This paper highlights three known identities, each of which involves sums
over alternating sign matrices. While proofs of all three are known, the only
known derivations are as corollaries of difficult results. The simplicity and
natural combinatorial interpretation of these identities, however, suggest that
there should be direct, bijective proofs

Bressoud, David M.

English

arXiv

AbstractThis paper highlights three known identities, each of which involves sums over alternating sign matrices. While proofs of all three are known, the only known derivations are as corollaries of difficult results. The simplicity and natural combinatorial interpretation of these identities, however, suggest that there should be direct, bijective proofs

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Three Alternating Sign Matrix Identities in Search of Bijective Proofs 

Advances in Applied Mathematics 27, 289–297 (2001)
doi:10.1006/aama.2001.0735, available online at http://www.idealibrary.com on
Three Alternating Sign Matrix Identities in Search of
Bijective Proofs
David M. Bressoud
Department of Mathematics, Macalester College, St. Paul, Minnesota 55105
Received July 18, 2000; accepted February 7, 2001
in honor of the master of bijective proofs, dominique foata
This paper highlights three known identities, each of which involves sums over
alternating sign matrices. While proofs of all three are known, the only known
derivations are as corollaries of difﬁcult results. The simplicity and natural com-
binatorial interpretation of these identities, however, suggest that there should be
direct, bijective proofs.  2001 Elsevier Science
1. INTRODUCTION
Alternating sign matrices (ASMs) are square matrices of 0’s, 1’s, and
−1’s with row and column-sums equal to 1 and with the restriction that the
non-zero entries alternate signs across each row and down each column.
An example is


0 1 0 0 0
1 −1 0 1 0
0 1 0 −1 1
0 0 0 1 0
0 0 1 0 0

 
These are rich combinatorial objects with connections to many problems in
algebraic combinatorics (see [2, 3, 13]). They also have many different rep-
resentations. The representation that was used in Kuperberg’s proof of the
counting function for alternating sign matrices [10] and Zeilberger’s proof
of the reﬁned alternating sign matrix conjecture [15] is the six-vertex model
of statistical mechanics. These are directed graphs in which each vertex has
in-degree two and out-degree two, and boundary conditions that the vertical
289
0196-8858/01 $35.00
 2001 Elsevier Science
All rights reserved.
290 david m. bressoud
arrows along the top and bottom are directed out, horizontal arrows along
the left and right are directed in, as in the following directed graph.
To actually make this a directed graph on 25 vertices, we can identify
the ith up-arrow along the top row with the ith right arrow along the left
edge, and similarly identify bottom and right arrows. This is called a six-
vertex model because there are six possible conﬁgurations at each vertex.
We shall describe a vertex as horizontal if both in-edges are horizontal,
vertical if both in-edges are vertical, and otherwise southwest, northwest,
northeast, or southeast, according to the direction of the sum of the four
vectors represented by the four adjacent edges.
It should be noted that the sum of all vertical vectors is zero, as is the sum
of all horizontal vectors. It follows that there will always be an equal number
of southwest and northeast vertices, and an equal number of southeast and
northwest vertices.
Our example of a six-vertex model corresponds to our example of an
alternating sign matrix. Each 1 in the ASM corresponds to a horizontal
vertex, each −1 to a vertical vertex, and the 0’s to the other vertices. This is
a bijection because once the positions of the horizontal and vertical vertices
are known, all other vertices are uniquely determined.
The six-vertex model is not the only insightful representation, but it
is very suggestive, especially because there is also a natural connection
between ASMs and complete directed graphs or tournaments. In response
to a preprint of this article, Chapman [4] has found such a direct bijec-
tive connection between ASMs and tournaments. We shall also present two
other related identities that cry out for bijective proofs.
2. THE λ-DETERMINANT
The ﬁrst two identities that I wish to present arise from the λ-
determinant of Robbins and Rumsey [14]. This is based on the Desnanot–
three asm identities 291
Jacobi adjoint matrix theorem [5, 8] that was used by Dodgson [6] to create
his algorithm for evaluating determinants. Given a square matrix M , we
let Mij denote M with row i and column j deleted. We then have that
detM = detM
1
1 · detMnn − detMn1 · detM1n
detM1n1n
 (1)
If we deﬁne the determinant of an empty matrix (0× 0) to be 1 and the
determinant of the 1× 1 matrix a to be a, then Eq. (1) can be used as a
recursive deﬁnition of the determinant. A natural one-parameter general-
ization of the determinant arises if we use the same initial conditions and
replace the minus sign in the numerator of the recursive step by +λ:
detλM =
detλ
(
M11
)
detλ
(
Mnn
)+ λ detλ(Mn1 )detλ(M1n)
detλ
(
M
1n
1n
)  (2)
The following generalization of the Vandermonde determinant evaluation
follows by induction.
Proposition 1.
detλ
(
xn−ij
) = ∏
1≤i<j≤n
xi + λxj (3)
If we expand a few λ-determinants, an interesting pattern emerges:
detλ

 a b cd e f
g h i

 = aei+ λ bdi+ afh + λ2 bfg + cdh
+λ3 ceg + λ1+ λbde−1fh
detλ


a b c d
e f g h
i j k l
m n o p

 = · · · + λ31+ λbef−1hkn
+λ31+ λ2cfg−1hij−1kn+ · · · 
The monomials in roman letters that correspond to permutation matrices
are each multiplied by λ raised to the inversion number of the permutation.
The other monomials in roman letters that appear, such as cfg−1hij−1kn,
correspond to alternating sign matrices, in this case


0 0 1 0
0 1 −1 1
1 −1 1 0
0 1 0 0

 
292 david m. bressoud
Each of these monomials is multiplied by a power of λ and a power of
1+ λ.
Let n be the set of n × n ASMs. Given A = aij ∈ n, we deﬁne its
inversion number,  A, to be
 A = ∑
i<k j>l
aij · akl
We deﬁne NA to be the number of −1’s in A. The following charac-
terization of the λ-determinant was published by Robbins and Rumsey in
1986 [14].
Proposition 2.
detλmij =
∑
A∈n
λ A−NA1+ λNA
n∏
ij=1
m
aij
ij  (4)
Zeilberger [16] has given a bijective proof of Eq. (1). It would be desir-
able to have a direct proof of Proposition 2 by ﬁnding a similar proof of
Eq. (2) when the λ-determinant is deﬁned by the right side of Proposition 2.
Problem 1. Find a direct, bijective proof of the following identity.
Within each summation, the range of indices for the alternating sign
matrices B and C is speciﬁed by the product term.
∑
BC∈n×n−2
λ B+ C−NB−NC1+ λNB+NC
n∏
ij=1
m
bij
ij
n−1∏
ij=2
m
cij
ij
= ∑
BC∈n−1×n−1
λ B+ C−NB−NC1+ λNB+NC
× ∏
1≤i≤n−1
1≤j≤n−1
m
bij
ij
∏
2≤i≤n
2≤j≤n
m
cij
ij
+λ ∑
BC∈n−1×n−1
λ B+ C−NB−NC1+ λNB+NC
× ∏
1≤i≤n−1
1≤j≤n−1
m
bij
ij
∏
2≤i≤n
2≤j≤n
m
cij
ij 
3. DIRECTED GRAPHS
If we combine Propositions 1 and 2, we get that
∏
1≤i<j≤n
xi + λxj =
∑
A∈n
λ A−NA1+ λNA
n∏
ij=1
x
n−iaij
j  (5)
three asm identities 293
It is worth noting that analogs of this identity for other root systems have
been found by Okada [12].
The left side of Eq. (5) can be interpreted as a sum over the set of
tournaments on n vertices, n. Each binomial xi + λxj corresponds to the
edge between vertices i and j. If the edge is directed from i to j, we choose
xi. If it is directed from j to i, we choose λxj . Each tournament corresponds
to a monomial in which the power of xi is ωi, the out-degree of vertex i,
and the power of λ is UT , the number of upsets in the tournament: j > i
and j → i,
∏
1≤i<j≤n
xi + λxj =
∑
T∈n
λUT 
n∏
i=1
x
ωi
i 
We shall use the six-vertex model to interpret the right side of Eq. (5).
We begin with the following observations which are explained below.
Proposition 3. Let A be an n× n ASM. In the corresponding six-vertex
model
• The number of horizontal vertices is n+NA,
• The number of vertical vertices is NA,
• The number of southwest vertices equals the number of northeast ver-
tices equals  A −NA,
• The number of southeast vertices equals the number of northwest ver-
tices equals
( n
2
)−  A.
The number of vertical vertices is immediate from the bijection, and there
must be one more 1 than −1 in each row. A southwest vertex corresponds
to a 0 of the ASM for which there is a 1 above it in its column (due north)
with no other non-zero entries in between, and a 1 to its left in its row (due
west) with no other non-zero entries in between.
1
0

0
1 0 · · · 0 0SW or − 1
The inversion number is the number of such pairs of 1’s: pairs of 1’s for
which there are only 0’s in the positions that are both due east of the lower
1 and strictly south and west of the upper 1, and there are only 0’s in the
positions that are both due south of the upper 1 and strictly north and east
of the lower 1. The entry in the unique position due east of the lower 1 and
due south of the upper 1 must be either a 0, corresponding to a southwest
294 david m. bressoud
vertex, or a −1. The remaining observations follow from the equality of the
number of southwest and northeast vertices, the equality of the number
of southeast and northwest vertices, and the fact that there are n2 vertices
in all.
If we let SW A, SEA, and V A denote, respectively, the number of
southwest, southeast, and vertical vertices in A and SWiA, SEiA, and
ViA the number of southwest, southeast, or vertical vertices, respectively,
in column i of A, then the right side of Eq. (5) can be written as
∑
A∈n
λSW A1+ λV A
n∏
i=1
x
SWiA+SEiA+ViA
i 
Equation (5) is equivalent to
∑
T∈n
λUT 
n∏
i=1
x
ωi
i =
∑
A∈n
λSW A1+ λV A
n∏
i=1
x
SWiA+SEiA+ViA
i  (6)
This suggests a natural bijection between tournaments on n vertices and
six-vertex models on n2 vertices in which we have chosen a direction (left
or right) at each vertical vertex. Each vertex in the six-vertex model that
has an in-edge from the north will deﬁne an out-edge of the tournament.
Call this vertex of the six-vertex model an initiating vertex. If an initiating
vertex is southwest, there is an out-edge to the left, and the corresponding
edge in the tournament will contribute to the upset number. If the initiating
vertex is southeast, there is an out-edge to the right, and the corresponding
edge in the tournament will not contribute to the upset number. If the
initiating vertex is vertical, we have a choice of taking either the left or
right out-edge. The left choice contributes one to the upset number of the
tournament; the right choice contributes nothing.
Problem 2. Find a bijective proof of Eq. (6).
Chapman [4] has now given such a bijection. His bijection implies the
following more general identity:
∏
1≤i<j≤n
xi + yj =
∑
A∈n
n∏
i=1
x
SWiA
i y
SEiA
i xi + yiViA (7)
4. THE IZERGIN–KOREPIN DETERMINANT EVALUATION
Kuperberg’s proof of the alternating sign matrix conjecture and Zeil-
berger’s proof of the reﬁned conjecture rest on the following determinant
evaluation of Izergin [7], described in Korepin, Bogoliubov, and Izergin’s
“Quantum Inverse Scattering Method” [9].
three asm identities 295
Proposition 4. Given A ∈ n, let i j be the vertex in row i, column j
of the corresponding six-vertex model, and let H, V SE SW NE NW be,
respectively, the sets of horizontal, vertical, southeast, southwest, northeast, and
northwest vertices. For indeterminants a, x1     xn, and y1     yn, we have
that
det
(
1
xi + yjaxi + yj
) ∏n
ij=1xi + yjaxi + yj∏
1≤i<j≤nxi − xjyi − yj
= ∑
A∈n
−1NA1− a2NAa n2 − A
× ∏
ij∈V
xiyj
∏
ij∈NE∪SW
axi + yj
∏
ij∈NW ∪SE
xi + yj (8)
As Lascoux has pointed out [11], the right way to understand this identity
is as an extension of Cauchy’s
det
(
1
xi + yj
) n∏
ij=1
xi + yj
∏
1≤i<j≤n
xi − xj−1yi − yj−1 = 1 (9)
This is true by inspection. The determinant times the product over i j is an
alternating polynomial in the xi and in the yj . Since any alternating polyno-
mial is divisible by the Vandermonde product, the left side of this equality
is a symmetric polynomial in the xi, and it is a symmetric polynomial in
the yj . The degree in x1 of this polynomial is zero, and the constant can be
checked by induction.
Applying this same reasoning to the left side of Eq. (8), we see that it is
a symmetric polynomial in the xi and in the yj . Its degree in x1 is n − 1.
On the right, we also have a polynomial in x1 of degree n − 1. We need
only check that these two sides agree for n values of x1. By induction, they
agree at x1 = −y1/a. If we can show that the right side is symmetric in the
yj , then the identity is proven.
Symmetry follows from Baxter’s triangle-to-triangle relation which was
used by Izergin to prove that∑
A∈n
∏
ij∈H
xi1−a
∏
ij∈V
yj1−a
∏
ij∈NE∪SW
axi+yj
∏
ij∈NW ∪SE
xi+yja1/2
is symmetric in the xi, and it is symmetric in the yj .
Among the corollaries of Proposition 4, we can set a = 1 to get
Borchardt’s permanent-determinant identity [1].
det
(
1
xi + yj2
) ∏n
ij=1xi + yj2∏
1≤i<j≤nxi − xjyi − yj
= perm
(
1
xi + yj
) n∏
ij=1
xi + yj (10)
296 david m. bressoud
where
permaij =
∑
σ∈n
n∏
i=1
aiσi
We can set a = ω = e2πi/3, xj = −ωqj and yj = q1−j , evaluate the deter-
minant, and then take the limit as q→ 1 to get the number of ASMs of a
given size:
n−1∏
j=0
3j + 1!
n+ j! = n (11)
If we set a = −1, then the matrix for which we take the determinant is
1/x2i − y2j , which can be evaluated using Cauchy’s formula, Eq. (9). The
left side of Eq. (8) simpliﬁes to
−1nn−1/2 ∏
1≤i<j≤n
xi + xjyi + yj
From Eq. (5), each of these Vandermonde-type products can be written
as a sum over alternating sign matrices. We let EiniA be the number of
vertices in row i with an in-edge from the left, NinjA be the number
of vertices in column j with an in-edge from below. Replacing yj by −yj
and multiplying each side by x1 · · ·xn, the case a = −1 is equivalent to the
identity
∑
BC∈n×n
2NB+NC
n∏
i=1
x
EiniB
i
n∏
j=1
y
NinjC
j
= ∑
A∈n
−1 A−NA4NA ∏
ij∈H
xi
∏
ij∈V
yj
∏
ij∈NE
xi + yj
× ∏
ij∈SW
−xi − yj
∏
ij∈NW
−xi + yj
∏
ij∈SE
xi − yj (12)
Problem 3. Find a bijective proof of Eq. (12).
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Three alternating sign matrix identities in search of bijective proofs

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