Quantum Serre Relations
Claus Michael RINGEL∗
Abstract
Let ⺪I be the free abelian group with basis I, let χ be a pair of integral
bilinear forms on ⺪I. We will endow the free K-algebra KI generated by I
with a comultiplication which depends on χ. This yields an associated bilinear
form on KI which may be called the Drinfeld form. We are going to show
that certain elements of KI which are similar to the well-known quantum
Serre relations belong to the left radical of the Drinfeld form, provided certain
integrality conditions are satisfied.
Résumé
Soit ⺪I le group abélien libre de base I et soit χ un couple de formes
bilinéaires entières sur ⺪I. Nous allons munir la K-algèbre libre KI
engendrée par I d’un coproduit qui dépend de χ. On obtient alors une forme
bilinéaire associée sur KI que l’on peut appeler forme de Drinfeld. Sous
l’hypothèse de certaines conditions d’intégralité, nous mettons en évidence
certains éléments de KI semblables aux relations de Serre quantiques bien
connues et nous montrons qu’ils appartiennent au radical à gauche de la forme
de Drinfeld.
Serre exhibited a presentation of the finite dimensional semisimple complex
Lie algebras by generators and relations in 1966. This presentation uses only the
data given by the Cartan matrix of the Lie algebra. Of great importance was the
introduction of correspondingly defined Lie algebras by Kac and Moody in 1968,
starting with a generalized Cartan matrix or, what is equivalent, with a suitable
bilinear form on a free abelian group ⺪I, where I is some finite set. Presentations
of a Lie algebra g by generators and relations are also presentations of the universal
enveloping algebra U (g); of course, one has to rewrite the Lie bracket operation in
terms of commutators.
The quantum groups as investigated by Drinfeld and Jimbo in 1985 are associative
algebras which are presented by relations similar to the Serre relations used in the
definition of a Kac-Moody algebra. Here, an additional parameter, say v, is involved:
the binomial coefficients are replaced by the corresponding Gauß polynomials in the
variable v. Even the most enthusiastic mathematician could not have foreseen the
AMS 1980 Mathematics Subject Classification (1985 Revision): 16W30, 17B37
für Mathematik, Universität Bielefeld, POBox 100 131, D-33 501 Bielefeld
∗ Fakultät
Société Mathématique de France
138
Claus Michael RINGEL
large variety of connections which have been encountered in the meantime relating
these algebras and quite different areas of mathematics and physics.
Now, let us consider in more detail the positive part U + of such a quantum group
U . Using the representation theory of finite dimensional hereditary algebras one can
realize U + in different ways: as a sort of Grothendieck ring of perverse sheaves [L]
or as a twisted generic composition algebra [G]. Closely related algebras such as
Hall algebras [R1,G] should be treated at the same time. Lusztig [L] has stressed
the importance of algebraic structures A which are similar to bialgebras, but where
multiplication and comultiplication are compatible only up to a twist on A ⊗ A.
Extending this approach we may start with a pair χ of bilinear forms on the free
abelian group ⺪I and consider what we call (K, v, χ)-bialgebras. Algebras such as
U + , the Hall algebras, and also the free K-algebra KI generated by I can then be
regarded as (K, v, χ)-bialgebras. Recall that U + can be defined as the factor algebra
of KI modulo the radical of the Drinfeld form, which is a bilinear form uniquely
determined by the (K, v, χ)-bialgebra structure of KI.
The present note will consider the Drinfeld form in the general case where χ is an
arbitrary pair of bilinear forms. Elements of KI which are similar to quantum Serre
relations will be shown to belong to the left radical of the Drinfeld form, provided
certain integrality conditions are satisfied. So we recover the result that U + , defined
as the factor algebra of KI modulo the radical of the Drinfeld form, satisfies the
quantum Serre relations ([L], Proposition 1.4.3). The general form of the quantum
Serre relations as exhibited below occurs quite naturally as the fundamental relations
of the Hall algebras [R2]. Similar considerations produce elements in the right radical
of the Drinfeld form.
1
(K, v, χ)-bialgebras
The data which are given are as follows: Let K be a commutative ring and v ∈ K
an invertible element. Let I be a set. We denote by ⺪I the free abelian group with
basis I, and χ , χ are two bilinear forms on ⺪I with values in ⺪.
Let A =
x∈⺪I Ax be a ⺪I-graded K-algebra. Given a bilinear form φ on ⺪I
and an invertible element v ∈ K, we denote by A[v,φ] the algebra obtained from
A by twisting the muliplication using v φ ; to be precise: the new multiplication ∗ is
defined on the same underlying K-module x∈⺪I Ax by
a ∗ b = v φ(x,y) ab,
where a, b are homogeneous elements of degree x, y ∈ ⺪I respectively. Note that
A[v,φ] is again a K-algebra [R3], and of course also ⺪I-graded. Let A be a ⺪ISéminaires et Congrès 2
139
Quantum Serre Relations
graded algebra. Given the pair χ = (χ , χ ) of bilinear forms on ⺪I and v ∈ K, we
consider a corresponding map (⺪I)4 → ⺪, which we also denote by χ and which is
defined as follows:
χ(x1 , x2 , x3 , x4 ) = χ (x1 , x4 ) + χ (x2 , x3 ).
Note that this map χ is a bilinear form on (⺪I)2 (the maps (⺪I)4 → ⺪ obtained
in this way may be characterized by certain bilinearity properties, see [R4]). We
may consider A ⊗ A as a (⺪I)2 -graded algebra, where for x, y ∈ ⺪I, we have
(A ⊗ A)(x,y) = Ax ⊗ Ay . Thus, given a pair χ = (χ , χ ) of bilinear forms on
⺪I, we may use the bilinear form χ on (⺪I)2 in order to twist the multiplication of
A ⊗ A and we obtain in this way the algebra (A ⊗ A)[v,χ] with multiplication
(a1 ⊗ a2 ) ∗ (b1 ⊗ b2 ) = v χ (a1 ,b2 )+χ
(a2 ,b1 )
a1 b 1 ⊗ a2 b 2 .
Here we see in which way the two bilinear forms are used: the form χ draws
attention to the interchange of the inner elements a2 , b1 , whereas the form χ is
called in for the outer elements a1 and b2 .
By definition, a (K, v, χ)-bialgebra is of the form A = (A, µ, δ), where A is a K
module with a direct decomposition A = x∈⺪I Ax , such that (A, µ) is a ⺪I-graded
algebra, (A, δ) is a ⺪I-graded coalgebra and such that on the one hand, the counit
satisfies (1) = 1, and, on the other hand,
δ : A → (A ⊗ A)[v,χ]
is an algebra homomorphism.
2
The free K-algebra KI as (K, v, χ)-bialgebra.
Let F = KI be the free K-algebra generated by I; we may consider it as
the semigroup algebra of the free semigroup I generated by I. The generator
corresponding to i ∈ I will be denoted by θi . Thus, the elements of I are words in
the letters θi (i ∈ I): there is the empty word which is denoted by 1, and there are
the words θi1 θi2 · · · θin of length n ≥ 1, with i1 , i2 , . . . , in ∈ I. The multiplication
in I is just the concatenation of words. We consider F as a ⺪I-graded algebra,
with the generator θi being of degree i and we denote by : F → F0 = K the
canonical projection. In order to define a comultiplication δ, we consider the algebra
(F ⊗ F )[v,χ] . Let δ be the algebra homomorphism F → (F ⊗ F )[v,χ] defined by
δ(θi ) = θi ⊗ 1 + 1 ⊗ θi .
With this comultiplication δ the K-algebra F becomes a (K, v, χ)-bialgebra, see
[R4]; its counit is .
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Claus Michael RINGEL
Given a ⺪I-graded K-module A = x∈⺪I Ax , a bilinear form −, − : A⊗A → K
is said to respect the grading provided we have a, b = 0 for a ∈ Ax , b ∈ Ay and
x = y in ⺪I.
Proposition 2.1 — There exists a unique bilinear form −, − on the (K, v, χ)bialgebra F = KI with the following properties:
(0) The bilinear form −, − respects the grading.
(1) We have θi , θi = 1 for all i ∈ I,
(2) If a, b1 , b2 ∈ F and δ(a) = ai1 ⊗ ai2 , with ai1 , ai2 ∈ F , then
a, b1 b2 =
ai1 , b1 ai2 , b2 .
The last condition may be rewritten as follows: We may introduce a bilinear form
on F ⊗ F which works componentwise and we denote it again by −, −; to be
precise: for a1 , a2 , b1 , b2 ∈ F, let a1 ⊗ a2 , b1 ⊗ b2 = a1 , b1 a2 , b2 . Then we can
write:
a, b1 b2 = δ(a), b1 ⊗ b2 ,
The bilinear form −, − on KI has been introduced by Drinfeld [D] in the case
where KI is a quantum group; thus we will call −, − the Drinfeld form of the
(K, v, χ)-bialgebra KI. A proof of the proposition can be found in the book of
Lusztig [L], at least in the special case of a quantum group. For the general case, we
refer to [R4]. We are interested in elements of KI which are in the left radical of
the Drinfeld form; by definition, the left radical is the set of all elements a ∈ KI
which satisfy a, b = 0 for all b ∈ KI.
3
The main result
For 0 ≤ t ≤ n, let
v n − v −n
,
[n] =
v − v −1
[n]! =
n
t=1
[t] ,
and
n
t
=
[n]!
.
[t]![n−t]!
These are Laurent polynomials in v with integer coefficients. Given such a Laurent
polynomial ϕ = ϕ(v), and an integer d, one denotes by ϕd the Laurent polynomial
which is obtained from ϕ by inserting v d , thus ϕd (v) = ϕ(v d ). For example,
dn
−dn
[n]d = v vd −v
.
−v −d
Theorem 3.1 — Let i = j be elements of I. Let us define η(i, j) = χ (i, j) − χ (j, i).
Assume that m is a natural number with
m · η(i, i) = −η(i, j) − η(j, i).
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Quantum Serre Relations
Define
1
−η(i, j) + η(j, i) ,
2
1
d = η(i, i).
2
e=
and assume that v e , v d are defined. Then the element
m+1
p
(−1)
m+1
p
p=0
v pe θip θj θim+1−p
d
belongs to the left radical of the Drinfeld form.
We may reformulate the assumptions as follows: First of all, if η(i, i) = 0, then
we have to assume that η(i, j) + η(j, i) = 0 and then we can choose m arbitrarily. If
η(i, i) = 0, then we have to assume that
m=−
η(i, j) + η(j, i)
η(i, i)
is a natural number. The condition that v e , v d are defined, is always satisfied in case
e, d both are integers; otherwise, we have to assume the existence of a square root
of v in K.
4
Preliminary calculations.
We fix an element i ∈ I. Let θ = θi and define a = χ (θ, θ) and a = χ (θ, θ). Then
d = 12 (a − a ), and we consider also c = 12 (a + a ).
Lemma 4.1 —
p
δ(θ ) =
p
t+t =p
t
v tt c θt ⊗ θt .
d
p
= 1 and tt = 0
t
for all t, and thus the right hand side is just θ ⊗ 1 + 1 ⊗ θ. Let p ≥ 1. We use the
Proof. For p = 0, we have δ(1) = 1 ⊗ 1. For p = 1, we have
multiplication rule in F ⊗ F
(θs ⊗ θs )(θt ⊗ θt ) = v st a +s ta θs+t ⊗ θs +t .
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Claus Michael RINGEL
and induction:
δ(θp+1 ) = δ(θ)δ(θp )
p
p
= (θ⊗ 1 + 1 ⊗ θ)
t
t=0
=
p
v
(p−t)a p
t
t=0
+
p
v
=
v
(p−t+1)a
t=1
+
p
t=0
=
p
v
d
v t(p−t)c θt+1 ⊗ θp−t
d
ta p
t=0
p+1
v t(p−t)c θt ⊗ θp−t
t
p
v t(p−t)c θt ⊗ θp−t+1
d
t−1
ta p
t
v (t−1)(p−t+1)c θt ⊗ θp−t+1
d
v t(p−t)c θt ⊗ θp−t+1
d
γ(t) θt ⊗ θp+1−t .
t=0
where the coefficients γ(t) are given by γ(0) = 1, and, for t ≥ 1, by
p
p
γ(t) = v (p−t+1)a +(t−1)(p−t+1)c
+ v ta +t(p−t)c
.
t−1
t
d
d
Note that we can rewrite the two exponents as follows:
(p − t + 1)a + (t − 1)(p − t + 1)c = (p − t + 1)d + t(p − t + 1)c
and
ta + t(p − t)c = −td + t(p − t + 1)c.
As a consequence, we can continue the calculation


p
p

γ(t) = v t(p+1−t)c v (p−t+1)d
+ v −td
t−1
t
d
d
p
p
+ v −t
= v t(p+1−t)c v p−t+1
t−1
t
d
p+1
= v t(p+1−t)c
.
t
d
This completes the proof.
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Quantum Serre Relations
Lemma 4.2 —
p
θp , θp = [p]!d · v (2)c .
Proof. Let p ≥ 1. We use Lemma 4.1 and induction:
θp+1 , θp+1 = δ(θp+1 ), θp ⊗ θ
p+1
=
v tt c · θt ⊗ θt , θp ⊗ θ
t
t+t =p+1
d
p+1
=
v pc · θp ⊗ θ, θp ⊗ θ
p
d
p
= [p+1]d · v pc · [p]!d · v (2)c
= [p+1]!d · v (
Here, we have used that p +
5
p
2
=
p+1
2
)c .
p+1
2 . This completes the proof.
The calculation of θip θj θip , θiq θj θiq .
Consider now a pair θi , θj where i = j. We want to calculate θip θj θip , θiq θj θiq in
case p + p = q + q . Again, we write θ = θi and we use the previous notation.
Proposition 5.1 — Let i = j and p + p = q + q . Then
θp θj θp , θq θj θq = h · [p]!d [p ]!d
f (s)g(t)
J
where the summation is over the set J of all quadrupels (s, t, s , t ) such that s+t = p,
s + t = p , s + s = q, t + t = q and where
q q 1
1 h = [q]!d [q ]!d · v (2)+( 2 ) c+q 2 q a +χ (j,i) +q 2 qa +χ (j,i) ,
1
f (s) =
v s(q d+χ (i,j)−χ (j,i)) ,
[s]!d [s ]!d
1
g(t) =
v t(−qd+χ (i,j)−χ (j,i)) .
[t]!d [t ]!d
We consider first the special case q = 0. Thus, let p + p = q. We have
δ(θp ) = θp ⊗ 1 + x, and δ(θp ) = θp ⊗ 1 + x , where x, x are linear combinations of
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Claus Michael RINGEL
elements of the form θs ⊗ θt with t ≥ 1. Therefore
θp θj θp , θq θj = δ(θp )δ(θj )δ(θp ), θq ⊗ θj = (θp ⊗ 1 + x)(θj ⊗ 1 + 1 ⊗ θj )(θp ⊗ 1 + x ), θq ⊗ θj = (θp ⊗ 1)(1 ⊗ θj )(θp ⊗ 1), θq ⊗ θj = v pχ (i,j)+p χ
= v pχ (i,j)+p χ
(j,i)
θq , θq θj , θj (j,i)
θq , θq .
This will be used in the next calculation.
We turn to the general case p + p = q + q . We observe that
(θs ⊗ θt )δ(θj )(θs ⊗ θt ), θq θj ⊗ θq = (θs ⊗ θt )(θj ⊗ 1)(θs ⊗ θt ), θq θj ⊗ θq ,
and this element is non-zero only in case s + s = q and t + t = q . As a consequence
θp θj θp , θq θj θq = δ(θp )δ(θj )δ(θp ), θq θj ⊗ θq p
p
=
v stc+s t c (θs ⊗ θt )δ(θj )(θs ⊗ θt ), θq θj ⊗ θq s+t=p s +t =p s d s d
p
p
=
v stc+s t c (θs ⊗ θt )(θj ⊗ 1)(θs ⊗ θt ), θq θj ⊗ θq s
s
J
d
d
p
p
=
v stc+s t c v st a +s ta +tχ (i,j)+t χ (j,i) θs θj θs ⊗ θt+t , θq θj ⊗ θq s
s
J
d
d
p
p
v r v tχ (i,j)+t χ (j,i) v sχ (i,j)+s χ (j,i) θq , θq θq , θq =
s
s
J
d
d
p
q
q
p
=
v r+r [q]!d · v (2)c · [q ]!d · v ( 2 )c
s
s
J
d
d
[p]![p ]![q]![q ]! (q)+(q )c
2
=
v 2
v r+r
[s]![s ]![t]![t ]! d
J
where
r = stc + s t c + st a + s ta
1
1
1
1
= sta + sta + s t a + s t a + st a + s ta
2
2
2
2
1
1
1
1
= s(t + t )a + s (t + t )a + t(s + s )a + t (s + s )a
2
2
2
2
1
= (sq a + s q a + tqa + t qa ).
2
Séminaires et Congrès 2
Quantum Serre Relations
145
and
r = tχ (i, j) + t χ (j, i) + sχ (i, j) + s χ (j, i).
Altogether we have
1 1 q a + χ (i, j) + s
q a + χ (j, i)
r+r =s
2
2
1 1 +t
qa + χ (i, j) + t
qa + χ (j, i)
2
2
1 1 =s
q a + χ (i, j) + (q − s)
q a + χ (j, i)
2
2
1 1 +t
qa + χ (i, j) + (q − t)
qa + χ (j, i)
2
2
= s q d + χ (i, j)−χ (j, i) + t (−qd + χ (i, j) − χ (j, i))
1 1 q a + χ (j, i) + q qa + χ (j, i) .
+q
2
2
We distribute the various terms in order to form h, f (s) and g(t). This completes
the proof.
6
Proof of Theorem.
Let p = m + 1 − p. Let q, q be natural numbers such that q + q = m + 1. We claim
that
1
v se f (s) =
v s(−q+1)d
[s]!d [s ]!d
1
v te g(t) =
v t(q −1)d
[t]!d [t ]!d
We have to consider the exponent of v s in [s]!d [s ]!d v se f (s); it is given by:
1
1
q d + χ (i, j) − χ (j, i) + e = q η(i, i) + η(i, j) + (−η(i, j) + η(j, i))
2
2
1 = (q η(i, i) + η(i, j) + η(j, i))
2
1 = (q η(i, i) − m · η(i, i))
2
1 = (q η(i, i) − (q + q − 1)η(i, i))
2
1
= (−q + 1)η(i, i)
2
= (−q + 1)d
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Claus Michael RINGEL
Similarly, we calculate the exponent of v t in [t]!d [t ]!d v te g(t), it is of the form:
1
1
−qd + χ (i, j) − χ (j, i) + e = −q η(i, i) − η(j, i) + (−η(i, j) + η(j, i))
2
2
1
= (−qη(i, i) − η(i, j) − η(j, i))
2
1
= (−qη(i, i) + m · η(i, i))
2
1
= (−qη(i, i) + (q + q − 1)η(i, i))
2
1 = (q − 1)η(i, i)
2
= (q − 1)d
We calculate (and we set p = m + 1 − p in the definition of J):
m+1
p
(−1)
p=0
m+1
p
v pe θip θj θim+1−p , θiq θj θiq d
m+1
=
(−1)p
v pe θip θj θim+1−p , θiq θj θiq p
p=0
d
m+1
[m+1]!
p
=
(−1)
v pe h [p]!d [m+1−p]!d
f (s) · g(t)
[p!][m+1−p]! d
p=0
m+1
J
=
[m+1]!d h
(−1)s v se f (s) · (−1)t v te g(t)
m+1
p=0
J
= [m+1]!d · h ·
q
q
(−1) v f (s) ·
(−1)t v te g(t) ,
s se
s=0
t=0
since any pair (s, t) with 0 ≤ s ≤ q and 0 ≤ t ≤ q gives a unique pair (p, (s, t, s , t ))
with 0 ≤ p ≤ m + 1 and (s, t, s , t ) ∈ J (namely p = s + t, s = q − s, t = q − t)
and any pair is obtained in this way. If q > 0, then the factor
q
(−1)s v se f (s) =
s=0
q
(−1)s
s=0
1
v ds(−q+1)
[s]!d [s ]!d
is zero, if q > 0, then the factor
q
t te
(−1) v g(t) =
t=0
Séminaires et Congrès 2
q
t=0
(−1)t
1
v dt(q −1)
[t]!d [t ]!d
Quantum Serre Relations
147
is zero. Here we use the following well-known formula: for q ≥ 1,
q
q
s s(q−1) q
s −s(q−1) q
=0=
,
(−1) v
(−1) v
s
s
s=0
s=0
(see, for example [L], 1.3.1). Since q + q = m + 1 > 0, at least one of the two factors
is zero.
The algebra KI is ⺪I-graded and the element
m+1
p m+1
a=
(−1)
v pe θip θj θim+1−p
p
p=0
d
has degree x = (m + 1)i + j ∈ ⺪I. Since the Drinfeld form respects the grading, and
since the elements θiq θj θiq with q + q = m + 1 form a basis of KIx , the calculation
above shows that a belongs to the left radical of the Drinfeld form. This completes
the proof.
7
Applications
7.1 The quantum group case
This is the special case when χ = 0 and χ is a Cartan datum as defined in [L];
thus, we assume that χ is a symmetric bilinear form on ⺪I, that χ (i, i) is an even
positive integer, for any i ∈ I, and finally that m(i, j) = −2 χχ(i,j)
(i,i) is a non-negative
integer, for i = j in I. In this case, we see that
η(i, j) = −χ (j, i),
so that η is again a symmetric bilinear form. It follows that e = 0. Also, d =
− 12 χ (i, i) = 0 is an integer, thus v d is defined, and the m considered above is equal
to m(i, j):
−η(i, j) − η(j, i)
χ (j, i) + χ (i, j)
χ (i, j)
=
= −2 = m(i, j).
η(i, i)
−χ (i, i)
χ (i, i)
n
n
We note that
=
. It follows that the elements exhibited above are just
t
t
d
−d
the usual quantum Serre relations.
m=
7.2 Skew commutation
Let us assume that we have
χ (i, j) + χ (j, i) = χ (i, j) + χ (j, i)
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Claus Michael RINGEL
for some pair i = j in I. Then
e = χ (j, i) − χ (i, j)
is an integer and the considerations above show that the element
θj θi − v e θi θj
belongs to the radical of the Drinfeld form (this is just the case m = 0).
References
[D]
V. G. Drinfeld, Quantum groups. In: Proc. Int. Congr. Math. Berkeley 1986,
vol.1. Amer. Math. Soc. (1988), 798-820.
[G]
J. A. Green, Hall algebras, hereditary algebras and quantum groups.
Inventiones math. 120 (1995), 361-377.
[L]
G. Lusztig, Introduction to Quantum Groups. Birkhäuser Progress Math.
(1993).
[R1]
C. M. Ringel, Hall algebras. In: Topics in Algebra. Banach Center Publ. 26.
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