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Flag Descents and P-Partitions
Ira M. Gessel
Department of MathematicsBrandeis University
2010 Joint Mathematics MeetingSpecial Session on Permutations
San Francisco, CAJanuary 16, 2010
Ehrhart polynomials
Let K be convex polytope in Rn with lattice points as vertices.Then for any positive integer m, #(mK ∩ Zn) is a polynomialLK (m) in m, called the Ehrhart polynomial of K .
For example, if K is the unit square in R2
(0, 0) (1, 0)
(0, 1) (1, 1)
then LK (m) = (m + 1)2.
Ehrhart polynomials
Let K be convex polytope in Rn with lattice points as vertices.Then for any positive integer m, #(mK ∩ Zn) is a polynomialLK (m) in m, called the Ehrhart polynomial of K .
For example, if K is the unit square in R2
(0, 0) (1, 0)
(0, 1) (1, 1)
then LK (m) = (m + 1)2.
If the interior of K is nonempty then LK (m) has degree n, so
∞∑m=0
LK (m)tm =A(t)
(1− t)n+1
for some polynomial A(t) of degree at most n. It is known thatA(t) has positive coefficients.
In our example
∞∑m=0
LK (m)tm =∞∑
m=0
(m + 1)2 =1 + t
(1− t)3
If the interior of K is nonempty then LK (m) has degree n, so
∞∑m=0
LK (m)tm =A(t)
(1− t)n+1
for some polynomial A(t) of degree at most n. It is known thatA(t) has positive coefficients.
In our example
∞∑m=0
LK (m)tm =∞∑
m=0
(m + 1)2 =1 + t
(1− t)3
If K is a convex polytope with rational points as vertices thenLK (m) = #(mK ∩ Zn) is a quasi-polynomial in m.
For example, if K is
(0, 0)
(0, 12 ) ( 1
2 , 12 )
( 12 , 0)
then LK (m) = (⌊m
2
⌋+ 1)2 and
∞∑m=0
LK (m)tm =1 + t2
(1− t)(1− t2)2 .
If K is a convex polytope with rational points as vertices thenLK (m) = #(mK ∩ Zn) is a quasi-polynomial in m.
For example, if K is
(0, 0)
(0, 12 ) ( 1
2 , 12 )
( 12 , 0)
then LK (m) = (⌊m
2
⌋+ 1)2 and
∞∑m=0
LK (m)tm =1 + t2
(1− t)(1− t2)2 .
We will also consider convex polytopes in which part of theboundary is missing. So the Ehrhart polynomial of
(0, 0) (1, 0)
(0, 1) (1, 1)
is m(m + 1).
Richard Stanley’s theory of P-partitions connects certainEhrhart polynomials with permutation enumeration.
We consider polytopes in the unit cube in Rn cut out byhyperplanes xi = xj . More precisely, these polytopes aredefined by inequalities of the form xi ≤ xj or xi > xj for i < j ,together with 0 ≤ xi ≤ 1.
Richard Stanley’s theory of P-partitions connects certainEhrhart polynomials with permutation enumeration.
We consider polytopes in the unit cube in Rn cut out byhyperplanes xi = xj . More precisely, these polytopes aredefined by inequalities of the form xi ≤ xj or xi > xj for i < j ,together with 0 ≤ xi ≤ 1.
For example, in R2, we might take x1 > x2 and we would havethe triangle
(0, 0) (1, 0)
(0, 1) (1, 1)
The Ehrhart polynomial counts integers i1, i2 satisfying0 ≤ i2 < i1 ≤ m and so it is equal to
(m+12
).
Sometimes a set of inequalities will be inconsistent, but when itis consistent it can be represented by a poset on[n] = 1,2, . . . ,n. For example, the set of inequalities x2 < x1,x2 ≤ x3 corresponds to the poset
1
2
3
Let P be a partial order on [n]. A P-partition is a point(x1, . . . , xn) in Rn such that
1. If i <P j then xi ≤ xj
2. If i <P j and i > j then xi < xj
Let P(P) be the set of P-partitions.
The Fundamental Theorem of P-partitions. (Stanley, Knuth,Kreweras, MacMahon)
P(P) =⊔
π∈L (P)
P(π)
Example:
1
2
3
L (P ) =
22
1
1 3
3
P =
x2 < x1, x2 ≤ x3 x2 < x1 ≤ x3 " x2 ≤ x3 < x1
Let P(P) be the set of P-partitions.
The Fundamental Theorem of P-partitions. (Stanley, Knuth,Kreweras, MacMahon)
P(P) =⊔
π∈L (P)
P(π)
Example:
1
2
3
L (P ) =
22
1
1 3
3
P =
x2 < x1, x2 ≤ x3 x2 < x1 ≤ x3 " x2 ≤ x3 < x1
The order polynomial of P, denoted by ΩP(m), is the Ehrhartpolynomial1 of the polytope associated to P. In other words, it isthe number of P-partitions with entries in the set 0,1, . . . ,m.
For example, if P is a disjoint union of n points, then ΩP(m) is(m + 1)n.
By the fundamental theorem of P-partitions
ΩP(m) =∑
π∈L (P)
Ωπ(m).
What is Ωπ(m)?
1Actually this is what Stanley calls ΩP(m + 1).
The order polynomial of P, denoted by ΩP(m), is the Ehrhartpolynomial1 of the polytope associated to P. In other words, it isthe number of P-partitions with entries in the set 0,1, . . . ,m.
For example, if P is a disjoint union of n points, then ΩP(m) is(m + 1)n.
By the fundamental theorem of P-partitions
ΩP(m) =∑
π∈L (P)
Ωπ(m).
What is Ωπ(m)?
1Actually this is what Stanley calls ΩP(m + 1).
The order polynomial of P, denoted by ΩP(m), is the Ehrhartpolynomial1 of the polytope associated to P. In other words, it isthe number of P-partitions with entries in the set 0,1, . . . ,m.
For example, if P is a disjoint union of n points, then ΩP(m) is(m + 1)n.
By the fundamental theorem of P-partitions
ΩP(m) =∑
π∈L (P)
Ωπ(m).
What is Ωπ(m)?
1Actually this is what Stanley calls ΩP(m + 1).
The order polynomial of P, denoted by ΩP(m), is the Ehrhartpolynomial1 of the polytope associated to P. In other words, it isthe number of P-partitions with entries in the set 0,1, . . . ,m.
For example, if P is a disjoint union of n points, then ΩP(m) is(m + 1)n.
By the fundamental theorem of P-partitions
ΩP(m) =∑
π∈L (P)
Ωπ(m).
What is Ωπ(m)?
1Actually this is what Stanley calls ΩP(m + 1).
First we note that linear orders on [n] may be identified withpermutations of [n]:
22
1
1 3
3
2 1 3 2 3 1
If π is the permutation 1 2 · · · n, then ΩP(m) =(m+n
n
).
What about an arbitrary permutation?
A descent of a permutation π of n is an i for whichπ(i) > π(i + 1). We denote by des(π) the number of descentsof π.
First we note that linear orders on [n] may be identified withpermutations of [n]:
22
1
1 3
3
2 1 3 2 3 1
If π is the permutation 1 2 · · · n, then ΩP(m) =(m+n
n
).
What about an arbitrary permutation?
A descent of a permutation π of n is an i for whichπ(i) > π(i + 1). We denote by des(π) the number of descentsof π.
First we note that linear orders on [n] may be identified withpermutations of [n]:
22
1
1 3
3
2 1 3 2 3 1
If π is the permutation 1 2 · · · n, then ΩP(m) =(m+n
n
).
What about an arbitrary permutation?
A descent of a permutation π of n is an i for whichπ(i) > π(i + 1). We denote by des(π) the number of descentsof π.
First we note that linear orders on [n] may be identified withpermutations of [n]:
22
1
1 3
3
2 1 3 2 3 1
If π is the permutation 1 2 · · · n, then ΩP(m) =(m+n
n
).
What about an arbitrary permutation?
A descent of a permutation π of n is an i for whichπ(i) > π(i + 1). We denote by des(π) the number of descentsof π.
Lemma. For any permutation π of n, Ωπ(m) =(m+n−des(π)
n
).
Proof by example. Consider the permutation π = 1 4•2 5•3.The π-partitions f with parts in [m] satisfy
0 ≤ x1 ≤ x4 < x2 ≤ x5 < x3 ≤ m
This is the same as
0 ≤ x1 ≤ x4 ≤ x2 − 1 ≤ x5 − 1 ≤ x3 − 2 ≤ m − 2
and the number of solutions of these inequalities is((m−2)+5
5
).
Lemma. For any permutation π of n, Ωπ(m) =(m+n−des(π)
n
).
Proof by example. Consider the permutation π = 1 4•2 5•3.The π-partitions f with parts in [m] satisfy
0 ≤ x1 ≤ x4 < x2 ≤ x5 < x3 ≤ m
This is the same as
0 ≤ x1 ≤ x4 ≤ x2 − 1 ≤ x5 − 1 ≤ x3 − 2 ≤ m − 2
and the number of solutions of these inequalities is((m−2)+5
5
).
As a consequence we have the fundamental result for countingpermutations by descents:
∞∑m=0
ΩP(m)tm =
∑π∈L (P) tdes(π)
(1− t)n+1
Proof. By linearity it is enough to consider the case in which Pis the total order corresponding to a permutation π for which wehave
∞∑m=0
Ωπ(m)tm =∞∑
m=0
(m + n − des(π)
n
)tm
=tdes(π)
(1− t)n+1
So for the case of an antichain, where L (P) is the set Sn of allpermutations of [n], we have
∞∑m=0
(m + 1)ntm =En(t)
(1− t)n+1 ,
whereEn(t) =
∑π∈Sn
tdes(π).
Signed P-partitions
We now consider some more general polytopes, associated withroot systems of type B (Reiner, C. Chow, Stembridge). In addi-tion to the hyperplanes xi = xj we also take the hyperplanesxi = −xj and xi = 0. More precisely we take the inequalitiesxi ≥ 0, xi < 0, xi + xj ≥ 0 and xi + xj ≤ 0, and we also restrict tothe cube [−1,1]n in Rn.
Signed P-partitions
We now consider some more general polytopes, associated withroot systems of type B (Reiner, C. Chow, Stembridge). In addi-tion to the hyperplanes xi = xj we also take the hyperplanesxi = −xj and xi = 0. More precisely we take the inequalitiesxi ≥ 0, xi < 0, xi + xj ≥ 0 and xi + xj ≤ 0, and we also restrict tothe cube [−1,1]n in Rn. So, for example, in R2 we are looking atpart of
Signed P-partitions
We now consider some more general polytopes, associated withroot systems of type B (Reiner, C. Chow, Stembridge). In addi-tion to the hyperplanes xi = xj we also take the hyperplanesxi = −xj and xi = 0. More precisely we take the inequalitiesxi ≥ 0, xi < 0, xi + xj ≥ 0 and xi + xj ≤ 0, and we also restrict tothe cube [−1,1]n in Rn.
We would like to represent these inequalities by posets of somekind.
Let [n]± = −n,−(n − 1), . . . ,−1,0,1, . . . ,n. A signed posetof order n is a partial order P on [n]± with the property thati <P j if and only if −j <P −i .
2
0
-1
1
-2
We would like to represent these inequalities by posets of somekind.
Let [n]± = −n,−(n − 1), . . . ,−1,0,1, . . . ,n. A signed posetof order n is a partial order P on [n]± with the property thati <P j if and only if −j <P −i .
2
0
-1
1
-2
If P is a signed poset of order n, a P-partition is a (2n + 1)-tuple(x−n, . . . , x−1, x0, x1, . . . , xn) such such that for all i ∈ [n]±,x−i = −xi (which implies that x0 = 0), together with the usualproperties for a P-partition: if i <P j then xi ≤ xj and if i <P jand i > j then xi < xi .
Since (x−n, . . . , xn) is determined by (x1, . . . xn), we can think ofa P-partition as a point in either R2n+1 or Rn.
If P is a signed poset of order n, a P-partition is a (2n + 1)-tuple(x−n, . . . , x−1, x0, x1, . . . , xn) such such that for all i ∈ [n]±,x−i = −xi (which implies that x0 = 0), together with the usualproperties for a P-partition: if i <P j then xi ≤ xj and if i <P jand i > j then xi < xi .
Since (x−n, . . . , xn) is determined by (x1, . . . xn), we can think ofa P-partition as a point in either R2n+1 or Rn.
For the poset
2
0
-1
1
-2
a P-partition is a 5-tuple (x−2, x−1, x0, x1, x2) satisfyingx−i = −xi for all i , 0 < x−2, and x−1 < x−2.
The set L (P) of signed linear extensions of P is the set oflinear extensions of P that are signed posets.
2
0
-1
1
-2
P =
L (P )
-2
-2
-2
-1
-1
-1
0 0 0
1
1
1
2 2
2
Just as for ordinary P-partitions, we have
P(P) =⊔
π∈L (P)
P(π)
and we can identify the linear extensions of P with signedpermutations.
-2
-1
0
1
2
-1 2 0 -2 1
or
-2 1
We define the order polynomial ΩP(m) for a signed poset P tobe the Ehrhart polynomial of the corresponding polytope, whichis the number of P-partitions with values in the set [m]±. Thenas before we have
∞∑m=0
ΩP(m)tm =
∑π∈L (P) tdes(π)
(1− t)n+1 ,
where a descent of a signed permutation π of [n] is ani ∈ 0,1, . . . ,n − 1 such that π(i) > π(i + 1). For example
• 4 1 2 • 5 3
writing i for −i .
As an example, if P is an antichain, then L (P) is the set Bn ofall signed permutations of [n] and thus
∞∑m=0
(2m + 1)ntm =
∑π∈Bn
tdes(π)
(1− t)n+1 .
(Steingrímsson)
Flag Descents
In 2001, Adin, Brenti, and Roichman introduced a variation ofthe descent number of a signed permutation. They defined theflag descent number fdes(π) by
fdes(π) =
2 des(π), if π(1) > 02 des(π)− 1, if π(1) < 0
In other words, a descent at the beginning of a signedpermutation contributes 1 to the flag descent number, but adescent anywhere else contributes 2.
Adin, Brenti, and Roichman showed that
∞∑m=0
(m + 1)ntm =
∑π∈Bn
t fdes(π)
(1− t)(1− t2)n ,
which implies that∑
π∈Bnt fdes(π) = (1 + t)nEn(t).
(Recently proved nicely by Lai and Petersen.)
We will see that flag descents arise from Ehrhart polynomials.
Adin, Brenti, and Roichman showed that
∞∑m=0
(m + 1)ntm =
∑π∈Bn
t fdes(π)
(1− t)(1− t2)n ,
which implies that∑
π∈Bnt fdes(π) = (1 + t)nEn(t).
(Recently proved nicely by Lai and Petersen.)
We will see that flag descents arise from Ehrhart polynomials.
Given a signed poset P, let KP be the polytope in [−1,1]n
associated to P, and let K flagP be obtained by shifting KP to
[0,1]n, i.e.
K flagP =
(12(x1 + 1), . . . , 1
2(xn + 1))
: (x1, . . . , xn) ∈ KP
and we define the flag order quasi-polynomial ΩflagP (m) to be
the Ehrhart quasi-polynomial of K flagP .
I If P is an antichain, then ΩflagP (m) = (m + 1)n,
I In general ΩflagP (m) is the number of P-partitions with
values from −m to m, all with the same parity as m.
Given a signed poset P, let KP be the polytope in [−1,1]n
associated to P, and let K flagP be obtained by shifting KP to
[0,1]n, i.e.
K flagP =
(12(x1 + 1), . . . , 1
2(xn + 1))
: (x1, . . . , xn) ∈ KP
and we define the flag order quasi-polynomial ΩflagP (m) to be
the Ehrhart quasi-polynomial of K flagP .
I If P is an antichain, then ΩflagP (m) = (m + 1)n,
I In general ΩflagP (m) is the number of P-partitions with
values from −m to m, all with the same parity as m.
What is the flag order quasi-polynomial for a total order, i.e., fora signed permutation?
Lemma. For any signed permutation π in Bn,
∞∑m=0
Ωflagπ (m)tm =
t fdes(π)
(1− t)(1− t2)n .
What is the flag order quasi-polynomial for a total order, i.e., fora signed permutation?
Lemma. For any signed permutation π in Bn,
∞∑m=0
Ωflagπ (m)tm =
t fdes(π)
(1− t)(1− t2)n .
Proof by example.Let’s consider the permutation π = 1342 with flag descentnumber 5 (• 1• 34•2). We want to count 5-tuples(x1, x3, x4, x2,m) satisfying
0 < x1 < x3 ≤ x4 < x2 ≤ m,
where all entries have the same parity. Each such 5-tuple hasweight tm. There is a minimal 5-tuple, (1,3,3,5,5) in whichm = fdes(π). We get all such 5-tuples by adding any multiplesof (1,1,1,1,1), (0,2,2,2,2), (0,0,2,2,2), (0,0,0,2,2), or(0,0,0,0,2). So there’s one way to increase m by 1 and fourways to increase m by 2.
Therefore∞∑
m=0
Ωflagπ (m)tm =
t fdes(π)
(1− t)(1− t2)n .
Proof by example.Let’s consider the permutation π = 1342 with flag descentnumber 5 (• 1• 34•2). We want to count 5-tuples(x1, x3, x4, x2,m) satisfying
0 < x1 < x3 ≤ x4 < x2 ≤ m,
where all entries have the same parity. Each such 5-tuple hasweight tm. There is a minimal 5-tuple, (1,3,3,5,5) in whichm = fdes(π). We get all such 5-tuples by adding any multiplesof (1,1,1,1,1), (0,2,2,2,2), (0,0,2,2,2), (0,0,0,2,2), or(0,0,0,0,2). So there’s one way to increase m by 1 and fourways to increase m by 2.
Therefore∞∑
m=0
Ωflagπ (m)tm =
t fdes(π)
(1− t)(1− t2)n .