Euler–Mahonian Statistics Via Polyhedral Geometry
Matthias Beck San Francisco State University Benjamin Braun University of Kentucky arXiv:1109.3353
- Adv. Math. (2013)
EulerMahonian Statistics Via Polyhedral Geometry [ n ] q ! n ! - - PowerPoint PPT Presentation
EulerMahonian Statistics Via Polyhedral Geometry [ n ] q ! n ! Matthias Beck San Francisco State University Benjamin Braun University of Kentucky arXiv:1109.3353 Adv. Math. (2013) Permutation Statistics S n permutation of { 1 ,
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 2
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 2
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 2
q tk =
j=0 (1 − tqj)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 3
q tk =
j=0 (1 − tqj)
j=0 (1 − tqj)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 3
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 3
q tk =
j=0 (1 − tqj)
Matthias Beck 4
q tk =
j=0 (1 − tqj)
Matthias Beck 4
q tk =
j=0 (1 − tqj)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 4
q tk =
j=0 (1 − tqj)
0 zm1 1
n
n
j + · · · + zk j
n
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 5
q tk =
j=0 (1 − tqj)
n
0 =
j=0
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 5
j=0 (1 − tqj)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 6
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 7
j=0 (1 − tqj)
n
−j[k]z−1
−j
0 =
−jw−j
ǫ1π(1)zǫ2 ǫ2π(2) · · · z ǫj ǫjπ(j) n
ǫ1π(1)zǫ2 ǫ2π(2) · · · z ǫj ǫjπ(j)
−1 = · · · =
−n = q, w−1 = · · · = w−n = s, and w1 = · · · = wn = 1
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 8
q tk =
j=1(1 − t2q2i)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 9
q tk =
j=1(1 − t2q2i)
j := 1 if ǫj = ǫj+1 and 0 otherwise. Then
n
0 =
aǫ
j=0
0z2 π(1)z2 π(2) · · · z2 π(j)
j=1
n
0 z2 π(1)z2 π(2) · · · z2 π(j)
Matthias Beck 10
n
0 =
aǫ
j=0
0z2 π(1)z2 π(2) · · · z2 π(j)
j=1
n
0 z2 π(1)z2 π(2) · · · z2 π(j)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 11
q tk =
j=1(1 − t2q2i)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 12
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 13
j=0 (1 − tqj)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 14
q tk =
j=1 (1 − t2q2i)
q tk =
j=1(1 − trqri)
q tk =
j=0 (1 − tqrj)
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 15
Euler–Mahonian Statistics Via Polyhedral Geometry Matthias Beck 16