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Proofs & Confirmations The story of the alternating sign matrix conjecture David M. Bressoud Macalester College Santa Clara University April 11, 2006 Bill Mills 1 IDA/CCR 0 0 –1 Howard Rumsey 1 0 –1 0 1 Dave Robbins Charles L. Dodgson 1 aka Lewis Carroll 0 0 –1 1 0 –1 “Condensation of Determinants,” 0 Proceedings of the Royal Society, London 1 1866 n An 1 1 1 2 2 0 3 7 0 4 42 –1 5 429 1 6 7436 0 –1 7 218348 0 8 10850216 1 9 911835460 n An How many n n alternating sign 1 1 matrices? 2 2 1 3 7 0 0 4 42 = 2 3 7 –1 5 429 = 3 11 13 1 6 7436 = 22 11 132 0 7 218348 = 22 132 17 19 –1 8 10850216 = 23 13 172 192 0 1 9 911835460 = 22 5 172 193 23 n An 1 1 2 2 1 3 7 0 0 4 42 = 2 3 7 –1 5 429 = 3 11 13 1 6 7436 = 22 11 132 0 7 218348 = 22 132 17 19 –1 8 10850216 = 23 13 172 192 0 1 9 911835460 = 22 5 172 193 23 There is exactly one 1 n An in the first row 1 1 1 2 2 0 3 7 0 4 42 –1 5 429 1 6 7436 0 –1 7 218348 0 8 10850216 1 9 911835460 There is exactly one 1 n An in the first row 1 1 1 2 1+1 0 3 2+3+2 0 4 7+14+14+7 –1 5 42+105+… 1 6 0 –1 7 0 8 1 9 1 1 1 1 0 2 3 2 0 –1 7 14 14 7 1 42 105 135 105 42 0 429 1287 2002 2002 1287 429 –1 0 1 1 1 1 1 0 2 3 2 0 –1 7 + 14 + 14 + 7 1 42 105 135 105 42 0 429 1287 2002 2002 1287 429 –1 0 1 1 1 1 1 0 2 3 2 0 –1 7 + 14 + 14 + 7 1 42 105 135 105 42 0 429 1287 2002 2002 1287 429 –1 0 1 1 1 1 2/2 1 0 2 2/3 3 3/2 2 0 –1 7 2/4 14 14 4/2 7 1 42 2/5 105 135 105 5/2 42 0 429 2/6 1287 2002 2002 1287 6/2 429 –1 0 1 1 1 1 2/2 1 0 2 2/3 3 3/2 2 0 –1 7 2/4 14 5/5 14 4/2 7 1 42 2/5 105 7/9 135 9/7 105 5/2 42 0 429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429 –1 0 1 2/2 1 0 2/3 3/2 0 2/4 5/5 4/2 –1 1 2/5 7/9 9/7 5/2 0 –1 2/6 9/14 16/16 14/9 6/2 0 1 Numerators: 1+1 1 0 1+1 1+2 0 –1 1+1 2+3 1+3 1 1+1 3+4 3+6 1+4 0 –1 1+1 4+5 6+10 4+10 1+5 0 1 Numerators: 1+1 1+1 1+2 1 0 1+1 2+3 1+3 0 1+1 3+4 3+6 1+4 –1 1+1 4+5 6+10 4+10 1+5 1 0 n 2 n 1 An, k k 1 k 1 –1 Conjecture 1: An, k 1 n 2 n 1 0 n k 1 n k 1 1 n 2 n 1 An, k k 1 k 1 Conjecture 1: 1 An, k 1 n 2 n 1 0 n k 1 n k 1 0 –1 Conjecture 2 (corollary of Conjecture 1): 1 0 n 1 An 3 j 1! 1!4!7!L 3n 2 ! j 0 n j ! n!n 1!L 2n 1! –1 0 1 1 0 0 –1 Richard Stanley 1 0 –1 0 1 1 0 0 –1 Richard Stanley 1 Andrews’ Theorem: the number 0 of descending plane partitions –1 of size n is 0 n 1 An 3 j 1! 1!4!7!L 3n 2 ! 1 j 0 n j ! n!n 1!L 2n 1! George Andrews All you have to do is find a 1-to-1 1 correspondence between n by n 0 alternating sign matrices and 0 descending plane partitions of size –1 n, and conjecture 2 will be proven! 1 0 –1 0 1 All you have to do is find a 1-to-1 1 correspondence between n by n 0 alternating sign matrices and 0 descending plane partitions of size –1 n, and conjecture 2 will be proven! 1 0 –1 What is a descending plane 0 partition? 1 Percy A. MacMahon 1 Plane Partition 0 0 –1 1 0 –1 0 Work begun in 1 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 1 pp j q j 1 q 3q 2 6q 3 13q 4 K 0 j 1 0 1 1 q –1 k k k 1 1 0 1 –1 1 q 1 q 1 q L 2 2 3 3 0 1 1912 MacMahon proves that the generating function for plane partitions in an n n n box is 1 1 qi j k 1 0 n 1 qi j k 2 1i, j,k 0 –1 At the same time, he conjectures that the 1 generating function for symmetric plane 0 partitions is –1 1 qi j k 1 1 q 2 i j k 1 0 1 n 1 qi j k 2 1i j n 1 q2i j k 2 1i j 1 k n 1 k n Symmetric Plane Partition 4 3 2 1 1 3 2 2 1 1 0 2 2 1 0 1 1 –1 1 1 “The reader must be warned that, although there 0 is little doubt that this result is correct, … the –1 result has not been rigorously established. … 0 Further investigations in regard to these matters 1 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 1977 George Andrews and Ian Macdonald –1 independently prove general case 1 0 –1 0 1 1912 MacMahon proves that the generating function for plane partitions in an n n n box is 1 1 qi j k 1 0 n 1 qi j k 2 1i, j,k 0 –1 At the same time, he conjectures that the 1 generating function for symmetric plane 0 partitions is –1 1 qi j k 1 1 q 2 i j k 1 0 1 n 1 qi j k 2 1i j n 1 q2i j k 2 1i j 1 k n 1 k n Macdonald’s observation: both generating functions are special cases of the following B r, s,t i, j, k 1 i r,1 j s,1 k t 1 ht i, j, k i j k 2 0 0 1 q 1 ht –1 generating function 1 B /G 1 q ht 0 where G is a group acting on the points in B and –1 B/G is the set of orbits. If G consists of only the 0 identity, this gives all plane partitions in B. If G is 1 the identity and (i,j,k) (j,i,k), then get generating function for symmetric plane partitions. Does this work for other groups of symmetries? 1 0 G = S3 ? 0 –1 1 0 –1 0 1 Does this work for other groups of symmetries? 1 0 G = S3 ? No 0 –1 1 0 –1 0 1 Does this work for other groups of symmetries? 1 0 G = S3 ? No 0 G = C3 ? (i,j,k) (j,k,i) (k,i,j) –1 1 It seems to work. 0 –1 0 1 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 1 ht 0 1 q 0 –1 1 q ht B /C3 1 0 “If I had to single out the most interesting open –1 problem in all of enumerative combinatorics, this 0 would be it.” Richard Stanley, review of 1 Symmetric Functions and Hall Polynomials, Bulletin of the AMS, March, 1981. 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 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 1979, Andrews counts cyclically symmetric plane partitions 1 0 L1 = W1 > L2 = W2 > L3 = W3 > … 0 –1 1 width 0 –1 0 1 length 1979, Andrews counts descending plane partitions 1 0 L1 > W1 ≥ L2 > W2 ≥ L3 > W3 ≥ … 0 –1 6 6 6 4 3 1 width 0 3 3 –1 0 2 1 length 6 X 6 ASM DPP with largest part ≤ 6 1 What are the corresponding 6 0 subsets of DPP’s? 0 –1 6 6 6 4 3 1 width 0 3 3 –1 0 2 1 length ASM with 1 at top of first column DPP with no parts of size n. 1 ASM with 1 at top of last column DPP with 0 n–1 parts of size n. 0 –1 6 6 6 4 3 1 width 0 3 3 –1 0 2 1 length Mills, Robbins, Rumsey Conjecture: # of n n ASM’s with 1 at top of column j equals # of 1 DPP’s ≤ n with exactly j–1 parts of size n. 0 0 –1 6 6 6 4 3 1 width 0 3 3 –1 0 2 1 length 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 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 Discovered an easier proof of Andrews’ 0 formula, using induction on j and n. 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 Discovered an easier proof of Andrews’ 0 formula, using induction on j and n. 0 Used this inductive argument to prove –1 Macdonald’s conjecture 1 0 “Proof of the Macdonald Conjecture,” Inv. Math., –1 1982 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 Discovered an easier proof of Andrews’ 0 formula, using induction on j and n. 0 Used this inductive argument to prove –1 Macdonald’s conjecture 1 0 “Proof of the Macdonald Conjecture,” Inv. Math., –1 1982 0 But they still didn’t have a 1 proof of their conjecture! Totally Symmetric Self-Complementary Plane Partitions 1 0 0 1983 –1 1 0 –1 David Robbins (1942–2003) 0 1 Vertical flip of ASM complement of DPP ? Totally Symmetric Self-Complementary Plane Partitions 1 0 0 –1 1 0 –1 0 1 1 0 0 –1 1 0 –1 0 1 Robbins’ Conjecture: The number of TSSCPP’s in a 2n X 2n X 2n box is 1 n 1 3 j 1! 1!4!7!L 3n 2 ! 0 n j ! n!n 1!L 2n 1! j0 0 –1 1 0 –1 0 1 Robbins’ Conjecture: The number of TSSCPP’s in a 2n X 2n X 2n box is 1 n 1 3 j 1! 1!4!7!L 3n 2 ! 0 n j ! n!n 1!L 2n 1! j0 0 –1 1989: William Doran shows equivalent to 1 counting lattice paths 0 1990: John Stembridge represents the counting –1 function as a Pfaffian (built on insights of 0 Gordon and Okada) 1 1992: George Andrews evaluates the Pfaffian, proves Robbins’ Conjecture December, 1992 Doron Zeilberger 1 announces a proof that 0 0 # of ASM’s of size n –1 equals of TSSCPP’s in 1 box of size 2n. 0 –1 0 1 December, 1992 Doron Zeilberger 1 announces a proof that 0 0 # of ASM’s of size n –1 equals of TSSCPP’s in 1 box of size 2n. 0 –1 0 1995 all gaps removed, published as “Proof of 1 the Alternating Sign Matrix Conjecture,” Elect. J. of Combinatorics, 1996. Zeilberger’s proof is an 84-page tour de force, but it still left open 1 the original conjecture: 0 0 n 2 n 1 –1 An, k k 1 k 1 1 An, k 1 n 2 n 1 n k 1 n k 1 0 –1 0 1 1996 Kuperberg announces a simple proof 1 “Another proof of the alternating 0 sign matrix conjecture,” 0 International Mathematics –1 Research Notices Greg Kuperberg 1 UC Davis 0 –1 0 1 1996 Kuperberg announces a simple proof 1 “Another proof of the alternating 0 sign matrix conjecture,” 0 International Mathematics –1 Research Notices Greg Kuperberg 1 UC Davis 0 –1 Physicists have been studying ASM’s for 0 decades, only they call them square ice 1 (aka the six-vertex model ). H O H O H O H O H O H H H H H H 1 H O H O H O H O H O H 0 H H H H H 0 –1 H O H O H O H O H O H 1 H H H H H 0 –1 H O H O H O H O H O H 0 H H H H H 1 H O H O H O H O H O H 1 0 0 –1 1 0 –1 0 1 Horizontal 1 1 0 0 –1 1 0 –1 0 Vertical –1 1 southwest 1 0 northwest 0 –1 1 southeast 0 –1 0 northeast 1 1 0 N = # of vertical 0 I = inversion x2, y3 –1 number 1 = N + # of NE 0 –1 0 1 1960’s Rodney Baxter’s 1 Triangle-to- 0 triangle relation 0 –1 1 0 –1 0 1 1960’s Rodney Baxter’s 1 Triangle-to- 0 triangle relation 0 –1 1980’s 1 0 –1 0 1 Anatoli Izergin Vladimir Korepin det 1 i, j 1 xi y j axi y j n xi y j axi y j 1i j n xi x j yi y j 1 1 a 2 N A a n(n 1)/2 InvA AA n 0 xi y j ax i yj x y i j 0 vert SW, NE NW, SE –1 1 0 –1 0 1 det 1 i, j 1 xi y j axi y j n xi y j axi y j 1i j n xi x j yi y j 1 1 a 2 N A a n(n 1)/2 InvA AA n 0 xi y j ax i yj x y i j 0 vert SW, NE NW, SE –1 a z 4 , xi z 2 , yi 1 1 0 RHS z z 1 n n 1 z z 1 2 N A AA n –1 z e i / 3 0 n n 1/2 1 RHS 3 An 1996 Doron Zeilberger 1 uses this 0 determinant to 0 prove the original –1 conjecture 1 0 –1 0 “Proof of the refined alternating sign matrix 1 conjecture,” New York Journal of Mathematics 1 0 The End 0 –1 (which is really just the beginning) 1 0 –1 0 1 1 0 The End 0 –1 (which is really just the beginning) 1 0 –1 This Power Point presentation can be 0 downloaded from 1 www.macalester.edu/~bressoud/talks