Mind your P and Q-symbols:
Why the Kazhdan-Lusztig basis of the
Hecke algebra of type A is cellular
Geordie Williamson
An essay submitted in partial fulfillment of
of the requirements for the degree of
B.A. (Honours)
D
ERE·M
E
N
S·E
E
SI
O
Pure Mathematics
University of Sydney
M
·MUTA
T
AD
October 2003
To my mother
Acknowledgements
First and foremost I would like to thank my supervisor Dr. Gus Lehrer. I would not have progressed
very far without his inspiration and support. I would also like to thank Anthony Henderson, Bob
Howlett and Andrew Mathas for giving up their time for me during the year. On a more personal
note, I would like to thank my family and friends (both on the fourth floor of Carslaw and at 8
Northumberland Ave.) for their support and friendship this year. Lastly, I would like to thank
John Lascelles, Alan Stapledon and Steve Ward for proof reading sections of the essay.
i
ii
Contents
Acknowledgements
i
Introduction
v
Preliminaries
vii
1 The Symmetric Group
1
1.1
The Length Function and Exchange Condition . . . . . . . . . . . . . . . . . . . . .
1
1.2
Generators and Relations for Symn . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3
The Bruhat Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.4
Descent Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.5
Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2 Young Tableaux
10
2.1
Diagrams, Shapes and Standard Tableaux . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2
The Row Bumping Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3
The Robinson-Schensted Correspondence . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4
Partitions Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.5
Growth Diagrams and the Symmetry Theorem . . . . . . . . . . . . . . . . . . . . . 14
2.6
Knuth Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.7
Tableau Descent Sets and Superstandard Tableaux . . . . . . . . . . . . . . . . . . . 20
2.8
The Dominance Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.9
Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 The Hecke Algebra and Kazhdan-Lusztig Basis
26
3.1
The Hecke Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.2
The Linear Representations of Hn (q) . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.3
Inversion in Hn (q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.4
An Involution and an anti-Involution . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.5
The Kazhdan-Lusztig Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.6
Multiplication Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.7
Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
iii
4 Cells
39
4.1
Cell Orders and Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.2
Representations Associated to Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4.3
Some Properties of the Kazhdan-Lusztig Polynomials . . . . . . . . . . . . . . . . . . 42
4.4
New Multiplication Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.5
New Definitions of the Cell Preorders . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.6
Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5 The Kazhdan-Lusztig Basis as a Cellular Basis
48
5.1
Cellular Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5.2
Elementary Knuth Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
5.3
The Change of Label Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.4
Left Cells and Q-Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
5.5
Property A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
5.6
The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
5.7
Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
A Appendix
65
A.1 An Alternative Proof of Kazhdan and Lusztig’s Basis Theorem . . . . . . . . . . . . 65
A.2 Two-Sided Cells and the Dominance Order . . . . . . . . . . . . . . . . . . . . . . . 67
Bibliography
69
iv
Introduction
The Hecke algebras emerge when one attempts to decompose an induced representation of certain
finite matrix groups. In particular, the Hecke algebra of the symmetric group is isomorphic to the
algebra of intertwining operators of the representation of GL(n, q) obtained by inducing the trivial
representation from the subgroup of upper triangular matrices. By Schur’s Lemma, the restriction
of any intertwining operator to an irreducible representation must be scalar. Hence the problem of
decomposing the induced representation is equivalent to decomposing the associated Hecke algebra.
However, the representation theory of the Hecke algebra is difficult. In 1979 Kazhdan and Lusztig
[17] introduced a special basis upon which a generating set for the Hecke algebra acts in an easily
described manner. In order to construct the representations afforded by this new basis the notion of
cells was introduced and it was shown that each cell corresponds to a representation of the algebra.
In general these representations are not irreducible. However Kazhdan and Lusztig showed that,
in the case of the Hecke algebra of the symmetric group, their construction does yield irreducible
representations. In order to prove this Kazhdan and Lusztig introduced and studied the so-called
‘star operations’ on elements of the symmetric group.
It is implicit in Kazhdan and Lusztig’s paper that these star operations (and the equivalence classes
which they generate) have a deep combinatorial significance. This significance can be explained
in terms of the Robinson-Schensted correspondence. In 1938 Robinson [25] showed that, to every
permutation, one can associate a pair of ‘standard tableaux’ of the same shape. Then in 1961
Schensted [26] showed that this map was a bijection. In 1970 Knuth [20] studied the equivalence
classes of the symmetric group corresponding to a fixed left or right tableau and showed that two
permutations are equivalent if and only if they can be related by certain basic rearrangements known
as ‘Knuth transformations’. The amazing thing is that these Knuth transformations are precisely
the star operations of Kazhdan and Lusztig. Hence, much of the algebraic theory developed by
Kazhdan and Lusztig can be described combinatorially, using the language of standard tableaux.
Since a multiplication table for the Hecke algebra of a Weyl group was first written down by Iwahori
[16] in 1964, many algebras similar to the Hecke algebra have been discovered. In almost all cases
their representation theory is approached by using analogous techniques to the original methods
of Kazhdan and Lusztig. The similarities between these algebras, in particular the multiplicative
properties of a distinguished basis, led Graham and Lehrer [12], in 1996, to define a ‘cellular algebra’.
This definition provides an axiomatic framework for a unified treatment of algebras which possess
a ‘cellular basis’.
In the paper in which Graham and Lehrer defined a cellular algebra, the motivating example was
the Hecke algebra of the symmetric group. However, there is no one source that explains why the
Hecke algebra of the symmetric group is a cellular algebra. The goal of this essay is to fill this gap.
However, we are not entirely successful in this goal. We must appeal to a deep theorem of Kazhdan
and Lusztig to show that a certain degenerate situation within a cell cannot occur (Kazhdan and
Lusztig prove this using geometric machinery beyond the scope of this essay).
The structure of this essay is as follows. In Chapter 1 we gather together some fundamental concepts
associated to the symmetric group including the length function, the Bruhat order and descent sets.
v
We also discover a set of generators and relations for the symmetric group. This is intended to
motivate the introduction of the Hecke algebra by generators and relations in Chapter 3.
Chapter 2 provides a self-contained introduction to the calculus of tableaux. Most of the results,
including the Robinson-Schensted correspondence, the Symmetry Theorem and Knuth equivalence
are fundamental to the subject and are present in any text on standard tableaux (possibly with
different proofs). Towards the end of the chapter we introduce the tableau descent set and superstandard tableau. These are less well-known but are fundamental to our arguments in Chapter 5.
Chapters 3 and 4 are an introduction to Kazhdan-Lusztig theory in the special case of the Hecke
algebra of the symmetric group. In Chapter 3 we define the Hecke algebra and then prove the
existence and uniqueness of the Kazhdan-Lusztig basis. In Chapter 4 we introduce the cells associated to the basis of an algebra and show how they lead to representations. The rest of the
chapter is then dedicated to deriving the original formulation of the cell preorders due to Kazhdan
and Lusztig. Although we will not make it explicit, all of the material of Chapters 3 and 4 can be
proved without alteration in the more general case of the Hecke algebra of a Coxeter group.
In Chapter 5 we define a cellular algebra and then combine the results of Chapters 2, 3 and 4 with
the aim of showing that the Kazhdan-Lusztig basis is cellular. We first show how the elementary
Knuth transformations can be realised algebraically and then work towards a complete description
of the cells in terms of the left and right tableau of the Robinson-Schensted correspondence. We also
show that the representations afforded by left cells within a given two-sided cell are isomorphic.
However, as mentioned above, we cannot complete our proof that the Kazhdan-Lusztig basis is
cellular entirely via elementary means: we must appeal to a theorem of Kazhdan and Lusztig to
show that all left cells within a two-sided cell are incomparable in the left cell preorder.
The appendix contains two sections. The first gives an elegant alternative proof of the existence
and uniqueness of the Kazhdan-Lusztig basis due to Lusztig [19]. In the second section we discuss
the relationship between the dominance order and two-sided cell order.
vi
Preliminaries
All rings and algebras are assumed to be associative and have identity. All homomorphisms between rings or algebras should preserve the identity. A representation of an R-algebra H is a
homomorphism φ : H → End M for some R-module M . Since homomorphisms must preserve the
identity all representations map the identity of H to the identity endomorphism of M . We will
refer without comment to the equivalence between H-modules and representations.
A preorder is a reflexive and transitive relation. That is, if ≤ is a preorder on a set X then x ≤ x
for all x ∈ X and if x ≤ y ≤ z then x ≤ z. A preorder need not satisfy antisymmetry and so
it is possible to have x ≤ y and y ≤ x with y 6= x. If ≤ is a preorder or partial order on a set
X, a subset of the relations is said to generate the order ≤ if all relations are either consequences
of reflexivity or follow via transitivity from the relations of the subset. For example, the relations
{i ≤ i + 1|i ∈ Z} generate the usual order on Z.
vii
viii
1
The Symmetric Group
This is an introductory chapter in which we gather results about the symmetric group. We prove
the strong exchange condition and give a set of generators and relations for the symmetric group.
We also introduce the Bruhat order and descent sets.
1.1
The Length Function and Exchange Condition
Let Symn be the symmetric group on {1, 2, . . . , n} acting on the left. Let id be the identity,
S = {(i, i + 1)|1 ≤ i < n} the set of simple transpositions and T = {(i, j)|1 ≤ i < j ≤ n} be the
transpositions. Throughout, si will always denote the simple transposition (i, i + 1) interchanging
i and i + 1 whereas r, u and v will be used to denote arbitrary simple transpositions. If w ∈ Symn ,
we can write w = w1 w2 . . . wn where w(i) = wi for all i. This is the string form of w. We will use
this notation more frequently than cycle notation.
Given w ∈ Symn the length of w, denoted `(w), is the number of pairs i < j such that w(i) > w(j).
Thus `(w) = 0 if and only if w is the identity. Our first result shows how the length of an element
w ∈ Symn is effected by multiplication by a simple transposition:
Lemma 1.1.1. If w ∈ Symn and sk = (k, k + 1) ∈ S then:
(i) `(wsk ) = `(w) + 1 if w(k) < w(k + 1)
(ii) `(wsk ) = `(w) − 1 if w(k) > w(k + 1)
Proof. Write N (w) = {(i, j)|i < j, w(i) > w(j)} so that `(w) = |N (w)|. Now assume that w(k) <
w(k + 1). If (p, q) ∈ N (w) then wsk (sk (p)) = w(p) > w(q) = wsk (sk (q)) and so (sk (p), sk (q)) ∈
N (wsk ) so long as sk (p) < sk (q). But if sk (p) > sk (q) then we must have (p, q) = (k, k + 1)
contradicting w(k) < w(k + 1). Hence, if (p, q) ∈ N (w) then (sk (p), sk (q)) ∈ N (wsk ). Similarly, if
(p, q) ∈ N (wsk ) and (p, q) 6= (k, k + 1) then (sk (p), sk (q)) ∈ N (w). By assumption, w(k) < w(k + 1)
and so (k, k + 1) ∈ N (wsk ). Hence |N (wsk )| = |N (w)| + 1. Thus `(wsk ) = `(w) + 1 and hence (i).
For (ii) note that if w(k) > w(k + 1) then wsk (k) < wsk (k + 1) and we can apply (i) to wsk to
conclude that `(w) = `(ws2k ) = `(wsk ) + 1 and hence (ii).
We will soon see that the simple transpositions generate Symn and so, given w ∈ Symn , we can
write w = r1 r2 . . . rm with ri ∈ S. This is called an expression for w. The expression is reduced if
the number of simple transpositions used is minimal.
Proposition 1.1.2. The simple transpositions generate Symn . Moreover, if w ∈ Symn then an
expression for w is reduced if and only if it contains `(w) simple transpositions.
Proof. We first show, by induction on `(w), that it is possible to express w using `(w) simple
transpositions. If `(w) = 0 then w is the identity and the result is clear (using the convention
that the empty word is the identity). Now, if `(w) > 0 then there exists i, j such that i < j but
w(i) > w(j). Hence there exists k such that w(k) > w(k + 1). Now from Lemma 1.1.1 above we
have `(wsk ) = `(w) − 1 and so we can apply induction to write wsk = r1 r2 . . . rm for some ri ∈ S
with m = `(w)−1. Hence w = r1 r2 . . . rm sk is an expression for w using `(w) simple transpositions.
Now let r1 r2 . . . rm be a reduced expression for w. Then, from above, we know that m ≤ `(w).
However, by repeated application of Lemma 1.1.1 we have `(w) = `(r1 r2 . . . rm ) ≤ m and so
1
`(w) = m. Conversely, we have seen that there exists a reduced expression for w using `(w) simple
transpositions and hence any expression with `(w) simple transpositions is reduced.
Using this new interpretation of the length function we have the following:
Proposition 1.1.3. If w ∈ Symn then `(w) = `(w−1 ). In particular, if r1 r2 . . . rm is a reduced
expression for w then rm rm−1 . . . r1 is a reduced expression for w−1 .
Proof. Let r1 r2 . . . rm be a reduced expression for w so that `(w) = m. Then rm rm−1 . . . r1 is an
expression for w−1 and so `(w−1 ) ≤ `(w). But `(w) = `((w−1 )−1 ) ≤ `(w−1 ) and so `(w) = `(w−1 ).
Hence rm rm−1 . . . r1 is a reduced expression for w−1 since it is an expression for w−1 and contains
`(w−1 ) terms.
The following are some useful congruences:
Lemma 1.1.4. Let x, y ∈ Symn , t ∈ T be a transposition and r, ri ∈ S be simple transpositions:
(i) `(xr) ≡ `(x) + 1 (mod 2)
(ii) `(r1 r2 . . . rm ) ≡ m (mod 2)
(iii) `(xy) ≡ `(x) + `(y) (mod 2)
(iv) `(t) ≡ 1 (mod 2)
In particular `(xt) 6= `(x) for all x ∈ Symn and t ∈ T .
Proof. We get (i) upon reduction of Lemma 1.1.1 modulo 2, and (ii) follows by repeated application
of (i) and the fact that `(id) = 0. If u1 u2 . . . up is an expression for x and v1 v2 . . . vq is an expression
for y then, by (ii), `(xy) = `(u1 . . . up v1 . . . vq ) ≡ p + q ≡ `(x) + `(y) (mod 2) and hence (iii). For
(iv) note that if t = (i, j) ∈ T then t = si si+1 . . . sj−2 sj−1 sj−2 . . . si and so we can express t using an
odd number of simple transpositions. Hence `(t) ≡ 1 (mod 2) by (ii). The last statement follows
since `(xt) ≡ `(x) + 1 (mod 2) by (iii) and (iv).
We now come to a theorem of fundamental importance:
Theorem 1.1.5. (Strong Exchange Condition) Let w ∈ Symn and choose an expression r1 r2 . . . rm
for w. If t = (i, j) ∈ T is such that `(wt) < `(w) then there exists a k such that wt =
r1 r2 . . . rbk . . . rm (where b denotes ommision).
Proof. We first show that `(wt) < `(w) implies that w(i) > w(j). As in Lemma 1.1.1 define N (w) =
{(p, q)|p < q, w(p) > w(q)}. Now assume that w(i) < w(j) and define a function ϕ : N (w) → N×N
by:
(
(t(p), t(q)) if t(p) < t(q)
ϕ(p, q) =
(p, q)
if t(p) > t(q)
We claim that Im(ϕ) ⊂ N (wt) and ϕ is injective. It then follows that `(w) = |N (w)| ≤ |N (wt)| =
`(wt).
We first verify that Im(ϕ) ⊂ N (wt). If t(p) < t(q) then wt(t(p)) = w(p) > w(q) = wt(t(q)) and
hence (t(p), t(q)) ∈ N (wt) if t(p) < t(q). On the other hand if t(p) > t(q) then since t = (i, j) and
w(i) < w(j) (so (i, j) ∈
/ N (w)) we must have either p = i and q = j but not both. If p = i then
we have wt(p) = w(j) > w(i) = w(p) > w(q) = wt(q) and hence (p, q) ∈ N (wt). Similarly, if q = j
then wt(p) = w(p) > w(q) = w(j) > w(i) = wt(q) and so (p, q) ∈ N (wt).
2
To see that ϕ is injective we argue by contradiction. So assume that (p, q) 6= (p0 , q 0 ) and that
ϕ(p, q) = ϕ(p0 , q 0 ). We may assume without loss of generality that t(p) < t(q) and that t(p0 ) > t(q 0 ).
Since p0 < q 0 and t(p0 ) > t(q 0 ) we must have either p0 = i and i < q 0 < j or q 0 = j and i < p0 < j.
If p0 = i and i < q 0 < j then since t(p) = p0 = i we have p = j. Hence j < q. But q 0 < j and
so we have a contradiction since q = q 0 . On the other hand if q 0 = j and i < p0 < j then since
t(q) = q 0 = j we have q = i. Hence p < q = i and we obtain a similar contradiction.
The above arguments show that if w(i) < w(j) then `(w) ≤ `(wt). Hence w(i) > w(j) since
`(w) > `(wt). Now write up = rp rp+1 . . . rm . Then if um (i) > um (j) then rm (i) > rm (j) and
2 = 1. So
so j = i + 1 and rm = t = (i, i + 1). The result then follows with k = m since rm
assume that um (i) < um (j). Then, since u1 (i) = w(i) > w(j) = u1 (j) there exists a k such that
uk (i) > uk (j) but uk+1 (i) < uk+1 (j). Hence uk+1 (i) < uk+1 (j) but rk (uk+1 (i)) > rk (uk+1 (j)) and
so rk = (uk+1 (i), uk+1 (j)) and we have:
wt = r1 r2 . . . rk−1 (uk+1 (i), uk+1 (j))uk+1 (i, j) = r1 r2 . . . rk−1 uk+1 = r1 r2 . . . rbk . . . rm
We also have a left-hand version of the exchange condition: If w = r1 r2 . . . rm ∈ Symn and t ∈ T
with `(tw) < `(w) then we have `(w(w−1 tw)) < `(w) and so we can apply the exchange condition
to w and w−1 tw (since w−1 tw ∈ T ) to get that tw = r1 r2 . . . rbk . . . rm for some k.
The exchange condition has two important corollaries:
Corollary 1.1.6. (The Deletion Condition) Let w ∈ Symn and let r1 r2 . . . rm be an expression
for w with `(w) < m. Then there exists p and q such that w = r1 . . . rbp . . . rbq . . . rm . Moreover, a
reduced expression for w can be obtained from r1 r2 . . . rm by deleting an even number of terms.
Proof. Since `(w) < m we have `(r1 r2 . . . rq−1 ) > `(r1 r2 . . . rq ) for some q. Choose such a q. Then
the exchange condition applies to the pair r1 r2 . . . rq−1 and rq so that there exists a p such that
r1 r2 . . . rq = r1 r2 . . . rbp . . . rq−1 . Hence w = r1 . . . rbp . . . rbq . . . rm . If w is unreduced `(w) is less than
the number of terms. We also have that the number of terms is congruent to `(w) modulo 2 by
Lemma 1.1.4(ii). Hence we can keep deleting two terms at a time until we have `(w) terms. By
Proposition 1.1.2 such an expression is reduced.
Corollary 1.1.7. If w ∈ Sym and r ∈ S then `(wr) < `(w) if and only if w has a reduced
expression ending in r.
Proof. Let w = r1 r2 . . . rm be a reduced expression for w. Then if `(wr) < `(w) there exists a k
such that wr = r1 r2 . . . rbk . . . rm by the exchange condition. Hence w = r1 r2 . . . rbk . . . rm r. Since
this expression has m = `(w) terms it is reduced by Proposition 1.1.2. The other implication is
clear.
1.2
Generators and Relations for Symn
The aim of this section is to write down a set of generators and relations for the symmetric group.
We have already seen that the simple transpositions generate the symmetric group. It is easily
verified that the following relations hold:
s2i = 1
si si+1 si = si+1 si si+1
if |i − j| ≥ 2
si sj = sj si
3
The second two relations are called the braid relations. It turns out that these relations determine
all others in the symmetric group, however this will take a little while to prove.
We first prove what is in fact a slightly stronger “universal property” of the symmetric group. Our
proof follows Dyer [4] closely.
Proposition 1.2.1. Suppose ϕ is a function from the set of simple transpositions S ⊂ Symn to a
monoid M such that:
ϕ(si )ϕ(si+1 )ϕ(si ) = ϕ(si+1 )ϕ(si )ϕ(si+1 )
ϕ(si )ϕ(sj ) = ϕ(sj )ϕ(si )
if |i − j| ≥ 2
Then there is a unique ϕ̃ extending ϕ to a map from Symn to M such that:
ϕ̃(w) = ϕ(r1 )ϕ(r2 ) . . . ϕ(rm )
whenever r1 r2 . . . rm is a reduced expression for w.
The key to the proof is the following lemma:
Lemma 1.2.2. Suppose u1 u2 . . . um and v1 v2 . . . vm are reduced expressions for w ∈ Symn such that
ϕ(u1 )ϕ(u2 ) . . . ϕ(um ) 6= ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ). Suppose further that, for any two reduced expressions
r1 r2 . . . rk = t1 t2 . . . tk with k < m we have ϕ(r1 )ϕ(r2 ) . . . ϕ(rk ) = ϕ(t1 )ϕ(t2 ) . . . ϕ(tk ). Then u1 6= v1
and v1 u1 . . . um−1 = u1 u2 . . . um but ϕ(v1 )ϕ(u1 ) . . . ϕ(um−1 ) 6= ϕ(u1 )ϕ(u2 ) . . . ϕ(um ).
Proof. Since u1 u2 . . . um = v1 v2 . . . vm we have:
v1 u1 . . . um = v2 . . . vm
(1.2.1)
Now the right hand side of (1.2.1) has m−1 terms and hence `(v1 u1 . . . um ) ≤ m−1 < m+1. Hence
the exchange condition applies to v1 and u1 u2 . . . um and there exists a j such that v1 u1 . . . um =
u1 . . . ubj . . . um . Some cancellation and rearranging yields:
v1 u1 . . . uj−1 = u1 u2 . . . uj
(1.2.2)
ϕ(v1 )ϕ(u1 ) . . . ϕ(uj−1 ) = ϕ(u1 ) . . . ϕ(uj )
(1.2.3)
Now, assume for contradiction that:
Then, by (1.2.2) v1 v2 . . . vm = u1 u2 . . . um = v1 u1 . . . ubj . . . um (again b denotes omission) and so:
v2 v3 . . . vm = u1 . . . ubj . . . um
(1.2.4)
Now both sides of (1.2.4) are reduced and have length less than m and so, by the conditions of the
lemma:
\
ϕ(v2 )ϕ(v3 ) . . . ϕ(vm ) = ϕ(u1 ) . . . ϕ(u
(1.2.5)
j ) . . . ϕ(um )
But left multiplying by ϕ(v1 ) and using (1.2.3) yields:
ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ) = ϕ(u1 )ϕ(u2 ) . . . ϕ(um )
This contradicts our assumption that ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ) 6= ϕ(u1 )ϕ(u2 ) . . . ϕ(um ). Hence:
ϕ(v1 )ϕ(u1 ) . . . ϕ(uj−1 ) 6= ϕ(u1 )ϕ(u2 ) . . . ϕ(uj )
(1.2.6)
Now if j < m then both sides of v1 u1 . . . uj−1 = u1 . . . uj have length less than m and so the
conditions of the lemma force the opposite of (1.2.6). Hence j = m and the result follows by (1.2.2)
and (1.2.6). Note that u1 6= v1 since v1 u1 . . . um−1 = u1 . . . um and u1 . . . um is reduced.
4
We can now give the proof:
Proof of Proposition 1.2.1. To show that ϕ̃ exists and is unique it is enough to show that
ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ) = ϕ(u1 )ϕ(u2 ) . . . ϕ(um )
whenever u1 u2 . . . um and v1 v2 . . . vm are two reduced expressions for some w ∈ Symn . We prove
this by contradiction. So assume that there exists some w ∈ Symn with two reduced expressions
u1 u2 . . . um = v1 v2 . . . vm but ϕ(u1 )ϕ(u2 ) . . . ϕ(um ) 6= ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ). Moreoever, assume
that w has minimal length amongst all such elements (so that if `(v) < `(w) then ϕ̃(v) is well
defined). Since w has minimal length the conditions of the above lemma are satisfied and so we
have v1 6= u1 and
v1 u1 u2 . . . um−1 = u1 u2 . . . um
but
ϕ(v1 )ϕ(u1 ) . . . ϕ(um−1 ) 6= ϕ(u1 )ϕ(u2 ) . . . ϕ(um )
(1.2.7)
Now the lemma applies to (1.2.7) to yield:
u1 v1 u1 u2 . . . um−2 = v1 u1 . . . um−1
but
ϕ(u1 )ϕ(v1 )ϕ(u1 )ϕ(u2 ) . . . ϕ(um−2 ) 6= ϕ(v1 )ϕ(u1 ) . . . ϕ(um−1 )
Continuing in this fashion yields:
u1 v1 u1 . . . = v1 u1 v1 . . .
| {z } | {z }
(1.2.8)
ϕ(u )ϕ(v1 )ϕ(u1 ) . . . 6= ϕ(v1 )ϕ(u2 )ϕ(v1 ) . . .
| 1
{z
} |
{z
}
(1.2.9)
m
m
But:
m
m
It is easily verified that if u1 = (i, i + 1) and v1 = (j, j + 1) then u1 v1 has order 3 if |i − j| = 1 and
2 if |i − j| ≥ 2. Hence if |i − j| = 1 by (1.2.8) we must have that m is a multiple of 3 so that (1.2.9)
contradicts the first relation in the proposition. On the other hand if |i − j| ≥ 2 we must have that
m is a multiple of 2 which contradicts the second relation. Hence if u1 u2 . . . um and v1 v2 . . . vm are
two reduced expressions for w then
ϕ(u1 )ϕ(u2 ) . . . ϕ(um ) = ϕ(v1 )ϕ(v2 ) . . . ϕ(vm )
and so we can define ϕ̃(w) = ϕ(u1 )ϕ(u2 ) . . . ϕ(um ) and the result follows.
This result has the following important corollary:
Corollary 1.2.3. If u1 u2 . . . um and v1 v2 . . . vm are two reduced expressions for some element
w ∈ Symn then it is possible to obtain one from the other using only the braid relations.
Proof. Let M be the monoid on generators mi with 1 ≤ i < n subject to the relations:
mi mi+1 mi = mi+1 mi mi+1
mi mj = mj mi
if |i − j| ≥ 2
(1.2.10a)
(1.2.10b)
Now define ϕ : S → M by ϕ(si ) = mi . By construction ϕ satisfies the hypotheses of the proposition
and hence ϕ̃ : Symn → M exists. Now let u1 u2 . . . um and v1 v2 . . . vm be two reduced expressions
for w ∈ Symn . Then:
ϕ(u1 )ϕ(u2 ) . . . ϕ(um ) = ϕ̃(w) = ϕ(v1 )ϕ(v2 ) . . . ϕ(vm )
5
Hence it is possible to obtain ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ) from ϕ(u1 )ϕ(un ) . . . ϕ(um ) using only the relations in (1.2.10). However, the relations in (1.2.10) are direct copies of the braid relations in Symn
and so ϕ(u1 )ϕ(u2 ) . . . ϕ(um ) = ϕ(v1 )ϕ(v2 ) . . . ϕ(vm ) if and only if it is possible to obtain v1 v2 . . . vm
from u1 u2 . . . um in Symn using only the braid relations.
As promised, we now have:
Theorem 1.2.4. The symmetric group is generated by the simple transpositions subject to the
relations:
s2i = 1
(1.2.11a)
si si+1 si = si+1 si si+1
if |i − j| ≥ 2
si sj = sj si
(1.2.11b)
(1.2.11c)
Proof. We argue that any relation of the form r1 r2 . . . rm = 1 with ri ∈ S is a consequence of the
given relations. We induct on m, with the case m = 0 being obvious. So assume that r1 r2 . . . rm = 1
holds in Symn . Then since `(1) = 0 we cannot have `(r1 r2 . . . rk ) < `(r1 r2 . . . rk rk+1 ) for all
k. Fix the smallest k with `(r1 r2 . . . rk ) > `(r1 r2 . . . rk+1 ) so that r1 r2 . . . rk is reduced. Letting
x = r1 r2 . . . rk we can apply Corollary 1.1.7 to conclude that x has a reduced expression u1 u2 . . . uk
with uk = rk+1 . Now r1 r2 . . . rk and u1 u2 . . . uk are both reduced expressions for x and so, from
the above corollary, we can obtain u1 u2 . . . uk from r1 r2 . . . rk using only the braid relations. Hence
we can use the relations to write:
r1 r2 . . . rm = r1 r2 . . . rk rk+1 . . . rm = u1 u2 . . . uk rk+1 . . . rm = u1 u2 . . . uk−1 rk+2 . . . rm
The last step follows since uk = rk+1 and so uk rk+1 = u2k = 1 by relation (i). Hence, using the
relations, we have reduced the number of terms in the relation by 2 and we can conclude that
r1 r2 . . . rm = 1 is a consequence of the given relations by induction on m.
1.3
The Bruhat Order
In this section we introduce a useful partial order on Symn . If v, w ∈ Symn with `(v) < `(w)
write v . w if there exists t ∈ T such that v = wt. We then write u < w if there exists a chain
u = v1 . v2 . · · · . vm = w. We write u ≤ w if u < w or u = w. The resulting relation is clearly
reflexive, anti-symmetric and transitive and hence is a partial order. It is called the Bruhat Order.
It is an immediate consequence of the definition that if r ∈ S then either wr < w or wr > w. The
following is another useful property of the Bruhat order:
Lemma 1.3.1. If v, w ∈ Symn with v ≤ w and r ∈ S, then either vr ≤ w or vr ≤ wr (or both).
Proof. First assume that v . w with `(wr) > `(vr). Since v . w there exists t ∈ T with v = wt.
Hence vr = wr(rtr) with `(vr) < `(wr) (by assumption) and rtr ∈ T . Hence vr . wr.
The other possibility if v . w is `(wr) ≤ `(vr). Since v = wt for some t ∈ T , `(w) − `(v) ≡ 1
(mod 2) (by Lemma 1.1.4). Since `(wr) = `(w) ± 1 and `(vr) = `(v) ± 1 the only way we can have
`(wr) ≤ `(vr) is if `(w) = `(v) + 1, `(wr) = `(w) − 1 and `(vr) = `(v) + 1. Now since `(wr) < `(w)
Corollary 1.1.7 applies and we can conclude that w has a reduced expression si1 si2 . . . sim such that
sim = r. We have v = wt with `(v) < `(w) and so the exchange condition applies and there exists
6
a k such that v = wt = si1 si2 . . . sc
ik . . . sim . But `(vr) > `(v) and sim = r and so we must have
k = m. Hence v = si1 si2 . . . sim−1 . But then vr = w and so vr ≤ w.
Now if v ≤ w then either v = w or there exists a chain v = u1 . u2 . · · · . un = w. If v = w then
vr = wr and so the result is obvious. On the other hand if there exists such a chain then from
above we have either u1 r ≤ u2 or u1 r ≤ u2 r. If u1 r ≤ u2 then vr = u1 r ≤ u2 . · · · . un = w and
so vr ≤ w. If u1 r ≤ u2 r then either u2 r ≤ u3 or u2 r ≤ u3 r. Continuing in this manner we obtain
a new chain vr ≤ u2 r ≤ . . . which either ends in w or wr. Hence vr ≤ w or vr ≤ wr.
The definition of the Bruhat order given above makes it immediately clear that the resulting relation
is a partial order. However it is possible to give a more practical criterion. If r1 r2 . . . rm is a reduced
expression for some w ∈ Symn , a subexpression of r1 r2 . . . rm is an expression ri1 ri2 . . . ris where
i1 , i2 , . . . is is a subsequence of 1, 2, . . . , m satisfying i1 < i2 · · · < is . In other words, ri1 ri2 . . . ris is
a subexpression of r1 r2 . . . rm if it is possible to obtain ri1 ri2 . . . ris from r1 r2 . . . rm by ‘crossing out’
various terms. The following proposition shows that the Bruhat order can be entirely characterised
in terms of subexpressions:
Proposition 1.3.2. Let u, v ∈ Symn . Then u ≤ v in the Bruhat order if and only if an expression
for u can be obtained as a subexpression of some reduced expression for v.
Proof. If u = v then the if and only if condition is clear. So assume that u < v. Then there exists
a sequence u = w1 . w2 . · · · . wm = v with `(wi ) < `(wi+1 ) for all i and wi = wi+1 ti for some
ti ∈ T . Now let si1 si2 . . . sim be a reduced expression for v and tm ∈ T such that wm−1 = vtm .
Then the exchange condition applies and so there exists a k such that wm−1 = si1 si2 . . . sc
ik . . . sim .
We can repeat this argument with wm−1 in place of wm to conclude that there exists a k 0 such
that wm−2 = si1 si2 . . . sc
c
ik . . . sc
ik0 . . . sim (or si1 si2 . . . sc
ik0 . . . s
ik . . . sim ). Continuing in this fashion
we obtain an expression for u as a subexpression of si1 si2 . . . sim .
For the other implication assume that uj1 uj2 . . . ujp is a subexpression of a reduced expression
u1 u2 . . . um (so that uk = sik for some sequence ik ). We use induction on m to show that
uj1 uj2 . . . ujp ≤ u1 u2 . . . um (the cases m = 0 and m = 1 being obvious). If jp < m then by induction
uj1 uj2 . . . ujp ≤ u1 u2 . . . um−1 . Now `(u1 u2 . . . um−1 ) < `(u1 u2 . . . um ) since u1 u2 . . . um is reduced
and u1 u2 . . . um−1 = (u1 u2 . . . um )um and so u1 u2 . . . um−1 ≤ u1 u2 . . . um . Hence uj1 uj2 . . . ujp ≤
u1 u2 . . . um . If jp = m then uj1 uj2 . . . ujp−1 ≤ u1 u2 . . . um−1 by induction and by Lemma 1.3.1 there
are two possibilities: uj1 uj2 . . . ujp ≤ u1 u2 . . . um−1 ≤ u1 u2 . . . um or uj1 uj2 . . . ujp ≤ u1 u2 . . . um . In
either case we have uj1 uj2 . . . ujp ≤ u1 u2 . . . um .
This alternative description yields a useful corollary:
Corollary 1.3.3. Let u, v ∈ Symn . Then u ≤ v if and only if u−1 ≤ v −1 .
Proof. If u ≤ v then, from above, there exists a reduced expression r1 r2 . . . rm for v with u =
ri1 ri2 . . . rim occurring as a subexpression. But u−1 = rim rim−1 . . . ri1 and rm rm−1 . . . r1 is a reduced
expression for v −1 by Proposition 1.1.3 and so u−1 occurs as a subexpression of a reduced expression
for v −1 . Hence u−1 ≤ v −1 .
We finish this section with a useful lemma:
Lemma 1.3.4. Suppose that x, y ∈ Symn and that there exists r ∈ S such that xr < x and yr < y.
Then x ≤ y if and only if xr ≤ yr.
7
Proof. If xr < yr then either x ≤ yr ≤ y or x ≤ y by Lemma 1.3.1. On the other hand if
x ≤ y we can fix a reduced expression r1 r2 . . . rm for y such that rm = r (since yr < y). Now,
since x ≤ y we can use the above characterisation of the Bruhat order to conclude that there
exists i1 < i2 < · · · < ik such that x = ri1 ri2 . . . rik . Now, if ik = m then xr = ri1 ri2 . . . rik−1
is a subexpression or yr = r1 r2 r3 . . . rm−1 and so xr ≤ yr. On the other hand if ik 6= m then
ri1 ri2 . . . rik = x is a subexpression of r1 r2 r3 . . . rm−1 = yr and so xr ≤ x ≤ yr.
1.4
Descent Sets
We finish this chapter with a discussion of the left and right descent sets of a permutation. Though
straightforward, this is a concept that will emerge repeatedly in what follows.
Given a permutation w ∈ Symn we define the left descent set and right descent set of a permutation
w ∈ Symn to be the sets L(w) = {r ∈ S |rw < w} and R(w) = {r ∈ S |wr < w} respectively.
For example, in Sym3 we have L(id) = R(id) = ∅, L(s1 s2 ) = s1 , R(s1 s2 ) = s2 and L(s1 s2 s1 ) =
R(s1 s2 s1 ) = {s1 , s2 }. We can give an alternative characterisation of the right descent set in terms
of the string form:
Lemma 1.4.1. Let w = w1 w2 . . . wn ∈ Symn . Then si ∈ R(w) if and only if wi > wi+1 .
Proof. This is immediate from Lemma 1.1.1: we have that `(wsi ) < `(w) (and so wsi < w) if and
only if wi = w(i) > w(i + 1) = wi+1 .
The following gives a relation between the left and right descent sets:
Lemma 1.4.2. Let w ∈ Symn . Then R(w) = L(w−1 ).
Proof. Note that if wr ≤ w then wr < w since wr 6= w. Hence the statement wr ≤ w is equivalent
to wr < w. Now let r ∈ S. Then, by Corollary 1.3.3, wr ≤ w if and only if (wr)−1 = rw−1 ≤ w−1 .
Hence r ∈ R(w) if and only if r ∈ L(w−1 ).
The last lemma of this section is useful in later arguments:
Lemma 1.4.3. Suppose that r ∈ S and w ∈ Symn .
(i) If rw > w then R(rw) ⊃ R(w).
(ii) If wr > w then L(wr) ⊃ L(w).
Proof. For (i) note that if si ∈ R(w) then w has a reduced expression ending in si by Corollary
1.1.7. Since rw > w left multiplying such an expression by r yields a reduced expression for rw.
Hence si ∈ R(rw). For, (ii) note that rw−1 > w−1 by Corollary 1.3.3 and hence L(w) = R(w−1 ) ⊂
R(rw−1 ) = L(wr) (using Lemma 1.4.2 above).
1.5
Notes
1. All the proofs of the first section are, as far as we are aware, original. They are motivated by
interpreting the symmetric group as a group of diagrams on 2n dots which group multiplication given by concatenation. For example, under this interpretation, our original definition
of the length of a permutation w counts the number of crossings in a diagram of w.
8
2. The “universal property” of the symmetric group proved in Section 1.2 is stated in the more
general case of Coxeter groups in Dyer [4]. We follow Dyer’s argument closely.
3. The definition of the Bruhat order, as well as the proof of the equivalence to the more familiar
subexpression condition, is adapted from an elegant argument of Humphreys [15].
4. Almost all the results in this section, including the strong exchange condition, the universal
property and the definition of the Bruhat order occur in the much more general setting of
Coxeter groups. See, for example, Bourbaki [3] or Humphreys [15].
9
2
Young Tableaux
In this section we introduce Young Tableaux, which offer a means to discuss the combinatorics of
the symmetric group. The central result of the section is the Robinson-Schensted correspondence,
which gives a bijection between elements of the symmetric group and pairs of standard tableaux of
the same shape. We also prove the symmetry theorem and introduce Knuth equivalence, tableau
descent sets and the dominance order on partitions.
2.1
Diagrams, Shapes and Standard Tableaux
A partition is a weakly decreasing sequence λ = (λ1 , λ2 , . . . ) with finitely many non-zero terms.
The λi ’s are the parts of λ. The weight of a partition, denoted |λ|, is the sum of the parts. If
|λ| = n we say that λ is a partition of n. The length of a partition λ, denoted l(λ), is the number of
non-zero parts. We will often denote a partition λ by (λ1 , λ2 , . . . , λm ) where m = l(λ) and λi = 0
for i > l(λ). Given a partition λ of n there is an associated Ferrer’s diagram (or simply diagram)
of λ consisting of l(λ) rows of boxes in which the ith row has λi boxes. For example, λ = (5, 4, 2)
is a partition of 11 and its associated diagram is:
If λ is a partition of n, a tableau of shape λ is a filling of the diagram of λ with positive integers
without repetition such that the entries increase from left to right along rows and from top to
bottom down columns. If T is a tableau we write Shape(T ) for the underlying partition. The
tableau is standard if the entries are precisely {1, 2, . . . n}. For example, the following are tableaux,
with T standard:
1 2 7 9
1 3 4 5
3
8
S=
T = 2 6
4
7
6
We have Shape(S) = (4, 2, 1, 1) and Shape(T ) = (4, 2, 1). The size of a tableau is the number of
boxes. Entries of a tableau are indexed by their row and column numbers (with rows indexed from
top to bottom and columns from left to right). For example, with T as above we have T13 = 4.
2.2
The Row Bumping Algorithm
Given a tableau T and a positive integer x not in T the Row Bumping Algorithm provides a means
of inserting x into T to produce a new tableau denoted T ← x. The algorithm is as follows: If x is
greater than all the elements of the first row then a new box is created right of the first row and x
is entered. Otherwise, x “bumps” the first entry greater than x into the second row. That is, the
smallest r is located such that T1r > x and then the entry x2 of T1r is replaced by x. The same
procedure is repeated in the second row with x2 in place of x. This process is repeated row by row
until a new box is created.
For example, with S as above, inserting 5 into S bumps the 7 from the first row, which in turn
10
bumps the 8 into the third row in which a new box is created:
1 2 7 9
3 8
←5=
4
6
1 2 5 9
3 7
4 8
6
A row insertion T ← x determines a bumping sequence, bumping route and new box : the bumping
sequence is the sequence x = x1 < x2 < · · · < xp of entries which are bumped from row to row; the
bumping route is the set of locations in which bumping takes place (or alternatively the location
of elements of the bumping sequence in T ← x); and the new box is the location of the box
created by inserting x. In the example above the bumping sequence is 5, 7, 8, the bumping route is
{(1, 3), (2, 2), (3, 2)} and the new box is (3, 2).
If (i, j) and (k, l) are locations in a tableau, (i, j) is strictly left of (k, l) if j < l and weakly left if
j ≤ l. Similarly (i, j) is strictly below (k, l) if i < k and weakly below if i ≤ k. If x and y are entries
in a tableau we say that x is left (or below) y if this is true of their locations. The following is a
technical but important lemma:
Lemma 2.2.1. (Row Bumping Lemma) The bumping route of T ← x moves to the left. That is,
if xi and xi+1 are elements in the bumping sequence then xi+1 is weakly left of xi . Furthermore,
if x < y then the bumping route of T ← x is strictly left of the bumping route of (T ← x) ← y
and the bumping route of (T ← y) ← x is weakly left of the bumping route of T ← y. Hence the
new box of T ← x is strictly left and weakly below the new box of (T ← x) ← y and the new box of
(T ← y) ← x is strictly below and weakly left of the new box of T ← y.
Proof. If the new box of T ← x is in the first row then there is nothing to prove. So assume that
the new box of T ← x is not in the first row. Let (i, j) be an element of the bumping route of
T ← x such that (i, j) is not equal to the new box of T ← x. Let y be the element bumped from
row i. Then, if Ti+1,y is an entry of the tableau then y < Ti+1,y (since Ti+1,j is immediately below
y in T) and so y bumps an element to the left of (i + 1, j). On the other hand, if Ti+1,j is not an
entry of the tableau then y either creates a new box left of (i, j) or bumps an entry left of (i, j)
since, in this case, all locations of the (i + 1)st row are left of (i, j). Hence y bumps an element or
creates a new box to the left of (i, j). Repeating this argument row by row shows that the bumping
route moves to the left.
Let x = x1 , x2 , . . . , xm be the bumping sequence of T ← x, let y = y1 , y2 , . . . , yn be the bumping
sequence of (T ← x) ← y and let p be the minimum of m and n. Since x < y the entry bumped by
x must be less than the entry bumped by y. Hence x2 < y2 . Repeating this row by row shows that
xi < yi for all i ≤ p. Since xi and yi are in the same row for all i ≤ p the bumping route of T ← x
is strictly left of the bumping route of (T ← x) ← y. This implies that the new box of T ← x is
weakly below and strictly left of the new box of (T ← x) ← y.
Now let y1 , y2 , . . . , yk be the bumping sequence of T ← y, let x1 , . . . xj be the bumping sequence of
(T ← y) ← x and let p be as above. Again, assume p > 1. Now x either bumps an entry to the left
of y (in which case x2 < y1 < y2 ) or x bumps y (in which case x2 = y < y2 ). We can repeat this
argument from row to row to conclude that xi < yi for all i ≤ p. Since we might have xi+1 = yi
for some i ≤ p we can only conclude that the bumping sequence of T (← y) ← x is weakly left of
T ← y. Hence the new box of (T ← y) ← x is strictly below and weakly left of the new box of
T ← y.
11
Since the bumping route moves to the left we know that T ← x has the shape of a partition. Also,
if x = x1 < x2 < · · · < xn is the bumping sequence and xi is in the (i, j)th position in T then we
know that xi is greater than all the entries to the left of (i, j) in the ith row. Since the bumping
route moves to the left we know, in T ← x, that xi lies to the left of (i, j) in the (i + 1)st row.
Hence the entry immediately above xi in T ← x is less than xi . These observations confirm that if
T is a tableau then so is T ← x.
We now state another technical property of the row bumping algorithm. If T is a tableau (or a
partition) an outside corner of T is a box (i, j) such that neither (i + 1, j) nor (i, j + 1) are boxes
of T . Deleting an outside corner produces another tableau. If the largest entry of T is n then n
must lie at the end of a row and at the base of a column and hence must occupy an outside corner.
Hence deleting the largest entry from a tableau always produces a tableau.
Lemma 2.2.2. Let T be a tableau containing a maximal element n and let x < n be such that x
is not in T . Then inserting x into T and then deleting n yields the same tableau as deleting n and
then inserting x.
Proof. If n is not an element of the bumping sequence x = x1 < x2 < · · · < xm of T ← x then the
result is clear. If n is an an element of the bumping sequence then we must have n = xm since n is
maximal. Now let T 0 be the tableau obtained from T by deleting n. Then the bumping sequence
of T 0 ← x is x1 < x2 < · · · < xm−1 . Hence T ← x and T 0 ← x only differ by the box containing n
and the result follows.
2.3
The Robinson-Schensted Correspondence
A word without repetitions (from now on simply word ) is a sequence of positive integers without
repetition. If w = w1 w2 . . . wi is a word there is a corresponding tableau, denoted ∅ ← w, obtained
by successively inserting the elements of w starting with the empty tableau. In symbols:
∅ ← w = (. . . ((∅ ← w1 ) ← w2 ) . . . ) ← wn
For example if w = 4125 then:
∅ ← w = (((∅ ← 4) ← 1) ← 2) ← 5 = (( 4 ← 1) ← 2) ← 5 = ( 1 ← 2) ← 5 = 1 2 ← 5 = 1 2 5
4
4
4
The Robinson-Schensted Correspondence gives a correspondence between permutations and pairs of
standard tableaux of the same shape. Given w = w1 w2 . . . wn ∈ Symn the algorithm is as follows:
Let P (0) = Q(0) = ∅ and recursively define P (i) and Q(i) by:
1) P (i+1) = P (i) ← wi+1
2) Q(i+1) is obtained from Q(i) by adding a box containing i + 1 in the location of the new box
of P (i) ← wi+1 .
Thus P (n) = ∅ ← w and Q(n) records the order in which the new boxes of P (n) are created. We then
set P (w) = P (n) and Q(w) = Q(n) . P (w) is referred to as the P -symbol or insertion tableau and
Q(w) is referred to as the Q-symbol or recording tableau. We write w ∼ (P (w), Q(w)) to indicate
the w corresponds to (P (w), Q(w)) under the Robinson-Shensted correspondence.
12
For example if w = 45132 we have:
P (1) = 4
P (2) = 4 5
P (3) = 1 5
4
P (4) = 1 3
4 5
Q(1) = 1
Q(2) = 1 2
Q(3) = 1 2
3
Q(4) = 1 2
3 4
Hence w = 45132 ∼ (
1 2
3 5
4
,
1 2
3 4
5
1
P (5) = 3
4
1
Q(5) = 3
5
2
5
2
4
).
Note from the construction Q(i) always has the same shape as P (i) . Also, since Q(i+1) is obtained
from Q(i) by adding i + 1 either in a new row at the bottom or at the end of a row it is clear
that Q(i) is always a standard tableau. Lastly, since w is a permutation w = w1 w2 . . . wn consists
of all the integers {1, 2, . . . , n} and so P is standard. Hence, given a permutation w ∈ Symn
the Robinson-Schensted correspondence does indeed yield a pair of standard tableaux of the same
shape.
Theorem 2.3.1. The Robinson-Schensted correspondence between elements w ∈ Symn and pairs
(P, Q) of standard tableaux of size n and the same shape is a bijection.
Proof. Notice that, if we are given T ← x together with the location of the new box of T ← x we
can recover T and x uniquely. We start at the new box (i, j) and label its entry xi . If i = 1 then
x = x1 and T is the tableau obtained by deleting the last box of the first row of T ← x. Otherwise
we have i > 1 and so xi must have been bumped by an element in the row above. Now the only
element which could have bumped xi is the largest entry smaller than xi . Label this entry xi−1 .
Continuing in this fashion we regain the bumping sequence x1 < x2 < · · · < xi . Hence we have
x = x1 and T is the tableau formed by shifting each xi up one row into the location previously
occupied by xi−1 . This process is known as the reverse bumping algorithm.
Hence, given (P, Q) we fix the location (i, j) of n in Q. Then the remarks above show that there is
only one possible P (n−1) and xn such that both P = P (n−1) ← xn and the new box of P (n−1) ← xn
is (i, j). Deleting the (i, j)th element of Q we obtain Q(n−1) . Repeating this argument with P (n−1)
and Q(n−1) in place of P and Q we uniquely obtain P (n−2) , Q(n−2) and xn−1 . Each time we repeat
this procedure we remove one of the elements from P (i) and obtain a new pair of tableau P (i−1) ,
Q(i−1) and a uniquely determined integer xi . Upon completion we are left with a string x1 x2 . . . xn
of entries of P in some order. Since at every iteration the xi is uniquely determined we conclude that
x1 x2 . . . xn is the only possible permutation corresponding to (P, Q). Hence we have demonstrated
an inverse with domain all pairs of standard tableaux of size n and the same shape and so the
Robinson-Schensted correspondence is a bijection.
2.4
Partitions Revisited
One of the most beautiful properties of the Robinson-Schensted correspondence is the symmetry
theorem which we will prove using the concept of the growth diagram. However, before we can
introduce the growth diagram we need to develop some notation surrounding partitions.
There are a number of set theoretic concepts associated with partitions which can be interpreted intuitively by considering the corresponding diagrams. If λ = (λ1 , λ2 , . . . , λn ) and µ =
13
(µ1 , µ2 , . . . , µm ) are partitions then we define (λ ∪ µ)i = max{λi , µi } for 1 ≤ i ≤ max{m, n}. For
example:
if λ =
and µ =
then λ ∪ µ =
We say that µ ⊆ λ if m ≤ n and µi ≤ λi for all i ≤ m. Geometrically, µ ⊆ λ means that the
diagram of µ “sits inside” λ. Clearly if λ ⊆ µ and λ and µ both partition the same n then λ = µ.
If µ ⊆ λ then the skew diagram, denoted λ\µ, is the diagram obtained by ommiting the squares of
µ from λ:
if µ =
and λ =
then λ\µ =
Now call a chain of partitions ∅ = λ0 ⊂ λ1 ⊂ · · · ⊂ λn saturated if each skew diagram λi+1 \λi contains precisely one square. There is an obvious bijection between saturated chains of partitions ∅ =
λ0 ⊂ λ1 ⊂ · · · ⊂ λn and standard tableaux of size n: we simply number the squares of λn in the order
in which they appear in the chain. For example the chain ∅ ⊂ ⊂
⊂
⊂
⊂
⊂
corresponds to the tableau
1 2 4
3 6
5
. Also, since the largest element of a tableau must occupy an out-
side corner (see the remarks prior to Lemma 2.2.2) it is clear that the shape of T with the largest
element removed is also the diagram of a partition. Hence all tableaux can be associated with a
saturated chain of partitions by taking the shapes of tableaux obtained by successively deleting the
largest entry.
With the correspondence between chains of partitions and standard tableaux we can give alternate
descriptions of the P and Q-symbols of a permutation:
Lemma 2.4.1. Let w = w1 w2 . . . wn ∈ Symn . Let P (i) = ∅ ← w1 w2 . . . wi be as in Section 2.3.
Also, let w(j) be the word obtained from w by omitting elements greater than j. Then:
(i) Shape(∅ ← w(1)) ⊂ Shape(∅ ← w(2)) ⊂ · · · ⊂ Shape(∅ ← w(n)) corresponds to P (w).
(ii) Shape(P (1) ) ⊂ Shape(P (2) ) ⊂ · · · ⊂ Shape(P (n) ) corresponds to Q(w).
Proof. Let i be the location of n in w so that w = w1 w2 . . . wi−1 nwi+1 . . . wn . Then an immediate
corollary of Lemma 2.2.2 is that if T = ∅ ← w1 w2 . . . wi−1 n then deleting n from T and then
inserting wi+1 . . . wn produces the same result as inserting wi+1 . . . wn and then deleting n. Hence
the P -symbol of w with n deleted is the same as ∅ ← w(n − 1). Hence the box containing n in P is
the unique box in Shape(∅ ← w(n))\Shape(∅ ← w(n − 1)). Hence we have (i) by induction on n.
In Section 2.3 we defined we defined Q(i+1) to be obtained from Q(i) be adding i + 1 in the
new box of P (i) ← wi+1 . Hence Q(1) corresponds to ∅ ⊂ Shape(P (1) ), Q(2) corresponds to ∅ ⊂
Shape(P (1)) ⊂ Shape(P (2) ) etc. Hence Q(w) corresponds to ∅ ⊂ Shape(P (1) ) ⊂ Shape(P (2) ) ⊂
· · · ⊂ Shape(P (n) ).
2.5
Growth Diagrams and the Symmetry Theorem
We now describe the construction of the growth diagram. Given a permutation w = w1 w2 . . . wn ∈
Symn we can associate an n × n array of boxes in which we label the box in the ith column and
with row with an X (with rows indexed from bottom to top and columns from left to right). For
example if w = 452613 ∈ Sym6 then the associated array looks like:
14
X
X
X
X
X
4
5
2
6
X
1
3
Now define the (i, j)th partial permutation, denoted w(i, j), to be the word formed by reading off
the column number of each X below and left of (i, j) from left to right. By convention, if either i
or j is 0 then w(i, j) = ∅. In our example the (5, 4)th partial permutation is 452:
X
X
X
X
X
4
5
2
6
X
1
3
Now define the growth diagram of w to be the n × n array with X’s inserted as above, in which
the (i, j)th entry contains the shape of the tableau formed by row-bumping the (i, j)th partial
permutation into the empty set. It is useful to also include the base and left hand side of the array
(that is, those locations with i = 0 or j = 0) in the growth diagram and label them with the empty
set. Using the convention mentioned above (that w(0, j) = w(i, 0) = ∅) we have that the (i, j)th
entry is Shape(∅ ← w(i, j)) for 0 ≤ i, j ≤ n.
In our example we have ∅ ← 452 =
The full growth diagram looks like:
2 5
4
and so the (5, 4)th entry of the growth diagram of w is
∅
X
∅
X
∅
X
∅
∅
∅
X
∅
∅
∅
X
∅
∅
∅
∅
∅
X
∅
∅
∅
∅
15
∅
∅
.
Given the growth diagram of a permutation w ∈ Symn we can immediately recover the P and Qsymbols of w. If r(i, j) denotes the diagram in the ith row and j th column then by Lemma 2.4.1 we
have that r(0, n) ⊂ r(1, n) ⊂ · · · ⊂ r(n, n) corresponds to P (w) and r(n, 0) ⊂ r(n, 1) ⊂ . . . r(n, n)
corresponds to Q(w).
Notice that, in our example, the partitions “grow” upwards and rightwards. That is, if λ is above
and right of µ then µ ⊆ λ. This is true in general and is the subject of the following lemma:
Lemma 2.5.1. Let r(i, j) denote the partition appearing in the ith row and j th column of the growth
diagram of w ∈ Symn . Then if i ≤ k and j ≤ l then r(i, j) ⊆ r(k, l).
Proof. It is enough to show that r(i, j) ⊆ r(i + 1, j) and r(i, j) ⊆ r(i, j + 1) for all 1 ≤ i, j ≤ n.
If w(i, j) = w1 w2 . . . wk then w(i, j + 1) = w(i, j) or w(i, j + 1) = w1 w2 . . . wk s for some s ≤ i
depending on whether there is an X in the same column and below (i, j + 1). In the first case
r(i, j) = r(i, j + 1) and in the second case we have:
r(i, j) = Shape(∅ ← w1 w2 . . . wk ) ⊂ Shape(∅ ← w1 w2 . . . wk s) = r(i, j + 1)
Hence, r(i, j) ⊆ r(i, j + 1) for all 1 ≤ i, j ≤ n.
As above, if w(i, j) = w1 w2 . . . wk then w(i+1, j) = w(i, j) or w(i+1, j) = w1 w2 . . . wl (i+1)wl+1 . . . wk
for some l ≤ j depending on whether there is an X in the same row and left of (i + 1, j). In the first
case we have r(i + 1, j) = r(i, j). In the second case an immediate corollary to Lemma 2.2.2 gives
us that ∅ ← w1 w2 . . . wk is equal to ∅ ← w1 w2 . . . wl (i + 1)wl+1 . . . wk with i + 1 deleted. Hence
r(i, j) can be obtained from r(i + 1, j) by removing a box and so r(i + 1, j) ⊂ r(i, j).
Now consider the diagram r(i, j) which appears in the ith row and j th column. We want to develop
rules which relate r(i, j) to the diagrams to the left and below r(i, j) so that the growth diagram
can be constructed inductively. To simplify notation we will let ρ = r(i, j), µ = r(i, j − 1), λ =
r(i − 1, j − 1) and ν = r(i − 1, j):
i
···
µ
ρ
···
λ
ν
..
.
..
.
j
Assume first that the (i, j)th square does not contain an X. Then there are four possibilities
(throughout “column below” refers to the set of boxes below and in the same column as ρ and “row
to the left” refers to those boxes to the left and in the same row as ρ):
1) There are no X’s in the column below or the row to the left. Clearly λ = ν = µ and ρ = λ.
2) There is an X in the column below but not in the row to the left. In this case clearly ρ = ν.
Here the (i, j − 1)th partial permutation is the (i − 1, j)th partial permutation truncated by
one and so µ ⊂ ν. Hence we can equally well write ρ = ν ∪ µ.
3) There is an X in the row to the left but not in the column below. Here the (i − 1, j)th partial
permutation is the (i, j − 1)th word truncated by one and so ν ⊂ µ. As in (2) we have
ρ = µ ∪ ν.
16
4) There is an X both in the column below and in the row to the left. First assume that µ 6= ν.
Then we know that µ ⊂ ρ and ν ⊂ ρ and hence ν ∪ µ ⊂ ρ. But we also know that ρ has only
two more boxes than λ (by considering the size of the corresponding partial permutations).
Since µ 6= ν, µ ∪ ν has two more squares than λ and so we must have ρ = µ ∪ ν.
The most difficult case is when µ = ν. Let T (µ), T (λ) and T (ν) be the tableaux given by
∅ ← w(i, j − 1), ∅ ← w(i − 1, j − 1) and ∅ ← w(i − 1, j) respectively (that is, they are the
tableaux used to construct the growth diagram). We know (from Lemma 2.2.2) that T (µ)
with i deleted is equal to T (λ) and, in this case, the shape of T (ν) is equal to the shape of
T (µ). Hence the new box of T (λ) ← wk must be in the same location as i in T (µ). Hence
inserting wk into T (µ) places wk in the box previously occupied by i and bumps i into the
next row. Hence if s is the unique integer such that µs = λs + 1 (the row number of i in µ)
then ρt = µt if t 6= s + 1 and ρs+1 = µs+1 + 1 (since i is bumped into the next row).
If (i, j) contains an X then clearly µ = λ = ν. If w(i − 1, j − 1) = w1 w2 . . . wk then w(i, j) =
w1 w2 . . . wk i and hence the shape of ∅ ← w1 w2 . . . wk i is the shape of ∅ ← w1 w2 . . . wk with one box
added to the first row (since i is greater than all the wi ). Hence ρi = λi when i 6= 1 and ρ1 = λ1 + 1.
Hence we have the following “local rules” for ρ based only on λ, µ and ν:
1) If the (i, j)th square does not contain an X and λ = µ = ν then ρ = λ.
2) If the (i, j)th square does not contain an X and µ 6= ν then ρ = µ ∪ ν.
3) If the (i, j)th square does not contain an X and λ 6= µ = ν then let i be the unique integer
such that µi = λi + 1. Then ρj = µj if j 6= i + 1 and ρi+1 = µi+1 + 1.
4) If the (i, j)th square does contain an X then λ = µ = ν and ρj = λj if j 6= 1 with ρ1 = λ1 + 1.
The above discussion shows that these rules are equivalent to our previous definition of the growth
diagram. Hence, given a permutation we could equally well construct the array of X’s, label the
base and left hand side with ∅’s and use the local rules to construct the growth diagram.
The proof of this important theorem is now straightforward:
Theorem 2.5.2. (Symmetry Theorem) If w ∼ (P, Q) under the Robinson-Schensted correspondence then w−1 ∼ (Q, P ).
Proof. The importance of the local rules is that they are perfectly symmetrical in µ and ν. So if G
is the growth diagram of w, then reflecting G along the line y = x yields the growth diagram of w−1 .
Hence if r(i, j) corresponds to the (i, j)th partition in G then the P -symbol of w−1 corresponds to
r(n, 0) ⊂ r(n, 1) ⊂ · · · ⊂ r(n, n) which is the Q-symbol of w. Similarly the Q-symbol of w−1 is
equal to the P -symbol of w.
2.6
Knuth Equivalence
Using the language of tableaux it is possible to split elements of the symmetric group into equivalence classes based on their P or Q-symbol. These equivalence classes turn out to be of fundamental
importance when we come to look at representations of the Hecke algebras. For this reason it is
essential to be able to decide when two elements of the symmetric group share a P or Q-symbol
without calculating their tableaux explicitly. The answer to this involves the notion of Knuth
equivalence. Throughout this section we will only concern ourselves with the problem of deciding
whether two permutations share a P -symbol. The theory that we develop below combined with
the symmetry theorem answers the analogous question for Q-symbols.
Let w = w1 w2 . . . wm be a word and let {x, y, z} with x < y < z be a set of three adjacent elements
17
in some order. An elementary Knuth transformation is a reordering of w using one of the following
rules:
. . . zxy . . . ↔ . . . xzy . . .
. . . yxz . . . ↔ . . . yzx . . .
(where the rest of w remains unchanged). We say that two words v and w are Knuth equivalent,
and write u ≡ w, if one can be transformed into the other by a series of the elementary Knuth
transformations. The following proposition shows that Knuth equivalent words share the same
P -symbol.
Proposition 2.6.1. Let u and v be words with u ≡ v. Then P (u) = P (v).
Proof. It is enough to show that elementary Knuth transformation does not alter the P -symbol.
This is equivalent to showing that if T is a tableau not containing x, y or z then T ← zxy = T ← xzy
and T ← yzx = T ← yxz. We show this by induction on the number of rows of T .
So assume that T has one row. We show that T ← zxy = T ← xzy. One can show T ← yxz =
T ← yzx in the case when T has one row by a similar examination of cases. Now, label the entries
of T as t1 < t2 < · · · < tm . There are seven possibilities which we examine case by case:
1) tm < x < y < z:
. . . tm x y
T ← zxy = t1 t2
= T ← xzy
ti z
2) ti−1 < x < ti < · · · < tm < y < z:
. . . x . . . tm y
= T ← xzy
T ← zxy = t1 t2
ti z
3) ti−1 < x < ti < · · · < tj−1 < y < tj < · · · < tm < z:
T ← zxy =
t1 t 2 . . . x . . . y . . . tm z
= T ← xzy
ti t j
4) ti−1 < x < ti < · · · < tj−1 < y < tj < · · · < tk−1 < z < tk < · · · < tm :
t 1 t2 . . . x . . . y . . . z . . . tm
T ← zxy = ti tj
= T ← xzy
tk
5) ti−1 < x < y < ti < · · · < tk−1 < z < tk < · · · < tm :
t1 t2 . . . x y . . . z . . . tm
T ← zxy = ti tj
= T ← xzy
tk
(tj := ti+1 )
6) ti−1 < x < ti < · · · < tk−1 < y < z < tk < · · · < tm :
t1 t 2 . . . x . . . y . . . tm
T ← zxy = ti z
= T ← xzy
tk
18
7) ti−1 < x < y < z < ti :
t1 t2 . . . x y . . . tm
= T ← xzy
T ← zxy = z tj
ti
(tj := ti+1 )
Hence the result is true if T has one row.
Now assume that T has r rows. Let R be the first row and S be the tableau obtained from T by
deleting the first row. Let w be the word bumped into the next row by R ← zxy and w0 be the
corresponding word for R ← xzy. Then, by examining the above cases there are three possibilities.
In cases 1, 2 and 3 we have w = w0 and so S ← w = S ← w0 and hence T ← zxy = T ← xzy. In
cases 4, 5 and 6, w has the form z 0 x0 y 0 with x0 < y 0 < z 0 and w0 = x0 z 0 y 0 and hence, by induction,
S ← w = S ← w0 forcing T ← zxy = T ← xzy. Finally, in case 7, w has the form y 0 x0 z 0
with x0 < y 0 < z 0 and w0 = y 0 z 0 x0 . Hence we still have S ← w = S ← w0 by induction. That
T ← yxz = T ← yzx in the general case follows by a similar induction.
Given a tableau T we form the tableau word, denoted w(T ), by reading off the entries of the tableau
from left to right and bottom to top. For example if:
1 3 4 5
T = 2 7 9
6 8
Then w(T ) = 682791345. Given a tableau word w1 w2 . . . wn we can recover the rows of the tableau
by splitting the word into its increasing sequences. In the example above we split up w(T ) as
68|279|1345 from which we regain the tableau’s rows in order from bottom to top. The following
lemma justifies our choice of the order in which the tableau word is formed:
Lemma 2.6.2. If T is a tableau then T = ∅ ← w(T ).
Proof. This is just a matter of examining the row bumping algorithm as w(T ) is inserted. Let
w(T ) = v1 v2 . . . va |va+1 . . . vb | . . . | . . . vm be the tableau word of T broken up into increasing sequences. Then:
∅ ← v1 v2 . . . va = v1 v2 . . . va
Now, va+1 < v1 , va+2 < v2 , etc. and so (with vi = va+1 and vj = v2a ):
. . . v . . . vb
∅ ← v1 v2 . . . va va+1 . . . vb = vi . . . j
v1
va
This process continues: each successive insertion of an increasing sequence shifts all the rows down
and places the new row on top, until the original tableau is regained.
Lemma 2.6.3. If T is a tableau then w(T ← v) ≡ w(T )v.
Proof. First consider the case when T has one row so that w(T ) = w1 w2 . . . wm with w1 < w2 <
· · · < wm . If v is larger than all of the first row then w(T ← v) = w1 w2 . . . wm v = w(T )v and so
19
w(T ← v) ≡ w(T )v. So assume that v bumps wi so that wj > v for all j ≥ i and wj < v if j < i.
Then:
w(T )v = w1 w2 . . . wi−1 wi wi+1 . . . wm−1 wm v
≡ w1 w2 . . . wi−1 wi wi+1 . . . wm−1 vwm
(yzx ↔ yxz)
≡ w1 w2 . . . wi−1 wi vwi+1 . . . wm−1 wm
(yzx ↔ yxz)
≡ w1 w2 . . . wi wi−1 vwi+1 . . . wm−1 wm
(xzy ↔ zxy)
≡ wi w1 w2 . . . wi−1 vwi+1 . . . wm−1 wm
(xzy ↔ zxy)
= w(T ← v)
Now if T has more than one row, label the words associated with the rows of T as r1 , r2 . . . rp so
that w(T ) = rp rp−1 . . . r2 r1 . Label the rows of T ← v as r10 , r20 , . . . , rq0 (with q = p or q = p + 1) and
let v = v1 < v2 · · · < vr be the bumping sequence of T ← v. From above we have that ri vi ≡ vi+1 ri0
for all i ≤ r. Hence:
w(T )v = rp rp−1 . . . r2 r1 v
≡ rp rp−1 . . . r2 v2 r10
≡ rp rp−1 . . . v3 r20 r10
0
≡ rq0 rq−1
. . . r20 r10
= w(T ← v)
The above Lemma allows us to prove:
Theorem 2.6.4. Let u, v ∈ Symn . Then u ≡ v if and only if their P -symbols coincide.
Proof. We have already seen (Proposition 2.6.1) that u ≡ v implies that P (u) = P (v). So assume
that P (u) = P (v). Then repeated application of the above lemma shows that u ≡ w(∅ ← u).
Hence u ≡ w(∅ ← u) = w(∅ ← v) ≡ v and so u and v are Knuth equivalent.
2.7
Tableau Descent Sets and Superstandard Tableaux
Recall that in Chapter 1 we defined the left and right descent sets of a permutation w = w1 w2 . . . wn ∈
Symn as the sets L(w) = {r ∈ S |rw < w} and R(w) = {r ∈ S |wr < w}. We also showed in
Lemma 1.4.1 that si ∈ R(w) if and only if wi > wi+1 . We wish to introduce a similar concept for
tableaux. If P is a tableau, let D(P ) be the set of i for which i + 1 lies strictly below and weakly
left of i in P . We call this the tableau descent set. For example:
1 3 7 8
if P = 2 5 9
4 6
then D(P ) = {1, 3, 5, 8}
The following proposition shows that the left and right descent sets of a permutation w ∈ Symn
can be characterised entirely in terms of the descent sets of the P and Q-symbols of w:
Proposition 2.7.1. Let w ∈ Symn and suppose that w ∼ (P, Q):
(i) We have si ∈ L(w) if and only if i ∈ D(P ).
(ii) We have si ∈ R(w) if and only if i ∈ D(Q).
20
Proof. We prove (ii) first. This is a matter of reinterpreting the Row Bumping Lemma (Lemma
2.2.1). Fix i and let R = ∅ ← w1 w2 . . . wi−1 (with R = ∅ if i = 1). If si ∈
/ R(w) then wi < wi+1
(Lemma 1.4.1) and by the Row Bumping Lemma the new box of R ← wi is strictly left and weakly
below the new box of (R ← wi ) ← wi+1 . Thus i is strictly left and weakly below i + 1 in Q and so
i∈
/ D(Q). On the other hand if si ∈ R(w) then wi > wi+1 (Lemma 1.4.1 again) and the row Row
Bumping Lemma gives us that the new box of (R ← wi ) ← wi+1 is weakly left and strictly below
the new box of R ← wi and so i ∈ D(Q). Hence si ∈ R(w) if and only if i ∈ D(Q).
For (i), we have si ∈ L(w) if and only if si ∈ R(w−1 ) (Lemma 1.4.2) which, by (ii), occurs if and
only if i ∈ D(Q(w−1 )) = D(P ) (since Q(w−1 ) = P (w) by the Symmetry Theorem).
It will be an important question in what follows as to what extent a tableau is determined by its
descent set. It is easy to see that two tableaux can have the same descent set and be unequal: for
example 13 24 and 13 2 4 both have descent set {2}. One might hope that the shape of a tableau
together with its descent set determines it uniquely. This also turns out to be false as can be seen
by considering the following two tableaux:
1 3 4
P = 2 7
5 8
6
1 3 4
Q= 2 5
6 7
8
Clearly P 6= Q but both tableaux have descent set {1, 4, 5, 7}.
Because of these problems we seek a suitable ‘test’ tableau P which has the property that any
tableau with the same shape and descent set must be equal to P . To this end, let λ be a partition
and define the column superstandard tableau of shape λ, denoted Sλ , to be the tableau obtained
from λ by filling the diagram of λ with 1, 2, . . . successively down each column. For example, if
λ = (3, 3, 2, 1) then the column superstandard tableau of shape λ is:
1 5 8
Sλ = 2 6 9
3 7
4
The following lemma shows that Sλ has our required ‘test’ property:
Lemma 2.7.2. Suppose Shape(P ) = λ and D(Sλ ) ⊂ D(P ) then P = Sλ .
Proof. We claim that there is only one way in which to fill a diagram of λ with {1, 2, . . . , n} in
order to satisfy the descent set condition. Let c1 ≥ c2 ≥ · · · ≥ cm be the column lengths of Sλ .
Then {1, 2, . . . , c1 − 1} ⊂ D(Sλ ) ⊂ D(P ) and so, in filling the diagram of λ, 2 must lie below 1, 3
must lie below 2, etc. So the first column must consist of 1, 2, . . . , c1 . We have now filled the first
column of λ and so c1 + 1 must lie in the first box of the second column. We can then repeat the
same argument to get that the second column of P must consist of c1 + 1, c1 + 2, . . . , c2 . Proceeding
in this fashion we see that there is only one way to fill a diagram of λ so that D(Sλ ) is a subset of
the descent set. Hence any tableau of shape λ and descent set contained in D(Sλ ) must be equal
to Sλ . Thus P = Sλ .
21
2.8
The Dominance Order
If λ and µ are partitions we say that λ is dominated by µ, and write λ E µ, if λ1 + λ2 + · · · + λk ≤
µ1 + µ2 + · · · + µk for all k ∈ N. Clearly, λ E λ for all partitions λ. If λ E µ E λ then
λ1 ≤ µ1 ≤ λ1 and so λ1 = µ1 . Similarly, λ1 + λ2 ≤ µ1 + µ2 ≤ λ1 + λ2 and so λ2 = µ2 . We can
continue in this fashion to see that λi = µi for all i and so λ = µ. Lastly, if λ E µ E π then
λ1 + · · · + λk E µ1 + · · · + µk E π1 + · · · + πk for all k and so λ E π. These calculations verify that
E is a partial order. We call E the dominance order. In general the dominance order is not a total
order: for example
and
are incomparable partitions of 6.
We give a more intuitive interpretation of the dominance order as follows. Recall that a box in a
partition or tableau is indexed by a pair (i, j) (where i is the number of rows from the top and j
is the number of columns from the left) and that an outside corner is a box (i, j) such that neither
(i + 1, j) nor (i, j + 1) are boxes of λ. Now define an inside corner as a location (i, j) which is not
a box of λ, such that either j = 1 and (i − 1, j) is a box of λ, i = 1 and (i, j − 1) is a box of λ,
or (i − 1, j) and (i, j − 1) are boxes of λ. For example, in the following diagram (of the partition
(5, 4, 2, 1, 1)) the outside corners are marked with an ‘o’ and the inside corners with an ‘i’:
o i
o i
o i
i
o
i
Given a partition λ, an inside corner (i, j) and an outside corner (k, l) strictly below (i, j) we let
λ0 be obtained from λ by deleting the box at (k, l) and inserting a box at (i, j). We say that
λ0 is obtained from λ by a raising operation. Intuitively, raising operations correspond to sliding
outside corners upwards into inside corners. For example, we can apply two raising operations to
the partition
and the outside corner (3, 1): corresponding to the inside corner (2, 2) we get
and corresponding to the inside corner (1, 3) we get
. The following diagram illustrates
that all partitions of 6 can be obtained from the partition (1, 1, 1, 1, 1, 1) = (16 ) by applying raising
operations:
/
/
:
vv
vvv
III
II
$
ss9
sss
LLL
LL&
KK
KK
K%
9
rr
rrr
MMM
MM&
pp7
ppp
/
/
It turns out that the dominance order can be entirely characterised in terms of raising operations:
Lemma 2.8.1. Let µ and λ be partitions of the same weight. Then µ E λ if and only if λ can be
obtained from µ by a sequence of raising operations.
Proof. If µ0 is obtained from µ by a raising operations then µ0i = µi + 1 and µ0j = µj − 1 for some
i < j and hence µ E µ0 . This shows that if λ can be obtained from µ by raising operations then
µ E λ.
22
P
For the opposite implication we induct on n =
|λk − µk |. If n = 0 then λ = µ and there is
nothing to prove. So assume that n > 0 and fix the first i for which µi < λi . Then µi < µi−1
(otherwise µi = µi−1 = λi−1 ≥ λi ) and so (i, µi + 1) is an inside corner. Since µi < λi and |µ| = |λ|
we must have either µj > λj for some j or l(µ) > l(λ).
If µj > λj for some j, fix the largest j for which this holds. Then µj > µj+1 (otherwise µj = µj+1 ≤
λj+1 ≤ λj ) and so (j, µj ) is an outside corner. Now, let µ0 be obtained from µ by performing a
raising operation from the outside corner (j, µj ) to the inside corner (i, µi + 1). Then:


if k 6= i, j
µk
0
µk = µk + 1 if k = i


µk − 1 if k = j
P
Since µi < λi and µj > λj we have
|λk − µ0k | < n and we can conclude, by induction, that λ can
be obtained from µ by raising operations.
On the other hand if l(µ) > l(λ) then (l(µ), µl(µ) ) is an outside corner and so we let µ0 be obtained
from µ by performing a raising operation from the outside corner (l(µ), µl(µ) ) to the inside corner
(i, µi + 1). Then µ0 is given by:


if k 6= i, l(µ)
µk
0
µk = µk + 1 if k = i


µk − 1 if k = l(µ)
P
Since µi < λi and λl(µ) = 0 we have
|λk − µ0k | < n and so the result follows by induction as
above.
Recall that in the previous section we introduced the column superstandard tableau Sλ as a useful
‘test’ tableau. The following proposition shows that Sλ also has useful properties with respect to
the dominance order:
Proposition 2.8.2. Let λ be a partition of n and P a standard tableau of size n satisfying D(P ) ⊇
D(Sλ ) . Then Shape(P ) E λ, with equality of shapes if and only if P = Sλ .
Proof. The statement that Shape(P ) = λ if and only if P = Sλ is Lemma 2.7.2. It remains to
show that Shape(P ) E λ. The proof is by induction on n with the case n = 1 being obvious. Now,
fix n and let r and s be the row number of n in Sλ and P respectively. Also, let P 0 be the tableau
obtained from P by deleting the box containing n. Define λ0 by:
(
λi − 1 if i = r
0
λi =
λi
if i 6= r
Thus, Sλ0 is obtained from Sλ by deleting the box containing n. Lastly, define µ0 = Shape(P 0 )
and µ = Shape(P ). We have D(Sλ0 ) = D(Sλ )\{n − 1} and D(P 0 ) = D(P )\{n − 1} and so
D(Sλ0 ) ⊂ D(P 0 ). Thus we can apply induction to conclude that Shape(P 0 ) E λ0 . There are two
possibilities:
Case 1: n − 1 ∈
/ D(Sλ ). In this case n does not lie below n − 1 in Sλ and so must belong to the
first row. Hence λ1 > λ01 and so:
µ01 + µ02 + · · · + µ0k < λ1 + λ2 + · · · + λk
23
for all k
(2.8.1)
Now, for some s, µs = µ0s + 1 and µi = µ0i if i 6= s. Hence:
µ1 + µ2 + · · · + µk ≤ λ1 + λ2 + · · · + λk
for all k
Hence µ E λ.
Case 2: n−1 ∈ D(Sλ ). By definition of Sλ we know that {n−1, n−2, . . . , n−r+1} ⊆ D(Sλ ) ⊆ D(P ).
Hence, in P , n must occur below n − 1, n − 1 must occur below n − 2, . . . , and n − r + 2 must occur
below n − r + 1. Hence s ≥ r. But λi = λ0i and µi = µ0i except when i = r and i = s respectively in
which case λi = λ0i + 1 and µi = µ0i + 1. Given that µ0 E λ0 the fact that s ≥ r immediately implies
µ E λ.
2.9
Notes
1. The Robinson-Schensted correspondence (as well as the more general Robinson-SchenstedKnuth correspondence) is well known. A good general reference is Fulton [10].
2. Schützenberger has developed an entirely different framework for viewing the combinatorics
of tableaux known as jeu de taquin (which is French for ‘teasing game’). In this framework
we define a skew-tableau as a set of tiles in N × N which can be manipulated by performing
certain ‘slides’. We allow four slides (viewing empty squares as having the value ∞):
← x
y
↑ y
x
x y
↓
x →
y
x<y
Schützenberger describes the construction of the P -symbol of a permutation as follows. Let
λn = (n, n−1, . . . , 2, 1) be the ‘staircase diagram’. Now, given a permutation w = w1 w2 . . . wm
insert the permutation into λn \λn−1 from bottom to top and perform slides until a non-skew
tableau is obtained. For example, if w = 4312 ∈ Sym4 this process yields:
2
1
3
4
1 2
∼
3
∼
4
1 2
1 2
1 2
∼ 3
∼ 3
3
4
4
4
Note that this is the same tableau as that obtained by row inserting 4312 into ∅. Similarly
define the Q-symbol of the permutation as the tableau obtained by sliding into shape the
‘staircase tableau’ of w−1 . To see that these two alternative definitions are equivalent one
defines the word of a skew tableau and shows that it is Knuth equivalent to any other word
obtained from the skew-tableau by jeu de taquin slides. See Shützenberger [27].
3. Most authors refer to a very different proof of the symmetry theorem due to Knuth [21].
However this proof gives little insight into why the theorem is true. The method of proof
via growth diagrams was discovered by Fomin [9]. Most authors refer to Stanley [30] for an
understanding of growth diagrams. Stanley first introduces the “local rules” and then shows
inductively that the entries of the growth diagram can be given an alternate description in
terms of partial permutations. Our approach in deriving the local rules from the intuitive
definition seems more motivated.
4. Knuth equivalence is treated in most books on tableaux. Our proof of the main theorem
(that two permutations share the same P -symbol if and only if they are Knuth equivalent)
is unique in that it uses only two basic lemmas (that w(T ← x) ≡ w(T )x and that u ≡ v
implies P (u) = P (v)) to derive the result.
24
5. The term ‘raising operation’ is due to MacDonald [23]. He treats them in the more general
setting of vectors in Zn . Here partitions of a fixed length emerge as a fundamental domain
for the natural action of Symn .
6. Our treatment of the descent of a tableau and superstandard tableaux is based on GarsiaMcLarnan [11]. Their proof of Proposition 2.8.2 uses an elegant combinatorial argument.
We could not reproduce it here because it relies on the concept of semi-standard tableaux
(tableaux in which it is possible to have repeated entries) which we have not treated.
25
3
The Hecke Algebra and Kazhdan-Lusztig Basis
In this chapter we introduce the Hecke algebra of the symmetric group. As outlined in the introduction, the Hecke algebra provides a useful structure to discuss the representation theory of both
the symmetric group and the general linear group over a finite field. However, the representation
theory of the Hecke algebra is difficult. The goal of this chapter is to introduce the Kazhdan-Lusztig
basis (and the associated Kazhdan-Lusztig polynomials) as a means of making the representation
theory of the Hecke algebra more tractable.
3.1
The Hecke Algebra
Let A = Z[q, q −1 ] be the ring of Laurent polynomials in the indeterminate q. The Hecke algebra
of the symmetric group, denoted Hn (q), is the algebra of A with identity element Tid generated by
elements Ti for 1 ≤ i < n, subject to the relations:
Ti Tj = Tj Ti
if |i − j| ≥ 2
Ti Ti+1 Ti = Ti+1 Ti Ti+1
Ti2
for 1 ≤ i < n − 1
= (q − 1)Ti + qTid
(3.1.1a)
(3.1.1b)
(3.1.1c)
The reader may notice the similarity between the above relations and those of the symmetric group
in Theorem 1.2.11. In fact, if we set q = 1, then Hn (q) is isomorphic to the group algebra of Symn .
It is for this reason that Hn (q) is often referred to as a deformation of the symmetric group algebra;
as in the case of the symmetric group (3.1.1a) and (3.1.1b) are known as the braid relations. The
last relation (3.1.1c) is known as the quadratic relation.
It will be useful to fix a set of elements which span Hn (q). If si1 si2 . . . sim and sj1 sj2 . . . sjm are
reduced expression for w ∈ Symn then, by Corollary 1.2.3, it is possible to obtain sj1 sj2 . . . sjm from
si1 si2 . . . sim using only the braid relations. Hence, we must have Ti1 Ti2 . . . Tim = Tj1 Tj2 . . . Tjm since
the braid relations also hold in Hn (q). So if we define Tw = Ti1 Ti2 . . . Tim we get a well defined
element of Hn (q) for each w ∈ Symn . In particular, Ti = Tsi for all si ∈ S.
Now, fix w ∈ Symn and let sj ∈ S be arbitrary. If wsj > w and si1 si2 . . . sim is a reduced expression
for w then si1 si2 . . . sim sj is a reduced expression for wsj . So, by the way that we have defined Tw ,
we have Tw Tsj = Twsj . On the other hand, if wsj < w then, by Corollary 1.1.7, w has a reduced
expression si1 si2 . . . sim ending in sj (so that wsj = si1 si2 . . . sim−1 ). Hence:
Tw Tsj = Ti1 Ti2 . . . Tim Tj
= Ti1 Ti2 . . . Ti2m
(since im = j)
= Ti1 Ti2 . . . Tim−1 ((q − 1)Tim + qTid )
= (q − 1)Tw + qTwsj
Therefore we have:
T w T sj =
Twsj
(q − 1)Tw + qTwsj
if wsj > w
if wsj < w
(3.1.2)
An identical argument yields a similar identity for Tsj Tw (see (3.4.1)). These identities show that
left and right multiplication by Tj ( = Tsj ) map the A–span of Tw for all w ∈ Symn into itself.
Since Hn (q) is generated by the Ti we can conclude that the Tw span Hn (q).
26
We will not prove the following theorem. For the proof see, for example, Humphreys [15] or Mathas
[24].
Theorem 3.1.1. Hn (q) is free as an A–module with basis Tw for w ∈ Symn .
We will refer to {Tw |w ∈ Symn } as the standard basis.
3.2
The Linear Representations of Hn (q)
Recall that a representation of Hn (q) is a ring homomorphism ρ : Hn (q) → EndA (M ) where M
is some A-module. We consider the representations afforded by Hn (q) when the module M is as
simple as possible—the ring A itself. Such representations are called linear representations. We
have that EndA A ∼
= A (since every element ϕ ∈ EndA A is uniquely determined by ϕ(1)) and so
representations φ : Hn (q) → EndA A are equivalent to homomorphisms ρ : Hn (q) → A. Now,
recalling our convention that homomorphisms must preserve the identity, we have that ρ(Tid ) = 1.
Using the relation (3.1.1c) we get:
(ρ(Ti ))2 = ρ(Ti2 ) = ρ((q − 1)Ti + qTid ) = (q − 1)ρ(Ti ) + q
Hence (ρ(Ti ) + 1)(ρ(Ti ) − q) = 0. Thus either ρ(Ti ) = q or ρ(Ti ) = −1. Now, if ρ(Ti ) = q then
(3.1.1b) forces ρ(Tj ) = q for all j. Similarly, if ρ(Ti ) = −1 then we have ρ(Tj ) = −1 for all j.
If w ∈ Symn is arbitrary we can choose a reduced expression w = si1 si2 . . . sim and since ρ is a
homomorphism we have:
ρ(Tw ) = ρ(Ti1 Ti2 . . . Tim ) = ρ(Ti1 )ρ(Ti2 ) . . . ρ(Tim )
Hence we have either ρ(Tw ) = q `(w) or ρ(Tw ) = (−1)`(w) .
It is straightforward to verify that both possibilities preserve the relations in (3.1.1) and therefore
define representations of Hn (q). These two representations occur throughout Kazhdan-Lusztig
theory and are given special notation: we write qw = q `(w) and w = (−1)`(w) . These are the
q-analogues of the trivial and sign representations of the symmetric group.
3.3
Inversion in Hn (q)
The involution ι which we will introduce in the next section is central to the definition of the
Kazhdan-Lusztig basis for Hn (q). However, before we can introduce ι, we need to better understand
how inversion works in Hn (q). We start with a technical lemma:
Lemma 3.3.1. Let r1 r2 . . . rm be a (possibly unreduced) subexpression of a reduced expression for
w. Then for some ax ∈ A we have:
X
Tr1 Tr2 . . . T rm =
ax Tx
x≤w
Proof. If r1 r2 . . . rm is reduced then Tr1 Tr2 . . . Trm = Tr1 r2 ...rm and the result follows since r1 r2 . . . rm ≤
w by Proposition 1.3.2. So assume that r1 r2 . . . rm is unreduced and fix the first i for which
`(r1 r2 . . . ri ) > `(r1 r2 . . . ri+1 ) so that r1 r2 . . . ri is reduced. Also, `(r1 r2 . . . ri+1 ) = i − 1 (since
27
`(r1 r2 . . . ri ) = i) and so, by the deletion condition, there exists p and q such that r1 r2 . . . rbp . . . rbq . . . ri+1
is a reduced expression for r1 r2 . . . ri+1 . Hence:
Tr1 Tr2 . . . Trm = Tr1 r2 ...ri Tri+1 Tri+2 . . . Trm
(since r1 r2 . . . ri is reduced)
= ((q − 1)Tr1 r2 ...ri + qTr1 ...ri+1 )Tri+2 . . . Trm
= (q − 1)Tr1 Tr2 . . . T[
ri+1 . . . Trm +
c
+ qTr . . . Tc
rp . . . Trq . . . T rm
(since r1 r2 . . . ri+1 < r1 r2 . . . ri )
(since r1 . . . rbp . . . rbq . . . ri+1 is reduced)
1
Now r1 r2 . . . rd
i+1 . . . rm and r1 r2 . . . rbp . . . rbq . . . rm are both subexpressions of a reduced expression
for w (since r1 r2 . . . rm is) and both have less than m terms. Hence the result follows by induction
on m.
This allows us to prove:
Proposition 3.3.2. For all w ∈ Symn the element Tw is invertible. Moreover, for all w ∈ Symn
there exists Rx,w ∈ A with Rw,w = q −`(w) such that:
Tw−1
−1 =
X
Rx,w Tx
x≤w
Proof. A simple calculation shows that, for all r ∈ S, Tr is invertible with inverse
Tr−1 = q −1 Tr + (q −1 − 1)Tid
(3.3.1)
If w ∈ Symn is arbitrary, fix a reduced expression r1 r2 . . . rm for w. Then:
−1
Tw−1
−1 = T
(r r
1 2 ...rm )
−1
= (Trm Trm−1 . . . Tr1 )−1
= (Tr1 )−1 (Tr2 )−1 . . . (Trm )−1
= (q −1 Tr1 + (q −1 − 1)Tid )(q −1 Tr2 + (q −1 − 1)Tid ) . . . (q −1 Trm + (q −1 − 1)Tid )
Now, expanding the right hand side we see that every term is of the form Tri1 Tri2 . . . Trik where
ri1 ri2 . . . rik is a subexpression of r1 r2 . . . rm . By the above lemma, for all such subexpressions we
can write Tri1 Tri2 . . . Trik as a linear combination of Tx with x ≤ w. Hence we can write the right
hand side above as a linear combination of Tx with x ≤ w. Since r1 r2 . . . rm is reduced the only
way that Tw can emerge is (q −1 Tr1 )(q −1 Tr1 ) . . . (q −1 Trm ) = q −`(w) Tw and hence Rw,w = q −`(w) .
3.4
An Involution and an anti-Involution
Let R be a ring and ϕ : R → R a function. We say that ϕ is an involution if it is a homomorphism
and has order two (that is, ϕ2 is the identity on R). We say that ϕ is an anti-involution if it has
order two and satisfies ϕ(a + b) = ϕ(a) + ϕ(b) and ϕ(ab) = ϕ(b)ϕ(a) for all a, b ∈ R (note the
reversal of order). In this chapter we introduce an involution ι and an anti-involution ∗ which are
fundamental tools in what follows: the involution ι is crucial to the definition of the KazhdanLusztig basis and ∗ will be useful in relating the many left and right identities as well as proving
crucial when we come to discuss cellular algebras in Chapter 5.
28
We begin with the anti-involution ∗ . Define Ts∗i = Tsi and extend so that ∗ is A-linear and (Tx Ty )∗ =
Ty∗ Tx∗ for all x, y ∈ Symn . Note that this implies that (Ti1 Ti2 . . . Tim )∗ = Tim Tim−1 . . . Ti1 . To verify
that ∗ is well defined it is enough to verify that ∗ preserves the relations in (3.1.1). However this is
straightforward:
(Ti Tj )∗ = Tj Ti = Ti Tj = (Tj Ti )∗
if |i − j| ≥ 2
∗
(Ti Ti+1 Ti ) = Ti Ti+1 Ti = Ti+1 Ti Ti+1 = (Ti+1 Ti Ti+1 )∗
(Ti2 )∗ = Ti2 = (q − 1)Ti + qTid = ((q − 1)Ti + qTid )∗
If w ∈ Symn fix a reduced expression si1 si2 . . . sim . Then sim sim−1 . . . si1 is a reduced expression
for w−1 by Lemma 1.1.3 and Tw∗ = (Ti1 . . . Tim )∗ = Tim Tim−1 . . . Ti1 = Tw−1 . Thus we could have
defined ∗ by:
X
X
∗
ay Ty =
ay Ty−1
y∈Symn
y∈Symn
∗
This makes it clear that has order 2. Since, by definition, ∗ satisfies (Tx Ty )∗ = Ty∗ Tx∗ we have
(ab)∗ = b∗ a∗ for all a, b ∈ Hn (q) since ∗ is A-linear. Thus ∗ is an anti-involution.
To illustrate the usefulness of
yields:
∗
we derive a left multiplication formula using (3.1.2). Applying
∗
∗
if wr > w
Twr
∗
∗
(q − 1)Tw + qTwr if wr < w
Trw−1
(q − 1)Tw−1 + qTrw−1
(Tw Tr ) = Tr Tw−1 =
=
∗
if wr > w
if wr < w
Now Lemma 1.4.2 shows that wr > w if and only if (wr)−1 = rw−1 > w−1 and similarly for
wr < w. Substituting w for w−1 yields:
Trw
if rw > w
Tr Tw =
(3.4.1)
(q − 1)Tw + qTrw if rw < w
This is our desired left-handed relation. An argument similar to the one used above is often used
to gain identities when only a left-hand or right-hand identity is known.
We now introduce the involution ι. Recall that the Hecke algebra is defined over the ring A =
Z[q, q −1 ] of Laurent polynomials in q. Define a function – : A → A by F (q) = F (q −1 ) for all
F (q) ∈ A. If is straightforward to verify that this defines a homomorphism of A of order 2 and
hence is an involution. In fact, this involution extends to Hn (q):
Proposition 3.4.1. The involution
defining:
X
ι
–
: A → A extends to an involution ι of the whole of Hn (q) by
Fw (q)Tw =
w∈Symn
X
Fw (q)Tw−1
−1
w∈Symn
Proof. Define ι(Ti ) = Ti−1 and extend ι multiplicatively by letting ι(Ti1 Ti2 . . . Tim ) = ι(Ti1 )ι(Ti2 ) . . . ι(Tim ).
In particular ι(Tw ) = ι(Ti1 )ι(Ti2 ) . . . ι(Tik ) if si1 si2 . . . sik is a reduced expression for w. Lastly, extend ι additively by defining:
X
X
ι
Fw (q)Tw =
Fw (q)ι(Tw )
w∈Symn
w∈Symn
29
To show that ι is a homomorphism it is enough to verify that ι preserves the relations of (3.1.1).
That is, that:
ι(Ti )ι(Tj ) = ι(Tj )ι(Ti ) if |i − j| ≥ 2
(3.4.2a)
ι(Ti )ι(Ti+1 )ι(Ti ) = ι(Ti+1 )ι(Ti )ι(Ti+1 )
(3.4.2b)
2
ι(Ti ) = (q − 1)ι(Ti ) + qι(Tid ) = (q
−1
− 1)ι(Ti ) + (q
−1
)Tid
(3.4.2c)
−1 −1 −1
By taking inverses in (3.1.1a) and (3.1.1b) we get Ti−1 Tj−1 = Tj−1 Ti−1 if |i−j| ≥ 2 and Ti+1
Ti Ti+1 =
−1 −1 −1
Ti+1 Ti Ti+1 for 1 ≤ i < n − 1 which are (3.4.2a) and (3.4.2b). We get (3.4.2c) by substituting
the expression for Ti−1 in (3.3.2). Hence ι is a homomorphism. To verify that ι is an involution it
is enough to verify that ι2 (Ti ) = Ti for all 1 ≤ i < n since ι is a homomorphism. Again, this is a
straightforward calculation using (3.3.2) and the definition of ι. Lastly, note that if s1 s2 . . . sm is a
reduced expression for w then
ι(Tw ) = ι(Ti1 )ι(Ti2 ) . . . ι(Tim ) = Ti−1
Ti−1
. . . Ti−1
= (Tim Tim−1 . . . Ti1 )−1 = Tw−1
−1
m
1
2
since sim sim−1 . . . si1 is a reduced expression for w−1 (see Lemma 1.1.3).
Our last result of this section relates the action of ι and
Proposition 3.4.2. The functions ι and
∗
∗
on Hn (q):
on Hn (q) commute.
Proof. Let y ∈ Symn be arbitrary and let si1 si2 . . . sim be a reduced expression for y (so that
sim sim−1 . . . si1 is a reduced expression for y −1 by Lemma 1.1.3). Then:
ι(Ty )∗ = (ι(Ti1 )ι(Ti2 ) . . . ι(Tim ))∗ = ι(Tim )ι(Tim−1 ) . . . ι(Ti1 ) = ι(Ty−1 ) = ι(Ty∗ )
Hence ι and ∗ commute on the standard basis. Now let
X
∗
ι
ay Ty =
y∈Symn
3.5
X
X
ay ι(Ty )∗ =
y∈Symn
P
y∈Symn
ay Ty ∈ Hn (q) be arbitrary. Then:
X
ay ι(Ty∗ ) = ι (
ay Ty )∗
y∈Symn
y∈Symn
The Kazhdan-Lusztig Basis
In the previous section the involution ι of Hn (q) was introduced. One of the major breakthroughs
of Kazhdan and Lusztig [17] was to notice that a special set of elements fixed by ι form a basis for
Hn (q). It turns out that this basis is parametrised by w ∈ Symn and that each element is a linear
combination of Tx where x is less than or equal to w in the Bruhat order. Some investigation shows
that we can achieve much simpler coefficients of Tx for x ≤ w if we allow our coefficients to lie in
1
1
the larger ring Z[q 2 , q − 2 ]. For example if r ∈ S, and we want A(q)Tr + B(q)Tid to be fixed by ι the
simplest solution with A(q), B(q) ∈ Z[q, q −1 ] is A(q) = q −1 + 1 and B(q) = −q. On the other hand,
1
1
1
1
1
1
if we allow polynomials in the larger ring Z[q 2 , q − 2 ] we can have A(q 2 ) = q − 2 and B(q 2 ) = −q 2 :
1
1
1
1
ι(q − 2 Tr − q 2 Tid ) = q 2 (q −1 Tr + (q −1 − 1)Tid ) − q − 2 Tid
=q
− 21
=q
− 12
Tr + q
− 12
1
2
Tid − q Tid − q
1
2
Tr − q Tid
30
− 12
Tid
(by (3.3.1))
For this reason (and others mentioned in the notes to this chapter) we redefine A to be the larger
1
1
1
ring Z[q 2 , q − 2 ]. Also, for the sake of clarity, we will often omit the parenthesised q 2 when referring
1
to a polynomial in A. Thus we will write F rather than F (q 2 ).
For the rest of this section we will be concerned with offering a proof of the following fundamental
theorem of Kazhdan and Lusztig [17]:
Theorem 3.5.1. For all w ∈ Symn there exists a unique element Cw such that ι(Cw ) = Cw and
Cw =
X
1
y w qw2 qy−1 Py,w Ty
y≤w
where Py,w ∈ Z[q] ⊂ A is of degree at most 12 (`(w) − `(y) − 1) for y < w and Pw,w = 1.
Before we prove the theorem notice that, assuming the validity of the theorem, the Cw certainly
form a basis. For, if we fix a total ordering of Symn compatible with the Bruhat order, then
1
the matrix of the linear map sending Cw to Tw is upper triangular with powers of q − 2 down the
1
diagonal (since Pw,w = 1). Hence the determinant of the map is a power of q − 2 which is a unit
in A and so the map is invertible. The Cw basis is referred to as the Kazhdan-Lusztig basis. The
polynomials Px,w are the Kazhdan-Lusztig polynomials.
We begin our proof by showing that, for each w ∈ Symn there is at most one possible Cw which
satisfies the conditions of the theorem.
Proof of Uniqueness. For fixed w ∈ Symn we induct on `(w) − `(x) for x ≤ w and argue that
there is at most one possible choice for Px,w . This immediately implies the uniqueness of Cw . If
`(w) − `(x) = 0 then x = w and so Px,w = 1 by assumption. Now for x < w assume that Py,w is
known for all x < y ≤ w. Using that ι(Cw ) = Cw we obtain:
X
X
X
1
1
−1
Cw =
z w qw2 qz−1 Pz,w Tz = ι
y w qw2 qy−1 Py,w Ty =
y w qw 2 qy Py,w Ty−1
−1
z≤w
y≤w
y≤w
Substituting our expression for Ty−1
−1 in Section 3.3 yields:
X
1
z w qw2 qz−1 Pz,w Tz =
z≤w
X
−1
y w qw 2 qy Py,w
X
Rz,y Tz
z≤y
y≤w
Now {Tw } forms a basis for Hn (q) and so we can fix x ≤ w and equate coefficients of Tx to obtain:
1
x w qw2 qx−1 Px,w =
−1
X
y w qw 2 qy Py,w Rx,y
y
x≤y≤w
−1
= x w qw 2 qx Px,w Rx,x +
X
−1
y w qw 2 qy Py,w Rx,y
y
x<y≤w
Rearranging and using the fact that Rx,x = qx−1 (Proposition 3.3.2) yields:
X
1
−1
−1
y w qw 2 qy Py,w Rx,y
x w qw2 qx−1 Px,w − qw 2 Px,w =
y
x<y≤w
31
1
Finally, multiplying both sides by x w qx2 yields:
1
−1
−1
1
qw2 qx 2 Px,w − qw 2 qx2 Px,w =
X
−1
−1
y x qw 2 qy qx 2 Py,w Rx,y
(3.5.1)
y
x<y≤w
By induction the right hand side is known. Now, by assumption Px,w has degree at most 21 (`(w) −
1
−1
1
`(x)−1) and so qw2 qx 2 Px,w is a polynomial in q 2 in which all terms have degree at least 1. Similarly
−1
1
1
−1
all terms in qw 2 qx2 Px,w have degree at most − 21 . Hence no cancellation occurs between qw2 qx 2 Px,w
−1
1
and qw 2 qx2 Px,w and so (3.5.1) has at most one solution for Px,w .
We now turn to the existence of the Cw . However, before we begin, some further notation is needed.
If Cw exists write x ≺ w if Px,w has degree 21 (`(w) − `(x) − 1) (the largest allowed by the theorem).
Since 12 (`(w) − `(x) − 1) is only an integer if `(w) − `(x) ≡ 1 (mod 2) we have w = −x if x ≺ w.
1
Define µ(x, w) to be the coefficient of q 2 (`(w)−`(x)−1) in Px,w . Thus µ(x, w) is only defined if x ≤ w
and µ(x, w) 6= 0 if and only if x ≺ w. If x w it is conventional (and sensible) to define Px,w = 0.
It will be seen that the relation ≺ and the function µ are of fundamental importance.
Proof of Existence. Setting Cid = Tid clearly satisfies the conditions of the theorem. Now consider
the case when r ∈ S. We have seen in the discussion prior to the statement of the theorem that
1
1
q − 2 Tr − q 2 Tid is an element fixed by ι. It is also routine to verify that it satisfies the conditions of
the theorem. Hence:
1
1
Cr = q − 2 Tr − q 2 Tid
(3.5.2)
Thus the theorem is verified in the case when w lies in S.
We proceed by induction on `(w). Assume that Cz is known for all z < w (so that x ≺ v and
µ(x, v) makes sense for x, v ∈ Symn such that x ≤ v < w). Since `(w) > 0 there exists r ∈ R such
that rw < w. Set v = rw so that Cv is known and rv = w. Now define:
X
Cw = Cr Cv −
µ(z, v)Cz
(3.5.3)
z≺v
rz<z
Since ι is a homomorphism and ι(Cz ) = Cz for all z < w (by induction) we have ι(Cw ) = Cw . Also,
1
1
since Cr = q 2 Tr − q 2 Tid it is clear that Cw is a A-linear combination of elements Ty satisfying
y ≤ w. It remains to show that the polynomials Px,w for x ≤ w lie in Z[q] and have the required
degree. This requires a careful examination of the terms which arise in right hand side of (3.5.3).
We can rewrite (3.5.3) using our inductive information and (3.5.2) as:
1
X
X
1
1
1
Cw = q − 2 Tr − q 2 Tid
y v qv2 qy−1 Py,v Ty −
µ(z, v)x z qz2 qx−1 Px,z Tx
(3.5.4)
x,z
x≤z≺v
rz<z
y≤v
For fixed x we want to obtain an expression for Px,w by considering the right hand side of (3.5.4).
First assume that rx > x. Then Tx emerges on the right hand side in three ways:
1
1
−1 P
1) In the first sum as Tr Trx = (q−1)Trx +qTx . In this case the coefficient is q 2 rx v qv2 qrx
rx,v .
1
1
2) In the first sum as Tid Tx . Here the coefficient is −q 2 x v qv2 qx−1 Px,v .
32
1
P
3) In the second sum. Here the coefficient is − µ(z, v)x z qz2 qx−1 Px,z with the sum over
those z satisfying x ≤ z ≺ v and rz < z.
−1 = q −1 q −1 . Also, = and so the coefficient of
Since rx > x, `(rx) = `(x) + 1 and so qrx
rx v
x w
x
Tx can be written as:
1
1
1
1
X
q − 2 x w qv2 qx−1 Prx,v − q 2 x v qv2 qx−1 Px,v −
1
µ(z, v)x z qz2 qx−1 Px,z
z
x≤z≺v
rz<z
1
1
1
Using that qw2 = q 2 qv2 , x v = −x w and z = w (since z ≺ v) the coefficient of Tx becomes:
X
1
1
−1
µ(z, v)qz2 qw 2 Px,z
w x qw2 qx−1 Px,v + q −1 Prx,v −
z
x≤z≺v
rz<z
1
1
Equating with w x qw2 qx−1 Px,w , cancelling w x qw2 qx−1 and applying ι we obtain:
Px,w = Px,v + qPrx,v −
−1
X
1
µ(z, v)qz 2 qw2 Px,z
if rx > x
(3.5.5)
z
x≤z≺v
rz<z
Now assume that rx < x. This time Tx emerges on the right hand side of (3.5.4) in four ways:
1
1
1) In the first sum as Tid Tx . In this case the coefficient is −q 2 x v qv2 qx−1 Px,v .
1
1
2) In the first sum as Tr Tx = (q−1)Tx +qTrx . Here the coefficient is q − 2 (q−1)x v qv2 qx−1 Px,v .
1
1
−1 P
3) In the first sum as Tr Trx = Tx . In this case the coefficient is q − 2 rx v qv2 qrx
rx,v .
1
P
4) In the second sum. As above the coefficient is − µ(z, v)x z qz2 qx−1 Px,z with the sum
over those z satisfying x ≤ z ≺ v and rz < z.
−1 = qq −1 and Since rx < x we have `(rx) = `(x) − 1 and so qrx
rx = −x . Adding the above four
x
coefficients and substituting these identities yields the following coefficient of Tx :
X
1
1
1
1
1
1
v qv2 x qx−1 − q 2 Px,v + (q 2 − q − 2 )Px,v − q 2 Prx,v −
µ(z, v)x z qz2 qx−1 Px,z
z
x≤z≺v
rz<z
By a similar process to that used in the first case (simplifying and then equating coefficients of Tx
in (3.5.3)) we get:
Px,w = qPx,v + Prx,v −
−1
X
1
µ(z, v)qz 2 qw2 Px,z
if rx < x
(3.5.6)
z
x≤z≺v
rz<z
If we define c to be 1 if rx > x and 0 if rx < x we can combine (3.5.5) and (3.5.6) into the following
expression for Px,w :
Px,w = q 1−c Px,v + q c Prx,v −
X
z
x≤z≺v
rz<z
33
−1
1
µ(z, v)qz 2 qw2 Px,z
(3.5.7)
−1
1
Since z ≺ v = rw we have that `(w) − `(z) ≡ 0 (mod 2) and so qz 2 qw2 ∈ Z[q]. Hence (3.5.7) shows
that Px,w is indeed a polynomial in q. It remains to show that Px,w has the required bounded
degree.
1 1
P
We first examine the terms in the sum
µ(z, v)qz2 qw2 Px,z . If x < z, then, by induction deg Px,z ≤
1
2 (`(z) − `(x) − 1) and so:
1
1
−1 1
deg µ(z, v)qz 2 qw2 Px,z ≤ (`(w) − `(z)) + (`(z) − `(x) − 1)
2
2
1
= (`(w) − `(x) − 1)
2
On the other hand, if x = z then Px,z = Px,x = 1 and so:
1
−1 1
deg µ(z, v)qz 2 qw2 Px,z = (`(w) − `(x))
2
1
Hence, if x ≺ v and rx < x, a term of the form −µ(x, v)q 2 (`(w)−`(x)) occurs in the sum. Moreover,
this is the only occurrence of a term of degree greater than 12 (`(w) − `(x) − 1).
We now consider the term q c Prx,v . If c = 1 then rx > x and so `(rx) = `(x) + 1. On the other
hand if c = 0 then rx < x and so `(rx) = `(x) − 1. Hence we may write `(rx) = `(x) − 1 + 2c. By
induction:
1
deg q c Prx,v ≤ (`(v) − `(rx) − 1) + c
2
1
= (`(w) − 1 − `(x) + 1 − 2c − 1) + c
2
1
= (`(w) − `(x) − 1)
2
(since rw = v)
Thus q c Prx,v never contributes a term of degree greater than 12 (`(w) − `(x) − 1).
We now consider the term q 1−c Px,v . In this case we have:
1
deg q 1−c Px,v ≤ (`(v) − `(x) − 1) + (1 − c)
2
1
= (`(w) − `(x)) − 1 + (1 − c)
2
Hence we could have deg q 1−c Px,v = 21 (`(w) − `(x)) if Px,v has maximal degree and c = 0. This is
slightly larger than that permitted in the statement of the theorem. However, in this case x ≺ v
(since Px,v has the maximal possible degree) and rx < x (since c = 0). So, from above, we have
1
a term of the form −µ(x, v)q 2 (`(w)−`(x)) occurring in the sum. This cancels the offending term in
1
q 1−c Px,v since µ(x, v) is the coefficient of q 2 (`(w)−`(x)) in q 1−c Px,v .
3.6
Multiplication Formulae
As mentioned in the introduction, the main motivation of Kazhdan and Lusztig in [17] was to
better understand the representation theory of Hn (q). As such, it is vital to know how Hn (q) acts
on the Kazhdan-Lusztig basis via left and right multiplication. The following theorem gives the
first indication of the unique properties of the Kazhdan-Lusztig basis:
34
Theorem 3.6.1. Let Cw be the Kazhdan-Lusztig basis element corresponding to w ∈ Symn and
let r ∈ S be a simple transposition.
Then:
(
−Cw
if rw < w
Tr Cw =
(3.6.1)
1
1 X
2
2
q Crw + qCw + q
µ(z, w)Cz if rw > w
z≺w
rz<z
Proof. If rw > w then, by (3.5.3), we have:
Crw = Cr Cw −
X
µ(z, w)Cz
z≺w
rz<z
1
1
But we know (see (3.5.2)) that Cr = q − 2 Tr − q 2 Tid and so we have:
1
1
X
Crw = q − 2 Tr Cw − q 2 Cw −
µ(z, w)Cz
z≺w
rz<z
1
Rearranging and multiplying by q 2 yields the case rw > w.
The first identity (the case rw < w) is not so straightforward. We have (by (3.5.2)):
1
1
1
1
1
1
Tr Cr = q − 2 ((q − 1)Tr + qTid ) − q 2 Tr = q 2 Tr − q − 2 Tr + q 2 Tid − q 2 Tr = −Cr
And so we may assume, for induction, that the first identity in (3.6.1) is known for all y < w
satisfying ry < y. Now:
1
1
X
Tr Cw = Tr (q − 2 Tr − q 2 Tid )Crw −
µ(z, rw)Cz
z≺rw
rz<z
1
1
1
X
= (q − 2 (q − 1)Tr + q 2 Tid − q 2 Tr )Crw +
µ(z, rw)Cz
(by induction)
z≺rw
rz<z
1
1
1
1
= (q 2 Tr − q − 2 Tr + q 2 Tid − q 2 Tr )Crw +
X
µ(z, rw)Cz
z≺rw
rz<z
1
1
= −(q − 2 Tr − q 2 Tid )Crw +
X
µ(z, rw)Cz
z≺rw
rz<z
= −Cr Crw +
X
µ(z, rw)Cz
z≺rw
rz<z
= −Cw
We would like to develop a right-hand version of Theorem 3.6.1. First we need to know how ∗ acts
on the Kazhdan-Lusztig basis. This provides some extra information about the Kazhdan-Lusztig
polynomials.
Proposition 3.6.2. Let w ∈ Symn . Then Cw∗ = Cw−1 . Hence if y ≤ w then Py,w = Py−1 ,w−1 and
µ(y, w) = µ(y −1 , w−1 ).
35
Proof. We know that ∗ and ι commute and so ι(Cw∗ ) = ι(Cw )∗ = Cw∗ . Hence Cw∗ is an ι-invariant.
Now Cw has the form:
X
1
Cw =
y w qw2 qy−1 Py,w Ty
y≤w
Hence, by the definition of ∗ , Cw∗ has the form:
Cw∗ =
X
1
y w qw2 qy−1 Py,w Ty−1
y≤w
Now noting that y ≤ w if and only if y −1 ≤ w−1 , `(y −1 ) = `(y) and `(w−1 ) = `(w) we can rewrite
this as:
X
1
Cw∗ =
y−1 w−1 qw2 −1 qy−1 Py,w Ty−1
y −1 ≤w−1
Now deg Py,w ≤ 21 (`(w)−`(y)−1) = 12 (`(w−1 )−`(y −1 )−1) and hence Cw∗ satisfies all the properties
of Cw−1 . By uniqueness we must have Cw∗ = Cw−1 . Now we can write Cw−1 as:
X
Cw−1 =
1
y−1 w−1 qw2 −1 qy−1 Py−1 ,w−1 Ty−1
y −1 ≤w−1
Comparing coefficients yields Py,w = Py−1 ,w−1 and hence µ(y, w) = µ(y −1 , w−1 ).
It is now straightforward to derive the corresponding right-hand multiplication formula:
Corollary 3.6.3. Let r ∈ S and
( w ∈ Symn . Then
−Cw
Cw Tr =
1
1 X
q 2 Cwr + qCw + q 2
µ(z, w)Cz
if wr < w
if wr > w
(3.6.2)
z≺w
zr<z
Proof. Applying
∗
to (3.6.1) yields:
∗
−Cw∗
if rw < w
1
1 P
∗
∗
∗
q 2 Crw + qCw + q 2
µ(z, w)Cz if rw > w
−Cw−1
1
1 P
q 2 Cw−1 r + qCw−1 + q 2
µ(z, w)Cz −1
(Tr Cw ) = Cw−1 Tr =
=
if rw < w
if rw > w
Now if z ≺ w then µ(z −1 , w−1 ) = µ(z, w) by the previous Proposition and so z −1 ≺ w−1 . Also, if
rz < z then z −1 r < z −1 (by Corollary 1.3.3) and similarly for w. Hence, replacing z −1 by z and
w−1 by w we get the result.
3.7
Notes
1. There are two very different proofs that the standard basis is indeed a basis for Hn (q). Most
authors define Hn (q) as the associative algebra generated by either Ti or Tw subject to the
relations (3.1.1) or (3.1.2) respectively. They then demonstrate a large endomorphism algebra
of Hn (q) which implies that the Tw are indeed a basis. For this approach see, for example,
Mathas [24] or Humphreys [15].
36
An outline of the second proof is as follows. Let G denote the general linear group of n × n
matrices over the finite field Fq . Let B denote the subgroup of upper triangular matrices.
Now, inside the group algebra CG of G we consider the ‘double coset algebra’ [B]CG[B] (where
[B] denotes the sum of all elements in B). If we write N for the subgroup of permutation
matrices it is not hard to show that G has a ‘Bruhat decomposition’ as the disjoint union
of double cosets of the form BwB with w ∈ N . Hence, [B]CG[B] has a basis consisting
of elements of the form [BwB] where w ∈ Symn (in which we regard a permutation as a
permutation matrix in the natural way). Now, if we normalise by defining eid = (1/|B|)[B]
and ew = (1/|B|)[BwB] then it can be shown that the multiplication of these basis elements
is given by:
(
erw
if `(rw) > `(w)
er ew =
(q − 1)ew + qerw if `(rw) < `(w)
Thus, for all primes powers q we have a homomorphism from Hn (q̃) onto [B]CG[B] P
by sending
Tw to ew and evaluatingP
f ∈ Z[q̃, q̃ −1 ] at q. Hence if a non-trivial linear relation
aw Tw =
0 holds in Hn (q̃) then
aw (q)ew = 0 also holds between the corresponding elements of
[B]CG[B] (which we know form a basis). Hence (q̃ − q) divides aw for all aw . But this holds
for infinitely many prime powers q. Hence all the coefficients are 0 and so the {Tw } form a
basis. A more detailed outline as to the derivation of generators and relations for [B]CG[B]
can be found in Exercise 24 of Bourbaki [3].
2. The above realisation of the Hecke algebra as a ‘double coset algebra’ is the origin of the term
‘Hecke algebra’. It is easy to see that, with G and [B] as above EndG (CG[B]) ∼
= [B]CG[B].
This is how the Hecke algebra first arose: during attempts to decompose the representation
obtained by inducing the trivial representation from B to G. See Iwahori [16].
3. In Section 3.1 it was mentioned that when q = 1 the Hecke algebra is isomorphic to the group
algebra of the symmetric group. The formal term for ‘assigning a value to q’ is specialisation.
1
1
If ϕ : Z[q 2 , q − 2 ] → R is a homomorphism (most often an evaluation homomorphism onto a
subring of C) then Hn (q) ⊗ϕ R is the algebra obtained by specialising at ϕ (or ϕ(q)). Hence,
our statement is that Hn (q) ⊗q7→1 C ∼
= CSymn where CSymn denotes the symmetric group
algebra over C. Tits has shown that a much stronger statement is true: that if a specialisation
q 7→ z is semi-simple (over C this means that the specialisation is isomorphic to a direct sum
of matrix algebras) then, in fact the specialisation remains isomorphic to CSymn . This is
known as the Tits deformation theorem. Tits’ proof is given in Steinberg [31].
4. Although Tits proved that semi-simple specialisations of Hn (q) are isomorphic to CSymn , no
explicit isomorphism is constructed in the proof. In [22], Lusztig uses the Kazhdan-Lusztig
basis to show that a certain left action of Hn (q) on CSymn commutes with the natural right
1
action of CSymn . This establishes a canonical isomorphism between Hn (q) ⊗ Q(q 2 ) and
1
Q(q 2 )Symn known as Lusztig’s isomorphism. The need to introduce a square root of q to
realise the isomorphism is used by many authors as further justification for enlarging A to
1
1
the ring Z[q − 2 , q 2 ].
5. The question of which values of z ∈ C yield semi-simple specialisations q 7→ z is answered by
Green [13]. If we define e(z) to be the first n such that z n = 1 (with e(z) = ∞ if no such n
exists) then Hn (q) ⊗q7→z C is semi-simple if and only if e(z) > n or n ≥ 3 and z = 0. To give
37
some suggestion as to why this result should be true, it can be shown that:
X
y∈Symn
Ty
2
=
n
Y
q i − 1 X
i=1
q−1
Ty
y∈Symn
P
Hence
Ty is a central nilpotent element if e(z) < n. The implication the other way is more
difficult. Gyoja and Uno [14] offer an elegant proof using characters.
6. The systematic use of the anti-involution ∗ is motivated by its importance in the cellular
algebra approach to the Kazhdan-Lusztig basis. See Chapter 5.
7. The first proof of the existence and uniqueness of the Kazhdan-Lusztig basis follows Kazhdan and Lusztig [17] very closely. However, Kazhdan and Lusztig are very terse in their
treatment—their proof takes only one journal page! Our approach (which is similar to Dyer
[4] and Humphreys [15]) is to try to elaborate on their argument.
8. The alternative proof, given in the appendix, of the existence and uniqueness of the KazhdanLusztig basis is due to Lusztig [19]. There are other, more abstract proofs which only use
basic properties of the involution to deduce the existence and uniqueness of a special basis.
See for example Chapter 8 of Du, Parshall and Wang [7].
38
4
Cells
In this chapter we introduce the cells associated to a basis of an algebra. We first give the abstract
definition in terms of a ‘cell preorder’ and explain how each cell corresponds to a representation of
the algebra. The rest of the chapter is devoted to deriving a more concrete (and, as it happens,
the original) definition of the cells in the Hecke algebra. We also derive some technical properties
essential to the further study of cells.
4.1
Cell Orders and Cells
Let H be an R-algebra which is free as a module over R and let {Ci |i ∈ I } be a basis for H. For
arbitrary a and j ∈ I we can express aCj uniquely as a linear combination of the basis elements.
Define a relation ← on I by declaring that i ← j if there exists an a ∈ H such that Ci appears
L
L
with non-zero coefficient in aCj . We have 1Ci = Ci and so i ← i for all i ∈ I. In other words, the
L
relation ← is reflexive. We similarly define i ← j if there exists an a ∈ H such that Ci appears
L
R
with non-zero coefficient in Cj a. We further define i ← j if either i ← j or i ← j.
LR
L
R
We might hope that the relations ←, ← and ← are transitive. However, in general this is not the
L
R
LR
case. Instead we take the transitive closure of ← by declaring that i ≤ j if there exists a sequence
L
L
i = i1 ← i2 ← . . . ← im = j. The resulting preorder is called the left cell preorder. We similarly
L
L
L
define the right cell preorder ≤ and the two-sided cell preorder ≤ .
R
LR
We can use the preorders ≤, ≤ and ≤ to define equivalence classes on I. We define i ∼ j if both
L
R
L
LR
i ≤ j and j ≤ i. Similarly, we write i ∼ j if i ≤ j ≤ i and i ∼ j if i ≤ j ≤ i. The equivalence
L
R
L
R
R
LR
LR
LR
classes of ∼ are called the left cells of I, those of ∼ are the right cells and those of ∼ are the
L
R
LR
two-sided cells. Since ≤, ≤ and ≤ are preorders they induce partial orders on the corresponding
L
R
LR
cells. These partial orders constitute the left cell poset, the right cell poset and the two-sided cell
poset respectively.
It is often difficult to decide whether i ≤ j. The following proposition shows that if we fix a set G
L
which generates H as an algebra, and consider gCk for each k ∈ I and g ∈ G then we obtain a set
of relations which generate ≤:
L
Proposition 4.1.1. Assume that G is a subset which generates H as an algebra. Then i ≤ j if
L
and only if there exists a chain i = i1 , i2 , . . . , im = j such that Cik appears with nonzero coefficient
in gCik+1 for some g ∈ G.
Proof. It is enough to show the existence of such a chain when i ← j. Since i ← j there exists an
L
L
a∈H
P such that Ci appears with non-zero coefficient in aCj . Since G generates H we can write
a=
λk Gk where each Gk is a finite product of elements of G. Now since Ci appears in aCj we
must have that Ci appears in Gk Cj for some k. Now, Gk = g1 g2 . . . gm for some gi ∈ G. Hence
there exists a chain i = i1 , i2 , . . . im = j with Ci1 appearing in g1 Ci2 , Ci2 appearing in g2 Ci3 etc.
This shows the existence of such a chain. The other implication is by definition of ≤.
L
Of course we get similar conditions for i ≤ j and i ≤ j by allowing only right multiplication by
R
LR
39
g ∈ G for ≤ and multiplication on either side for ≤ .
R
LR
We now give some examples of cells. First consider the cells of the standard basis of H2 (q).
Multiplication of the standard basis elements is given by:
Tid
Tid
Tid
Tid
T s1
T s1
T s1
(q − 1)Ts1 + qTid
By considering T1 Tid we have s1 ≤ id and by considering Ts21 we have that id ≤ s1 . Hence we have
L
L
only one left cell consisting of all of Sym2 . It is not too hard to see that we only ever get one left
cell when considering Hn (q) with respect to the standard basis. For if we fix w ∈ Symn we have
Tw Tid = Tw and so w ≤ id. On the other hand, by Proposition 3.3.2, each standard basis element is
L
invertible and so we have Tw−1 Tw = Tid which yields id ≤ w. Hence w ≤ id ≤ w for all w ∈ Symn .
L
L
L
This implies that all of Symn lies in the same left-cell.
However, when we consider Hn (q) with respect to the Kazhdan-Lusztig basis we obtain a rich
cell structure. First consider the Kazhdan-Lusztig basis of H2 (q). We have Cid = Tid and Cs1 =
1
1
q − 2 Ts1 − q 2 Tid . A simple calculation (or use of the multiplication formulae in Section 3.6) yields
the following multiplication table:
Cid
Cs1
Cid
Cid
Cs1
Cs1
Cs1
1
1
(−q 2 − q − 2 )Cs1
Since {Cid , Cs1 } is a basis it generates H2 (q) as an algebra and so, by Proposition 4.1.1, the relation
s1 ≤ id generates ≤. Hence there are two left-cells: {id} and {s1 }. The left cell poset looks like:
L
L
{id}
{s1 }
We now consider the cells of H3 (q) with respect to the Kazhdan-Lusztig basis. Using the recurrence
(3.5.7) for the Kazhdan-Lusztig polynomials it is routine to verify that Px,w = 1 for all x ≤ w in
Sym3 . Hence x ≺ w if and only if x ≤ w and `(w) − `(x) = 1 and in this case µ(x, w) = 1.
We now have all the information we need in order to use the multiplication formulae in Section
3.6. The following two tables (calculated using the multiplication formulae) show the effect of left
multiplying the Kazhdan-Lusztig basis elements by Ts1 and Ts2 :
T s1
T s2
T s1
T s2
1
2
q Cs1 s2 s1
Cid
q Cs1 + qCid
1
q 2 Cs2 + qCid
1
2
Cs1
−Cs1
1
q 2 Cs2 s1 + qCs1
Cs1 s2
−Cs1 s2
1
+ qCs1 s2 + q 2 Cs2
1
2
q Cs1 s2 s1
40
Cs2
q Cs1 s2 + qCs2
−Cs2
1
2
Cs2 s1
1
+ qCs2 s1 + q 2 Cs1
−Cs2 s1
Cs1 s2 s1
−Cs1 s2 s1
−Cs1 s2 s1
Hence ≤ is generated by s1 ≤ id, s2 ≤ id, s1 s2 ≤ s2 , s2 s1 ≤ id, s1 s2 s1 ≤ s2 s1 , s2 ≤ s1 s2 and
L
L
L
L
L
L
L
s1 ≤ s2 s1 . Hence the left cells are {id}, {s1 , s2 s1 }, {s2 , s1 s2 } and {s1 s2 s1 }. The left cell poset
L
looks like:
qq
qqq
q
q
q
qqq
{id} M
MMM
MMM
MMM
M
{s1 , s2 s1 }
{s2 , s1 s2 }
MMM
MMM
MMM
M
q
qqq
q
q
q
qqq
{s1 s2 s1 }
4.2
Representations Associated to Cells
Let H be an R-algebra with fixed basis {Cw |w ∈ I } and let ≤ and ∼ be the relations introduced
L
L
in the previous section. We will show that to every left cell we can associate a left H-module and
hence a representation of H. Although we will not make it explicit, a similar process yields right
H-modules to every right cell and (H, H)-bimodules to every two-sided cell.
Fix x ∈ I and consider aCx for arbitrary a ∈ H. If Cy appears with non-zero coefficient then, by
definition, we must have y ≤ x. So we can always write:
L
aCx =
X
ra (y, x)Cy
(4.2.1)
y≤x
for some ra (y, x) ∈ R. Hence, if we define H(≤ w) to be the linear span of those Cy satisfying
L
y ≤ w then (4.2.1) shows that H(≤ w) is a left ideal of H.
L
L
If w ∈ I, write x < w if x ≤ w and w x. Then {y ∈ I |y < w} is the set of elements less than
L
L
L
L
w in the left cell preorder which are not in the same left cell. Now define H(< w) to be the linear
L
span of those Cy satisfying y < w. Now if x < w and a ∈ H is arbitrary then, by (4.2.1), we can
L
L
P
write aCx = y≤x ra (y, x)Cy . Now, if y ≤ x then y < w (otherwise we would have w ≤ y ≤ x
L
L
L
L
L
contradicting x < w). Hence, left multiplication by H maps H(< w) into itself and so H(< w) is
L
L
L
also a left ideal of H.
Clearly H(< w) is contained inside H(≤ w) and so we can consider the quotient H(≤ w)/H(< w).
L
L
L
L
Now, H(≤ w) has a basis consisting of Cx with x ≤ w. Similarly H(< w) has a basis consisting of
L
L
L
those Cy with y < w. Hence H(≤ w)/H(< w) has a basis consisting of the images of those Cx with
L
L
L
x ≤ w but x ≮ w. But if x ≤ w and x ≮ w then w ≤ x and so x ∼ w. Hence H(≤ w)/H(< w)
L
L
L
L
L
L
L
L
has a natural basis consisting of the images of those Cx with x ∼ w. In the quotient module
L
H(≤ w)/H(< w) the multiplication in (4.2.1) becomes:
L
L
aCx ≡
X
ra (y, w)Cy
y∼w
L
41
(mod H(< w))
L
This is the cell module associated to the cell {x|x ∼ w}. Clearly the cell module has rank equal
L
to the number of elements in the cell. The representation afforded by the cell module is the cell
representation.
For example, we consider some of the representation afforded by the cells of H3 (q) with respect to
the Kazhdan-Lusztig basis. To simplify notation in this example we will write H(≤ w) in place of
L
H3 (q)(≤ w). In the last section we saw that the left cells of H3 (q) with respect to the KazhdanL
Lusztig basis were {id}, {s1 , s2 s1 }, {s2 , s1 s2 } and {s1 s2 s1 }. Let us calculate the cell representation
associated to {id}. Reducing the calculations of the previous section modulo H(< w) we have:
L
Ts1 Cid = qCid
(mod H(< w))
Ts2 Cid = qCid
(mod H(< w))
L
L
Thus the cell {id} affords the representation Tw 7→ qw . It is not too hard to see from the multiplication formulae in Section 3.6 that this is a general fact: id always lies in a cell by itself and affords
the ‘q-trivial’ representation Tw 7→ qw .
For a slightly more complicated example consider the cell {s1 , s2 s1 }. Our calculations of the
previous section yield:
Ts1 Cs1
Ts1 Cs2 s1
= −Cs
1
1
= q 2 Cs1 + qCs2 s1 + q 2 Cs1 s2 s1
Ts2 Cs1
Ts2 Cs2 s1
1
= qCs1 + q 2 Cs2 s1
= −Cs2 s1
When we pass to the quotient H(≤ s1 )/H(< s1 ) the only basis element which lies in H(< s1 ) is
L
L
L
Cs1 s2 s1 . Thus the representing matrices of Ts1 and Ts2 are:
1 q
−1 q 2
Ts1 7→
Ts1 7→
1
0
q
q2
0
−1
We multiply these to get the matrices of the other standard basis elements of Hn (q):
!
!
!
3
3
3
−q q 2
0 q2
0
−q 2
Ts1 s2 7→
Ts2 s1 7→
Ts1 s2 s1 7→
1
3
3
−q 2 0
0
q 2 −q
−q 2
4.3
Some Properties of the Kazhdan-Lusztig Polynomials
We now explore the preorders ≤, ≤ and ≤ introduced in the last two sections in the context
L
R
LR
of the Hecke algebras. The goal of this chapter is to derive a more concrete formulation of the
relations ≤, ≤ and ≤ . To get started, however, we need to develop some technical properties of
L
R
LR
the Kazhdan-Lusztig polynomials.
The following is a formalisation of the inductive formula for the Kazhdan-Lusztig polynomials
developed during the proof of Theorem 3.5.1:
Proposition 4.3.1. If x ≤ w and w 6= id we can find r ∈ S such that rw < w. If we then define
cx by:
1 if rx > x
cx =
0 if rx < x
42
Then we have an inductive identity for Px,w :
Px,w = q 1−cx Px,rw + q cx Prx,rw −
X
−1
1
qz 2 qw2 µ(z, rw)Px,z
(4.3.1)
z
x≤z≺rw
rz<z
We use this inductive formula to prove the following lemma:
Lemma
(i)
(ii)
(iii)
4.3.2. Let x, w ∈ Symn and r ∈ S.
If rw < w then Prw,w = 1 and so rw ≺ w and µ(rw, w) = 1.
If rx < x and x rw then Px,w = Prx,rw .
If x < w and rw < w then Px,w = Prx,w .
Proof. For (i) note that if rw < w then r(rw) = w > rw and so crw = 1. The inductive formula
(4.3.1) gives:
X
− 1 12
Prw,w = q 0 Prw,rw + qPw,rw −
qz 2 qrw
µ(z, rw)Px,z
z
rw≤z≺rw
rz<z
Now z ≺ rw implies z < rw and so the sum is empty. Also, we know Prw,rw = 1 and Pw,rw = 0
since w > rw. Hence Prw,w = q 0 Prw,rw = 1.
For (ii) note that rx < x and so cx = 0. Since x rw we have Px,rw = 0 and there are no z
satisfying x ≤ z ≺ rw. Hence the inductive formula gives Px,w = Prx,rw .
For (iii) we use induction on `(w). If `(w) = 0 then w = id and so w does not satisfy the conditions
of the theorem. If `(w) = 1 then w = t for some t ∈ S and rt < t forces r = t. Hence, in this
case the statement is that Px,w = Pid,t = Pt,t = 1 but this is a special case of (i). So assume,
for induction that Px,z = Prx,z for all z satisfying `(z) < `(w) and rz < z. We want to show
that Px,w = Prx,w under the assumption that rw < w. First note that if z satisfies rz < z and
x ≤ z ≺ rw then, by Lemma 1.3.1, either rx ≤ z or rx ≤ rz. Hence in either case rx ≤ z ≺ rw
(since rz < z). Conversely, if z satisfies rz < z and rx ≤ z ≺ rw then x ≤ z ≺ rw by an identical
argument. We have therefore shown:
{z ∈ Symn |rz < z, x ≤ z ≺ rw} = {z ∈ Symn |rz < z, rx ≤ z ≺ rw}
(4.3.2)
Now, by the inductive identity:
X
Px,w = q 1−cx Px,rw + q cx Prx,rw −
−1
1
qz 2 qw2 µ(z, rw)Px,z
z
x≤z≺rw
rz<z
X
= q cx Prx,rw + q 1−cx Px,rw −
1
−1
qz 2 qw2 µ(z, rw)Prx,z
(by induction and (4.3.2))
z
rx≤z≺rw
rz<z
= q 1−crx Prx,rw + q crx Px,rw −
X
−1
1
qz 2 qw2 µ(z, rw)Prx,z
z
rx≤z≺rw
rz<z
= Prx,w
This lemma allows us to prove:
43
(since crx = 1 − cx )
Proposition 4.3.3. Let w ∈ Symn and r ∈ S such that rw > w. Then the only element x ∈ Symn
satisfying w ≺ x and rx < x is rw, and in this case µ(w, x) = 1.
Proof. Assume that x 6= rw. By assumption rx < x and so, by the lemma above we have Pw,x =
Prw,x . Now:
1
1
1
deg Pw,x = deg Prw,x ≤ (`(x) − `(rw) − 1) = (`(x) − `(w) − 2) < (`(x) − `(w) − 1)
2
2
2
Hence w ⊀ x. On the other hand if x = rw then the above lemma applies to yield w ≺ x and
µ(w, x) = µ(w, rw) = 1.
4.4
New Multiplication Formulae
Recall that in Chapter 1 we defined the left and right descent sets of a permutation w ∈ Symn
as the sets L(w) = {r ∈ S |rw < w} and R(w) = {r ∈ S |wr < w}. Using this notation we can
rewrite the multiplication formula
of Theorem 3.6.1 as:
(
−Cw
if r ∈ L(w)
X
Tr Cw =
(4.4.1)
1
1
q 2 Crw + qCw + q 2
µ(z, w)Cz if r ∈
/ L(z)
z≺w
r∈L(w)
We want to simplify this further. Recall that µ(x, y) is only defined if x ≤ y. We extend the
definition of µ by defining µ(x, y) = µ(y, x) if x ≥ y and µ(x, y) = 0 if x and y are incomparable.
Thus µ(x, y) is defined, and symmetric in x and y, for all x 6= y. We say that x is joined to y, and
write x—y, if x 6= y and µ(x, y) 6= 0.
Now, fix r ∈ S and w ∈ Symn such that rw > w and consider the sum:
X
µ(z, w)Cz
z—w
r∈L(z)
Now z—w if and only if µ(z, w) 6= 0 and so z ≺ w or w ≺ z. Hence, by symmetry of µ we have:
X
X
X
µ(z, w)Cz =
µ(w, z)Cz +
µ(z, w)Cz
z—w
r∈L(z)
w≺z
r∈L(z)
z≺w
r∈L(z)
Now consider which terms emerge in the first sum on the right hand side. If Cz appears we have that
rw > w (by assumption), rz < z (since r ∈ L(z)) and w ≺ z. Hence we can apply Proposition 4.3.3
of the previous section to conclude that the only term in the first sum is rw and that µ(w, rw) = 1.
Hence:
X
X
µ(z, w)Cz = Crw +
µ(w, z)Cz
(4.4.2)
z—w
r∈L(z)
z≺w
r∈L(z)
Substituting (4.4.2) into (4.4.1) we have:
(
−Cw
Tr Cw =
1 X
µ(z, w)Cz
qCw + q 2
if r ∈ L(w)
if r ∈
/ L(w)
(4.4.3)
z—w
r∈L(z)
By applying ∗ (noting that µ(z −1 , w−1 ) = µ(z, w) and L(w−1 ) = R(w)) we obtain the right-handed
identity:
44
(
−Cw
X
Cw Tr =
1
qCw + q 2
µ(z, w)Cz
if r ∈ R(w)
if r ∈
/ R(w)
(4.4.4)
z—w
r∈R(z)
4.5
New Definitions of the Cell Preorders
Equipped with the new multiplication formulae of the previous section we can give explicit conditions in terms of descent sets and the joins relation for the preorders ≤, ≤ and ≤ . We start by
L
R
LR
asking: under what conditions does a non-zero coefficient of Cz emerge in Tr Cw for some r ∈ S? If
r ∈ L(w) then Tr Cw = −Cw and we only get a non-zero coefficient of Cw . If r ∈
/ L(w) then (4.4.3)
shows that we get a non-zero coefficient of Cz if and only if z—w and r ∈ L(z). Hence, we can
get a non-zero coefficient of Cz by multiplying Cw on the left by some Tr if and only if z—w and
L(z) * L(w).
Similarly, (4.4.4) shows that we can get a non-zero coeffcient of Cz by multiplying Cw on the
right by some Tr if and only if z—w and R(z) * R(w). Since {Tr |r ∈ S } generates Hn (q) we
can apply Proposition 4.1.1 to conclude that the relations {x ≤ y|x—y and L(x) * L(y)} and
L
{x ≤ y|x—y and R(x) * R(y)} generate ≤ and ≤ respectively. Lastly, we have x ≤ y if and only
R
L
R
LR
if there exists a chain x = x1 , x2 , . . . , xm = y such that either xi ≤ xi+1 or xi ≤ xi+1 for all
L
R
1 ≤ i < m. We have therefore shown:
Proposition 4.5.1. Let x, y ∈ Symn and consider the preorders ≤, ≤ and ≤ with respect to the
L
R
LR
Kazhdan-Lusztig basis. Then:
1. We have x ≤ y if and only if there exists a chain x = x1 —xw — . . . —xm = y such that
L
L(xi ) * L(xi+1 ) for all 1 ≤ i < m.
2. We have x ≤ y if and only if there exists a chain x = x1 —xw — . . . —xm = y such that
R
R(xi ) * R(xi+1 ) for all 1 ≤ i < m.
3. We have x ≤ y if and only if there exists a chain x = x1 —xw — . . . —xm = y such that either
LR
L(xi ) * L(xi+1 ) or R(xi ) * R(xi+1 ) for all 1 ≤ i < m.
If x ≤ y then the above Proposition shows there exists a chain x = x1 —xw — . . . —xm = y
L
such that L(xi ) * L(xi+1 ) for all 1 ≤ i < m. Now µ(x, y) = µ(x−1 , y −1 ) and so x—y if and
−1
−1
−1 such that
only if x−1 —y −1 . Hence we also have a chain x−1 = x−1
1 —xw — . . . —xm = y
−1
−1
−1
−1
L(xi ) = R(xi ) * R(xi+1 ) = L(xi+1 ) for all 1 ≤ i < m. In other words, x ≤ y . An identical
R
argument shows that if x ≤ y then x−1 ≤ y −1 . Hence:
R
L
Corollary 4.5.2. Let x, y ∈ Symn . Then x ≤ y if and only if x−1 ≤ y −1 . Hence x ∼ y if and only
L
L
R
if x−1 ∼ y −1 .
R
The last result of this section (which uses the above characterisation of ≤ and ≤) will be important
L
in the next Chapter:
45
R
Proposition 4.5.3. Let x, y ∈ Symn .
(i) If x ≤ y then R(x) ⊃ R(y). Hence if x ∼ y then R(x) = R(y).
L
L
(i) If x ≤ y then L(x) ⊃ L(y). Hence if x ∼ y then L(x) = L(y).
R
R
Proof. If x ≤ y then there exists a chain x = x1 —x2 — . . . —xm = y such that L(xi ) * L(xi+1 ) for
L
all 1 ≤ i < m. Hence, if we can show that xi —xi+1 and L(xi ) * L(xi+1 ) implies R(xi ) ⊃ R(xi+1 )
we will have (i). So assume x—y and L(x) * L(y). Then, there are two possibilities:
Case 1: x ≺ y. Suppose, for contradiction, that r ∈ R(y)\R(x). Then r ∈ L(y −1 )\L(x−1 ) and so
x−1 ≺ y −1 and rx−1 > x−1 but ry −1 < y −1 . By Proposition 4.3.3 this forces x−1 = ry −1 and so
xr = y. But then, by Lemma 1.4.3, we have L(x) ⊂ L(xr) = L(y). This contradicts L(x) * L(y).
Hence R(y)\R(x) = ∅. In other words R(x) ⊃ R(y).
Case 2: y ≺ x. Choose r ∈ L(x)\L(y). Then ry > y and rx < x and so, by Proposition 4.3.3, we
must have x = ry. Hence R(x) = R(ry) ⊃ R(y) by Lemma 1.4.3.
Hence (i) is proven. For (ii) note that if x ≤ y then x−1 ≤ y −1 and so R(x−1 ) = L(x) ⊃ L(y) =
R
L
R(y −1 ).
4.6
Notes
1. Our approach to the definition of the cells in the Hecke algebra is not the standard one.
Most authors use the conditions of Proposition 4.5.1 to define the cell preorders and then
remark that the multiplication formulae show that left multiplication by arbitrary a ∈ Hn (q)
maps Cw into the A-span of Cx satisfying x ≤ w. Our approach (which was suggested by
L
Fishel-Grojnowski [8]) takes longer to develop but is more motivated.
2. In the first two sections we have tried to use as much of the language of cellular algebras as
possible; hence the terms ‘cell module’ and ‘cell representation’ and the notation H(≤ w).
This is intended to motivate their introduction in the next chapter.
3. The elegant proof of part (ii) of Lemma 4.3.2 (that Px,w = Prx,w if rw < w) is due to Shi
[28]. Most authors use a cumbersome expansion of the identity Tr Cw = −Cw if rw < w to
derive the result.
4. The multiplication formulae developed in Section 4.4 suggest that only a basic set of information is needed to determine the action of Tr on Cw . We only need to know the appropriate
descent sets of each element, which elements are joined (by —), and what integer value corresponds to each joined pair (the µ function). This suggests that a graph might be constructed
to carry all the necessary information. This is the approach taken by Kazhdan-Lusztig [17].
They define a W -graph as a graph with vertices X and edge set Y such that each vertex x ∈ X
is labelled with a subset Ix of the simple transpositions and each edge {x, y} is labeled with
an integer µ(x, y). This graph is subject to the requirement that, if M is the free A-module
on the vertices of the graph, then defining:
(
−x
if r ∈ Ix
X
τr (x) =
1
2
qx + q
µ(y, x)y if r ∈
/ Ix
{x,y}∈Y
r∈Iy
46
yields a representation of Hn (q) on M via Tr 7→ τr . The results of this chapter show that if
we let X = Symn , Y = {{x, y}|x—y}, Ix = L(x) or R(x) and label each edge {x, y} with
µ(x, y) then we obtain a W -graph. It is customary to place the integers corresponding to the
()*+
elements of the descent set in circles. Hence, if L(w) = {s1 } we would represent this as /.-,
1.
For example, in the case n = 3 our W -graph looks like:
89:;
id u?>=<
∅ II
II 1
II
II
uu
u
u
?>=<
89:;
?>=<
2 s2
s1 1 TTT1T
1 89:;
TTTT jjjjjjj
T
j
T
j
T
j
1
1
TTTT
jjj
TT
jjjj
?>=<
89:;
89:;
?>=<
s1 s2 1 II
2 s2 s1
u
III1
1 uuu
III
u
uuu
HIJK
ONML
1, 2 s1 s2 s1
1 uuu
(Once one becomes accustomed to W -graphs much unnecessary information can be omitted).
If we consider the full subgraph consisting of vertices belonging to a particular left cell we get
a W -graph. The representation which it affords is the cell representation.
47
5
The Kazhdan-Lusztig Basis as a Cellular Basis
In this chapter we start by defining a cellular algebra and then spend the rest of the chapter
completing our study of the cells in Hn (q) with the goal of showing that Hn (q) is a cellular algebra.
We cannot complete this entirely via elementary means: in showing that no non-trival relation can
hold between two distinct left cells within a two-sided cell we must appeal to a theorem of Kazhdan
and Lusztig which we are unable to prove.
5.1
Cellular Algebras
In the previous chapter we introduced the cell preorders and showed how we can use them to define
cells which, in turn, lead to representations of the algebra. However, given an arbitrary algebra
with basis the concept of cells is far too general to be any use. We have seen, for example, that
different bases of the same algebra can yield very different cell structures: with respect to the
standard basis of Hn (q) there is only one left cell; whereas with respect to the Kazhdan-Lusztig
basis the left cell structure gives a partitioning at least as fine as that given by considering right
descent sets (Proposition 4.5.3). Also, even if a fixed basis yields an interesting cell structure, we
cannot be certain that we fully understand the representation theory of the algebra. We would like
to know, for example, whether the representations afforded by the cells are irreducible and which
cells afford isomorphic representations.
It is for this reason that we seek a class of algebras with fixed basis such that the resulting cell
structure is interesting and regular enough that questions of irreducibility and equivalence can be
addressed. Such a class of algebras is provided by Graham and Lehrer’s [12] definition of a ‘cellular
algebra’ which we now define. Let H be an R-algebra that is free as an R-module. Suppose that Λ is
a finite poset such that, for each λ ∈ Λ, we are given a finite set M (λ). Suppose further, that for each
λ |λ ∈ Λ and P, Q ∈ M (λ)}
λ
∈ H such that {CP,Q
λ ∈ Λ and P, Q ∈ M (λ), there is an element CP,Q
λ } a cellular basis if:
form an R-basis for H. We call {CP,Q
λ )∗ = C λ
(C1) The R-linear map ∗ defined by (CP,Q
Q,P is an anti-involution of H.
µ
(C2) Let H(< λ) be the R-span of those elements CU,V
with µ < λ in Λ and U, V ∈ M (µ). Then,
for all a ∈ H we have:
X
λ
aCP,Q
≡
ra (P 0 , P )CPλ 0 ,Q (mod H(< λ))
P 0 ∈M (λ)
where ra (P 0 , P ) ∈ R is independent of Q.
A cellular algebra is an algebra which has a cellular basis.
The simplest example of a cellular algebra is provided by the algebra R of n × n matrices over a
n
th
field k. If we let Λ = {n}, M (n) = {1, 2, . . . , n} and
P define Ci,j = eij where eij is the (i, j) matrix
unit, then a simple calculation shows that if a = λij eij ∈ R is arbitrary then (C2) is satisfied:
X
X
X
n
n
aCk,l
=
λij eij ekl =
λik eil =
λik Ci,l
(5.1.1)
i,j
i
i
n )∗ = C n sends e to e and hence is the transpose map. The identity (ab)T = bT aT
The map (Ci,j
ij
ji
j,i
shows that this is an anti-involution and hence (C1) is satisfied.
Let us consider what the axioms mean for the left, right and two-sided cells of H. If we define
µ
H(≤ λ) to be the R-submodule generated by CU,V
with µ ≤ λ then (C2) shows that H(≤ λ) is a
48
left ideal of H. Applying
∗
to (C2) yields:
X
λ
CQ,P
a∗ ≡
λ
ra (P 0 , P )CQ,P
0
(mod H(< λ))
P 0 ∈M (λ)
µ
λ
Hence H(≤ λ) is also a right ideal and so H(≤ λ) is a two sided ideal of H. Thus, if CU,V
and CP,Q
µ
λ
lie in the same two-sided cell then CU,V ∈ H(≤ λ) and CP,Q
∈ H(≤ µ). Hence λ ≤ µ ≤ λ and so
λ = µ. Hence, we can think of Λ as indexing the two-sided cells.1
λ
Now fix λ ∈ Λ and P, Q ∈ M (λ). Then (C2) shows that if we left multiply CP,Q
by a ∈ H we get
λ
0
λ
a linear combination of CP 0 ,Q for P ∈ M (λ) as well as some terms in H(< λ). Hence, if CP,Q
and
λ
λ
λ
CU,V lie in the same left cell then we must have Q = V . Similarly, if CP,Q and CU,V lie in the same
right cell we must have P = U . Hence we can think of pairs λ ∈ Λ and Q ∈ M (λ) as indexing the
left cells (and similarly for right cells).
This is made a little clearer by again considering the case when R is the algebra of n × n matrices
n is the cellular basis introduced above. It is easily seen that C n ∼ C n if and
over a field k and Ci,j
i,j L
k,l
n ∼ C n if and only if i = k. If 1 ≤ i, j ≤ n then we can depict the left cell
only if j = l and Ci,j
k,l
R
indexed by i and the right cell indexed by j as:
n
C1,1
n
C1,2
...
n
C2,1
n
C2,2
...
..
.
..
.
..
n
C1,i
...
n
C1,n
...
n
C2,n
∼
L
n
Cj,1
∼
R
n
Cj,2
..
.
..
.
n
Cn,1
...
n
C2,i
..
.
.
∼
R
n
Cj,n
∼
L
n
Cn,i
...
..
.
...
..
.
n
Cn,n
This situation is typical in a cellular algebra. We can imagine each quotient module H(≤ λ)/H(< λ)
as a ‘deformed matrix algebra’ which looks something like the above.
Now fix λ ∈ Λ and Q ∈ M (λ). If W is the free module on {CP,Q |P ∈ M (λ)} then (C2) shows that
we get a representation of H on W by defining:
X
aCP,Q =
ra (P, P 0 )CP 0 ,Q
P 0 ∈M (λ)
Now, axiom (C2) states that ra (P, P 0 ) is independent of Q. Thus we get isomorphic representations for all choices of Q ∈ M (λ). Hence it makes sense to define the cell representation of H
corresponding to λ as the the left module W (λ) with free R-basis {CP |P ∈ M (λ)} and H action
given by:
X
aCP =
ra (P 0 , P )CP 0
P 0 ∈M (λ)
1
λ
λ
Note that it is not a consequence of the axioms that CP,Q
and CU,V
lie in the same two-sided cell.
49
Due to limitations of space we cannot discuss cellular algebras in any more depth. However, it can
be shown that, over a field, certain quotients of W (λ) for all λ ∈ Λ constitute a full set of pairwise
inequivalent irreducible representations of H. Thus, the notion of a cellular algebra does indeed
address (and answer) the questions raised during the introduction to this section. The reader is
referred to Graham-Lehrer [12] for a more detailed account of the properties of cellular algebras as
well as a number of interesting examples.
5.2
Elementary Knuth Transformations
Our goal for the rest of this chapter it to show that the Kazhdan-Lusztig basis is a cellular basis for
Hn (q). It is perhaps surprising that standard tableaux provide a useful combinatorial framework
in which to discuss the cellular structure. The first indication of the usefulness of tableaux in this
context is in a result that we will show in this section: if x and y have the same Q-symbol then x
and y lie in the same left cell.
Recall from Chapter 2 that if x and y have the same Q-symbol, then, by the Symmetry Theorem x−1
and y −1 have the same P -symbol. Furthermore, if x−1 and y −1 have the same P -symbol then they
are Knuth equivalent and hence can be related by a sequence of elementary Knuth transformations
(Theorem 2.6.4). It is these elementary Knuth transformations that provide the key to the further
study of the cells in terms of tableaux.
In this section we develop functions between subsets of Symn which realise, algebraically, the elementary Knuth transformations. However, before we introduce these functions we need a technical
lemma which gives us information about certain cosets that arise repeatedly. If si , si+1 ∈ S are
simple transpositions let hsi , si+1 i denote the subgroup of Symn generated by si and si+1 .
Lemma 5.2.1. For all w ∈ Symn there exists a unique element w0 of minimal length in the coset
whsi , si+1 i. Moreover, w0 satisfies w0 (i) < w0 (i + 1) < w0 (i + 2) and we can depict the coset, with
the induced order, as:
w0 si si+1 si
NNN
N
p
ppp
w0 si si+1 VVV
w0 si+1 si
VVhVhVhVhhh
h
VVVV
h
hhhhh
w0 si+1
w0 si NN
NNN
N
w0
pp
ppp
Proof. Let w0 be an element of minimal length in whsi , si+1 i. Now if w0 (i) < w0 (i + 1) < w0 (i + 2)
does not hold then we can right multiply by si or si+1 to reduce the length (by Lemma 1.1.1). This
is a contradiction. Hence w0 (i) < w0 (i + 1) < w0 (i + 2). Since w0 (k) = w(k) if k ∈
/ {i, i + 1, i + 2},
0
0
0
0
w is uniquely determined by the condition w (i) < w (i + 1) < w (i + 2).
Since w0 is of minimal length we have w0 < w0 si < w0 si si+1 and w0 < w0 si+1 < w0 si+1 si
(otherwise we would have another element of length equal to w0 ). Also w0 si+1 < w0 si si+1 and
w0 si < w0 si+1 si by considering reduced expressions (Proposition 1.3.2). Now w0 si+1 < w0 si si+1
implies either w0 si+1 si ≤ w0 si si+1 or w0 si+1 si ≤ w0 si si+1 si (Lemma 1.3.1). But w0 si+1 si w0 si si+1 because they have the same length but are not equal. Hence w0 si si+1 < w0 si si+1 si .
Hence w0 si+1 si < w0 si si+1 si (again by considering reduced expressions).
50
We will denote by w0 the unique element of minimal length in whsi , si+1 i and call it the distinguished
coset representative.
Recall from Chapter 2 that if w = w1 w2 . . . wn ∈ Symn an elementary Knuth transformation of w
is a reordering of w according to one of the following patterns (where x < y < z):
. . . zxy . . . ↔ . . . xzy . . .
. . . yxz . . . ↔ . . . yzx . . .
We showed (in Theorem 2.6.4) that the P -symbols of x and y are equal if and only if x and y can
be linked by a sequence of elementary Knuth transformations.
Let us make some observations about the elementary Knuth transformations. Notice that, if
wi wi+1 wi+2 are three consecutive letters of w ∈ Symn then it is not always possible to perform
an elementary Knuth transformation: if wi < wi+1 < wi+2 or wi > wi+1 > wi+2 then no elementary Knuth transformations apply. Also note that, if it is possible to perform an elementary
Knuth transformation on wi wi+1 wi+2 then there is only one possibility. Hence, given an element
w ∈ Symn and a sequence wi wi+1 wi+2 upon which an elementary Knuth transformation can be
applied, the result of performing the elementary Knuth transformation is well-defined.
Suppose now that w = w1 w2 . . . wn ∈ Symn and that we wish to perform a Knuth transformation
on the subsequence wi wi+1 wi+2 for some 1 ≤ i < n − 1. We cannot have wi > wi+1 > wi+2
or wi < wi+1 < wi+2 and hence we must have either wi < wi+1 > wi+2 or wi > wi+1 < wi+2 .
Now if wi < wi+1 > wi+2 then wsi > w and wsi+1 < w (Lemma 1.1.1). On the other hand, if
wi > wi+1 < wi+2 then wsi < w and wsi+1 > w. In other words, if w ∈ Symn is such that we can
perform an elementary Knuth transformation on the i, i+1 and i+2 positions then R(w)∩{si , si+1 }
contains one element. On the other hand, R(w) ∩ {si , si+1 } contains one element then we must
have wi < wi+1 > wi+2 or wi > wi+1 < wi+2 (by Lemma 1.1.1 again) and hence an elementary
Knuth transformation is applicable. Hence, for 1 ≤ i < n − 1 we define:
Di = {w ∈ Symn |R(w) ∩ {si , si+1 } contains one element}
Then w ∈ Di if and only if it is possible to perform an elementary Knuth transformation on the
i, i + 1 and i + 2 positions of w.
Now, if w ∈ Di consider the coset whsi , si+1 i and let w0 be the distinguished coset representative.
As above we can depict the coset as:
w0 si si+1 si
vv
vv
vv
HH
HH
HH
II
II
II
u
uu
uu
u
u
w0 si si+1 T
w0 si+1 si
TTTT kkkk
T
kkkkTTTTTT
kkkk
k
w0 si+1
w0 si I
w0
Now consider the elements w0 si , w0 si+1 , w0 si si+1 and w0 si+1 si . Right multiplication by si lifts
w0 si+1 and w0 si si+1 and lowers w0 si and w0 si+1 si . On the other hand right multiplication by
si+1 lifts w0 si and w0 si+1 si and lowers w0 si+1 and w0 si si+1 . Hence all of these elements are in
Di . Now right multiplication by both si and si+1 lifts w0 and lowers w0 si si+1 si and so neither w0
51
nor w0 si si+1 si are in Di . Hence w must be one of the ‘middle’ elements w0 si , w0 si+1 , w0 si si+1
or w0 si+1 si . The above comments also make it clear that either wsi ∈ Di or wsi+1 ∈ Di but not
both. In other words Di ∩ {wsi , wsi+1 } contains precisely one element and so we can define a map
Ki : Di → Di by:
Ki (w) = the unique element of Di ∩ {wsi , wsi+1 }
Note that, if w ∈ Di then Ki (w) = wr for some r ∈ {si , si+1 }. Then wr2 = w ∈ Di and so
Ki (Ki (w)) = w. We have therefore shown:
Lemma 5.2.2. Ki is an involution on Di .
The following proposition shows that the functions Ki for 1 ≤ i < n − 1 realise the elementary
Knuth transformations:
Proposition 5.2.3. Suppose that w ∈ Symn and that it is possible to perform an elementary
Knuth transformation on the i, i + 1 and i + 2 positions of w. Then w ∈ Di and Ki (w) is the
permutation obtained from w by performing the only possible elementary Knuth transformation on
the subsequence wi wi+1 wi+2 of w.
Proof. We have already seen that w ∈ Di if and only if it is possible to perform an elementary Knuth
transformation on the i, i + 1 and i + 2 positions and that, in this case, only one elementary Knuth
transformation is possible. Now, consider the coset whsi , si+1 i and let w0 be the distinguished coset
representative. Let x = w0 (i), y = w0 (i + 1) and z = w0 (i + 2) so that w0 has the form . . . xyz . . .
with x < y < z. Now, if wsi < w and wsi+1 > w then either w = w0 si or w = w0 si+1 si . In the
first case Ki (w) = w0 si si+1 and so Ki affords the elementary Knuth transformation . . . yxz . . . 7→
. . . yzx . . . . In the second case Ki (w) = w0 si+1 and so Ki affords the transformation . . . zxy . . . 7→
. . . xzy . . . . Now Ki is an involution and so Ki maps . . . xzy . . . to . . . zxy . . . and . . . zxy . . . to
. . . xzy . . . . Hence, the action of Ki agrees with the the elementary Knuth transformations for
all arrangements of x, y, z in the i, i + 1 and i + 2 positions of w for which elementary Knuth
transformations are possible.
The following is immediate:
Corollary 5.2.4. Suppose that x, y ∈ Symn . Then the P -symbols of x and y are equal if and
only if there exists a sequence i1 , i2 , . . . im such that Kik−1 Kik−2 . . . Ki1 (x) ∈ Dik for all k and
y = Kim Kim−1 . . . Ki1 (x)
Proof. We have seen (in Theorem 2.6.4) that the P -symbols of x and y are equal if and only if
there exists a chain x = x1 , x2 , . . . , xm = y in which xk+1 is obtained from xk by performing an
elementary Knuth transformation in the ik , ik + 1 and ik + 2 positions of xk for some ik . Now,
from above xk ∈ Dik and xk+1 = Kik xk for all k. Hence Kik−1 Kik−2 . . . Kik1 (x) ∈ Dik for all k and
y = Kim Kim−1 . . . Ki1 (x).
We can now begin to apply the function Ki to the cells in the Hecke algebra:
Lemma 5.2.5. If x ∈ Di then x ∼ Ki (x).
R
52
Proof. Since Ki (x) = xsi or Ki (x) = xsi+1 we have µ(x, Ki (x)) 6= 0 by Lemma 4.3.2(i)2 and so
x—Ki (x). If Ki (x) = xsi then si ∈ R(x) if and only if si ∈
/ R(Ki (x)). Similarly, if Ki (x) = xsi+1
then si+1 ∈ R(x) if and only if si+1 ∈
/ R(Ki (x)). But, since x and Ki (x) are in Di only one
of si and si+1 are elements of R(x) and similarly for R(Ki (x)). Hence R(x) * R(Ki (x)) and
R(Ki (x)) * R(x). Hence x ∼ Ki (x).
R
This allows us to prove:
Proposition 5.2.6. Suppose x, y ∈ Symn have the same Q-symbol. Then x ∼ y.
L
Proof. If x and y have the same Q-symbol then, by the Symmetry Theorem (Theorem 2.5.2), x−1
and y −1 have the same P -symbol. Hence, by Corollary 5.2.4, there exists a sequence i1 , i2 , . . . ,
im such that Kik−1 . . . Ki1 x−1 ∈ Dik for all k and y −1 = Kim Kim−1 . . . Ki1 x−1 . Now from above
x−1 ∼ Ki1 x−1 ∼ Ki2 Ki1 x−1 ∼ . . . ∼ y −1 . Hence x−1 ∼ y −1 and so x ∼ y (Corollary 4.5.2).
R
5.3
R
R
R
R
L
The Change of Label Map
In the previous section we saw that if x and y have the same Q-symbol then they lie in the same
left cell. In the next section we will prove the remarkable converse, thus establishing that x and y
lie in the same left cell if and only if their Q-symbols are the equal. It then follows easily that x
and y are in the same two-sided cell if and only if P (x) and P (y) have the same shape.
Assuming this result, we can label the left cells within any two-sided cell by standard tableaux
of shape λ. The next step in showing that the Kazhdan-Lusztig basis is cellular is to show that
the representations afforded by the left cells within a two-sided cell are isomorphic. To do this we
need, for fixed Q1 and Q2 of shape λ, a map between the sets {(P, Q1 )|P standard of shape λ}
and {(P, Q2 )|P }. The obvious map (which sends (P, Q1 ) to (P, Q2 )) will be shown to yield an
isomorphism of representations. We call this the change of label map.
To examine the effect of the change of label map on the left cells we need a way of realising the
map algebraically. However, at this point it is not at all obvious how this might be achieved. It
turns out that if x, y ∈ Di , x ∼ (P, Q) and y ∼ (P 0 , Q) then Ki (x) ∼ (P, R) and Ki (y) ∼ (P 0 , R)
for some standard tableau R. Hence, we can use chains of elementary Knuth transformations to
realise the change of label map. For the moment, however, we will be content to investigate the
effects of elementary Knuth transormations on the function µ and relation ∼. The fact that the
L
elementary Knuth tranformation realise the change of label map will emerge as a corollary in the
next section.
If Ki does indeed realise the change of label map then it would send x ∼ (P, Q) and y ∼ (P 0 , Q)
to Ki (x) ∼ (P, R) and Ki (y) ∼ (P 0 , R) for some standard tableau R. Hence, by the results of
the previous section, we would have Ki (x) ∼ Ki (y). The aim of this section is to prove that this
L
holds in general. To show that if x, y ∈ Di and x ∼ y implies Ki (x) ∼ Ki (y) we need two facts:
L
L
that L(x) = L(Ki (x)) and that µ(x, y) = µ(Ki (x), Ki (y)). The first is easy: we have seen that
2
Up until this chapter most identities involving Kazhdan-Lusztig polynomials have been developed with simple
transpositions etc. acting on the left. This reflects the conventional focus in the literature on left cells. However,
due to the nature of the Knuth transformations, it will be more convenient to work on the right during this chapter.
We will refer without comment to left-hand results from previous chapters. The conversion is always straightforward
using the anti-involution ∗ and the fact that Px,w = Px−1 ,w−1 (Proposition 3.6.2).
53
x ∼ Ki (x) and hence L(x) = L(Ki (x)) by Proposition 4.5.3. However, the second requires a long
R
and intricate proof. In the proof, we follow the original argument of Kazhdan-Lusztig [17] closely.
Proposition 5.3.1. Suppose that x, y ∈ Di .
(i) If x−1 y ∈ hsi , si+1 i then x ≺ y if and only if Ki (y) ≺ Ki (x). In this case µ(x, y) =
µ(Ki (y), Ki (x)) = 1.
(ii) If x−1 y ∈
/ hsi , si+1 i then x ≺ y if and only if Ki (x) ≺ Ki (y). In this case µ(x, y) =
µ(Ki (y), Ki (x)).
Proof. Let us fix some notation which we will use throughout the proof. If x ≺ y then Px,y has
degree 12 (`(y) − `(x) − 1). If P 0 ∈ Z[q] is another polynomial write Px,y ∼ P 0 if Px,y − P 0 has degree
strictly less than 21 (`(y) − `(x) − 1). Thus Px,y ∼ P 0 if and only if P 0 has degree 21 (`(y) − `(x) − 1)
and leading coefficient µ(x, y).
Since x ∈ Di we have either xsi < x and xsi+1 > x or xsi+1 < x and xsi > x. In order to deal
with these two configurations simultaneously we will use r and t to denote either si and si+1 or
si+1 and si . In each case we will define r and then assume that t is defined such that t ∈ {si , si+1 }
with t 6= r.
For (i) we have x ≺ y and x and y are in the same left coset of hsi , si+1 i. Now, let w0 be the
distinguished coset representative of xhsi , si+1 i. Since there are only four elements of xhsi , si+1 i in
Di and x < y (since x ≺ y) we must have x = w0 r for r ∈ {si , si+1 } and hence either y = w0 rt or
y = w0 tr. Thus µ(x, y) = 1 by Lemma 4.3.2(i). Now, since x = w0 r we have Ki (x) = w0 rt and
Ki (y) = w0 r or Ki (y) = w0 t. In either case Lemma 4.3.2(i) applies again to yield Ki (y) ≺ Ki (x)
and µ(Ki (y), Ki (x)) = 1. The converse follows since Ki is an involution.
The proof of (ii) is more difficult. The proof is by cases:
Case 1: x−1 Ki (x) = y −1 Ki (y).
Hence Ki (x) = xr and Ki (y) = yr for some r ∈ {si , si+1 }. Now, if xr > x and yr < y then
Proposition 4.3.3 applies to give that x = yr and so x and y lie in the same left coset of hsi , si+1 i.
This is a contradiction. On the other hand if xr < x and yr > r then xt > x and yt < t (since
x, y ∈ Di ) and so Proposition 4.3.3 applies again to give the same contradiction. Hence, either
xr < x and yr < y or xr > x and yr > y. Throughout, we will argue the equivalence of x ≺ y
and Ki (x) ≺ Ki (y). Hence we can assume without loss of generality that xr < x and yr < y. (If
xr > x and yr > y then we can replace x by xr and y by yr to get that xr = Ki (x) ≺ yr = Ki (y)
if and only if Ki (xr) = x ≺ Ki (yr) = y.) So we are in the following situation:
yt
xt
w DDD
ww
DD
w
DD
ww
w
D
w
x
xtr RRR
l
RRR
lll
RRlRllll
l R
Ki
lll RRRRR
R
lll
xr
xrtrG
GG
zz
z
GG
zz
GG
G zzz
y BBB
yy
BB
y
BB
yy
B
yy
y
ytr QQ
m
m
QQQ
m
QQQ mmmmm
Ki
mmQQ
mmm QQQQQ
Q mmm
yr
yrtrE
EE
||
|
EE
|
EE
||
E
||
xrt
yrt
54
If x yr then Lemma 4.3.2(ii) gives Px,w = Pxr,wr and hence µ(x, w) = µ(xr, wr) and the result
follows. So assume that x ≤ yr. Since xr < x we have cx = 0 in the inductive formula (4.3.1) and
we have:
X −1 1
Px,y ∼ qPx,yr + Pxr,yr −
qz 2 qy2 µ(z, yr)Px,z
z
x≤z≺yr
zr<z
If either x ≺ y or xr ≺ yr then x ⊀ yr (since both x ≺ y and xr ≺ yr imply x = −y whereas
x ≺ yr forces x = y ) and hence we can rewrite the sum over all those z satisfying x < z ≺ yr and
zr < z. Also, if either x ≺ y or xr ≺ yr then there is a term on one side of the equation with degree
at least 21 (`(y) − `(x) − 1). Hence any terms of degree less than 12 (`(y) − `(x) − 1) can be ignored.
1
1
Now, if x ⊀ z then deg Px,z < 12 (`(z) − `(x) − 1) and so deg qz2 qy2 Px,y < 12 (`(y) − `(x) − 1). Hence
we can ignore any z in the sum which do not satisfy x ≺ z. Also, if x < z then the only term of
−1
1
−1
1
degree 12 (`(y) − `(x) − 1) in qz 2 qy2 Px,z is qz 2 qy2 multiplied by the leading coefficient of Px,z (which
1
has coefficient µ(x, z)). Hence we can replace Px,z with µ(x, z)q 2 (`(z)−`(x)−1) . Thus we have:
Px,y ∼ qPx,yr + Pxr,yr −
−1
X
1
1
qz 2 qy2 µ(z, yr)µ(x, z)q 2 (`(z)−`(x)−1)
z
x≺z≺yr
zr<z
∼ qPx,yr + Pxr,yr −
1
X
q 2 (`(y)−`(x)−1) µ(z, yr)µ(x, z)
z
x≺z≺yr
zr<z
Now assume that z appears in the sum. Then, if t ∈ R(z) then we have zt < z, xt > x and x ≺ z.
Then Proposition 4.3.3 applies to yield that z = xt. On the other hand if t ∈
/ R(z) then zt > z,
yrt < yr and y ≺ yrt. Again, Proposition 4.3.3 applies forcing z = yrt. Hence z = xt or z = yrt.
But we require r ∈ R(z). Now r ∈ R(xt) but r ∈
/ R(yrt). Hence the only possibility is z = xt.
Now, µ(x, xt) = 1 by Lemma 4.3.2(i) and so our expression becomes:
1
Px,y ∼ Pxr,yr + qPx,yr − q 2 (`(y)−`(x)−1) µ(xt, yr)
We have yrt < yr and so, by Lemma 4.3.2(iii), we have Px,yr = Pxt,yr . Hence:
1
qPx,yr = qPxt,yr ∼ q 2 (`(yr)−`(xt)−1)+1 µ(xt, yr)
1
= q 2 (`(y)−`(x)−1) µ(xt, yr)
1
Thus qPx,yr − q 2 (`(y)−`(x)−1) µ(xt, yr) is a polynomial of degree less than 12 (`(y) − `(x) − 1). Hence
Px,y ∼ Pxr,yr and so µ(x, y) = µ(Ki (x), Ki (y)).
Case 2: x−1 Ki (x) 6= y −1 Ki (y).
Hence Ki (x) = xr and Ki (y) = xt for r, t ∈ {si , si+1 } with r 6= t. Now, if xr < x and yt < x
then xt > x (since x ∈ Di ) and so Proposition 4.3.3 applies to yield x = yt. This contradicts
the fact that x and y lie in different left cosets of hsi , si+1 i. Interchanging r and t yields a similar
contradiction if xr > x and yt > y. Hence either xr < x and yt > y or xr < x and yt > y. As in
the previous case, we can assume without loss of generality that xr < x and yt > y since we argue
55
that x ≺ y if and only if Ki (x) ≺ Ki (y). So we are in the following situation:
xt
w DDD
ww
DD
w
DD
ww
w
D
w
xtr RRR
lx
RRR
lll
RRlRllll
l R
Ki
lll RRRRR
R
lll
xr
xrtrG
GG
zz
z
GG
zz
GG
G zzz
yy
yy
y
y
yy
ytr B
BB
BB
BB
yrtr QQ
yt [
mmm
QQQ
QQmmQmmmm
Ki
m QQQ
QQQ
mmm
Q
mmm
y
yrt F
FF
FF
FF
F
xrt
yr
|
||
||
|
||
If xr ≮ yt then x ≮ y (otherwise xr < x < y < yt) and so µ(Ki (x)), Ki (y)) = µ(x, y) = 0. So
we may assume that xr < yt. Now t ∈ R(xr) ∩ R(yt) and so xrt < y (Lemma 1.3.4). Similarly
xr < yt implies x < yt or x < ytr (Lemma 1.3.1) and so x < ytr.
Now assume that xr y. Then xr (yt)t = y and so, by Lemma 4.3.2(ii), Pxr,yt = Pxrt,y . If
xrt ≺ y then (xrt)r > xrt but yr < y forcing xrt = yr (Proposition 4.3.3). This contradicts the
fact that x and y do not lie in the same left coset of hsi , si+1 i and so we must have xrt ⊀ y. Hence
deg Pxrt,y < 12 (`(y) − `(xrt) − 1) = 12 (`(yt) − `(xr) − 1) and so xr ⊀ yt (since Pxr,yt = Pxrt,y ). On
the other hand xr y implies x y (otherwise xr < x ≤ y) and so x ⊀ y. Thus if xr y only
one case can occur: µ(x, y) = µ(Ki (x), Ki (y)) = 0.
Now assume that xr ≤ y. Then (xr)t < xr and so cxr = 0 in the inductive formula (4.3.1). This
yields:
X
− 1 12
Pxr,yt ∼ qPxr,y + Pxrt,y −
µ(z, y)qz 2 qyt
Pxr,z
z
xr≤z≺y
zt<z
As in Case 1, Pxr,z does not have a large enough degree to contribute if xr ⊀ z and if xr ≺ z we
−1
1
1
2
can replace qz 2 qyt
Pxr,z with q 2 (`(yt)−`(xr)−1) µ(xr, z). We have also seen that xrt ⊀ y and so Pxrt,y
has degree less than 21 (`(yt) − `(xr) − 1). Thus, we can rewrite our expression as:
Pxr,yt ∼ qPxr,y −
X
1
µ(z, y)µ(xr, z)q 2 (`(yt)−`(xr)−1)
z
xr≺z≺y
zt<z
Let us consider which z emerge in the sum. If r ∈ R(z) then (xr)r > xr, zr < z and xr ≺ z forcing
z = x. If r ∈
/ R(z) then zr > z, yr < y and z ≺ y forcing z = yr (both by Proposition 4.3.3). But
neither x not yr have t in their right descent set. Hence the sum is empty and we can conclude
that Pxr,yt ∼ qPxr,y . Now yr < y and hence Pxr,y = Px,y by Lemma 4.3.2(iii). Hence Pxr,yt ∼ qPx,y .
Hence xr ≺ yt if and only if x ≺ y and µ(Ki (x), Ki (y)) = µ(x, y).
As promised, we have:
56
Corollary 5.3.2. Suppose that x, y ∈ Di . Then x ≤ y if and only if Ki (x) ≤ Ki (y). Hence, if
L
L
x ∼ y then Ki (x) ∼ Ki (y).
L
L
Proof. Suppose first that x—y with L(x) * L(y). Then the above Proposition shows that Ki (x)—Ki (y).
Also L(x) = L(Ki (x)) and L(y) = L(Ki (y)) since x ∼ Ki (x) and y ∼ Ki (y) (Lemma 5.2.5 and
R
R
Proposition 4.5.3). Hence Ki (x)—Ki (y) and L(Ki (x)) * L(Ki (y)). Now, if x ≤ y then there exists
L
a chain x = x1 —x2 — . . . —xm = y with L(xj ) * L(xj+1 ) for all j < m. Now, by assumption, x
and y lie in Di and hence R(x) ∩ {si , si+1 } and R(y) ∩ {si , si+1 } contain precisely one element. But
since R(x) ⊃ R(y) we must have R(x) ∩ {si , si+1 } = R(y) ∩ {si , si+1 }. Furthermore, since R(x) ⊃
R(xj ) ⊃ R(y) for all j (since x ≤ xj ≤ y) we must have R(x) ∩ {si , si+1 } = R(xj ) ∩ {si , si+1 } and
L
L
hence xj ∈ Di for all j. Hence, by the above arguments, Ki (x) = Ki (x1 )—Ki (x2 )— . . . —Ki (xm ) =
Ki (y) with L(Ki (xj−1 )) * L(Ki (xj )) for all i < m and so Ki (x) ≤ Ki (y). If x ∼ y then x ≤ y and
L
L
L
y ≤ x. Thus, Ki (x) ≤ Ki (y) and Ki (y) ≤ Ki (x) and so Ki (x) ∼ Ki (y).
L
5.4
L
L
L
Left Cells and Q-Symbols
Equipped with the results of the previous section we can give a complete description of the left
cells in terms of the Q-symbol of the permutation.
Theorem 5.4.1. Let x, y ∈ Symn . Then x ∼ y if and only if Q(x) = Q(y).
L
Before we commence the proof we recall some definitions and results from Sections 2.7 and 2.8
which are central to the proof. If λ is a partition of n, we defined the column superstandard
tableau, Sλ , as the tableau obtained from a diagram of λ by filling it with 1, 2, . . . n successively
down columns. We defined the descent of a tableau P , denoted D(P ), as the set of i for which i + 1
occurs strictly below and weakly left of i in P and showed that, if w ∈ Symn , then si ∈ R(w) if
and only if i ∈ D(Q(w)) (Proposition 2.7.1). Lastly, we showed that if the tableau descent of P
contains the tableau descent of Sλ then the shape of P is dominated by λ and that Shape(P ) = λ
if and only if P = Sλ (Proposition 2.8.2). We will use these results without reference during the
proof. The argument is based on Ariki [1].
Proof. We have already seen in Proposition 5.2.6 that if x and y have the same Q-symbol then
they lie in the same left cell. It remains to show the converse.
Assume first that x ∼ y with x ∼ (Sλ , Q) and y ∼ (Sµ , Q0 ) where Sλ and Sµ are column superstanL
dard tableau. Now, define x0 by x0 ∼ (Sλ , Sλ ). Then x and x0 have the same P -symbol and hence
are Knuth equivalent. Hence, by Corollary 5.2.4, there exists a sequence i1 , i2 , . . . im such that:
Kik−1 Kik−2 . . . Ki1 (x) ∈ Dik for all 1 ≤ k ≤ m
x0 = Kim Kim−1 . . . Ki1 (x)
(5.4.1)
Now, x ∼ y and hence R(x) = R(y). Hence, y ∈ Di1 since x ∈ Di1 . Hence Ki (y) is a well defined
L
element and Ki (x) ∼ Ki (y) by Corollary 5.3.2. Repeating this argument we see that Ki2 Ki1 (y) is
L
a well-defined element satisfying Ki2 Ki1 (x) ∼ Ki2 Ki1 (y). Thus we may define y 0 by:
L
0
y = Kim Kim−1 . . . Ki1 (y)
57
(5.4.2)
We have y 0 ∼ x0 and hence R(y 0 ) = R(x0 ). Hence D(Q(y 0 )) = D(Q(x0 )) = D(Sλ ) and hence µ E λ.
L
Now the above argument is perfectly symmetrical in x and y and so we can repeat it with x and y
interchanged to get λ E µ. Hence λ = µ. Now D(Q(y 0 )) = D(Sλ ) and Q(y 0 ) has shape λ forcing
Q(y 0 ) = Sλ . Hence y 0 ∼ (Sλ , Sλ ) and so y 0 = x0 . Applying Ki1 Ki2 . . . Kim to (5.4.1) and (5.4.2) we
get x = y since Ki is an involution.
Now let x ∼ y with x ∼ (P (x), Q(x)) and y ∼ (P (y), Q(y)) be arbitrary. Let λ and µ be the shape
L
of P (x) and P (y) respectively. Define x̂ and ŷ by x̂ ∼ (Sλ , Q(x)) and ŷ ∼ (Sµ , Q(y)). Now, x̂ ∼ x
L
since x and x̂ have the same Q-symbol and similarly ŷ ∼ y. Hence x̂ ∼ ŷ and the above argument
L
L
applies to force x̂ = ŷ. Hence Q(x) = Q(x̂) = Q(ŷ) = Q(y).
Using the Symmetry Theorem it is straightforward to extend this to the right and two-sided cells:
Corollary 5.4.2. Let x, y ∈ Symn .
(i) Then x ∼ y if and only if P (x) = P (y).
R
(ii) Then x ∼ y if and only if P (x) and P (y) have the same shape.
LR
Proof. For (i) we have x ∼ y if and only if x−1 ∼ y −1 (Corollary 4.5.2) which, from above, occurs if
R
L
and only if Q(x−1 ) = Q(y −1 ). By the Symmetry Theorem (Theorem 2.5.2) we have Q(x−1 ) = P (x)
and Q(y −1 ) = P (y).
For (ii) note that (i) combined with the above theorem yields that if x ∼ y or x ∼ y then
L
Shape(P (x)) = Shape(P (y)).
R
If x ∼ y then there exists a chain x = x1 , x2 , . . . , xm such
LR
that xi ∼ xi+1 or xi ∼ xi+1 for 1 ≤ i < m. Hence Shape(P (x)) = Shape(P (x1 )) = · · · =
L
R
Shape(P (xm )) = Shape(P (y)). On the other hand if x ∼ (P, Q) and y ∼ (P 0 , Q0 ) and x and y have
the same shape then x ∼ (P 0 , Q) ∼ y and so x ∼ y.
L
R
LR
The following result shows that the elementary Knuth transformations realise the change of label
map:
Corollary 5.4.3. Assume that x ∼ (P, Q) and y ∼ (P 0 , Q) and that R is an arbitrary standard
tableau of the same shape as Q. Then there exists a sequence i1 , i2 , . . . im and well-defined elements Kim Kim−1 . . . Ki1 (x) and Kim Kim−1 . . . Ki1 (y) satisfying Kim Kim−1 . . . Ki1 (x) ∼ (P, R) and
Kim Kim−1 . . . Ki1 (y) ∼ (P 0 , R).
Proof. Define x̂ by x̂ ∼ (P, R). Then x and x̂ have the same P -symbol and hence are Knuth equivalent. Hence, by Corollary 5.2.4, there exists a sequence i1 , i2 , . . . , im such that Kik−1 Kik−2 . . . Ki1 (x) ∈
Dik for all k and x̂ = Kim Kim−1 . . . Ki1 (x). Now, x and y have the same Q-symbol and hence are
in the same left cell. Hence R(x) = R(y) and so y ∈ Di1 and, by Corollary 5.3.2, Ki1 (x) ∼
L
Ki1 (y). Now, since Ki1 (x) and Ki1 (y) lie in the same left cell we have, by Proposition 4.5.3,
that R(Ki1 (x)) = R(Ki1 (y)). Hence Ki1 (y) ∈ Di2 and we have a well defined Ki2 Ki1 (y) satisfying Ki2 Ki1 (x) ∼ Ki2 Ki1 (y). Repeating this argument we see that we have a well-defined
L
ŷ = Kim Kim−1 . . . Ki1 (y) and x̂ ∼ ŷ. Now P (ŷ) = P 0 since Knuth equivalent permutations have
L
the same P -symbol. Also, x̂ ∼ ŷ implies (from the theorem above) that R = Q(x̂) = Q(ŷ). Hence
x̂ ∼ (P, R) and ŷ ∼ (P 0 , R).
L
58
The following can be used to show that representations afforded by left cells corresponding to
Q-symbols of the same shape are isomorphic:
Corollary 5.4.4. Suppose that P , P 0 , Q and R are standard tableaux of the same shape and that
x ∼ (P, Q), y ∼ (P 0 , Q), x̂ ∼ (P, R) and ŷ ∼ (P 0 , R). Then µ(x, y) = µ(x̂, ŷ).
Proof. By the above Corollary we can find i1 , i2 , . . . , im such that Kim Kim−1 . . . Ki1 (x) = x̂ and
Kim Kim−1 . . . Ki1 (y) = ŷ. By repeated application of Proposition 5.3.1 we have:
µ(x̂, ŷ) = µ(Kim Kim−1 . . . Ki1 (x), Kim Kim−1 . . . Ki1 (y))
= µ(Kim−1 Kim−2 . . . Ki1 (x), Kim−1 Kim−2 . . . Ki1 (y))
..
.
= µ(x, y)
5.5
Property A
In the previous section we obtained a complete description, in terms of P and Q-symbols, for the
left, right and two-sided cells. Given this characterisation we might suspect that our cellular basis
λ
has the form Cw = CP,Q
where w ∼ (P, Q) and λ is the shape of P . For this basis to be cellular,
λ
axiom (C2) states that, in left multiplying CP,Q
by arbitrary a ∈ Hn (q) the only elements in the
same two-sided cell which appear with non-zero coefficient are of the form CPλ 0 ,Q with P 0 another
standard tableau of shape λ. In other words, in left multiplying Cw by a ∈ Hn (q) the only elements
in the same two-sided cell as w which appear are in the same left cell as w.
If Cx appears with non-zero coefficient in aCw for some a ∈ H then (recalling our original definition
of the cell preorders given in Section 4.1) we have x ≤ w. Hence we must show that if x and w are
L
in the same two-sided cell and x ≤ w then x and w are in the same left cell. Unfortunately, this
L
statement is equivalent to a deep result of Kazhdan-Lusztig theory known as “Property A”. This
is proved by Kazhdan and Lusztig using intersection cohomology in [18].3 We will be content to
offer a proof in a special case and refer the courageous reader to Kazhdan-Lusztig.
Property A. Suppose that x, y ∈ Symn satisfy x—y, L(x) * L(y) and R(x) * R(y). Then x and
y do not lie in the same two-sided cell.
Proof in a special case: Assume that y ∼ (P, Sλ ) where λ is the shape of P and Sλ is the column
superstandard tableau of shape λ. Then, since x—y and L(x) * L(y) we have x ≤ y. Hence
L
R(x) ⊃ R(y) (Proposition 4.5.3). If P (x) and P (y) have the same shape then Lemma 2.7.2 forces
Q(x) = Sλ . This contradicts the fact that R(x) * R(y) (since si ∈ R(x) if and only if i ∈ D(Q(x))
by Proposition 2.7.1). Hence P (x) and P (y) do not have the same shape and so lie in different
two-sided cells.
We use this to prove:
Lemma 5.5.1. Suppose x ≤ y and P (x) and P (y) have the same shape. Then R(x) = R(y).
L
3
Actually, Kazhdan and Lusztig prove that, in the Hecke algebra of a Weyl group (of which the symmetric group
is an example), the coefficients of the Kazhdan-Lusztig polynomials are all positive. See the notes to this chapter.
59
Proof. Since x ≤ y there exists a chain x = x1 —x2 — . . . —xm = y such that L(xi ) * L(xi+1 ) for
L
all 1 ≤ i < m. In particular x ≤ xi ≤ y for all i. By assumption x and y have the same shape and
L
L
so x ∼ y. Hence for all i we have x ≤ xi ≤ y ≤ x and so all of the xi lie in the same two-sided
LR
LR
LR
LR
cell.
Now, assume for contradiction that R(xi ) 6= R(xi+1 ) for some i. Since xi ≤ xi+1 and so R(xi ) ⊃
L
R(xi+1 ). Hence, if R(xi ) 6= R(xi+1 ) then R(xi ) * R(xi+1 ). But then we have xi —xi+1 , L(xi ) *
L(xi+1 ) and R(xi ) * R(xi+1 ) and we can apply Property A above to conclude that xi and xi+1 do
not lie in the same two sided cell. This is a contradiction. Hence R(xi ) = R(xi+1 ) for all 1 ≤ i < m
and so R(x) = R(y).
We can now prove that the situation described at the start of this section cannot occur:
Proposition 5.5.2. Let x, y ∈ Symn .
(i) If x ≤ y and x ∼ y then x ∼ y.
LR
L
L
(ii) If x ≤ y and x ∼ y then x ∼ y.
LR
R
R
Proof. For (i) assume that x ∼ (P, Q) and y ∼ (P 0 , Q0 ). Let λ by the shape of y and define ŷ
by ŷ ∼ (P 0 , Sλ ). Since y and ŷ have the same P -symbol there exists a sequence i1 , i2 , . . . im such
that Kik−1 Kik−2 . . . Ki1 (y) ∈ Dik for all k and ŷ = Kim Kim−1 . . . Ki1 (y) by Corollary 5.2.4. We
have x ≤ y and x ∼ y and so R(x) = R(y) by Lemma 5.5.1 above. Hence x ∈ Di1 if and only
LR
L
if y ∈ Di1 . Thus, we can apply Corollary 5.3.2 to get that Ki1 (x) ≤ Ki1 (y). Now, from above
L
R(Ki1 (x)) = R(Ki1 (y)) and so Ki1 (x) ∈ Di2 . Hence, Ki2 Ki1 (x) ≤ Ki2 Ki1 (y). Continuing in this
L
fashion we see that we get a well defined x̂ by defining x̂ = Kim Kim−1 . . . Ki1 (x) and we have x̂ ≤ ŷ.
L
Thus R(x̂) ⊃ R(ŷ) and so D(Q(x̂)) ⊃ D(Q(ŷ)) = D(Sλ ) (Proposition 2.7.1). But, by assumption x
and y have the same shape. Hence x̂ and ŷ have the same shape and Lemma 2.7.2 forces Q(x̂) = Sλ .
Hence, by Corollary 5.4.3, we see that Ki1 Ki2 . . . Kim (x̂) and Ki1 Ki2 . . . Kim (ŷ) have the same Qsymbol. But Ki is an involution and hence Ki1 Ki2 . . . Kim (x̂) = x and Ki1 Ki2 . . . Kim (ŷ) = y. Thus
x and y have the same Q-symbol and so x ∼ y (Proposition 5.2.6).
L
For (ii), if x ≤ y and x ∼ y then
LR
R
x−1
≤
y −1
and x−1 ∼ y −1 by Corollary 4.5.2. By (i) we have
LR
L
x−1 ∼ y −1 implying x ∼ y (again by Corollary 4.5.2).
L
5.6
R
The Main Theorem
Let Λ denote the set of partitions of n. If λ ∈ Λ write M (λ) for the set of standard tableau of shape
λ
λ. If x ∈ Symn and x ∼ (P, Q) under the Robinson-Schensted correspondence write CP,Q
= Cx
where λ is the shape of P . If λ, µ ∈ Λ write λ ≤ µ if there exists x, y ∈ Symn such that x ≤ y with
LR
Shape(P (x)) = λ and Shape(P (y)) = µ. Since ≤ is a preorder ≤ is a preorder. Now if λ ≤ µ ≤ λ
LR
then there exists x and y such that Shape(P (x)) = λ and Shape(P (y)) = µ with x ≤ y ≤ x.
LR
LR
Hence x ∼ y and so λ = µ by Corollary 5.4.2. Hence ≤ is a partial order on partitions. Using this
LR
new notation we can show that the Kazhdan-Lusztig basis is cellular.
60
As in the definition of a cellular algebra define H(< λ) and H(≤ λ) to be the A-span of those
µ
with U, V ∈ M (µ) satisfying µ < λ and µ ≤ λ respectively. We start by
basis elements CU,V
reformulating the multiplication formulae of Section 4.4 in the quotient module H(≤ λ)/H(< λ):
λ
Lemma 5.6.1. The action of Ti on CP,Q
in H(≤ λ)/H(< λ) is given by:
(
λ
−CP,Q
if i ∈ D(P )
λ
X
Ti CP,Q
≡
(mod H(< λ))
1
λ
0
λ
2
qCP,Q + q
µ(P , P )CP 0 ,Q if i ∈
/ D(P )
P 0 ∈M (λ)
i∈D(P 0 )
Where µ(P 0 , P ) ∈ A is independent of Q.
Proof. First note that H(≤ λ) and H(< λ) are certainly ideals of Hn (q) because if w ∈ Symn is
such that P (w) has shape λ then H(≤ λ) = H(≤ w) and H(< λ) = H(< w) by the way that we
LR
LR
have defined ≤. Now, recall the multiplication
formula given in Section 4.4:
(
−Cw
if si ∈ L(w)
Tsi Cw =
1 X
qCw + q 2
µ(z, w)Cz if si ∈
/ L(w)
(5.6.1)
z—w
r∈L(z)
We want to reduce (5.6.1) modulo H(< w). If Cz appears with non-zero coefficient we have z ≤ w
LR
L
(recalling the original definition of the cell preorders given in Section 4.1). If Cz ∈
/ H(< w) then
LR
z ∼ w. Hence, by Proposition 5.5.2, the only Cz not in H(< w) which emerge with non-zero
LR
LR
coefficient in (5.6.1) satisfy z ∼ w. Also, since we have defined µ(z, w) = 0 if the relation z—w
L
does not hold, we can omit
the
requirement that z—w:
(
−Cw
if si ∈ L(w)
X
Tsi Cw ≡
(mod H(< w))
(5.6.2)
1
LR
qCw + q 2
µ(z, w)Cz if si ∈
/ L(w)
z
si ∈L(w)
z∼w
L
We want to reinterpret (5.6.2) in terms of our new notation for the basis. Fix w ∼ (P, Q). Then,
summing over µ(z, w)Cz with z ∼ w is equivalent to summing over µ(z, w)CPλ 0 ,Q with P 0 ∈ M (λ)
L
by Proposition 5.4.1. Also, if si ∈ L(w) then i ∈ D(P ) by Lemma 2.7.1. Lastly, by the remarks at
the start of the proof H(< w) = H(< λ). Hence (5.6.2) becomes:
LR
(
λ
−CP,Q
if i ∈ D(P )
λ
X
Ti CP,Q
≡
(mod H(< λ))
(5.6.3)
1
λ
λ
qCP,Q + q 2
µ(z, w)CP 0 ,Q if i ∈
/ D(P )
P 0 ∈M (λ)
z∼(P 0 ,Q)
i∈D(P 0 )
Now, by Corollary 5.4.4, if R is another standard tableau of shape λ and if ŵ ∼ (P, R) and
ẑ ∼ (P 0 , R) then µ(ẑ, x̂) = µ(z, w). Hence, if we define µ(P 0 , P ) = µ(z, w) we obtain a well-defined
integer independent of Q. Thus, (5.6.3) becomes:
(
λ
−CP,Q
if i ∈ D(P )
λ
X
Ti CP,Q
≡
(mod H(< λ))
1
λ
0
λ
qCP,Q + q 2
µ(P , P )CP 0 ,Q if i ∈
/ D(P )
P 0 ∈M (λ)
i∈D(P 0 )
We can now prove:
61
λ |λ ∈ Λ, P, Q ∈ M (λ)} is a cellular basis for
Theorem 5.6.2. The Kazhdan-Lusztig basis {CP,Q
Hn (q).
λ
λ
Proof. The A-linear map given by CP,Q
7 CQ,P
→
sends Cw to Cw−1 by the Symmetry Theorem
∗
(Theorem 2.5.2) and hence is the map . We have seen in Section 3.4 that ∗ is an anti-involution
and hence (C1) is satisfied.
Define ri (P 0 , P ) by:


−1



0
ri (P 0 , P ) =

q



q 12 µ(P 0 , P )
if
if
if
if
i ∈ D(P )
i ∈ D(P )
i∈
/ D(P )
i∈
/ D(P )
Lemma 5.6.1 shows that, for all i we have:
X
λ
Ti CP,Q
≡
ri (P 0 , P )CPλ 0 ,P
and
and
and
and
P0
P0
P0
P0
=P
6= P
=P
6= P
(mod H(< λ))
(5.6.4)
P 0 ∈M (λ)
Now Hn (q) is generated by {Ti |1 ≤ i < n} and hence (5.6.4) shows that, for all a ∈ Hn (q), we can
find ra (P 0 , P ), independent of Q, such that:
X
λ
aCP,Q
≡
ra (P 0 , P )CPλ 0 ,P (mod H(< λ))
P 0 ∈M (λ)
λ |λ ∈ Λ, P, Q ∈ M (λ)} is a cellular basis.
Hence (C2) is satisfied and so {CP,Q
5.7
Notes
1. The subgroup hsi , si+1 i is known as a parabolic subgroup. More generally, in a Coxeter group
a (standard) parabolic subgroup is any subgroup generated by a subset of simple reflections.
See, for example, Bourbaki [3] or Humphreys [15].
2. The functions Ki on the subsets Di are the ‘right star operations’ of Kazhdan-Lusztig [17].
In Kazhdan-Lusztig Di is denoted by DR (si , si+1 ) and Ki (x) is denoted by x∗ . We have not
introduced the ‘left star operations’ (denoted ∗ x) or DL (si , si+1 ). These correspond to the
‘dual Knuth transformations’. See, for example, Stanley [30] for the definition of dual Knuth
transformations.
3. The idea of viewing an element of x ∈ Di as embedded in the coset xhsi , si+1 i was suggested
by Shi [28].
4. Corollary 5.4.3 is a purely combinatorial result which we prove using the connection between
Q-symbols and left cells. For a combinatorial proof see Schützenberger [27].
5. As we mentioned in the text, the proof of Proposition 5.3.1 follows the original argument of
Kazhdan-Lusztig [17] closely.
6. The proof that x ∼ y if and only if Q(x) = Q(y) is based on Ariki [1] who, in turn, cites
L
Garsian-McLarnan [11].
62
7. In [18], Kazhdan and Lusztig prove that, in the Hecke algebra of a Weyl group (of which the
symmetric group is an example), the coefficients of the Kazhdan-Lusztig polynomials are all
non-negative integers. In fact this implies Property A. For a proof of this implication see
Dyer [4].
63
64
A
Appendix
A.1
An Alternative Proof of Kazhdan and Lusztig’s Basis Theorem
The proof of Theorem 3.5.1 given in Chapter 3 follows the original proof of Kazhdan and Lusztig
[17] closely. We gave it in its entirety for two reasons: it reveals the intricacy of the KazhdanLusztig basis and yields explicit inductive information. In this appendix we present a briefer proof
originally due to Lusztig [19] and presented elegantly in Soergel [29].
1
−
It is convenient to renormalise, and define a new basis of Hn (q) by Tew = qw 2 Tw . A simple
calculation shows that the multiplication in (3.1.2) becomes:
(
Tewr
if wr > w
Tew Ter =
(A.1.1)
1
1
−
Tewr + (q 2 − q 2 )Tew if wr < w
Under the new basis the identity in (3.3.1) becomes:
1
1
Ter−1 = Ter + (q − 2 − q 2 )Teid
(A.1.2)
Also, making use of Lemma 3.3.2 we have:
−1
1
ι(Tew ) = ι(qw 2 Tw ) = qw2 (
X
1
Ry,w Ty ) = qw2 Rw,w Tw +
X
Ry,w Ty
y<w
y≤w
Using the fact that Rw,w = q −`(w) (Proposition 3.3.2) and the definition of Tey yields:
ι(Tew ) = Tew +
X
1
qy2 Ry,w Tew
(A.1.3)
y<w
Now, in terms of this new basis Theorem 3.5.1 states that, for all w ∈ Symn there is a unique element
1
P
−1
Cw =
y w qw2 qy 2 P y,w Tew with Px,w ∈ Z[q], Pw,w = 1 and deg Px,w ≤ 1 (`(w) − `(x) − 1) for
y≤w
2
x < w. Now if Px,w has degree at most
1
2 (`(w) − `(x) − 1)
1
then
1
2
−1
qw qx 2 P x,w
1
= q 2 (`(w)−`(x)) P x,w is a
1
−1
1
1
polynomial in Z[q 2 ] without constant term. In other words, if x < w then qw2 qx 2 P x,w ∈ q 2 Z[q 2 ].
Hence, Theorem 3.5.1 is equivalent to:4
Theorem 3.5.1 (Restatement). For all w ∈ Symn there exists a unique element Cw such that
P
1
1
ι(Cw ) = Cw and Cw ∈ Tew + y<w q 2 Z[q 2 ]Tey .
P
P
Proof. We first show uniqueness. So assume that Cw = Tew + y<w hy Tey and Cw0 = Tew + y<w h0y Tey
both satisfy the conditions of the theorem. Then Cw − Cw0 is also ι-invariant since ι is a homomorphism. If Cw 6= Cw0 then there exists x maximal (with respect to the Bruhat order) so that
hx 6= h0x . Then Cw − Cw0 = ι(Cw − Cw0 ) implies:
X
X
(hy − h0y )Tey =
ι(hy − h0y )ι(Tey )
y<w
y<w
4
It is actually not true that the two statements are entirely equivalent: if x < w the original theorem states
1
that Px,w is a polynomial in q whereas the restatement only guarantees that Px,w is a polynomial in q 2 . Once the
restatement is proved it is straightforward to prove inductively that Px,w indeed lies in Z[q].
65
Since x is maximal we see from (A.1.3) that the coefficient of Tex on the right hand side is ι(hx −h0x ).
1
1
Equating coefficients yields hx − h0x = ι(hx − h0x ) which is impossible since hx − h0x ∈ q 2 Z[q 2 ] and
1
1
so ι(hx − h0x ) ∈ q − 2 Z[q − 2 ]. Hence Cw = Cw0 and there is at most one choice of the Cw satisfying
the conditions of the theorem.
We now show the existence of the Cw . Clearly ι(Tid ) = Tid and so setting Cid = Tid satisfies
1
1
the conditions of the theorem. Now if r ∈ S then ι(Ter ) = Ter + (q 2 − q − 2 )Teid by (A.1.2) and so
1
1
1
1
1
1
ι(Ter − q 2 Teid ) = Ter + (q − 2 − q 2 )Teid − q − 2 Teid = Ter − q 2 Teid . Hence we have Cr = Ter − q 2 Teid for all
simple transpositions r ∈ S. A simple calculation yields the following multiplication formula for
Cr :
(
1
Tewr − q 2 Tew
if wr > w
e
Tw Cr =
(A.1.4)
− 12 e
e
Twr − q Tw if wr < w
Now, for the inductive step assume that the Cy are known and satisfy the conditions of the theorem
for all y < w. Choose r ∈ S such that wr < w (so that Cwr is known). Then by assumption
P
1
1
Cwr = Tewr + y<wr h0y Tey for some h0y ∈ q 2 Z[q 2 ]. Now:
Cwr Cr = Tewr Cr +
X
h0y Tey Cr
y<wr
1
Since wr2 = w > wr we have Tewr Cr = Tew − q 2 Tewr by (A.1.4). Also by (A.1.4), Tey Cr is either
1
1
1
equal to Teyr + q 2 Tey or Teyr − q − 2 Tey . In either case the coefficients in h0y Tey Cr are in Z[q 2 ] since
1
1
h0y ∈ q 2 Z[q 2 ]. Also, since y < wr then by Lemma 1.3.1 either yr < w or yr < wr (and so yr < w
since wr < w). Hence we may write:
X
Cwr Cr = Tew +
hy Tey
(A.1.5)
y<w
1
for some hy ∈ Z[q 2 ]. Now Cwr Cr is certainly ι-invariant (again since ι is a homomorphism and
Cwr and Cr are ι-invariant). However, the hy are not necessarily without constant term. Notice,
however, that for y < w the coefficient of Tey in hy Tey − hy (0)Cy is hy − hy (0) which is certainly
without constant term. And, by our inductive assumptions on Cy , the coefficient of Tex for x ≤ y is
1
1
in q 2 Z[q 2 ]. Therefore, we may form:
X
Cw = Tew +
(hy Ty − hy (0)Cy )
(A.1.6)
y<w
P
1
1
Now, from the observations above Cw ∈ Tew + y<w q 2 Z[q 2 ]Tey . Note also that Cw = Cwr Cr −
P
y<w hy (0)Cy and hence ι(Cw ) = Cw . Hence the existence of Cw is proven.
66
A.2
Two-Sided Cells and the Dominance Order
In Section 5.4 we saw that the partitions of n index the two-sided cells of Symn . In order to
show that the Kazhdan-Lusztig basis is cellular we needed to define a partial order on partitions
compatible with the two-sided cell order. We did this in the obvious way: we defined λ ≤ µ if there
exists an x and y in Symn such that Shape(P (x)) = λ, Shape(P (y)) = µ and x is less than or
equal to y in the two-sided cell preorder.
From a combinatorial point of view it would be desirable to have an easily described partial order
on partitions compatible with the two-sided cell order. It is part of the folklore of this subject that
the dominance order provides such an order. That is, that x ≤ y if and only if Shape(P (x)) E
LR
Shape(P (y)). However, we are only able to prove one implication:
Proposition A.2.1. Suppose that x and y are in Symn with Shape(P (x)) E Shape(P (y)). Then
x ≤ y.
LR
The difficulty in the proof is more notational than conceptual and so we illustrate the method with
an example before giving the proof. Consider the partitions λ =
and µ =
of 6. The
partition µ is obtained from λ by performing a raising operation from the outside corner (3,1) to
the outside corner (2,3). We want to show that there exists x and y in Symn such that x ≤ y and
LR
Shape(P (x)) = λ and Shape(P (y)) = µ.
We start with the permutation w(λ) = 321546. Clearly w(λ) ∼ (Sλ , Sλ ). Now we define w by
replacing the largest number in the column in which we wish to move a box from (column 1) by
the largest entry in the column in which we wish to move a box to (column 3). To make sure that
we still have a permutation we must decrement all the numbers in between the two columns by 1.
In our example, we want to move a box from the first row to the third and so we replace the 3 with
a 6 and then decrement the second two rows. Thus we obtain w = 621435. It is easily verified that
this procedure does not change the shape and so Shape(P (w)) = λ.
We then form a chain of elements:
x1 = x
R(x1 ) = {s1 , s4 }
= 621435
x2 = x1 s1 = 261435
R(x2 ) = {s1 , s2 , s4 }
x3 = x2 s2 = 216435
R(x3 ) = {s1 , s3 , s4 }
x4 = x3 s3 = 214635
R(x3 ) = {s1 , s4 }
x5 = x4 s4 = 214365
R(x4 ) = {s1 , s3 , s5 }
(Note that each permutation is obtained from the previous one by moving the 6 one place to the
right.) For 1 < i ≤ 4 we have xi−1 —xi by Lemma 4.3.2(i). We also have si ∈ R(xi ) for all i but
si ∈
/ R(xi+1 ) and hence R(xi ) * R(xi+1 ) for all 1 ≤ i < 5. Hence x = x1 ≤ x5 . Now, P (x5 ) = 12 34 56
R
and hence if we set y = x5 we have x ≤ y with the shape of P (y) obtained from the shape of P (x)
R
by performing a raising operation.
Proof (Sketch): Suppose that µ is obtained from λ by performing a raising operation from the
outside corner at the end of the j th column to the inside corner at the end of the k th column. Let
67
c1 , c2 , . . . cm be the column lengths of λ. Define c0i =
wλ =
P
c01 c01 + 1 c01 + 2 . . .
1
c02
c02 − 1 . . .
1
2
...
c01 c01 − 1 . . .
j≤i cj
and let wλ be the permutation:
c02
...
0
c1 + 1 . . .
c0m−1 + 1 . . .
c0m
...
c0m
0
cm−1 + 1
It is straightforward to verify that wλ ∼ (Sλ , Sλ ).
Now, define a new permutation w by:


wλ (i)
w(i) = c0k


wλ (i) − 1
if i < c0j−1 + 1 or i > c0k
if i = c0j−1 + 1
if c0j−1 + 1 < i ≤ c0k
Then w is a permutation with the same shape as wλ . Let p = c0j−1 + 1 and q = c0k−1 − 1. Define a
family of elements by:
wp = w
wp+1 = wp sp
wp+2 = wp+1 sp+1
..
.
wq = wq−1 sq−1
Now, for all p ≤ i ≤ q we have wi (i) = c0k which is the greater than or equal to wi (l) for all
p ≤ i, l ≤ q. Hence si ∈ R(wi ) but si ∈
/ R(wi+1 ) for all p ≤ i < q. Also, wi —wi+1 for all p ≤ i < q
by Lemma 4.3.2(i). Hence w ≤ wq . It can be verified that wq = wµ and so we have w ≤ wµ with
R
R
Shape(w) = λ and Shape(wµ ) = µ.
Now, if λ E µ we have seen in Lemma 2.8.1, that there exists a chain λ = λ1 , λ2 , . . . λm = µ in
which each λi+1 is obtained from λi by performing a raising operation. Now, from above, for each
i we can find a sequence of elements xi ≤ yi with Shape(xi ) = λi and Shape(yi ) = λi+1 . Since yi
R
and xi+1 have the same shape we have yi ∼ xi+1 for all 1 ≤ i < m (Corollary 5.4.2). We thus have
LR
a chain:
x1 ≤ y1 ∼ x2 ≤ y2 ∼ x3 ≤ . . . ≤ ym−1 ∼ xm ≤ ym
R
LR
R
LR
R
LR
R
R
In which Shape(P (x1 )) = λ and Shape(P (ym )) = µ. Now, if x and y are any permutations
satisfying Shape(P (x)) = λ and Shape(P (y)) = µ then x ∼ x1 ≤ ym ∼ y and hence x ≤ y.
L
68
LR
LR
LR
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