Generic Symmetries of the Laurent Extension of Quantum Plane
A list of generic Uq (sl)-module algebra structures on the Laurent polynomial algebra over the quantum plane with uncountably many isomorphism classes is produced. Also, a complete list of these structures is presented in which the action of Cartan generator of Uq (sl) is not reduced to multiplying...
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irk-123456789-1198842017-06-11T03:02:46Z Generic Symmetries of the Laurent Extension of Quantum Plane Sinel'shchikov, S. A list of generic Uq (sl)-module algebra structures on the Laurent polynomial algebra over the quantum plane with uncountably many isomorphism classes is produced. Also, a complete list of these structures is presented in which the action of Cartan generator of Uq (sl) is not reduced to multiplying x and y (the generators of quantum plane) by constants. Построен список структур Uq (sl2) -модульной алгебры общего положения на алгебре полиномов Лорана над квантовой плоскостью, содержащий несчетное количество классов изоморфизма. Предъявлен полный список подобных структур, в которых действие картановской образующей квантовой универсальной обертывающей Uq (sl2) не сводится к умножению x и y (образующих квантовой плоскости) на константы. 2015 Article Generic Symmetries of the Laurent Extension of Quantum Plane / S. Sinel'shchikov // Журнал математической физики, анализа, геометрии. — 2015. — Т. 11, № 4. — С. 333-358. — Бібліогр.: 8 назв. — англ. 1812-9471 DOI: doi.org/10.15407/mag11.04.333 MSC2000: 81R50, 17B37. http://dspace.nbuv.gov.ua/handle/123456789/119884 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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A list of generic Uq (sl)-module algebra structures on the Laurent polynomial algebra over the quantum plane with uncountably many isomorphism classes is produced. Also, a complete list of these structures is presented in which the action of Cartan generator of Uq (sl) is not reduced to multiplying x and y (the generators of quantum plane) by constants. |
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Sinel'shchikov, S. Generic Symmetries of the Laurent Extension of Quantum Plane Журнал математической физики, анализа, геометрии |
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Generic Symmetries of the Laurent Extension of Quantum Plane |
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Generic Symmetries of the Laurent Extension of Quantum Plane |
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Generic Symmetries of the Laurent Extension of Quantum Plane |
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Generic Symmetries of the Laurent Extension of Quantum Plane |
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Generic Symmetries of the Laurent Extension of Quantum Plane |
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generic symmetries of the laurent extension of quantum plane |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Generic Symmetries of the Laurent Extension of Quantum Plane / S. Sinel'shchikov // Журнал математической физики, анализа, геометрии. — 2015. — Т. 11, № 4. — С. 333-358. — Бібліогр.: 8 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT sinelshchikovs genericsymmetriesofthelaurentextensionofquantumplane |
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2025-07-08T16:50:46Z |
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2025-07-08T16:50:46Z |
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1837098298074202112 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2015, vol. 11, No. 4, pp. 333–358
Generic Symmetries of the Laurent Extension
of Quantum Plane
S. Sinel’shchikov
B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkiv, 61103, Ukraine
E-mail: sinelshchikov@ilt.kharkov.ua
Received September 16, 2014, revised April 3, 2015
A list of generic Uq(sl2)-module algebra structures on the Laurent poly-
nomial algebra over the quantum plane with uncountably many isomorphism
classes is produced. Also, a complete list of these structures is presented in
which the action of Cartan generator of Uq(sl2) is not reduced to multiplying
x and y (the generators of quantum plane) by constants.
Key words: quantum universal enveloping algebra, Hopf algebra, Verma
module, Laurent polynomial, weight.
Mathematics Subject Classification 2010: 81R50, 17B37.
1. Introduction
It is well known that the quantum plane Cq[x, y] admits a structure of Uq(sl2)-
module algebra (see, e.g., [5]). In fact, it was a single selected structure which
was implicit in various related topics and applications. The question on to what
extent this structure is unique was initially raised in [4]. It was established
that there exists an uncountable family of non-isomorphic structures of Uq(sl2)-
module algebra on quantum plane, and a complete classification of them was
presented. This result was extended by S. Duplij, Y. Hong, and F. Li in [3],
where the structures of Uq(slm)-module algebra on a generalized quantum plane,
a polynomial algebra in n quasi-commuting variables, m,n > 2, are considered. In
all the above cases, once m,n are fixed, the structures in question, or symmetries
for short, belong to finitely many series. Every such series is labeled by a pair (in
the simplest case [4]; in the more general context of [3] their quantity is (m−1)n)
of the so-called weight constants, which determine the action of Cartan generators
of the quantum universal enveloping algebra on the generators of (generalized)
quantum plane.
c© S. Sinel’shchikov, 2015
S. Sinel’shchikov
In our opinion, a somewhat different generalization of the results of [4], as
compared to [3], is of a separate interest. Namely, instead of increasing either
the number m or n of generators, we suggest to retain m = n = 2, but to add
the inverse elements x−1 and y−1 for the generators of the ”standard” quantum
plane. This way we obtain the Laurent polynomial algebra Cq[x±1, y±1] over the
quantum plane. Our research shows that this newly formed quantum algebra con-
stitutes a much more symmetric object than the standard quantum plane. More
precisely, the symmetries listed in [4] can be produced by separating those sym-
metries on the extended algebra Cq[x±1, y±1] which leave the subalgebra Cq[x, y]
invariant. Our list of symmetries looks more regular than that of [4].
Our approach anticipates passing through some additional difficulties. The
latter are related to the fact that the action of the Cartan generator on the
extended algebra Cq[x±1, y±1], we consider in this work, is not reduced in general
to multiplying x and y by weight constants, as it was the case in [4, 3]. This
new context, after introducing some Preliminaries, is a subject of Sec. 3. The
complete list of symmetries in which monomials are not weight vectors, is given
by Theorem 3.5.
Contrary to the latter rather poor collection of symmetries, we use Sec. 4
to present the so-called generic symmetries in which all monomials are weight
vectors (see Theorem 4.1). The collection of generic series is abundant in the
sense that it splits into uncountably many isomorphism classes of symmetries. In
contrast to [4, 3], the collection of pairs of weight constants involved in generic
symmetries is also uncountable. These pairs of weight constants, coming from
the generic symmetries, contain all but a countable family of weight constants
(some rational powers of q) which appear as weight constants for (non-generic)
symmetries. The latter symmetries are going to be a subject of a subsequent
work.
The author would like to express his gratitude to the referee for pointing out
a large number of inconsistencies in an earlier version of this work.
2. Preliminaries
Let H be a Hopf algebra whose comultiplication is ∆, counit is ε, and antipode
is S [1]. Also, let A be a unital algebra whose unit is 1. We will also use the
Sweedler notation ∆(h) =
∑
i h
′
i ⊗ h′′i [8].
Definition 2.1. By a structure of H-module algebra on A we mean a homo-
morphism of algebras π : H → EndCA such that
(i) π(h)(ab) =
∑
i π(h′i)(a) · π(h′′i )(b) for all h ∈ H, a, b ∈ A;
(ii) π(h)(1) = ε(h)1 for all h ∈ H.
The structures π1, π2 are said to be isomorphic if there exists an automorphism
Ψ of the algebra A such that Ψπ1(h)Ψ−1 = π2(h) for all h ∈ H.
334 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
Throughout the paper we assume that q ∈ C \ {0} is not a root of 1 (qn 6= 1
for all non-zero integers n). Consider the quantum plane which is a unital algebra
Cq[x, y] with two generators x, y and a single relation
yx = qxy. (2.1)
Let us complete the list of generators with two more elements x−1, y−1, and
the list of relations with
xx−1 = x−1x = yy−1 = y−1y = 1. (2.2)
The extended unital algebra Cq[x±1, y±1] defined this way is called the Laurent
extension of quantum plane, more precisely, the algebra of Laurent polynomials
over quantum plane.
Given an integral matrix σ =
(
k l
m n
)
∈ SL(2,Z) and a pair of non-zero com-
plex numbers (α, β) ∈ (C∗)2, we associate an automorphism ϕσ,α,β of Cq[x±1, y±1]
determined on the generators x and y by
ϕσ,α,β(x) = αxkym, ϕσ,α,β(y) = βxlyn. (2.3)
A well-known result claims that every automorphism of Cq[x±1, y±1] has the form
(2.3), and the group Aut(Cq[x±1, y±1]) of automorphisms of Cq[x±1, y±1] is just
the semidirect product of its subgroups SL(2,Z) and (C∗)2 determined by setting
σ(α, β)σ−1 = (α, β)σ def= (αkβm, αlβn), (2.4)
[6] (see also [2, 7]).
The quantum universal enveloping algebra Uq (sl2) is a unital associative al-
gebra defined by its (Chevalley) generators k, k−1, e, f and the relations
k−1k = 1, kk−1 = 1, (2.5)
ke = q2ek, (2.6)
kf = q−2fk, (2.7)
ef − fe =
k− k−1
q − q−1
. (2.8)
The standard Hopf algebra structure on Uq(sl2) is determined by
∆(k) = k⊗ k, (2.9)
∆(e) = 1⊗ e + e⊗ k, (2.10)
∆(f) = f ⊗ 1 + k−1 ⊗ f, (2.11)
S(k) = k−1, S(e) = −ek−1, S(f) = −kf, (2.12)
ε(k) = 1, ε(e) = ε(f) = 0. (2.13)
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 335
S. Sinel’shchikov
3. The Symmetries with Non-Trivial σ
It should be observed that, given a Uq(sl2)-module algebra structure on
Cq[x±1, y±1] (to be referred to as a symmetry or merely an action for brevity), the
generator k acts via an automorphism of Cq[x±1, y±1], as one can readily deduce
from invertibility of k, Definition 2.1(i) and (2.9). In particular, every symmetry
determines uniquely a matrix σ ∈ SL(2,Z) as in (2.3).
R e m a r k 3.1. It turns out that there exists a one-to-one correspondence
between the Uq(sl2)-symmetries that leave the subalgebra Cq[x, y] invariant and
the Uq(sl2)-symmetries on Cq[x, y]. One can readily restrict the symmetry of
Cq[x±1, y±1] with the above invariance property to Cq[x, y].
On the other hand, suppose we are given an arbitrary symmetry π
on Cq[x±1, y±1], not necessarily leaving Cq[x, y] invariant. One has the following
relations:
π(k)(x−1)=(π(k)x)−1, π(k)(y−1)=(π(k)y)−1, (3.1)
π(e)(x−1)=−x−1(π(e)x)(π(k)x)−1, π(e)(y−1)=−y−1(π(e)y)(π(k)y)−1, (3.2)
π(f)(x−1)=−(π(k−1)x)−1(π(f)x)x−1, π(f)(y−1)=−(π(k−1)y)−1(π(f)y)y−1. (3.3)
Here (3.1) is straightforward since π(k) is an automorphism; by ‘differentiating’,
i.e., applying e and f to (2.2), we can get (3.2) and (3.3), respectively. Certainly,
these relations remain true when x or y is replaced by an arbitrary invertible
element.
Thus, given a symmetry on Cq[x, y], relations (3.1)–(3.3) determine a well-
defined extension of it to the additional generators x−1, y−1, hence to Cq[x±1, y±1].
It might look as if there should be a sharp difference with the picture dis-
covered in [4] due to the (conjectured) abundance of non-weight actions related
to non-trivial matrices σ. However, it turns out that there exists only a small
collection of such symmetries. Let us start with describing those actions.
Suppose we are given a matrix σ =
(
k l
m n
)
∈ SL(2,Z), and let λ and µ be
its eigenvalues. We consider two series of assumptions on σ:
(i) λ = µ and (since λµ = 1) |λ| = |µ| = 1,
(ii) λ, µ ∈ R and λ, µ /∈ {−1, 1}.
One observes first that λ + µ ∈ Z. In the case (i), this, together with other
restrictions of (i), implies that the only possible values for λ + µ = tr σ could be
0, ±1, ±2. Thus, being intended to find the symmetries corresponding to the
assumptions of (i), we need to consider separately five subcases.
(i-1) Suppose that trσ = 2. This means that λ = µ = 1.
The case when σ is just the identity matrix I is postponed till the next section.
336 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
Let us consider the case of (matrix conjugate to) Jordan block, that is, the
eigenspace of σ is one dimensional. We need the following
Lemma 3.2. The complete list of Uq(sl2)-symmetries on the Laurent polyno-
mial algebra C[z±1] of a single variable z is as follows.
1) Let γ ∈ C \ {0} be such that γr−1 = q2 for some r ∈ Z. There exists a
one-parameter (a ∈ C \ {0}) family of Uq(sl2)-symmetries on C[z±1] given by
π(k)(z) = γz; π(e)(z) =
a
q2 − 1
zr; π(f)(z) = q3(γ − 1)a−1z2−r.
Additionally, there exist two more symmetries
π(k)(z) = ±z; π(e)(z) = π(f)(z) = 0.
All the symmetries with fixed γ are isomorphic, e.g., to that with a = 1. There
exists an isomorphism between the symmetries with γ and γ−1. In all other cases
the symmetries are non-isomorphic.
2) Let γ ∈ C \ {0}. There exists a Uq(sl2)-symmetry on C[z±1] given by
π(k)(z) = γz−1; π(e)(z) = π(f)(z) = 0.
All these symmetries are isomorphic, e.g., to that with γ = 1.
The symmetries from 1) are non-isomorphic to those from 2).
P r o o f. Since, with π being a symmetry, π(k) is an automorphism of
C[z±1], and any automorphism of C[z±1] is given by either z 7→ γz or z 7→ γz−1,
γ ∈ C \ {0} [7], we need to consider two cases.
1) Let π(k)(z) = γz.
Assume first π(e)(z) 6= 0. As a consequence of (2.6), we have π(ke)(z) =
q2γπ(e)(z), and with π(e)(z) =
∑
i aiz
i, the assumption ar 6= 0 implies arγ
rzr =
q2γarz
r, hence
γr−1 = q2. (3.4)
Since q is not a root of 1, such r ∈ Z is unique.
Thus we establish that π(e)(z) = azr, a ∈ C \ {0}.
A completely similar argument allows one to deduce that π(f)(z) = bz2−r.
Here b ∈ C \ {0}, because otherwise π(f) is identically zero on C[z±1]. With the
latter assumption, we observe that (2.8), being applied to z, fails, as its l.h.s.
vanishes, while its r.h.s. is non-zero, since γ, due to (3.4), is not a root of 1
together with q. This contradiction shows that π(f)(z) 6= 0.
It remains to use our formulas for π(e)(z) and π(f)(z) when applying (2.8) to
z in order to compute the relation between a and b. This requires two additional
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 337
S. Sinel’shchikov
formulas:
π(e)(zp) =
γp − 1
γ − 1
azp+r−1 =
γp − 1
γ − 1
zp−1π(e)(z), (3.5)
π(f)(zp) =
γ−p − 1
γ − 1
bzp+1−r =
γ−p − 1
γ − 1
zp−1π(f)(z), (3.6)
p ∈ Z, whose proof is completely routine. We have
π(ef)(z) = π(e)(bz2−r) = b
γ2−r − 1
γ − 1
z1−rπ(e)(z) = ab
γ2−r − 1
γ − 1
z,
π(fe)(z) = π(f)(azr) = a
γ−r − 1
γ − 1
zr−1π(f)(z) = ab
γ−r − 1
γ − 1
z,
whence
π(ef − fe)(z) = ab
γ2−r − γ−r
γ − 1
z = abγ−r(γ + 1)z.
On the other hand,
π(k− k−1)
q − q−1
(z) =
γ − γ−1
q − q−1
z =
γ−1(γ2 − 1)
q − q−1
z.
We thus obtain an equation
abγ−r(γ + 1) =
γ−1(γ2 − 1)
q − q−1
,
whence, in view of (3.4), one deduces
ab =
q2(γ − 1)
q − q−1
.
At this we obtain the first family of symmetries in the statement of Lemma
(1). The above argument thus demonstrates that in the case we consider now,
(π(k)(z) = γz, π(e)(z) 6= 0), there can be no other symmetries. On the other
hand, a routine verification shows that those formulas pass through all the rela-
tions (2.1), (2.2), (2.5)–(2.8), hence determine a family of well-defined Uq (sl2)-
symmetries on C[z±1].
Now consider separately the case π(e)(z) = 0. This means π(e) is identically
zero on C[z±1]. Apply again (2.8) to z to deduce that now γ = ±1. Of course,
π(f) ≡ 0 in this case because (a relation similar to) (3.4) fails. We thus obtain
two additional symmetries as in the claim of Lemma (1); these can be readily
verified to be well-defined.
One can verify that the isomorphisms of actions with fixed γ and different a are
given by Ψ(z) = αz for a suitable α. The isomorphism of actions corresponding
338 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
to γ, γ−1 is determined by Ψ(z) = z−1. This exhausts the action of the group
of automorphisms on the space of parameters of symmetries, hence there are no
other isomorphisms between the symmetries.
2) Let π(k)(z) = γz−1.
It is a matter of direct computation that π(k)2 = id, whence
π(e) = π(k2e) = q4π(ek2) = q4π(e),
that is, π(e) ≡ 0. A similar argument proves that π(f) ≡ 0. Thus (2.8) is satisfied.
The isomorphism between the symmetries with different γ is given by Ψ(z) =
αz for a suitable α.
Proposition 3.3. There exist no Uq (sl2)-symmetries on Cq[x±1, y±1] with k
acting via an automorphism ϕσ,α,β such that the matrix σ has eigenvalues λ =
µ = 1 and a one-dimensional eigenspace.
P r o o f . Suppose that such symmetry π exists. Clearly, an eigenvector
(
v1
v2
)
of σ can be chosen so that v1, v2 are coprime integers. Let u1, u2 be integers with
the property u1v1 + u2v2 = 1. Consider the matrix θ =
(
v1 −u2
v2 u1
)
∈ SL(2,Z)
together with the automorphism Φ = ϕθ,1,1 of Cq[x±1, y±1] as in (2.3). In the
associated isomorphic symmetry, the Cartan generator k acts via Φ−1◦π(k)◦Φ as
in (2.3) with the matrix θ−1σθ, which is of the form
(
1 l
0 1
)
for some l ∈ Z\{0}.
Thus we may assume that σ itself has this form.
Now, by (2.3), we have that π(k)(x) = αx, π(k)(y) = βxly for some α, β ∈
C\{0}. Let π(e)(x) =
∑
i,j aijx
iyj . A direct computation which uses the relation
(xrys)i = q
i(i−1)
2
rsxriysi, i, r, s ∈ Z,
shows that
q2π(ek)(x) = q2α
∑
i,j
aijx
iyj , (3.7)
π(ke)(x) =
∑
i,j
aijα
iβjq
j(j−1)
2
lxi+ljyj . (3.8)
Since l 6= 0, a comparison of (3.7) and (3.8) demonstrates that if aij 6= 0 for
some i, j with j 6= 0, then (2.6) fails. So we deduce that π(e)(x) ∈ C[x±1]. One
can prove that π(f)(x) ∈ C[x±1] in a similar way. It follows that the action π of
Uq (sl2) leaves the subalgebra C[x±1] invariant, thus defining a Uq (sl2)-symmetry
on C[x±1]. By Lemma 3.2, we need to consider two cases.
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 339
S. Sinel’shchikov
(A) Let αr−1 = q2 for some r ∈ Z and a ∈ C \ {0} be such that
π(e)(x) =
a
q2 − 1
xr; π(f)(x) = q3(α− 1)a−1x2−r.
With π(f)(y) =
∑
i,j dijx
iyj we compute using (3.6):
q−2π(fk)(y) = βq−2π(f)(xly) = βq−2π(f)(xl)y + βq−2π(k)−1(xl)π(f)(y) =
= β(α−l − 1)qa−1xl−r+1y + βα−lq−2xlπ(f)(y) =
= β(α−l − 1)qa−1xl−r+1y + βα−lq−2
∑
i,j
dijx
i+lyj ; (3.9)
π(kf)(y) =
∑
i,j
dijα
iβjq
j(j−1)
2
lxi+ljyj . (3.10)
Since l 6= 0, a comparison of (3.9) and (3.10) demonstrates that if dij 6= 0 for
some i, j with j 6= 1, then (2.7) fails. This implies that
π(f)(y)x = qxπ(f)(y). (3.11)
Let us use (3.11) and (3.6) to compute
0 = π(kf − q−2fk)(y) = π(kf)(y)− q−2βπ(f)(xl)y − q−2βπ(k)−1(xl)π(f)(y) =
= π(kf)(y)− q−2β
α−l − 1
α− 1
xl−1π(f)(x)y − q−2βα−lxlπ(f)(y),
whence
π(kf)(y)− q−2α−lβxlπ(f)(y)− q−2β
α−l − 1
α− 1
xl−1π(f)(x)y = 0. (3.12)
Furthermore, an application of (3.11) and the explicit form of π(f)(x) in the
case we consider now yields
0 = π(f)(yx− qxy) =
= π(f)(y)x + π(k)−1(y)π(f)(x)− qπ(f)(x)y − qπ(k)−1(x)π(f)(y) =
= q(1− α−1)xπ(f)(y) + q2−rαlβ−1x−lπ(f)(x)y − qπ(f)(x)y,
whence
π(f)(y) +
α
α− 1
(
q1−rαlβ−1x−l−1 − x−1
)
π(f)(x)y = 0. (3.13)
We need two derived relations. The first one is just −π(k) applied to (3.13):
−π(kf)(y) +
αq−2
α− 1
(
βxl−1 − q1−rx−1
)
π(f)(x)y = 0. (3.14)
340 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
The next derived relation is nothing more than (3.13) multiplied on the left by
q−2α−lβxl:
q−2α−lβxlπ(f)(y) +
αq−2
α− 1
(
q1−rx−1 − α−lβxl−1
)
π(f)(x)y = 0. (3.15)
Finally, sum up (3.12), (3.14), and (3.15) to obtain
βq−2
α− 1
(1 + α)
(
1− α−l
)
xl−1π(f)(x)y = 0.
Since Cq[x±1, y±1] is a domain, we conclude that some constant multiplier in this
product should be zero. However, in the special case (A) we consider now this
can not happen. Thus we obtain a contradiction.
(B) Let α = ±1, π(e)(x) = π(f)(x) = 0.
We have
0 = π(e)(yx− qxy) = π(e)(y)π(k)(x)− qxπ(e)(y) = απ(e)(y)x− qxπ(e)(y),
whence
π(e)(y)x = α−1qxπ(e)(y).
This quasi-commutation relation is possible only if π(e)(y) = ϕyp for some ϕ ∈
C[x±1], p ∈ Z.
In the case α = −1, this can not happen if one has qp−1 = −1. The latter
implies that p − 1 6= 0 and q2(p−1) = 1, which contradicts to our assumptions
on q.
It remains to assume that α = 1. In this case, p = 1, and one has that
π(e)(y) = ϕy, π(k)(ϕ) = ϕ, hence
0 = π(ke− q2ek)(y) = π(k)(ϕ)βxly − q2βπ(e)(xly) = βxlϕy − q2βxlϕy,
whence
β(1− q2)xlϕy = 0.
This implies ϕ = 0, hence π(e)(y) = 0. Therefore π(e) is identically zero on
Cq[x±1, y±1]. Since π(k)(y) 6= π(k)−1(y), we observe that (2.8) being applied to
y fails. Thus we obtain the final contradiction, which completes the proof of
Proposition.
(i-2) Suppose that trσ = 1. This means that λ = 1
2 + i
√
3
2 , µ = 1
2 − i
√
3
2 . In
particular, the matrix σ has a finite order, more precisely, σ6 = I. Hence the
same is true for the corresponding automorphism of Cq[x±1, y±1] as in (2.3) with
α = β = 1. However, we need a more subtle claim.
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 341
S. Sinel’shchikov
Lemma 3.4. Assume we are given an arbitrary pair (α, β) ∈ (C∗)2 and a
matrix σ ∈ SL(2,Z) with the properties listed in the subcase (i-2) as well as also
in the subcases (i-3), (i-4) below. Then the automorphism ϕσ,α,β of Cq[x±1, y±1]
determined by (2.3) has a finite order, the latter being larger than 2.
P r o o f . One readily computes that det(σ − I) = 2 − trσ = 1, hence the
inverse matrix (σ − I)−1 is integral. Thus we have a well-defined pair (α′, β′) =
(σ− I)(α, β)(σ− I)−1 ∈ (C∗)2 as in (2.4). Now a simple computation shows that
in the group Aut(Cq[x±1, y±1]) one has the conjugation
(α′, β′)−1σ(α′, β′) = (α, β)σ.
It follows that ϕ6
σ,α,β = id, which proves the lemma in the present subcase (i-2).
It will become clear below that in the subcases (i-3), (i-4) this proof requires only
minor modifications.
Now assume that π is a symmetry with σ possessing the properties listed
in the subcase (i-2). It follows from (2.6) that k6ek−6 = q12e. By Lemma 3.4,
π(k)6 = id, hence q12π(e) = π(e). Since q is not a root of 1, this implies that π(e)
is identically zero. A similar argument establishes also that π(f) ≡ 0.
Now let us apply π to (2.8). The above observations show that we obtain
identically zero in the l.h.s. But this is not the case with the r.h.s, because π(k) 6≡
π(k)−1, as, by Lemma 3.4, the order of π(k) is larger than 2. The contradiction
we obtain this way proves that there exist no symmetries corresponding to σ as
in the subcase (i-2).
(i-3) Suppose that trσ = 0. This means that λ = i, µ = −i. The matrix σ
has order 4, σ4 = I, hence ϕ4
σ,1,1 = id.
To prove Lemma 3.4 in this subcase, observe first that the subgroup
T ⊂ (C∗)2 ⊂ Aut(Cq[x±1, y±1]), formed by (r, s) with r, s = ±1, is normal
in Aut(Cq[x±1, y±1]), as one can see from (2.4). Therefore, the subgroup of
Aut(Cq[x±1, y±1]) generated by T and σi, i = 0, 1, 2, 3, is finite.
In this subcase one has det(σ − I) = 2 − trσ = 2, hence (σ − I)σ′ = 2I for
some integral matrix σ′. As the second equality of (2.4) determines a well-defined
action of the semigroup of integral matrices on (C∗)2, we conclude that, given
arbitrary (α, β) ∈ (C∗)2, one has
(α, β)(σ−I)σ′ =
(
(α, β)σ′
)(σ−I)
= (α, β)2.
After passing to square roots, we obtain
(
(α′, β′)σ′
)(σ−I)
= (r, s)(α, β)
342 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
for some α′, β′, r, s such that α′2 = α, β′2 = β; r, s ∈ {−1, 1}. Now set
(α′′, β′′) def= (α′, β′)σ′ . A routine computation shows that one has the conjugation
(α′′, β′′)−1(r, s)σ(α′′, β′′) = (α, β)σ.
By the above argument, (r, s)σ has finite order as an element of a finite subgroup,
hence the same is true for its conjugate (α, β)σ. This order can not be less than
the order of the projection σ to SL(2,Z), which is 4. This proves Lemma 3.4. It
remains to proceed the same way as in the subcase (i-2) to conclude that there
exist no symmetries corresponding to σ as in the subcase (i-3).
(i-4) Suppose that trσ = −1. This means that λ = −1
2 + i
√
3
2 , µ = −1
2 − i
√
3
2 .
The matrix σ has order 3, σ3 = I, hence ϕ3
σ,1,1 = id.
To prove Lemma 3.4 in this subcase, observe first that the subgroup T ⊂
(C∗)2 ⊂ Aut(Cq[x±1, y±1]), formed by (r, s) with r, s = ζi, ζ = −1
2 + i
√
3
2 , i =
0, 1, 2, is normal in Aut(Cq[x±1, y±1]), as one can see from (2.4). Therefore, the
subgroup of Aut(Cq[x±1, y±1]) generated by T and σi, i = 0, 1, 2, is finite.
In this subcase one has det(σ − I) = 2 − trσ = 3, hence (σ − I)σ′ = 3I for
some integral matrix σ′. As the second equality of (2.4) determines a well-defined
action of the semigroup of integral matrices on (C∗)2, we conclude that, given
arbitrary (α, β) ∈ (C∗)2, one has
(α, β)(σ−I)σ′ =
(
(α, β)σ′
)(σ−I)
= (α, β)3.
After passing to cubic roots, we obtain
(
(α′, β′)σ′
)(σ−I)
= (r, s)(α, β)
for some α′, β′, r, s such that α′3 = α, β′3 = β; r, s ∈ {ζi, i = 0, 1, 2}. Now set
(α′′, β′′) def= (α′, β′)σ′ . A routine computation shows that one has the conjugation
(α′′, β′′)−1(r, s)σ(α′′, β′′) = (α, β)σ.
By the above argument, (r, s)σ has finite order as an element of a finite subgroup,
hence the same is true for its conjugate (α, β)σ. This order can not be less than
the order of the projection σ to SL(2,Z), which is 3. This proves Lemma 3.4. It
remains to proceed the same way as in the subcase (i-2) to conclude that there
exist no symmetries corresponding to σ as in the subcase (i-4).
(i-5) Suppose that trσ = −2. This means that λ = µ = −1, hence either
σ = −I or σ is a matrix conjugate to Jordan block, that is, the eigenspace of σ
is one-dimensional.
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 343
S. Sinel’shchikov
Theorem 3.5. There exists a two-parameter (α, β ∈ C∗) family of Uq(sl2)-
symmetries on Cq[x±1, y±1] that correspond to σ = −I :
π(k)(x) = α−1x−1, π(k)(y) = β−1y−1, (3.16)
π(e)(x) = 0, π(e)(y) = 0, (3.17)
π(f)(x) = 0, π(f)(y) = 0. (3.18)
These are all the symmetries with σ = −I. These symmetries are all isomorphic,
in particular, to the symmetry with α = β = 1.
P r o o f . A routine verification demonstrates that the action given by (3.16)–
(3.18) passes through all the relations (2.1), (2.2) in Cq[x±1, y±1] and (2.5)–(2.13)
in Uq(sl2). Therefore this action is really a Uq(sl2)-symmetry.
To see that there are no more symmetries with σ = −I, observe that α and β
are arbitrary non-zero complex numbers, hence (3.16) exhausts all the possibilities
for the action of k by an automorphism (see (2.3)). As for the action of e or f,
note first that π(k)2 = id, as it is resulted from a straightforward computation.
Hence, by (2.6), one has π(e) = π(k2ek−2) = q4π(e). Since q is not a root of 1,
we deduce that π(e) ≡ 0. Similarly, π(f) ≡ 0.
Let us verify that all the symmetries listed in the theorem are isomorphic.
Given any such symmetry with π(k) being (α, β)(−I) ∈ Aut(Cq[x±1, y±1]) and
an arbitrary automorphism A = (µ, ν)τ , τ ∈ SL(2,Z), (µ, ν) ∈ (C∗)2, one readily
computes that
A(α, β)(−I)A−1 = (µ2, ν2)(α, β)τ (−I).
Clearly, an appropriate choice of (µ, ν), τ makes this conjugate automorphism
(−I).
R e m a r k 3.6. Although the action of k in the symmetries of Theorem 3.5 is
not reduced to multiplying the generators x, y by weight constants as in [4], the
associated Uq(sl2)-actions are weight modules. Namely, a basis of weight vectors
in Cq[x±1, y±1] is given by
{1} ∪ {uij = αiβjxiyj + x−iy−j | i, j > 0} ∪ {vij = αiβjxiyj − x−iy−j | i, j > 0},
so that π(k)(1) = 1, π(k)(uij) = uij , π(k)(vij) = −vij .
Proposition 3.7. There exist no Uq (sl2)-symmetries on Cq[x±1, y±1] with k
acting via an automorphism ϕσ,α,β such that the matrix σ has eigenvalues λ =
µ = −1 and a one-dimensional eigenspace.
P r o o f. Suppose that such symmetry π exists. One can apply the same
argument as at the beginning of the proof of Proposition 3.3 in order to reduce
matters to the case σ =
(−1 l
0 −1
)
for some l ∈ Z \ {0}.
344 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
Now, by (2.3), we have that π(k)(x) = αx−1, π(k)(y) = βxly−1 for some
α, β ∈ C \ {0}. Let π(e)(x) =
∑
i,j aijx
iyj . A direct computation which uses
(3.2) shows that
q2π(ek)(x) = −
∑
i,j
qj+2aijx
iyj , (3.19)
π(ke)(x) =
∑
i,j
aijα
iβjq−
j(j−1)
2
lx−i+ljy−j . (3.20)
Let us compare (3.19) and (3.20). Equate the coefficients at xi, and x−i, respec-
tively, i ∈ Z, j = 0:
−q2ai0 = a−i,0α
−i, −q2a−i,0 = ai0α
i.
The product of two latter relations is q4ai,0a−i,0 = ai,0a−i,0. This clearly implies
that ai,0a−i,0 = 0. Due to the above relations, it follows that ai,0 = 0 for all i.
Now let j 6= 0. Let us compare again (3.19), (3.20) and equate the coefficients
at x−i+ljy−j to obtain
a−i+lj,−j = −q−
j(j−1)
2
l+j−2αiβjaij .
One more iteration of this relation yields
ai−2lj, j = −q−
(−j)(−j−1)
2
l−j−2α−i+ljβ−ja−i+lj,−j = q−j2l−4αljaij .
Since l 6= 0, this implies that, once some aij with j 6= 0 is assumed to be non-
zero, then infinitely many of other aij ’s appear to be non-zero. This is certainly
impossible, because the element π(e)(x) =
∑
i,j aijx
iyj of the algebra Cq[x±1, y±1]
is anyway a finite sum of monomials. This proves that with j 6= 0, aij = 0 for all
i ∈ Z.
Thus we conclude that π(e)(x) = 0. Now let us compute
0 = π(e)(yx− qxy) = απ(e)(y)x−1 − qxπ(e)(y),
whence απ(e)(y)x−1 = qxπ(e)(y). The latter can not happen with non-zero
π(e)(y) because the maximum degree in x of monomials in the l.h.s. is lower than
the similar degree in the r.h.s. Therefore, π(e)(y) = 0, hence π(e) is identically
zero on Cq[x±1, y±1]. On the other hand, π(k)(y) 6= π(k)−1(y), hence (2.8) being
applied to y fails. This contradiction completes the proof of proposition.
Let us consider the case (ii).
Proposition 3.8. There exist no Uq(sl2)-symmetries on Cq[x±1, y±1] with k
acting via an automorphism ϕσ,α,β such that the matrix σ has eigenvalues λ, µ ∈
R \ {−1, 1}.
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 345
S. Sinel’shchikov
To prove the proposition, we need some observations. Firstly, given a sym-
metry π, let us introduce the notation
π(e)(x) =
∑
i,j
aijx
iyj , π(e)(y) =
∑
i,j
bijx
iyj . (3.21)
Here, of course, only finitely many of aij , bij are non-zero. Let us denote D ={(
i
j
) ∈ Z2 : aij 6= 0
}
, E =
{(
i
j
) ∈ Z2 : bij 6= 0
}
.
A straightforward induction argument allows one to deduce the relation
π(e)(xp) =
p−1∑
r=0
xp−1−rπ(e)(x)π(k)(x)r (3.22)
for integers p > 0, together with a similar but slightly different formula for p < 0
(just due to (3.2)):
π(e)(xp) = −
−p−1∑
r=0
xp+rπ(e)(x)π(k)(x)−r−1. (3.23)
One can combine these two formulas in order to obtain a universal formula for
all p ∈ Z \ {0}, which, however, ignores the specific values of constant multipliers
at monomials. More precisely, we have
π(e)(xp) =
∑
(i
j)∈D
max{0,p}−1∑
r=min{0,p}
const(i, j, p, r, k, m)xp−1−r+i+rkyj+rm, (3.24)
where k and m determine the action of k as in (2.3), the constants aij are as in
(3.21), and all the constants const(. . .) are non-zero. This formula is going to
be useful in the sequel where we are about to compute the sums of monomials
modulo constant multipliers.
In a similar way, one has
π(e)(yp) =
p−1∑
r=0
yp−1−r(π(e)y)(π(k)y)r (3.25)
for integers p > 0, and
π(e)(yp) = −
−p−1∑
r=0
yp+r(π(e)y)(π(k)y)−r−1 (3.26)
for p < 0.
346 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
Again, the latter two formulas can be combined in order to obtain a universal
formula for all p ∈ Z \ {0} similar to (3.24):
π(e)(yp) =
∑
(i
j)∈E
max{0,p}−1∑
r=min{0,p}
const(i, j, p, r, l, n)xi+rlyp−1−r+j+rn, (3.27)
where l and n determine the action of k as in (2.3), the constants bij are as in
(3.21), and all the constants const(. . .) are non-zero.
Another observation is related to the specific form of the (integral) entries of
the matrix
σN =
(
a(N) b(N)
c(N) d(N)
)
.
Let λ, λ−1 be the real eigenvalues of σ as under the assumptions of Proposi-
tion 3.8, |λ| > 1. Then
σN = Φ
(
λN 0
0 λ−N
)
Φ−1
for some invertible complex matrix Φ and N ∈ Z. A routine computation shows
that
σN =
(
aλN + a′λ−N bλN + b′λ−N
cλN + c′λ−N dλN + d′λ−N
)
with some a, a′, b, b′, c, c′, d, d′ ∈ C. Substitute now N = 0; as σ0 = I, we deduce
that in fact
σN =
(
aλN + (1− a)λ−N b(λN − λ−N )
c(λN − λ−N ) dλN + (1− d)λ−N
)
.
Finally, computing detσN , which is just 1, we find that ad−bc = 0 and d = 1−a.
Note that under the assumptions of Proposition 3.8 b 6= 0. In fact, if b = 0,
then ad = 0, hence either a = 0 or a = 1. In both cases σ becomes a triangular
matrix whose diagonal entries are λ, λ−1. Since these are integers, one has
λ = ±1, which is not our case.
In a similar way, one deduces that c 6= 0, a 6= 0, a 6= 1.
Thus we have c = a(1−a)
b , so
σN =
(
a(N) b(N)
c(N) d(N)
)
=
(
aλN + (1− a)λ−N b(λN − λ−N )
a(1−a)
b (λN − λ−N ) (1− a)λN + aλ−N
)
. (3.28)
P r o o f of Proposition 3.8. Assume the contrary, that is, there exists a
symmetry π with the properties as in Proposition 3.8. We restrict our considera-
tions to the case where k acts via an automorphism ϕσ,α,β with α = β = 1. It will
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 347
S. Sinel’shchikov
become clear in what follows that this extra assumption is not really restrictive.
However, this restriction will be implicit unless the contrary is stated explicitly.
Let us use (2.3), (2.10), (3.24), (3.27), to compute (modulo non-zero constant
multipliers at monomials)
π
(
ekN
)
(x) = π(e)
(
xa(N)yc(N)
)
= xa(N)π(e)
(
yc(N)
)
+ π(e)
(
xa(N)
)
π(k)
(
yc(N)
)
= xa(N)
∑
(i
j)∈E
max{0,c(N)}−1∑
r=min{0,c(N)}
constxi+rb(1)yj+c(N)−1+r(d(1)−1)
+
∑
(i
j)∈D
max{0,a(N)}−1∑
r=min{0,a(N)}
constxi+a(N)−1+r(a(1)−1)yj+rc(1)
(
xb(1)yd(1)
)c(N)
=
∑
(i
j)∈E
max{0,c(N)}−1∑
r=min{0,c(N)}
constxi+a(N)+rb(1)yj+c(N)−1+r(d(1)−1)
+
∑
(i
j)∈D
max{0,a(N)}−1∑
r=min{0,a(N)}
constxi+a(N)+b(1)c(N)−1+r(a(1)−1)yj+d(1)c(N)+rc(1). (3.29)
The constant multipliers here are all non-zero. In the case where either a(N) = 0
or c(N) = 0, the corresponding sum in (3.29) is totally absent (equals 0), which
agrees with π(e)(1) = 0. On the other hand,
π
(
kNe
)
(x) = π
(
kN
)
∑
(i
j)∈D
constxiyj
=
∑
(i
j)∈D
const
(
xa(N)yc(N)
)i (
xb(N)yd(N)
)j
=
∑
(i
j)∈D
constxia(N)+jb(N)yic(N)+jd(N) (3.30)
with all the constant multipliers being non-zero.
Now let ξ =
(
ξ1
ξ2
)
, η =
(
η1
η2
)
be the eigenvectors of σ corresponding to the
eigenvalues λ and λ−1, with |λ| > 1 and |λ−1| < 1, respectively. One may assume
that ξi, ηi are real, so that everything is embedded into the real vector space R2,
348 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
together with the action of SL(2,Z) on it, which leaves the integral lattice Z2
invariant.
Note that
η2
η1
is irrational. In fact, if one assumes the contrary, one can
normalize η so that it has integral coordinates, together with all σNη, N ∈ Z.
Since the latter vectors are just λ−Nη, this makes a contradiction. In a similar
way, one observes that
ξ2
ξ1
is irrational.
Consider the linear functionals Φ, Ψ on the real vector space R2 given by
Φ(sξ + tη) = s, Ψ(sξ + tη) = t, s, t ∈ R.
Set also
Lε =
{
w ∈ R2 : |Ψ(w)| < ε
}
for ε > 0.
We are going to use the one-to-one correspondence constxiyj 7→ (
i
j
)
between
the monomials in Cq[x±1, y±1] modulo non-zero constant multipliers and the pairs
of integers in Z2. In the latter picture, as a consequence of (2.6), the (finite)
collections of pairs of integers coming from (3.29) and (3.30) must coincide. But
the collections themselves look very different. As for (3.30), the corresponding
subset in Z2 ⊂ R2 is nothing more than σND, which is inside a narrow stripe Lε,
ε > 0 being as small as desired with N ∈ Z+ big enough. On the other hand,
the set of pairs coming from (3.29) is formed by the two collections of arithmetic
progressions
{(
i + a(N) + rb(1)
j + c(N)− 1 + r(d(1)− 1)
)
:
r ∈ Z, min{0, c(N)} ≤ r < max{0, c(N)}
}
(3.31)
with
(
i
j
) ∈ E, and
{(
i + a(N) + b(1)c(N)− 1 + r(a(1)− 1)
j + d(1)c(N) + rc(1)
)
:
r ∈ Z, min{0, a(N)} ≤ r < max{0, a(N)}
}
(3.32)
with
(
i
j
) ∈ D. In the case where either a(N) = 0 or c(N) = 0, the corresponding
set of progressions is treated as void because the associated sum in (3.29) is totally
absent (equals 0). The steps of progressions (3.31), (3.32), namely hE =
( b(1)
d(1)−1
)
and hD =
(a(1)−1
c(1)
)
, respectively, do not depend on N . These vectors are certainly
linearly independent, being the columns of the non-degenerate matrix σ − I.
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 349
S. Sinel’shchikov
So, the only problem in producing a desired contradiction is in observing
that some pairs in (3.31), (3.32) may fail to be distinct, hence the corresponding
monomials in (3.29) may fail to survive after possible reductions.
Lemma 3.9. Each of the sets D and E contains at most two elements.
P r o o f. We prove this lemma for D. A completely similar argument can be
used to prove it for E.
The above argument on irrationality implies that all the values, Φ
(
i
j
)
,
(
i
j
) ∈ D,
are pairwise different, hence
dΦ = min
{∣∣∣∣Φ
(
i
j
)
− Φ
(
i′
j′
)∣∣∣∣ :
(
i
j
)
6=
(
i′
j′
)
∈ D
}
> 0,
and
|λN |dΦ = min
{∣∣∣∣Φ
(
i
j
)
− Φ
(
i′
j′
)∣∣∣∣ :
(
i
j
)
6=
(
i′
j′
)
∈ σND
}
for N ∈ Z. Of course, the latter inequality (dΦ > 0) anticipates cardD > 1, while
the opposite assumption already implies the claim of the lemma.
In a similar way, set
dΨ = max
{∣∣∣∣Ψ
(
i
j
)
−Ψ
(
i′
j′
)∣∣∣∣ :
(
i
j
)
,
(
i′
j′
)
∈ D
}
.
Let A = max
{∣∣∣Φ
(
i
j
)− Φ
(
i′
j′
)∣∣∣ :
(
i
j
)
,
(
i′
j′
) ∈ D
}
. As Ψ(hD) 6= 0, one can choose
ε > 0 such that ε < 1
2 |Ψ(hD)|, and then choose N > 0 such that σND ⊂ Lε and
|λN |dΦ > A + (dΨ + 2ε)
∣∣∣∣
Φ(hD)
Ψ(hD)
∣∣∣∣ + 1. (3.33)
Consider the second sum in (3.29) corresponding to
(
i
j
) ∈ D together with
the associated set of progressions (3.32) in Z2. By our choice of ε, every such
progression has at most one intersection point with Lε and thus also with σND.
We claim that in fact all the progressions in (3.32) together produce at most
one intersection point with σND. To see this, assume the contrary, that is,
two progressions as in (3.32) meet σND in two different points. As the steps
of these progressions are the same, these progressions are disjoint. To be more
precise, we have some
(
i1
j1
)
,
(
i2
j2
) ∈ (a(N)+b(1)c(N)−1
d(1)c(N)
)
+D and some r1, r2 ∈ Z with
min{0, a(N)} ≤ r < max{0, a(N)} such that
(
i1
j1
)
+ r1hD,
(
i2
j2
)
+ r2hD ∈ σND.
350 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
In these settings we have
dΨ ≥
∣∣∣∣Ψ
(
i1
j1
)
−Ψ
(
i2
j2
)∣∣∣∣
≥ |r1 − r2||Ψ(hD)| −
∣∣∣∣Ψ
((
i1
j1
)
+ r1hD
)
−Ψ
((
i2
j2
)
+ r2hD
)∣∣∣∣
≥ |r1 − r2||Ψ(hD)| − 2ε,
hence
|r1 − r2| ≤ dΨ + 2ε
|Ψ(hD)| .
This estimate, together with (3.33), implies
A ≥
∣∣∣∣Φ
(
i1
j1
)
− Φ
(
i2
j2
)∣∣∣∣
≥
∣∣∣∣Φ
((
i1
j1
)
+ r1hD
)
− Φ
((
i2
j2
)
+ r2hD
)∣∣∣∣− |r1 − r2||Φ(hD)|
≥ |λN |dΦ − dΨ + 2ε
|Ψ(hD)| |Φ(hD)| > A + 1.
We thus obtain a contradiction which proves the existence of at most one intersec-
tion point of progressions (3.32) and σND for N chosen above and all bigger N .
A similar argument, possibly after decreasing ε and increasing N , allows one
to establish the existence of additionally at most one intersection point of pro-
gressions (3.31) and σND. This proves the claim of the lemma, because, due to
(2.6), the union of points in (3.32) and (3.31) must contain σND.
Lemma 3.10. cardD = cardE = 2, and D =
{(
i
j
)
;
(
i
j
)
+ hD
}
, E ={(
i′
j′
)
;
(
i′
j′
)
+ hE
}
for some integers i, j, i′, j′.
P r o o f. In view of Lemma 3.9, we have to consider finitely many cases to
be discarded.
(a) Let us suppose cardD = 0. This means that π(e)(x) = 0. If one also has
cardE = 0, i.e., π(e)(y) = 0, then π(e) is identically zero. Let us apply (2.8)
via the action π to x. As the left-hand side is zero, the right-hand side being
zero is equivalent to π(k)2(x) = x. This already implies (even without assuming
α = β = 1 as at the beginning of the proof) that the matrix σ2 has eigenvalue 1,
which is not our case. Thus we get a contradiction.
Suppose now in this case cardE = 1, that is, (3.31) contains just one progres-
sion (note that c(N) 6= 0), while (3.32) is void. As the corresponding monomials
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 351
S. Sinel’shchikov
in (3.29) are linearly independent, we conclude that (3.29) is non-zero, while
(3.30) is zero. This contradicts to (2.6).
The last subcase here is cardE = 2. Again, since c(N) 6= 0, (3.31) contains
just two progressions that can not coincide as sets of points, because they corre-
spond to different points of E and have the same step hE . There exists a point
that belongs to one of the progressions but not to the other one. The correspond-
ing monomial in (3.29) will survive under possible reductions, thus making (3.29)
non-zero, while (3.30) is zero. This contradicts to (2.6).
(b) Consider the case cardD = 1. There exists a single progression in (3.32).
Let us choose ε > 0 and N > 0 as in the proof of Lemma 3.9, which provides
that the progression in (3.32) contains at most one intersection point with σND.
Additionally, it can contain at most two intersection points with progressions in
(3.31). The latter is due to Lemma 3.9 and the fact that the steps hD and hE are
linearly independent. As for the rest of points in the progression in (3.32) (which
are certainly present with N > 0 big enough), they are all outside of σND, and
the corresponding monomials in (3.29) survive under possible reductions. This
contradicts to (2.6).
(c) Consider the case cardD = 2, that is, D is formed by two distinct points(
i1
j1
)
,
(
i2
j2
)
. Respectively, (3.32) is formed by two progressions whose step is hD
and length is |a(N)|. As one observes from (3.28), the latter value is non-zero
and grows for N > 0 big enough, whose choice is at our hand.
Suppose that these two progressions are disjoint. In this case one can choose
each to apply the argument in (b) in order to get the desired contradiction. Of
course, this argument can be also applied to the progressions of (3.31).
Finally, suppose that the two progressions in (3.32) are not disjoint (at least
when their length |a(N)| is big enough). More precisely, one has to assume that(
i2
j2
)
=
(
i1
j1
)
+ shD for some integer s. In fact, s = 1, because otherwise the
symmetric difference of these two progressions contains 2s (at least 4) points.
After discarding at most one intersection point with progressions of (3.31), we
obtain at least 3 points in (3.32) corresponding to at least 3 monomials in (3.29)
which survive after all the reductions in (3.29). Since (3.30) contains at most
2 monomials, we get a contradiction with (2.6). Thus s = 1, and a similar
argument works also for E.
Turn back to the P r o o f of Proposition 3.8. At this step, we need to diverge
from our previous approach based on disregarding the specific non-zero constant
multipliers at monomials. Also, we now assume that k acts via an automorphism
ϕσ,α,β of the general form. Namely, we have
π(k)(x) = αxa(1)yc(1), π(k)(y) = βxb(1)yd(1). (3.34)
352 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
Also, by Lemma 3.10, we have
π(e)(x) = a1x
iyj + a2x
i+a(1)−1yj+c(1), (3.35)
π(e)(y) = b1x
i′yj′ + b2x
i′+b(1)yj′+d(1)−1 (3.36)
for some a1, a2, b1, b2 ∈ C \ {0}. Let us compute
π(e)(yx) = yπ(e)(x) + π(e)(y)π(k)(x)
= a1q
ixiyj+1 + a2q
i+a(1)−1xi+a(1)−1yj+c(1)+1 + b1αqa(1)j′xi′+a(1)yj′+c(1)
+ b2αqa(1)(j′+d(1)−1)xi′+a(1)+b(1)yj′+c(1)+d(1)−1, (3.37)
qπ(e)(xy) = qxπ(e)(y) + qπ(e)(x)π(k)(y)
= b1qx
i′+1yj′ + b2qx
i′+b(1)+1yj′+d(1)−1 + a1βqb(1)j+1xi+b(1)yj+d(1)
+ a2βqb(1)(j+c(1))+1xi+a(1)+b(1)−1yj+c(1)+d(1). (3.38)
Equating (3.37) and (3.38), we obtain a relation with a sum of 8 non-zero
monomials equal to zero. Let us consider the picture appearing via the one-
to-one correspondence constxiyj 7→ (
i
j
)
between the monomials in Cq[x±1, y±1]
modulo non-zero constant multipliers and Z2. Among the 8 monomials, those 4,
which contain i, j in their exponents, correspond to the following 4 points in Z2:
u =
(
i
j+1
)
, u + hD, u + hE , u + hD + hE . These 4 points are pairwise distinct,
hence the corresponding monomials are linearly independent.
In a similar way, the 4 monomials, which contain i′, j′ in their exponents,
correspond to the following 4 distinct points in Z2: v =
(
i′+1
j′
)
, v + hD, v + hE ,
v + hD + hE ; hence the corresponding monomials are linearly independent. Note
that in both cases we have parallelograms whose sides are determined by the
same pair of vectors hD, hE . In view of our observations, we conclude that the
sum of 8 monomials can be zero only in the case when the two parallelograms
coincide. This implies, in particular, that i = i′ + 1, j + 1 = j′.
With the latter conclusion being taken into account, our next step is to equate
the coefficients at the corresponding monomials in (3.37), (3.38). This leads to
the following system of equations:
a1q
i = b1q,
a2q
i+a(1)−1 = −b1αqa(1)j+a(1),
a1βqb(1)j+1 = −b2q,
a2βqb(1)(j+c(1))+1 = b2αqa(1)(j+d(1)).
The general solution of this system has the form
a1 = g, a2 = −gαqa(1)j , b1 = gqi−1, b2 = −gβqb(1)j , g ∈ C \ {0}.
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 353
S. Sinel’shchikov
This allows one to rewrite (3.35) and (3.36) as follows:
π(e)(x) = gxiyj − gαqa(1)jxi+a(1)−1yj+c(1), (3.39)
π(e)(y) = gqi−1xi−1yj+1 − gβqb(1)jxi+b(1)−1yj+d(1). (3.40)
Now let us turn back to (3.29), (3.30), together with the associated pictures in
Z2 (3.31), (3.32), both in the case N = 1, in order to extract more consequences
from (2.6) applied to x. Looking at (3.32), in view of Lemma 3.10, we note that
the entire picture is formed by two progressions, both of the same length |a(1)|
and with the same step hD. These progressions are non-disjoint; more precisely,
they are translates of each other in such a way that the initial point of one of two
progressions coincides to the second point of another one. This means that the
endpoints of the union of these two progressions correspond to the monomials in
(3.29) which survive after possible reductions within the second sum in (3.29).
To write these two points, u1, u2 ∈ Z2, explicitly, one has to substitute r = 0
and r = a(1) to (3.32), respectively, and this is clearly independent of the sign of
a(1). Namely, we have
u1 =
(
i + a(1)− 1 + b(1)c(1)
j + c(1)d(1)
)
, u2 =
(
i + a(1)2 − 1 + b(1)c(1)
j + a(1)c(1) + c(1)d(1)
)
.
It might look all this assumes a(1) 6= 0. However, this is true only in the above
part of the picture. It will become clear in what follows that the case a(1) = 0, in
spite of vanishing the above two progressions, does not break the entire argument.
In a similar way, let us consider the progressions of (3.31), whose length is
|c(1)| and the step is hE , and write explicitly the endpoints v1, v2 ∈ Z2 explicitly
substituting r = 0 and r = c(1) to (3.31) and using the relations between i, j, i′, j′:
v1 =
(
i + a(1)− 1
j + c(1)
)
, v2 =
(
i + a(1)− 1 + b(1)c(1)
j + c(1)d(1)
)
.
Note that c(1) 6= 0, because the matrix σ can not be triangular under our as-
sumptions. Again, the corresponding monomials in (3.29) survive after possible
reductions within the first sum in (3.29). However, all the four monomials should
somehow vanish when equating (3.29) to (3.30) due to (2.6). In view of this, let us
compare u1, u2, v1, v2 with the points of σD which correspond to the monomials
of (3.30). By (3.39),
(
i
j
) ∈ D, hence the two points of σD are just w1 = σ
(
i
j
)
and
w2 = σ
((
i
j
)
+ hD
)
.
Note that u1 = v2. It follows that {u2, v1} = σD (obviously, there is no other
opportunity). More precisely, since u2 = u1 + a(1)hD = u1 + a(1)(σ − I)
(
1
0
)
and
v1 = v2 − c(1)hE = v2 − c(1)(σ − I)
(
0
1
)
, one has
u2 − v1 = a(1)(σ − I)
(
1
0
)
+ c(1)(σ − I)
(
0
1
)
= (σ − I)
(
a(1)
c(1)
)
= (σ − I)σ
(
1
0
)
.
354 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
Also one has
w2 − w1 = σhD = σ(σ − I)
(
1
0
)
,
and since the matrices σ, σ − I commute, u2 − v1 = w2 − w1. It follows that
v1 = w1, u2 = w2. The first of these equalities can be written as
σ
(
i
j
)
=
(
i + a(1)− 1
j + c(1)
)
,
hence
(σ − I)
(
i
j
)
= hD = (σ − I)
(
1
0
)
,
and since σ − I is an invertible matrix, we conclude that i = 1, j = 0, and this
solution is unique. Certainly, the same result can be deduced from u2 = w2.
This, together with (3.34), now makes possible one more adjustment of (3.39),
(3.40):
π(e)(x) = gx− gαxa(1)yc(1) = g(id−π(k))(x), (3.41)
π(e)(y) = gy − gβxb(1)yd(1) = g(id−π(k))(y) (3.42)
for some g ∈ C \ {0}.
A routine verification shows that the linear map Φ = g(id−π(k)) on
Cq[x±1, y±1] is subject to the same rule Φ(ξη) = ξΦ(η) + Φ(ξ)π(k)(η) as π(e).
This allows one to extend the relations (3.41), (3.42) from the generators to the
entire algebra Cq[x±1, y±1], so that π(e) = g(id−π(k)) identically on Cq[x±1, y±1].
Since this map clearly commutes with π(k), we conclude that (2.6) fails. This
contradiction completes the proof of the proposition.
R e m a r k 3.11. In view of the results of this section, the complete list of
Uq (sl2)-symmetries on Cq[x±1, y±1] such that k acts by an automorphism ϕσ,α,β
with a non-unit matrix σ, is given by Theorem 3.5.
4. σ = I: the Generic Case
In the case σ = I, the action of the Cartan element k is given by multiplication
of the generators x, y by the weight constants. We follow (2.3) in denoting these
weight constants by α and β, respectively. Certainly, monomials form a basis of
weight vectors (eigenvectors for π(k)), and the associated eigenvalues are called
weights.
Let us consider the case where either π(e) or π(f) is not identically zero. In
this setting we describe a series of symmetries which we call generic.
A pair of non-zero complex constants α and β which could appear as weight
constants for some Uq(sl2)-symmetry of Cq[x±1, y±1], can not be arbitrary. In
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 355
S. Sinel’shchikov
fact, an obvious consequence of (2.6) claims that π(e) sends a vector whose weight
is γ to a vector whose weight is q2γ. In particular, π(e)(x), if non-zero, is a sum
of monomials with (the same) weight q2α. Since the weight of the monomial
axiyj (with a 6= 0) is αiβj , one has that αuβv = q2 for some integers u, v. Of
course, similar conclusions can be also derived by applying (2.6), (2.7) to x and
y. Under our assumptions, this argument must work at least once.
The following theorem covers all but a countable family of admissible pairs
of weight constants.
Theorem 4.1. Let α, β ∈ C \ {0} be such that αuβv = q2 for some u, v ∈ Z
and αm 6= βn for non-zero integers m, n. Then there exists a one-parameter
(a ∈ C \ {0}) family of Uq(sl2)-symmetries of Cq[x±1, y±1]:
π(k)(x) = αx, π(k)(y) = βy, (4.1)
π(e)(x) = aquv+3 1− αqv
(1− q2)2
xu+1yv, π(e)(y) = aquv+3 qu − β
(1− q2)2
xuyv+1, (4.2)
π(f)(x) = −α−1 − q−v
a
x−u+1y−v, π(f)(y) = −β−1q−u − 1
a
x−uy−v+1. (4.3)
There exist no other symmetries with the weight constants α and β.
Lemma 4.2. Let π be a Uq(sl2)-symmetry on Cq[x±1, y±1] such that
π(k)(x) = αx, π(k)(y) = βy, (4.4)
π(e)(x) =
∑
i,j
ai,jx
iyj , π(e)(y) =
∑
i,j
bi,jx
iyj , (4.5)
π(f)(x) =
∑
i,j
ci,jx
iyj , π(f)(y) =
∑
i,j
di,jx
iyj , (4.6)
with α, β ∈ C \ {0}, ai,j , bi,j , ci,j , di,j ∈ C, and the above sums being finite. Then
ai+1,j
(
qi − β
)
= bi,j+1
(
1− αqj
)
, (4.7)
ci+1,j
(
1− β−1qi
)
= di,j+1
(
qj − α−1
)
. (4.8)
P r o o f. This is a consequence of (2.1). We have
π(e)(yx) = yπ(e)(x) + π(e)(y)π(k)(x) =
∑
i,j
ai,jq
ixiyj+1 +
∑
i,j
bi,jαqjxi+1yj ,
qπ(e)(xy) = qxπ(e)(y) + qπ(e)(x)π(k)(y) =
∑
i,j
bi,jqx
i+1yj +
∑
i,j
ai,jβqxiyj+1.
356 Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4
Generic Symmetries of the Laurent Extension of Quantum Plane
With this, we project the relation π(e)(yx) = qπ(e)(xy) to the one-dimensional
subspace Cxi+1yj+1 parallel to the linear span of all other monomials to obtain
(4.7). The proof of (4.8) is similar.
P r o o f of Theorem 4.1. We are about to apply the following relations valid
under the assumption σ = I of the present section:
π(e)(xp) =
∑
i,j
ai,j
αpqjp − 1
αqj − 1
xp−1+iyj ,
π(e)(yp) =
∑
i,j
bi,j
βp − qip
β − qi
xiyp−1+j ,
π(f)(xp) =
∑
i,j
ci,j
α−p − qjp
α−1 − qj
xp−1+iyj ,
π(f)(yp) =
∑
i,j
di,j
β−pqip − 1
β−1qi − 1
xiyp−1+j ,
where p ∈ Z, ai,j , bi,j , ci,j , di,j ∈ C are as in (4.5), (4.6). These relations are due
to a straightforward induction argument.
A direct computation that applies the above relations, allows one to verify
that the extended action (4.1)–(4.3) from the generators to the entire algebras
Uq(sl2) and Cq[x±1, y±1] passes through all the relations (2.1), (2.2), (2.5)–(2.8).
Hence (4.1)–(4.3) determine a well-defined Uq(sl2)-symmetry on Cq[x±1, y±1].
Let us prove that there are no other symmetries. Observe first that the
assumptions of the theorem on the weight constants α and β imply that the pair of
integers u, v with αuβv = q2 is unique. Therefore the monomials modulo a (non-
zero) constant multiplier are in one-to-one correspondence with their weights.
Since, in view of (2.6), (2.7), π(e) and π(f) ‘multiply’ the weight of a weight
vector by q2 and q−2, respectively, one deduces that π(e)(x), π(e)(y), π(f)(x),
π(f)(y) are monomials which, up to constant multipliers, should be just as in
(4.2), (4.3).
Observe that, under the assumptions of the theorem on weight constants α
and β, no differences in (4.2), (4.3) can be zero. Thus, it follows from Lemma
4.2 that the ratio of coefficients at the monomials π(e)(x) and π(e)(y) should be
just as in (4.2). Of course, a similar claim is also true for (4.3).
It remains to establish the ratio of coefficients in (4.2) and (4.3). This is done
by applying (2.8) to x and y. The computation in question, which is left to the
reader, in both cases leads to the same result reflected in (4.2), (4.3).
R e m a r k 4.3. Theorem 4.1 describes an uncountable family of isomorphism
classes of Uq(sl2)-symmetries of Cq[x±1, y±1].
Journal of Mathematical Physics, Analysis, Geometry, 2015, vol. 11, No. 4 357
S. Sinel’shchikov
In fact, one clearly has an uncountable family of admissible (i.e., those that
are subject to the assumptions of Theorem 4.1) pairs of weight constants α, β. On
the other hand, the action of the group of automorphisms of Cq[x±1, y±1] (which
is the semidirect product of its subgroups SL(2,Z) and (C∗)2) on the space of
parameters of generic symmetries is such that the action of normal subgroup
(C∗)2 remains every pair of weight constants (α, β) intact. It follows that (the
projection of) each orbit of the automorphism group on the space of admissible
pairs of weight constants α, β is only countable.
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