Proofs & Confirmations The story of the alternating sign matrix conjecture

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

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

3. 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

4. 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

5. 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?

6. 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

7. 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

8. 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

9. 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

10. 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 + + +

11. 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 + + +

12. 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

13. 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

14. 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

15. 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

16. 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:

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

18. 1 0 0 –1 1 0 –1 0 1 Richard Stanley

19. 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

20. 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

21. 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?

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

23. 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)

24. 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

25. Generating function: 1 0 0 –1 1 0 –1 0 1

26. 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

27. 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)

28. 1971 Basil Gordon proves case for n = infinity 1 0 0 –1 1 0 –1 0 1

29. 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

30. 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

31. 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.

32. Does this work for other groups of symmetries? 1 0 0 –1 1 0 –1 0 1 G = S3 ?

33. Does this work for other groups of symmetries? 1 0 0 –1 1 0 –1 0 1 G = S3 ? No

34. 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.

35. Cyclically Symmetric Plane Partition 1 0 0 –1 1 0 –1 0 1

36. 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.

37. 1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1

38. 1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1

39. 1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1

40. 1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1

41. 1979, Andrews counts cyclically symmetric plane partitions 1 0 0 –1 1 0 –1 0 1 L1 = W1 > L2 = W2 > L3 = W3 > … width length

42. 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

43. 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

44. 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

45. 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

46. 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 0 –1 1 0 –1 0 1

47. 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 0 –1 1 0 –1 0 1 Discovered an easier proof of Andrews’ formula, using induction on j and n.

48. 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 0 –1 1 0 –1 0 1 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

49. 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 0 –1 1 0 –1 0 1 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!

50. Totally Symmetric Self-Complementary Plane Partitions 1 0 0 –1 1 0 –1 0 1 1983 David Robbins (1942–2003) Vertical flip of ASM  complement of DPP ?