Character Polynomials

Transcription

1 Character Polynomials

2 Problem From Stanley s Positivity Problems in Algebraic Combinatorics Problem : Give a combinatorial interpretation of the row sums of the character table for S n (combinatorial proof of non-negativity)

3 Symmetric Group S n = permutations of n things Contains n! elements S 3 =permutations of {,,3} (3, 3, 3, 3, 3, 3) Permutations can be represented with n n matrices Character: trace of a matrix representation Character Table: table of all irreducible characters of a group

4 Representations of S 3 vertices of an equilateral triangle

5 Representations of S 3 vertices of an equilateral triangle pick a permutation:

6 Representations of S 3 vertices of an equilateral triangle pick a permutation:

7 Representations of S is 0 CW rotation 3 3 Character = Trace =

8 Character Table for S 3,,, 3 3, 0 -,, -

9 Character Table for S 4 4, 3,

10 Character Polynomials compute characters without matrices depend only on small parts of the cycle type connections to Murnaghan-Nakayama rule, Schur functions

11 Character Table for S 4 4, 3, Sum

12 Character Polynomials Partition Polynomial n n-, a n-, n-, n-3,3 n-3,, n-3, 3 n-4, 4

13 Character Table for S 4 4, 3,

14 Character Polynomials Partition Polynomial n n-, a n-, n-, n-3,3 n-3,, n-3, 3 n-4, 4 a a a (a ) a a a (a )

15 Character Polynomials Partition Polynomial n n-, a n-, n-, n-3,3 n-3,, n-3, 3 n-4, 4 a a a (a ) a a a (a ) a 3 a a a a (a ) a (a )(a ) 6 a 3 a (a ) a (a )(a ) a 3 a 3 a a a a (a ) a (a )(a ) 6 a

16 Character Polynomials Partition Polynomial n n-, a n-, n-, n-3,3 n-3,, n-3, 3 n-4, 4 a a a (a ) a a a (a ) a 3 a a a a (a ) a (a )(a ) 6 a 3 a (a ) a (a )(a ) a 3 a 3 a a a a (a ) a (a )(a ) 6 a a( a ) a( a ) a4 aa3 a a3 aa a ( a )( a )( a 3) a ( a )( a ) a ( a ) a 4 6

17 Generating Functions and Row Sums n0 p(n)x n i x i x (+x+x +x3 +x4 + )(+x +x4 + )(+x3 +x6 + )(+x4 +x8 + )+ i i Can get x 4 from:. x 4,,,. x x 3 3, 3. x 4, 4. x x,, 5. x 4 4 p(4)=5

18 Example: n-, Character Polynomial: a 3 3 ux u x u x ux u ux 3 0 x ux 3u x u 0 xx 3x ux u 3 counts number of s!

19 Example: n-, u ( ux) x( ux) u ( ux) u x ( x) n0 p(n)x n i x i x u ux x x x i i u i i

20 Example: n-, x x x x x n 0 i i 3 p( n) x n ( x x x ) p( n) x n x n a n0 n x p( n ) p( n ) p( n 3) Row Sum= p( n ) p( n ) p( n 3) p( n)

21 Row n p(n) Row Sum Rows Sums n-, p( n ) p( n ) p( n 3) p( n 4) p( n 5) p( n) n-, p( n ) p( n 3) 3 p( n 4) 3 p( n 5) 5 p( n 6) p( n ) n-, p( n) p( n ) p( n 3) p( n 4) 3 p( n 5) 3 p( n 6) p( n ) n-3,3 n-3,, p( n 3) 4 p( n 5) 7 p( n 6) p( n 7) p( n ) p( n ) p( n 4) 5 p( n 5) 0 p( n 6) p( n ) p( n 3) n-3, 3 p( n ) p( n ) p( n 4) p( n 5) 6 p( n 6) p( n)

22 Growth of p(n) p(n-) p(n) p(n-)+p(n-)

23 Row n pn ( ) Row Sum Positivity Rows Sums n-, p( n ) p( n ) p( n 3) p( n 4) p( n 5) p( n) n-, p( n ) p( n 3) 3 p( n 4) 3 p( n 5) 5 p( n 6) p( n ) n-, p( n) p( n ) p( n 3) p( n 4) 3 p( n 5) 3 p( n 6) p( n ) n-3,3 n-3,, p( n 3) 4 p( n 5) 7 p( n 6) p( n 7) p( n ) p( n ) p( n 4) 5 p( n 5) 0 p( n 6) p( n ) p( n 3) n-3, 3 p( n ) p( n ) p( n 4) p( n 5) 6 p( n 6) p( n)

24 Growth of p(n) p(n-) p(n) p(n-)+p(n-) super-polynomial, sub-exponential n0 Q( x) p( n) x n asymptotics good enough to show that finitely many subtracted terms guaranteed to cancel out for n sufficiently large

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

26 Character Table for S 4 4, 3,

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 n n s( n) x Q( x) s( nx ) 0 ( ) i i n i x n0 For every row sum formula in terms of p(n), the conjugate row has the same formula in terms of s(n). s(n-) s(n) s(n-)+s(n-) for n >

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