or
by
Massachusetts Institute of Technology
June 26, 2004 on the occasion of Richard P. Stanley’s Birthday
1

P n
( x
1
, . . . , x n +1
) := ConvexHull(( x w (1)
, . . . , x w ( n +1)
) | w ∈ S n +1
)
This is a convex n -dimensional polytope in H ⊂ R n +1 .
Example: n = 2 (type A
2
)
( x
1
, x
2
, x
3
)
P
2
( x
1
, x
2
, x
3
) =
More generaly, for a Weyl group W , P
W
( x ) := ConvexHull( w ( x ) | w ∈ W ) .
2

Question: What is the volume V n
:= Vol P n
?
Volume form is normalized so that the volume of a parallelepiped formed by generators of the lattice Z n +1 ∩ H is 1.
Question: What is the number of lattice points N n
:= P n
∩ Z n +1 ?
We will see that polynomial of P n
V
The polynomial V n is n and
E ( t
N
) = n
N are polynomials in x
1
, . . . , x n +1 is the top homogeneous part of n
( tx
1
, . . . , tx n
), and V n
N n of degree n
. The Ehrhart is its top coefficient.
.
We will give 3 totally different formulas for these polynomials.
3

P n
( n + 1 , n, . . . , 1) = ConvexHull(( w (1) , ..., w ( n + 1)) | w ∈ S n +1
) is the most symmetric permutohedron.
regular hexagon subdivided into 3 rhombi
It is a zonotope = Minkowsky sum of line intervals.
Well-known facts:
➠ V n
( n + 1 , . . . , 1) = ( n + 1) n − 1 is the number of trees on n + 1 labelled vertices.
P n
( n + 1 , . . . , 1) can be subdivided into parallelepipeds of unit volume associated with trees. This works for any zonotope.
➠ N n
( n + 1 , . . . , 1) is the number of forests on n + 1 labelled vertices.
4

Fix any distinct numbers λ
1
, . . . , λ n +1 such that λ
1
+ · · · + λ n +1
= 0.
V n
( x
1
, . . . , x n +1
) =
1
X n !
w ∈ S n +1
( λ w (1) x
1
+ · · · + λ w ( n +1) x n +1
) n
( λ w (1)
− λ w (2)
)( λ w (2)
− λ w (3)
) · · · ( λ w ( n )
− λ w ( n +1)
)
Notice that the symmetrization in RHS does not depends on λ i
’s.
Idea of Proof Use Khovansky-Puchlikov’s method:
➠ Instead of just counting the number of lattice points in P , define [ P ] = sum of formal exponents e a over lattice points a ∈ P ∩ Z n .
➠ Now we can work with unbounded polytopes. For example, for a simplicial cone C , the sum [ C ] is given by a simple rational expression.
➠ Any polytope P can be explicitly presented as an alternating sum of simplicial cones: [ P ] = [ C
1
] ± [ C
2
] ± · · · .
Applying this procedure to the permutohedron, we obtain . . .
5

Let α
1
, . . . , α n be a system of let L be the root lattice .
simple roots for Weyl group W , and
Theorem: For a dominant weight µ ,
[ P
W
( µ )] :=
X a ∈ P
W
( µ ) ∩ ( L + µ )
X e w ( µ ) e a = w ∈ W
(1 − e − w ( α
1
) ) · · · (1 − e − w ( α n
) )
Compare this with Weyl’s character formula!
Note: LHS is obtained from the character ch V
µ of the irrep V
µ by replacing all nonzero coefficients with 1. In type A , ch V
µ
= Schur polynomial s
µ
.
From this expression, one can deduce the First Formula and also its generalizations to other Weyl groups .
6

Let us use the coordinates y
1
, . . . , y n +1 related x
1
, . . . , x n +1 by









 y y
1
2
= − x
= − x
2
1
+ x
1 y
3
= − x
3
+ 2 x
2
− x
1









· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · y n +1
= − n
0 x n
+ n
1 x n − 1
− · · · ± n n x
1 and write V n
= Vol P n
( x
1
, . . . , x n +1
) as a polynomial in y
1
, . . . , y n +1
.
Examples:
V
1
= Vol ([( x
1
, x
2
) , ( x
2
, x
1
)]) = x
1
− x
2
= y
2
V
2
= · · · = 3 y 2
2
+ 3 y
2 y
3
+ 1
2 y 2
3
7

Theorem:
V n
( x
1
, . . . , x n +1
) =
1
X n !
( S
1
,...,S n
) y
| S
1
|
· · · y
| S n
|
, where the sum is over ordered collections of subsets S
1
, . . . , S n
⊂ such that either of the following equivalent conditions is satisfied:
[ n + 1]
➠ For any distinct i
1
, . . . , i k
, we have | S i
1
∪ · · · ∪ S i k
| ≥ k + 1
(cf.
Hall’s Marriage Theorem )
➠ For any j ∈ [ n + 1], there is a system of distinct representatives in
S
1
, . . . , S n that avoids j .
Thus n !
V n is a polynomial in y
2
, . . . , y n +1 with positive integer coefficients.
8

This formula can be extended to generalized permutohedra a generalized permutohedron
Generalized permutohedra are obtained from usual permutohedra by moving faces while preserving all angles.
this is also a generalized permutohedron
9

Coordinate simplices in R n +1 : ∆
I
= ConvexHull(
Let Y = { Y
I
} be the collection of variables Y
I e i
| i ∈ I ) , for I ⊆ [ n +1].
≥ 0 associated with all subsets I ⊂ [ n + 1]. Define
P n
( Y ) :=
X
I ⊂ [ n +1]
Y
I
· ∆
I
(Minkowsky sum)
Its combinatorial type depends only on the set of I ’s for which Y
I
= 0.
Examples:
➠ If Y
I
= y
| I |
, then P n
( Y ) is a usual permutohedron .
➠ If Y
I
= 0 iff I is a consecutive interval , then P n
( Y ) is an associahedron .
➠ If Y
I
= 0 iff I is a cyclic interval , then P n
( Y ) is a cyclohedron .
➠ If Y
I
= 0 iff I is a connected set in Dynkin diagram , then P n
( Y generalized associahedron related to DeConcini-Procesi’s work.
) is a
(Do not confuse with Fomin-Zelevinsky’s generalized associahedra!)
➠ If Y
I
= 0 iff I is an initial interval
Stanley-Pitman polytope .
{ 1 , . . . , i } , then P n
( Y ) is the
10

Theorem: The volume of the generalized permutohedron is given by
Vol P n
( Y ) =
1
X n !
( S
1
,...,S n
)
Y
S
1
· · · Y
S n
, where S
1
, . . . , S n satisfy the same condition.
Theorem: The # of lattice points in the generalized permutohedron is
P n
( Y ) ∩ Z n +1 =
1
X n !
( S
1
,...,S n
)
{ Y
S
1
· · · Y
S n
} ,
(
Y
Y
I a
I
I
)
:= ( Y
[ n +1]
+1) { a
[ n +1]
} Y
I =[ n +1]
Y
I
{ a
I
}
, where Y
{ a } = Y ( Y +1) · · · ( Y + a − 1) .
This extends a formula from [Stanley-Pitman] for the volume of their polytope. In this case, the above summation is over parking functions .
11

We also have a combinatorial description of face structure of generalized permutohedra in terms of nested collections of subsets in [ n + 1]. This is related to DeConcini-Procesi’s wonderful arrangements .
Not enough time for this now.
The most interesting part of the talk is . . .
12

Let use the coordinates z
1
, . . . , z n related to x
1
, . . . , x n +1 by z
1
= x
1
− x
2
, z
2
= x
2
− x
3
, · · · , z n
= x n
− x n +1
These coordinates are canonically defined for an arbitrary Weyl group.
Then the permutohedron P n is the Minkowsky sum
P n
= z
1
∆
1 n
+ z
2
∆
2 n
+ · · · + z n
∆ nn of hypersimplices ∆ kn
= P n
(1 , . . . , 1 , 0 , . . . , 0) (with k 1’s).
+ =
13

This implies
Vol P n
=
X c
1
,...,c n
A c
1
,...,c n z c
1
1 c
1
!
· · · z c n n
, c n
!
where the sum is over c
1
, . . . , c n
≥ 0, c
1
+ · · · + c n
= n , and
A c
1
,...,c n
= MixedVolume(∆ c
1
1 n
, . . . , ∆ c n nn
) ∈ Z
> 0
In particular, n !
V n is a positive integer polynomial in z
1
, . . . , z n
.
Let us call the integers A c
1
,...,c n the Mixed Eulerian numbers .
Examples:
V
1
V
2
V
3
= 1 z
1
= 1
= 1 z
2
2
1 z
3
1
3!
+ 2 z
1 z
2
+ 2 z
2
1
2 z
2
+ 1 z
2
2
2
+ 4 z
1 z
2
2
+ 4 z
3
2
3!
+ 3
+ 4 z
2
2
2 z
2
1
2 z
3 z
3
+ 6
+ 3 z
1 z
1 z
2
2
3 z
2 z
3
+
+ 2 z
2 z
2
2
3 + 1 z
3
3
3!
(The mixed Eulerian numbers are marked in red .)
14

➠ A c
1
,...,c n are positive integers defined for c
1
, . . . , c n
≥ 0, c
1
+ · · · + c n
= n .
➠ P 1 c
1
!
··· c n
!
A c
1
,...,c n
= ( n + 1) n − 1 .
➠ A
0 ,..., 0 ,n, 0 ,..., 0
( n is in k -th position) is the usual Eulerian number A kn
= # permutations in S n with k descents = n ! Vol ∆ kn
.
➠ A
1 ,..., 1
= n !
➠ A k, 0 ,..., 0 ,n − k
= n k
➠ A c
1
,...,c n
= 1 c
1 2 c
2 · · · n c n
There are exactly C n
= if c
1
1 n +1
+ · · · + c i
2 n n
≥ i , for i = 1 such sequences ( c
1
, . . . , n
, . . . , c n
.
).
When I showed these numbers to Richard Stanley, he conjectured that
➠ P A c
1
,...,c n
= n !
C n
.
Moreover, he conjectured that . . .
15

One can subdivide all sequences ( c
1
, . . . , c n
) into C n classes such that the sum of mixed Eulerian numbers for each class is n !. For example, A
1 ,..., 1
= n !
and A n, 0 ,..., 0
+ A
0 ,n, 0 ,..., 0
+ A
0 , 0 ,n,..., 0
+ · · · + A
0 ,..., 0 ,n
= n !
, because the sum of Eulerian numbers P k
A kn is n !.
Let us write ( c
1
, . . . , c n
) ∼ ( c 0
1
, . . . , c 0 n
) iff ( c
1
, . . . , c n
, 0) is a cyclic shift of
( c 0
1
, . . . , c 0 n
, 0). Stanley conjectured that, for fixed ( c
1
, . . . , c n
), we have
X
( c 0
1
,...,c 0 n
) ∼ ( c
1
,...,c n
)
A c 0
1
,...,c 0 n
= n !
Exercise: Check that there are exactly C n equivalence classes of sequences.
Every equivalence class contains exactly one sequence ( c
1
, . . . , c n
) such that c
1
+ · · · + c i
≥ i , for i = 1 , . . . , n . (For this sequence, A c
1
,...,c n
= 1 c
1 · · · n c n .)
These conjectures follow from . . .
16

Theorem: Let U n
( z
1
, . . . , z n +1
) = Vol P n
. (It does not depend on z n +1
.)
U n
( z
1
, . . . , z n +1
) + U n
( z n +1
, z
1
, . . . , z n
) + · · · + U n
( z
2
, . . . , z n +1
, z
1
) =
= ( z
1
+ · · · + z n +1
) n
This theorem has a simple geometric proof. It extends to any Weyl group.
Cyclic shifts come from symmetries of type A extended Dynkin diagram.
Idea of Proof:
The area of blue triangle is 1
6 sum of the areas of three hexagons .
17

Corollary: Fix z
1
, . . . , z n +1
, λ
1
, . . . , λ n +1
Symmetrizing the expression such that λ
1
+ · · · + λ n +1
= 0.
1 n !
( λ
1 z
1
+ ( λ
1
+ λ
2
) z
2
+ · · · ( λ
1
+ · · · + λ n +1
) z n +1
) n
( λ
1
− λ
2
) · · · ( λ n
− λ n +1
) with respect to ( n + 1)! permutations of λ
1
, . . . , λ n +1 permutations of z
1
, . . . z n +1
, we obtain and ( n + 1) cyclic
( z
1
+ · · · + z n +1
) n
.
Problem: Find a direct proof.
18

A c
1
,...,c n
1 z
3
5 a plane binary tree on n nodes
2 z
4
2
T = z
1
5
1 z
T = z
3 z
4 z
8 z
7 z
1 z
7 z
4 z
3
(the order is given by green labels)
8 z
3
3 z
4
4
7
6 z
8
3
7 6 z
7 z
7
4
8
➠ The nodes are labelled by 1 , . . . , n such that, for a node labelled l , labels of all in the left (right) branch are less (greater) than l . The labels of all descendants of a node form a consecutive interval I = [ a, b ] .
➠ We have an increasing labelling of the nodes by 1 , . . . , n .
➠ Each node is labeled by z i such that i ∈ I ; z T := product of all z i
’s.
➠ The weight of a node labelled by if i ≤ l , and b − i +1 b − l +1 l and z i with interval [ a, b ] is i − a +1 l − a +1 if i ≥ l . The weight wt ( T ) of tree is the product of weights of its nodes.
19

Theorem: The volume of the permutohedron is
V n
=
X wt ( T ) · z
T
T where the sum is over plane binary trees with blue , red , and green labels.
Combinatorial interpretation for the mixed Eulerian numbers:
Theorem: Let z i
1
· · · z i n
= z c
1
1
· · · z c n n
. Then
A c
1
,...,c n
=
X n !
wt ( T )
T over same kind of trees T such that z T = z i
1
· · · z i n
(in this order).
Note that all terms n !
wt ( T ) in this formula are positive integer.
Comparing different formulas for V n
, we obtain a lot of interesting combinatorial identities. For example . . .
20

Corollary:
( n + 1) n − 1 =
X
2 n
T n !
Y v ∈ T
1 +
1 h ( v )
, where is sum is over unlabeled plane binary trees T on n nodes, and h ( v ) denotes the “hook-length” of a node v in T , i.e., the number of descendants of v (including v ).
Example: n = 3
1
2
3
2
1
3 3 3
2
2
1 1 hook-lengths of binary trees
1
The identity says that
(3 + 1) 2 = 3 + 3 + 3 + 3 + 4 .
3
1
Problem: Prove this identity combinatorially.
21