Proofs and Confirmations: the Story of the Alternating Sign Matrix Conjecture

Full text

Proofs & Confirmations!
The story of the !
alternating sign matrix conjecture!
David M. Bressoud
Macalester College
Arizona State University
Tempe, AZ
March 19, 2009!
These slides are available at !
www.macalester.edu/~bressoud/talks!

Artwork by Greg Kuperberg!
0 0 1 0 0
0 1 −1 0 1
1 −1 0 1 0
0 1 0 0 0
0 0 1 0 0
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
Square matrix:!
• Entries are 0, 1, –1!
• Row and column sums
are +1!
• Non-zero entries
alternate in sign in each
row!

Bill Mills!
Howard Rumsey!
IDA-CCR!
David Robbins
(1942–2003)!
MAA Robbins Prize in
algebra, combinatorics, or
discrete math!

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

Desnanot-Jacobi adjoint matrix theorem (Desnanot for
n ≤ 6 in 1819, Jacobi for general case in 1833!
M
j
i
is matrix M with row i and column j removed.!
det M =
det M
1
1
⋅ det M
n
n
− det M
n
1
⋅ det M
1
n
det M
1,n
1,n
Given that the determinant of the empty
matrix is 1 and the determinant of a 1×1
is the entry in that matrix, this uniquely
defines the determinant for all square
matrices.!
Carl Jacobi (1804–1851)!

det M =
det M
1
1
⋅ det M
n
n
− det M
n
1
⋅ det M
1
n
det M
1,n
1,n
det
λ
M =
det
λ
M
1
1
⋅ det
λ
M
n
n
+
λ
det
λ
M
n
1
⋅ det
λ
M
1
n
det
λ
M
1,n
1,n
det
−1
M = det M
( )
det
λ
a
j
i −1
( )
i, j =1
n
= a
i
+
λ
a
j
( )
1≤i < j ≤n
∏

det
λ
a b
c d
⎛
⎝
⎜
⎞
⎠
⎟
= ad +
λ
bc
det
λ
a b c
d e f
g h j
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
= aej +
λ
bdj + afh
( )
+
λ
2
bfg + cdh
( )
+
λ
3
ceg
+
λ
1 +
λ
( )
bde
−1
fh
det M =
det M
1
1
⋅ det M
n
n
− det M
n
1
⋅ det M
1
n
det M
1,n
1,n
det
λ
M =
det
λ
M
1
1
⋅ det
λ
M
n
n
+
λ
det
λ
M
n
1
⋅ det
λ
M
1
n
det
λ
M
1,n
1,n

det
λ
a
1,1
a
1,2
a
1,3
a
1,4
a
2,1
a
2,2
a
2,3
a
2,4
a
3,1
a
3,2
a
3,3
a
3,4
a
4,1
a
4,2
a
4,3
a
4,4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
= a
1,1
a
2,2
a
3,3
a
4,4
+
λ
a
1,2
a
2,1
a
3,3
a
4,4
+ a
1,1
a
2,3
a
3,2
a
4,4
+ a
1,1
a
2,2
a
3,4
a
4,3
( )
+  sums over other permutations ×
λ
inversion number
+
λ
3
1 +
λ
−1
( )
a
1,2
a
2,1
a
2,2
−1
a
2,3
a
3,4
a
4,2
+ 
+
λ
3
1 +
λ
−1
( )
2
a
1,2
a
2,1
a
2,2
−1
a
2,3
a
3,2
a
3,3
−1
a
3,4
a
4,3
+ 

det
λ
a
1,1
a
1,2
a
1,3
a
1,4
a
2,1
a
2,2
a
2,3
a
2,4
a
3,1
a
3,2
a
3,3
a
3,4
a
4,1
a
4,2
a
4,3
a
4,4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
= a
1,1
a
2,2
a
3,3
a
4,4
+
λ
a
1,2
a
2,1
a
3,3
a
4,4
+ a
1,1
a
2,3
a
3,2
a
4,4
+ a
1,1
a
2,2
a
3,4
a
4,3
( )
+  sums over other permutations ×
λ
inversion number
+
λ
3
1 +
λ
−1
( )
a
1,2
a
2,1
a
2,2
−1
a
2,3
a
3,4
a
4,2
+ 
+
λ
3
1 +
λ
−1
( )
2
a
1,2
a
2,1
a
2,2
−1
a
2,3
a
3,2
a
3,3
−1
a
3,4
a
4,3
+ 
0 1 0 0
1 −1 1 0
0 1 −1 1
0 0 1 0
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
0 1 0 0
1 −1 1 0
0 0 0 1
0 1 0 0
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟

det
λ
a
1,1
a
1,2
a
1,3
a
1,4
a
2,1
a
2,2
a
2,3
a
2,4
a
3,1
a
3,2
a
3,3
a
3,4
a
4,1
a
4,2
a
4,3
a
4,4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
= a
1,1
a
2,2
a
3,3
a
4,4
+
λ
a
1,2
a
2,1
a
3,3
a
4,4
+ a
1,1
a
2,3
a
3,2
a
4,4
+ a
1,1
a
2,2
a
3,4
a
4,3
( )
+  sums over other permutations ×
λ
inversion number
+
λ
3
1 +
λ
−1
( )
a
1,2
a
2,1
a
2,2
−1
a
2,3
a
3,4
a
4,2
+ 
+
λ
3
1 +
λ
−1
( )
2
a
1,2
a
2,1
a
2,2
−1
a
2,3
a
3,2
a
3,3
−1
a
3,4
a
4,3
+ 
det
λ
x
i, j
( )
=
λ
Inv A
( )
A= a
i , j
( )
∑
1 +
λ
−1
( )
N A
( )
x
i, j
a
i , j
i, j
∏
Sum is over all alternating sign matrices, N(A) = # of –1’s!

n!
1!
2!
3!
4!
5!
6!
7!
8!
9!
A
n!
1!
2!
7!
42!
429!
7436!
218348!
10850216!
911835460!
0 0 1 0 0
0 1 −1 0 1
1 −1 0 1 0
0 1 0 0 0
0 0 1 0 0
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟

n!
1!
2!
3!
4!
5!
6!
7!
8!
9!
A
n!
1!
2!
7!
42 !
429!
7436!
218348!
10850216!
911835460!
= 3 × 11 × 13!
= 2
2
× 11 × 13
2!
= 2
2
× 13
2
× 17 × 19!
= 2
3
× 13 × 17
2
× 19
2!
= 2
2
× 5 × 17
2
× 19
3
× 23!
= 2 × 3 × 7!
How many n × n alternating sign
matrices?!

n!
1!
2!
3!
4!
5!
6!
7!
8!
9!
A
n!
1!
2!
7!
42!
429!
7436!
218348!
10850216!
911835460!
0 0 1 0 0
0 1 −1 0 1
1 −1 0 1 0
0 1 0 0 0
0 0 1 0 0
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
There is exactly one 1
in the first row!

n!
1!
2!
3!
4!
5!
6!
7!
8!
9!
A
n!
1!
1+1!
2+3+2!
7+14+14+7!
42+105+…!
0 0 1 0 0
0 1 −1 0 1
1 −1 0 1 0
0 1 0 0 0
0 0 1 0 0
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
There is exactly one 1
in the first row!

1!
1 1!
2 3 2!
7 14 14 7!
42 105 135 105 42!
429 1287 2002 2002 1287 429!

1!
1 1!
2 3 2!
7 14 14 7!
42 105 135 105 42!
429 1287 2002 2002 1287 429!
+! +! +!

1!
1 1!
2 3 2!
7 14 14 7!
42 105 135 105 42!
429 1287 2002 2002 1287 429!
+! +! +!

1!
1 2/2 1!
2 2/3 3 3/2 2!
7 2/4 14 14 4/2 7!
42 2/5 105 135 105 5/2 42!
429 2/6 1287 2002 2002 1287 6/2 429!

1!
1 2/2 1!
2 2/3 3 3/2 2!
7 2/4 14 5/5 14 4/2 7!
42 2/5 105 7/9 135 9/7 105 5/2 42!
429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429!

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!

1+1 !
1+1 1+2 !
1+1 2+3 1+3 !
1+1 3+4 3+6 1+4 !
1+1 4+5 6+10 4+10 1+5 !
Numerators:!

1+1 !
1+1 1+2 !
1+1 2+3 1+3 !
1+1 3+4 3+6 1+4 !
1+1 4+5 6+10 4+10 1+5 !
Conjecture 1:!
Numerators:!
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 1:!
Conjecture 2 (corollary of 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
⎛
⎝
⎜
⎞
⎠
⎟
A
n
=
3 j + 1
( )
!
n + j
( )
!
j = 0
n−1
∏
=
1!⋅4!⋅7! 3n − 2
( )
!
n!⋅ n + 1
( )
! 2n − 1
( )
!

Richard Stanley!

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

What is a
descending
plane partition?!

Percy A. MacMahon!
Plane Partition
Work begun in!
1897!

6 5 5 4 3 3!
Plane partition of 75
# of pp’s of 75 = pp(75)

6 5 5 4 3 3!
Plane partition of 75
# of pp’s of 75 = pp(75) = 37,745,732,428,153

Generating function:!
1 + pp j
( )
j =1
∞
∑
q
j
= 1 + q + 3q
2
+ 6q
3
+ 13q
4
+ …
=
1
1 − q
k
( )
k
k =1
∞
∏
=
1
1 − q
( )
1 − q
2
( )
2
1 − q
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
1 − q
i + j + k − 2
1≤i, j, k ≤n
∏

Symmetric Plane Partition
4 3 2 1 1
3 2 2 1
2 2 1
1 1
1

At the same time, he conjectures that the
generating function for symmetric plane
partitions is
1 − q
i + j + k −1
1 − q
i + j + k − 2
1≤i = j ≤n
1≤k ≤n
∏
1 − q
2 i+ j + k −1
( )
1 − q
2 i+ j + k − 2
( )
1≤i < j ≤n
1≤k ≤n
∏
“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

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

Cyclically Symmetric Plane Partition

Macdonald’s Conjecture (1979): The
generating function for cyclically
symmetric plane partitions in B(n,n,n) is!
“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.!
1 − q
η
1+ ht
η
( )
( )
1 − q
η
ht
η
( )
η
∈B /C
3
∏

1979, Andrews counts cyclically symmetric
plane partitions

1979, Andrews counts cyclically symmetric
plane partitions

1979, Andrews counts cyclically symmetric
plane partitions

1979, Andrews counts cyclically symmetric
plane partitions

1979, Andrews counts cyclically symmetric
plane partitions
length!
width!
L
1
= W
1
> L
2
= W
2
> L
3
= W
3
> …!

1979, Andrews counts descending plane
partitions
length!
width!
L
1
> W
1
≥ L
2
> W
2
≥ L
3
> W
3
≥ …!
6 6 6 4 3!
3 3!
2!

length!
width!
6 6 6 4 3!
3 3!
2!
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.!

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!!

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

Totally Symmetric Self-Complementary
Plane Partitions!

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

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

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

December, 1992!
Doron Zeilberger
announces a proof that
# of ASM’s of size n
equals of TSSCPP’s in
box of size 2n.!
1995 all gaps removed, published as “Proof of
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
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
⎛
⎝
⎜
⎞
⎠
⎟

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

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.!

Horizontal = 1!
Vertical = –1!
0 0 1 0 0
0 1 −1 0 1
1 −1 0 1 0
0 1 0 0 0
0 0 1 0 0
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟

1960’s
Anatoli Izergin
Vladimir Korepin
1980’s
Rodney Baxter’s
Triangle-to-
triangle relation!

det
1
x
i
− y
j
( )
ax
i
− y
j
( )
⎛
⎝
⎜
⎞
⎠
⎟
x
i
− y
j
( )
ax
i
− y
j
( )
i, j =1
n
∏
x
i
− x
j
( )
y
i
− y
j
( )
1≤i < j ≤n
∏
= 1 − a
( )
2 N A
( )
a
n(n−1)/2 −Inv A
( )
A∈
A
n
∑
× x
i
vert
∏
y
j
ax
i
− y
j
( )
SW, NE
∏
x
i
− y
j
( )
NW, SE
∏

a = z
−4
, x
i
= z
2
, y
i
= 1
RHS = z − z
−1
( )
n n −1
( )
z + z
−1
( )
2 N A
( )
A∈
A
n
∑
z = e
π
i /3
RHS = −3
( )
n n −1
( )
/2
A
n
det
1
x
i
− y
j
( )
ax
i
− y
j
( )
⎛
⎝
⎜
⎞
⎠
⎟
x
i
− y
j
( )
ax
i
− y
j
( )
i, j =1
n
∏
x
i
− x
j
( )
y
i
− y
j
( )
1≤i < j ≤n
∏
= 1 − a
( )
2 N A
( )
a
n(n−1)/2 −Inv A
( )
A∈
A
n
∑
× x
i
vert
∏
y
j
ax
i
− y
j
( )
SW, NE
∏
x
i
− y
j
( )
NW, SE
∏

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

!e End"
(which is really just the beginning)!
These slides can be downloaded from
www.macalester.edu/~bressoud/talks!