From Eulerian Polynomials and Chromatic Polynomials to Hessenberg - PowerPoint PPT Presentation

From Eulerian Polynomials and Chromatic Polynomials to Hessenberg Varieties Michelle Wachs University of Miami 1 1 1 1 4 1 1 11 11 1 1 26 66 26 1

Eulerian polynomials - Euler’s definition t � t i = 1 − t i ≥ 1

Eulerian polynomials - Euler’s definition t � t i = 1 − t i ≥ 1 t � it i = (1 − t ) 2 i ≥ 1

Eulerian polynomials - Euler’s definition t � t i = 1 − t i ≥ 1 t � it i = (1 − t ) 2 i ≥ 1 t (1 + t ) � i 2 t i = (1 − t ) 3 i ≥ 1

Eulerian polynomials - Euler’s definition t � t i = 1 − t i ≥ 1 t � it i = (1 − t ) 2 i ≥ 1 t (1 + t ) � i 2 t i = (1 − t ) 3 i ≥ 1 t (1 + 4 t + t 2 ) � i 3 t i = (1 − t ) 4 i ≥ 1

Eulerian polynomials - Euler’s definition Euler’s triangle t � t i = 1 − t 1 i ≥ 1 1 1 t � it i = 1 4 1 (1 − t ) 2 i ≥ 1 1 11 11 1 t (1 + t ) � 1 26 66 26 1 i 2 t i = (1 − t ) 3 i ≥ 1 Euler’s definition t (1 + 4 t + t 2 ) � i 3 t i = (1 − t ) 4 t A n ( t ) � i ≥ 1 i n t i = (1 − t ) n +1 i ≥ 1 Leonhard Euler (1707-1783)

Eulerian polynomials - Euler’s definition Euler’s triangle t � t i = 1 − t 1 i ≥ 1 1 1 t � it i = 1 4 1 (1 − t ) 2 i ≥ 1 1 11 11 1 t (1 + t ) � 1 26 66 26 1 i 2 t i = (1 − t ) 3 i ≥ 1 Euler’s definition t (1 + 4 t + t 2 ) � i 3 t i = (1 − t ) 4 t A n ( t ) � i ≥ 1 i n t i = (1 − t ) n +1 i ≥ 1 Euler’s exponential generating function formula A n ( t ) z n 1 − t � n ! = e ( t − 1) z − t n ≥ 0 Leonhard Euler (1707-1783)

Eulerian polynomials - Euler’s definition Euler’s triangle t � t i = 1 − t 1 i ≥ 1 1 1 t � it i = 1 4 1 (1 − t ) 2 i ≥ 1 1 11 11 1 t (1 + t ) � 1 26 66 26 1 i 2 t i = (1 − t ) 3 i ≥ 1 Euler’s definition t (1 + 4 t + t 2 ) � i 3 t i = (1 − t ) 4 t A n ( t ) � i ≥ 1 i n t i = (1 − t ) n +1 i ≥ 1 Euler’s exponential generating function formula A n ( t ) z n e ( t − 1) z − t = (1 − t ) e z 1 − t � n ! = e tz − te z n ≥ 0 Leonhard Euler (1707-1783)

Eulerian polynomials - combinatorial interpretation For σ ∈ S n , Descent set: DES ( σ ) := { i ∈ [ n − 1] : σ ( i ) > σ ( i + 1) } σ = 3 . 25 . 4 . 1 DES ( σ ) = { 1 , 3 , 4 } Define des ( σ ) := | DES ( σ ) | . So des (32541) = 3

Eulerian polynomials - combinatorial interpretation For σ ∈ S n , Descent set: DES ( σ ) := { i ∈ [ n − 1] : σ ( i ) > σ ( i + 1) } σ = 3 . 25 . 4 . 1 DES ( σ ) = { 1 , 3 , 4 } Define des ( σ ) := | DES ( σ ) | . So des (32541) = 3 Excedance set: EXC ( σ ) := { i ∈ [ n − 1] : σ ( i ) > i } σ = 32541 EXC ( σ ) = { 1 , 3 } Define exc ( σ ) := | EXC ( σ ) | . So exc (32541) = 2

Eulerian polynomials - combinatorial interpretation des exc S 3 � 123 0 0 t des ( σ ) = 1+4 t + t 2 1 132 1 1 1 1 σ ∈ S 3 213 1 1 1 4 1 � t exc ( σ ) = 1+4 t + t 2 231 1 2 1 11 11 1 312 1 1 1 26 66 26 1 σ ∈ S 3 321 2 1

Eulerian polynomials - combinatorial interpretation des exc S 3 � 123 0 0 t des ( σ ) = 1+4 t + t 2 1 132 1 1 1 1 σ ∈ S 3 213 1 1 1 4 1 � t exc ( σ ) = 1+4 t + t 2 231 1 2 1 11 11 1 312 1 1 1 26 66 26 1 σ ∈ S 3 321 2 1 Eulerian polynomial n − 1 � � � � � t j = t des ( σ ) = n t exc ( σ ) A n ( t ) = j j =0 σ ∈ S n σ ∈ S n

Eulerian polynomials - combinatorial interpretation des exc S 3 � 123 0 0 t des ( σ ) = 1+4 t + t 2 1 132 1 1 1 1 σ ∈ S 3 213 1 1 1 4 1 � t exc ( σ ) = 1+4 t + t 2 231 1 2 1 11 11 1 312 1 1 1 26 66 26 1 σ ∈ S 3 321 2 1 Eulerian polynomial n − 1 � � � � � t j = t des ( σ ) = n t exc ( σ ) A n ( t ) = j j =0 σ ∈ S n σ ∈ S n MacMahon (1905) showed equidistribution of des and exc . Carlitz and Riordin (1955) showed these are Eulerian polynomials.

Mahonian Permutation Statistics - q-analogs Let σ ∈ S n . Inversion Number: inv ( σ ) := |{ ( i , j ) : 1 ≤ i < j ≤ n , σ ( i ) > σ ( j ) }| . inv (3142) = 3 Major Index: � maj ( σ ) := i i ∈ DES ( σ ) maj (3142) = maj (3 . 14 . 2) = 1 + 3 = 4 Major Percy Alexander MacMahon (1854 - 1929)

Mahonian Permutation Statistics - q-analogs inv maj S 3 123 0 0 132 1 2 213 1 1 231 2 2 312 2 1 321 3 3

Mahonian Permutation Statistics - q-analogs inv maj S 3 123 0 0 � � q inv ( σ ) q maj ( σ ) = 132 1 2 σ ∈ S 3 σ ∈ S 3 213 1 1 1 + 2 q + 2 q 2 + q 3 = 231 2 2 312 2 1 321 3 3

Mahonian Permutation Statistics - q-analogs inv maj S 3 123 0 0 � � q inv ( σ ) q maj ( σ ) = 132 1 2 σ ∈ S 3 σ ∈ S 3 213 1 1 1 + 2 q + 2 q 2 + q 3 = 231 2 2 (1 + q + q 2 )(1 + q ) 312 2 1 = 321 3 3

Mahonian Permutation Statistics - q-analogs inv maj S 3 123 0 0 � � q inv ( σ ) q maj ( σ ) = 132 1 2 σ ∈ S 3 σ ∈ S 3 213 1 1 1 + 2 q + 2 q 2 + q 3 = 231 2 2 (1 + q + q 2 )(1 + q ) 312 2 1 = 321 3 3 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

q-Eulerian polynomials � A inv , des q inv ( σ ) t des ( σ ) ( q , t ) := n σ ∈ S n � A maj , des q maj ( σ ) t des ( σ ) ( q , t ) := n σ ∈ S n � A inv , exc q inv ( σ ) t exc ( σ ) ( q , t ) := n σ ∈ S n � A maj , exc q maj ( σ ) t exc ( σ ) ( q , t ) := n σ ∈ S n

q-Eulerian polynomials � A inv , des q inv ( σ ) t des ( σ ) ( q , t ) := n σ ∈ S n � q maj ( σ ) t des ( σ ) A maj , des ( q , t ) := n σ ∈ S n � A inv , exc q inv ( σ ) t exc ( σ ) ( q , t ) := n σ ∈ S n � q maj ( σ ) t exc ( σ ) A maj , exc ( q , t ) := n σ ∈ S n Theorem (MacMahon, 1916; Carlitz, 1954) q t i = tA maj , des ( q , t ) � n [ i ] n � n i =0 (1 − tq i ) i ≥ 1

q-analogs of Euler’s exp. generating function formula Theorem (Stanley, 1976) ( q , t ) z n 1 − t � A inv , des [ n ] q ! = n Exp q ( z ( t − 1)) − t n ≥ 0 where q ( n 2 ) z n � Exp q ( z ) := [ n ] q ! n ≥ 0

q-analogs of Euler’s exp. generating function formula Theorem (Stanley, 1976) ( q , t ) z n 1 − t � A inv , des [ n ] q ! = n Exp q ( z ( t − 1)) − t n ≥ 0 where q ( n 2 ) z n � Exp q ( z ) := [ n ] q ! n ≥ 0 Theorem (Shareshian-W., 2006) ( q , t ) z n (1 − tq ) exp q ( z ) � A maj , exc [ n ] q ! = n exp q ( ztq ) − tq exp q ( z ) n ≥ 0 where z n � exp q ( z ) := [ n ] q ! n ≥ 0

q -Eulerian polynomials and q -Eulerian numbers Theorem (Shareshian-W., 2006) ( q , tq − 1 ) z n (1 − t ) exp q ( z ) � A maj , exc [ n ] q ! = n exp q ( zt ) − t exp q ( z ) n ≥ 0 We use symmetric function theory and bijective combinatorics to prove this.

q -Eulerian polynomials and q -Eulerian numbers Theorem (Shareshian-W., 2006) ( q , tq − 1 ) z n (1 − t ) exp q ( z ) � A maj , exc [ n ] q ! = n exp q ( zt ) − t exp q ( z ) n ≥ 0 We use symmetric function theory and bijective combinatorics to prove this. From now on the q -Eulerian polynomials and the q -Eulerian numbers are � A n ( q , t ) := A maj , exc ( q , tq − 1 ) = q maj ( σ ) − exc ( σ ) t exc ( σ ) n σ ∈ S n � n � � q maj ( σ ) − exc ( σ ) q := j σ ∈ S n exc ( σ )= j So the result with Shareshian becomes (1 − t ) exp q ( z ) A n ( q , t ) z n � [ n ] q ! = exp q ( zt ) − t exp q ( z ) n ≥ 0

Palindromicity and unimodality of the q -Eulerian numbers n \ j 0 1 2 3 4 1 1 2 1 1 2 + q + q 2 3 1 1 3 + 2 q + 3 q 2 + 2 q 3 + q 4 3 + 2 q + 3 q 2 + 2 q 3 + q 4 4 1 1 4 + 3 q + 5 q 2 + ... 6 + 6 q + 11 q 2 + ... 4 + 3 q + 5 q 2 + ... 5 1 1 Theorem (Shareshian-W., 2006) � � The q-Eulerian polynomial A n ( q , t ) = � n − 1 q t j is n t =0 j � � � � n n palindromic in the sense that q = q for j n − 1 − j 0 ≤ j ≤ n − 1 2 � � � � n n q − q ∈ N [ q ] for q-unimodal in the sense that j j − 1 1 ≤ j ≤ n − 1 2

A symmetric function analog of the Eulerian polynomials Let ω be the involution on the ring of symmetric functions that takes the elementary symmetric functions e n to the complete homogeneous symmetric functions h n . For a homogeneous symmetric function f ( x 1 , x 2 , . . . ) of degree n with coefficients in ring R , the stable principal specialization of f is n � ps q ( f ( x 1 , x 2 , . . . )) = f (1 , q , q 2 , . . . ) (1 − q i ) ∈ R [ q ] . i =1

A symmetric function analog of the Eulerian polynomials Let ω be the involution on the ring of symmetric functions that takes the elementary symmetric functions e n to the complete homogeneous symmetric functions h n . For a homogeneous symmetric function f ( x 1 , x 2 , . . . ) of degree n with coefficients in ring R , the stable principal specialization of f is n � ps q ( f ( x 1 , x 2 , . . . )) = f (1 , q , q 2 , . . . ) (1 − q i ) ∈ R [ q ] . i =1 Let W n := { w ∈ Z n > 0 : w i � = w i +1 ∀ i } (Smirnov words) and let � t des ( w ) x w 1 · · · x w n . W n ( x , t ) := w ∈ W n Example: 37572 ∈ W 5 contributes t 2 x 2 x 3 x 5 x 2 7 to W 5 ( x , t ).