Proofs & Confirmations The story of the alternating sign matrix conjecture

Transcription

1 Proofs & Confirmations The story of the alternating sign matrix conjecture David M. Bressoud Macalester College University of Nebraska Omaha, NE These slides are available at March 18, 2015 MAA

2 IDA-CCR Bill Mills Howard Rumsey David Robbins ( ) MAA Robbins Prize in algebra, combinatorics, or discrete math

3 Charles L. Dodgson aka Lewis Carroll Condensation of Determinants, Proceedings of the Royal Society, London 1866

4 Square matrix: Entries are 0, 1, 1 Row and column sums are +1 Artwork by Greg Kuperberg UC Davis Non-zero entries alternate in sign in each row

5 n A n

6 n A n How many n n alternating sign matrices? = = = = = =

7 There is exactly one 1 in the first row n 1 A n

8 There is exactly one 1 in the first row n 1 A n

9

10

11 ???? 0???? 0???? 0????

12 1 1 2/ /3 3 3/ / / / / / /2 429

13 1 1 2/ /3 3 3/ /4 14 5/5 14 4/ / / / / / / / / /2 429

14 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

15 Numerators:

16 Numerators: Conjecture 1: A n,k A n,k +1 = n 2 k 1 + n 1 k 1 n 2 n k 1 + n 1 n k 1

17 Conjecture 1: A n,k A n,k +1 = n 2 k 1 + n 1 k 1 n 2 n k 1 + n 1 n k 1 Conjecture 2 (corollary of Conjecture 1): A n = n 1 ( 3j + 1)! = ( n + j)! j =0 1! 4! 7! ( 3n 2 )! n! ( n + 1)! ( 2n 1)!

18 Richard Stanley

19 Richard Stanley George Andrews A n = Andrews Theorem: the number of descending plane partitions of size n is n 1 ( 3j + 1)! 1! 4! 7! = ( 3n 2 )! ( n + j)! n! ( n + 1)! ( 2n 1)! j =0

20 What is a descending plane partition?

21 Percy A. MacMahon Plane Partition Work begun in 1897

22 Plane partition of # of pp s of 75 = pp(75)

23 Plane partition of MacMahon finds a simple recursive algorithm for computing these numbers (based on generating function). # of pp s of 75 = pp(75) = 37,745,732,428,153

24 Symmetric Plane Partition MacMahon conjectures a simple recursive algorithm for computing these numbers (based on generating function).

25 1971 Basil Gordon proves case for n = infinity 1977 George Andrews and Ian Macdonald independently prove general case

26 Cyclically Symmetric Plane Partition

27 1979: Ian Macdonald conjectures the generating function for cyclically symmetric plane partitions, producing an efficient recursive algorithm for computing these numbers. 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.

28 1979, Andrews counts the number of cyclically symmetric plane partitions in an n n n box.

29 1979, Andrews counts the number of cyclically symmetric plane partitions in an n n n box.

30 1979, Andrews counts the number of cyclically symmetric plane partitions in an n n n box.

31 1979, Andrews counts the number of cyclically symmetric plane partitions in an n n n box.

32 1979, Andrews counts the number of cyclically symmetric plane partitions in an n n n box. L 1 = W 1 > L 2 = W 2 > L 3 = W 3 > width length

33 1979, Andrews counts descending plane partitions L 1 > W 1 L 2 > W 2 L 3 > W 3 width length

34 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. width length

35 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!

36 Totally Symmetric Self-Complementary Plane Partitions 1983 David Robbins Vertical flip of ASM = complement of DPP?

37 Totally Symmetric Self-Complementary Plane Partitions

38

39 Robbins Conjecture: The number of TSSCPP s in a 2n X 2n X 2n box is n 1 ( 3j + 1)! = ( n + j)! j =0 1! 4! 7! ( 3n 2 )! n! ( n + 1)! ( 2n 1)!

40 Robbins Conjecture: The number of TSSCPP s in a 2n X 2n X 2n box is n 1 ( 3j + 1)! = ( n + j)! j =0 1! 4! 7! ( 3n 2 )! n! ( n + 1)! ( 2n 1)! 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

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

42 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.

43 Zeilberger s proof is an 84-page tour de force, but it still left open the original conjecture: A n,k A n,k +1 = n 2 k 1 + n 1 k 1 n 2 n k 1 + n 1 n k 1

44 1996 Kuperberg announces a simple proof Another proof of the alternating sign matrix conjecture, International Mathematics Research Notices Greg Kuperberg UC Davis

45 1996 Kuperberg announces a simple proof Another proof of the alternating sign matrix conjecture, International Mathematics Research Notices Greg Kuperberg UC Davis Physicists had been studying ASM s for decades, only they called them the six-vertex model.

46 Horizontal = 1 Vertical = 1

47 1960 s Rodney Baxter s Triangle-totriangle relation 1980 s Anatoli Izergin Vladimir Korepin

48 1996 Doron Zeilberger uses this determinant to prove the original conjecture Proof of the refined alternating sign matrix conjecture, New York Journal of Mathematics

49 The End (which is really just the beginning) These slides can be downloaded from