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
∆ =
s ss
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
s s
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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
s s s
ss ss
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QQQQQQQ
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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not flag
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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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