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
Howard Rumsey
–1
1
0
–1
0
1
Dave Robbins
Charles L. Dodgson
1
aka Lewis Carroll
0
0
–1
1
0
–1
0
1
“Condensation of Determinants,”
Proceedings of the Royal Society, London
1866
n
An
1
1
1
2
2
0
3
7
0
4
42
5
429
6
7436
–1
7
218348
0
8
10850216
1
9
911835460
–1
1
0
1
0
n
An
1
1
2
2
3
7
How many n  n alternating sign
matrices?
0
4
42 = 2  3  7
–1
5
429 = 3  11  13
1
6
7436 = 22  11  132
0
7
–1
0
1
8
9
218348 = 22  132  17  19
10850216 = 23  13  172  192
911835460 = 22  5  172  193  23
1
0
n
An
1
1
2
2
3
7
0
4
42 = 2  3  7
–1
5
429 = 3  11  13
1
6
7436 = 22  11  132
0
7
–1
0
1
8
9
218348 = 22  132  17  19
10850216 = 23  13  172  192
911835460 = 22  5  172  193  23
n
An
1
1
1
2
2
0
3
7
0
4
42
5
429
6
7436
–1
7
218348
0
8
10850216
1
9
911835460
There is exactly one 1
in the first row
–1
1
0
n
An
1
1
1
2
1+1
0
3
2+3+2
0
4 7+14+14+7
There is exactly one 1
in the first row
–1
1
0
5
6
–1
7
0
8
1
9
42+105+…
1
1
1
0
2
0
7
–1
42
1
0
–1
0
1
429
1287
1
3
14
105
2002
2
14
135
7
105
2002
42
1287
429
1
1
1
0
2
0
42
1
–1
0
1
3
2
7 + 14 + 14 + 7
–1
0
1
429
1287
105
2002
135
105
2002
42
1287
429
1
1
1
0
2
0
42
1
–1
0
1
3
2
7 + 14 + 14 + 7
–1
0
1
429
1287
105
2002
135
105
2002
42
1287
429
1
1
1 2/2 1
0
2 2/3 3 3/2 2
0
7 2/4 14
–1
1
0
–1
0
1
42 2/5 105
429 2/6 1287
2002
14 4/2 7
135
105 5/2 42
2002
1287 6/2 429
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
–1
0
1
429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429
2/2
1
2/3
0
2/4
0
–1
2/5
1
0
–1
0
1
2/6
9/14
3/2
5/5
7/9
4/2
9/7
16/16
5/2
14/9
6/2
Numerators:
1+1
1
1+1
0
0
–1
1
0
–1
0
1
1+2
1+1 2+3
1+1
1+3
3+4 3+6 1+4
1+1 4+5 6+10 4+10 1+5
1+1
Numerators:
1+1
1
1+1
0
0
–1
1
1+1
0
1
2+3
3+4
1+3
3+6
1+4
1+1 4+5 6+10 4+10
0
–1
1+2
Conjecture 1:
An, k
An,k 1
1+5
 n  2   n  1
 k  1    k  1

 n  2   n 1 
 n  k  1   n  k  1
1
Conjecture 1:
0
An, k
An,k 1
 n  2   n  1
 k  1    k  1

 n  2   n 1 
 n  k  1   n  k  1
0
–1
Conjecture 2 (corollary of Conjecture 1):
1
0
–1
0
1
3 j  1! 1!4!7!L 3n  2 !

An  

n!n  1!L 2n  1!
j 0 n  j !
n1
1
0
0
–1
1
0
–1
0
1
Richard Stanley
1
0
0
Richard Stanley
–1
1
0
–1
0
1
Andrews’ Theorem: the number
of descending plane partitions
of size n is
n1
3 j  1! 1!4!7!L 3n  2 !

An  

n!n  1!L 2n  1!
j 0 n  j !
George Andrews
1
0
0
–1
1
0
–1
0
1
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
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
–1
0
1
What is a descending plane
partition?
Percy A. MacMahon
Plane Partition
1
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
0
0
–1

1   pp  j q j  1  q  3q 2  6q 3  13q 4  K
j 1


k 1
1
0
–1
0
1

1
1  q 
k k
1  q 1  q
1
 1  q  L
2 2
3 3
1912 MacMahon proves that the generating
function for plane partitions in an n  n  n
box is
1  q i  j  k 1

i jk2
1i, j,k n 1  q
1
0
0
–1
1
0
–1
0
1
At the same time, he conjectures that the
generating function for symmetric plane
partitions is
i  j  k 1
1 q

i jk2
1

q
1i  j n
1 k n
2 i  j  k 1
1 q

2 i  j  k  2 
1

q
1i  j n
1 k n
Symmetric Plane Partition
4 3 2 1 1
1
3 2 2 1
0
2 2 1
0
1 1
–1
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)
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  q i  j  k 1

i jk2
1i, j,k n 1  q
1
0
0
–1
1
0
–1
0
1
At the same time, he conjectures that the
generating function for symmetric plane
partitions is
i  j  k 1
1 q

i jk2
1

q
1i  j n
1 k n
2 i  j  k 1
1 q

2 i  j  k  2 
1

q
1i  j 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
0
0
–1
1
0
–1
0
1
ht i, j, k   i  j  k  2
1  q   
generating function  
 ht  
1

q
 B /G
 1 ht 
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 get generating
function for symmetric plane partitions.
Does this work for other groups of
symmetries?
1
0
0
–1
1
0
–1
0
1
G = S3 ?
Does this work for other groups of
symmetries?
1
0
0
–1
1
0
–1
0
1
G = S3 ? No
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.
Cyclically Symmetric Plane Partition
1
0
0
–1
1
0
–1
0
1
1
0
0
–1
Macdonald’s Conjecture (1979): The
generating function for cyclically
symmetric plane partitions in B(n,n,n) is
 1 ht  
1 q

 ht  
 B /C3 1  q
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.
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
L 1 = W1 > L 2 = W2 > L 3 = W3 > …
0
–1
1
width
0
–1
0
1
length
1979, Andrews counts descending plane
partitions
1
0
L 1 > W1 ≥ L 2 > W2 ≥ L 3 > W3 ≥ …
0
–1
1
6 6 6 4 3
width
3 3
0
–1
2
0
1
length
6 X 6 ASM  DPP with largest part ≤ 6
What are the corresponding 6
subsets of DPP’s?
1
0
0
–1
1
6 6 6 4 3
width
3 3
0
–1
2
0
1
length
ASM with 1 at top of first column DPP with
no parts of size n.
1
0
ASM with 1 at top of last column DPP with
n–1 parts of size n.
0
–1
1
6 6 6 4 3
width
3 3
0
–1
2
0
1
length
1
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.
0
0
–1
1
6 6 6 4 3
width
3 3
0
–1
2
0
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
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.
Discovered an easier proof of Andrews’
formula, using induction on j and n.
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.
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
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.
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!
Totally Symmetric Self-Complementary
Plane Partitions
1
0
0
1983
–1
1
0
–1
0
1
David Robbins
(1942–2003)
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
0
0
–1
1
0
–1
0
1
3 j  1!  1!4!7!L 3n  2 !

n!n  1!L 2n  1!
j 0 n  j !
n1
Robbins’ Conjecture: The number of
TSSCPP’s in a 2n X 2n X 2n box is
1
0
0
–1
1
0
–1
0
1
3 j  1!  1!4!7!L 3n  2 !

n!n  1!L 2n  1!
j 0 n  j !
n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
December, 1992
1
0
0
–1
1
0
–1
0
1
Doron Zeilberger
announces a proof that
# of ASM’s of size n
equals of TSSCPP’s in
box of size 2n.
December, 1992
1
0
0
–1
1
Doron Zeilberger
announces a proof that
# of ASM’s of size n
equals of TSSCPP’s in
box of size 2n.
0
–1
0
1
1995 all gaps removed, published as “Proof of
the Alternating Sign Matrix Conjecture,” Elect. J.
of Combinatorics, 1996.
1
0
Zeilberger’s proof is an 84-page
tour de force, but it still left open
the original conjecture:
0
–1
1
0
–1
0
1
An, k
An,k 1
 n  2   n  1
 k  1    k  1

 n  2   n 1 
 n  k  1   n  k  1
1996 Kuperberg
announces a simple proof
1
0
0
–1
1
0
–1
0
1
“Another proof of the alternating
sign matrix conjecture,”
International Mathematics
Research Notices
Greg Kuperberg
UC Davis
1996 Kuperberg
announces a simple proof
1
0
0
–1
1
0
–1
0
1
“Another proof of the alternating
sign matrix conjecture,”
International Mathematics
Research Notices
Greg Kuperberg
UC Davis
Physicists have been studying ASM’s for
decades, only they call them square ice
(aka the six-vertex model ).
H O H O H O H O H O H
H
1
0
0
–1
1
0
–1
0
1
H
H
H
H
H O H O H O H O H O H
H
H
H
H
H
H O H O H O H O H O H
H
H
H
H
H
H O H O H O H O H O H
H
H
H
H
H
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
1
Vertical  –1
southwest
1
0
northwest
0
–1
1
southeast
0
–1
0
1
northeast
1
0
0
–1
1
0
–1
0
1
N = # of vertical
I = inversion
number
= N + # of NE
x2, y3
1960’s
Rodney Baxter’s
1
0
0
–1
1
0
–1
0
1
Triangle-totriangle relation
1960’s
Rodney Baxter’s
Triangle-totriangle relation
1
0
0
–1
1980’s
1
0
–1
0
1
Anatoli Izergin
Vladimir Korepin






  i, j 1 xi  y j axi  y j
1
det 

 xi  y j axi  y j   1i  j n xi  x j yi  y j
n

1
0
0
–1
1
0
–1
0
1




2 N A  n(n1)/2 Inv A 
1

a
a



AA n
  xi y j
vert
 ax
i
SW, NE
 yj
  x  y 
i
NW, SE
j






  i, j 1 xi  y j axi  y j
1
det 

 xi  y j axi  y j   1i  j n xi  x j yi  y j
n

1
0
0
–1
1
0
–1
0
1




2 N A  n(n1)/2 Inv A 
1

a
a



AA n
  xi y j
vert
 ax
i
 yj
  x  y 
SW, NE
i
j
NW, SE
a  z 4 , xi  z 2 , yi  1

RHS  z  z

 z  z 
1 n n 1
1 2 N A 
AA n
z  e i / 3
n n 1/2
RHS  3
An
1996
1
0
0
–1
Doron Zeilberger
uses this
determinant to
prove the original
conjecture
1
0
–1
0
1
“Proof of the refined alternating sign matrix
conjecture,” New York Journal of Mathematics
1
0
The End
0
–1
1
0
–1
0
1
(which is really just the beginning)
1
0
The End
0
–1
1
(which is really just the beginning)
0
–1
0
1
This Power Point presentation can be
downloaded from
www.macalester.edu/~bressoud/talks