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Character Polynomials Problem From Stanleys Positivity Problems in - - PowerPoint PPT Presentation

Character Polynomials Problem From Stanleys Positivity Problems in Algebraic Combinatorics Problem 12: Give a combinatorial interpretation of the row sums of the character table for S n (combinatorial proof of non-negativity)


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SLIDE 1

Character Polynomials

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SLIDE 2

Problem

  • From Stanley’s Positivity Problems in

Algebraic Combinatorics

  • Problem 12: Give a combinatorial

interpretation of the row sums of the character table for Sn (combinatorial proof

  • f non-negativity)
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SLIDE 3

Symmetric Group

  • Sn = permutations of n things
  • Contains n! elements
  • S3=permutations of {1,2,3}

(123, 132, 213, 231, 312, 321)

  • Permutations can be represented with n × n

matrices

  • Character: trace of a matrix representation
  • Character Table: table of all irreducible

characters of a group

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SLIDE 4

Representations of S3

  • vertices of an equilateral triangle

1 1        3 2 2 3 1                3 2 1 2 2              

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SLIDE 5

Representations of S3

  • vertices of an equilateral triangle
  • pick a permutation: 123  312

3 2 1 2 2               3 2 2 3 1                1 1       

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SLIDE 6

Representations of S3

  • vertices of an equilateral triangle
  • pick a permutation: 123  312

3 2 2 2 1                3 2 1 2 1               3 1       

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SLIDE 7
  • 123  312 is 120° CW rotation
  • Character = Trace =

1 3 2 2 3 1 2 2               

 1 2  1 2  1

Representations of S3

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SLIDE 8

Character Table for S3

1,1,1 2,1 3 3 1 1 1 2,1 2

  • 1

1,1,1 1

  • 1

1

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SLIDE 9

Character Table for S4

14 2,12 22 3,1 4 1 1 1 1 1 3 1

  • 1
  • 1

2 2

  • 1

3

  • 1
  • 1

1 1

  • 1

1 1

  • 1
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SLIDE 10

Character Polynomials

  • compute characters without matrices
  • depend only on small parts of the cycle

type

  • connections to Murnaghan-Nakayama

rule, Schur functions

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SLIDE 11

Character Table for S4

14 2,12 22 3,1 4 1 1 1 1 1 3 1

  • 1
  • 1

2 2

  • 1

3

  • 1
  • 1

1 1

  • 1

1 1

  • 1

Sum

5 2 3 2 1

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SLIDE 12

Character Polynomials

Partition Polynomial n 1 n-1,1 n-2,2 n-2,12 n-3,3 n-3,2,1 n-3,13 n-4,14 a1 1

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SLIDE 13

Character Table for S4

14 2,12 22 3,1 4 1 1 1 1 1 3 1

  • 1
  • 1

2 2

  • 1

3

  • 1
  • 1

1 1

  • 1

1 1

  • 1
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SLIDE 14

Character Polynomials

Partition Polynomial n 1 n-1,1 n-2,2 n-2,12 n-3,3 n-3,2,1 n-3,13 n-4,14 a2  a1 1 a1(a1 1) 2 a1 1 a2  a1  a1(a1 1) 2

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SLIDE 15

Character Polynomials

Partition Polynomial n 1 n-1,1 n-2,2 n-2,12 n-3,3 n-3,2,1 n-3,13 n-4,14 a2  a1 1 a1(a1 1) 2 a1 1 a2  a1  a1(a1 1) 2 a3  a1a2  a2  a1(a1 1) 2  a1(a1 1)(a1  2) 6 a3  a1(a1 1)  a1(a1 1)(a1  2) 3  a1 a3  a1a2  a2  a1(a1 1) 2  a1(a1 1)(a1  2) 6  a1 1

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SLIDE 16

Character Polynomials

Partition Polynomial n 1 n-1,1 n-2,2 n-2,12 n-3,3 n-3,2,1 n-3,13 n-4,14 a2  a1 1 a1(a1 1) 2 a1 1 a2  a1  a1(a1 1) 2 a3  a1a2  a2  a1(a1 1) 2  a1(a1 1)(a1  2) 6 a3  a1(a1 1)  a1(a1 1)(a1  2) 3  a1 a3  a1a2  a2  a1(a1 1) 2  a1(a1 1)(a1  2) 6  a1 1

2 2 2 2 4 1 3 3 1 2 1 1 1 1 1 1 1 1 1 2 1

( 1) ( 1) 2 2 ( 1)( 2)( 3) ( 1)( 2) ( 1) 1 24 6 2 a a a a a a a a a a a a a a a a a a a a a                   

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SLIDE 17

Generating Functions and Row Sums

p(n)xn  1 1 xi

i1

n0

1 1 xi 

i1

(1+x+x2+x3+x4+···)(1+x2+x4+···)(1+x3+x6+···)(1+x4+x8+···)+···

  • 1. x4 · 1 · 1 · 1  1,1,1,1
  • 2. x · 1 · x3 · 1  3,1
  • 3. 1 · x4 · 1 · 1  2,2
  • 4. x2 · x2 · 1 · 1  2,1,1
  • 5. 1 · 1 · 1 · x4  4
  • p(4)=5

Can get x4 from:

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SLIDE 18

Example: n-1,1

a1 1

Character Polynomial:

2 2 3

1 2 3 1 x ux u x u ux        

2 2 3 3

1 1 1 ux u x u x ux      

2 3 1

1 2 3 1

u

x x u u x x

       

counts number of 1s!

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SLIDE 19

Example: n-1,1

 u (1 ux)1  x(1 ux)2

1 2 1

(1 (1 ) )

u

u x x u x

 

    

p(n)xn  1 1 xi

i1

n0

1 2 1

1 1 1 1 1 1 1

i i i u i

x x u ux x x

  

      

 

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SLIDE 20

Example: n-1,1

( 1) ( 2) ( 3)

n

x p n p n p n           

Row Sum= ( 1) ( 2) ( 3) ( ) p n p n p n p n       

1

1 ( ) 1 1 1

n i n i

x x p n x x x x

 

    

2 3

( ) ( )

n n

x x x p n x

   

xn      a1

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SLIDE 21

Rows Sums

Row Row Sum

n n-1,1 n-2,2 n-2,12 n-3,3 n-3,2,1 n-3,13

p(n) ( 1) ( 2) ( 3) ( 4) ( 5) ( ) p n p n p n p n p n p n            ( 2) ( 3) 3 ( 4) 3 ( 5) 5 ( 6) ( 1) p n p n p n p n p n p n             ( ) ( 2) ( 3) ( 4) 3 ( 5) 3 ( 6) ( 1) p n p n p n p n p n p n p n              ( 3) 4 ( 5) 7 ( 6) 12 ( 7) 2 ( 2) p n p n p n p n p n           ( 1) ( 2) ( 4) ( 5) 6 ( 6) ( ) p n p n p n p n p n p n            ( 1) ( 4) 5 ( 5) 10 ( 6) ( 2) 2 ( 3) p n p n p n p n p n p n            

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SLIDE 22

Growth of p(n)

  • p(n-1) ≤ p(n) ≤ p(n-1)+p(n-2)
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SLIDE 23

Rows Sums

Row Row Sum Positivity

n n-1,1 n-2,2 n-2,12 n-3,3 n-3,2,1 n-3,13

( ) p n ( 1) ( 2) ( 3) ( 4) ( 5) ( ) p n p n p n p n p n p n            ( 2) ( 3) 3 ( 4) 3 ( 5) 5 ( 6) ( 1) p n p n p n p n p n p n             ( ) ( 2) ( 3) ( 4) 3 ( 5) 3 ( 6) ( 1) p n p n p n p n p n p n p n              ( 3) 4 ( 5) 7 ( 6) 12 ( 7) 2 ( 2) p n p n p n p n p n           ( 1) ( 2) ( 4) ( 5) 6 ( 6) ( ) p n p n p n p n p n p n            ( 1) ( 4) 5 ( 5) 10 ( 6) ( 2) 2 ( 3) p n p n p n p n p n p n            

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SLIDE 24

Growth of p(n)

  • p(n-1) ≤ p(n) ≤ p(n-1)+p(n-2)
  • super-polynomial, sub-exponential
  • asymptotics good enough to show that

finitely many subtracted terms guaranteed to cancel out for n sufficiently large

( ) ( )

n n

Q x p n x

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SLIDE 25

From the bottom up

  • The sum of the last row is the number of

self-conjugate partitions of n, call this s(n).

  • Conjugate row obtained by multiplying by

bottom row

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SLIDE 26

Character Table for S4

14 2,12 22 3,1 4 1 1 1 1 1 3 1

  • 1
  • 1

2 2

  • 1

3

  • 1
  • 1

1 1

  • 1

1 1

  • 1
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SLIDE 27

From the bottom up

  • The sum of the last row is the number of

self-conjugate partitions of n, call this s(n).

  • Conjugate row obtained by multiplying by

bottom row

  • For every row sum formula in terms of

p(n), the conjugate row has the same formula in terms of s(n).

  • s(n-1) ≤ s(n) ≤ s(n-1)+s(n-2) for n > 1

1

1 ( ) ( ) ( ) 1 ( 1)i

i n n n n i

s n x Q x s x n x

  

   

  

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SLIDE 28

Words of Wisdom

The worst thing you can do to a problem is to solve it completely… because then you have to find something else to work on. ̶ Dan Kleitman