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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