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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
                    1i, 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
         1i  j
                                       n 1  q2i  j  k  2
                                     1i  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
                    1i, 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
         1i  j
                                       n 1  q2i  j  k  2
                                     1i  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!
       j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!
         j0
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   1i  j n xi  x j yi  y j      
1
           1  a 2 N A a n(n 1)/2 InvA
           AA 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   1i  j n xi  x j yi  y j                   
1
           1  a 2 N A a n(n 1)/2 InvA
           AA 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 

                                                       AA 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

								
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