Proofs & Confirmations The story of the alternating sign matrix conjecture David M. Bressoud Macalester College New Jersey Section, MAA November 6, 2005 Monmouth University

1 0 –1 1 0 –1 0 1 Bill Mills Dave Robbins Howard Rumsey IDA/CCR

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

1 0 –1 1 0 –1 0 1 n n A n

n n A n = 2  3  7 = 3  11  13 = 2 2  11  13 2 = 2 2  13 2  17  19 = 2 3  13  17 2  19 2 = 2 2  5  17 2  19 3  23 How many n  n alternating sign matrices? 1 0 –1 1 0 –1 0 1

n n A n = 2  3  7 = 3  11  13 = 2 2  11  13 2 = 2 2  13 2  17  19 = 2 3  13  17 2  19 2 = 2 2  5  17 2  19 3  –1 1 0 –1 0 1

0 –1 1 0 –1 0 1 n n A n There is exactly one 1 in the first row

1 0 –1 1 0 –1 0 1 n n A n … There is exactly one 1 in the first row

1 0 –1 1 0 –

1 0 –1 1 0 –

1 0 –1 1 0 –

1 0 –1 1 0 – / /3 3 3/ / / / / / /2 429

1 0 –1 1 0 – / /3 3 3/ /4 14 5/5 14 4/ / / / / / / / / /2 429

1 0 –1 1 0 – /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

1 0 –1 1 0 – Numerators:

1 0 –1 1 0 – Conjecture 1: Numerators:

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

1 0 –1 1 0 –1 0 1 Richard Stanley

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

1 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 –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! What is a descending plane partition?

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

1 0 –1 1 0 – Plane partition of 75 # of pp’s of 75 = pp(75)

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

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

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

1 0 –1 1 0 –1 0 1 Symmetric Plane Partition “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)

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

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

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

1 0 –1 1 0 –1 0 1 Macdonald’s observation: both generating functions are special cases of the following 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 get generating function for symmetric plane partitions.

1 0 –1 1 0 –1 0 1 Does this work for other groups of symmetries? G = S 3 ?

1 0 –1 1 0 –1 0 1 Does this work for other groups of symmetries? G = S 3 ? No

1 0 –1 1 0 –1 0 1 Does this work for other groups of symmetries? G = S 3 ? No G = C 3 ? (i,j,k)  (j,k,i)  (k,i,j) It seems to work.

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

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

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

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

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

1 0 –1 1 0 – , Andrews counts cyclically symmetric plane partitions length width L 1 = W 1 > L 2 = W 2 > L 3 = W 3 > …

1 0 –1 1 0 – , Andrews counts descending plane partitions length width L 1 > W 1 ≥ L 2 > W 2 ≥ L 3 > W 3 ≥ …

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

1 0 –1 1 0 –1 0 1 length width 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 –1 1 0 –1 0 1 length width 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 –1 1 0 –1 0 1 Mills, Robbins, & Rumsey proved that # of DPP’s ≤ n with j parts of size n was given by their conjectured formula for ASM’s.

1 0 –1 1 0 –1 0 1 Mills, Robbins, & Rumsey proved that # of DPP’s ≤ n with j parts of size n was given by their conjectured formula for ASM’s. Discovered an easier proof of Andrews’ formula, using induction on j and n.

1 0 –1 1 0 –1 0 1 Mills, Robbins, & Rumsey proved that # of DPP’s ≤ n with j parts of size n was given by their conjectured formula for ASM’s. Discovered an easier proof of Andrews’ formula, using induction on j and n. Used this inductive argument to prove Macdonald’s conjecture “Proof of the Macdonald Conjecture,” Inv. Math., 1982

1 0 –1 1 0 –1 0 1 Mills, Robbins, & Rumsey proved that # of DPP’s ≤ n with j parts of size n was given by their conjectured formula for ASM’s. Discovered an easier proof of Andrews’ formula, using induction on j and n. Used this inductive argument to prove Macdonald’s conjecture “Proof of the Macdonald Conjecture,” Inv. Math., 1982 But they still didn’t have a proof of their conjecture!

1 0 –1 1 0 – Totally Symmetric Self-Complementary Plane Partitions Vertical flip of ASM  complement of DPP ? David Robbins (1942–2003)

1 0 –1 1 0 –1 0 1 Totally Symmetric Self-Complementary Plane Partitions

1 0 –1 1 0 –1 0 1

0 –1 1 0 –1 0 1 Robbins’ Conjecture: The number of TSSCPP’s in a 2n X 2n X 2n box is

1 0 –1 1 0 –1 0 1 Robbins’ Conjecture: The number of TSSCPP’s in a 2n X 2n X 2n box is 1989: William Doran shows equivalent to counting lattice paths 1990: John Stembridge represents the counting function as a Pfaffian (built on insights of Gordon and Okada) 1992: George Andrews evaluates the Pfaffian, proves Robbins’ Conjecture

1 0 –1 1 0 –1 0 1 December, 1992 Doron Zeilberger announces a proof that # of ASM’s of size n equals of TSSCPP’s in box of size 2n.

1 0 –1 1 0 –1 0 1 December, 1992 Doron Zeilberger announces a proof that # of ASM’s of size n equals of TSSCPP’s in box of size 2n all gaps removed, published as “Proof of the Alternating Sign Matrix Conjecture,” Elect. J. of Combinatorics, 1996.

1 0 –1 1 0 –1 0 1 Zeilberger’s proof is an 84-page tour de force, but it still left open the original conjecture:

1996 Kuperberg announces a simple proof “Another proof of the alternating sign matrix conjecture,” International Mathematics Research Notices Greg Kuperberg UC Davis 1 0 –1 1 0 –1 0 1

1996 Kuperberg announces a simple proof “Another proof of the alternating sign matrix conjecture,” International Mathematics Research Notices Greg Kuperberg UC Davis Physicists have been studying ASM’s for decades, only they call them square ice (aka the six-vertex model ). 1 0 –1 1 0 –1 0 1

0 –1 1 0 –1 0 1 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

1 0 –1 1 0 –1 0 1

0 –1 1 0 –1 0 1 Horizontal  1 Vertical  –1

1 0 –1 1 0 –1 0 1 southwest northeast northwest southeast

1 0 –1 1 0 –1 0 1 N = # of vertical I = inversion number = N + # of NE x 2, y 3

1 0 –1 1 0 – ’s Rodney Baxter’s Triangle-to- triangle relation

1 0 –1 1 0 – ’s Anatoli Izergin Vladimir Korepin 1980’s Rodney Baxter’s Triangle-to- triangle relation

1 0 –1 1 0 –1 0 1

0 –1 1 0 –1 0 1

0 –1 1 0 – Doron Zeilberger uses this determinant to prove the original conjecture “Proof of the refined alternating sign matrix conjecture,” New York Journal of Mathematics

1 0 –1 1 0 –1 0 1 The End (which is really just the beginning)

1 0 –1 1 0 –1 0 1 The End (which is really just the beginning) This Power Point presentation can be downloaded from