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 L (P) be the set of extensions of P to a total order.

1

2

3

L (P ) =

22

1

1 3

3

P =

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 .

Theorem. For any signed poset P on [n]±,

∞∑m=0

ΩflagP (m)tm =

∑π∈L (P) t fdes(π)

(1− t)(1− t2)n .

The result of Adin, Brenti, and Roichman is the special case inwhich P is an antichain.