ASM matrices and Grothendieck
Polynomials
Alain Lascoux
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Institut Gaspard Monge, Université Paris-Est
Alain.Lascoux@univ-mlv.fr
phalanstere.univ-mlv.fr/∼al
’ ’ ’ ’
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Alain Lascoux
ASM & Grothendieck
1 / 24
Permutations can be viewed in many different ways. I shall
consider today only two interpretations :
♠ as sequences of integers, as configurations of points in the
plane (non attacking rooks), as flags of vector spaces,
♠ as elements of the symmetric group, interpreted as a Coxeter
group generated by simple transpositions.
The first point of view leads to order properties, inversion
diagrams, Bruhat order, lattice operations on permutations, &c.
The second forces to use the group algebra, or its deformations
( Hecke algebra), Yang-Baxter relations, &c.
I shall connect these two points of view using polynomials.
Alain Lascoux
ASM & Grothendieck
2 / 24
Permutations have inversions. Given a permutation σ, an
inversion is a pair i < j such that σi > σj . We prefer
polynomials, and associate to such inversion the weight
yi xσ−1
−1
j
Let
Inv (σ) =
Y
−
1
yi xσ−1
j
(j,i) inversion
be the inversion polynomial associated σ.
This one more family of polynomials associated to permutations,
apart from Schubert, Grothendieck and key polynomials !
Alain Lascoux
ASM & Grothendieck
3 / 24
What to do with inversion polynomials ?
Summing over S2 , one obtains
(y1 /x1 − 1) + 1 = y1 /x1 .
GOOD
S3 instead gives
y2 y2 y1 2 y2 y1
+
−
x2 x1 2 x2 x1 x2
that one can cure into y12 y2 x1−2 x2−1 by adding y2 x2−1 (y1 x1−1 − 1).
However, S4 seems hopeless.
Alain Lascoux
ASM & Grothendieck
4 / 24
The solution comes from a two-dimensional object well adapted
to the season:
square ice .
For a mathematician, it is made of 6 types of molecules:
H
H −O−H
H –O
H
H
H –O
O–H
H
H
O–H
O
H
Alain Lascoux
ASM & Grothendieck
5 / 24
Here is an example of an ice configuration :
H −− O
H −− O −− H
O −− H
O −− H
O −− H
H
H
H
H
H
H −− O −− H
O
H −− O
H
H
H
H
H
H −− O
O −− H
O −− H
O
H −− O −− H
H
H
H
H
H
H −− O
H −− O
H −− O
H
H
H
H −− O
H −− O
Alain Lascoux
H −− O −− H
H −− O −− H
H −− O −− H
H −− O −− H
ASM & Grothendieck
O −− H
H
H
O −− H
O −− H
6 / 24
Ice configurations are in bijection with alternating-sign matrices
(ASM in short),
replacing horizontal molecules by 1,
vertical molecules by −1 ,
and the others by 0.
Such matrices of 0, 1, −1 are characterized by the property that
non zero entries alternate in each row and column, always
starting and finishing with a 1.
Continuing with the same example, we get the following ASM :


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
Alain Lascoux
ASM & Grothendieck
7 / 24
No information has been lost. One recovers the Ice from the
ASM thanks to Rothe (1800!), who described inversions by a
diagram. A matrix has four corners, and four inversion diagrams:
Given a 0-entry in an ASM, ignore all the other zeroes. Then the
current 0 is next to a 1 in its column and its row. Replace now
this 0 by a box that will be attributed to one of 4 diagrams,
depending on the orientation :
0 →1
↓
1
1← 0
↓
1
NW
gives
↓
1
1←
gives
Alain Lascoux
→1
,
NE
↓
1
,
1
↑
0 →1
1
↑
1← 0
ASM & Grothendieck
gives
1
↑
→1
SW
1
↑
gives
1←
.
SE
8 / 24
The preceding ASM gives the four diagrams :

 
1 · · ·
· 1 ·
 1 −1 1 ·   1 −1 ·

 
 · 1 · −1 1 ,  1 ·

 
 · · 1 ·   ·
· · 1 · ·
1

·
1

·

·
·
Alain Lascoux
 
1 ·


−1
· 1  1

1 −1 1 
 , ·
· · 1 ·  ·
· 1 · ·
·
1
−1
1
·
·
ASM & Grothendieck
·
1
−1
1
·

·
·

1

·
·

· · ·
· 1 ·

· −1 1 
 .
· 1 
1 9 / 24
As is frequent in combinatorics, and as most of you know, we
need another object in bijection with the ASM, which are the
monotone triangles
(Young tableaux with weakly decreasing diagonals) :
Read the successive rows of an ASM, from right to left,
a 1 in column i meaning that the letter i appears in the tableau,
a −1 meaning that it disappears.
This gives a sequence of sets of numbers, or, equivalently, a
sequence of columns of a triangle.
Alain Lascoux
ASM & Grothendieck
10 / 24
.
5
.
.
.
4
b
4
4
.
.
.
.
.
.
3
.
2
b
2
2
.
.
.

1
.
.
←→ {4}, {2, 5}, {1, 4, 5}, {1, 2, 4, 5}, {1, 2, 3, 4, 5}
5

4
5
←→
3
4
5
2
2
4
5
1
1
1
2
Alain Lascoux





·
1
·
·
·
1
−1
1
·
·
·
·
·
·
1
·
1
−1
1
o
·
·
1
o
o



 .


4
ASM & Grothendieck
11 / 24
Ehresmann defined an order (called Bruhat order) on the
symmetric group, which plays a fundamental role in geometry,
algebra and representation theory:
two permutations are consecutive if they have consecutive
length and differ by multiplication
123
by a transposition
(on the right or left).
thick arrows =
permutohedron
213
132
231
312
321
Alain Lascoux
ASM & Grothendieck
12 / 24
M.P. Schützenberger and I transformed the interpretation
of the Bruhat order by embedding
123
the symmetric group into the
distributive lattice of ASM.
213
132
•
231
312
321
Alain Lascoux
ASM & Grothendieck
13 / 24
Instead of ASM, let us use staircase Young tableaux which are
more directly related to Demazure characters
and
Grothendieck polynomials .
Recall the definition of isobaric divided differences acting on
polynomials:
si transpose xi , xi+1 ,
πi is the operator f → (xi f − xi+1 f si ) (xi − xi+1 )−1
π
bi = πi −1 is the operator f → (f − f si ) (xi /xi+1 − 1)−1
Alain Lascoux
ASM & Grothendieck
14 / 24
The starting points are dominant polynomials (i.e. indexed by
decreasing partitions λ. Here we need only λ = ω := [n, . . . , 1],
the maximal permutation).
Y
Kω = x ω = x1n . . . xn & Gω =
(1 − yj xi−1 )
i+j≤n+1
The Demazure characters Kσ and the Grothendieck polynomials
Gσ are ALL the images of the dominant ones by products of πi .
bσ are all the images of K
bω := x ω under
The key polynomials K
products of π
bi .
Alain Lascoux
ASM & Grothendieck
15 / 24
However, the operators si , πi , π
bi lift into operators
acting on the free algebra
b F = . . . 2λ2 1λ1 , and the
Given a partition λ, one defines KλF = K
λ
other key polynomials by recursion. For i, v such that vi > vi+1 ,
then
bF = K
bF π
KvFsi = KvF πi & K
v si
v bi
Facts.
b F is the sum of all tableaux with right key v .
K
v
X
bF
KF =
K
v
u≤v
u
The tableaux in KvF index a basis of a Demazure module.
Alain Lascoux
ASM & Grothendieck
16 / 24
Tableaux generated by π
b1 , π
b2 , starting from a Yamanouchi
tableau (tableau with 1 in the bottom row, 2 in the row above,
etc)
bF = 2
K
210
1 1 O
o
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bF = 2
K
120
1 2
bF = 3
K
201
1 1
bF = 2 + 3
K
102
1 3
1 3
bF = 3 + 3
K
021
1 2
2 2
OOO
OOO
O
o
ooo
ooo
bF = 3
K
012
2 3
Alain Lascoux
ASM & Grothendieck
17 / 24
Instead of ASM, let us use staircase tableaux. The weight which
extends the inversions of a permutation matrix is defined as a
product on elementary staircases.
The weight of a subtableau of shape [1, 2] on columns j, j +1 and
consecutive rows is, for three integers a < b < c,
ϕ weight
c
ab
yj xb −1
c
b b
yj xb −1 − 1
c
b c
1
b
a c .
0
ϕ(T ) is the product of these elementary weights
( the staircases which are not monotone have weight 0).
Alain Lascoux
ASM & Grothendieck
18 / 24
For example, pointing out the rightmost box of the elementary
4
tableaux contributing to the weight, the tableau t = 2 3 has
1 1 2
weight
4
2
3
1
1
Alain Lascoux
→
2
•
−1
ϕ(t) = • y1 x3
• y1 x1−1 −1 y2 x2−1
ASM & Grothendieck
19 / 24
Theorem. Let σ ∈ Sn . Then
(−1)`(σ) G(σ) (x, y) =
X
bF
T ∈K
σ
ϕ(T )
Proof. Grothendieck polynomials, as well as key polynomials,
Schubert polynomials and Macdonald polynomials satisfy a
transition equation
which allows to cut them into smaller polynomials , and this
gives a recursion to prove the theorem.
(this property is stronger than mere triangularity in the basis of
monomials for an appropriate order, that these four bases
satisfy.)
Alain Lascoux
ASM & Grothendieck
20 / 24
For example, for σ = [6, 4, 8, 3, 1, 7, 2, 5], one has
G64831527 − G64831725
x
6
y5
= G64831527 − G65831427 − G64851327 − G64835127
+ G65841327 + G65834127 + G64853127 − G65843127 ,
this recursion corresponding to a boolean lattice ...
Alain Lascoux
ASM & Grothendieck
21 / 24
[65843127]
[65841327]
[65834127]
XXX
X
X
X
XXX
[65831427]
XXX
X
XXX
XXX
XXX
[64851327]
XXX
XX
[64853127]
XXX
X
X
X
XXX
[64835127]
[64831527]
Going back to the theorem, one has for example ...
Alain Lascoux
ASM & Grothendieck
22 / 24
bF.
For σ = [4, 2, 1, 5, 3], there are fifteen tableaux in K
σ
Only 4 of them are monotone triangles.
5
4 5
y3 y1
( −1)( yx22 −1)( xy11 −1)( yx21 −1)( yx31 −1)
3 4 5
x4 x2
2 2 2 4
1 1 1 1 4
5
4 5
y3 y1
( −1)( yx12 −1)( xy22 −1)( yx11 −1)( yx21 −1)( yx31 −1)
3 3 5
x4 x3
2 2 2 4
1 1 1 1 4
5
4 5
( xy21 −1)( yx34 −1)( yx11 −1)( yx21 −1)( xy13 −1)
3 4 5
2 2 4 4
1 1 1 1 4
5
4 5
y3 y1
( −1)( yx12 −1)( xy11 −1)( yx21 −1)( yx31 −1)
3 3 5
x4 x3
2 2 3 4
1 1 1 1 4
Alain Lascoux
ASM & Grothendieck
23 / 24
Therefore, the Grothendieck polynomial G(42153) is equal to the
sum of these four weights:
G(42153) (x, y) = (1−
y1
y1
y2
y3
y1 y2 y3
)(1− )(1− )(1− )(1−
),
x2
x1
x1
x1
x2 x3 x4
More interesting polynomials are to be expected, starting with
the tableaux corresponding to a Demazure module, when taking
another shape, or another weight !
Alain Lascoux
ASM & Grothendieck
24 / 24