(n − 1).



An(t) =

b n+12c∑

m=1

∑k1,...,km≥2

(n

k1 − 1, k2, . . . , km

)tm−1

m∏i=1

[ki − 1]t

where[k]t := 1 + t + · · ·+ tk−1.




Center of symmetry:

(m − 1) +m∑i=1

ki − 2

2=

1

2(n − 1).

Mahonian Permutation Statistics - q-analogs

Let σ ∈ Sn.

Inversion Number:

inv(σ) := |{(i , j) : 1 ≤ i < j ≤ n, σ(i) > σ(j)}|.

inv(3142) = 3

Major Index:

maj(σ) :=∑

i∈DES(σ)

i

maj(3142) = maj(3.14.2) = 1 + 3 = 4

Major Percy Alexander MacMahon

(1854 - 1929)


S3 inv maj

123 0 0

132 1 2

213 1 1

231 2 2

312 2 1

321 3 3

∑σ∈S3

qinv(σ) =∑σ∈S3

qmaj(σ)

= 1 + 2q + 2q2 + q3

= (1 + q + q2)(1 + q)

Theorem (MacMahon 1905)∑σ∈Sn

qinv(σ) =∑σ∈Sn

qmaj(σ) = [n]q!

where [n]q := 1 + q + · · ·+ qn−1 and [n]q! := [n]q[n − 1]q · · · [1]q


S3 inv maj

123 0 0

132 1 2

213 1 1

231 2 2

312 2 1

321 3 3

∑σ∈S3


qmaj(σ)

= 1 + 2q + 2q2 + q3

= (1 + q + q2)(1 + q)



qmaj(σ) = [n]q!



S3 inv maj

123 0 0

132 1 2

213 1 1

231 2 2

312 2 1

321 3 3

∑σ∈S3


qmaj(σ)

= 1 + 2q + 2q2 + q3

= (1 + q + q2)(1 + q)



qmaj(σ) = [n]q!



S3 inv maj

123 0 0

132 1 2

213 1 1

231 2 2

312 2 1

321 3 3

∑σ∈S3


qmaj(σ)

= 1 + 2q + 2q2 + q3

= (1 + q + q2)(1 + q)



qmaj(σ) = [n]q!


q-Eulerian polynomials

Ainv,desn (q, t) :=

∑σ∈Sn

qinv(σ)tdes(σ)

Amaj,desn (q, t) :=

∑σ∈Sn

qmaj(σ)tdes(σ)

Ainv,excn (q, t) :=

∑σ∈Sn

qinv(σ)texc(σ)

Amaj,excn (q, t) :=

∑σ∈Sn

qmaj(σ)texc(σ)


Ainv,desn (q, t) :=

∑σ∈Sn

qinv(σ)tdes(σ)

Amaj,desn (q, t) :=

∑σ∈Sn

qmaj(σ)tdes(σ)

Ainv,excn (q, t) :=

∑σ∈Sn

qinv(σ)texc(σ)

Amaj,excn (q, t) :=

∑σ∈Sn

qmaj(σ)texc(σ)

Theorem (Carlitz 1954)∑i≥1

[i ]nq t i =tAmaj,des

n (q, t)∏ni=0(1− tqi )


Ainv,desn (q, t) :=

∑σ∈Sn

qinv(σ)tdes(σ)

Amaj,desn (q, t) :=

∑σ∈Sn

qmaj(σ)tdes(σ)

Ainv,excn (q, t) :=

∑σ∈Sn

qinv(σ)texc(σ)

Amaj,excn (q, t) :=

∑σ∈Sn

qmaj(σ)texc(σ)

Theorem (Carlitz 1954 MacMahon 1916)∑i≥1

[i ]nq t i =tAmaj,des

n (q, t)∏ni=0(1− tqi )

q-analogs of Euler’s exp. generating function formula

Theorem (Stanley 1976)∑n≥0

Ainv,desn (q, t)

zn

[n]q!=

1− t

Expq(z(t − 1))− t

where

Expq(z) :=∑n≥0

q(n2)zn

[n]q!

Theorem (Shareshian & MW 2006)∑n≥0

Amaj,excn (q, t)

zn

[n]q!=

(1− tq) expq(z)

expq(ztq)− tq expq(z)

where

expq(z) :=∑n≥0

zn

[n]q!

q-analogs of Euler’s exp. generating function formula

Theorem (Stanley 1976)∑n≥0

Ainv,desn (q, t)

zn

[n]q!=

1− t

Expq(z(t − 1))− t

where

Expq(z) :=∑n≥0

q(n2)zn

[n]q!


Amaj,excn (q, t)

zn

[n]q!=

(1− tq) expq(z)

expq(ztq)− tq expq(z)

where

expq(z) :=∑n≥0

zn

[n]q!

q-Eulerian polynomials and q-Eulerian numbers


Amaj,excn (q, tq−1)

zn

[n]q!=

(1− t) expq(z)

expq(zt)− t expq(z)

where

expq(z) :=∑n≥0

zn

[n]q!




zn

[n]q!=

(1− t) expq(z)


where

expq(z) :=∑n≥0

zn

[n]q!

Specialization of (quasi)symmetric function identity∑n≥0

n∑j=0

Qn,j(x)t jzn =(1− t)H(z)

H(zt)− tH(z),

the Qn,j(x) are what we call Eulerian quasisymmetric functions(a sum of certain fundamental quasisymmetric functions)

H(z) :=∑

n≥0 hn(x)zn (complete homogeneous symmetricfunctions)




zn

[n]q!=

(1− t) expq(z)


where

expq(z) :=∑n≥0

zn

[n]q!

From now on

An(q, t) := Amaj,excn (q, tq−1) =

∑σ∈Sn

qmaj(σ)−exc(σ)texc(σ)

and ⟨nj

⟩q

:=∑σ∈Sn

exc(σ)=j

qmaj(σ)−exc(σ)

Palindromicity and unimodality of the q-Eulerian numbers

n\j 0 1 2 3 4

1 1

2 1 1

3 1 2 + q + q2 1

4 1 3 + 2q + 3q2 + 2q3 + q4 3 + 2q + 3q2 + 2q3 + q4 1

5 1 4 + 3q + 5q2 + ... 6 + 6q + 11q2 + ... 4 + 3q + 5q2 + ... 1

Theorem (Shareshian and MW)

The q-Eulerian polynomial An(q, t) =∑n−1

t=0

⟨nj

⟩qt j is

palindromic in the sense that⟨

nj

⟩q

=⟨

nn − 1− j

⟩qfor

0 ≤ j ≤ n−12

q-unimodal in the sense that⟨

nj

⟩q−⟨

nj − 1

⟩q∈ N[q] for

1 ≤ j ≤ n−12

Palindromicity and unimodality of the q-Eulerian numbers

Proof: We use our q-analog of Euler’s exponential generatingfunction formula to prove

An(q, t) =

b n+12c∑

m=1

∑k1,...,km≥2

[n

k1 − 1, k2, . . . , km

]q

tm−1m∏i=1

[ki − 1]t ,

where [n

k1, . . . , km

]q

=[n]q!

[k1]q! · · · [km]q!

Then apply the Sum & Product Lemma.

Geometric Interpretation of the Eulerian numbers

(⟨n0

⟩,⟨

n1

⟩, . . . ,

⟨n

n − 1

⟩)is the h-vector of the type An−1 Coxeter

complex ∆n.

Stanley (1980): The h-vector of every simplicial convex polytope isunimodal (and palindromic).

This is part of the celebrated g -theorem of Billera, Lee and Stanley.

Proof idea: Let P be a d-dimensional convex polytope in Rd withinteger vertices. Let VP be the toric variety associated with P.

Danilov (1978): The h-vector of P equals

(dimH0(VP), dimH2(VP), . . . , dimH2d(VP))

(Hence⟨

nj

⟩= dimH2j(V∆n).)


(⟨n0

⟩,⟨

n1

⟩, . . . ,

⟨n

n − 1


complex ∆n.






(Hence⟨

nj



(⟨n0

⟩,⟨

n1

⟩, . . . ,

⟨n

n − 1


complex ∆n.






(Hence⟨

nj



Theorem (Hard Lefschetz Theorem)

Let V be a “smooth” irreducible complex projective variety of(complex) dimension m. Then for some ω ∈ H2(V) and alli = 0, . . . ,m, the map H i (V)→ H2m−i (V), given by multiplicationby ωm−i in the singular cohomology ring H∗(V), is a vector spaceisomorphism.

H2j(V)ω−→ H2(j+1)(V)

ω−→ · · · ω−→ H2m−2j(V)

=⇒ H2j(V)ω−→ H2(j+1)(V) is injective.

=⇒ dimH2j(V) ≤ dimH2(j+1)(V)

Consequently the Poincare polynomial∑m

j=0 dimH2j(V)t j isunimodal and palindromic.

Hence An(t) =∑n−1

j=0 dimH2j(V∆n)t j is unimodal and palindromic




H2j(V)ω−→ H2(j+1)(V)

ω−→ · · · ω−→ H2m−2j(V)










H2j(V)ω−→ H2(j+1)(V)

ω−→ · · · ω−→ H2m−2j(V)







Geometric interpretation of the q-Eulerian numbers

There is a natural action of Sn on the toric variety V∆n whichinduces a representation on each H2j(V∆n).

{Representations of Sn}ch−→ Λn

Zpsq−−→ Z[q]

ch: Frobenius characteristicΛnZ: homogeneous symmetric functions over Z of degree n

psq: stable principal specialization.

psq(f (x1, x2, . . . )) = f (1, q, q2, . . . )n∏

i=1

(1− qi )

Theorem (Shareshian & MW)⟨nj

⟩q

= psq(chH2j(V∆n))

Follows from∑n≥0

n∑j=0


H(zt)− tH(z)=∑n≥0

n∑j=0

chH2j(V∆n)t jzn

Shareshian and MW Procesi and Stanley

Geometric interpretation of the q-Eulerian numbersThere is a natural action of Sn on the toric variety V∆n whichinduces a representation on each H2j(V∆n).


Zpsq−−→ Z[q]



psq(f (x1, x2, . . . )) = f (1, q, q2, . . . )n∏

i=1

(1− qi )


⟩q

= psq(chH2j(V∆n))


n∑j=0



n∑j=0

chH2j(V∆n)t jzn


Geometric interpretation of the q-Eulerian numbersThere is a natural action of Sn on the toric variety V∆n whichinduces a representation on each H2j(V∆n).


Zpsq−−→ Z[q]



psq(f (x1, x2, . . . )) = f (1, q, q2, . . . )n∏

i=1

(1− qi )


⟩q

= psq(chH2j(V∆n))


n∑j=0



n∑j=0

chH2j(V∆n)t jzn


Geometric interpretation of the q-Eulerian numbers

The hard Lefschetz map ω commutes with the action of Sn onV = V∆n . This gives Sn-module maps for j < m

2

H2j(V)ω−→ H2(j+1)(V)

ω−→ · · · ω−→ H2m−2j(V)

=⇒ H2j(V)ω−→ H2(j+1)(V) is an Sn-module injection

=⇒ chH2(j+1)(V)− chH2j(V) is Schur-positive

=⇒⟨

nj + 1

⟩q−⟨

nj

⟩q∈ N[q]

=⇒ An(q, t) is q-unimodal and palindromic.

Rawlings’ Mahonian permutation statistic (1981)

Let 1 ≤ r ≤ n. For σ ∈ Sn, set

inv<r (σ) := |{(i , j) : 1 ≤ i < j ≤ n, 0 < σ(i)− σ(j)< r}|.DES≥r (σ) := {i ∈ [n − 1] : σ(i)− σ(i + 1)≥ r}maj≥r (σ) :=

∑i∈DES≥r

i

r = 1: inv<1(σ) = 0 maj≥1(σ) = maj(σ)

r = n: inv<n(σ) = inv(σ) maj≥n(σ) = 0

Theorem (Rawlings, 1981)∑σ∈Sn

qmaj≥r (σ)+inv<r (σ) = [n]q!

A(r)n (q, t) :=

∑σ∈Sn

qmaj≥r (σ)t inv<r (σ)

Rawlings’ Mahonian permutation statistic (1981)

Let 1 ≤ r ≤ n. For σ ∈ Sn, set

inv<r (σ) := |{(i , j) : 1 ≤ i < j ≤ n, 0 < σ(i)− σ(j)< r}|.DES≥r (σ) := {i ∈ [n − 1] : σ(i)− σ(i + 1)≥ r}maj≥r (σ) :=

∑i∈DES≥r

i

r = 1: inv<1(σ) = 0 maj≥1(σ) = maj(σ)

r = n: inv<n(σ) = inv(σ) maj≥n(σ) = 0

Theorem (Rawlings, 1981)∑σ∈Sn

qmaj≥r (σ)+inv<r (σ) = [n]q!

A(r)n (q, t) :=

∑σ∈Sn


A(r)n (q, t) :=

∑σ∈Sn


A(1)n (q, t) =

∑σ∈Sn

qmaj(σ) = [n]q!

A(n)n (q, t) =

∑σ∈Sn

t inv(σ) = [n]t !

A(2)n (q, t) =?

The (< 2)-inversions are

635412 635412 635412

(635142)−1 = 46.25.3.1

inv<2(σ) = des(σ−1) A(2)n (1, t) =

∑σ∈Sn

tdes(σ−1) = An(t)


A(2)n (q, t) = An(q, t)

Proof involves Stanley’s theory of P-partitions, Gessel’s theory ofquasisymmetric functions, our Eulerian quasisymmetric functions.

A(r)n (q, t) :=

∑σ∈Sn


A(1)n (q, t) =

∑σ∈Sn

qmaj(σ) = [n]q!

A(n)n (q, t) =

∑σ∈Sn

t inv(σ) = [n]t !

A(2)n (q, t) =?


635412 635412 635412

(635142)−1 = 46.25.3.1

inv<2(σ) = des(σ−1) A(2)n (1, t) =

∑σ∈Sn



A(2)n (q, t) = An(q, t)


A(r)n (q, t) :=

∑σ∈Sn


A(1)n (q, t) =

∑σ∈Sn

qmaj(σ) = [n]q!

A(n)n (q, t) =

∑σ∈Sn

t inv(σ) = [n]t !

A(2)n (q, t) =?


635412 635412 635412

(635142)−1 = 46.25.3.1

inv<2(σ) = des(σ−1) A(2)n (1, t) =

∑σ∈Sn



A(2)n (q, t) = An(q, t)


A(r)n (q, t) :=

∑σ∈Sn


A(1)n (q, t) =

∑σ∈Sn

qmaj(σ) = [n]q!

A(n)n (q, t) =

∑σ∈Sn

t inv(σ) = [n]t !

A(2)n (q, t) =?


635412 635412 635412

(635142)−1 = 46.25.3.1

inv<2(σ) = des(σ−1) A(2)n (1, t) =

∑σ∈Sn



A(2)n (q, t) = An(q, t)


A(r)n (q, t) :=

∑σ∈Sn


A(1)n (q, t) =

∑σ∈Sn

qmaj(σ) = [n]q!

A(n)n (q, t) =

∑σ∈Sn

t inv(σ) = [n]t !

A(2)n (q, t) =?


635412 635412 635412

(635142)−1 = 46.25.3.1

inv<2(σ) = des(σ−1) A(2)n (1, t) =

∑σ∈Sn



A(2)n (q, t) = An(q, t)


A(r)n (q, t) :=

∑σ∈Sn


So A(r)n (1, t) is a generalized Eulerian polynomial and A

(r)n (q, t) is

a generalized q-Eulerian polynomial.

r A(r)n (q, t)

1 [n]q!

2

b n+12c∑

m=1

∑k1,...,km≥2

[n

k1 − 1, k2, . . . , km

]q

tm−1m∏i=1

[ki − 1]t

... ???

n − 2 [n]t [n − 3]t ![n − 3]2t + [n]q tn−3[n − 4]t ![n − 2]t([n − 3]t + [2]t [n − 4]t)

+

[n

n − 2, 2

]q

t3n−10[n − 4]![n − 2]t [2]t

n − 1 [n]t [n − 2]t ![n − 2]t + [n]qtn−2[n − 2]t !

n [n]t !

Conjecture (Shareshian and MW)

A(r)n (q, t) is q-unimodal (and palindromic).

A(r)n (q, t) :=

∑σ∈Sn


So A(r)n (1, t) is a generalized Eulerian polynomial and A

(r)n (q, t) is

a generalized q-Eulerian polynomial.

r A(r)n (q, t)

1 [n]q!

2

b n+12c∑

m=1

∑k1,...,km≥2

[n

k1 − 1, k2, . . . , km

]q

tm−1m∏i=1

[ki − 1]t

... ???

n − 2 [n]t [n − 3]t ![n − 3]2t + [n]q tn−3[n − 4]t ![n − 2]t([n − 3]t + [2]t [n − 4]t)

+

[n

n − 2, 2

]q

t3n−10[n − 4]![n − 2]t [2]t

n − 1 [n]t [n − 2]t ![n − 2]t + [n]qtn−2[n − 2]t !

n [n]t !


A(r)n (q, t) is q-unimodal (and palindromic).

A(r)n (q, t) :=

∑σ∈Sn


Exercise (Stanley EC1, 1.50 f): Prove that A(r)n (1, t) is palindromic

and unimodal.Solution:

Theorem (De Mari and Shayman - 1988)

Let Hn,r be the type An−1 regular semisimple Hessenberg varietyof degree r . Then

A(r)n (1, t) =

d(n,r)∑j=0

dimH2j(Hn,r )t j

Consequently by the hard Lefschetz theorem, A(r)n (1, t) is

palindromic and unimodal.

Stanley: Is there a more elementary proof of unimodality?Shareshian and MW: Is there a q-analog of this result?(Hn,2 is the toric variety V∆n)

A(r)n (q, t) :=

∑σ∈Sn






A(r)n (1, t) =

d(n,r)∑j=0

dimH2j(Hn,r )t j



Stanley: Is there a more elementary proof of unimodality?

Shareshian and MW: Is there a q-analog of this result?(Hn,2 is the toric variety V∆n)

A(r)n (q, t) :=

∑σ∈Sn






A(r)n (1, t) =

d(n,r)∑j=0

dimH2j(Hn,r )t j



Stanley: Is there a more elementary proof of unimodality?Shareshian and MW: Is there a q-analog of this result?(Hn,2 is the toric variety V∆n)

Hessenberg Varieties (De Mari and Shayman - 1988)

Let Fn be the set of all flags

F : V1 ⊂ V2 ⊂ · · · ⊂ Vn = Cn

where dimVi = i . Fix X ∈ GLn(C) with n distinct eigenvalues.

The type A regular semisimple Hessenberg variety of degree r is

Hn,r := {F ∈ Fn | XVi ⊆ Vi+r−1 for all i}

For q-unimodality we want an action of Sn on Hn,r .

The Representation

Tymoczko (2008) used a theory of Goresky, Kottwitz andMacPherson (GKM theory) to define a representation of Sn

on H2j(Hn,r ).MacPherson & Tymoczko show that the hard Lefschetz mapcommutes with the action of Sn on H2j(Hn,r ) .


A(r)n (q, t) =

d(n,r)∑j=0

psq(chH2j(Hn,r ))t j

Consequently A(r)n (q, t) is q-unimodal (and palindromic).

The r = 2 case is the toric variety case.

A(2)n (q, t) = An(q, t) =

n−1∑j=0

psq(chH2j(V∆n))t j =n−1∑j=0

psq(chH2j(Hn,2))t j

Also true for r = 1, n − 2, n − 1, n

The Representation

Tymoczko (2008) used a theory of Goresky, Kottwitz andMacPherson (GKM theory) to define a representation of Sn

on H2j(Hn,r ).MacPherson & Tymoczko show that the hard Lefschetz mapcommutes with the action of Sn on H2j(Hn,r ) .


A(r)n (q, t) =

d(n,r)∑j=0

psq(chH2j(Hn,r ))t j

Consequently A(r)n (q, t) is q-unimodal (and palindromic).

The r = 2 case is the toric variety case.

A(2)n (q, t) = An(q, t) =

n−1∑j=0

psq(chH2j(V∆n))t j =n−1∑j=0

psq(chH2j(Hn,2))t j

Also true for r = 1, n − 2, n − 1, n

Chromatic symmetric functions

1

23

4

5

67 8

Wednesday, December 7, 2011

1

23

4

5

67 8

77

15 15

23

23

23

35

23

23

Let V (G ) = {1, 2, . . . , n}. Let C (G ) be set of proper colorings ofG , where a proper coloring is a map c : V (G )→ P such thatc(i) 6= c(j) if {i , j} ∈ E (G ).

Chromatic symmetric function (Stanley, 1995)

XG (x) :=∑

c∈C(G)

xc(1)xc(2) . . . xc(n)

XG (1, 1, . . . , 1︸︷︷︸, 0, 0, . . . ) = χG (m)

m


1

23

4

5

67 8

1

23

4

5

67 8

77

15 15

23

23

23

35

23

23



XG (x) :=∑

c∈C(G)

xc(1)xc(2) . . . xc(n)

XG (1, 1, . . . , 1︸︷︷︸, 0, 0, . . . ) = χG (m)

m


1

23

4

5

67 8

77

15 15

23

23

23

35

23

23



XG (x) :=∑

c∈C(G)

xc(1)xc(2) . . . xc(n)

XG (1, 1, . . . , 1︸︷︷︸, 0, 0, . . . ) = χG (m)

m


1

23

4

5

67 8

77

15 15

23

23

23

35

23

23



XG (x) :=∑

c∈C(G)

xc(1)xc(2) . . . xc(n)

XG (1, 1, . . . , 1︸︷︷︸, 0, 0, . . . ) = χG (m)

m


1

23

4

5

67 8

77

15 15

23

23

23

35

23

23



XG (x) :=∑

c∈C(G)

xc(1)xc(2) . . . xc(n)

XG (1, 1, . . . , 1︸︷︷︸, 0, 0, . . . ) = χG (m)

m

Chromatic quasisymmetric function

1

23

4

5

67 8

77

15 15

23

23

23

35

23

23

Chromatic quasisymmetric function (Shareshian and MW)

XG (x, t) :=∑

c∈C(G)

tdes(c)xc(1)xc(2) . . . xc(n)

where

des(c) := |{{i , j} ∈ E (G ) : i < j and c(i) > c(j)}|.

Chromatic quasisymmetric function

eλ(x1, x2, . . . ) := elementary symmetric function indexed bypartition λ

Fµ(x1, x2, . . . ) := fundamental quasisymmetric functionindexed by composition µ

1 2 3G =

XG (x, t) = e3 + (e3 + e2,1)t + e3t2

1 3 2G =

XG (x, t) = (e3 + F1,2) + 2e3t + (e3 + F2,1)t2

Formulae

Let Gn,r be the graph with vertex set {1, 2, . . . , n} and edge set{{i , j} | 0 < |j − i | < r}.

r XGn,r (x, t)

1 en1

2

b n+12c∑

m=1

∑k1,...,km≥2∑

ki=n+1

ek1−1,k2,...,km tm−1m∏i=1

[ki − 1]t

... ???

n − 2 en[n]t [n − 3]t ![n − 3]2t + en−1,1 tn−3[n − 4]t ![n − 2]t([n − 3]t + [2]t [n − 4]t)

+en−2,2t3n−10[n − 4]![n − 2]t [2]t

n − 1 en[n]t [n − 2]t ![n − 2]t + en−1,1tn−2[n − 2]t !

n en[n]t !


A(r)n (q, t) = psq(ωXGn,r (x, t))

Formulae


r XGn,r (x, t)

1 en1

2

b n+12c∑

m=1

∑k1,...,km≥2∑

ki=n+1

ek1−1,k2,...,km tm−1m∏i=1

[ki − 1]t

... ???


+en−2,2t3n−10[n − 4]![n − 2]t [2]t


n en[n]t !


A(r)n (q, t) = psq(ωXGn,r (x, t))

The Main Conjecture


Let G be the incomparability graph of a natural unit interval order.Then

ωXG (x, t) =

d(n,r)∑j=0

chH2j(HG )t j

where HG is the Hessenberg variety associated with G .

⇒ q-unimodality of A(r)n (q, t) (set G = Gn,r )

⇒ Schur-positivity and Schur-unimodality of XG (x, t)

Theorem (Shareshian and MW - t-analog of Gasharov)

Let G be the incomparability graph of a natural unit interval orderP. Then

XG (x, t) =∑T∈TP

t invG (T )sλ(T ),

where TP is the set of P-tableaux, invG is an inversion statistic ontableaux, and λ(T ) is the shape of T .

The Main Conjecture



ωXG (x, t) =

d(n,r)∑j=0

chH2j(HG )t j







t invG (T )sλ(T ),


The Main Conjecture



ωXG (x, t) =

d(n,r)∑j=0

chH2j(HG )t j







t invG (T )sλ(T ),


Refinement of Stanley-Stembridge Conjecture


Let G be the incomparability graph of a natural unit interval order.Then XG (x, t) is e-positive and e-unimodal.


r XGn,r

1 e1n

2

b n+12c∑

m=1

∑k1,...,km≥2∑

ki=n+1

ek1−1,k2,...,km tm−1m∏i=1

[ki − 1]t

... ???


+en−2,2t3n−10[n − 4]![n − 2]t [2]t


n en[n]t !

Documents

Michelle Wachs University of Miami Based on joint … › files › wam › mwachs.pdfMichelle Wachs University of Miami Based on joint work with John Shareshian Binomial Coe cients