Primitive ideals and automorphisms of quantum matrices

S Launois and T H Lenagan ^{∗}

Abstract

Let q be a nonzero complex number that is not a root of unity. We give a criterion for ⟨0⟩ to be a primitive ideal of the algebra O_{q}(M_{m,n}) of quantum matrices. Next, we describe all height one primes of O_{q}(M_{m,n}); these two problems are actually interlinked since it turns out that ⟨0⟩ is a primitive ideal of O_{q}(M_{m,n}) whenever O_{q}(M_{m,n}) has only finitely many height one primes. Finally, we compute the automorphism group of O_{q}(M_{m,n}) in the case where m ≠ n. In order to do this, we first study the action of this group on the prime spectrum of O_{q}(M_{m,n}). Then, by using the preferred basis of O_{q}(M_{m,n}) and PBW bases, we prove that the automorphism group of O_{q}(M_{m,n}) is isomorphic to the torus (ℂ^{∗})^{m+n−1} when m ≠ n and (m, n) ≠ (1, 3), (3, 1).

2000 Mathematics subject classification: 16W35, 16W20, 20G42, 81R50.

Key words: Quantum matrices, quantum minors, prime ideals, primitive ideals, automorphisms.

Introduction

The automorphism group of a polynomial algebra ℂ[X_{1}, …, X_{n}], for n ≥ 2, is in general far from being understood. In the case n = 2 the group was described explicitly by Jung, [12]. However, it is only recently that Shestakov and Umirbaev, [19], have proved that the wellknown Nagata automorphism of ℂ[X_{1}, X_{2}, X_{3}] is wild; that is, the Nagata automorphism cannot be written as a product of elementary automorphisms.

∗This research was supported by a Marie Curie Intra-European Fellowship within the 6th European Community Framework Programme and by Leverhulme Research Interchange Grant F/00158/X

In this paper, we are interested in the quantum case. More precisely, when q is a nonzero complex number that is not a root of unity, we study the automorphism group of the algebra O_{q}(M_{m,n}) of m × n quantum matrices which in turn is a non-commutative deformation of a polynomial ring in m × n indeterminates. We denote by Y_{i,α}, (i, α) ∈ [[1, m]] × [[1, n]], the canonical generators of O_{q}(M_{m,n}). It is well-known that the group H := (ℂ^{∗})^{m+n} acts on O_{q}(M_{m,n}) by ℂ-automorphisms via:

(a_{1}, …, a_{m}; b_{1}, …, b_{n}).Y_{i,α} = a_{i}b_{α}Y_{i,α} ((i, α) ∈ [[1, m]] × [[1, n]]).

(Note that this is not a faithful action; for example, (a_{,} …, a; 1, …, 1) and (1, …, 1; a, …, a) have the same action on O_{q}(M_{m,n}), multiplying each Y_{i,α} by a. This explains why the automorphism group is an (m + n − 1)-torus rather than an (m + n)-torus.)

It was observed by Alev and Chamarie, [1], that quantization implies rigidity and so puts limits on the automorphism group of quantized algebras. This explains why it has been possible to compute the automorphism group of several quantum algebras at least in the generic case. For instance, Alev and Chamarie, [1], have described the automorphism group of the algebra of 2 × 2 quantum matrices, Alev and Dumas, [2], the automorphism group of the positive part of the quantized enveloping algebra of a complex simple Lie algebra of type A_{2}, the first author, [13], the automorphism group of the positive part of the quantized enveloping algebra of a complex simple Lie algebra of type B_{2}, ... .

One method to study the automorphism group of an algebra is to use the invariance of the set of height one primes under the action of the automorphism group. This method was used successfully by Rigal, [18], to compute the automorphism group of quantized Weyl algebras and next by Gomez-Torrecillas and El Kaoutit, [6], to calculate the automorphism group of the coordinate ring of quantum symplectic spaces. We also use this method in the present paper. However, in the cases of quantized Weyl algebras and coordinate ring of quantum symplectic spaces, the number of height one primes is finite (because of the choice of parameters) and so the restrictions on the automorphisms are very strong. In the case of the algebra of quantum matrices, the set of height one primes is in general not finite; so the situation is substantially more complicated.

We start by considering when the algebra O_{q}(M_{m,n}) has only finitely many height one primes. It turns out that, because of the stratification theorem of Goodearl and Letzter, see [3] for example, this situation arises exactly when ⟨0⟩ is a primitive ideal of O_{q}(M_{m,n}). Thus, the first section of this paper is devoted to this question: we establish a criterion for the ideal ⟨0⟩ to be primitive in O_{q}(M_{m,n}). More precisely, we prove that ⟨0⟩ is a primitive ideal of O_{q}(M_{m,n}) if and only if v_{2}(m) ≠ v_{2}(n), where v_{2}(k) denotes the 2-adic valuation of a positive integer k. This shows, for example, that ⟨0⟩ is a primitive ideal of O_{q}(M_{2,3}).

This criterion together with the stratification theorem of Goodearl and Letzter shows that, if v_{2}(m) ≠ v_{2}(n), then O_{q}(M_{m,n}) has only finitely many height one primes and it turns out that they are all H-invariant. On the other hand, if v_{2}(m) = v_{2}(n), then O_{q}(M_{m,n}) has infinitely many height one primes. A finite number (those that are H-invariant) are already known from previous work of the authors and Rigal, [14]. In the second part of this paper, we provide an explicit generator for every height one prime of O_{q}(M_{m,n}). (Note that, since O_{q}(M_{m,n}) is a Noetherian (non-commutative) UFD, every height one prime of O_{q}(M_{m,n}) is generated by a normal element, [14].) Hence, in this second part, we describe all normal elements of O_{q}(M_{m,n}) that generate a prime ideal.

Finally, in the third section, we investigate the automorphism group of O_{q}(M_{m,n}) when m ≠ n. Using the description of the height one primes of O_{q}(M_{m,n}) together with graded arguments, we show that all height one primes that are H-invariant, except possibly one, are invariant under every automorphism of O_{q}(M_{m,n}). Next, by using the preferred basis of O_{q}(M_{m,n}) and certain PBW bases, we are able to prove that the automorphism group of O_{q}(M_{m,n}) is isomorphic to the torus (ℂ^{∗})^{m+n−1} when m ≠ n and (m, n) ≠ (1, 3), (3, 1). This latter restriction occurs because of the existence of a rogue automorphism for quantum 3space (which may be viewed as 1×3 quantum matrices), see [1]. This is the only exception, and our analysis recovers the Alev-Chamarie result for this exceptional case.

In the case where m = n, the algebra O_{q}(M_{n}) has a homogeneous central element of degree n, the quantum determinant. The existence of this element considerably complicates the computation of the automorphism group, since the graded arguments that we use in the non-squared case do not put strong limits on the automorphism group and its action on the height one primes in the square case. In addition, transposition provides an automorphism that also clouds the analysis. The conjecture is that the automorphim group of O_{q}(M_{n}) is generated by a torus and the transposition automorphism. This has been verified by Alev and Chamarie, [1], in the 2 × 2 case. Our present methods can recover the Alev-Chamarie result, but, as yet, we are unable to deal with the general case, although we do have partial results. We intend to return to this question in a subsequent paper.

1 A criterion for quantum matrices to be primitive.

In this section, we use the H-stratification theory of Goodearl and Letzter together with the deleting derivations theory of Cauchon in order to characterize the integers m and n such that ⟨0⟩ is a primitive ideal in O_{q}(M_{m,n}).

1.1 The H-stratification of the prime spectrum of O_{q}(M_{m,n}).

Throughout this paper, we use the following conventions.

• If I is a finite set, |I| denotes its cardinality.

• [[a, b]] := {i ∈ ℕ | a ≤ i ≤ b}.

• ℂ denotes the field of complex numbers and we set ℂ^{∗} := ℂ \ {0}.

• q ∈ ℂ^{∗} is not a root of unity.

• m, n denote positive integers.

• R = O_{q}(M_{m,n}) is the quantization of the ring of regular functions on m×n matrices with entries in ℂ; it is the ℂ-algebra generated by the m × n indeterminates Y_{i,α}, 1 ≤ i ≤ m and 1 ≤ α ≤ n, subject to the following relations:

\begin{matrix}Y_{i,β}Y_{i,α} = q^{−1}Y_{i,α}Y_{i,β}, & (α < β); \\ Y_{j,α}Y_{i,α} = q^{−1}Y_{i,α}Y_{j,α}, & (i < j); \\ Y_{j,β}Y_{i,α} = Y_{i,α}Y_{j,β}, & (i < j, α > β); \\ Y_{j,β}Y_{i,α} = Y_{i,α}Y_{j,β} − (q − q^{−1})Y_{i,β}Y_{j,α}, & (i < j, α < β). \end{matrix}

It is well-known that R can be presented as an iterated Ore extension over ℂ, with the generators Y_{i,α} adjoined in lexicographic order. Thus the ring R is a Noetherian domain; we denote by F its skew-field of fractions. Moreover, since q is not a root of unity, it follows from [10, Theorem 3.2] that all prime ideals of R are completely prime. We denote by Spec(R) the set of (completely) prime ideals of R.

• It is well-known that the algebras O_{q}(M_{m,n}) and O_{q}(M_{n,m}) are isomorphic. Hence, we assume that m ≤ n.

• It is easy to check that the group H := (ℂ^{∗})^{m+n} acts on R by ℂ-algebra automorphisms via:

(a_{1}, …, a_{m}, b_{1}, …, b_{n}).Y_{i,α} = a_{i}b_{α}Y_{i,α} for all (i, α) ∈ [[1, m]] × [[1, n]].

An H-eigenvector x of R is a nonzero element x ∈ R such that h.x ∈ ℂ^{∗}x for each h ∈ H. An ideal I of R is said to be H-invariant if h.I = I for all h ∈ H. We denote by H-Spec(R) the set of H-invariant prime ideals of R. Since q is not a root of unity, it follows from [9, 5.7] that H-Spec(R) is a finite set. Note that ⟨0⟩ is an H-invariant prime ideal of R, since R is a domain.

The action of H on R allows us to use the H-stratification theory of Goodearl and Letzter, see [3, II.2], to constuct a partition of Spec(R) as follows. If J is an H-invariant prime ideal of R, we denote by Spec_{J}(R) the H-stratum of Spec(R) associated to J. Recall that Spec_{J}(R) := {P ∈ Spec(R) | ⋂_{h∈H} h.P = J}. Then the H-strata Spec_{J}(R), with J ∈ H-Spec(R), form a partition of Spec(R), see [3]:

Spec(R) = ⨆_{J∈H-Spec(R)} Spec_{J}(R).  (1)

Naturally, this partition induces a partition of the set Prim(R) of all (left) primitive ideals of R as follows. For all J ∈ H-Spec(R), we set Prim_{J}(R) := Spec_{J}(R) ∩ Prim(R). Then it is obvious that the H-strata Prim_{J}(R) (J ∈ H-Spec(R)) form a partition of Prim(R):

Prim(R) = ⨆_{J∈H-Spec(R)} Prim_{J}(R).

One of the reasons that makes the H-stratification interesting is that it provides a powerful tool for recognizing primitive ideals. Indeed, since ℂ is uncountable and since the Noetherian domain R is generated as an algebra by a finite number of elements, it follows from [3, Proposition II.7.16] that the algebra R satisfies the Nullstellensatz over ℂ, see [3, II.7.14]. Further the set of H-invariant prime ideals of R is finite. Thus [3, Theorem II.8.4] implies that Prim_{J}(R) (J ∈ H-Spec(R)) coincides with the set of those primes in Spec_{J}(R) that are maximal in Spec_{J}(R). Hence, we deduce the following criterion for R to be a primitive ring.

Proposition 1.1 R is a primitive ring if and only if the ⟨0⟩-stratum of Spec(R) is reduced to ⟨0⟩; that is, ⟨0⟩ is the only prime ideal in Spec_{⟨0⟩}(R).

1.2 Dimension of the ⟨0⟩-stratum of Spec(R).

In this section, we show that the ⟨0⟩-stratum Spec_{⟨0⟩}(O_{q}(M_{m,n})) is Zariski-homeomorphic to the prime spectrum of a commutative Laurent polynomial ring in dim(ker(B)) indeterminates, where B is a mn × mn matrix with entries in ℂ. First, we describe the matrix B explicitly and then we will compute the dimension of the kernel of B.

First recall, see [5], that the theory of deleting derivations can be applied to the iterated Ore extension R = ℂ[Y_{1,1}] … [Y_{m,n}; σ_{m,n}, δ_{m,n}] (where the indices are increasing for the lexicographic order ≤). The corresponding deleting derivations algorithm is called the standard deleting derivations algorithm. Before recalling its construction, we need to introduce some notation.

• We denote by ≤_{s} the lexicographic ordering on ℕ^{2}. We often call it the standard ordering on ℕ^{2}. Recall that (i, α) ≤_{s} (j, β) if and only if [(i < j) or (i = j and α ≤ β)].

• We set E_{s} = ([[1, m]] × [[1, n]] ∪ {(m, n + 1)}) \ {(1, 1)}.

• Let (j, β) ∈ E_{s}. If (j, β) ≠ (m, n + 1), then (j, β)^{+} denotes the least element (relative to ≤_{s}) of the set {(i, α) ∈ E_{s} | (j, β) <_{s} (i, α)}.

As described in [5], the standard deleting derivations algorithm constructs, for each r ∈ E_{s}, a family (Y_{i,α}^{(r)})_{(i,α)∈[[1,m]]×[[1,n]]} of elements of F := Frac(R), defined as follows.

1. \underline{If r = (m, n + 1)}, then Y_{i,α}^{(m,n+1)} = Y_{i,α} for all (i, α) ∈ [[1, m]] × [[1, n]].

2. \underline{Assume that r = (j, β) <_{s} (m, n + 1)} and that the Y_{i,α}^{(r^{+})} ((i, α) ∈ [[1, m]] × [[1, n]]) are already constructed. Then, it follows from [4, Théorème 3.2.1] that Y_{j,β}^{(r^{+})} ≠ 0 and, for all (i, α) ∈ [[1, m]] × [[1, n]], we have:

Y_{i,α}^{(r)} = \{ \begin{matrix}Y_{i,α}^{(r^{+})} − Y_{i,β}^{(r^{+})} (Y_{j,β}^{(r^{+})})^{−1} Y_{j,α}^{(r^{+})} & if i < j and α < β \\ Y_{i,α}^{(r^{+})} & otherwise. \end{matrix}

As in [4], we denote by \overline{R} the subalgebra of Frac(R) generated by the indeterminates obtained at the end of this algorithm, that is, we denote by \overline{R} the subalgebra of Frac(R) generated by the T_{i,α} := Y_{i,α}^{(1,2)} for each (i, α) ∈ [[1, m]] × [[1, n]].

Let N ∈ ℕ^{∗} and let Λ = (Λ_{i,j}) be a multiplicatively antisymmetric N × N matrix over ℂ^{∗}; that is, Λ_{i,i} = 1 and Λ_{j,i} = Λ_{i,j}^{−1} for all i, j ∈ [[1, N]]. We denote by ℂ_{Λ}[T_{1}, …, T_{N}] the corresponding quantum affine space; that is, the ℂ-algebra generated by the N indeterminates T_{1}, …, T_{N} subject to the relations T_{i}T_{j} = Λ_{i,j}T_{j}T_{i} for all i, j ∈ [[1, N]]. In [5, Section 2.2], Cauchon has shown that \overline{R} can be viewed as the quantum affine space generated by the indeterminates T_{i,α} for (i, α) ∈ [[1, n]]^{2}, subject to the following relations.

\begin{matrix}T_{i,β}T_{i,α} = q^{−1}T_{i,α}T_{i,β}, & (α < β); \\ T_{j,α}T_{i,α} = q^{−1}T_{i,α}T_{j,α}, & (i < j); \\ T_{j,β}T_{i,α} = T_{i,α}T_{j,β}, & (i < j, α > β); \\ T_{j,β}T_{i,α} = T_{i,α}T_{j,β}, & (i < j, α < β). \end{matrix}

Hence \overline{R} = ℂ_{Λ}[T_{1,1}, T_{1,2}, …, T_{m,n}], where Λ denotes the mn × mn matrix defined as follows. We set

A := ( \begin{matrix}0 & 1 & 1 & … & 1 \\ −1 & 0 & 1 & … & 1 \\ ⋮ & ... & ... & ... & ⋮ \\ −1 & … & −1 & 0 & 1 \\ −1 & … & … & −1 & 0 \end{matrix} ) ∈ M_{m}(ℂ),

and

B := ( \begin{matrix}A & I_{m} & I_{m} & … & I_{m} \\ −I_{m} & A & I_{m} & … & I_{m} \\ ⋮ & ... & ... & ... & ⋮ \\ −I_{m} & … & −I_{m} & A & I_{m} \\ −I_{m} & … & … & −I_{m} & A \end{matrix} ) ∈ M_{mn}(ℂ),

where I_{m} denotes the identity matrix of M_{m}. Then Λ is the mn×mn matrix whose entries are defined by Λ_{k,l} = q^{b_{k,l}} for all k, l ∈ [[1, mn]].

It follows from [4, Théorèmes 5.1.1, 5.5.1, 5.5.2] that Spec_{⟨0⟩}(R) is Zariski-homeomorphic to the prime spectrum of the quantum torus P (Λ) := \overline{R}Σ^{−1}, where Σ denotes the multiplicative system of \overline{R} generated by the normal elements T_{i,α} with (i, α) ∈ [[1, m]] × [[1, n]]. Next, Spec(P (Λ)) is Zariski-homeomorphic via extension and contraction to the prime spectrum of the centre Z(P (Λ)) of P (Λ), by [8, Corollary 1.5]. Further, Z(P (Λ)) turns out to be a Laurent polynomial ring. To make this result precise, we need to introduce the following notation.

If \underline{s} = (s_{1,1}, s_{1,2}, …, s_{m,n}) ∈ ℤ^{mn}, then we set T \underline{s} := T_{1,1}^{s_{1,1}}T_{1,2}^{s_{1,2}} … T_{m,n}^{s_{m,n}} ∈ P (Λ).

As in [8], we denote by σ : ℤ^{mn} × ℤ^{mn} → ℂ^{∗} the antisymmetric bicharacter defined by

σ(\underline{s}, \underline{t}) := ∏_{k,l=1}^{mn} Λ_{k,l}^{s_{k}t_{l}} for all \underline{s}, \underline{t} ∈ ℤ^{mn}.

Then it follows from [8, 1.3] that the centre Z(P (Λ)) of P (Λ) is a Laurent polynomial ring in the variables (T \underline{b_{1}})^{±1}, …, (T \underline{b_{r}})^{±1}, where (\underline{b_{1}}, …, \underline{b_{r}}) is any basis of S := \{\underline{s} ∈ ℤ^{mn} | σ(\underline{s}, −) ≡ 1\}. Since q is not a root of unity, easy computations show that \underline{s} ∈ S if and only if B^{t}\underline{s}^{t} = 0. Hence the centre Z(P (Λ)) of P (Λ) is a Laurent polynomial ring in ker(B^{t}) indeterminates; so we have just proved the following statement.

Proposition 1.2 Spec_{⟨0⟩}(O_{q}(M_{m,n})) is Zariski-homeomorphic to the prime spectrum of a commutative Laurent polynomial ring in dim(ker(B^{t})) = dim(ker(B)) indeterminates.

We deduce from Propositions 1.1 and 1.2 the following criterion for R to be a primitive ring.

Corollary 1.3 R is a primitive ring if and only if B is invertible.

We now compute the dimension of the kernel of B. First, straightforward computations show that

dim(ker(B)) = m − rk [(A + I)^{n} + (A − I)^{n}] .

If i is a positive integer greater than or equal to 2, we denote by v_{2}(i) the 2-adic valuation of i; that is, m is the largest integer such that 2^{m}|i.

One can easily check the following result.

Proposition 1.4

dim(ker(B)) = \{ \begin{matrix}0 & if v_{2}(m) ≠ v_{2}(n) \\ m ∧ n & otherwise \end{matrix}

where m ∧ n denotes the greatest common divisor of m and n.

Proof. Set e_{k} := exp (i (\frac{(2k+1)π}{m})) for all k ∈ [[0, m − 1]]; that is, the e_{k} are the mth root of −1. It is easy to show that there exists U ∈ GL_{m}(ℂ) such that U^{−1}AU = diag(\frac{e_{0}+1}{e_{0}−1}, …, \frac{e_{m−1}+1}{e_{m−1}−1}). Hence we have

U^{−1} [(A + I)^{n} + (A − I)^{n}] U =

diag ((\frac{2e_{0}}{e_{0} − 1})^{n} + (\frac{2}{e_{0} − 1})^{n} , …, (\frac{2e_{m−1}}{e_{m−1} − 1})^{n} + (\frac{2}{e_{m−1} − 1})^{n}) .

Since U ∈ GL_{m}(ℂ),

rk ((A + I)^{n} + (A − I)^{n}) = rk (U^{−1} [(A + I)^{n} + (A − I)^{n}] U)
= \{k ∈ [[0, m − 1]] | (\frac{2e_{k}}{e_{k}−1})^{n} + (\frac{2}{e_{k}−1})^{n} ≠ 0\} ,

so that

dim(ker(B)) = \{k ∈ [[0, m − 1]] | (\frac{2e_{k}}{e_{k}−1})^{n} + (\frac{2}{e_{k}−1})^{n} = 0\} .

Now, easy calculations show that (\frac{2e_{k}}{e_{k}−1})^{n} + (\frac{2}{e_{k}−1})^{n} = 0 if and only if e_{k}^{n} = −1; that is, if and only if exp (i (\frac{(2k+1)nπ}{m})) = −1, that is, if and only if m|(2k + 1)n and \frac{(2k+1)n}{m} is an odd integer. Hence,

dim(ker(B)) = \{k ∈ [[0, m − 1]] | m|(2k + 1)n and \frac{(2k+1)n}{m}is odd\} .

We now distinguish between two cases.

• First, assume that v_{2}(m) ≠ v_{2}(n). Suppose that there exists k ∈ [[0, m − 1]] such that m|(2k + 1)n. Then v_{2}(m) ≤ v_{2}(n). This implies that v_{2}(m) < v_{2}(n), since v_{2}(m) ≠ v_{2}(n); and so \frac{(2k+1)n}{m} is even. Hence there does not exist k ∈ [[0, m − 1]] such that m|(2k + 1)n and \frac{(2k+1)n}{m} is odd . This proves that, in this case,

dim(ker(B)) = \{k ∈ [[0, m − 1]] | m|(2k + 1)n and \frac{(2k+1)n}{m} is odd \} = 0,

as desired.

• Next, assume that v_{2}(m) = v_{2}(n) = α. Write n = 2^{α}v and m = 2^{α}u with u, v odd. It follows that

dim(ker(B)) = \{k ∈ [[0, m − 1]] | u|(2k + 1)v and \frac{(2k+1)v}{u} is odd \}
= {k ∈ [[0, m − 1]] | u|(2k + 1)v} .

Set d′ = u ∧ v and u = d′u′. Note that d′ and u′ are odd, since u and v are odd. This implies that dim(ker(B)) = {k ∈ [[0, m − 1]] | u′|(2k + 1)} , the number of odd multiples of u′ less that 2m − 1 = 2^{α+1}d′u′ − 1. Hence dim(ker(B)) = 2^{α}d′ = m ∧ n, as required. □

Proposition 1.4 together with Proposition 1.2 and Corollary 1.3 yields the following results.

Theorem 1.5 Spec_{⟨0⟩}(O_{q}(M_{m,n})) is Zariski-homeomorphic to the prime spectrum of a commutative Laurent polynomial ring in α_{m,n} indeterminates, where

α_{m,n} = \{ \begin{matrix}0 & if v_{2}(m) ≠ v_{2}(n) \\ m ∧ n & otherwise. \end{matrix}

Theorem 1.6 O_{q}(M_{m,n}) is a primitive ring if and only if v_{2}(m) ≠ v_{2}(n).

For example, O_{q}(M_{2}) is not primitive, while O_{q}(M_{2,3}) is primitive.

2 Height one primes in quantum matrices.

In this section, we investigate height one primes of R = O_{q}(M_{m,n}). Every height one prime ideal of R is generated by a normal element, since R is a Noetherian UFD, [14]. In this section, we describe explicitly the normal elements that generate the height one prime ideals. Some of them are already known. Indeed, the height one primes that are H-invariant have been described in [14]. Hence, we mainly focus our attention on the other height one primes: these belong to the ⟨0⟩-stratum of Spec(O_{q}(M_{m,n})).

2.1 Height one primes of O_{q}(M_{m,n}) that are H-invariant.

The algebra O_{q}(M_{n}) has a special element, det_{q}, the quantum determinant, defined by

det_{q} := ∑_{σ} (−q)^{l(σ)}Y_{1,σ(1)} ⋯ Y_{n,σ(n)},

where the sum is taken over the permutations of {1, …, n} and l(σ) is the usual length function on such permutations. The quantum determinant is a central element of O_{q}(M_{n}), see, for example, [17, Theorem 4.6.1]. If I is a t-element subset of {1, …, m} and Γ is a telement subset of {1, …, n}, then the quantum determinant of the subalgebra of O_{q}(M_{m,n}) generated by Y_{i,α}, with i ∈ I and α ∈ Γ, is denoted by [I | Γ]. The elements [I | Γ] are the quantum minors of O_{q}(M_{m,n}). Note that the quantum minors are H-eigenvectors; and so every ideal generated by a family of quantum minors is H-invariant.

It follows from [14, Proposition 4.2] that there are exactly m + n − 1 H-invariant prime ideals in R that have height one. To make this result precise, let us introduce some notation.

For 1 ≤ i ≤ n + m − 1, let b_{i} be the quantum minor defined as follows.

b_{i} := \{ \begin{matrix}[1, …, i | n − i + 1, …, n] & if 1 ≤ i ≤ m \\ [1, …, m | n − i + 1, …, n + m − i] & if m < i ≤ n \\ [i − n + 1, …, m | 1, …, m + n − i] & if n < i ≤ m + n − 1 \end{matrix}

Note that the b_{i} with m ≤ i ≤ n are precisely the m × m minors of O_{q}(M_{n}) that have consecutive column indices. They are homogeneous of degree m. This fact will be used several times later.

Proposition 2.1 ([14], Proposition 4.2) There are precisely m+n−1 height one prime ideals that are H-invariant in O_{q}(M_{m,n}). They are the ideals generated by b_{1}, …, b_{m+n−1}.

The quantum minors b_{i} are normal elements of R. Moreover they belong to the algebra \overline{R} obtained from R by the standard deleting derivations algorithm, see Section 1.2. Indeed, every quantum minor b_{i} can be expressed as a product of the canonical generators T_{i,α} of \overline{R} as follows.

Lemma 2.2 For 1 ≤ i ≤ m + n − 1, we have

b_{i} = \{ \begin{matrix}T_{1,n−i+1}T_{2,n−i+2} … T_{i,n} & if 1 ≤ i ≤ m \\ T_{1,n−i+1}T_{2,n−i+2} … T_{m,n+m−i} & if m < i ≤ n \\ T_{i−n+1,1}T_{i−n+2,2} … T_{m,m+n−i} & if n < i ≤ n + m − 1 \end{matrix}

Proof. This lemma is a consequence of [5, Proposition 5.2.2]. □

Recall that two elements a, b of Frac(R) = Frac(\overline{R}) are said to q-commute if there exists an integer • such that ab = q^{•}ba. Since the T_{i,α} q-commute pairwise, it follows from the previous Lemma 2.2 that the b_{j} q-commute with the T_{i,α} and that the b_{j} also q-commute pairwise. Sometimes, it will not be necessary to know exactly the integers that appear in the power of q. However, at some points, we will need the following commutation relations that can be easily deduced from Lemma 2.2 and from the commutation relations between the T_{i,α}.

Corollary 2.3 Assume that m < n. Then,

1. for all m ≤ i < j ≤ n,

b_{i}b_{j} = q^{α_{ij}}b_{j}b_{i},

where α_{ij} := {n − i + 1, …, m + n − i} ∩ {n − j + 1, …, m + n − j} − m,

2. for all i ∈ {1, …, m − 1},

b_{n}b_{i} = q^{α_{i}}b_{i}b_{n} and b_{n}b_{m+n−i} = b_{m+n−i}b_{n},

where α_{i} := {n − i + 1, …, n} ∩ {m + 1, …, n} .

2.2 The ⟨0⟩-stratum of Spec(O_{q}(M_{m,n})).

First, it follows from Theorem 1.5 that the ⟨0⟩-stratum of Spec(O_{q}(M_{m,n})) is reduced to ⟨0⟩ when v_{2}(m) ≠ v_{2}(n). Thus, throughout this section, we assume that v_{2}(m) = v_{2}(n); so that Spec_{⟨0⟩}(O_{q}(M_{m,n})) is Zariski-homeomorphic to the prime spectrum of a commutative Laurent polynomial ring in m ∧ n indeterminates. We set d := m ∧ n and we denote by m′ and n′ the positive integers such that m = dm′ and n = dn′. Observe that m′ and n′ are odd, since v_{2}(m) = v_{2}(n). This observation will be crucial in what follows.

In Section 1.2, we have shown that Spec_{⟨0⟩}(R) is Zariski-homeomorphic to the prime spectrum of the quantum torus P (Λ) = \overline{R}Σ^{−1}, where Σ denotes the multiplicative system of \overline{R} generated by the normal elements T_{i,α} with (i, α) ∈ [[1, m]]×[[1, n]]. Moreover, Spec(P (Λ)) is Zariski-homeomorphic via extension and contraction to the prime spectrum of the centre Z(P (Λ)) of P (Λ), by [8, Corollary 1.5]. Before describing the ⟨0⟩-stratum of R, we calculate the centre Z(P (Λ)) of the quantum torus P (Λ)

2.2.1 The centre of the quantum torus P (Λ) = \overline{R}Σ^{−1}

Recall that the quantum minors b_{i} belong to \overline{R}. Moreover, the b_{i} are invertible in the quantum torus P (Λ) = \overline{R}Σ^{−1},

For j ∈ {1, …, d}, set

Δ_{j} := ∏_{i=0}^{m′+n′−1} b_{id+j}^{(−1)^{i}}.

(Here we set b_{m+n} := 1.)

Theorem 2.4 Z(P (Λ)) = ℂ[Δ_{1}^{±1}, …, Δ_{d}^{±1}]

Proof. First, straightforward computations (see also [16, Theorem 2.13]) show that Δ_{1}, …, Δ_{d} are central in P (Λ), so that Z(P (Λ)) ⊇ ℂ[Δ_{1}^{±1}, …, Δ_{d}^{±1}].

If \underline{s} = (s_{1,1}, s_{1,2}, …, s_{m,n}) ∈ ℤ^{mn}, we set T \underline{s} := T_{1,1}^{s_{1,1}}T_{1,2}^{s_{1,2}} … T_{m,n}^{s_{m,n}} ∈ P (Λ). Since the T_{i,α} q-commute, we deduce from Lemma 2.2 that the central elements Δ_{i} can be expressed as follows:

Δ_{i} = q^{γ_{i}}T \underline{u^{(i)}},

where γ_{i} ∈ ℤ and \underline{u^{(i)}} = (u_{1,1}^{(i)}, u_{1,2}^{(i)}, …, u_{m,n}^{(i)}) ∈ ℤ^{mn} with (u_{1,m−d+1}^{(i)}, u_{1,m−d+2}^{(i)}, …, u_{1,m}^{(i)}) = (0, …, 0, 1, 0, …, 0) (the one being in the i-th position).

As in [8], we denote by σ : ℤ^{mn} × ℤ^{mn} → ℂ^{∗} the antisymmetric bicharacter defined by

σ(\underline{s}, \underline{t}) = ∏_{k,l=1}^{mn} Λ_{k,l}^{s_{k}t_{l}}

for all \underline{s}, \underline{t} ∈ ℤ^{mn}.

Then it follows from [8, 1.3] that the centre Z(P (Λ)) of P (Λ) is a Laurent polynomial ring in the variables (T \underline{b_{1}})^{±1}, …, (T \underline{b_{r}})^{±1}, where (\underline{b_{1}}, …, \underline{b_{r}}) is any basis of S := \{\underline{s} ∈ ℤ^{mn} | σ(\underline{s}, −) ≡ 1\}.

Now, because of Theorem 1.5, we know r = rk(S) = d. Moreover, the Δ_{i} are central, so that the \underline{u^{(i)}} belong to S. To conclude, observe that, since we have

(u_{1,m−d+1}^{(i)}, u_{1,m−d+2}^{(i)}, …, u_{1,m}^{(i)}) = (0, …, 0, 1, 0, …, 0),

the group ℤ^{mn}/ ∑_{i=1}^{d} ℤ\underline{u^{(i)}} is torsionfree. Hence, the \underline{u^{(i)}} form a basis of S, and so the centre Z(P (Λ)) of P (Λ) is a Laurent polynomial ring in the variables Δ_{1}^{±1}, …, Δ_{d}^{±1}, as desired. □

2.2.2 Height one primes in O_{q}(M_{m,n}).

Let P a height one prime of R = O_{q}(M_{m,n}). Because of the H-stratification, see (1) in Section 1.1, there exists an H-invariant prime ideal J of R such that P belongs to the H-stratum associated to J. In particular, we have J ⊆ P. Since P has height one, this implies that the height of J is at most 1. Thus, two cases arise.

1. First, suppose that J has height one. In this case, P = J, since P has height one; so that P is an H-invariant height one prime ideal of R. Hence it follows from Proposition 2.1 that there exists i ∈ {1, …, m + n − 1} such that P = ⟨b_{i}⟩.

2. Next, suppose that J = ⟨0⟩. In this case, P is a height one prime ideal that belongs to the ⟨0⟩-stratum of Spec(R). Note that this case can only arise when v_{2}(m) = v_{2}(n), since it follows from Theorem 1.5 that the ⟨0⟩-stratum of Spec(O_{q}(M_{m,n})) is reduced to ⟨0⟩ when v_{2}(m) ≠ v_{2}(n).

Let Spec^{1}(R) denote the set of all height one primes of R. The previous discussion proves the following statement.

Proposition 2.5

1. If v_{2}(m) ≠ v_{2}(n), then

Spec^{1}(R) = {⟨b_{i}⟩ | i ∈ {1, …, m + n − 1}} .

2. If v_{2}(m) = v_{2}(n), then

Spec^{1}(R) = {⟨b_{i}⟩ | i ∈ {1, …, m + n − 1}} ∪ Spec_{⟨0⟩}^{1}(R),

where Spec_{⟨0⟩}^{1}(R) denotes the set of all height one primes of R that belong to the ⟨0⟩-stratum of Spec(R).

In order to complete the previous result, we now describe, in the case where v_{2}(m) = v_{2}(n), the height one primes of O_{q}(M_{m,n}) that belongs to Spec_{⟨0⟩}(O_{q}(M_{m,n})).

Proposition 2.6 Assume that v_{2}(m) = v_{2}(n). Then, for any height one prime ideal P of O_{q}(M_{m,n}) that belongs to Spec_{⟨0⟩}(O_{q}(M_{m,n})), there exists a unique (up to scalar) irreducible polynomial V = ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}}X_{1}^{i_{1}} … X_{d}^{i_{d}} ∈ ℂ[X_{1}, …, X_{d}] (where r_{i} = deg_{X_{i}} V) with V ≠ X_{i} for all i ∈ {1, …, d} such that P = ⟨u⟩ where

u := ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}} ∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}} .

Moreover, u is normal in R.

Proof. We proceed in three steps.

• Step 1. A generator for the extension of P in a localisation of R.

First, observe that the prime ideals in Spec_{⟨0⟩}(R) do not contain any b_{i}. Indeed, assume that this is not the case; that is, assume that there exists P ∈ Spec_{⟨0⟩}(R) with b_{i} ∈ P for a certain i. Then, since b_{i} is an H-eigenvector, we have b_{i} ∈ ⋂_{h∈H} h.P = ⟨0⟩. This is a contradiction; and so Spec_{⟨0⟩}(R) ⊆ {P ∈ Spec(R) | b_{i} ∉ P for all i}. On the other hand, if P is a prime ideal of R such that b_{i} ∉ P for all i, then ⋂_{h∈H} h.P is an H-invariant prime ideal of R that does not contain any b_{i}. However, because of [14, Proposition 2.9], every nonzero H-invariant prime ideal of R contains a height one prime that is Hinvariant. In other words, every nonzero H-invariant prime ideal of R contains a b_{i}. Thus ⋂_{h∈H} h.P = ⟨0⟩ and so P ∈ Spec_{⟨0⟩}(R). To sum up, we have shown that Spec_{⟨0⟩}(R) = {P ∈ Spec(R) | b_{i} ∉ P for all i}.

Denote by T the localisation of R with respect of the multiplicatively closed set B generated by the normal elements b_{i}. Note that the torus H still acts rationally by automorphisms on T, since the b_{i} are H-eigenvectors, see [3, Exercise II.3.A]. Moreover, it follows from the previous study, and from classical results of non-commutative localisation theory, that the map φ : P → P B^{−1} is an increasing bijection from Spec_{⟨0⟩}(R) onto Spec (T).

Next, T is H-simple; that is, the only H-invariant ideals in T are ⟨0⟩ and T, by [14, Proposition 3.5]. The action of H on T is rational, see [3, Exercise II.3.A], and T is H-simple; so [3, Corollary II.3.9] applies to T. Thus extension and contraction provide mutually inverse bijections between Spec(T) and Spec(Z(T)).

Now, recall that the b_{i} can be expressed as products of T_{i,α}, see Lemma 2.2; so the b_{i} belong to the quantum torus P (Λ) = \overline{R}Σ^{−1}. Moreover, it follows from [4, Théorème 3.3.1] that there exists a multiplicative system S of R such that R ⊆ RS^{−1} = P (Λ) = \overline{R}Σ^{−1}. Hence T is a subalgebra of the quantum torus P (Λ) = RS^{−1}, and R ⊆ T ⊆ P (Λ) = \overline{R}Σ^{−1} = RS^{−1}. Thus Z(T) ⊆ Z(P (Λ)) = ℂ[Δ_{1}^{±1}, …, Δ_{d}^{±1}]. But the Δ_{i} are just products of b_{i}^{±1}, so that they belong to T. Hence Z(T) = ℂ[Δ_{1}^{±1}, …, Δ_{d}^{±1}].

Observe that Δ_{j} can be written as follows.

Δ_{j} = q^{•} ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j} ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}^{−1},

where as usual • denotes an integer, since the b_{i} q-commute. For j ∈ {1, …, d}, we set

Δ′_{j} = ( ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}) ( ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j})^{−1} .  (2)

Since Z(T) = ℂ[Δ_{1}^{±1}, …, Δ_{d}^{±1}], the centre of T is also the (commutative) Laurent polynomial ring in the indeterminates Δ′_{1}, …, Δ′_{d}, that is,

Z(T) = ℂ[Δ_{1}^{′±1}, …, Δ_{d}^{′±1}].

Now, let P be a height one prime of R that belongs to Spec_{⟨0⟩}(R). It follows from the previous study that φ(P) ∩ Z(T) is a height one prime of Z(T) = ℂ[Δ_{1}^{′±1}, …, Δ_{d}^{′±1}]. Hence there exists an irreducible polynomial V = ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}}X_{1}^{i_{1}} … X_{d}^{i_{d}} ∈ ℂ[X_{1}, …, X_{d}], with r_{i} = deg_{X_{i}} V, and with V ≠ X_{i} for all i ∈ {1, …, d}, such that

φ(P) ∩ Z(T) = ⟨ ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}}Δ_{1}^{′i_{1}} … Δ_{d}^{′i_{d}}⟩_{Z(T)} .

Thus,

φ(P) = ⟨ ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}}Δ_{1}^{′i_{1}} … Δ_{d}^{′i_{d}} ⟩_{T} ,

since extension and contraction provide mutually inverse bijections between Spec(T) and Spec(Z(T)). Since the b_{i} are invertible in T, (2) leads to φ(P) = ⟨u⟩, where

u := ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}} ∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}} .

Note that u is a normal element in both R and T.

• Step 2. We prove that u is not contained in any ⟨b_{k}⟩.

Set

u_{i_{1},...,i_{d}} := ∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}}

for all (i_{1}, …, i_{d}) ∈ [[0, r_{1}]] × ⋯ × [[0, r_{d}]]; so that

u = ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}}u_{i_{1},...,i_{d}}.

Since the b_{k} are H-eigenvectors, it is clear that the u_{i_{1},...,i_{d}} are also H-eigenvectors. Hence, for all (i_{1}, …, i_{d}) ∈ [[0, r_{1}]] × ⋯ × [[0, r_{d}]], there exists a (unique) character f_{i_{1},...,i_{d}} : H → ℂ^{∗} such that h.u_{i_{1},...,i_{d}} = f_{i_{1},...,i_{d}}(h)u_{i_{1},...,i_{d}} for all h ∈ H. The character f_{i_{1},...,i_{d}} is called the H-eigenvalue of the H-eigenvector u_{i_{1},...,i_{d}}.

Claim 2.7 The u_{i_{1},...,i_{d}} are H-eigenvectors with pairwise distinct H-eigenvalues.

Proof of Claim 2.7. Let (i_{1}, …, i_{d}) ∈ [[0, r_{1}]]×⋯×[[0, r_{d}]] and h := (g_{1}, …, g_{m}; h_{1}, …, h_{n}) ∈ H. Then,

h.u_{i_{1},...,i_{d}} = g_{1}^{•} … g_{m}^{•}h_{1}^{•} … h_{n−m}^{•}h_{n−m+1}^{α_{m−1}} … h_{n}^{α_{0}}u_{i_{1},...,i_{d}},

where • ∈ ℤ and α_{k} is the number of quantum minors involving the (n − k)-th column in the product that defines u_{i_{1},...,i_{d}}.

Now one can observe that, for all k ∈ {0, …, d − 1}, the (n − k)-th column appears in b_{id+j} if and only if either (i) k + 1 ≤ j ≤ d and 0 ≤ i ≤ m′ − 1, or (ii) 1 ≤ j ≤ k and 1 ≤ i ≤ m′. Hence, since m′ is odd,

α_{k} = ∑_{j=1}^{k} (\frac{m′ − 1}{2}i_{j} + \frac{m′ + 1}{2}(r_{j} − i_{j})) + ∑_{j=k+1}^{d} (\frac{m′ + 1}{2}i_{j} + \frac{m′ − 1}{2}(r_{j} − i_{j}))
= ∑_{j=1}^{k} (\frac{m′ + 1}{2}r_{j} − i_{j}) + ∑_{j=k+1}^{d} (\frac{m′ − 1}{2}r_{j} + i_{j})

Now, let (i_{1}, …, i_{d}), (i′_{1}, …, i′_{d}) ∈ [[0, r_{1}]] × ⋯ × [[0, r_{d}]], and assume that u_{i_{1},...,i_{d}} and u_{i′_{1},...,i′_{d}} are associated to the same H-eigenvalue. Then, since ℂ is infinite, it follows from the previous study that,

∑_{j=1}^{k} (\frac{m′ + 1}{2}r_{j} − i_{j})+ ∑_{j=k+1}^{d} (\frac{m′ − 1}{2}r_{j} + i_{j}) = ∑_{j=1}^{k} (\frac{m′ + 1}{2}r_{j} − i′_{j})+ ∑_{j=k+1}^{d} (\frac{m′ − 1}{2}r_{j} + i′_{j}) ,

for all k ∈ {0, …, d − 1}. Hence,

∑_{j=1}^{k} (i′_{j} − i_{j}) + ∑_{j=k+1}^{d} (i_{j} − i′_{j}) = 0,

for all k ∈ {0, …, d−1}. This forces i_{j} = i′_{j} for all j ∈ {1, …, d}; so the claim is proved. □

Next, we prove that u does not belong to any ⟨b_{k}⟩. Indeed, assume that u ∈ ⟨b_{k}⟩ for a certain k ∈ {1, …, m + n − 1}. Then, since ⟨b_{k}⟩ is an H-invariant prime ideal of R, it follows from the previous claim and from [3, II.2.10] that:

If a_{i_{1},...,i_{d}} ≠ 0, then u_{i_{1},...,i_{d}} = ∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}} ∈ ⟨b_{k}⟩ .  (3)

Set k := rd + s with r ∈ {0, …, m′ + n′ − 1} and s ∈ {1, …, d}. We distinguish between two cases. First, assume that r is even. Since the ⟨b_{i}⟩ are pairwise distinct height one completely prime ideals of R, we deduce from (3) that a_{i_{1},...,i_{d}} = 0 if i_{s} = 0. This implies that V = ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}}X_{1}^{i_{1}} … X_{d}^{i_{d}} = X_{s}V′, with V′ ∈ ℂ[X_{1}, …, X_{d}]. This contradicts the facts that V is irreducible and V ≠ X_{s}.

Next, assume that r is odd. Since the ⟨b_{i}⟩ are pairwise distinct height one completely prime ideals of R, we deduce from (3) that a_{i_{1},...,i_{d}} = 0 if i_{s} = r_{s}. Now this contradicts deg_{X_{s}} V = r_{s}.

To sum up: u does not belong to any ⟨b_{k}⟩.

• Step 3. We prove that P is generated by u.

Recall from step one that φ(P) is generated by u. Hence, it is clear that P = φ(P)∩R ⊇ ⟨u⟩. Let now x ∈ P. It remains to prove that x ∈ ⟨u⟩. There exists (α_{1}, …, α_{m+n−1}) ∈ ℕ^{m+n−1} such that xb_{1}^{α_{1}} … b_{m+n−1}^{α_{m+n−1}} = ur with r ∈ R. Choose such a (α_{1}, …, α_{m+n−1}) ∈ ℕ^{m+n−1} minimal (for the lexicographic order). If (α_{1}, …, α_{m+n−1}) ≠ 0, there exists k such that α_{k} ≠ 0. Then ur = xb_{1}^{α_{1}} … b_{m+n−1}^{α_{m+n−1}} belongs to the completely prime ideal of R generated by b_{k}. Hence, u ∈ ⟨b_{k}⟩ or r ∈ ⟨b_{k}⟩. Because of step two, the first possibility can not happen. Hence, r ∈ ⟨b_{k}⟩. Since b_{k} is normal, we can write r = r′b_{k} with r′ ∈ R. Thus xb_{1}^{α_{1}} … b_{m+n−1}^{α_{m+n−1}} = ur = ur′b_{k}; and so xb_{1}^{α_{1}} … b_{k}^{α_{k}−1} … b_{m+n−1}^{α_{m+n−1}} = ur′. This contradicts the minimality of (α_{1}, …, α_{m+n−1}). Hence (α_{1}, …, α_{m+n−1}) = 0; and so x = ur ∈ ⟨u⟩, as desired. □

3 Automorphisms of quantum matrices

In this section, we investigate the group of automorphisms of O_{q}(M_{m,n}). Using graded arguments together with the results of the previous sections, we show that, in the non-square case, every H-invariant height one prime of O_{q}(M_{m,n}), except possibly one, is invariant under every automorphism. Next, by using the preferred basis of O_{q}(M_{m,n}) introduced in [7], we show that the group of automorphisms of O_{q}(M_{m,n}), with 2 ≤ m < n, is isomorphic to the torus (ℂ^{∗})^{m+n−1}.

In the sequel, we will use several times the following well-known result concerning normal elements of R.

Lemma 3.1 Let u and v two nonzero normal elements of R such that ⟨u⟩ = ⟨v⟩. Then there exist λ, μ ∈ ℂ^{∗} such that u = λv and v = μu.

3.1 q-commutation, gradings and automorphisms.

Let A = ⊕_{i∈ℕ}A_{i} be a ℕ-graded ℂ-algebra with A_{0} = ℂ. Assume that A is a domain generated as an algebra by x_{1},..., x_{n}, and that A_{1} = ℂx_{1}⊕⋯⊕ℂx_{n}. We set A_{≥d} := ⊕_{i≥d}A_{i}. The following result was inspired by a result in [1].

Proposition 3.2 Assume that, for all i ∈ {1, …, n}, there exist j ≠ i and q_{ij} ≠ 1 such that x_{i}x_{j} = q_{ij}x_{j}x_{i}. Let σ be an automorphism of A and x be a nonzero homogeneous element of degree d of A. Then σ(x) = y_{d} + y_{>d}, where y_{d} ∈ A_{d} \ {0} and y_{>d} ∈ A_{>d}.

Proof. First, observe that it is sufficient to prove that σ(A_{d}) ⊆ A_{≥d}, for every automorphism σ of A. Indeed, assume that this is the case, and let x be a nonzero homogeneous element of degree d of A. Then we can write σ(x) = y_{d} + y_{>d}, where y_{d} ∈ A_{d} and y_{>d} ∈ A_{>d}. If y_{d} = 0, then σ(x) = y_{>d} ∈ A_{>d}, and thus σ^{−1}(A_{>d}) is not contained in A_{>d}. This is a contradiction.

Hence it just remains to prove that, for every automorphism σ of A, we have σ(A_{d}) ⊆ A_{≥d}. Naturally, it is sufficient to prove this result when x = x_{i} is one of the canonical generators of A. So let σ be an automorphism σ of A and i ∈ {1, …, n}. We can write

σ(x_{i}) = α_{i} + f_{i},

where α_{i} ∈ ℂ and f_{i} ∈ A_{≥1}. We have to prove that α_{i} = 0.

Now, by hypothesis, there exist j ∈ {1, …, n} and q_{ij} ≠ 1 such that x_{i}x_{j} = q_{ij}x_{j}x_{i}. Set

σ(x_{j}) = f_{j} + g_{j},

where f_{j} ∈ A_{t}, f_{j} ≠ 0 and g_{j} ∈ A_{>t}. Applying σ to the equality x_{i}x_{j} = q_{ij}x_{j}x_{i}, and next identifying the homogeneous part of degree t yields α_{i}f_{j} = q_{ij}f_{j}α_{i}. Thus, α_{i}f_{j} = 0, since q_{ij} ≠ 1. Now, since f_{j} ≠ 0, this forces α_{i} = 0, as desired. □

Note that the commutativity hypothesis of the Proposition 3.2 is satisfied by the algebra R = O_{q}(M_{m,n}), provided that n ≥ 2. Indeed, the relations that define R are all quadratic, so that R = ⊕_{i∈ℕ}R_{i} is a ℕ-graded algebra, the canonical generators Y_{i,α} of R having degree one.

Next, for all (i, α) ∈ [[1, m]] × [[1, n]] and β ≠ α, we have Y_{i,α}Y_{i,β} = q^{•}Y_{i,β}Y_{i,α} with • = ±1. Thus, the commutativity hypothesis of Proposition 3.2 is satisfied; and so one can state:

Corollary 3.3 Let σ be an automorphism of R = O_{q}(M_{m,n}) and x an homogeneous element of degree d of R. Then σ(x) = y_{d} + y_{>d}, where y_{d} ∈ R_{d} \ {0} and y_{>d} ∈ R_{>d}.

Note, for later use, that a t×t quantum minor of R is a homogeneous element of degree t with respect to this grading of R. In the sequel, R will always be endowed with this grading.

3.2 The action of Aut(O_{q}(M_{m,n})) on the set of height one primes: the non-square case.

Throughout this section, we assume that m < n. Our aim in this section is to show that every H-invariant height one prime of O_{q}(M_{m,n}), except possibly one, is invariant under every automorphism. In order to do this, we will distinguish between two cases.

3.2.1 The case where n ≠ 3m.

Throughout this section, we assume that m < n and n ≠ 3m. In this case, we first show that the set of all H-invariant height one primes of R is invariant under every automorphism of R, that is, we have:

Lemma 3.4 Assume that n ≠ 3m. Let σ be an automorphism of R and i ∈ {1, …, m + n − 1}. Then there exists j ∈ {1, …, m + n − 1} such that σ(⟨b_{i}⟩) = ⟨b_{j}⟩.

Proof. First, assume that v_{2}(m) ≠ v_{2}(n). It follows from Proposition 2.5 that {⟨b_{i}⟩ | i ∈ [[1, m + n − 1]]} is exactly the set of all height one primes of R and so the result is obvious in this case.

Next, assume that v_{2}(m) = v_{2}(n). We use the notation of Section 2.2. In particular, d denotes the greatest common divisor of m and n, and we set m = dm′ and n = dn′. Observe that m′ and n′ are odd. Note further that, since n ≠ 3m, we have (m′, n′) ≠ (1, 3). Moreover, since m < n, we have m′ < n′. Since m′ and n′ are both odd, this forces n′ ≥ 5.

Since ⟨b_{i}⟩ is a height one prime ideal of R, its image under σ is also a height one prime ideal of R. We distinguish between two cases.

First, if ⟨σ(b_{i})⟩ is H-invariant, then it follows from Proposition 2.1 that ⟨σ(b_{i})⟩ = ⟨b_{j}⟩ for some j and so the proof is complete in this case.

Next, assume that ⟨σ(b_{i})⟩ is not H-invariant. In this case, ⟨σ(b_{i})⟩ is a height one (completely) prime ideal of R which is not H-invariant. Thus, ⟨σ(b_{i})⟩ belongs to the ⟨0⟩- stratum of the prime spectrum of R, by Proposition 2.5; and so we deduce from Proposition 2.6 that there exist (r_{1}, …, r_{d}) ∈ ℕ^{d} \ {0} and scalars a_{i_{1},...,i_{d}} ∈ ℂ such that ⟨σ(b_{i})⟩ = ⟨u⟩, where

u := ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{d}=0}^{r_{d}} a_{i_{1},...,i_{d}} ∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}} .

It follows from Proposition 2.6 that u is a normal element of R. On the other hand, σ(b_{i}) is normal, since b_{i} is normal. Thus, we deduce from Lemma 3.1 that u ∈ ℂ^{∗}σ(b_{i}).

Now b_{i} is a quantum minor of R = O_{q}(M_{m,n}), and so b_{i} is a homogeneous element of degree less than or equal to m. So Corollary 3.3 implies that

u ∉ R_{>m}.  (4)

On the other hand, since (r_{1}, …, r_{d}) ≠ 0, there exists k such that r_{k} ≥ 1. Now, recalling that m′ and n′ are odd and that n′ ≥ 5, we have:

deg (∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}} ) ≥ \{ \begin{matrix}deg (b_{k}b_{(m′+1)d+k}) & if i_{k} ≥ 1 \\ deg (b_{m′d+k}b_{(m′+n′−3)d+k}) & if i_{k} = 0 \end{matrix}

Since m < n, we have m′ < n′. Further, m′ and n′ are odd, and so m′ + 1 < n′. Hence m = m′d ≤ m′d + k ≤ (m′ + 1)d + k ≤ n = n′d. This implies that b_{(m′+1)d+k} and b_{m′d+k} are m × m quantum minors. Thus, in both cases, we get

deg (∏_{j=1}^{d} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{m′+n′−1} b_{id+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{m′+n′−1} b_{id+j}]^{r_{j}−i_{j}} ) ≥ m + 1.

Thus u is a linear combination of terms of degree greater than m. This implies that u ∈ R_{>m}, contradicting (4); and so the proof is complete. □

In fact, more is true: each H-invariant height one prime is left invariant by any automorphism, as the following result shows.

Proposition 3.5 Assume that n ≠ 3m and let σ be an automorphism of R. Then, for each i ∈ {1, …, m + n − 1}, there exists λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{i}.

Proof. Let i ∈ {1, …, m + n − 1}. It follows from Lemma 3.4 that there exists j ∈ {1, …, m + n − 1} such that σ(⟨b_{i}⟩) = (⟨b_{j}⟩). Since σ(b_{i}) and b_{j} are normal, it follows from Lemma 3.1 that there exists λ_{j} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}.

Thus, there exist scalars λ_{1}, …, λ_{m+n−1} ∈ ℂ^{∗} and a permutation s ∈ S_{m+n−1} such that σ(b_{i}) = λ_{i}b_{s(i)} for all i ∈ {1, …, m + n − 1}.

We will now prove that s is the identity, that is, s(i) = i for all i ∈ {1, …, m + n − 1}. First, let i ∈ {m, …, n}, so that b_{i} is a m × m quantum minor. Then b_{i} ∈ R_{m} and it

follows from Corollary 3.3 that σ(b_{i}) ∈ R_{≥m}. Hence b_{s(i)} ∈ R_{≥m}. This implies that b_{s(i)} is also a m × m quantum minor, so that s(i) ∈ {m, …, n}. Thus s induces a permutation of {m, …, n}.

We now prove with the help of a decreasing induction that s(i) = i for all i ∈ {m, …, n}. It follows from Corollary 2.3 that b_{n}b_{j} = q^{•}b_{j}b_{n} with • ≥ 0 for all j ∈ {m, …, n}.

Then, applying σ leads to b_{s(n)}b_{s(j)} = q^{•}b_{s(j)}b_{s(n)} with • ≥ 0 for all j ∈ {m, …, n}. Since s is a permutation of {m, …, n}, this implies that b_{s(n)}b_{j} = q^{•}b_{j}b_{s(n)} with • ≥ 0 for all j ∈ {m, …, n}. Now, if s(n) ≠ n, then we have s(n) + 1 ∈ {m, …, n} and b_{s(n)}b_{s(n)+1} = q^{−1}b_{s(n)+1}b_{s(n)}. This is a contradiction and so s(n) = n.

We now assume that m ≤ i < n. It follows from the induction hypothesis that s induces a permutation of {m, …, i}; so that s(i) ≤ i. By using a similar argument to that in the previous paragraph, we obtain s(i) = i.

Hence, s(i) = i for all i ∈ {m, …, n}.

Now let i ∈ {1, …, m − 1}. Then b_{i} ∈ R_{i}; and so it follows from Proposition 3.2 that σ(b_{i}) ∈ R_{≥i} \ R_{>i}. Hence b_{s(i)} ∈ R_{≥i} \ R_{>i} so that b_{s(i)} is also a i × i quantum minor. This implies that either s(i) = i or s(i) = m + n − i. Note that similar arguments show that either s(m + n − i) = i or s(m + n − i) = m + n − i; so s induces a permutation of {i, m + n − i}. Observe that it follows from Corollary 2.3 that b_{n}b_{m+n−i} = b_{m+n−i}b_{n} and b_{n}b_{i} = q^{•}b_{i}b_{n} with • > 0. Since we have already proved that s(n) = n, applying σ leads to b_{n}b_{s(m+n−i)} = b_{s(m+n−i)}b_{n} and b_{n}b_{s(i)} = q^{•}b_{s(i)}b_{n} with • > 0. This forces s(i) = i and s(m + n − i) = m + n − i. □

3.2.2 The case where n = 3m.

Throughout this section, we assume that n = 3m. In this case, we are not able to prove directly that the set of all H-invariant height one primes is invariant under every automorphism of R = O_{q}(M_{m,n}). However, by using arguments similar to those developed in the proof of Lemma 3.4, one can establish the following weaker result.

Lemma 3.6 Assume that n = 3m. Let σ be an automorphism of R and i ∈ {1, …, m + n − 1}. Then, either

1. there exists j ∈ {1, …, m + n − 1} such that σ(⟨b_{i}⟩) = ⟨b_{j}⟩, or,

2. there exist λ ∈ ℂ^{∗} and μ ∈ ℂ such that σ(b_{i}) = λb_{2m} + μb_{m}b_{3m}.

Proof. Note that v_{2}(m) = v_{2}(n), since n = 3m. Also, observe that the greatest common divisor d of m and n is equal to m and that, if we set m = dm′ and n = dn′, then m′ = 1 and n′ = 3.

Since ⟨b_{i}⟩ is a height one prime ideal of R, its image under σ is also a height one prime ideal of R. We distinguish between two cases.

If ⟨σ(b_{i})⟩ is H-invariant, then it follows from Proposition 2.1 that σ(⟨b_{i}⟩) = ⟨b_{j}⟩ for some j and so the proof is complete in this case.

Assume now that ⟨σ(b_{i})⟩ is not H-invariant. In this case, ⟨σ(b_{i})⟩ is a height one (completely) prime ideal of R which is not H-invariant. Thus it follows from Proposition 2.5 that ⟨σ(b_{i})⟩ belongs to the ⟨0⟩-stratum of the prime spectrum of R; and so, recalling that d = m, we deduce from Proposition 2.6 that there exist (r_{1}, …, r_{m}) ∈ ℕ^{m} \{0} and scalars a_{i_{1},...,i_{m}} ∈ ℂ such that ⟨σ(b_{i})⟩ = ⟨u⟩ where

u := ∑_{i_{1}=0}^{r_{1}} ⋯ ∑_{i_{m}=0}^{r_{m}} a_{i_{1},...,i_{m}} ∏_{j=1}^{m} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{3} b_{im+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{3} b_{im+j}]^{r_{j}−i_{j}} .

It follows that u is a normal element of R, by Proposition 2.6. On the other hand, σ(b_{i}) is normal in R, since b_{i} is normal in R. . Hence, u ∈ ℂ^{∗}σ(b_{i}), by Lemma 3.1.

Now b_{i} is a quantum minor of R = O_{q}(M_{m,n}); and so b_{i} is a homogeneous element of degree less than or equal to m. Thus, Corollary 3.3 implies that

u ∉ R_{>m}.  (5)

On the other hand, since (r_{1}, …, r_{m}) ≠ 0, there exists k such that r_{k} ≥ 1. We consider three separate cases.

• First, suppose that k < m, then

deg (∏_{j=1}^{m} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{3} b_{im+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{3} b_{im+j}]^{r_{j}−i_{j}} ) ≥ \{ \begin{matrix}deg (b_{k}b_{2m+k}) & if i_{k} ≥ 1 \\ deg (b_{m+k}b_{3m+k}) & if i_{k} = 0 \end{matrix}

Now, m + k and 2m + k are both between m and n, since k < m and n = 3m. Thus, b_{m+k} and b_{2m+k} are m×m quantum minors. Also, b_{k} and b_{3m+k} are homogeneous of degree greater than or equal to 1. Thus, in both cases, we get

deg (∏_{j=1}^{m} [ ∏_{\begin{matrix}i=0 \\ i even \end{matrix}}^{3} b_{im+j}]^{i_{j}} [ ∏_{\begin{matrix}i=0 \\ i odd \end{matrix}}^{3} b_{im+j}]^{r_{j}−i_{j}} ) ≥ m + 1.

Hence u is a linear combination of terms of degree greater than m. This implies that u ∈ R_{>m}. This contradicts (5).

• Next, assume that k = m and r_{m} ≥ 2. In this case, one can prove, by using similar arguments, that u is a linear combination of terms of degree greater than m; so this case cannot happen.

• Finally, assume that k = m and (r_{1}, …, r_{m}) = (0, …, 0, 1). In this case, there exist λ′, μ′ ∈ ℂ such that

u = λ′b_{2m} + μ′b_{m}b_{3m}.

If λ′ = 0, then once again u is a linear combination of terms of degree greater than m, contradicting (5). Hence λ′ ≠ 0. Since we have already proved that u ∈ ℂ^{∗}σ(b_{i}), we see that there exist λ ∈ ℂ^{∗} and μ ∈ ℂ such that σ(b_{i}) = λb_{2m} + μb_{m}b_{3m}, as desired. □

The following commutation relations can be easily deduced from Corollary 2.3.

Lemma 3.7 For all m ≤ i ≤ 3m:

b_{i}(λb_{2m} + μb_{m}b_{3m}) = \{ \begin{matrix}q^{α_{i,2m}}(λb_{2m} + μb_{m}b_{3m})b_{i} & if i < 2m \\ q^{−α_{i,2m}}(λb_{2m} + μb_{m}b_{3m})b_{i} & if i > 2m, \end{matrix}

where α_{i,2m} := {3m − i + 1, …, 4m − i} ∩ {m + 1, …, 2m} − m.

We can now obtain the analogous (but slightly weaker) result to Proposition 3.5 in the n = 3m case.

Proposition 3.8 Assume that n = 3m and let σ be an automorphism of R. Then, for all i ∈ {1, …, 4m − 1}, i ≠ 2m, there exists λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{i}.

Proof. Let i ∈ {1, …, 4m − 1} and assume that there exists j ∈ {1, …, 4m − 1} such that σ(⟨b_{i}⟩) = ⟨b_{j}⟩. Then, it follows from Lemma 3.1 that there exists λ_{j} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}, since σ(b_{i}) and b_{j} are normal.

Thus, we deduce from Lemma 3.6 that, for all i ∈ {1, …, 4m − 1}, either there exist j ∈ {1, …, 4m − 1} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}, or there exist λ_{i}, μ_{i} ∈ ℂ with λ_{i} ≠ 0 such that σ(b_{i}) = λ_{i}b_{2m} + μ_{i}b_{m}b_{3m}.

We distinguish between two cases.

• If, for each i ∈ {1, …, 4m − 1}, there exist j ∈ {1, …, 4m − 1} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}, then, by using similar arguments to those in the proof of Proposition 3.5, we show that σ(b_{i}) = λ_{i}b_{i} for all i ∈ {1, …, 4m − 1}.

• Now, assume that there exist k ∈ {1, …, 4m − 1}, and λ_{k}, μ_{k} ∈ ℂ with λ_{k} ≠ 0, such that σ(b_{k}) = λ_{k}b_{2m} + μ_{k}b_{m}b_{3m}.

•• First, we show that k = 2m.

Observe that, for all i ∈ {1, …, m − 1} ∪ {3m + 1, …, 4m − 1}, the quantum minor b_{i} is a homogeneous element of degree less than m. Hence, σ(b_{i}) ∉ R_{≥m}, by Corollary 3.3. Since σ(b_{k}) = λ_{k}b_{2m} + μ_{k}b_{m}b_{3m} ∈ R_{≥m}, this shows that i ≠ k. Thus, k ∈ {m, …, 3m}.

Next, let i ∈ {m, …, 3m}, so that b_{i} is a m × m quantum minor. Then b_{i} ∈ R_{m} and it follows from Corollary 3.3 that σ(b_{i}) ∈ R_{≥m}. Hence, either

there exist j ∈ {m, …, 3m} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j},  (6)

or

there exist λ_{i}, μ_{i} ∈ ℂ with λ_{i} ≠ 0 such that σ(b_{i}) = λ_{i}b_{2m} + μ_{i}b_{m}b_{3m}.  (7)

We now prove that (7) can not happen, for all i ∈ {2m + 1, …, 3m}. Indeed, assume that there exists i ∈ {2m + 1, …, 3m} such that σ(b_{i}) = λ_{i}b_{2m} + μ_{i}b_{m}b_{3m}. Note that b_{i}b_{j} = q^{•}b_{j}b_{i} with • > 0, for all m ≤ j < i, by Corollary 2.3.

Hence, by applying σ, we obtain:

(λ_{i}b_{2m} + μ_{i}b_{m}b_{3m})σ(b_{j}) = q^{•}σ(b_{j})(λ_{i}b_{2m} + μ_{i}b_{m}b_{3m})  (8)

with • > 0 for all m ≤ j < i. This implies that σ(b_{j}) can not be equal to λb_{2m}+μb_{m}b_{3m} with λ, μ ∈ ℂ, since λb_{2m} + μb_{m}b_{3m} commutes with λ_{i}b_{2m} + μ_{i}b_{m}b_{3m} by Lemma 3.7. Hence we deduce from (6) and (7) that, for each j such that m ≤ j < i, there exist l ∈ {m, …, 3m−1} and λ_{j} ∈ ℂ^{∗} such that σ(b_{j}) = λ_{j}b_{l}. Moreover, we deduce from (8) and Lemma 3.7 that we must have l ∈ {m, …, 2m − 1}. Thus, since i > 2m, there exist j ≠ j′ with m ≤ j, j′ < i with σ(b_{j}) ∈ ℂ^{∗}σ(b_{j′}). This is impossible since σ is an automorphism and ⟨b_{j}⟩ ≠ ⟨b_{j′}⟩, by Proposition 2.1.

Hence, for all i ∈ {2m + 1, …, 3m}, there exist j ∈ {m, …, 3m} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}. In other words, k ≠ i for all i ∈ {2m + 1, …, 3m}.

A similar argument shows that, for all i ∈ {m, …, 2m−1}, there exist j ∈ {m, …, 3m} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}, and so k ≠ i for all i ∈ {m, …, 2m − 1}.

In conclusion, the only possibility is k = 2m. Hence, we have already proved that:

1. There exist λ, μ ∈ ℂ, with λ ≠ 0, such that σ(b_{2m}) = λb_{2m} + μb_{m}b_{3m}.

2. For all i ∈ {m, …, 3m} with i ≠ 2m, there exist j ∈ {m, …, 3m} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}.

3. For all i ∈ {1, …, m − 1} ∪ {3m + 1, …, 4m − 1}, there exist j ∈ {1, …, 4m − 1} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}.

•• We prove by induction that σ(b_{i}) = λ_{i}b_{i}, for all i ∈ {2m + 1, …, 3m}.

First, we know that there exist λ_{2m+1} ∈ ℂ^{∗} and j ∈ {m, …, 3m} such that σ(b_{2m+1}) = λ_{2m+1}b_{j}. It follows from Corollary 2.3 that b_{2m}b_{2m+1} = qb_{2m+1}b_{2m}. Hence, applying σ yields:

(λb_{2m} + μb_{m}b_{3m})b_{j} = qb_{j}(λb_{2m} + μb_{m}b_{3m}).

In view of Lemma 3.7, this forces j = 2m + 1, as desired.

Next, let i ∈ {2m + 2, …, 3m}. It follows from the previous study that there exist j ∈ {m, …, 3m} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j}. Moreover we deduce from the induction hypothesis that σ(b_{i−1}) = λ_{i−1}b_{i−1}. Now, Corollary 2.3 shows that b_{i}b_{i−1} = qb_{i−1}b_{i}. Applying σ yields

b_{j}b_{i−1} = qb_{i−1}b_{j}.

In view of Corollary 2.3, this implies that j = i, as desired.

Hence, σ(b_{i}) = λ_{i}b_{i}, for all i ∈ {2m + 1, …, 3m}. A similar argument shows that σ(b_{i}) = λ_{i}b_{i}, for all i ∈ {m, …, 2m − 1}.

•• Now, let i ∈ {1, …, m − 1}. Then b_{i} ∈ R_{i}; and so it follows from Proposition 3.2 that σ(b_{i}) ∈ R_{≥i} \ R_{>i}. Now, there exist j ∈ {1, …, 4m − 1} and λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{j} because of the previous study. Hence we have b_{j} ∈ R_{≥i} \ R_{>i}, so that b_{j} is also a i × i quantum minor. This implies that either j = i or j = 4m − i. Now, it follows from Corollary 2.3 that b_{3m}b_{i} = q^{•}b_{i}b_{3m} with • > 0. Since we have already proved that σ(b_{3m}) = λ_{3m}b_{3m}, composing by σ leads to b_{3m}b_{j} = q^{•}b_{j}b_{3m} with • > 0. On the other hand, it follows Corollary 2.3 that b_{3m}b_{4m−i} = b_{4m−i}b_{3m}. So j can not be equal to 4m − i. Hence j = i and so σ(b_{i}) = λ_{i}b_{i}, as desired.

A similar argument shows that there exist λ_{i} ∈ ℂ^{∗} such that σ(b_{i}) = λ_{i}b_{i}, for all i ∈ {3m + 1, …, 4m − 1}. □

3.3 The automorphism group of non-square quantum matrices.

Theorem 3.9 Assume that m < n and (m, n) ≠ (1, 3). Let σ be an automorphism of O_{q}(M_{m,n}). Then there exist μ_{i,α} ∈ ℂ^{∗} such that σ(Y_{i,α}) = μ_{i,α}Y_{i,α}, for all (i, α) ∈ [[1, m]] × [[1, n]].

Proof. We proceed by induction on n. The case n = 2 easily follows from Proposition 3.5. So we assume that n ≥ 3. If m = 1, then n > 3, and so once again the result easily follows from Proposition 3.5. So we assume that m ≥ 2. We need to see that σ acts on each generator Y_{i,α} by multiplication by a scalar. We do this by using preferred basis arguments. We use the language and notation of [7].

Note that, because of [7, Proposition 5.3], the quantum minor b_{n} = [1 … m|1 … m] commutes with Y_{i,α}, for 1 ≤ i, α ≤ m, and that b_{n}[R | C] = q^{•}[R | C]b_{n} with • ≥ 0 for all other quantum minors [R | C] of O_{q}(M_{m,n}). Thus, b_{n} q^{•}-commutes with each monomial in the preferred basis where • is ≥ 0, and [7, Proposition 5.3] shows that • is equal to zero if and only if the only quantum minors that occur in the monomial are those of (Y_{i,α})_{i,α∈{1,...,m}}. Observe further that, if b_{n} = [1 … m|1 … m] commutes with an element y ∈ O_{q}(M_{m,n}), then it must commute with each of the monomials in the expression for y in the preferred basis.

Now, let σ be an automorphism of O_{q}(M_{m,n}). We will show first that σ(Y_{i,α}) = μ_{i,α}Y_{i,α} for all 1 ≤ i, α ≤ m.

Set y_{i,α} := σ(Y_{i,α}). Since Y_{i,α} commutes with b_{n} and since σ acts on b_{n} by multiplication by a scalar (see Proposition 3.5 if n ≠ 3m or Proposition 3.8 if n = 3m), then y_{i,α} must commute with b_{n}. Thus, any monomial z in the expression for y_{i,α} in the preferred basis must also commute with b_{n}. This means that the only quantum minors that can occur in z are those of (Y_{i,α})_{i,α∈{1,...,m}}. In particular, σ(Y_{i,α}) belongs to the subalgebra R_{m,m} of R = O_{q}(M_{m,n}) generated by Y_{i,α}, 1 ≤ i, α ≤ m (which is a copy of O_{q}(M_{m,m})). Hence σ induces an automorphism of R_{m,m}.

Moreover, σ acts on the quantum minors b_{n+i} = [i + 1, …, m|1, …, m − i] by multiplication by scalars, by Proposition 3.5 (if n ≠ 3m) or Proposition 3.8 (if n = 3m). In particular, there exists μ_{m,1} ∈ ℂ^{∗} such that σ(Y_{m,1}) = μ_{m,1}Y_{m,1}. Let j ∈ {1, …, m − 1}. Then Y_{j,1}Y_{m,1} = qY_{m,1}Y_{j,1}; so that σ(Y_{j,1})Y_{m,1} = qY_{m,1}σ(Y_{j,1}). Write σ(Y_{j,1}) in the PBW basis of R_{m,m}:

σ(Y_{j,1}) = ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}}Y_{1,2}^{γ_{1,2}} … Y_{m,m}^{γ_{m,m}},

where Γ is a finite subset of ℕ^{m^{2}}. Hence σ(Y_{j,1})Y_{m,1} = qY_{m,1}σ(Y_{j,1}) implies that, for all \underline{γ} ∈ Γ such that c\underline{γ} ≠ 0, we have ∑_{i=1}^{m−1} γ_{i,1} − ∑_{i=2}^{m} γ_{m,i} = 1. Thus, for all \underline{γ} ∈ Γ such that c\underline{γ} ≠ 0, there exists i ∈ \{1, …, m − 1\} with γ_{i,1} ≠ 0. Denote by J, the prime ideal of R generated by the Y_{i,1}, with i ∈ {1, …, m}, and, similarly, denote by J′ the prime ideal of R_{m,m} generated by the Y_{i,1}, with i ∈ {1, …, m}. Observe that J = Y_{1,1}R +⋯+Y_{m,1}R and J′ = Y_{1,1}R_{m,m}+⋯+Y_{m,1}R_{m,m}. We have just proved that σ(Y_{j,1}) ∈ J′ for all j ∈ {1, …, m}. Hence σ(J′) = J′ and σ(J) = J.

Thus, σ induces an automorphism of R_{m,m}/J′ ≃ O_{q}(M_{m,m−1}). Now, because of the induction hypothesis (and the isomorphism O_{q}(M_{m,m−1}) ≃ O_{q}(M_{m−1,m})), we obtain σ(Y_{i,α}) = μ_{i,α}Y_{i,α} + ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}}Y_{1,2}^{γ_{1,2}} … Y_{m,m}^{γ_{m,m}} with μ_{i,α} ≠ 0 and at least one of γ_{k,1} > 0 for each nonzero term of the sum. (Observe that we can apply the inductive hypothesis since (m − 1, m) ≠ (1, 3).)

On the other hand, it follows from Proposition 3.5 or Proposition 3.8 that σ(b_{n−1}) = λb_{n−1} with λ ∈ ℂ^{∗}. Moreover, recalling that b_{n−1} = [1 … m|2 … m + 1], we deduce from a transposed version of [7, Lemma 5.1] that, for all 1 ≤ i, α ≤ m, we have

b_{n−1}Y_{i,α} = q^{•}Y_{i,α}b_{n−1}

with • = −1 if i = 1 and 0 otherwise. Let 1 ≤ i, α ≤ m with α ≠ 1. Thus we must have b_{n−1}σ(Y_{i,α}) = σ(Y_{i,α})b_{n−1}. Since σ(Y_{i,α}) = μ_{i,α}Y_{i,α} + ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}}Y_{1,2}^{γ_{1,2}} … Y_{m,m}^{γ_{m,m}} with at least one of γ_{k,1} ≠ 0 for each nonzero term of the sum, this implies that the sum is empty and so σ(Y_{i,α}) = μ_{i,α}Y_{i,α} for all 1 ≤ i, α ≤ m with α ≠ 1.

Now σ induces an automorphism of R_{m,m} that acts on the indeterminates Y_{i,α} (1 ≤ i ≤ m and 1 < α ≤ m) by multiplication by scalars. Moreover, it follows from Proposition 3.5 or Proposition 3.8 that σ also acts on the b_{i} = [i − n + 1, …, m | 1, …, m + n − i] (n ≤ i ≤ m + n − 1) by multiplication by scalars. This forces σ(Y_{i,1}) = μ_{i,1}Y_{i,1} for all 1 ≤ i ≤ m.

This establishes the following claim:

Claim 3.10 For all (i, α) ∈ [[1, m]]^{2}, there exists μ_{i,α} ∈ ℂ^{∗} such that σ(Y_{i,α}) = μ_{i,α}Y_{i,α}.

It remains to consider the case that α > m. Let us now distinguish between two cases. If n = m + 1, then σ is an automorphism of R = O_{q}(M_{m,m+1}) that acts on the indeter-

minates Y_{i,α} (1 ≤ i, α ≤ m) by multiplication by scalars. Further, because of Proposition 3.5 or Proposition 3.8, σ also acts on the quantum minors b_{i} = [1, …, i | n − i + 1, …, n] (1 ≤ i ≤ m) by multiplication by scalars. This forces σ(Y_{i,m+1}) = μ_{i,m+1}Y_{i,m+1} for all 1 ≤ i ≤ m. This finishes the proof of Theorem 3.9 in the case where n = m + 1.

Now, assume that m + 1 < n. Recall that J denotes the two-sided ideal generated by the Y_{k,1}, 1 ≤ k ≤ m. Note that J = Y_{1,1}R + ⋯ + Y_{m,1}R. It is well-known that the monomials

Y_{1,1}^{γ_{1,1}}Y_{2,1}^{γ_{2,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}}

form a PBW basis of R. Observe that x = ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}} belongs to J if and only if at least one of the γ_{k,1} ≥ 1 for all γ ∈ Γ such that c\underline{γ} ≠ 0.

It follows from Claim 3.10 that σ(J) = J. Hence σ induces an automorphism of R/J ≃ O_{q}(M_{m,n−1}). Since m < n − 1 and m ≥ 2, it follows from the inductive hypothesis that, for all (i, α) with m + 1 ≤ α ≤ n, we can write

σ(Y_{i,α}) = μ_{i,α}Y_{i,α} + ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}}  (9)

where Γ is a finite subset of ℕ^{mn} and at least one of the γ_{k,1} ≥ 1 for all γ ∈ Γ such that c\underline{γ} ≠ 0.

Let K the two-sided ideal generated by the Y_{k,2}, 1 ≤ k ≤ m. Note that an element

x = ∑_{\underline{γ}∈Γ′} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}} ∈ R

belongs to K if and only if at least one of the γ_{k,2} ≥ 1 for all γ ∈ Γ′ such that c\underline{γ} ≠ 0.

It follows from Claim 3.10 that σ(K) = K. Hence σ induces an automorphism of R/K ≃ O_{q}(M_{m,n−1}). Since m < n − 1 and m ≥ 2, the inductive hypothesis applies and, for all (i, α) with m + 1 ≤ α ≤ n, we can write

σ(Y_{i,α}) = μ′_{i,α}Y_{i,α} + ∑_{\underline{γ}∈Γ′} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}}  (10)

where Γ′ is a finite subset of ℕ^{mn} and at least one of the γ_{k,2} ≥ 1 for all γ ∈ Γ′ such that c\underline{γ} ≠ 0.

Let (i, α) with m+1 ≤ α ≤ n. It remains to prove that σ acts on Y_{i,α} by multiplication by a scalar. First, identifying the two expressions (9) and (10) of σ(Y_{i,α}) in the PBW basis of R leads to:

σ(Y_{i,α}) = μ_{i,α}Y_{i,α} + ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}}  (11)

where Γ is a finite subset of ℕ^{mn} such that at least one of the γ_{k,1} ≥ 1 and at least one of the γ_{l,2} ≥ 1 for all γ ∈ Γ such that c\underline{γ} ≠ 0.

By Proposition 3.5 (if n ≠ 3m) or Proposition 3.8 (if n = 3m), the automorphism σ acts on b_{m} = [1 … m | n − m + 1 … n] by multiplication by a scalar. Further, it follows from a transposed version of [7, Lemma 5.1] that

b_{m}Y_{i,α} = \{ \begin{matrix}q^{−1}Y_{i,α}b_{m} & if α ≤ n − m \\ Y_{i,α}b_{m} & if α ≥ n − m + 1. \end{matrix}

Hence

b_{m}σ(Y_{i,α}) = \{ \begin{matrix}q^{−1}σ(Y_{i,α})b_{m} & if α ≤ n − m \\ σ(Y_{i,α})b_{m} & if α ≥ n − m + 1, \end{matrix}

Thus

∑_{\underline{γ}∈Γ} c\underline{γ}q^{−γ_{1,1}−···−γ_{m,1}−···−γ_{1,n−m}−···−γ_{m,n−m}}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}}

= q^{•} ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}},

where q^{•} = −1 if α ≤ n − m, and q^{•} = 0 otherwise.

Consequently,

γ_{1,1} + ⋯ + γ_{m,1} + ⋯ + γ_{1,n−m} + ⋯ + γ_{m,n−m} = \{ \begin{matrix}1 & if α ≤ n − m \\ 0 & if α ≥ n − m + 1, \end{matrix} (12)

for all \underline{γ} ∈ Γ such that c\underline{γ} ≠ 0, since q is not a root of unity.

On the other hand, recall that

σ(Y_{i,α}) = μ_{i,α}Y_{i,α} + ∑_{\underline{γ}∈Γ} c\underline{γ}Y_{1,1}^{γ_{1,1}} … Y_{m,1}^{γ_{m,1}} … Y_{1,n}^{γ_{1,n}} … Y_{m,n}^{γ_{m,n}}

where, for each nonzero term of the sum, at least one of the γ_{k,1} is a positive integer and at least one of the γ_{l,2} is a positive integer. Hence, γ_{1,1}+⋯+γ_{m,1}+⋯+γ_{1,n−m}+⋯+γ_{m,n−m} ≥ 2 for all \underline{γ} ∈ Γ with c\underline{γ} ≠ 0, since n − m ≥ 2. This contradicts (12). Thus Γ must be empty for all (i, α) with α ≥ m and so σ(Y_{i,α}) = μ_{i,α}Y_{i,α} for all α > m. This finishes the proof. □

Corollary 3.11 Assume that m < n and (m, n) ≠ (1, 3). Let σ be an automorphism of O_{q}(M_{m,n}). Then there exist unique nonzero complex numbers h_{1}, …, h_{m}, h′_{1}, …, h′_{n−1} such that σ(Y_{i,α}) = h_{i}h′_{α}Y_{i,α} (with the convention h′_{n} = 1). Thus, Aut(O_{q}(M_{m,n})) is isomorphic to the torus (ℂ^{∗})^{m+n−1}.

Proof. Let σ be an automorphism of O_{q}(M_{m,n}). By Theorem 3.9, there exists a family (μ_{i,α})_{(i,α)∈[[1,m]]×[[1,n]]} of elements of ℂ^{∗} such that σ(Y_{i,α}) = μ_{i,α}Y_{i,α} for all (i, α) ∈ [[1, m]] × [[1, n]]. Recall that, if (i, α), (j, β) ∈ [[1, m]] × [[1, n]] with i < j and α < β, then Y_{j,β}Y_{i,α} = Y_{i,α}Y_{j,β} − (q − q^{−1})Y_{i,β}Y_{j,α}. Hence, since σ is an automorphism, we must have μ_{i,α}μ_{j,β} = μ_{i,β}μ_{j,α} for all (i, α), (j, β) ∈ [[1, m]] × [[1, n]] with i < j and α < β. In other words, the matrix (μ_{i,α})_{(i,α)∈[[1,m]]×[[1,n]]} has rank 1. Hence, there exist unique nonzero complex numbers h_{1}, …, h_{m}, h′_{1}, …, h′_{n−1}, h′_{n} = 1 such that μ_{i,α} = h_{i}h′_{α} for all i, α, as desired. □

Note that, in the exceptional case where (m, n) = (1, 3), the automorphism group of O_{q}(M_{1,3}) has been computed by Alev and Chamarie, [1, Théorème 1.4.6]; in this case, the group is not isomorphic to the torus (ℂ^{∗})^{3}, since the second case of Lemma 3.6 does arise. Indeed, Alev and Chamarie show that any automorphism of O_{q}(M_{1,3}) is of the form

σ(Y_{11}) = μ_{1}Y_{11}, σ(Y_{12}) = μ_{2}Y_{12} + λY_{11}Y_{13}, σ(Y_{13}) = μ_{3}Y_{13},

where μ_{i} ≠ 0 and λ are complex numbers; so that the automorphism group is isomorphic to the semidirect product ℂ × (ℂ^{∗})^{3}. This result can easily be obtained from our analysis.

Acknowledgments. We thank Jacques Alev, Ken Goodearl and Laurent Rigal for helpful conversations and comments.

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T H Lenagan: School of Mathematics, University of Edinburgh, James Clerk Maxwell Building, King’s Buildings, Mayfield Road, Edinburgh EH9 3JZ, Scotland E-mail: tom@maths.ed.ac.uk