Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 43 (2002), No. 2, 597-601.

The Group of Automorphisms of the Coordinate Ring of Quantum Symplectic Space

J. Gómez-Torrecillas L. EL Kaoutit^{∗}

Departamento de Algebra, Facultad de Ciencias Universidad de Granada, E18071-Granada, Spain

Abstract. The group of automorphisms of the coordinate ring of quantum symplectic space O_{q}(𝔰𝔭𝕜^{2×n}) is isomorphic to the algebraic torus (𝕜^{×})^{n+1}, when q is not a root of unity.

Introduction.

In contrast with the classical commutative case, the rigidity of some quantized coordinate algebras in the ‘generic case’ allows an explicit description of their groups of automorphisms. Thus, J. Alev and M. Chamarie showed that the group of automorphisms of the quantum plane is a torus of rank two in the non root of unit case [1]. Under the same condition on the quantum parameters, J. Alev and F. Dumas [2] found later that the group of automorphisms of the first quantum Weyl algebra is an algebraic torus. This result was extended by L. Rigal to the n^{th} quantum Weyl algebra [12].

The aim of this paper is to describe explicitly the group of automorphisms of the coordinate ring of quantum space O_{q}(𝔰𝔭𝕜^{2×n}) (q is not a root of unity), which will be denoted by Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})). We use Rigal’s method, namely, the fact that the set of height one prime ideals (explicitly determined in [5]) is invariant under the action of Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})).

∗The author wishes to express his thanks to A.E.C.I for financial support.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2002 Heldermann Verlag

598 J. Gómez-Torrecillas, L. EL Kaoutit: The Group of Automorphisms …

1. The quantum symplectic space

Let 𝕜 be a field, and q a non-zero element in 𝕜 which is not a root of unity. I. M. Musson found [8, §1.1] that the coordinate ring O_{q}(𝔰𝔭𝕜^{2×n}) of the quantum symplectic space (cf. [11, Definition 14] or [13, §4]) is the 𝕜-algebra generated by y_{1}, x_{1}, …, y_{n}, x_{n} satisfying the following relations

\begin{matrix}y_{j}x_{i} = q^{−1}x_{i}y_{j}, & y_{j}y_{i} = qy_{i}y_{j} & (1 ≤ i < j ≤ n) \\ x_{j}x_{i} = q^{−1}x_{i}x_{j}, & x_{j}y_{i} = qy_{i}x_{j} & (1 ≤ i < j ≤ n) \\ x_{i}y_{i} − q^{2}y_{i}x_{i} = (q^{2} − 1)∑^{i−1} ; l=1q^{i−l}y_{l}x_{l} & & (1 ≤ i ≤ n) \end{matrix}  (1)

By [9, Proposition 1.10] or [10, Example 6], O_{q}(𝔰𝔭𝕜^{2×n}) can be written as an iterated Ore extension

R_{0} ⊆ R_{1} ⊆ ⋯ ⊆ R_{n} = O_{q}(𝔰𝔭𝕜^{2×n})

where R_{0} = 𝕜 and R_{k} = R_{k−1}[y_{k}; α_{k}][x_{k}; β_{k}, δ_{k}] for k ≥ 1, with

\begin{matrix}α_{k}(x_{l}) = q^{−1}x_{l} = β_{k}(x_{l}) & (1 ≤ l < k ≤ n) \\ α_{k}(y_{l}) = qy_{l} = β_{k}(y_{l}) & (1 ≤ l < k ≤ n) \\ β_{k}(y_{k}) = q^{2}y_{k} & (1 ≤ k ≤ n) \\ δ_{k}(R_{k−1}) = 0, δ_{k}(y_{k}) = (q^{2} − 1)k−1 ; ∑_{l=1}q^{k−l}y_{l}x_{l} & (1 ≤ k ≤ n) \end{matrix}

Consider the elements Ω_{i} = ∑_{l=1}^{i} q^{i−l}y_{l}x_{l} (i ≥ 1). From [9, Lemma 1.3] we get

Ω_{i}y_{k} = q^{2}y_{k}Ω_{i}, Ω_{i}x_{k} = q^{−2}x_{k}Ω_{i} (k ≤ i)
Ω_{i}x_{k} = x_{k}Ω_{i}, Ω_{i}y_{k} = y_{k}Ω_{i}  (i < k)
Ω_{i}Ω_{k} = Ω_{k}Ω_{i}  (for all i, k)

(2)

\begin{matrix}Ω_{i} = ∑_{j<l≤i} q^{i−l}y_{l}x_{l} + q^{i−j}Ω_{j}, & (j ≤ i) \\ x_{i}y_{i} − q^{2}y_{i}x_{i} = (q^{2} − 1)qΩ_{i−1} \\ x_{i}y_{i} − y_{i}x_{i} = (q^{2} − 1)Ω_{i}. \end{matrix}  (3)

By the relations (1), O_{q}(𝔰𝔭𝕜^{2×n}) is a P.B.W 𝕜-algebra in the sense of [3, 4] with respect to the lexicographical order ≤_{lex} on ℕ^{2n} with (1, 0, ⋯, 0) <_{lex} (0, 1, ⋯, 0) <_{lex} ⋯ <_{lex} (0, ⋯, 0, 1). This implies that every f ∈ O_{q}(𝔰𝔭𝕜^{2×n}) is uniquely written as

f = ∑_{α∈ℕ^{2n}} c_{α}X^{α}

with respect to the 𝕜-basis

B = {X^{α} = y_{1}^{ν_{1}}x_{1}^{μ_{1}} ⋯ y_{n}^{ν_{n}}x_{n}^{μ_{n}}| α = (ν_{1}, μ_{1}, ⋯, ν_{n}, μ_{n}) ∈ ℕ^{2n}} .

For f ≠ 0, define

exp(f) = max_{≤_{lex}}{α | c_{α} ≠ 0}.

By [4, Proposition 1.3(i)],

exp(fg) = exp(f) + exp(g) for any f, g ∈ O_{q}(𝔰𝔭𝕜^{2×n}) \ {0}.

In [5] we found an explicit description of the prime spectrum (the set of prime ideals) of O_{q}(𝔰𝔭𝕜^{2×n}). From [5, Proposition 3.2, Corollary 3.5], we can deduce that the set of all height one prime ideals is

P = {⟨y_{1}⟩, ⟨x_{1}⟩, ⟨Ω_{2}⟩, ⋯, ⟨Ω_{n}⟩} .

2. The group of automorphisms

To determine this group, we first show that the 𝕜-vector space generated by an element v ∈ {y_{1}, x_{1}, Ω_{2}, ⋯, Ω_{n}} is invariant by Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})). Then we compute the image of the rest of generators, by a fixed element of Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})). Let V_{n} be the 𝕜-vector space generated by {Ω_{1}, ⋯, Ω_{n}}, by (3) the set {y_{i}x_{i}}_{1≤i≤n} is a 𝕜-basis of this vector space.

Lemma 2.1. Let σ ∈ Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})). Then for any i ∈ {2, ⋯, n} there exists j ∈ {2, ⋯, n}, such that

σ(Ω_{i}) = λ_{ij}Ω_{j}, and σ(x_{1}) = μ_{1}x_{1}, σ(y_{1}) = ν_{1}y_{1},

where λ_{ij}, ν_{1}, μ_{1} ∈ 𝕜^{∗}.

Proof. By the Theorem of the principal ideal [7, Theorem 4.1.11] (or [6]), P is invariant by Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})). Fix σ ∈ Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})), and let x ∈ {y_{1}, x_{1}, Ω_{2}, ⋯, Ω_{n}}, then there exist h, h′ ∈ O_{q}(𝔰𝔭𝕜^{2×n}) \ {0} and y ∈ {y_{1}, x_{1}, Ω_{2}, ⋯, Ω_{n}}, such that σ(x) = hy and σ^{−1}(y) = h′x. Hence hσ(h′) = σ(h′)h = 1, so h, h′ are invertible thus h, h′ ∈ 𝕜^{∗}. Suppose now that there exist i ∈ {2, ⋯, n} and j ∈ {2, ⋯, n}, such that σ(Ω_{i}) = αy_{1}, σ(Ω_{j}) = βx_{1} for some α, β ∈ 𝕜^{∗}. From (2), we have Ω_{i}Ω_{j} = Ω_{j}Ω_{i}. Applying σ to this equality we get αβ = 0, because q is not a root of unity. So for each i ∈ {2, ⋯, n} there exists j ∈ {2, ⋯, n}, such that σ(Ω_{i}) = λ_{ij}Ω_{j}, λ_{ij} ∈ 𝕜^{∗}. By the same way, we show that x_{1} (resp, y_{1}) can not be the image of y_{1} (resp, x_{1}) by σ. Therefore σ(x_{1}) = μ_{1}x_{1}, σ(y_{1}) = ν_{1}y_{1}, ν_{1}, μ_{1} ∈ 𝕜^{∗}. □

Remark 2.2. By the previous lemma, for each σ ∈ Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})) and i ∈ {2, ⋯, n}, there exist k ∈ ℕ, such that σ^{k}(Ω_{i}) = λΩ_{i}, λ ∈ 𝕜^{∗}.

Lemma 2.3. Let a, b ∈ O_{q}(𝔰𝔭𝕜^{2×n}) \ 𝕜, such that ab = ∑_{1≤k_{i}≤n} k_{i}y_{i}x_{i} ∈ V_{n} with k_{i} ∈ 𝕜, and k_{n} ≠ 0. Then there exist λ, λ′ ∈ 𝕜^{∗}, such that a = λy_{n}, and b = λ′x_{n} (or a = λ′x_{n}, and b = λy_{n}).

Proof. It is clear that exp(ab) = (0, 0, ⋯, 0, 1, 1), so we have

\{ \begin{matrix}exp(a) = (0, ⋯, 0, 1, 0), & exp(b) = (0, ⋯, 0, 0, 1) \\ or \\ exp(a) = (0, ⋯, 0, 0, 1), & exp(b) = (0, ⋯, 0, 1, 0). \end{matrix}

Then we have, for example, a = λy_{n} + a_{0}, b = λ′x_{n} + b_{0}, where λ, λ′ ∈ 𝕜^{∗} and a_{0} ∈ R_{n−1}, b_{0} ∈ R_{n/2} = R_{n−1}[y_{n}; α_{n}]. So ab = λλ′y_{n}x_{n} + λ′a_{0}x_{n} + λy_{n}b_{0} + a_{0}b_{0}, and u = λ′a_{0}x_{n} + λy_{n}b_{0} + a_{0}b_{0} ∈ V_{n}. But exp(λ′a_{0}x_{n}) = exp(u) = (ν, 0, 1), with ν ∈ ℕ^{2(n−1)}. Hence a_{0} = b_{0} = 0. By the same way we get the other case. □

Proposition 2.4. Let σ ∈ Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})). Then for each i ∈ {1, ⋯, n} we have σ(Ω_{i}) = λ_{i}Ω_{i}, λ_{i} ∈ 𝕜^{∗}, and σ(x_{i}) = μ_{i}x_{i}, σ(y_{i}) = ν_{i}y_{i}, ν_{i}, μ_{i} ∈ 𝕜^{∗}.

Proof. For i = 1, we know by the Lemma 2.1 that σ(x_{1}) = μ_{1}x_{1}, σ(y_{1}) = ν_{1}y_{1}, ν_{1}, μ_{1} ∈ 𝕜^{∗}, and so σ(Ω_{1}) = λ_{1}Ω_{1}, λ_{1} = ν_{1}μ_{1}. Suppose that there exist i ≠ j ∈ {2, ⋯, n}, such that σ(Ω_{i}) = λΩ_{j}, λ ∈ 𝕜^{∗}. If σ(Ω_{n}) = λΩ_{n} then there does not exist k ∈ ℕ^{∗} such that σ^{k}(Ω_{i}) = λ_{0}Ω_{n}, with λ_{0} ∈ 𝕜^{∗}. Let m be the maximal element in the set

{j | σ(Ω_{l}) = λΩ_{j} for some λ ∈ 𝕜^{∗} and l ≠ j}.

Let σ(Ω_{i}) = λ_{0}Ω_{m}, λ_{0} ∈ 𝕜^{∗}. Note that 1 < i < m ≤ n, σ(Ω_{i−1}) = λ′_{0}Ω_{r}, σ(Ω_{i+1}) = λ′′Ω_{s} for some λ′, λ′′ ∈ 𝕜^{∗} and r, s < m. Applying σ to Ω_{i} = y_{i}x_{i} + qΩ_{i−1}, we get σ(y_{i})σ(x_{i}) = ∑_{1≤l≤m} k_{l}y_{l}x_{l}, with k_{m} = λ_{0}. By Lemma 2.3 (with m = n) applied to σ(y_{i})σ(x_{i}), we have for example σ(y_{i}) = μ′x_{m}, μ′ ∈ 𝕜^{∗}. If we apply σ to Ω_{i+1} = y_{i+1}x_{i+1} +qΩ_{i}, then we get σ(y_{i+1}) = μ′_{0}x_{m}, μ′_{0} ∈ 𝕜^{∗} (or σ(x_{i+1}) = μ′_{0}x_{m}), which contradicts the injectivity of σ. In conclusion we have σ(Ω_{i}) = λ_{i}Ω_{i}, λ_{i} ∈ 𝕜^{∗}. Now applying σ, to Ω_{i} = y_{i}x_{i} + qΩ_{i−1}, i = 2, …, n, we get σ(y_{i})σ(x_{i}) ∈ V_{n}. By Lemma 2.3 (with i = n) applied to σ(y_{i})σ(x_{i}), we have σ(y_{i}) = ν_{i}y_{i}, σ(x_{i}) = μ_{i}x_{i}, ν_{i}, μ_{i} ∈ 𝕜^{∗}. Observe that we can not have, σ(x_{i}) = ν_{i}y_{i}, σ(y_{i}) = μ_{i}x_{i}. If it is the case, then one have μ_{i}ν_{i}y_{i}x_{i} − q^{2}μ_{i}ν_{i}x_{i}y_{i} = (q^{2} − 1)qλ_{i−1}Ω_{i−1}, so μ_{i}ν_{i} = 0. □

Theorem 2.5. Let O_{q}(𝔰𝔭𝕜^{2×n}), with q not a root of unity. Then

Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})) ≅ (𝕜^{×})^{n+1}.

Proof. Pick σ ∈ Aut_{𝕜}(O_{q}(𝔰𝔭𝕜^{2×n})), by the Proposition 2.4 we have for each i = 1, …, n, σ(x_{i}) = μ_{i}x_{i}, σ(y_{i}) = ν_{i}y_{i}, μ_{i}, ν_{i} ∈ 𝕜^{∗} and σ(Ω_{i}) = λ_{i}Ω_{i}, λ_{i} ∈ 𝕜^{∗}, with λ_{1} = μ_{1}ν_{1}. Applying σ to Ω_{2} = y_{2}x_{2} + (q^{2} − 1)qy_{1}x_{1}, we get λ_{2} = λ_{1} = μ_{1}ν_{1} = μ_{2}ν_{2}. We continue to applying σ to other equalities listed in (3), we get in the end, λ_{1} = ⋯ = λ_{n} = μ_{1}ν_{1} = ⋯ = μ_{n}ν_{n}. The theorem now is clear. □

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Received November 24, 2000