THE REPRESENTATION DIMENSION OF QUANTUM COMPLETE INTERSECTIONS

PETTER ANDREAS BERGH & STEFFEN OPPERMANN

Abstract. We give a lower and an upper bound for the representation dimension of a quantum complete intersection.

1. Introduction

In [Au1], Auslander introduced the representation dimension of an Artin algebra in order to study algebras of infinite representation type. He showed that a nonsemisimple algebra is of finite type if and only if its representation dimension is exactly two, whereas it is of infinite type if and only if the representation dimension is at least three.

For more than three decades no general method for computing the representation dimension was known, and it was even unclear if this dimension could exceed three. A negative answer to the latter would imply that the finitistic dimension conjecture holds (cf. [IgT]). However, in 2006 Rouquier showed in [Ro2] that the representation dimension of the exterior algebra on a d-dimensional vector space is d + 1, using the notion of the dimension of a triangulated category (cf. [Ro1]). In particular, there exist finite dimensional algebras of arbitrarily large representation dimension. Other examples illustrating this were subsequently given in [KrK] and [Op1].

In this paper, we study the representation dimension of quantum complete intersections, a class of algebras originating from work by Manin (cf. [Man]) and Avramov, Gasharov and Peeva (cf. [AGP]). In particular, under some assumptions we show that the representation dimension of such an algebra is strictly greater than its codimension.

2. Representation dimension

Let Λ be an Artin algebra, and denote by mod Λ the category of finitely generated left Λ-modules. The representation dimension of Λ, denoted repdim Λ, is defined as

repdim Λ =^{def} inf{gldim End_{Λ}(M) | M generates and cogenerates mod Λ},

where gldim denotes the global dimension of an algebra. A priori we see that the representation dimension might be infinite, but Iyama showed in [Iya] that it is finite for every Artin algebra. Auslander himself showed that the representation dimension of a selfinjective algebra is at most its Loewy length.

When computing the representation dimension of the exterior algebras, Rouquier used the notion of dimensions of triangulated categories, a concept he introduced in [Ro1]. As we will also use this notion when computing a lower bound for the representation dimension of a quantum complete intersection, we recall the definitions. Let T be a triangulated category, and let C and D be subcategories of T .

2000 Mathematics Subject Classification. 16G60, 16S80, 16U80, 81R50.

Key words and phrases. Representation dimension, quantum complete intersections.

The first author was supported by NFR Storforsk grant no. 167130, and the second author by a EU LieGrits grant.

We denote by ⟨C⟩ the full subcategory of T consisting of all the direct summands of finite direct sums of shifts of objects in C. Furthermore, we denote by C ∗ D the full subcategory of T consisting of objects M such that there exists a distinguished triangle

C → M → D → C[1]

in T , with C ∈ C and D ∈ D. Finally, we denote the subcategory ⟨C ∗ D⟩ by C ⋄ D. Now define ⟨C⟩_{0} and ⟨C⟩_{1} to be 0 and ⟨C⟩, respectively, and for each n ≥ 2 define inductively ⟨C⟩_{n} to be ⟨C⟩_{n−1} ⋄ ⟨C⟩. The dimension of T , denoted dim T , is defined as

dim T =^{def} inf{d ∈ ℤ | there exists an object M ∈ T such that T = ⟨M⟩_{d+1}}.

From the definition, we see that the dimension of a triangulated category is not necessarily finite. However, the following elementary result provides a method for computing an upper bound in terms of dense subcategories.

Proposition 2.1. [Ro1, Lemma 3.4] If F: S → T is a functor of triangulated categories whose image is dense in T , then dim T ≤ dim S.

We will use this result to show that for any quantum complete intersection of codimension n, there exists a chain of n subalgebras and a corresponding chain of n−1 inequalities bounding the representation dimension from below. When proving both this and the main results, the triangulated category we use is the stable module category of the algebra. Recall therefore that when Λ is a selfinjective algebra, its stable module category, denoted \underline{mod} Λ, is defined as follows: the objects of \underline{mod} Λ are the same as in mod Λ, but two morphisms in mod Λ are equal in \underline{mod} Λ whenever their difference factors through a projective Λ-module. The cosyzygy functor Ω_{Λ}^{−1} : \underline{mod} Λ → \underline{mod} Λ is an equivalence of categories, and a triangulation of \underline{mod} Λ is given by using this functor as a shift and by letting short exact sequences in mod Λ correspond to triangles. Thus \underline{mod} Λ is a triangulated category, and its dimension is related to the representation dimension of Λ by the following result.

Proposition 2.2. [Ro2, Proposition 3.7] If Λ is a non-semisimple selfinjective algebra, then repdim Λ ≥ dim(\underline{mod} Λ) + 2.

We end this section with the following lemma. It gives a lower bound for the dimension of the stable module category of a selfinjective algebra, in terms of certain subalgebras.

Lemma 2.3. Let A and B be selfinjective Artin algebras. If there exist algebra homomorphisms

A ⟶^{i} B ⟶^{π} A

such that πi is the identity and such that B is a projective A-module, then dim(\underline{mod} A) ≤ dim(\underline{mod} B).

Proof. Every B-module is an A-module via the map i. Moreover, if a map in mod B factors through a projective B-module, then it factors through a projective Amodule since B is A-projective. Therefore the map i induces a functor F: \underline{mod} B → \underline{mod} A, and this is clearly a functor of triangulated categories. Now let M be an object in \underline{mod} A. Then M is a B-module via the map π, hence every object in \underline{mod} A belongs to the image of F. The result now follows from Proposition 2.1. □

3. Quantum complete intersections

Throughout the rest of this paper, let k be a field and n a positive integer. Let a_{1}, …, a_{n} be a sequence of integers with a_{u} ≥ 2 for all u, and for each pair (i, j) of integers with 1 ≤ i < j ≤ n, let q_{ij} be a nonzero element of k. Denote by Λ the algebra

Λ = k⟨X_{1}, …, X_{n}⟩/({X_{u}^{a_{u}}}_{u=1}^{n}, {X_{i}X_{j} − q_{ij}X_{j}X_{i}}_{1≤i<j≤n}),

and by x_{i} the image of X_{i} in Λ. This algebra is finite dimensional of dimension ∏_{u=1}^{n} a_{u} and Loewy length 1+∏_{u=1}^{n}(a_{u} −1), and a quantum complete intersection of codimension n. Namely, it is the quotient of the quantum symmetric algebra

k⟨X_{1}, …, X_{n}⟩/(X_{i}X_{j} − q_{ij}X_{j}X_{i} for 1 ≤ i < j ≤ n)

by the quantum regular sequence x_{1}^{a_{1}}, …, x_{n}^{a_{n}}. Furthermore, these algebras are Frobenius; the codimension two argument in the beginning of [BeE, Section 3] transfers to the general situation. In particular, these algebras are selfinjective.

The class of quantum complete intersections contains some well known algebras. For example, when q_{ij} = 1 for 1 ≤ i < j ≤ n we obtain the truncated polynomial ring

k[X_{1}, …, X_{n}]/(X_{1}^{a_{1}}, …, X_{n}^{a_{n}}),

which is a finite dimensional commutative complete intersection. Moreover, when a_{u} = 2 for all u and q_{ij} = −1 for 1 ≤ i < j ≤ n, then the resulting quantum complete intersection is just the exterior algebra on k^{n}.

As mentioned in the introduction, the notion of quantum complete intersections originates from work by Manin (cf. [Man]), who introduced the concept of quantum symmetric algebras. These algebras were used by Avramov, Gasharov and Peeva in [AGP] to study modules behaving homologically as modules over commutative complete intersections. In particular, they introduced quantum regular sequences of endomorphisms of modules, thus generalizing the classical notion of regular sequences. In [BEH] a rank variety theory was given for quantum complete intersections satisfying certain conditions, and in [BeE] the Hochschild cohomology and homology of these algebras were studied.

We now prove the first of the main results. Given any subset {i_{1}, …, i_{t}} of {1, …, n}, let Λ_{i_{1},...,i_{t}} denote the subalgebra of Λ generated by x_{i_{1}}, …, x_{i_{t}}. Thus Λ_{i_{1},...,i_{t}} is the codimension t quantum complete intersection we obtain when “forgetting” the variables X_{i} not in the sequence X_{i_{1}}, …, X_{i_{t}}.

Theorem 3.1. Let {i_{1}, …, i_{n−1}} be any subset of {1, …, n} of size n − 1. Then

dim(\underline{mod} Λ_{i_{1}}) ≤ dim(\underline{mod} Λ_{i_{1},i_{2}}) ≤ ⋯ ≤ dim(\underline{mod} Λ_{i_{1},...,i_{n−1}}) ≤ dim(\underline{mod} Λ).

Proof. For any t ≤ n there are algebra homomorphisms

Λ_{i_{1},...,i_{t−1}} → Λ_{i_{1},...,i_{t}} → Λ_{i_{1},...,i_{t−1}}

whose composition is the identity. Moreover, the middle term is free as a module over Λ_{i_{1},...,i_{t−1}}. Namely, it is isomorphic to ⨁_{j=0}^{a_{i_{t}} −1}(Λ_{i_{1},...,i_{t−1}})x_{i_{t}}^{j}. The inequalities now follow from Lemma 2.3. □

We end this section with a result giving an upper bound for the representation dimension of a quantum complete intersection.

Theorem 3.2. The representation dimension of Λ is at most 2n.

Proof. The algebra Λ is ℤ^{n}-graded, where the degree of x_{i} is the ith unit vector (0, …, 1, …, 0) in ℤ^{n}. We prove by induction on n that if Λ is a quantum complete intersection of codimension n, then there exists a finitely generated graded Λ-module M such that the following hold:

(i) Λ is a direct summand of M, (ii) the global dimension of End_{Λ}(M) is at most 2n, (iii) all the simple End_{Λ}(M)-modules are one-dimensional.

The theorem will then follow from (i) and (ii).

If n = 1, then Λ is of the form k[X]/(X^{a}). In this case, define M by M = ⨁_{i=1}^{a} k[X]/(X^{i}). Then it is well known that (i), (ii) and (iii) hold for M. Next let n ≥ 1, and suppose that the above claim holds for every quantum complete intersection of codimension at most n. Let Λ be a codimension n + 1 quantum complete intersection

Λ = k⟨X_{1}, …, X_{n+1}⟩/({X_{u}^{a_{u}}}_{u=1}^{n+1}, {X_{i}X_{j} − q_{ij}X_{j}X_{i}}_{1≤i<j≤n+1}),

and let Λ_{1,...,n} and Λ_{n+1} be the subalgebras generated by x_{1}, …, x_{n} and x_{n+1}, respectively. The algebra Λ_{1,...,n} is ℤ^{n}-graded, whereas Λ_{n+1} is ℤ-graded.

Define a map g: ℤ^{n} × ℤ → k \ {0} by

((z_{1}, …, z_{n}), z) ↦ ∏_{i=1}^{n} q_{i,n+1}^{−zz_{i}}.

It is straightforward to check that this is a homomorphism of groups. We now use this homomorphism to define a “twisted” tensor product algebra Λ_{1,...,n} ⊗_{k}^{g} Λ_{n+1} as follows. As a k-vector space Λ_{1,...,n} ⊗_{k}^{g} Λ_{n+1} is just the usual tensor product Λ_{1,...,n} ⊗_{k} Λ_{n+1}, but the multiplication is given by

(λ_{1} ⊗ γ_{1})(λ_{2} ⊗ γ_{2}) =^{def} g(|λ_{2}|, |γ_{1}|)λ_{1}λ_{2} ⊗ γ_{1}γ_{2},

where λ_{i} and γ_{i} are homogeneous elements of Λ_{1,...,n} and Λ_{n+1}, respectively. This algebra is ℤ^{n+1}-graded; the homogeneous elements of degree (z_{1}, …, z_{n+1}) are those of the form λ ⊗ γ for homogeneous elements λ ∈ Λ_{1,...,n} and γ ∈ Λ_{n+1} of degrees (z_{1}, … z_{n}) and z_{n+1}, respectively. With this ring structure we see that the graded algebras Λ_{1,...,n} ⊗_{k}^{g} Λ_{n+1} and Λ are isomorphic.

By induction, there exist graded modules M_{1} ∈ mod Λ_{1,...,n} and M_{2} ∈ mod Λ_{n+1} satisfying (i), (ii) and (iii) over Λ_{1,...,n} and Λ_{n+1}, respectively. Consider the graded module M_{1} ⊗_{k}^{g} M_{2} over Λ_{1,...,n} ⊗_{k}^{g} Λ_{n+1}, where the scalar action is defined in the natural way. It is not hard to see that Λ_{1,...,n} ⊗_{k}^{g} Λ_{n+1} is a direct summand of M_{1} ⊗_{k}^{g} M_{2}, and so this module satisfies (i). Now let Γ be its endomorphism ring over Λ_{1,...,n} ⊗_{k}^{g} Λ_{n+1}. Then Γ ≃ Γ_{1} ⊗_{k}^{g} Γ_{2}, where Γ_{1} = End_{Λ_{1,...,n}}(M_{1}) and Γ_{2} = End_{Λ_{n+1}}(M_{2}) (note that Γ_{1} and Γ_{2} are graded, being endomorphism rings of finitely generated graded modules). If S is a simple Γ-module, then it is a quotient of a module of the form S_{1} ⊗_{k}^{g} S_{2}, where S_{i} is a simple Γ_{i}-module for i = 1, 2. Since S_{1} and S_{2} are both one-dimensional, the module S must be isomorphic to S_{1} ⊗_{k}^{g} S_{2}. Now given graded modules X ∈ mod Γ_{1} and Y ∈ mod Γ_{2}, the minimal projective Γ_{1} ⊗_{k}^{g} Γ_{2}-resolution of X ⊗_{k}^{g} Y is just the tensor product of the minimal projective resolutions of X and Y over Γ_{1} and Γ_{2}, respectively. This can be seen by the same argument as in the classical case, i.e. when there is no twist. Therefore

pd_{Λ_{1,...,n}⊗_{k}^{g}Λ_{n+1}} S = pd_{Λ_{1,...,n}} S_{1} + pd_{Λ_{n+1}} S_{2} ≤ 2n + 2.

In particular, we see that (ii) and (iii) hold for the module M_{1}⊗_{k}^{g}M_{2}. This completes the proof. □

4. Homogeneous quantum complete intersections

We now turn our attention to “homogeneous” quantum complete intersections of codimension n. Throughout the rest of this paper, fix an integer a ≥ 2 and a primitive ath root of unity q ∈ k. We denote by Λ_{n}^{a} the quantum complete intersection

Λ_{n}^{a} = k⟨X_{1}, …, X_{n}⟩/({X_{u}^{a}}_{u=1}^{n}, {X_{i}X_{j} − qX_{j}X_{i}}_{1≤i<j≤n}),

that is, all the defining exponents and commutators are equal to a and q, respectively.

Our aim is to show that the representation dimension of Λ_{n}^{a} is at least n + 1. To do this, we show that given any object M ∈ \underline{mod} Λ_{n}^{a}, there are objects \{N_{i} ∈ \underline{mod} Λ_{n}^{a}\}_{i=1}^{n} and morphisms \{N_{i} ⟶^{f_{i}} N_{i+1}\}_{i=1}^{n−1} satisfying the following: for each i the map

\underline{Hom}_{Λ_{n}^{a}}(−, N_{i}) ⟶^{(f_{i})_{∗}} \underline{Hom}_{Λ_{n}^{a}}(−, N_{i+1})

vanishes on ⟨M⟩, whereas the composition (f_{n−1})_{∗} ∘ ⋯ ∘ (f_{1})_{∗} does not vanish on \underline{mod} Λ_{n}^{a}. We may then conclude from [Ro1, Lemma 4.11] that dim(\underline{mod} Λ_{n}^{a}) ≥ n−1, and then Proposition 2.2 gives repdim Λ_{n}^{a} ≥ n + 1.

The first step is the following lemma on the behavior of a linear combination of the generators. Whenever we have a tuple α = (α_{1}, …, α_{n}) ∈ k^{n}, we denote the corresponding element α_{1}x_{1} + ⋯ + α_{n}x_{n} ∈ Λ_{n}^{a} by σ_{α}.

Lemma 4.1. For any n-tuple α in k^{n}, the following hold.

(i) σ_{α}^{a} = 0.

(ii) σ_{α}^{a−1}x_{i} + σ_{α}^{a−2}x_{i}σ_{α} + ⋯ + σ_{α}x_{i}σ_{α}^{a−2} + x_{i}σ_{α}^{a−1} = 0 for all i.

Proof. The first part of the lemma is just [BEH, Lemma 2.3]. To prove the second part, assume that k is infinite, and let ϵ be any nonzero element of k. Then from (i) the element (σ_{α} + ϵx_{i})^{a} is zero, and expanding we get

0 = σ_{α}^{a} + ϵ (∑_{j=0}^{a−1} σ_{α}^{j}x_{i}σ_{α}^{a−1−j}) + ϵ^{2}z,

where z is some element in Λ_{n}^{a} depending on ϵ. Therefore the set of all elements ϵ in k for which the equality

0 = ϵz + ∑_{j=0}^{a−1} σ_{α}^{j}x_{i}σ_{α}^{a−1−j}

holds contains all the nonzero elements. However, this set is closed in the Zariski topology, and so since k is infinite the zero element must also belong to it. This proves the second part of the lemma in the case when k is infinite. If k is finite, let K/k be an infinite field extension. Then K ⊗_{k} Λ_{n}^{a} is the quantum complete intersection we obtain when allowing the scalars to be elements of K, with the same defining relations as in Λ_{n}^{a}. Since (ii) holds in K ⊗_{k} Λ_{n}^{a} by the above, it also holds in Λ_{n}^{a}. □

In order to prove the main result, we also need the following two lemmas. These are versions of [Op1, Lemma 9] and [Op1, Proposition 11], respectively.

Lemma 4.2. Suppose k is infinite, and let M ∈ mod Λ_{n}^{a} be a module. Then there exists a non-empty open subset U_{M} ⊆ k^{n} such that for any α ∈ U_{M} and any m ∈ M, the following implications hold:

σ_{α}m = 0 ⇒ σ_{β}m ∈ σ_{α}M for all β ∈ k^{n}
σ_{α}^{a−1}m = 0 ⇒ (∑_{i=0}^{a−2} σ_{α}^{i}σ_{β}σ_{α}^{a−2−i}) m ∈ σ_{α}^{a−1}M for all β ∈ k^{n}.

Proof. Let U_{1} be the set of all elements α ∈ k^{n} such that the matrix representing the linear map M ⟶^{·σ_{α}} M (multiplication with σ_{α} from the right) has maximal rank. Then U_{1} is non-empty and open, by an argument similar to that of the proof of [Op1, Lemma 9]. For an element α ∈ U_{1}, choose a basis

{m_{1}, …, m_{s}, w_{1}, …, w_{t}}

of M such that σ_{α}m_{i} = 0 and the set {σ_{α}w_{1}, …, σ_{α}w_{t}} is linearly independent. Then for any nonzero ϵ ∈ k and any β ∈ k^{n}, the set

{σ_{(α+ϵβ)}m_{i}, σ_{(α+ϵβ)}w_{1}, …, σ_{(α+ϵβ)}w_{t}}

is linearly dependent by the choice of α. Since σ_{α}m_{i} = 0, the set

{σ_{β}m_{i}, σ_{(α+ϵβ)}w_{1}, …, σ_{(α+ϵβ)}w_{t}}

is also linearly dependent for any nonzero ϵ ∈ k and any β ∈ k^{n}. However, the set of all ϵ ∈ k such that this set is linearly dependent is closed, and since it contains all the nonzero elements it must be k itself. Consequently, for any element β ∈ k^{n} the set

{σ_{β}m_{i}, σ_{α}w_{1}, …, σ_{α}w_{t}}

is linearly dependent, i.e. σ_{β}m_{i} ∈ σ_{α}M. Therefore, given any elements α ∈ U_{1} and m ∈ M, the first implication in the lemma holds.

Next define U_{2} to be the set of all elements α ∈ k^{n} such that the matrix repre-

senting the linear map M ⟶^{·σ_{α}^{a−1}} M has maximal rank. Then U_{2} is non-empty and open. An argument similar to that for U_{1} shows that given any elements α ∈ U_{2} and m ∈ M, the second implication in the lemma holds, since

σ_{α}^{a−1}σ_{β} + σ_{α}^{a−2}σ_{β}σ_{α} + ⋯ + σ_{α}σ_{β}σ_{α}^{a−2} + σ_{β}σ_{α}^{a−1} = 0

for any β ∈ k^{n} by Lemma 4.1(ii). Now since U_{1} and U_{2} are non-empty open sets in the Zariski topology, their intersection U_{M} = U_{1} ∩ U_{2} is also non-empty and open, and this is a set having the properties we are seeking. □

Lemma 4.3. Suppose k is infinite, and let M be a Λ_{n}^{a}-module. Then there exists a non-empty open subset U ⊆ k^{n} such that for any α ∈ U and any 1 ≤ p ≤ n, the following hold.

(i) For every j ∈ ℤ, any composition

Λ_{n}^{a}/(σ_{α}) \underrightarrow{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} Λ_{n}^{a}/(σ_{α}^{a−1}) → Ω_{Λ_{n}^{a}}^{j}(M)

of left Λ_{n}^{a}-homomorphisms is zero in \underline{mod} Λ_{n}^{a}. (ii) For every j ∈ ℤ, any composition

Λ_{n}^{a}/(σ_{α}^{a−1}) ⟶^{·x_{p}} Λ_{n}^{a}/(σ_{α}) → Ω_{Λ_{n}^{a}}^{j}(M)

of left Λ-homomorphisms is zero in \underline{mod} Λ_{n}^{a}.

Proof. For simplicity, we denote our algebra Λ_{n}^{a} by just Λ. First note that the maps

Λ/(σ_{α}) \underrightarrow{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} Λ/(σ_{α}^{a−1}) and Λ/(σ_{α}^{a−1}) ⟶^{·x_{p}} Λ/(σ_{α}) of left modules are

well defined by Lemma 4.1(ii). Choose two non-empty open subsets U_{M} and U_{Ω_{Λ}(M)} of k^{n} satisfying Lemma 4.2, and consider their intersection U = U_{M} ∩ U_{Ω_{Λ}(M)}. Then U is also non-empty and open. Take any α ∈ U and let Λ/(σ_{α}^{a−1}) ⟶^{g} N be a homomorphism of left Λ-modules, where N is either M or Ω_{Λ}(M). Furthermore, fix any 1 ≤ p ≤ n, and let m = g(1 + (σ_{α}^{a−1})). Then σ_{α}^{a−1}m = 0, and so by Lemma 4.2 there exists an element m′ ∈ N with the property that (∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}) m =

σ_{α}^{a−1}m′. This gives the factorization

\xymatrix{Λ/(σ_{α}) \ar[r]^{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} \ar[dr]_{1↦σ_{α}^{a−1}} & Λ/(σ_{α}^{a−1}) \ar[r]^{g} & N \\  & Λ \ar[ur]_{1↦m′}}

proving (i) in the case when j = 0, 1. A similar argument shows that (ii) holds for j = 0, 1.

Consider the two diagrams

\xymatrix{0 \ar[r] & Λ/(σ_{α}) \ar[d]^{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} \ar[r]^{·σ_{α}^{a−1}} & Λ \ar[d]^{·x_{p}} \ar[r] & Λ/(σ_{α}^{a−1}) \ar[d]^{·x_{p}} \ar[r] & 0 \\ 0 \ar[r] & Λ/(σ_{α}^{a−1}) \ar[r]^{·(−σ_{α})} & Λ \ar[r] & Λ/(σ_{α}) \ar[r] & 0 \\ 0 \ar[r] & Λ/(σ_{α}^{a−1}) \ar[d]^{·x_{p}} \ar[r]^{·(−σ_{α})} & Λ \ar[d]^{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} \ar[r] & Λ/(σ_{α}) \ar[d]^{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} \ar[r] & 0 \\ 0 \ar[r] & Λ/(σ_{α}) \ar[r]^{·σ_{α}^{a−1}} & Λ \ar[r] & Λ/(σ_{α}^{a−1}) \ar[r] & 0}

which are commutative by Lemma 4.1(ii). Since the rows are exact, the modules Λ/(σ_{α}) and Λ/(σ_{α}^{a−1}) are both 2-periodic with respect to the syzygy operator, i.e. Ω_{Λ}^{2u}(Λ/(σ_{α})) ≃ Λ/(σ_{α}) and Ω_{Λ}^{2u}(Λ/(σ_{α}^{a−1})) ≃ Λ/(σ_{α}^{a−1}) for any u ∈ ℤ. Moreover, the commutativity of the squares shows that the two maps

Λ/(σ_{α}) \underrightarrow{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} Λ/(σ_{α}^{a−1}) and Λ/(σ_{α}^{a−1}) ⟶^{·x_{p}} Λ/(σ_{α}) are also 2-periodic

with respect to the syzygy operator. Now let Λ/(σ_{α}^{a−1}) ⟶^{g} Ω_{Λ}^{j}(M) and Λ/(σ_{α}) ⟶^{f} Ω_{Λ}^{j}(M) be any maps, and choose an integer u ∈ ℤ such that Ω_{Λ}^{j+2u}(M) is isomorphic to either M or Ω_{Λ}(M) in \underline{mod} Λ. Then the diagrams

\xymatrix{Λ/(σ_{α}) \ar[r]^{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} \ar[d]^{≀} & Λ/(σ_{α}^{a−1}) \ar[r]^{g} \ar[d]^{≀} & Ω_{Λ}^{j}(M) \\ Λ/(σ_{α}) \ar[r]^{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{p}σ_{α}^{a−2−i}} & Λ/(σ_{α}^{a−1}) \ar[r]^{Ω_{Λ}^{2u}(g)} & Ω_{Λ}^{j+2u}(M) \\ Λ/(σ_{α}^{a−1}) \ar[r]^{·x_{p}} \ar[d]^{≀} & Λ/(σ_{α}) \ar[r]^{f} \ar[d]^{≀} & Ω_{Λ}^{j}(M) \\ Λ/(σ_{α}^{a−1}) \ar[r]^{·x_{p}} & Λ/(σ_{α}) \ar[r]^{Ω_{Λ}^{2u}(f)} & Ω_{Λ}^{j+2u}(M)}

are commutative in \underline{mod} Λ, where the vertical maps are isomorphisms. We have already shown that the bottom compositions are zero in \underline{mod} Λ, but then so are the top compositions, since these are shifts of the bottom compositions. □

We are now ready to prove that the representation dimension of Λ_{n}^{a} is at least n + 1 when n is even and the field k is infinite.

Proposition 4.4. If k is infinite and n is even, then repdim Λ_{n}^{a} ≥ n + 1.

Proof. As in the previous proof, we denote our algebra Λ_{n}^{a} by just Λ. Let α ∈ k^{n} be an n-tuple, and denote the element

x_{n−1} (∑_{i=0}^{a−2} σ_{α}^{i}x_{n}σ_{α}^{a−2−i}) x_{n−3} (∑_{i=0}^{a−2} σ_{α}^{i}x_{n−2}σ_{α}^{a−2−i}) ⋯ x_{3} (∑_{i=0}^{a−2} σ_{α}^{i}x_{4}σ_{α}^{a−2−i}) x_{2}

by w_{α}. In the first (and longest) part of this proof, we show that the set of all α ∈ k^{n} such that w_{α} does not belong to σ_{α}Λ + Λσ_{α} contains a non-empty open set.

Fix any n-tuple α = (α_{1}, …, α_{n}) ∈ k^{n} with α_{1} ≠ 0, and, for simplicity, denote the corresponding σ_{α} and w_{α} by σ and w, respectively. Every element λ ∈ Λ admits a unique decomposition λ = N_{x_{1}}(λ) + σR_{x_{1}}(λ), in which x_{1} does not occur in any of the monomials in N_{x_{1}}(λ). With this notation, we see that

w ∈ σΛ + Λσ ⇔ N_{x_{1}}(w) ∈ σΛ + Λσ
⇔ N_{x_{1}}(w) + hσ ∈ σΛ for some h ∈ Λ
⇔ N_{x_{1}}(w) + hσ ∈ σΛ for some h ∈ Λ with N_{x_{1}}(h) = h
⇔ N_{x_{1}} (N_{x_{1}}(w) + hσ) = 0 for some h ∈ Λ with N_{x_{1}}(h) = h,

where we may assume that the h occurring is homogeneous. Now note that the degree of w is (\frac{n}{2} − 1) a + 1. This implies that any homogeneous h satisfying the above implications is of degree (\frac{n}{2} − 1) a. If in addition N_{x_{1}}(h) = h, i.e. if x_{1} does not occur in any of the monomials in h, then hx_{1} = q^{−(\frac{n}{2} −1)a}x_{1}h = x_{1}h. Writing σ′ = σ − α_{1}x_{1} we then get

hσ = α_{1}hx_{1} + hσ′ = α_{1}x_{1}h + hσ′ = σh − σ′h + hσ′.

What we have so far gives

w ∈ σΛ + Λσ ⇔ N_{x_{1}} (N_{x_{1}}(w) + hσ) = 0 with N_{x_{1}}(h) = h
⇔ N_{x_{1}} (N_{x_{1}}(w) + σh − σ′h + hσ′) = 0 with N_{x_{1}}(h) = h
⇔ N_{x_{1}} (N_{x_{1}}(w) − σ′h + hσ′) = 0 with N_{x_{1}}(h) = h
⇔ N_{x_{1}}(w) − σ′h + hσ′ = 0 with N_{x_{1}}(h) = h,

where the last implication follows from the fact that x_{1} does not occur in any of the monomials in N_{x_{1}}(w) − σ′h + hσ′. Thus w belongs to σΛ + Λσ if and only if N_{x_{1}}(w) = σ′h − hσ′ for some homogeneous element h satisfying N_{x_{1}}(h) = h.

Let λ be the element x_{2}x_{3}^{a−1}x_{4} ⋯ x_{n−1}^{a−1}x_{n}, which has the same degree as w. For any homogeneous element h of degree one less than that of w, the coefficient of λ in σ′h − hσ′ is easily seen to be zero. Therefore w ∉ σΛ + Λσ if the coefficient of λ in N_{x_{1}}(w) is nonzero. Note that the set of tuples (α_{2}, …, α_{n}) ∈ k^{n−1} for which this holds is open.

Consider the tuple (α_{1}, 0, 1, …, 1, 0) ∈ k^{n}, that is, we take as σ the element

σ = α_{1}x_{1} + x_{3} + ⋯ + x_{n−1}.

Define the element w′ ∈ Λ by

w′ = x_{n−1} ∑_{i=0}^{a−2} σ^{i}x_{n}σ^{a−2−i},

i.e. w′ is the “first part” of w. Since x_{n}σ = q^{−1}σx_{n}, we see that

w′ = x_{n−1}βσ^{a−2}x_{n}
= β(q^{−1}α_{1}x_{1} + ⋯ + q^{−1}x_{n−3} + x_{n−1})^{a−2}x_{n−1}x_{n}

for some nonzero element β ∈ k. Using the equality N_{x_{1}}(zy) = N_{x_{1}} (N_{x_{1}}(z)y), which holds for any elements z, y ∈ Λ, an induction argument gives

N_{x_{1}} ((q^{−1}α_{1}x_{1} + ⋯ + q^{−1}x_{n−3} + x_{n−1})^{i}) = (∏_{j=1}^{i}(1 − q^{−j})) x_{n−1}^{i}.

Applying N_{x_{1}} to the above expression for w′ then gives

N_{x_{1}}(w′) = βN_{x_{1}} (N_{x_{1}} [(q^{−1}α_{1}x_{1} + ⋯ + q^{−1}x_{n−3} + x_{n−1})^{a−2}] x_{n−1}x_{n})
= βN_{x_{1}} ([∏_{j=1}^{a−2}(1 − q^{−j})] x_{n−1}^{a−1}x_{n})
= β′x_{n−1}^{a−1}x_{n},

where β′ is some nonzero element in k. Now define w′′ by

w′′ = x_{n−3} (∑_{i=0}^{a−2} σ^{i}x_{n−2}σ^{a−2−i}) ⋯ x_{3} (∑_{i=0}^{a−2} σ^{i}x_{4}σ^{a−2−i}) x_{2},

that is, the element w is given by w = w′w′′. By what we have just shown, the equality

N_{x_{1}}(w) = N_{x_{1}} (N_{x_{1}}(w′)w′′) = β′N_{x_{1}}(x_{n−1}^{a−1}x_{n}w′′)

holds. Now every monomial in w′′ containing x_{n−1} does not “contribute” to x_{n−1}^{a−1}x_{n}w′′, therefore we may replace every σ in w′′ by σ − x_{n−1}. Then if we repeat the above process, we see that

N_{x_{1}}(w) = δx_{2}x_{3}^{a−1}x_{4} ⋯ x_{n−1}^{a−1}x_{n} = δλ,

where δ is some nonzero element in k. This shows that the coefficient of the monomial λ in N_{x_{1}}(w) is nonzero, i.e. w ∉ σΛ + Λσ.

Define the set V ⊆ k^{n} by

V = {α ∈ k^{n} | w_{α} ∉ σ_{α}Λ + Λσ_{α}},

and consider the set

W = {(α_{2}, …, α_{n}) ∈ k^{n−1} | (α_{1}, α_{2}, …, α_{n}) ∈ V for all 0 ≠ α_{1} ∈ k}.

We have just shown that W is open and non-empty, hence the subset

{(α_{1}, α_{2}, …, α_{n}) ∈ k^{n} | α_{1} ≠ 0 and (α_{2}, …, α_{n}) ∈ W }

of V is non-empty and open in k^{n}.

Now let M be a Λ-module, and let U ⊆ k^{n} be a non-empty open set having the properties stated in Lemma 4.3. Since V contains a non-empty open subset, the intersection U ∩V is not empty, and therefore contains an n-tuple α ∈ k^{n}. For each 1 ≤ i ≤ n − 1, define maps f_{i} ∈ \underline{mod} Λ by

f_{i} = \{ \begin{matrix}Λ/(σ_{α}^{a−1}) ⟶^{·x_{n−i}} Λ/(σ_{α}) & when i is odd and i ≤ n − 3 \\ Λ/(σ_{α}) \underrightarrow{· ∑_{i=0}^{a−2} σ_{α}^{i}x_{n+2−i}σ_{α}^{a−2−i}} Λ/(σ_{α}^{a−1}) & when i is even \\ Λ/(σ_{α}^{a−1}) ⟶^{·x_{2}} Λ/(σ_{α}) & when i = n − 1. \end{matrix}

By Lemma 4.3 each induced map (f_{i})_{∗} vanishes on the subcategory ⟨M⟩ of \underline{mod} Λ. However, the composition (f_{n−1})_{∗}∘⋯∘(f_{1})_{∗} does not vanish: if it did then the map Λ/(σ_{α}^{a−1}) ⟶^{·w_{α}} Λ/(σ_{α}) would be zero in \underline{mod} Λ. However, the injective envelope of Λ/(σ_{α}^{a−1}) is Λ/(σ_{α}^{a−1}) ⟶^{·σ_{α}} Λ, whereas the projective cover of Λ/(σ_{α}) is the projection Λ ⟶^{π} Λ/(σ_{α}). Therefore, if the map Λ/(σ_{α}^{a−1}) ⟶^{·w_{α}} Λ/(σ_{α}) was zero in \underline{mod} Λ, then it would factor in a commutative diagram

\xymatrix{Λ/(σ_{α}^{a−1}) \ar[d]^{·σ_{α}} \ar[r]^{·w_{α}} & Λ/(σ_{α}) \\ Λ \ar[r] & Λ \ar[u]^{π}}

in mod Λ. Since any homomorphism Λ → Λ is right multiplication with an element in Λ, this would mean that there exists an element λ ∈ Λ such that w_{α} = σ_{α}λ in Λ/(σ_{α}). In other words, the element w_{α} would belong to σ_{α}Λ+Λσ_{α}, and this is not the case since α ∈ V. From [Ro1, Lemma 4.11] we see that dim(\underline{mod} Λ) ≥ n − 1, and so Proposition 2.2 gives repdim Λ ≥ n + 1. □

Remark. It is also possible to prove Proposition 4.4 using [Op2, Theorem 1(b)]. However, as in the proof we just gave, the key ingredient in such a proof is deciding whether or not w_{α} belongs to σ_{α}Λ + Λσ_{α}.

Having established the case when n is even and the field k is infinite, we prove the main theorem in this paper, namely the general case.

Theorem 4.5. The representation dimension of Λ_{n}^{a} satisfies

n + 1 ≤ repdim Λ_{n}^{a} ≤ 2n.

Proof. For the inequality n + 1 ≤ repdim Λ_{n}^{a}, we first assume k to be infinite. The case when n is even was treated in Proposition 4.4, so it remains to give a proof for odd n. To do so, we keep n even, and prove the inequality n ≤ repdim Λ_{n−1}^{a}. We denote our algebra Λ_{n}^{a} by Λ. Furthermore, denote by Λ̃ the algebra Λ_{2,3,...,n}, where the notation is the same as that used in Theorem 3.1. In other words, the algebra Λ̃ is the codimension n − 1 quantum complete intersection subalgebra of Λ generated by x_{2}, …, x_{n}, and therefore isomorphic to Λ_{n−1}^{a}. In this notation, our aim is to show that repdim Λ̃ ≥ n. We may assume n ≥ 4, since we know that the representation dimension of the truncated polynomial ring k[X]/(X^{a}), which is of finite representation type, is two.

Given an n-tuple α ∈ k^{n}, define as before the corresponding element σ_{α} ∈ Λ, and denote by w_{α} the element

x_{n−1} (∑_{i=0}^{a−2} σ_{α}^{i}x_{n}σ_{α}^{a−2−i}) x_{n−3} (∑_{i=0}^{a−2} σ_{α}^{i}x_{n−2}σ_{α}^{a−2−i}) ⋯ x_{3} (∑_{i=0}^{a−2} σ_{α}^{i}x_{4}σ_{α}^{a−2−i}) x_{2}.

We showed in the proof of the previous result that the set

V = {α ∈ k^{n} | w_{α} ∉ σ_{α}Λ + Λσ_{α}}

contains a non-empty open subset. Now for any (n − 1)-tuple α̃ = (α_{2}, …, α_{n}) ∈ k^{n−1}, define the element σ̃α̃ ∈ Λ̃ by σ̃α̃ = α_{2}x_{2} + ⋯ + α_{n}x_{n}, and denote by w̃α̃ the element

x_{n−1} (∑_{i=0}^{a−2} σ̃i ; α̃x_{n}σ̃a−2−i ; α̃) x_{n−3} (∑_{i=0}^{a−2} σ̃i ; α̃x_{n−2}σ̃a−2−i ; α̃) ⋯ x_{3} (∑_{i=0}^{a−2} σ̃i ; α̃x_{4}σ̃a−2−i ; α̃)

in Λ̃. Furthermore, consider the subset

Ṽ = {α̃ ∈ k^{n−1} | w̃α̃ ∉ σ̃α̃Λ̃ + Λ̃σ̃a−1 ; α̃}

of k^{n−1}. We show that this set contains a non-empty subset which is open. Recall from the previous proof that the set

W = {(α_{2}, …, α_{n}) ∈ k^{n−1} | (α_{1}, α_{2}, …, α_{n}) ∈ V for all 0 ≠ α_{1} ∈ k}

is non-empty and open in k^{n−1}. Furthermore, let W′ be the set of all (n − 1)- tuples (α_{2}, …, α_{n}) in which α_{2} is nonzero. Since W′ is non-empty and open, the intersection W̃ = W ∩ W′ is also non-empty and open. Pick therefore an element α̃ = (α_{2}, …, α_{n}) ∈ W̃, let α ∈ k^{n} be the n-tuple α = (α_{2}, α_{2}, …, α_{n}), and define a map f: Λ̃ → Λ by

x_{i} ↦ \{ \begin{matrix}x_{1} + x_{2} & when i = 2 \\ x_{i} & when i ≠ 2. \end{matrix}

Since f(x_{u}^{a}) = 0 = f(x_{i}x_{j} − qx_{j}x_{i}) for 2 ≤ u ≤ n and 2 ≤ i < j ≤ n, this is a well defined algebra homomorphism. Note that f(σ̃α̃) = σ_{α} and that α belongs to V. Moreover, note that f(w̃α̃)x_{2} = w_{α}, and so if w̃α̃ belongs to σ̃α̃Λ̃ + Λ̃σ̃a−1 ; α̃ then w_{α} belongs to σ_{α}Λ + Λσ_{α}^{a−1}x_{2}. However, from Lemma 4.1(ii) we see that σ_{α}Λ + Λσ_{α}^{a−1}x_{2} is contained in σ_{α}Λ + Λσ_{α}, and we know that w_{α} ∉ σ_{α}Λ + Λσ_{α} since α ∈ V. Therefore w̃α̃ cannot belong to σ̃α̃Λ̃ + Λ̃σ̃a−1 ; α̃, and this shows that W̃ is a subset of Ṽ. Consequently, the set Ṽ contains a non-empty subset which is open.

The arguments we applied at the end of the previous proof now shows that repdim Λ̃ ≥ n. We have therefore proved that repdim Λ_{n}^{a} ≥ n + 1 for any n, when the field k is infinite. However, when k is finite, the strategy applied in [Op1, Section 4] carries over to our algebra and shows that the inequality still holds. Therefore repdim Λ_{n}^{a} ≥ n + 1 regardless of whether the field k is infinite. This proves the first inequality in the theorem, the other is Theorem 3.2. □

Corollary 4.6. Let Λ be a general quantum complete intersection, i.e.

Λ = k⟨X_{1}, …, X_{n}⟩/({X_{u}^{a_{u}}}_{u=1}^{n}, {X_{i}X_{j} − q_{ij}X_{j}X_{i}}_{1≤i<j≤n})

where a_{u} ≥ 2 and q_{ij} is nonzero. If there exists a subset {i_{1}, …, i_{t}} of {1, …, n} such that the subalgebra Λ_{i_{1},...,i_{t}} is a homogeneous quantum complete intersection of the form Λ_{t}^{a}, then repdim Λ ≥ t + 1.

Proof. By Theorem 3.1 the inequality dim(\underline{mod} Λ_{t}^{a}) ≤ dim(\underline{mod} Λ) holds, and in the previous proof we showed that dim(\underline{mod} Λ_{t}^{a}) is at least t − 1. Proposition 2.2 now gives repdim Λ ≥ t + 1. □

Remarks. (i) In the main results we have only considered homogeneous quantum complete intersections of the form

k⟨X_{1}, …, X_{n}⟩/({X_{u}^{a}}_{u=1}^{n}, {X_{i}X_{j} − qX_{j}X_{i}}_{1≤i<j≤n}),

that is, algebras where the defining exponents are all equal to a and where q is a primitive ath root of unity. However, the proofs are also valid if we relax the requirement that the defining exponents of the “even” indeterminates X_{2}, X_{4}, … are equal to a, as long as we require that these exponents belong to {2, …, a}. That is, the main results apply to quantum complete intersections of the form

k⟨X_{1}, …, X_{n}⟩/({X_{u}^{a}}_{u odd}, {X_{v}^{a_{v}}}_{v even}, {X_{i}X_{j} − qX_{j}X_{i}}_{1≤i<j≤n}),

where q is a primitive ath root of unity and a_{v} ∈ {2, …, a}. This is because the key ingredient in the proof when n is even is to show that the coefficient of the element x_{2}x_{3}^{a−1}x_{4} ⋯ x_{n−1}^{a−1}x_{n} in

x_{n−1} (∑_{i=0}^{a−2} σ_{α}^{i}x_{n}σ_{α}^{a−2−i}) x_{n−3} (∑_{i=0}^{a−2} σ_{α}^{i}x_{n−2}σ_{α}^{a−2−i}) ⋯ x_{3} (∑_{i=0}^{a−2} σ_{α}^{i}x_{4}σ_{α}^{a−2−i}) x_{2}

is nonzero for every α belonging to a certain non-empty open subset of k^{n}.

(ii) Work in progress by Avramov and Iyengar (cf. [AvI]) shows that the dimension of the stable derived category of a commutative local complete intersection of codimension c is at least c − 1. Consequently, the representation dimension of an Artin complete intersection is strictly greater than its embedding dimension. In particular, they have shown that the representation dimension of k[X_{1}, …, X_{n}]/(X_{1}^{a_{1}}, …, X_{n}^{a_{n}}) is at least n + 1 when a_{i} ≥ 2.

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Institutt for matematiske fag, NTNU, N-7491 Trondheim, Norway

   E-mail address: bergh@math.ntnu.no
   E-mail address: opperman@math.ntnu.no