Counting permutations by congruence class of major index H el` - PowerPoint PPT Presentation

Counting permutations by congruence class of major index H´ el` ene Barcelo (Arizona State University), Bruce Sagan (Michigan State University), and Sheila Sundaram (Bard College) www.math.msu.edu/ ˜ sagan March 20, 2006

The major index The inversion number Shuffles The case k = ℓ

Outline The major index The inversion number Shuffles The case k = ℓ

Let [ n ] = { 1 , 2 , . . . , n } , and S n = symmetric group of permutations of [ n ] .

Let [ n ] = { 1 , 2 , . . . , n } , and S n = symmetric group of permutations of [ n ] . Then π = a 1 a 2 . . . a n ∈ S n has major index � maj π = i . a i > a i + 1

Let [ n ] = { 1 , 2 , . . . , n } , and S n = symmetric group of permutations of [ n ] . Then π = a 1 a 2 . . . a n ∈ S n has major index � maj π = i . a i > a i + 1 π = 1 2 3 4 5 6 Ex. If 5 > 3 6 > 1 2 4

Let [ n ] = { 1 , 2 , . . . , n } , and S n = symmetric group of permutations of [ n ] . Then π = a 1 a 2 . . . a n ∈ S n has major index � maj π = i . a i > a i + 1 π = 1 2 3 4 5 6 Ex. If then 5 > 3 6 > 1 2 4 maj π = 2 + 4 = 6 .

Let [ n ] = { 1 , 2 , . . . , n } , and S n = symmetric group of permutations of [ n ] . Then π = a 1 a 2 . . . a n ∈ S n has major index � maj π = i . a i > a i + 1 π = 1 2 3 4 5 6 Ex. If then 5 > 3 6 > 1 2 4 maj π = 2 + 4 = 6 . Theorem If q is an indeterminate then q maj π = 1 ( 1 + q )( 1 + q + q 2 ) · · · ( 1 + q + · · · + q n − 1 ) . � π ∈ S n

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5.

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5. Theorem If k , ℓ ≤ n and gcd ( k , ℓ ) = 1 then n ( i , j ) = n ! m k ,ℓ ∀ i , j . k ℓ

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5. Theorem If k , ℓ ≤ n and gcd ( k , ℓ ) = 1 then n ( i , j ) = n ! m k ,ℓ ∀ i , j . k ℓ History. Implicit in GORDON (1963) and ROSELLE (1974).

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5. Theorem If k , ℓ ≤ n and gcd ( k , ℓ ) = 1 then n ( i , j ) = n ! m k ,ℓ ∀ i , j . k ℓ History. Implicit in GORDON (1963) and ROSELLE (1974). Explicit in BARCELO, MAULE, and SUNDARAM (2002).

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5. Theorem If k , ℓ ≤ n and gcd ( k , ℓ ) = 1 then n ( i , j ) = n ! m k ,ℓ ∀ i , j . k ℓ History. Implicit in GORDON (1963) and ROSELLE (1974). Explicit in BARCELO, MAULE, and SUNDARAM (2002). Combinatorial Proof (BSS) m 1 ,ℓ (1) Prove the special case k = 1: n ( i , j ) = n ! /ℓ ∀ i , j .

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5. Theorem If k , ℓ ≤ n and gcd ( k , ℓ ) = 1 then n ( i , j ) = n ! m k ,ℓ ∀ i , j . k ℓ History. Implicit in GORDON (1963) and ROSELLE (1974). Explicit in BARCELO, MAULE, and SUNDARAM (2002). Combinatorial Proof (BSS) m 1 ,ℓ (1) Prove the special case k = 1: n ( i , j ) = n ! /ℓ ∀ i , j . m n ,ℓ (2) Use (1) to prove the case k = n : n ( i , j ) = n ! / ( n ℓ ) ∀ i , j .

Given k , ℓ we let m k ,ℓ be the k × ℓ matrix with ( i , j ) entry n n ( i , j ) = # { π ∈ S n : maj π ≡ i ( mod k ) , maj π − 1 ≡ j ( mod ℓ ) } . m k ,ℓ Ex. Suppose n = 5. Theorem If k , ℓ ≤ n and gcd ( k , ℓ ) = 1 then n ( i , j ) = n ! m k ,ℓ ∀ i , j . k ℓ History. Implicit in GORDON (1963) and ROSELLE (1974). Explicit in BARCELO, MAULE, and SUNDARAM (2002). Combinatorial Proof (BSS) m 1 ,ℓ (1) Prove the special case k = 1: n ( i , j ) = n ! /ℓ ∀ i , j . m n ,ℓ (2) Use (1) to prove the case k = n : n ( i , j ) = n ! / ( n ℓ ) ∀ i , j . (3) Use (2) and induction on n to prove the final case n > k .

Outline The major index The inversion number Shuffles The case k = ℓ

imaj π = maj π − 1 . Let

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 6 4 then π − 1 = 1 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 6 4 then π − 1 = 1 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4 ∴ imaj π = 1 + 4

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 4 then π − 1 = 1 6 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4 � ∴ imaj π = 1 + 4 = i . i + 1 left of i

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 4 then π − 1 = 1 6 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4 � ∴ imaj π = 1 + 4 = i . i + 1 left of i The inversion number of π is inv π = # { ( a i , a j ) : i < j and a i > a j } .

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 4 then π − 1 = 1 6 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4 � ∴ imaj π = 1 + 4 = i . i + 1 left of i inv π = # { 21 , 53 , 51 , 54 , 31 , 61 , 64 } = 7 . The inversion number of π is inv π = # { ( a i , a j ) : i < j and a i > a j } .

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 4 then π − 1 = 1 6 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4 � ∴ imaj π = 1 + 4 = i . i + 1 left of i inv π = # { 21 , 53 , 51 , 54 , 31 , 61 , 64 } = 7 . The inversion number of π is inv π = # { ( a i , a j ) : i < j and a i > a j } . Theorem q inv π = 1 ( 1 + q )( 1 + q + q 2 ) · · · ( 1 + q + · · · + q n − 1 ) . � π ∈ S n

imaj π = maj π − 1 . Let Ex. If π = 1 2 3 4 5 4 then π − 1 = 1 6 2 3 4 5 6 2 5 3 6 1 5 > 1 3 6 > 2 4 � ∴ imaj π = 1 + 4 = i . i + 1 left of i inv π = # { 21 , 53 , 51 , 54 , 31 , 61 , 64 } = 7 . The inversion number of π is inv π = # { ( a i , a j ) : i < j and a i > a j } . Theorem q inv π = 1 ( 1 + q )( 1 + q + q 2 ) · · · ( 1 + q + · · · + q n − 1 ) . � π ∈ S n We say maj and inv are equidistributed, i.e., have the same generating function. So are ( maj , imaj ) and ( inv , imaj ) .

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ Let M k ,ℓ n ( i , j ) = { π ∈ S n : inv π ≡ i ( mod k ) , imaj π ≡ j ( mod ℓ ) } .

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ Let M k ,ℓ n ( i , j ) = { π ∈ S n : inv π ≡ i ( mod k ) , imaj π ≡ j ( mod ℓ ) } . ∴ m k ,ℓ n ( i , j ) = # M k ,ℓ n ( i , j ) .

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ Let M k ,ℓ n ( i , j ) = { π ∈ S n : inv π ≡ i ( mod k ) , imaj π ≡ j ( mod ℓ ) } . ∴ m k ,ℓ n ( i , j ) = # M k ,ℓ n ( i , j ) . Let M ( i , j ) = M n ,ℓ n ( i , j ) and m ( i , j ) = # M ( i , j ) .

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ Let M k ,ℓ n ( i , j ) = { π ∈ S n : inv π ≡ i ( mod k ) , imaj π ≡ j ( mod ℓ ) } . ∴ m k ,ℓ n ( i , j ) = # M k ,ℓ n ( i , j ) . Let M ( i , j ) = M n ,ℓ n ( i , j ) and m ( i , j ) = # M ( i , j ) . It suffices to find a bijection M ( i , j ) ← → M ( i + 1 , j ) ∀ i , j

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ Let M k ,ℓ n ( i , j ) = { π ∈ S n : inv π ≡ i ( mod k ) , imaj π ≡ j ( mod ℓ ) } . ∴ m k ,ℓ n ( i , j ) = # M k ,ℓ n ( i , j ) . Let M ( i , j ) = M n ,ℓ n ( i , j ) and m ( i , j ) = # M ( i , j ) . It suffices to find a bijection M ( i , j ) ← → M ( i + 1 , j ) ∀ i , j since then m ( 1 , j ) = m ( 2 , j ) = . . . = m ( n , j ) .

n ( i , j ) = n ! n ( i , j ) = n ! Proof of (2): If m 1 ,ℓ m n ,ℓ then ( ∀ i , j ) . ℓ n ℓ Let M k ,ℓ n ( i , j ) = { π ∈ S n : inv π ≡ i ( mod k ) , imaj π ≡ j ( mod ℓ ) } . ∴ m k ,ℓ n ( i , j ) = # M k ,ℓ n ( i , j ) . Let M ( i , j ) = M n ,ℓ n ( i , j ) and m ( i , j ) = # M ( i , j ) . It suffices to find a bijection M ( i , j ) ← → M ( i + 1 , j ) ∀ i , j since then m ( 1 , j ) = m ( 2 , j ) = . . . = m ( n , j ) . Also, m ( 1 , j ) + m ( 2 , j ) + · · · + m ( n , j )