Enumerative Combinatorics from an Algebraic-Geometric ... · Enumerative Combinatorics from an Algebraic-Geometric point of view Christos Athanasiadis University of Athens March 22,

Enumerative Combinatorics from an Algebraic-Geometricpoint of view

Christos Athanasiadis

University of Athens

March 22, 2018

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Outline

1 Three examples from combinatorics

2 An algebraic-geometric perspective

3 Gamma-positivity

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Eulerian polynomials

We let

• Sn be the group of permutations of [n] := {1, 2, . . . , n}

and for w ∈ Sn

• des(w) := # {i ∈ [n − 1] : w(i) > w(i + 1)}• exc(w) := # {i ∈ [n − 1] : w(i) > i}

be the number of descents and excedances of w , respectively. Thepolynomial

An(x) :=∑w∈Sn

xdes(w) =∑w∈Sn

xexc(w)

is the nth Eulerian polynomial.

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Example

An(x) =

1, if n = 1

1 + x , if n = 2

1 + 4x + x2, if n = 3

1 + 11x + 11x2 + x3, if n = 4

1 + 26x + 66x2 + 26x3 + x4, if n = 5

1 + 57x + 302x2 + 302x3 + 57x4 + x5, if n = 6.

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Derangement polynomials

We let

• Dn be the set of derangements (permutations without fixed points)in the symmetric group Sn.

For instance,

• D3 = { (2, 3, 1), (3, 1, 2) }.

The polynomial

dn(x) :=∑w∈Dn

xexc(w)

is the nth derangement polynomial.

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Example

dn(x) =

0, if n = 1

x , if n = 2

x + x2, if n = 3

x + 7x2 + x3, if n = 4

x + 21x2 + 21x3 + x4, if n = 5

x + 51x2 + 161x3 + 51x4 + x5, if n = 6

x + 113x2 + 813x3 + 813x4 + 113x5 + x6, if n = 7.

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Binomial Eulerian polynomials

The polynomial

An(x) := 1 + xn∑

k=1

(n

k

)Ak(x) =

n∑k=0

(n

k

)xn−kAk(x)

is the nth binomial Eulerian polynomial.

Example

An(x) =

1 + x , if n = 1

1 + 3x + x2, if n = 2

1 + 7x + 7x2 + x3, if n = 3

1 + 15x + 33x2 + 15x3 + x4, if n = 4

1 + 31x + 131x2 + 131x3 + 31x4 + x5, if n = 5

1 + 63x + 473x2 + 883x3 + 473x4 + 63x5 + x6, if n = 6.

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Note: All these polynomials are symmetric and unimodal. There is anendless list of generalizations, refinements and variations with similarproperties.

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Symmetry and unimodality

Definition

A polynomial f (x) ∈ R[x ] is

• symmetric (or palindromic) and• unimodal

if for some n ∈ N,

f (x) = p0 + p1x + p2x2 + · · ·+ pnx

n

with

• pk = pn−k for 0 ≤ k ≤ n and• p0 ≤ p1 ≤ · · · ≤ pbn/2c.

The number n/2 is called the center of symmetry.

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Question: Can algebra or geometry shed light into such phenomena?

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Face enumeration of simplicial complexes

We let

• ∆ be a simplicial complex of dimension n − 1• fi (∆) be the number of i-dimensional faces.

Definition

The h-polynomial of ∆ is defined as

h(∆, x) =n∑

i=0

fi−1(∆) x i (1− x)n−i =n∑

i=0

hi (∆) x i .

The sequence h(∆) = (h0(∆), h1(∆), . . . , hn(∆)) is the h-vector of ∆.

Note: h(∆, 1) = fn−1(∆).

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Example

For the 2-dimensional complex

∆ =

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ss

ss

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we have f0(∆) = 8, f1(∆) = 15 and f2(∆) = 8 and hence

h(∆, x) = (1− x)3 + 8x(1− x)2 + 15x2(1− x) + 8x3

= 1 + 5x + 2x2.

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Example

The boundary complex Σn of the n-dimensional cross-polytope is a trian-gulation of the (n − 1)-dimensional sphere:

We have

h(Σn, x) = (1 + x)n

for every n ≥ 1.

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The face ring

We let

• m be the number of vertices of ∆• k be a field• S = k[x1, x2, . . . , xm]• I∆ = 〈

∏i∈F xi : F /∈ ∆ 〉 be the Stanley–Reisner ideal of ∆

•k[∆] = S/I∆

be the Stanley–Reisner ring (or face ring) of ∆.

Then k[∆] is a graded k-algebra with Hilbert series∑i≥0

dimk(k[∆]i ) ti =

h(∆, t)

(1− t)n,

where n − 1 = dim(∆).14 / 52

Theorem (Klee, Reisner, Stanley)

The polynomial h(∆, x):

• has nonnegative coefficients if ∆ triangulates a ball or a sphere,• is symmetric if ∆ triangulates a sphere,• is unimodal if ∆ is the boundary complex of a simplicial polytope.

Note: If ∆ triangulates a ball or a sphere (more generally, if ∆ is Cohen–Macaulay over k), then there exists a graded quotient k(∆) of k[∆] suchthat hi (∆) = dimk(k(∆)i ) for every i .

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Example

We let

• V be an n-element set,• 2V be the simplex on the vertex set V ,• ∆ be the first barycentric subdivision of the boundary complex of 2V .

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Then h(∆, x) = An(x).

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Example

For n = 3

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∆ =

h(∆, x) = (1− x)2 + 6x(1− x) + 6x2 = 1 + 4x + x2.

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The local h-polynomial

We let

• V be an n-element set,• Γ be a triangulation of the simplex 2V on the vertex set V .

Definition (Stanley, 1992)

The local h-polynomial of Γ (with respect to V ) is defined as

`V (Γ, x) =∑F⊆V

(−1)n−|F | h(ΓF , x),

where ΓF is the restriction of Γ to the face F of the simplex 2V .

Note: This polynomial plays a major role in Stanley’s theory of subdivisi-ons of simplicial (and more general) complexes.

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Example

For the 2-dimensional triangulation

Γ =

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we have

`V (Γ, x) = (1 + 5x + 2x2) − (1 + 2x) − (1 + x) − 1

+ 1 + 1 + 1 − 1 = 2x + 2x2.

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Theorem (Stanley, 1992)

The polynomial `V (Γ, x)

• is symmetric,• has nonnegative coefficients,• is unimodal for every regular triangulation Γ of 2V .

Note: Stanley showed that there exists a graded S-module LV (Γ) whoseHilbert series equals `V (Γ, x).

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Barycentric subdivision

For the barycentric subdivision Γ of the simplex 2V on the vertex set V

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Stanley showed that

`V (Γ, x) =n∑

k=0

(−1)n−k(n

k

)Ak(x) =

∑w∈Dn

xexc(w) = dn(x).

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The triangulation Σ(Γ)

We now turn attention to An(x). We let

• V be an n-element set,• Γ be a triangulation of the simplex 2V on the vertex set V .

Then, there exists a triangulation Σ(Γ) of Σn which restricts to Γ on onefacet of Σn and satisfies

h(Σ(Γ), x) =∑F⊆V

xn−|F | h(ΓF , x).

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Example

For the barycentric subdivision Γ of 2V we have

h(Σ(Γ), x) =n∑

k=0

(n

k

)xn−kAk(x) = An(x).

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h-vectors of triangulations of complexes

Recall that the link of a simplicial complex ∆ at a face F ∈ ∆ is definedas link∆(F ) := {G r F : F ⊆ G ∈ ∆}.

Proposition (Stanley, 1992)

For every triangulation ∆′ of a pure simplicial complex ∆,

h(∆′, x) =∑F∈∆

`F (∆′F , x) h(link∆(F ), x).

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An application

We let again

• V be an n-element set,• Γ be a triangulation of the simplex 2V on the vertex set V .

Theorem (Kubitzke–Murai–Sieg, 2017)

`V (Γ, x) =∑F⊆V

(h(ΓF , x)− h(∂(ΓF ), x)) · dn−|F |(x).

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Corollary (Kubitzke–Murai–Sieg, 2017)

We have

dn(x) =n−2∑k=0

(n

k

)dk(x)(x + x2 + · · ·+ xn−1−k)

for n ≥ 2. In particular, dn(x) is symmetric and unimodal for all n.

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Question: Are there nice (symmetric and unimodal) analogues of Eulerian,derangement and binomial Eulerian polynomials for the hyperoctahedralgroup?

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Eulerian polynomials of type Bn

We let

• Bn = {w = (w(1),w(2), . . . ,w(n)) : |w | ∈ Sn} be the group of sign-ed permutations of [n]

• ∆ be the first barycentric subdivision of the boundary complex of then-dimensional cube.

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Then, h(∆, 1) = 2nn! = #Bn.

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Moreover,

h(∆, x) = Bn(x) :=∑w∈Bn

xdesB(w)

where

• desB(w) := # {i ∈ {0, 1, . . . , n − 1} : w(i) > w(i + 1)}

for w ∈ Bn as before, with w(0) := 0.

Example

Bn(x) =

1 + x , if n = 1

1 + 6x + x2, if n = 2

1 + 23x + 23x2 + x3, if n = 3

1 + 76x + 230x2 + 76x3 + x4, if n = 4

1 + 237x + 1682x2 + 1682x3 + 237x4 + x5, if n = 5.

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More barycentric subdivisions

We let

• V be an n-element set,• K be the barycentric subdivision of the cubical barycentric subdivision

of 2V .

Example

n = 3

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Consider the polynomial

d+n (x) = `V (K , x).

Example

d+n (x) =

0, if n = 1

3x , if n = 2

7x + 7x2, if n = 3

15x + 87x2 + 15x3, if n = 4

31x + 551x2 + 551x3 + 31x4, if n = 5

63x + 2803x2 + 8243x3 + 2803x4 + 63x5, if n = 6.

Note: The sum of the coefficients of d+n (x) is equal to the number of even-

signed derangements (signed permutations without fixed points of positivesign) in Bn.

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For w = (w1,w2, . . . ,wn) ∈ Bn we let

fex(w) = 2 · excA(w) + neg(w),

where

• excA(w) := # {i ∈ [n − 1] : w(i) > i}• neg(w) := # {i ∈ [n] : w(i) < 0}.

Theorem (A, 2014)

We haved+n (x) =

∑w∈D+

n

x fex(w)/2

where D+n is the set of even-signed derangements in Bn.

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We now consider the polynomial

B+n (x) := h(Σ(K ), x).

Savvidou showed that

h(K , x) = B+n (x) :=

∑w∈B+

n

xdesB(w)

where B+n is the set of w ∈ Bn with positive last coordinate. Hence,

B+n (x) =

n∑k=0

(n

k

)xn−kB+

k (x) = h(Σ(K ), x)

is a symmetric and unimodal analogue of An(x) for the group Bn.

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Example

B+n (x) =

1 + x , if n = 1

1 + 5x + x2, if n = 2

1 + 19x + 19x2 + x3, if n = 3.

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Symmetric functions

We let

• x = (x1, x2, x3, . . . ) be a sequence of commuting indeterminates,• hn(x) be the complete homogeneous symmetric function in x of

degree n, defined by

H(x, z) :=∑n≥0

hn(x)zn =∏i≥1

1

1− xiz,

• sλ(x) be the Schur function in x corresponding to the partition λ.

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We define polynomials Rλ(t),Pλ(t),Qλ(t) by

1− t

H(x; tz)− tH(x; z)=∑λ

Rλ(t)sλ(x) z |λ|,

(1− t)H(x, z)

H(x; tz)− tH(x; z)=∑λ

Pλ(t)sλ(x) z |λ|

and

(1− t)H(x, z)H(x, tz)

H(x; tz)− tH(x; z)=∑λ

Qλ(t)sλ(x) z |λ|.

Note: The left-hand sides of these equations arise from algebraic-geome-tric and representation-theoretic considerations.

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Note: We have ∑λ`n

f λRλ(t) = dn(t),

∑λ`n

f λPλ(t) = An(t)

and ∑λ`n

f λQλ(t) = An(t),

where f λ is the number of standard Young tableaux of shape λ.

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Theorem (Brenti, Gessel, Shareshian–Wachs, Stanley)

The polynomials Rλ(t),Pλ(t) and Qλ(t) are symmetric and unimodal, withcenters of symmetry n/2, (n − 1)/2 and n/2, respectively, for every λ ` n.Moreover, there exist nonnegative integers ξλ,i , γλ,i and γλ,i such that

Rλ(t) =

bn/2c∑i=0

ξλ,i ti (1 + t)n−2i ,

Pλ(t) =

b(n−1)/2c∑i=0

γλ,i ti (1 + t)n−1−2i ,

Qλ(t) =

bn/2c∑i=0

γλ,i ti (1 + t)n−2i .

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Gamma-positivity

Proposition (Branden, 2004, Gal, 2005)

Suppose f (x) ∈ R[x ] has nonnegative coefficients and only real roots andthat it is symmetric, with center of symmetry n/2. Then,

f (x) =

bn/2c∑i=0

γi xi (1 + x)n−2i

for some nonnegative real numbers γ0, γ1, . . . , γbn/2c.

Definition

The polynomial f (x) is called γ-positive if there exist nonnegative real nu-mbers γ0, γ1, . . . , γbn/2c as above, for some n ∈ N.

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Thus, An(x), dn(x) and An(x) are γ-positive for all n.

Example

An(x) =

1, if n = 1

1 + x , if n = 2

(1 + x)2 + 2x , if n = 3

(1 + x)3 + 8x(1 + x), if n = 4

(1 + x)4 + 22x(1 + x)2 + 16x2, if n = 5

(1 + x)5 + 52x(1 + x)3 + 186x2(1 + x), if n = 6.

Note: Every γ-positive polynomial (even if it has nonreal roots) is symme-tric and unimodal.

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An index i ∈ [n] is called a double descent of a permutation w ∈ Sn if

w(i − 1) > w(i) > w(i + 1),

where w(0) = w(n + 1) = n + 1.

Theorem (Foata–Schutzenberger, 1970)

We have

An(x) =

b(n−1)/2c∑i=0

γn,i xi (1 + x)n−1−2i ,

where γn,i is the number of w ∈ Sn which have no double descent anddes(w) = i . In particular, An(x) is symmetric and unimodal.

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Recently, gamma-positivity attracted attention after the work of

• Branden (2004, 2008) on P-Eulerian polynomials,• Gal (2005) on flag triangulations of spheres.

Expositions can be found in:

• A, Gamma-positivity in combinatorics and geometry, arXiv:1711.05983.

• T.K. Petersen, Eulerian Numbers, Birkhauser, 2015.

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Note: Γ-positivity is known to hold for

• Bn(x), d+n (x) and B+

n (x)

for all n.

Example

Bn(x) =

1 + x , if n = 1

(1 + x)2 + 4x , if n = 2

(1 + x)3 + 20x(1 + x), if n = 3

(1 + x)4 + 72x(1 + x)2, if n = 4

(1 + x)5 + 232x(1 + x)3 + 976x2(1 + x), if n = 5

(1 + x)6 + 716x(1 + x)4 + 7664x2(1 + x)2, if n = 6.

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Flag complexes and Gal’s conjecture

Definition

A simplicial complex ∆ is called flag if it contains every simplex whose1-skeleton is a subcomplex of ∆.

Example

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flag44 / 52

Example

For a 1-dimensional sphere ∆ with m vertices we have

h(∆, x) = 1 + (m − 2)x + x2.

Note that h(∆, x) is γ-positive ⇔ m ≥ 4 ⇔ ∆ is flag.

Conjecture (Gal, 2005)

The polynomial h(∆, x) is γ-positive for every flag triangulation ∆ of thesphere.

Note: This extends a conjecture of Charney–Davis (1995).

Example

The complex Σn is a flag and h(Σn, x) = (1 + x)n for every n ≥ 1.

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Conjecture (A, 2012)

The polynomial `V (Γ, x) is γ-positive for every flag triangulation Γ of 2V .

Note: This is stronger than Gal’s conjecture. There is considerable eviden-ce for both conjectures.

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We recall our notation Σ(Γ) and note that the formula

h(Σ(Γ), x) =∑F⊆V

xn−|F | h(ΓF , x)

may be rewritten as

h(Σ(Γ), x) =∑F⊆V

`F (ΓF , x) (1 + x)n−|F |.

Corollary

The γ-positivity of h(Σ(Γ), x) is implied by that of the `F (ΓF , x). In par-ticular:

• The γ-positivity of An(x) follows from that of dn(x).

• The γ-positivity of B+n (x) follows from that of d+

n (x).

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The analogue

(1− t)H(x, z)H(x, tz)

H(x; tz)H(y; tz)− tH(x; z)H(y; tz)=∑λ,µ

Pλ,µ(t)sλ(x)sµ(y) z |λ|+|µ|

for the group Bn of

(1− t)H(x, z)

H(x; tz)− tH(x; z)=∑λ

Pλ(t)sλ(x) z |λ|

was found by Dolgachev–Lunts and Stembridge (1994). We then have

∑(λ,µ)`n

(n

|λ|

)f λf µPλ,µ(t) = Bn(t)

for every n ≥ 1.

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Open Problem:

• Find the analogues of the identites involving Rλ and Qλ for Bn.• Find a combinatorial interpretation for Pλ,µ(t).• Prove that the polynomials Pλ,µ(t) are γ-positive.

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Consider the second barycentric subdivision Γ2 of the simplex 2V .

One can show that

`V (Γ2, x) =∑(

n

r0, r1, . . . , rk

)dk(x) dr0(x)Ar1(x) · · ·Ark (x),

where the sum ranges over all k ≥ 0 and over all sequences (r0, r1, . . . , rk)of integers which satisfy r0 ≥ 0, r1, . . . , rk ≥ 1 and sum to n.

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Note: This implies the γ-positivity of `V (Γ2, x).

Exercise: The sum of the coefficients of `V (Γ2, x) is equal to the numberof pairs (u, v) ∈ Sn ×Sn of permutations with no common fixed point.

Open Problem: Find a combinatorial interpretation for:

• `V (Γ2, x),

• the corresponding γ-coefficients.

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Thank you for your attention!

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