66
David M. Bressoud Macalester College, St. Paul, MN Talk given at University of Florida October 29, 2004

Exploiting Symmetries

  • Upload
    gino

  • View
    66

  • Download
    4

Embed Size (px)

DESCRIPTION

Exploiting Symmetries. Alternating Sign Matrices and the Weyl Character Formulas. David M. Bressoud Macalester College, St. Paul, MN Talk given at University of Florida October 29, 2004. The Vandermonde determinant Weyl’s character formulae Alternating sign matrices - PowerPoint PPT Presentation

Citation preview

Page 1: Exploiting Symmetries

David M. BressoudMacalester College, St. Paul, MNTalk given at University of FloridaOctober 29, 2004

Page 2: Exploiting Symmetries

1. The Vandermonde determinant

2. Weyl’s character formulae

3. Alternating sign matrices

4. The six-vertex model of statistical mechanics

5. Okada’s work connecting ASM’s and character formulae

Page 3: Exploiting Symmetries

x1n−1 x2

n−1 L xnn−1

M M O M

x1 x2 L xn

1 1 L 1

= −1( )I σ( ) xin−σ i( )

i=1

n

∏σ∈Sn

Cauchy 1815

“Memoir on functions whose values are equal but of opposite sign when two of their variables are interchanged”

(alternating functions) Augustin-Louis

Cauchy (1789–1857)

Page 4: Exploiting Symmetries

Cauchy 1815

“Memoir on functions whose values are equal but of opposite sign when two of their variables are interchanged”

(alternating functions)

This function is 0 when so it is divisible by

xi =xjxi −xj( )

i< j∏

x1n−1 x2

n−1 L xnn−1

M M O M

x1 x2 L xn

1 1 L 1

= −1( )I σ( ) xin−σ i( )

i=1

n

∏σ∈Sn

Page 5: Exploiting Symmetries

Cauchy 1815

“Memoir on functions whose values are equal but of opposite sign when two of their variables are interchanged”

(alternating functions)

This function is 0 when so it is divisible by

xi =xjxi −xj( )

i< j∏

But both polynomials have same degree, so ratio is constant, = 1.

= xi − x j( )i< j∏

x1n−1 x2

n−1 L xnn−1

M M O M

x1 x2 L xn

1 1 L 1

= −1( )I σ( ) xin−σ i( )

i=1

n

∏σ∈Sn

Page 6: Exploiting Symmetries

Cauchy 1815

Any alternating function in divided by the Vandermonde determinant yields a symmetric function:

x1, x2 ,K , xn

x1λ1 +n−1 x2

λ1 +n−1 L xnλ1 +n−1

M M O M

x1λn−1 +1 x2

λn−1 +1 L xnλn−1 +1

x1λn x2

λn L xnλn

x1n−1 x2

n−1 L xnn−1

M M O M

x11 x2

1 L xn1

x10 x2

0 L xn0

=sλ x1,x2 ,K ,xn( )

Page 7: Exploiting Symmetries

Cauchy 1815

Any alternating function in divided by the Vandermonde determinant yields a symmetric function:

x1, x2 ,K , xn

Called the Schur function. I.J. Schur (1917) recognized it as the character of the irreducible representation of GLn indexed by λ.

x1λ1 +n−1 x2

λ1 +n−1 L xnλ1 +n−1

M M O M

x1λn−1 +1 x2

λn−1 +1 L xnλn−1 +1

x1λn x2

λn L xnλn

x1n−1 x2

n−1 L xnn−1

M M O M

x11 x2

1 L xn1

x10 x2

0 L xn0

=sλ x1,x2 ,K ,xn( )

Issai Schur (1875–1941)

Page 8: Exploiting Symmetries

sλ 1,1,K ,1( ) is the dimension of the representation

sλ 1,1,K ,1( ) =ρ + λ( )⋅rρ ⋅rr∈An−1

+∏

where ρ =n−12

,n−32

,K ,1−n2

⎛⎝⎜

⎞⎠⎟=12

rr∈An−1

+∑ ,

λ = λ1,λ2 ,K ,λn( ),

An−1+ = ei −ej 1≤i < j ≤n{ } ,

ei is the unit vector with 1 in the ith coordinate

Note that the symmetric group on n letters is the group of transformations of

An−1 = ei −ej 1≤i ≠ j ≤n{ }

Page 9: Exploiting Symmetries

Weyl 1939 The Classical Groups: their invariants and representations

Sp2n λ; rx( ) =

x1λ1 +n −x1

−λ1−n L xnλ1 +n −xn

−λ1−n

M O M

x1λn+1 −x1

−λn−1 L xnλn+1 −xn

−λn−1

x1n −x1

−n L xnn −xn

−n

M O M

x11 −x1

−1 L xn1 −xn

−1

Sp2n λ; rx( ) is the character of the irreducible representation, indexed by the partition λ, of the symplectic group (the subgoup of GL2n of isometries).

Hermann Weyl (1885–1955)

Page 10: Exploiting Symmetries

Sp2n λ;1r

( ) =ρ + λ( )⋅rρ ⋅rr∈Cn

+∏

where ρ = n,n−1,K ,1( ) =12

rr∈Cn

+∑ ,

λ = λ1,λ2 ,K ,λn( ),

Cn+ = ei ±ej 1≤i < j ≤n{ } U 2ei 1≤i ≤n{ } ,

ei is the unit vector with 1 in the ith coordinate

The dimension of the representation is

Page 11: Exploiting Symmetries

Weyl 1939 The Classical Groups: their invariants and representations

x1n −x1

−n L xnn −xn

−n

M O M

x11 −x1

−1 L xn1 −xn

−1

x1x2L xn( )n

xi −xj( )i< j∏

is a symmetric polynomial. As a polynomial in x1 it has degree n + 1 and roots at

±1, x j−1 for 2 ≤ j ≤ n

Page 12: Exploiting Symmetries

Weyl 1939 The Classical Groups: their invariants and representations

x1n −x1

−n L xnn −xn

−n

M O M

x11 −x1

−1 L xn1 −xn

−1

x1x2L xn( )n

xi −xj( )i< j∏

= xi2 −1( )

i∏ xixj −1( )

i< j∏

is a symmetric polynomial. As a polynomial in x1 it has degree n + 1 and roots at

±1, x j−1 for 2 ≤ j ≤ n

Page 13: Exploiting Symmetries

Weyl 1939 The Classical Groups: their invariants and representations: The Denominator Formulas

x1

n−12 −x1−n+ 12 L xn

n−12 −xn−n+ 12

M O M

x112 −x1

−12 L xn

12 −xn

−12

x1x2L xn( )n−12

xi −xj( )i< j∏

= xi −1( )i∏ xixj −1( )

i< j∏

x1n−1 + x1

−n+1 L xnn−1 + xn

−n+1

M O M

x10 + x1

−0 L xn0 + xn

−0

x1x2L xn( )n−1

xi −xj( )i< j∏

=2 xixj −1( )i< j∏

Page 14: Exploiting Symmetries
Page 15: Exploiting Symmetries

Desnanot-Jacobi adjoint matrix thereom (Desnanot for n ≤ 6 in 1819, Jacobi for general case in 1833M j

i is matrix M with row i and column j removed.

detM =detM1

1 ⋅detMnn −detMn

1 ⋅detM1n

detM1,n1,n

Given that the determinant of the empty matrix is 1 and the determinant of a 11 is the entry in that matrix, this uniquely defines the determinant for all square matrices.

Carl Jacobi (1804–1851)

Page 16: Exploiting Symmetries

detM =detM1

1 ⋅detMnn −detMn

1 ⋅detM1n

detM1,n1,n

detλ M =detM1

1 ⋅detMnn + λdetMn

1 ⋅detM1n

detM1,n1,n

det−1 M =detM( )

detλ aji−1( )

i, j=1

n= ai + λaj( )

1≤i< j≤n∏

David Robbins (1942–2003)

Page 17: Exploiting Symmetries

detM =detM1

1 ⋅detMnn −detMn

1 ⋅detM1n

detM1,n1,n

detλ M =detM1

1 ⋅detMnn + λdetMn

1 ⋅detM1n

detM1,n1,n

detλa bc d

⎛⎝⎜

⎞⎠⎟=ad+ λbc

detλ

a b cd e fg h j

⎜⎜

⎟⎟=aej + λ bdj + afh( ) + λ2 bfg+ cdh( ) + λ 3ceg

+ λ 1+ λ( )bde−1 fh

Page 18: Exploiting Symmetries

detλ

a1,1 a1,2 a1,3 a1,4

a2,1 a2,2 a2,3 a2,4

a3,1 a3,2 a3,3 a3,4

a4,1 a4,2 a4,3 a4,4

⎜⎜⎜⎜

⎟⎟⎟⎟

=a1,1a2,2a3,3a4,4

+ λ a1,2a2,1a3,3a4,4 + a1,1a2,3a3,2a4,4 + a1,1a2,2a3,4a4,3( )

+L sums over other permutations×λ inversion number

+ λ 3 1+ λ−1( )a1,2a2,1a2,2−1 a2,3a3,4a4,2 +L

+ λ 3 1+ λ−1( )2a1,2a2,1a2,2

−1 a2,3a3,2a3,3−1a3,4a4,3 +L

Page 19: Exploiting Symmetries

detλ

a1,1 a1,2 a1,3 a1,4

a2,1 a2,2 a2,3 a2,4

a3,1 a3,2 a3,3 a3,4

a4,1 a4,2 a4,3 a4,4

⎜⎜⎜⎜

⎟⎟⎟⎟

=a1,1a2,2a3,3a4,4

+ λ a1,2a2,1a3,3a4,4 + a1,1a2,3a3,2a4,4 + a1,1a2,2a3,4a4,3( )

+L sums over other permutations×λ inversion number

+ λ 3 1+ λ−1( )a1,2a2,1a2,2−1 a2,3a3,4a4,2 +L

+ λ 3 1+ λ−1( )2a1,2a2,1a2,2

−1 a2,3a3,2a3,3−1a3,4a4,3 +L

0 1 0 0

1 −1 1 00 1 −1 10 0 1 0

⎜⎜⎜⎜

⎟⎟⎟⎟

0 1 0 0

1 −1 1 00 0 0 10 1 0 0

⎜⎜⎜⎜

⎟⎟⎟⎟

Page 20: Exploiting Symmetries

detλ

a1,1 a1,2 a1,3 a1,4

a2,1 a2,2 a2,3 a2,4

a3,1 a3,2 a3,3 a3,4

a4,1 a4,2 a4,3 a4,4

⎜⎜⎜⎜

⎟⎟⎟⎟

=a1,1a2,2a3,3a4,4

+ λ a1,2a2,1a3,3a4,4 + a1,1a2,3a3,2a4,4 + a1,1a2,2a3,4a4,3( )

+L sums over other permutations×λ inversion number

+ λ 3 1+ λ−1( )a1,2a2,1a2,2−1 a2,3a3,4a4,2 +L

+ λ 3 1+ λ−1( )2a1,2a2,1a2,2

−1 a2,3a3,2a3,3−1a3,4a4,3 +L

0 1 0 0

1 −1 1 00 1 −1 10 0 1 0

⎜⎜⎜⎜

⎟⎟⎟⎟

0 1 0 0

1 −1 1 00 0 0 10 1 0 0

⎜⎜⎜⎜

⎟⎟⎟⎟

Page 21: Exploiting Symmetries

detλ

a1,1 a1,2 a1,3 a1,4

a2,1 a2,2 a2,3 a2,4

a3,1 a3,2 a3,3 a3,4

a4,1 a4,2 a4,3 a4,4

⎜⎜⎜⎜

⎟⎟⎟⎟

=a1,1a2,2a3,3a4,4

+ λ a1,2a2,1a3,3a4,4 + a1,1a2,3a3,2a4,4 + a1,1a2,2a3,4a4,3( )

+L sums over other permutations×λ inversion number

+ λ 3 1+ λ−1( )a1,2a2,1a2,2−1 a2,3a3,4a4,2 +L

+ λ 3 1+ λ−1( )2a1,2a2,1a2,2

−1 a2,3a3,2a3,3−1a3,4a4,3 +L

detλ xi, j( ) = λ Inv A( )

A= ai , j( )

∑ 1+ λ−1( )N A( )

xi, jai , j

i, j∏

Sum is over all alternating sign matrices, N(A) = # of –1’s

Page 22: Exploiting Symmetries

detλ

a1,1 a1,2 a1,3 a1,4

a2,1 a2,2 a2,3 a2,4

a3,1 a3,2 a3,3 a3,4

a4,1 a4,2 a4,3 a4,4

⎜⎜⎜⎜

⎟⎟⎟⎟

=a1,1a2,2a3,3a4,4

+ λ a1,2a2,1a3,3a4,4 + a1,1a2,3a3,2a4,4 + a1,1a2,2a3,4a4,3( )

+L sums over other permutations×λ inversion number

+ λ 3 1+ λ−1( )a1,2a2,1a2,2−1 a2,3a3,4a4,2 +L

+ λ 3 1+ λ−1( )2a1,2a2,1a2,2

−1 a2,3a3,2a3,3−1a3,4a4,3 +L

detλ xi, j( ) = λ Inv A( )

A= ai , j( )

∑ 1+ λ−1( )N A( )

xi, jai , j

i, j∏

xi + λxj( ) = λ Inv A( ) 1+ λ−1( )

N A( )xj

n−i( )ai , j

i, j∏

A∈An

∑1≤i< j≤n∏

Page 23: Exploiting Symmetries

n

1

2

3

4

5

6

7

8

9

An

1

2

7

42

429

7436

218348

10850216

911835460

= 2 3 7

= 3 11 13

= 22 11 132

= 22 132 17 19

= 23 13 172 192

= 22 5 172 193 23

How many n n alternating sign matrices?

Page 24: Exploiting Symmetries

n

1

2

3

4

5

6

7

8

9

An

1

2

7

42

429

7436

218348

10850216

911835460

= 2 3 7

= 3 11 13

= 22 11 132

= 22 132 17 19

= 23 13 172 192

= 22 5 172 193 23

Page 25: Exploiting Symmetries

n

1

2

3

4

5

6

7

8

9

An

1

2

7

42

429

7436

218348

10850216

911835460

There is exactly one 1 in the first

row

Page 26: Exploiting Symmetries

n

1

2

3

4

5

6

7

8

9

An

1

1+1

2+3+2

7+14+14+7

42+105+…

There is exactly one 1 in the first

row

Page 27: Exploiting Symmetries

1

1 1

2 3 2

7 14 14 7

42 105 135 105 42

429 1287 2002 2002 1287 429

Page 28: Exploiting Symmetries

1

1 1

2 3 2

7 14 14 7

42 105 135 105 42

429 1287 2002 2002 1287 429

+ + +

Page 29: Exploiting Symmetries

1

1 1

2 3 2

7 14 14 7

42 105 135 105 42

429 1287 2002 2002 1287 429

+ + +

1 0 0 0 0

0 ? ? ? ?

0 ? ? ? ?

0 ? ? ? ?

0 ? ? ? ?

⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟

Page 30: Exploiting Symmetries

1

1 2/2 1

2 2/3 3 3/2 2

7 2/4 14 14 4/2 7

42 2/5 105 135 105 5/2 42

429 2/6 1287 2002 2002 1287 6/2 429

Page 31: Exploiting Symmetries

1

1 2/2 1

2 2/3 3 3/2 2

7 2/4 14 5/5 14 4/2 7

42 2/5 105 7/9 135 9/7 105 5/2 42

429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429

Page 32: Exploiting Symmetries

2/2

2/3 3/2

2/4 5/5 4/2

2/5 7/9 9/7 5/2

2/6 9/14 16/16 14/9 6/2

Page 33: Exploiting Symmetries

2

2 3

2 5 4

2 7 9 5

2 9 16 14 6

Page 34: Exploiting Symmetries

1+1

1+1 1+2

1+1 2+3 1+3

1+1 3+4 3+6 1+4

1+1 4+5 6+10 4+10 1+5

Numerators:

Page 35: Exploiting Symmetries

1+1

1+1 1+2

1+1 2+3 1+3

1+1 3+4 3+6 1+4

1+1 4+5 6+10 4+10 1+5

Conjecture 1:

Numerators:

An,k

An,k+1

=

n−2k−1

⎛⎝⎜

⎞⎠⎟+

n−1k−1

⎛⎝⎜

⎞⎠⎟

n−2n−k−1

⎛⎝⎜

⎞⎠⎟+

n−1n−k−1

⎛⎝⎜

⎞⎠⎟

Page 36: Exploiting Symmetries

Conjecture 1:

Conjecture 2 (corollary of Conjecture 1):

An,k

An,k+1

=

n−2k−1

⎛⎝⎜

⎞⎠⎟+

n−1k−1

⎛⎝⎜

⎞⎠⎟

n−2n−k−1

⎛⎝⎜

⎞⎠⎟+

n−1n−k−1

⎛⎝⎜

⎞⎠⎟

An =

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!

n!⋅n+1( )!L 2n−1( )!

Page 37: Exploiting Symmetries

Conjecture 2 (corollary of Conjecture 1):

An =

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!

n!⋅n+1( )!L 2n−1( )!

Exactly the formula found by George Andrews for counting descending plane partitions.

George Andrews Penn State

Page 38: Exploiting Symmetries

Conjecture 2 (corollary of Conjecture 1):

An =

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!

n!⋅n+1( )!L 2n−1( )!

Exactly the formula found by George Andrews for counting descending plane partitions. In succeeding years, the connection would lead to many important results on plane partitions.

George Andrews Penn State

Page 39: Exploiting Symmetries

A n; x( ) = xN A( )

A∈An

∑A 1;x( ) =1,

A 2;x( ) =2,

A 3;x( ) =6 + x,

A 4;x( ) =24 +16x+ 2x2 ,

A 5;x( ) =120 + 200x+ 94x2 +14x3 + x4 ,

A 6;x( ) =720 + 2400x+ 2684x2 +1284x3 + 310x4 + 36x5 + 2x6

A 7;x( ) =5040 + 24900x+ 63308x2 + 66158x3 + 38390x4 +13037x5

+ 2660x6 + 328x7 + 26x8 + x9

Page 40: Exploiting Symmetries

A n; x( ) = xN A( )

A∈An

∑A 1;x( ) =1,

A 2;x( ) =2,

A 3;x( ) =6 + x,

A 4;x( ) =24 +16x+ 2x2 ,

A 5;x( ) =120 + 200x+ 94x2 +14x3 + x4 ,

A 6;x( ) =720 + 2400x+ 2684x2 +1284x3 + 310x4 + 36x5 + 2x6

A 7;x( ) =5040 + 24900x+ 63308x2 + 66158x3 + 38390x4 +13037x5

+ 2660x6 + 328x7 + 26x8 + x9

xi + λxj( ) = λ Inv A( ) 1+ λ−1( )

N A( )xj

n−i( )ai , j

i, j∏

A∈An

∑1≤i< j≤n∏

A n;0( ) =n!

A n;1( ) =An =3i +1( )!n+ i( )!i=0

n−1

∏⎛

⎝⎜⎞

⎠⎟

A n;2( ) =2n(n−1)/2

Page 41: Exploiting Symmetries

A n; x( ) = xN A( )

A∈An

∑A 1;x( ) =1,

A 2;x( ) =2,

A 3;x( ) =6 + x,

A 4;x( ) =24 +16x+ 2x2 ,

A 5;x( ) =120 + 200x+ 94x2 +14x3 + x4 ,

A 6;x( ) =720 + 2400x+ 2684x2 +1284x3 + 310x4 + 36x5 + 2x6

A 7;x( ) =5040 + 24900x+ 63308x2 + 66158x3 + 38390x4 +13037x5

+ 2660x6 + 328x7 + 26x8 + x9

A n; 3( ) =3n n−1( )

2n n−1( )

3 j −i( ) +13 j −i( )1≤i, j≤n

j−i odd

∏ Conjecture:

(MRR, 1983)

A n;0( ) =n!

A n;1( ) =An =3i +1( )!n+ i( )!i=0

n−1

∏⎛

⎝⎜⎞

⎠⎟

A n;2( ) =2n(n−1)/2

Page 42: Exploiting Symmetries

Mills & Robbins (suggested by Richard Stanley) (1991)

Symmetries of ASM’s

A n( ) =3 j +1( )!n+ j( )!j=0

n−1

AV 2n+1( ) = −3( )n2 3 j −i( ) +1

j −i + 2n+11≤i, j≤2n+12 j

∏⎛

⎜⎜⎜

⎟⎟⎟

A n( )

AHT 2n( ) = −3( )n n−1( )/2 3 j −i( ) + 2j −i +ni, j

∏⎛

⎝⎜⎞

⎠⎟A n( )

AQT 4n( ) =AHT 2n( )⋅A n( )2

Vertically symmetric ASM’s

Half-turn symmetric ASM’s

Quarter-turn symmetric ASM’s

Page 43: Exploiting Symmetries

December, 1992

Zeilberger announces a proof that # of ASM’s equals

3 j +1( )!n+ j( )!j=0

n−1

Doron Zeilberger

Rutgers University

Page 44: Exploiting Symmetries

December, 1992

Zeilberger announces a proof that # of ASM’s equals

3 j +1( )!n+ j( )!j=0

n−1

1995 all gaps removed, published as “Proof of the Alternating Sign Matrix Conjecture,” Elect. J. of Combinatorics, 1996.

Page 45: Exploiting Symmetries

Zeilberger’s proof is an 84-page tour de force, but it still left open the original conjecture:

An,k

An,k+1

=

n−2k−1

⎛⎝⎜

⎞⎠⎟+

n−1k−1

⎛⎝⎜

⎞⎠⎟

n−2n−k−1

⎛⎝⎜

⎞⎠⎟+

n−1n−k−1

⎛⎝⎜

⎞⎠⎟

Page 46: Exploiting Symmetries

1996 Kuperberg announces a simple proof

“Another proof of the alternating sign matrix conjecture,” International Mathematics Research Notices

Greg Kuperberg

UC Davis

Page 47: Exploiting Symmetries

“Another proof of the alternating sign matrix conjecture,” International Mathematics Research NoticesPhysicists have been

studying ASM’s for decades, only they call them square ice (aka the six-vertex model ).

1996 Kuperberg announces a simple proof

Page 48: Exploiting Symmetries

H O H O H O H O H O H

H H H H H

H O H O H O H O H O H

H H H H H

H O H O H O H O H O H

H H H H H

H O H O H O H O H O H

H H H H H

H O H O H O H O H O H

Page 49: Exploiting Symmetries
Page 50: Exploiting Symmetries

Horizontal 1

Vertical –1

Page 51: Exploiting Symmetries

southwest

northeast

northwest

southeast

Page 52: Exploiting Symmetries

N = # of verticalI = inversion number = N + # of SW

x2, y3

Page 53: Exploiting Symmetries

Anatoli Izergin

Vladimir Korepin

SUNY Stony Brook

1980’s

Page 54: Exploiting Symmetries

det1

xi −yj( ) axi −yj( )

⎝⎜

⎠⎟

xi −yj( ) axi −yj( )i, j=1

n∏xi −xj( ) yi −yj( )1≤i< j≤n∏

= 1−a( )2N A( ) an(n−1)/2−Inv A( )

A∈An

× xivert∏ yj axi −yj( )

SW, NE∏ xi −yj( )

NW, SE∏

Proof:LHS is symmetric polynomial in x’s and in y’s

Degree n – 1 in x1

By induction, LHS = RHS when x1 = y1

Sufficient to show that RHS is symmetric polynomial in x’s and in y’s

Page 55: Exploiting Symmetries

LHS is symmetric polynomial in x’s and in y’s

Degree n – 1 in x1

By induction, LHS = RHS when x1 = –y1

Sufficient to show that RHS is symmetric polynomial in x’s and in y’s — follows from Baxter’s triangle-to-triangle relation

Proof:

Rodney J. Baxter

Australian National University

Page 56: Exploiting Symmetries

a =z−4 , xi =z2 , yi =1

RHS= z−z−1( )n n−1( )

z+ z−1( )2N A( )

A∈An

det1

xi −yj( ) axi −yj( )

⎝⎜

⎠⎟

xi −yj( ) axi −yj( )i, j=1

n∏xi −xj( ) yi −yj( )1≤i< j≤n∏

= 1−a( )2N A( ) an(n−1)/2−Inv A( )

A∈An

× xivert∏ yj axi −yj( )

SW, NE∏ xi −yj( )

NW, SE∏

Page 57: Exploiting Symmetries

det1

xi −yj( ) axi −yj( )

⎝⎜

⎠⎟

xi −yj( ) axi + yj( )i, j=1

n∏xi −xj( ) yi −yj( )1≤i< j≤n∏

= 1−a( )2N A( ) an(n−1)/2−Inv A( )

A∈An

× xivert∏ yj axi −yj( )

SW, NE∏ xi −yj( )

NW, SE∏

z =eπ i /3 : RHS= −3( )n n−1( )/2 An ,

z=eπ i /4 : RHS= −2( )n n−1( )/2 2N A( )

A∈An

∑ ,

z=eπ i /6 : RHS= −1( )n n−1( )/2 3N A( )

A∈An

∑ .

a =z−4 , xi =z2 , yi =1

RHS= z−z−1( )n n−1( )

z+ z−1( )2N A( )

A∈An

Page 58: Exploiting Symmetries

1996

Doron Zeilberger uses this determinant to prove the original conjecture

“Proof of the refined alternating sign matrix conjecture,” New York Journal of Mathematics

Page 59: Exploiting Symmetries

2001, Kuperberg uses the power of the triangle-to-triangle relation to prove some of the conjectured formulas:

AV 2n +1( ) = −3( )n2 3 j −i( ) +1

j −i + 2n+1i, j≤2n+12 j

∏⎛

⎜⎜⎜

⎟⎟⎟

A n( )

AHT 2n( ) = −3( )n n−1( )/2 3 j −i( ) + 2j −i +ni, j

∏⎛

⎝⎜⎞

⎠⎟A n( )

AQT 4n( ) =AHT 2n( )⋅A n( )2

Page 60: Exploiting Symmetries

Kuperberg, 2001: proved formulas for counting some new six-vertex models:

AUU 2n( ) = −3( )n2

22n 3 j −i( ) + 2j −i + 2n+11≤i, j≤2n+1

2 j

1 −1 0 10 1 −1 00 0 0 00 0 1 0

⎜⎜⎜⎜

⎟⎟⎟⎟

Page 61: Exploiting Symmetries

Kuperberg, 2001: proved formulas for many symmetry classes of ASM’s and some new ones

1 −1 0 10 1 −1 00 0 0 00 0 1 0

⎜⎜⎜⎜

⎟⎟⎟⎟

AUU 2n( ) = −3( )n2

22n 3 j −i( ) + 2j −i + 2n+11≤i, j≤2n+1

2 j

Page 62: Exploiting Symmetries

Soichi Okada, Nagoya University

1993, Okada finds the equivalent of the λ-determinant for the other Weyl Denominator Formulas.

2004, Okada shows that the formulas for counting ASM’s, including those subject to symmetry conditions, are simply the dimensions of certain irreducible representations, i.e. specializations of Weyl Character formulas.

Page 63: Exploiting Symmetries

sλ 1,1,K ,1( ) =ρ + λ( )⋅rρ ⋅rr∈A2n−1

+∏ =

3 j +1( )2

⎢⎣⎢

⎥⎦⎥− 3i +1( )

2⎢⎣⎢

⎥⎦⎥

j −i1≤i< j≤2n∏

3−n(n−1)/2sλ 1,1,K ,1( ) =3i +1( )!n+ i( )!i=0

n−1

Number of n n ASM’s is 3–n(n–1)/2

times the dimension of the irreducible representation of GL2n indexed by λ = n −1,n −1,n − 2,n − 2,K ,1,1,0,0( )

A2n−1+ = ei −ej 1≤i < j ≤2n{ }

ρ = n−12,n−3

2,K ,−n+ 12( )

Page 64: Exploiting Symmetries

dim Sp4n λ( ) =ρC + λ( )⋅rρC ⋅rr∈C2n

+∏

=6n+ 2 −

3i +12

⎢⎣⎢

⎥⎦⎥−

3 j +12

⎢⎣⎢

⎥⎦⎥

4n+ 2 −i − j

⎜⎜⎜

⎟⎟⎟1≤i< j≤2n

∏3 j +12

⎢⎣⎢

⎥⎦⎥−

3i +12

⎢⎣⎢

⎥⎦⎥

j −i

⎜⎜⎜

⎟⎟⎟

3n+1−3i +12

⎢⎣⎢

⎥⎦⎥

2n+1−ii=1

2n

Number of (2n+1) (2n+1) vertically symmetric ASM’s is 3–

n(n–1) times the dimension of the irreducible representation of Sp4n indexed by λ = n −1,n −1,n − 2,n − 2,K ,1,1,0,0( )

C2n+ = ei ±ej 1≤i < j ≤2n{ } U 2ei 1≤i ≤2n{ }

ρC =12

r =r∈C2n

+∑ 2n,2n−1,K ,1( )

Page 65: Exploiting Symmetries

NEW for 2004:

Number of (4n+1) (4n+1) vertically and horizontally symmetric ASM’s is 2–2n 3–n(2n–1)

times

λ = n −1,n −1,n − 2,n − 2,K ,1,1,0,0( )

μ = n − 12,n − 3

2,n − 32,K , 3

2, 32, 1

2( )

dim Sp4n λ( )×dim %O4n μ( ) =

ρC + λ( )⋅rρC ⋅rr∈C2n

+∏

⎝⎜

⎠⎟

ρD + μ( )⋅rρD ⋅rr∈D2n

+∏

⎝⎜

⎠⎟

C2n+ = ei ±ej 1≤i < j ≤2n{ } U 2ei 1≤i ≤2n{ }

ρC = 2n,2n−1,K ,1( )

D2n+ = ei ±ej 1≤i < j ≤2n{ }

ρD = 2n−1,2n−2,K ,1,0( )

Page 66: Exploiting Symmetries

Proofs and Confirmations: The Story of the Alternating Sign Matrix Conjecture

Cambridge University Press & MAA, 1999

OKADA, Enumeration of Symmetry Classes of Alternating Sign Matrices and Characters of Classical Groups, arXiv:math.CO/0408234 v1 18 Aug 2004