Table Of ContentGraphical Calculus for the Hecke Algebra
Pablo Gonzalez Pagotto
Departamento de Matem´atica, IGCE–UNESP Campus Rio Claro
August 2013
Abstract
We develop a graphical calculus for the Hecke algebra of type A, when its parameter is nonzero and not a
nontrivial root of unity. Then we proceed to generalize the concepts of symmetrizer and antisymmetrizer to
thisalgebra. Wealsoexploresomeoftheirproperties,recursiverelationsandtheirinteractionwitheachother.
1 Introduction
The symmetric group S is the group of permutations of the set {1,2,...,n}. The study of permutations and
n
combinatorics dates back to the 6th century B.C. in India, where the first traces of this kind of thinking can
be found. Their importance grew substantially in 1770 when Lagrange published “R´eflexions sur la R´esolution
Alg´ebrique des E´quations”. This paper initiated the study of permutations in connection with the study of the
solution of polynomial equations. It was probably the first clear instance of implicit group-theoretic thinking in
mathematics. It led directly to the works of P. Ruffini, N. Abel and E. Galois during the first third of the 19th
century, and the concept of permutation group and, in particular, the symmetric group. Later in that century
A. Cayley defined the modern concept of a group and proved that every finite group is isomorphic to a subgroup
of the symmetric group, a result known as Cayley’s Theorem.
The finite dimensional irreducible representations of S in characteristic zero are very well understood, with
n
everythingfromtheirdegreesandcharacterformulaetoexplicitmatrixrepresentationbeingknownformanyyears.
In contrast, very little detailed information is known about the modular representation theory of the symmetric
groups.
In this work we are concerned with the Hecke algebra (or Iwahori-Hecke algebra) of type A which is a
q-deformation of the group algebra of the symmetric group. Loosely speaking, it is the quantum analogue of
the symmetric group that corresponds to passing from Lie algebras to quantum groups. When q is nonzero and
notanontrivialrootofunity,thenitsrepresentationtheoryisverysimilartotherepresentationtheoryofthesym-
metric group. At a nontrivial root of unity, its representation theory is more similar to the modular representation
theory of the symmetric group in characteristic zero.
The study of Hecke algebras was pioneered by R. Dipper and G. James in a series of landmark papers (see
[5] for references and a more detailed overview of the history of this subject). The main motivation for the study
of Hecke algebras is that they provide a bridge between the representation theory of the symmetric and general
linear groups. In addition, Hecke algebras play an important role in the theory of categorification. They also have
applications to many other areas of algebra, combinatorics and geometric representation theory.
Inthebook[2]wefindagraphicalcalculususingstringdiagramsforsymmetrizersandantisymmetrizersdefined
on the symmetric group. The goal of this work is to generalize these notions for the Hecke algebra together with
theirgraphicaldescription. Thegraphicalcalculusisapowerfultoolfortheoreticalandpracticalpurposes. Froma
practicalpointofview, sinceweareveryvisualbeings, ithelpsustovisualizerelationsandpropertiesmoreeasily,
whichmaynotevenbenoticedwithoutit,givingahandtointuition. Forinstance,inthegraphicalcalculus,oneof
the defining relations of the Hecke algebra (relation (b) in Definition 1) follows immediately from the fact that we
consider diagrams up to isotopy relative the boundary. Another feature of the graphical calculus is that, since the
groupalgebraofthesymmetricgroupandtheHeckealgebraarequotientsofthegroupalgebraofthebraidgroup,
their graphical calculi share deep similarities. From the theoretical point of view there is interest in diagrammatic
categorification in which Hecke algebras play an important role. For instance, the graphical calculus of the Hecke
algebra is of fundamental importance in the categorification of the Heisenberg algebra developed in [4]. In fact, in
the current paper we will provide proofs of some of the statements asserted there without proof.
Acknowledgments. I would like to thank the Mitacs Globalink Program, the University of Ottawa and the
Capes for providing such an amazing oportunity for undergraduate students around the world. I would also like to
thank A. Savage for guiding me through the subject and the staff of the mathematics department of UNESP - Rio
Claroforthehelpandincentive. Finally, IwouldliketothankN.Manningforhelpfuldiscussionsandsuggestions.
2 Graphical Description
In our work we are going to explore properties of the symmetrizer, antisymmetrizer and their relation. All these
entities are defined on the Hecke algebra.
Definition 1 (Hecke algebra.). Let F be a field of characteristic zero and consider q (cid:54)=0 an element of F, which is
not a nontrivial root of unity. For n ≥ 2, the Hecke algebra H (q) is the F-algebra generated by t ,...,t with
n 1 n−1
relations
(a) t2 =q+(q−1)t for i=1,2,...,n−1,
i i
(b) t t =t t for i,j =1,2,...,n−1 such that |j−i|>1,
i j j i
(c) t t t =t t t for i=1,2,...,n−2.
i i+1 i i+1 i i+1
By convention, we set H (q)=H (q)=F. We let 1 denote the identity element of H (q). The algebra H has
0 1 n n
a basis {t } where for σ ∈S , t =t ···t , where σ =s ···s is a reduced expression for σ.
σ σ∈Sn n σ i1 ik i1 ik
Throughout our work we shall make use of a graphical description of the Hecke algebra formalized below. The
graphical description of the Hecke algebra is the F-module generated by planar diagrams modulo local relations
(Heckealgebrarelations). Thediagramsareorientedcompact1-manifoldsimmersedinthestrip[0,1]×R, modulo
rel boundary isotopies. The endpoints of the 1-manifold are located at {0}×{1,2,...,n} and {1}×{1,2,...,n}.
In such a description, the element t is represent by crossing the ith and (i+1)th lines. There is also a natural
i
product defined by gluing diagrams. The local relations are as follows.
(cid:16) (cid:17) (cid:16) (cid:17)
i i i
(1) =q +(q−1)
i+1 i+1 i+1
i i
i+1 i+1
(2) =
j j
j+1 j+1
i i
(3) =
i+1 i+1
i+2 i+2
Notice that if we set q =1 in Definition 1, then H (1) is the group algebra of the symmetric group S .
n n
Definition 2. We define the homomorphism ρ : H (q) → H (q) by ρ(t ) = t . This homomorphism is in fact
n n i n−i
a automorphism called the ‘flipping’ automorphism. Diagrammatically, this automorphism performs a flip in the
horizontal axis. For example, the image of the element t t t t under the flipping automorphism is t t t t .
2 4 3 1 3 1 2 4
1 1
2 ρ 2
(4) (cid:55)−→ .
3 3
4 4
5 5
2
3 The Symmetrizer and The Antisymmetrizer
Definition 3. In H (q), let m,k be positive integers such that k ≤ k+m−1 ≤ n. Define the homomorphism
n
ιm :S →S by
k m n
(cid:40)
(k−1)+σ(t+(1−k)), t∈{k,k+1,...,k+(m−1)},
(5) ιm(σ)(t)=
k t, t∈{1,2,...,k−1,k+m,...,n}.
We define the symmetrizer to be
1 (cid:88)
(6) S = t .
k,k+1,...,k+(m−1) [m]q! ιmk(σ)
σ∈Sm
Where [n] !=[1] [2] ···[n−1] [n] and
q q q q q
n−1
(cid:88)
(7) [n] = qi.
q
i=0
It is depicted by a white box enveloping the strings from k to k+(m−1).
1
2
k−1
k
k+1
k+(m−2)
k+(m−1)
k+m
n−1
n
When k =1 and m=n we shall denote the symmetrizer S by S.
1,2,...,n
Remark 1. Given k,m,n positive integers such that k+m−1 ≤ n, there exists a natural embedding of H (q)
m
into H (q) with relation to the kth string given by
n
I :H (q)→H (q)
k m n
t (cid:55)→t .
i k+i−1
In view of this the following statements for the symmetrizer S and for the (to be defined) antisymmetrizer A are
also valid for S and A .
k,...,k+m−1 k,...,k+m−1
1
2
k−1
1 k
2 k+1
3
4 (cid:55)−I→k
5
k+(m−2)
m k+(m−1)
k+m
n−1
n
3
The next result tells us that crossings may be absorbed by the symmetrizer at the cost of a factor q.
Theorem 1. For all i∈{1,2,...,n−1} we have t S =qS =St .
i i
1 1 1
(8) i =q i = i .
i+1 i+1 i+1
n n n
Proof. Leti∈{1,2,...,n−1}beafixedinteger. LetA={σ ∈S ;(cid:96)(s σ)≥(cid:96)(σ)}andB ={σ ∈S ;(cid:96)(s σ)<(cid:96)(σ)}.
n i n i
Let f : A → B be the function defined by f(σ) = s σ. We must prove that f is a bijection. In fact, clearly f
i
is injective, to show that f is onto let τ ∈ B. The permutation s τ is in A since (cid:96)(s (s τ)) = (cid:96)(τ) > (cid:96)(s τ),
i i i i
furthermore f(s τ)=τ. In view of this we may rewrite the symmetrizer as
i
1 (cid:88)
(9) S = (t +t ).
[n] ! σ f(σ)
q
σ∈A
On the other hand,
(10) t (t +t )=t t +t t =t +t 2t =t +qt +(q−1)t t =q(t +t ),
i σ f(σ) i σ i f(σ) f(σ) i σ f(σ) σ i σ f(σ) σ
where we have used that t t =t , since (cid:96)(s σ)=(cid:96)(σ)+(cid:96)(s ) for σ ∈A. Therefore we have
i σ siσ i i
1 (cid:88) 1 (cid:88)
(11) t S = t (t +t )= q(t +t )=qS.
i [n] ! i σ f(σ) [n] ! f(σ) σ
q q
σ∈A σ∈A
The other equality follows by symmetry.
Lemma 1. (a) For a given i∈{1,2,...,n+1}, let S(i) ={τ ∈S ;τ(n+1)=i}. Then
n+1 n+1
(cid:88) (cid:88)
(12) q(cid:96)(τ) =qn+1−i q(cid:96)(τ).
τ∈S(i) τ∈Sn
n+1
(b) For a given positive integer n we have the following identity,
(cid:88)
(13) [n] != q(cid:96)(σ).
q
σ∈Sn
Proof. (a) In fact, it is easy to see that equation (12) holds for i = n+1. Since the correspondence φ (cid:55)→ ψφ,
where ψ =(i,i+1,...,n,n+1), between S(n+1) and S(i) is a bijection we can write
n+1 n+1
(cid:88) (cid:88)
(14) q(cid:96)(τ) = q(cid:96)(ψφ).
τ∈S(i) φ∈S(n+1)
n+1 n+1
We claim that (cid:96)(ψφ) = (cid:96)(ψ) + (cid:96)(φ). Consider the sets Φ of the inversions of the permutation φ and
Ψ={(j,n+1);n+1>φ(j)≥i}. These sets are disjoint since φ(n+1) = n+1 and it is a straightfor-
ward case-by-case analysis to check that both are subsets of the set of inversions of ψφ, whence we have
that
(cid:96)(ψφ)≥#(Ψ∪Φ)=(cid:96)(ψ)+(cid:96)(φ).
Since (cid:96)(ψφ)≤(cid:96)(ψ)+(cid:96)(φ) we conclude that (cid:96)(ψφ)=(cid:96)(ψ)+(cid:96)(φ). Then equation (14) becomes
(cid:88) (cid:88) (cid:88) (cid:88)
(15) q(cid:96)(τ) = q(cid:96)(ψ)+(cid:96)(φ) =q(cid:96)(ψ) q(cid:96)(φ) =qn+1−i q(cid:96)(φ).
τ∈S(i) φ∈S(n+1) φ∈S(n+1) φ∈Sn
n+1 n+1 n+1
4
(b) We shall prove by induction on n. For n=1 it is trivial. Suppose the equality holds for some positive integer
n and we shall prove it holds for n+1. Let S(i) = {τ ∈ S ;τ(n+1) = i}. It is easy to see that these
n+1 n+1
sets are disjoint and their union is equal to S . Thus we may write:
n+1
(cid:88) (cid:88) (cid:88) (cid:88) (cid:88)
(16) q(cid:96)(τ) = q(cid:96)(τ)+ q(cid:96)(τ)+···+ q(cid:96)(τ)+ q(cid:96)(τ).
τ∈Sn+1 τ∈S(1) τ∈S(2) τ∈S(n) τ∈S(n+1)
n+1 n+1 n+1 n+1
By Lemma 1(a) we have that
(cid:88) (cid:88)
(17) q(cid:96)(τ) =qn+1−i q(cid:96)(τ)
τ∈S(i) τ∈Sn
n+1
then, using our induction hypothesis that
(cid:88)
[n] != q(cid:96)(τ),
q
τ∈Sn
we can rewrite equation (16) to obtain
(cid:88)
(18) q(cid:96)(τ) =qn[n] !+qn−1[n] !+···+q[n] !+[n] !=(qn+qn−1+···+1)[n] !=[n+1] [n] !=[n+1] !.
q q q q q q q q
τ∈Sn+1
The next result tells us that bigger symmetrizers absorb smaller ones.
Theorem 2. If 1≤a<a+b≤n+1, then
(19) SS =S =S S.
a,...,a+(b−1) k,...,k+(m−1) a,...,a+(b−1)
1 1
a−1 a−1
a a
(20) =
a+(b−1) a+(b−1)
a+b a+b
n n
Proof. By definition of S we have,
a,...,a+(b−1)
(cid:32) (cid:33)
1 (cid:88) 1 (cid:88)
(21) S S = t S = (t S).
a,...,a+(b−1) [b]q! ιba(σ) [b]q! ιba(σ)
σ∈Sb σ∈Sb
Now, applying Theorem 1 we obtain,
1 (cid:88) 1 (cid:88)
(22) (t S)= (q(cid:96)(σ)S).
[b]q! ιba(σ) [b]q!
σ∈Sb σ∈Sb
Finally, by the distributive law and Lemma 1(b),
(cid:32) (cid:33)
1 (cid:88) 1 (cid:88)
(23) (q(cid:96)(σ)S)= q(cid:96)(σ) S =S.
[b] ! [b] !
q q
σ∈Sb σ∈Sb
5
Theorem 3. We have the following recursive relation
1
(24) S = [1+t +t t +t t t +···+t t ···t ]S .
[n] 1 2 1 3 2 1 n−1 n−2 1 2,...,n
q
1 1 1 1 1 1
2 2 2 2 2 2
(25) 543 = [n1]q 543 + 543 + 543 +···+ 3 543
n−1
n n n n n n
Proof. By definition of S we have that
2,...,n
1 1 (cid:88)
(26) [1+t +t t +···+t ···t ]S = [1+t +t t +···+t ···t ] t .
[n]q 1 2 1 n−1 1 2,...,n [n]q! 1 2 1 n−1 1 ιn2−1(σ)
σ∈Sn−1
As we have seen in the proof of Lemma 1(a), for i=0,1,...,n−2 and σ ∈S ,
n−1
(27) t t ···t t =t .
1+i i 1 ιn2−1(σ) s1+isi···s1ιn2−1(σ)
Furthermore, the sets ιn−1(S ),s ιn−1(S ),s s ιn−1(S ),...,s s ···s ιn−1(S ) are pairwise dis-
2 n−1 1 2 n−1 2 1 2 n−1 n−1 n−2 1 2 n−1
joint, since they each fix different subsets of {1,2,...,n}, and their union equals S since we have n! distinct
n
elements of S .
n
Then we can write equation (26) as
1 (cid:88)
[1+t +t t +···+t ···t ] t
[n]q! 1 2 1 n−1 1 ιn2−1(σ)
σ∈Sn−1
1 (cid:88) (cid:88) (cid:88)
= [n] ! tσ+ tσ+···+ tσ
q
σ∈ιn2−1(Sn−1) σ∈s1ιn2−1(Sn−1) σ∈sn−1···s1ιn2−1(Sn−1)
1 (cid:88)
= t
[n] ! σ
q
σ∈Sn
(28) =S.
By (26) and (28) we conclude that
1
(29) S = [1+t +t t +t t t +···+t t ···t ]S .
[n] 1 2 1 3 2 1 n−1 n−2 1 2,...,n
q
Corollary 1. We have the following recursive relation
1
(30) S = [1+t +t t +t t t +···+t t ···t ]S .
[n] n−1 n−2 n−1 n−3 n−2 n−1 1 2 n−1 1,...,n−1
q
1 1 1 1 1 1
2 2 2
(31) 543 = [n1]q 543 + nnn−−−234 + nnn−−−234 +···+ n−2 nnn−−−234
n−1 n−1 n−1 n−1
n n n n n n
Proof. The image of S under the flipping automorphism ρ is S. Thus applying ρ on both sides of equation (24) of
Theorem 3 we obtain equation (30).
Corollary 2. We have the following relations:
6
(i)
1 1 1
2 2 2
1
(32) 3 = 3 + 3
4 [n] 4 4
5 q 5 5
n n n
(ii)
1 1 1
2
1
(33) 3 = n−4 + n−4
4 [n] n−3 n−3
5 q n−2 n−2
n−1 n−1
n n n
1
2
Proof. (i) Multiply equation (25), on the left, by 3 and use Theorem 1.
4
5
n
1
n−4
(ii) Multiply equation (31), on the left, by and use Theorem 1.
n−3
n−2
n−1
n
Definition 4. Let m,k be positive integers such that k ≤k+m−1≤n. We define the antisymmetrizer to be
1 (cid:88)
(34) A = (−q)−(cid:96)(σ)t .
k,k+1,...,k+(m−1) [m]q−1! ιmk(σ)
σ∈Sm
It is depicted by a black box enveloping the strings from k to k+(m−1).
1
2
k−1
k
k+1
k+(m−2)
k+(m−1)
k+m
n−1
n
When k =1 and m=n we shall denote the antisymmetrizer A by A.
1,2,...,n
As noted in Remark 1 all the statements below about the antisymmetrizer A are also valid for A .
k,...,k+m−1
The next result tells us that crossings may be absorbed by the antisymmetrizer at the cost of a sign.
Theorem 4. For all i∈{1,2,...,n−1} we have t A=−A=At .
i i
1 1 1
(35) i =− i = i .
i+1 i+1 i+1
n n n
7
Proof. Leti∈{1,2,...,n−1}beafixedinteger. LetL={σ ∈S ;(cid:96)(s σ)≥(cid:96)(σ)}andB ={σ ∈S ;(cid:96)(s σ)<(cid:96)(σ)}.
n i n i
Let f :L→B be the function defined by f(σ)=s σ. As we have seen in the proof of Theorem 1, f is a bijection.
i
In view of this we may rewrite the antisymmetrizer as
1 (cid:88)
(36) A= ((−q)−(cid:96)(σ)t +(−q)−(cid:96)(f(σ))t ).
[n] ! σ f(σ)
q−1
σ∈A
On the other hand,
t ((−q)−(cid:96)(σ)t +(−q)−(cid:96)(f(σ))t )=(−q)−(cid:96)(σ)t t +(−q)−(cid:96)(f(σ))t t
i σ f(σ) i σ i f(σ)
=(−q)−(cid:96)(σ)t +(−q)−(cid:96)(f(σ))t 2t
f(σ) i σ
=−q(−q)−(cid:96)(f(σ))t +(−q)−(cid:96)(f(σ))(qt +(q−1)t t )
f(σ) σ i σ
=−q(−q)−(cid:96)(f(σ))t −(−q)−(cid:96)(σ)t −(−q)−(cid:96)(f(σ))(1−q)t t )
f(σ) σ i σ
(37) =−(t +t ),
f(σ) σ
where we have used that t t =t , since (cid:96)(s σ)=(cid:96)(σ)+(cid:96)(s ) for σ ∈A. Therefore we have
i σ siσ i i
1 (cid:88)
(38) t A= t ((−q)−(cid:96)(σ)t +(−q)−(cid:96)(f(σ))t )
i [n] ! i σ f(σ)
q−1
σ∈A
1 (cid:88)
= −((−q)−(cid:96)(σ)t +(−q)−(cid:96)(f(σ))t )=−A.
[n] ! σ f(σ)
q−1
σ∈A
The other equality follows by symmetry.
The next result tells us that bigger antisymmetrizers absorb smaller ones.
Theorem 5. If 1≤a<b+a≤n+1, then
(39) AA =A=A A.
a,...,a+(b−1) a,...,a+(b−1)
1 1
a−1 a−1
a a
(40) = .
a+(b−1) a+(b−1)
a+b a+b
n n
Proof. By definition of A we have,
a,...,a+(b−1)
(cid:32) (cid:33)
1 (cid:88) 1 (cid:88)
(41) A A= (−q)−(cid:96)(σ)t A= ((−q)−(cid:96)(σ)t A).
a,...,a+(b−1) [b]q−1! ιba(σ) [b]q−1! ιba(σ)
σ∈Sb σ∈Sb
Now, applying Theorem 4 we obtain,
1 (cid:88) 1 (cid:88) 1 (cid:88)
(42) ((−q)−(cid:96)(σ)t A)= ((−q)−(cid:96)(σ)(−1)(cid:96)(σ)A)= (q−(cid:96)(σ)A).
[b]q−1! ιba(σ) [b]q−1! [b]q−1!
σ∈Sb σ∈Sb σ∈Sb
Finally, by the distributive law, together with the fact that
(cid:88)
(43) [b] != q−(cid:96)(σ),
q−1
σ∈Sb
which is Lemma 1(b) after replacing q by q−1, we have
(cid:32) (cid:33)
1 (cid:88) 1 (cid:88)
(44) (q−(cid:96)(σ)A)= q−(cid:96)(σ) A=A.
[b] ! [b] !
q−1 q−1
σ∈Sb σ∈Sb
8
Theorem 6. We have the following recursive relation
1
(45) A= [1−q−1t +q−2t t −q−3t t t +···+(−q)1−nt t ···t ]A .
[n] 1 2 1 3 2 1 n−1 n−2 1 2,...,n
q−1
1 1 1 1 1 1
2 1 2 12 1 2 (−1)n−1 2 2
(46) 3 = 3 − 3 + 3 +···+ 3 3
4 [n] 4 q 4 q2 4 qn−1 4
5 q−1 5 5 5 5
n−1
n n n n n n
Proof. By definition of A we have that
2,...,n
1
(47) [1+(−q)−1t +(−q)−2t t +···+(−q)1−nt ···t ]A =
[n] 1 2 1 n−1 1 2,...,n
q−1
1 (cid:88)
[1+(−q)−1t +(−q)−2t t +···+(−q)1−nt ···t ] (−q)−(cid:96)(σ)t .
[n]q−1! 1 2 1 n−1 1 ιn2−1(σ)
σ∈Sn−1
As we have seen in the proof of Lemma 1(a), for i=0,1,...,n−2 and σ ∈S ,
n−1
(48) t t ···t t =t .
1+i i 1 ιn2−1(σ) s1+isi···s1ιn2−1(σ)
Furthermore, the sets ιn−1(S ),s ιn−1(S ),s s ιn−1(S ),...,s s ···s ιn−1(S ) are pairwise dis-
2 n−1 1 2 n−1 2 1 2 n−1 n−1 n−2 1 2 n−1
joint, since they each fix different subsets of {1,2,...,n}, and their union equals S since we have n! distinct
n
elements of S .
n
Then we can write equation (47) as
1 (cid:88)
[1+(−q)−1t +(−q)−2t t +···+(−q)1−nt ···t ] (−q)−(cid:96)(σ)t
[n]q−1! 1 2 1 n−1 1 ιn2−1(σ)
σ∈Sn−1
1 (cid:88) (cid:88) (cid:88)
= [n] ! (−q)−(cid:96)(σ)tσ+ (−q)−(cid:96)(σ)tσ+···+ (−q)−(cid:96)(σ)tσ
q−1
σ∈ιn2−1(Sn−1) σ∈s1ιn2−1(Sn−1) σ∈sn−1···s1ιn2−1(Sn−1)
1 (cid:88)
= (−q)−(cid:96)(σ)t
[n] ! σ
q−1
σ∈Sn
(49) =A.
By (47) and (49) we conclude that
1
(50) A= [1−q−1t +q−2t t −q−3t t t +···+(−q)1−nt t ···t ]A .
[n] 1 2 1 3 2 1 n−1 n−2 1 2,...,n
q−1
Corollary 3. We have the following recursive relation
1
(51) A= [1−q−1t +q−2t t −q−3t t t +···+(−q)1−nt t ···t ]A .
[n] n−1 n−2 n−1 n−3 n−2 n−1 1 2 n−1 1,...,n−1
q−1
1 1 1 1 1 1
2 1 2 1 1 (−1)n−1 2
(52) 3 = 3 − n−4 + n−4 +···+ n−4
4 [n] 4 qn−3 q2 n−3 qn−1 n−3
5 q−1 5 n−2 n−2 n−2 n−2
n−1 n−1 n−1 n−1
n n n n n n
Proof. It is easy to check that the image of A under the flipping automorphism ρ is A. Thus applying ρ on both
sides of equation (45) of Theorem 6 we obtain equation (51).
Corollary 4. We have the following relations:
9
(i)
1 1 1
2 1 2 1−[n] 2
(53) 3 = 3 + q−1 3
4 [n] 4 [n] 4
5 q−1 5 q−1 5
n n n
(ii)
1 1 1
2 1 1−[n]
(54) 3 = n−4 + q−1 n−4
4 [n] n−3 [n] n−3
5 q−1 n−2 q−1 n−2
n−1 n−1
n n n
1
2
Proof. (i) Multiply equation (46), on the left, by 3 and use Theorem 4.
4
5
n
1
n−4
(ii) Multiply equation (52), on the left, by and use Theorem 4.
n−3
n−2
n−1
n
Proposition 1. For i∈1,2,...,n−1 we have that A S =0=S A .
i,i+1 i,i+1 i,i+1 i,i+1
1 1
i−1 i−1
i i
(55) =0=
i+1 i+1
i+2 i+2
n n
Proof.
1 1 1 1 1
2 2 2 2
i−i1 (cid:18) 1 1 (cid:19)i−1 i−1 i−1 i−1
= i + i i −q−1 i
ii++21 [2]q![2]q−1! ii++21 ii++21 ii++21 ii++21
n n−1 n−1 n−1 n−1
n n n n
1 1 1
2 2 2
(cid:18) 1 1 (cid:19)i−1 i−1 i−1
= i +(1−q−1) i −q−1 i
[2] ![2] ! i+1 i+1 i+1
q q−1 i+2 i+2 i+2
n−1 n−1 n−1
n n n
=0.
The last equality holds because, by item (a) of Definition 1, we have that
1 1 1
2 2 2
i−1 i−1 i−1
q−1 i = i +(1−q−1) i .
i+1 i+1 i+1
i+2 i+2 i+2
n−1 n−1 n−1
n n n
10
