Eulerian polynomials, chromatic quasisymmetric functions, and - PowerPoint PPT Presentation

Eulerian polynomials, chromatic quasisymmetric functions, and Hessenberg varieties Michelle Wachs University of Miami Joint work with John Shareshian

Eulerian Polynomials For σ ∈ S n DES ( σ ) := { i ∈ { 1 , . . . , n − 1 } : σ ( i ) > σ ( i + 1) } des ( σ ) := | DES ( σ ) | For σ = 3 . 25 . 4 . 1 DES ( σ ) = { 1 , 3 , 4 } des ( σ ) = 3 Eulerian polynomial � t des ( σ ) A n ( t ) = σ ∈ S n A 1 ( t ) = 1 A 2 ( t ) = 1 + t palindromic A 3 ( t ) = 1 + 4 t + t 2 A 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 unimodal A 5 ( t ) = 1 + 26 t + 66 t 2 + 26 t 3 + t 4

Eulerian Polynomials For σ ∈ S n DES ( σ ) := { i ∈ { 1 , . . . , n − 1 } : σ ( i ) > σ ( i + 1) } des ( σ ) := | DES ( σ ) | For σ = 3 . 25 . 4 . 1 DES ( σ ) = { 1 , 3 , 4 } des ( σ ) = 3 Eulerian polynomial � t des ( σ ) A n ( t ) = σ ∈ S n A 1 ( t ) = 1 A 2 ( t ) = 1 + t palindromic A 3 ( t ) = 1 + 4 t + t 2 A 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 unimodal A 5 ( t ) = 1 + 26 t + 66 t 2 + 26 t 3 + t 4

Eulerian Polynomials For σ ∈ S n DES ( σ ) := { i ∈ { 1 , . . . , n − 1 } : σ ( i ) > σ ( i + 1) } des ( σ ) := | DES ( σ ) | For σ = 3 . 25 . 4 . 1 DES ( σ ) = { 1 , 3 , 4 } des ( σ ) = 3 Eulerian polynomial � t des ( σ ) A n ( t ) = σ ∈ S n A 1 ( t ) = 1 A 2 ( t ) = 1 + t palindromic A 3 ( t ) = 1 + 4 t + t 2 A 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 unimodal A 5 ( t ) = 1 + 26 t + 66 t 2 + 26 t 3 + t 4

Eulerian Polynomials A new formula? ⌊ n +1 2 ⌋ � � m n � � � t m − 1 A n ( t ) = [ k i − 1] t k 1 − 1 , k 2 , . . . , k m m =1 k 1 ,..., k m ≥ 2 i =1 where [ k ] t := 1 + t + · · · + t k − 1 .

Eulerian Polynomials A new formula? ⌊ n +1 2 ⌋ � � m n � � � t m − 1 A n ( t ) = [ k i − 1] t k 1 − 1 , k 2 , . . . , k m m =1 k 1 ,..., k m ≥ 2 i =1 where [ k ] t := 1 + t + · · · + t k − 1 . Sum & Product Lemma: Let A ( t ) and B ( t ) be positive, unimodal, palindromic with respective centers of symmetry c A and c B . Then A ( t ) B ( t ) is positive, unimodal, and palindromic with center of symmetry c A + c B . If c A = c B then A ( t ) + B ( t ) is positive, unimodal and palindromic with center of symmetry c A .

Eulerian Polynomials A new formula? ⌊ n +1 2 ⌋ � � m n � � � t m − 1 A n ( t ) = [ k i − 1] t k 1 − 1 , k 2 , . . . , k m m =1 k 1 ,..., k m ≥ 2 i =1 where [ k ] t := 1 + t + · · · + t k − 1 . Sum & Product Lemma: Let A ( t ) and B ( t ) be positive, unimodal, palindromic with respective centers of symmetry c A and c B . Then A ( t ) B ( t ) is positive, unimodal, and palindromic with center of symmetry c A + c B . If c A = c B then A ( t ) + B ( t ) is positive, unimodal and palindromic with center of symmetry c A . Center of symmetry: m k i − 2 = 1 � ( m − 1) + 2( n − 1) . 2 i =1

A n ( t ) z n 1 − t � 1 + = (Euler’s exponential generating function formula) e z ( t − 1) − t n ! n ≥ 1 (1 − t ) exp( z ) = exp( zt ) − t exp( z ) 1 − t = exp( z ) t k z k − tz k � k ≥ 0 k ! 1 − t = exp( z ) ( t k − t ) z k � k ≥ 0 k ! − 1     z r t [ k − 1] t z k � � =  1 −   r ! k ! r ≥ 0 k ≥ 2 m     z r t [ k − 1] t z k �  � � =  r ! k ! r ≥ 0 m ≥ 0 k ≥ 2 ⌊ n 2 ⌋ n m � � n � � � t m � A n ( t ) = [ k i − 1] t r , k 1 , . . . , k m m =0 r =0 k 1 ,..., k m ≥ 2 i =1 = further manipulations ⌊ n +1 2 ⌋ m � � n � � � t m − 1 = [ k i − 1] t k 1 − 1 , k 2 , . . . , k m m =1 k 1 ,..., k m ≥ 2 i =1

Mahonian Permutation Statistics - q-analog Let σ ∈ S n . Inversion Number: inv ( σ ) := |{ ( i , j ) : 1 ≤ i < j ≤ n , σ ( i ) > σ ( j ) }| . inv (32541) = 6 Major Index: � maj ( σ ) := i i ∈ DES ( σ ) maj (32541) = maj (3 . 25 . 4 . 1) = 1 + 3 + 4 = 8

Mahonian Permutation Statistics - q-analog Let σ ∈ S n . Inversion Number: inv ( σ ) := |{ ( i , j ) : 1 ≤ i < j ≤ n , σ ( i ) > σ ( j ) }| . inv (32541) = 6 Major Index: � maj ( σ ) := i i ∈ DES ( σ ) maj (32541) = maj (3 . 25 . 4 . 1) = 1 + 3 + 4 = 8 Theorem (MacMahon 1905) � � q inv ( σ ) = q maj ( σ ) = [ n ] q ! σ ∈ S n σ ∈ S n where [ n ] q := 1 + q + · · · + q n − 1 and [ n ] q ! := [ n ] q [ n − 1] q · · · [1] q

Rawlings major index Let 1 ≤ r ≤ n . For σ ∈ S n , set inv < r ( σ ) := |{ ( i , j ) : 1 ≤ i < j ≤ n , 0 < σ ( i ) − σ ( j ) < r }| . DES ≥ r ( σ ) := { i ∈ [ n − 1] : σ ( i ) − σ ( i + 1) ≥ r } � maj ≥ r ( σ ) := i i ∈ DES ≥ r � maj if r = 1 Note maj ≥ r + inv < r = inv if r = n

Rawlings major index Let 1 ≤ r ≤ n . For σ ∈ S n , set inv < r ( σ ) := |{ ( i , j ) : 1 ≤ i < j ≤ n , 0 < σ ( i ) − σ ( j ) < r }| . DES ≥ r ( σ ) := { i ∈ [ n − 1] : σ ( i ) − σ ( i + 1) ≥ r } � maj ≥ r ( σ ) := i i ∈ DES ≥ r � maj if r = 1 Note maj ≥ r + inv < r = inv if r = n Theorem (Rawlings, 1981) � q maj ≥ r ( σ )+ inv < r ( σ ) = [ n ] q ! σ ∈ S n

Rawlings major index Let 1 ≤ r ≤ n . For σ ∈ S n , set inv < r ( σ ) := |{ ( i , j ) : 1 ≤ i < j ≤ n , 0 < σ ( i ) − σ ( j ) < r }| . DES ≥ r ( σ ) := { i ∈ [ n − 1] : σ ( i ) − σ ( i + 1) ≥ r } � maj ≥ r ( σ ) := i i ∈ DES ≥ r � maj if r = 1 Note maj ≥ r + inv < r = inv if r = n Theorem (Rawlings, 1981) � q maj ≥ r ( σ )+ inv < r ( σ ) = [ n ] q ! σ ∈ S n � A ( r ) q maj ≥ r ( σ ) t inv < r ( σ ) n ( q , t ) := σ ∈ S n

n ( q , t ) := � A ( r ) σ ∈ S n q maj ≥ r ( σ ) t inv < r ( σ ) A (1) n ( q , t ) = � σ ∈ S n q maj ( σ ) = [ n ] q ! A ( n ) n ( q , t ) = � σ ∈ S n t inv ( σ ) = [ n ] t ! A (2) n ( q , t ) =?

n ( q , t ) := � A ( r ) σ ∈ S n q maj ≥ r ( σ ) t inv < r ( σ ) A (1) n ( q , t ) = � σ ∈ S n q maj ( σ ) = [ n ] q ! A ( n ) n ( q , t ) = � σ ∈ S n t inv ( σ ) = [ n ] t ! A (2) n ( q , t ) =? inv < 2 (635142) = 3 since ( < 2)-inversions are 635142 635142 635142

n ( q , t ) := � A ( r ) σ ∈ S n q maj ≥ r ( σ ) t inv < r ( σ ) A (1) n ( q , t ) = � σ ∈ S n q maj ( σ ) = [ n ] q ! A ( n ) n ( q , t ) = � σ ∈ S n t inv ( σ ) = [ n ] t ! A (2) n ( q , t ) =? inv < 2 (635142) = 3 since ( < 2)-inversions are 635142 635142 635142 (635142) − 1 = 46 . 25 . 3 . 1

n ( q , t ) := � A ( r ) σ ∈ S n q maj ≥ r ( σ ) t inv < r ( σ ) A (1) n ( q , t ) = � σ ∈ S n q maj ( σ ) = [ n ] q ! A ( n ) n ( q , t ) = � σ ∈ S n t inv ( σ ) = [ n ] t ! A (2) n ( q , t ) =? inv < 2 (635142) = 3 since ( < 2)-inversions are 635142 635142 635142 (635142) − 1 = 46 . 25 . 3 . 1 A (2) n (1 , t ) = � σ ∈ S n t des ( σ ) = A n ( t ) inv < 2 ( σ ) = des ( σ − 1 )

n ( q , t ) := � A ( r ) σ ∈ S n q maj ≥ r ( σ ) t inv < r ( σ ) A (1) n ( q , t ) = � σ ∈ S n q maj ( σ ) = [ n ] q ! A ( n ) n ( q , t ) = � σ ∈ S n t inv ( σ ) = [ n ] t ! A (2) n ( q , t ) =? inv < 2 (635142) = 3 since ( < 2)-inversions are 635142 635142 635142 (635142) − 1 = 46 . 25 . 3 . 1 A (2) n (1 , t ) = � σ ∈ S n t des ( σ ) = A n ( t ) inv < 2 ( σ ) = des ( σ − 1 ) So A ( r ) n (1 , t ) is a generalized Eulerian polynomial and A ( r ) n ( q , qt ) is a Mahonian q -analog.

Generalized Eulerian polynomial n (1 , t ) = � A ( r ) n ( t ) := A ( r ) σ ∈ S n t inv < r ( σ ) Eulerian polynomials are palindromic and unimodal. A (2) 3 ( t ) = 1 + 4 t + t 2 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 A (2) What about generalized Eulerian polynomials?

Generalized Eulerian polynomial n (1 , t ) = � A ( r ) n ( t ) := A ( r ) σ ∈ S n t inv < r ( σ ) Eulerian polynomials are palindromic and unimodal. A (2) 3 ( t ) = 1 + 4 t + t 2 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 A (2) What about generalized Eulerian polynomials? � t inv ( σ ) = [ n ] t ! A ( n ) n ( t ) = σ ∈ S n

Generalized Eulerian polynomial n (1 , t ) = � A ( r ) n ( t ) := A ( r ) σ ∈ S n t inv < r ( σ ) Eulerian polynomials are palindromic and unimodal. A (2) 3 ( t ) = 1 + 4 t + t 2 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 A (2) What about generalized Eulerian polynomials? n � � t inv ( σ ) = [ n ] t ! = A ( n ) (1 + t + · · · + t i ) n ( t ) = σ ∈ S n i =1

Generalized Eulerian polynomial n (1 , t ) = � A ( r ) n ( t ) := A ( r ) σ ∈ S n t inv < r ( σ ) Eulerian polynomials are palindromic and unimodal. A (2) 3 ( t ) = 1 + 4 t + t 2 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 A (2) What about generalized Eulerian polynomials? n � � t inv ( σ ) = [ n ] t ! = A ( n ) (1 + t + · · · + t i ) n ( t ) = σ ∈ S n i =1 A ( n − 1) � [ n ] t [ n − 2] t + nt n − 2 � ( t ) = [ n − 2] t ! n center of symmetry: 1 2 (( n − 1) + ( n − 3)) = n − 2

n ( t ) = � Generalized Eulerian polynomial A ( r ) σ ∈ S n t inv < r ( σ ) Problem (Stanley EC1, 1.50 f): Prove that A ( r ) n ( t ) is palindromic and unimodal. Solution: