Proofs & Confirmations The story of the alternating sign matrix conjecture

Proofs & Confirmations The story of the alternating sign matrix conjecture David M. Bressoud Macalester College MAA Northeastern Section November 20, 2004 Worcester, MA

Bill Mills Institute for Defense Analysis 1 0 0 –1 1 0 –1 0 1 Howard Rumsey Dave Robbins

Charles L. Dodgson aka Lewis Carroll 1 0 0 –1 1 0 –1 0 1 “Condensation of Determinants,” Proceedings of the Royal Society, London 1866

n 1 2 3 4 5 6 7 8 9 An 1 2 7 42 429 7436 218348 10850216 911835460 1 0 0 –1 1 0 –1 0 1

n 1 2 3 4 5 6 7 8 9 An 1 2 7 42 429 7436 218348 10850216 911835460 1 0 0 –1 1 0 –1 0 1 = 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?

n 1 2 3 4 5 6 7 8 9 An 1 2 7 42 429 7436 218348 10850216 911835460 1 0 0 –1 1 0 –1 0 1 very suspicious = 2  3  7 = 3  11  13 = 22 11  132 = 22 132  17  19 = 23 13  172  192 = 22 5  172  193  23

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 1 0 0 –1 1 0 –1 0 1

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 1 0 0 –1 1 0 –1 0 1

1 1 1 2 3 2 7 14 14 7 42 105 135 105 42 429 1287 2002 2002 1287 429 1 0 0 –1 1 0 –1 0 1

1 1 1 2 3 2 7 14 14 7 42 105 135 105 42 429 1287 2002 2002 1287 429 1 0 0 –1 1 0 –1 0 1 + + +

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 1 0 0 –1 1 0 –1 0 1

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 1 0 0 –1 1 0 –1 0 1

2/2 2/33/2 2/45/54/2 2/57/99/75/2 2/69/1416/1614/96/2 1 0 0 –1 1 0 –1 0 1

Numerators: 1+1 1+11+2 1+12+31+3 1+13+43+61+4 1+14+56+104+101+5 1 0 0 –1 1 0 –1 0 1

1+1 1+11+2 1+12+31+3 1+13+43+61+4 1+14+56+104+101+5 Numerators: 1 0 0 –1 1 0 –1 0 1 Conjecture 1:

Conjecture 1: 1 0 0 –1 1 0 –1 0 1 Conjecture 2 (corollary of Conjecture 1):

1 0 0 –1 1 0 –1 0 1 Richard Stanley

1 0 0 –1 1 0 –1 0 1 Richard Stanley Andrews’ Theorem: the number of descending plane partitions of size n is George Andrews

All you have to do is find a 1-to-1 correspondence between n by n alternating sign matrices and descending plane partitions of size n, and conjecture 2 will be proven! 1 0 0 –1 1 0 –1 0 1

All you have to do is find a 1-to-1 correspondence between n by n alternating sign matrices and descending plane partitions of size n, and conjecture 2 will be proven! 1 0 0 –1 1 0 –1 0 1 What is a descending plane partition?

Percy A. MacMahon Plane Partition 1 0 0 –1 1 0 –1 0 1 Work begun in 1897

Plane partition of 75 6 5 5 4 3 3 1 0 0 –1 1 0 –1 0 1 # of pp’s of 75 = pp(75)

Plane partition of 75 6 5 5 4 3 3 1 0 0 –1 1 0 –1 0 1 # of pp’s of 75 = pp(75) = 37,745,732,428,153

Generating function: 1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

1912 MacMahon proves that the generating function for plane partitions in an nnn box is 1 0 0 –1 1 0 –1 0 1 At the same time, he conjectures that the generating function for symmetric plane partitions is

Symmetric Plane Partition 4 3 2 1 1 3 2 2 1 2 2 1 1 1 1 1 0 0 –1 1 0 –1 0 1 “The reader must be warned that, although there is little doubt that this result is correct, … the result has not been rigorously established. … Further investigations in regard to these matters would be sure to lead to valuable work.’’ (1916)

1971 Basil Gordon proves case for n = infinity 1 0 0 –1 1 0 –1 0 1

1971 Basil Gordon proves case for n = infinity 1 0 0 –1 1 0 –1 0 1 1977 George Andrews and Ian Macdonald independently prove general case

Macdonald’s observation: both generating functions are special cases of the following 1 0 0 –1 1 0 –1 0 1 where G is a group acting on the points in B and B/G is the set of orbits. If G consists of only the identity, this gives all plane partitions in B. If G is the identity and (i,j,k)  (j,i,k), then getgenerating function for symmetric plane partitions.

Does this work for other groups of symmetries? 1 0 0 –1 1 0 –1 0 1 G = S3 ? No G = C3 ? (i,j,k)  (j,k,i)  (k,i,j) It seems to work.

Cyclically Symmetric Plane Partition 1 0 0 –1 1 0 –1 0 1

Macdonald’s Conjecture (1979): The generating function for cyclically symmetric plane partitions in B(n,n,n) is 1 0 0 –1 1 0 –1 0 1 “If I had to single out the most interesting open problem in all of enumerative combinatorics, this would be it.” Richard Stanley, review of Symmetric Functions and Hall Polynomials, Bulletin of the AMS, March, 1981.

1 0 0 –1 1 0 –1 0 1 Macdonald’s Second Conjecture (1979): For every subgroup G of S3, the product on the right counts the total number of plane partitions contained in B and invariant under G. G = C3: proved by Andrews, 1979

1 0 0 –1 1 0 –1 0 1 Macdonald’s Second Conjecture (1979): For every subgroup G of S3, the product on the right counts the total number of plane partitions contained in B and invariant under G. G = S3: proved by John Stembridge, 1995

1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1

1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1 L1 = W1 > L2 = W2 > L3 = W3 > … width length

1979, Andrews counts descending plane partitions 1 0 0 –1 1 0 –1 0 1 L1 > W1 ≥ L2 > W2 ≥ L3 > W3 ≥ … 6 6 6 4 3 3 3 2 width length

6 X 6 ASM  DPP with largest part ≤ 6 What are the corresponding 6 subsets of DPP’s? 1 0 0 –1 1 0 –1 0 1 6 6 6 4 3 3 3 2 width length

ASM with 1 at top of first column DPP with no parts of size n. ASM with 1 at top of last column DPP with n–1 parts of size n. 1 0 0 –1 1 0 –1 0 1 6 6 6 4 3 3 3 2 width length

Mills, Robbins, Rumsey Conjecture: # of nn ASM’s with 1 at top of column j equals # of DPP’s ≤ n with exactly j–1 parts of size n. 1 0 0 –1 1 0 –1 0 1 6 6 6 4 3 3 3 2 width length

Proofs & Confirmations The story of the alternating sign matrix conjecture