Nilpotent elements in Hochschild cohomology

Karin Erdmann

Abstract

We study the algebra A = K⟨x, y⟩/(x^{2}, y^{2}, (xy)^{k} + q(yx)^{k}) over the field K where k ≥ 1, and where 0 ≠ q ∈ K. We determine a minimal projective bimodule resolution of A. In the case when q is not a root of unity, we compute its Hochschild cohomology. In particular we show that for n ≥ 3, the n-th part HH^{n}(A) has dimension k − 1 if char(K) does not divide k. We also show that every element in HH^{n}(A) for n ≥ 1 is nilpotent. This is motivated by the problem of understanding why the finite generation condition (Fg) fails which is needed to ensure existence of support varieties.

MR Subject Classification: 16E40, 16S80, 20C20, 16G20. Keywords: Algebra of dihedral type, socle deformation, bimodule resolution, Hochschild cohomology, nilpotent elements in Hochschild cohomology.

To Dave

1 Introduction

Assume A is a finite-dimensional selfinjective algebra over some field. We would like to have support varieties for A-modules algebras, similar to those for representations of finite groups. If G is a finite group then such varieties are constructed using group cohomology, which is graded commutative and noetherian. These varieties have very good properties and have proved to be a powerful tool in many contexts.

In [12] this is extended to finite-dimensional algebras, by defining support varieties based on Hochschild cohomology. It is shown in [6] that with suitable finiteness conditions, much of the properties in the group setting generalize.

Following the account of [13], the finiteness condition needed is called (Fg), and it states that the Hochschild cohomology HH^{∗}(A) should be noetherian, and the extalgebra Ext_{A}(A/J, A/J) of A should be finitely generated as a module over HH^{∗}(A), here J is the radical of A.

Here we study the local selfinjective algebras of dimension 4k,

A = K⟨x, y⟩/(x^{2}, y^{2}, (xy)^{k} + q(yx)^{k})

where k ≥ 1 and 0 ≠ q ∈ K. When char(K) = 2 and k is a power of 2 and moreover q = −1, this is isomorphic to the group algebra of a dihedral 2-group. It is known that the indecomposable non-projective modules are independent of q, and their parametrisation is independent of the characteristic. One would therefore expect that its homological algebra should be similar to the group situation.

However this is not the case. Take k = 1 and q not a root of unity, then the algebra is precisely the famous example of [4] which is selfinjective but its Hochschild cohomology is finite-dimensional. In particular condition (Fg) fails. For general k, there is a problem as well. Namely as it is shown in [8], when q is not a root of unity, there are A-modules with complexity one (that is, with bounded periodic resolutions) which are not periodic. By Theorem 5.3 in [6] this implies that (Fg) cannot hold. One would like to understand why.

We determine explicitly a minimal projective bimodule resolution for A, for arbitrary characteristic, and arbitrary non-zero q. Furthermore, assume that q is not a root of unity. For such algebras, we compute the dimensions of the homogeneous parts of the Hochschild cohomology. We show that for n ≥ 3, the n-th Hochschild cohomology HH^{n}(A) has dimension k − 1 if the characteristic of the field does not divide k, and otherwise it has dimension k. Moreover, we show that every element of positive degree is nilpotent. This uses a general result of [12] (see 2.3). In particular this shows that the Hochschild cohomology algebra is not noetherian for k ≥ 2.

When k = 1 this shows again that the Hochschild cohomology is finite-dimensional, as in [4]: In this case the characteristic does not divide k and hence HH^{n}(A) = 0 for n ≥ 3.

In general we see that the dimension of HH^{n}(A) is bounded when q is not a root of unity. Given that the algebra A is of infinite representation type, this may be unexpected. It also is different from the group setting, see [10].

The second section collects relevant background. In the third section we determine a minimal projective bimodule resolution, and in the fourth section we compute the cohomology.

2 Preliminaries

2.1 The algebras

The algebra A with q = −1 is symmetric, and it is one of the algebras of dihedral type introduced in [5]. In general, the socle of A is spanned by (xy)^{k}, and the factor algebra A/soc(A) is independent of q. That is, the general algebra A is a socle deformation of an algebra of dihedral type.

Indecomposable non-projective A-modules are annihilated by soc(A), and hence they are independent of q. However the action of the Heller operator Ω does depend on q: As mentioned above, for q not a root of unity there are modules of complexity one which are not Ω-periodic. On the other hand, when q = −1, all modules of complexity one have Ω-period ≤ 2.

Recall that if R is a finite-dimensional algebra then every finite-dimensional Rmodule M has a projective cover π : P → M, and Ω(M) is the kernel of π. The kernels of the maps in a minimal projective resolution of M are the modules Ω^{n}(M) for n ≥ 1, called syzygies. We use this when R = A as above. As well, for Hochschild cohomology we use this when R is the enveloping algebra of A, that is R = A^{e} = A ⊗_{K} A^{op}, and when M = A. We write Ω_{A}(−) and we write just Ω for Ω_{A^{e}}.

2.2 Hochschild cohomology

The Hochschild cohomology HH^{∗}(A) is isomorphic to Ext_{A^{e}}^{∗}(A, A). Here A is viewed as a left module for the enveloping algebra A^{e}. We will compute this from a minimal projective resolution of A as an A^{e}-module, equivalently, as a a bimodule:

(ℙ)  … → ℙ_{n} →^{d_{n}} ℙ_{n−1} → … →^{d_{1}} ℙ_{0} →^{d_{0}} A → 0

Here the kernel of d_{n−1} is the bimodule Ω^{n}(A). We identify HH^{n}(A) with Ext_{A^{e}}^{n}(A, A), and to compute the dimension of HH^{n}(A) we will use the exact sequence

0 → Hom_{A^{e}}(Ω^{n−1}(A), A) → Hom_{A^{e}}(ℙ_{n−1}, A) → Hom_{A^{e}}(Ω^{n}(A), A) → HH^{n}(A) → 0.

The terms of a minimal projective bimodule resolution can be described more generally, by a result of Happel, see [9]. In our case, K is the only simple A-module, and A^{e} (≅ A^{e}(1 ⊗ 1)) is the only indecomposable projective A^{e}-module (up to isomorphism), so each ℙ_{n} is isomorphic to a direct sum of copies of A^{e}. Happel’s result becomes the following.

Proposition 2.1. The multiplicity of A^{e} as a direct summand of ℙ_{n} is equal to the dimension of Ext_{A}^{n}(K, K).

In fact, tensoring (ℙ) over A with K on right gives a minimal projective resolution of K as a left A-module which is perhaps implicit in Happel’s proof. The syzygies of K as a left A-module are well understood, we describe them below. This motivates the bimodule resolution we will construct in the next section.

2.2.1 The modules Ω_{A}^{n}(K)

The algebra A is tame, and the indecomposable A-modules are classified. It is important for our context that the indecomposable non-projective modules are independent of q as explained above.

There is a parametrisation of indecomposables, originally due to C. M. Ringel [11]. The description can also be found in the appendix of [2] (the assumption on the characteristic and on k is not needed). The non-projective indecomposables come in two types, now called string modules and band modules, originally called modules ’of the first kind’ and ’of the second kind’.

The syzygies of K are string modules. A string module for A can be visualized by a linear quiver where each vertex corresponds to a basis vector. Each arrow is labelled by either by x or by y alternating, and the length of a maximal path with no change of orientation is bounded by 2k − 1. We describe Ω_{A}(K) and Ω_{A}^{2}(K) when k = 2.

(a) The module Ω_{A}(K) may be written as:

Kv_{1} ⟶^{x} Kv_{2} ⟶^{y} Kv_{3} ⟶^{x} Kv_{4} ⟵^{y} Kv_{5} ⟵^{x} Kv_{6} ⟵^{y} Kv_{7}.

This means that x takes v_{1} → v_{2}, v_{3} → v_{4} and v_{6} → v_{5} and it maps all other basis vectors to zero. Moreover y acts by v_{2} → v_{3}, v_{7} → v_{6} and v_{5} → v_{4}, and all other basis vectors are mapped to zero.

This module is generated by v_{1}, v_{7} and can be identified with Ω_{A}(K), that is, Ax + Ay ⊂ A, by mapping x → v_{7} and y ↦ v_{1} which extends to an A-module isomorphism.

(b) Ω_{A}^{2}(K) is the following: We use shorthand notation, and write just v_{i} instead of Kv_{i}.

v_{1} ⟶^{x} v_{2} ⟶^{y} v_{3} ⟶^{x} v_{4} ⟵^{y} v_{5} ⟶^{x} v_{6} ⟵^{y} v_{7} ⟵^{x} v_{8} ⟵^{y} v_{9}.

One can turn this into an exact sequence of A-modules. The submodule generated by v_{9} is isomorphic to Ax, and the submodule generated by v_{1} is isomorphic to Ay. These two submodules have trivial intersection and span a maximal submodule. We have an exact sequence

0 → Ax ⊕ Ay → Ω_{A}^{2}(A) → K ≅ Ω_{A}^{0}(K) → 0.

(c) This generalizes. To obtain the description of Ω_{A}^{n+2}(A) one may start with Ω^{n}(A) and extend the quiver describing it at each end by a maximal linear subquiver of length 2k − 1. This reflects the fact that there is a short exact sequence of left A-modules

0 → Ax ⊕ Ay → Ω^{n+2}(A) → Ω^{n}(A) → 0.

Since K is simple, we have for each n that

Ext_{A}^{n}(K, K) ≅ Hom_{A}(Ω^{n}(K), K).

The non-zero homomorphisms from Ω^{n}(K) map a vector corresponding to a sink to some scalar and map any other basis vector to zero. Hence we see that

dim Ext_{A}^{n}(K, K) = n + 1.

For the group setting, see also [1].

2.2.2 Small degrees

The first two terms of a minimal bimodule resolution can be described explicitly, more generally, for a finite-dimensional algebra of the form A = KQ/I where Q is a finite quiver and I is an admissible ideal of KQ. Let Q_{0} be the set of vertices of the quiver, and Q_{1} be the set of arrows. If α : i → j is an arrow, write iα = i and tα = j. We have

ℙ_{0} = ⊕_{i∈Q_{0}} A^{e}(e_{i} ⊗ e_{i})
ℙ_{1} = ⊕_{α∈Q_{1}} A^{e}(e_{iα} ⊗_{α} e_{tα}).

We must label the generators of ℙ_{1} by arrows, so that we can distinguish equal idempotents in case of multiple arrows. The first two differentials are defined by

d_{0}(e_{i} ⊗ e_{i}) := e_{i}
d_{1}(e_{iα} ⊗_{α} e_{tα}) := α(e_{tα} ⊗ e_{tα}) − (e_{iα} ⊗ e_{iα})α,

and the images of the generators of ℙ_{1} labelled by arrows are minimal generators of Ω(A).

To describe minimal generators for Ω^{2}(A) the following notation is convenient.

Notation 2.2. (1) If a_{1}a_{2} … a_{m} is a monomial in KQ then define

ρ(a_{1} … a_{m}) := ∑_{j=1}^{m} a_{1}a_{2} … a_{j−1}(e_{ia_{j}} ⊗_{a_{j}} e_{ta_{j}})a_{j+1} … a_{m}.

This is an element in ℙ_{1}, and ρ extends to a linear map KQ → ℙ_{1}.

(2) Then minimal generators for Ω^{2}(A) can be written down, roughly speaking, by the images of minimal relations under ρ.

(3) We return to the algebra A. Its quiver has one vertex, with idempotent 1_{A}, and two loops, x and y, and minimal relations x^{2}, y^{2}, (xy)^{k} + q(yx)^{k}. We label the generator 1 ⊗ 1 of ℙ_{0} as [f_{0}^{0}]. The generators of ℙ_{1} are written as [f_{0}^{1}], corresponding to x, and [f_{1}^{1}], corresponding to y. Note that as elements of A^{e}, each of these is 1 ⊗ 1. With this,

d_{1}([f_{0}^{1}]) = x ⊗ 1 − 1 ⊗ x = x[f_{0}^{0}] − [f_{0}^{0}]x =: f_{0}^{1}

and

d_{1}([f_{1}^{1}]) = y ⊗ 1 − 1 ⊗ y = y[f_{0}^{0}] − [f_{0}^{0}]y =: f_{1}^{1}.

(4) The module ker(d_{1}) = Ω^{2}(A) has generators ρ(x^{2}) and ρ(y^{2}) and ρ((xy)^{k} +q(yx)^{k}). Here

f_{0}^{2} := ρ(x^{2}) = x[f_{0}^{1}] + [f_{0}^{1}]x and f_{2}^{2} := ρ(y^{2}) = y[f_{1}^{1}] + [f_{1}^{1}]y.

We will introduce a shorthand notation for the element ρ((xy)^{k} + q(yx)^{k}) of ℙ_{1}.

We define the following linear maps: If M is an A, A bimodule, define a linear map Tr_{xy} : M → M which substitutes for each occurance of x, and a linear map Tr_{yx} : M → M which substitutes for each occurance of y,

Tr_{xy}(m) := ∑_{t=0}^{k−1}(xy)^{t}(m)(yx)^{k−1−t}

Tr_{yx}(m) := ∑_{t=0}^{k−1}(yx)^{t}(m)(xy)^{k−1−t}.

With this, we have the formula

ρ((xy)^{k} + q(yx)^{k}) = (Tr_{xy}[f_{0}^{1}])y + qy(Tr_{xy}[f_{0}^{1}]) + x(Tr_{yx}[f_{1}^{1}]) + q(Tr_{yx}[f_{1}^{1}])x.

The maps Tr_{xy} and Tr_{yx} will appear again later.

2.3 Nilpotent elements

We wish to identify nilpotent elements of Hochschild cohomology. This can sometimes be done without knowing the algebra structure of HH^{∗}(A) explicitly, by exploiting a more general result, that is Proposition 4.4 of [12].

Proposition 2.3. Assume K is a field and A is a finite-dimensional K-algebra with radical J. Let η ∈ HH^{n}(A) such that η ⊗_{A} A/J is zero in Ext_{A}^{n}(A/J, A/J). Then η is nilpotent in HH^{∗}(A) with nilpotency index at most the radical length of A.

3 A minimal bimodule resolution

3.1 From now we assume that A is a local algebra of dihedral type as above. In this section we will construct a minimal resolution of A as an A^{e}-module. Here the parameter q is arbitrary, as well the field is arbitrary. As we have seen, the n-th term of a minimal bimodule resolution is free of rank n + 1.

Our aim is to find recursively generators for the kernel of Ω^{n}(A). By the above there will be n + 1 generators, and we will write these as f_{0}^{n}, …, f_{n}^{n}. Then we label generators of ℙ_{n} as

[f_{0}^{n}], …, [f_{n}^{n}]

( each of these is equal to 1⊗1 as an element in A^{e} but of course we need to distinguish different copies). With this, we define the differential d_{n} : ℙ_{n} → ℙ_{n−1} by

[f_{i}^{n}] ↦ f_{i}^{n} ∈ ℙ_{n−1}.

We will show that the f_{i}^{n} are in the kernel of d_{n−1}. To see that they are generators, there are several general arguments in the literature. In this case, one can even see directly that if U_{n} is the sub-bimodule generated by the f_{i}^{n} then U_{n} ⊗_{A} K ≅ Ω_{A}^{n}(K).

In Notation 2.2, we have already fixed the generators of ℙ_{0}, the differential d_{0}, generators for Ω(A), the differential d_{1} and generators for Ω^{2}(A). 3.2 Before giving the general definition, we will explain why Ω^{r}(A)⊗_{A} K is isomorphic to Ω_{A}^{r}(K) for r = 1, 2. Since (−) ⊗_{A} K annihilates the radical, we see

A(f_{0}^{1}) ⊗_{A} K = Ax(1 ⊗ 1) ⊗_{A} K ≅ Ax

and similarly A(f_{1}^{1}) ⊗_{A} K ≅ Ay and we see directly that Ker(d_{0}) ⊗_{A} K ≅ Ax + Ay = Ω_{A}(K).

Similarly consider Ker(d_{1}) ⊗_{A} K, this has submodules

A(f_{0}^{2}) ⊗_{A} K ≅ Ax, and A(f_{2}^{2}) ⊗_{A} K ≅ Ay.

Consider ζ := f_{1}^{2} ⊗_{A} K. This has only two terms, namely

ζ = (xy)^{k−1}x[f_{1}^{1}] ⊗_{A} K + q(yx)^{k−1}y[f_{0}^{1}] ⊗_{A} K.

This generates a 3-dimensional left A-submodule of ℙ_{1}, isomorphic to A/J^{2}. We can also see that the intersection of Aζ with A(f_{0}^{2})⊗_{A}K is equal to yζ, and the intersection with A(f_{2}^{2}) ⊗_{A} K is equal to xζ, both one-dimensional. We see that Ω^{2}(A) ⊗_{A} K ≅ Ω_{A}^{2}(K).

The formulae below for generators may similarly be thought of as being a lift of a one-sided minimal projective resolution.

3.3 We will inductively define explicit minimal generators of Ω^{n}(A), they will be denoted by f_{i}^{n} for 0 ≤ i ≤ n, they are elements of ℙ_{n−1}. Then we fix generators of ℙ_{n} and label them by [f_{i}^{n}] for 0 ≤ i ≤ n. With this, we define the differential d_{n} : ℙ_{n} → ℙ_{n−1}

d_{n} : [f_{i}^{n}] ⟶ f_{i}^{n} (0 ≤ i ≤ n).

Note that the elements f_{i}^{n} will be expressed in terms of [f_{i}^{n−1}].

3.3.1 For any n ≥ 1, we may set

f_{0}^{n} := x[f_{0}^{n−1}] + (−1)^{n}[f_{0}^{n−1}]x, f_{n}^{n} := y[f_{n−1}^{n−1}] + (−1)^{n}[f_{n−1}^{n−1}]y.

We can see directly, using the recipe for the differentials, that these elements are in the kernel of d_{n−1}.

These may be thought of a lift to A^{e} of the process of ’extending by Ax and Ay’. When n = 1, these two elements generate Ω(A). We have also already written down generators for Ω^{2}(A).

3.3.2 We write down the remaining generators for Ω^{3}(A), the kernel of d_{2} : ℙ_{2} → ℙ_{1}.

f_{1}^{3} := ([f_{0}^{2}](yx)^{k−1}y − q^{2}(yx)^{k−1}y[f_{0}^{2}]) − (x[f_{1}^{2}] − q[f_{1}^{2}]x)
f_{2}^{3} = ([f_{1}^{2}]y − qy[f_{1}^{2}]) − ((xy)^{k−1}x[f_{2}^{2}] − q^{2}[f_{2}^{2}](xy)^{k−1}x).

It is straightforward to check that these are in the kernel of d_{2}. By Proposition 2.1 we know that Ω^{3}(A) has four independent generators, and then by general arguments or directly it follows that the f_{i}^{3} for 0 ≤ i ≤ 3 generate Ω^{3}(A).

Now we state the answer for degrees n ≥ 4. Here the field is arbitrary, and q can be any non-zero element of K.

Proposition 3.1. Assume n ≥ 4. Then generators for Ω^{n}(A) may be taken as follows:

(a) Let 1 ≤ i < (n/2). Then

f_{i}^{n} = ([f_{i−1}^{n−1}](yx)^{k−1}y + (−1)^{n}q^{n−i}(yx)^{k−1}y[f_{i−1}^{n−1}]) + (−1)^{i}(x[f_{i}^{n−1}] + (−1)^{n}q^{i}[f_{i}^{n−1}]x).

(b) If n is even and i = (n/2) then we have a generator f_{i}^{n} equal to

f_{i}^{n} = (Tr_{xy}[f_{i−1}^{n−1}])y + q^{i}(yTr_{xy}[f_{i−1}^{n−1}]) + (−1)^{i}x(Tr_{yx}[f_{i}^{n−1}]) + (−1)^{i}q^{i}(Tr_{yx}[f_{i}^{n−1}])x

with the notation as in 2.2.

(c) If (n/2) < i ≤ n − 1 then

f_{i}^{n} = ([f_{i−1}^{n−1}]y + (−1)^{n}q^{n−i}y[f_{i−1}^{n−1}]) + (−1)^{i}((xy)^{k−1}x[f_{i}^{n−1}] + (−1)^{n}q^{i}[f_{i}^{n−1}](xy)^{k−1}x)

(d) For any n, we have

f_{0}^{n} = x[f_{n}^{n−1}] + (−1)^{n}[f_{0}^{n−1}]x, f_{n}^{n} = y[f_{n−1}^{n−1}] + (−1)^{n}[f_{n−1}^{n−1}]y.

Proof Part (d) has already been explained above. Parts (a) to (c) are proved by induction. We give details for part (b) and leave the proof for (a) and (c) to the reader, it is fairly straightforward. Assume i = n/2 and n ≥ 4, and we have the formulae for degree n − 1 as the inductive hypothesis.

We take the stated expression f_{i}^{n}, and we must show that d_{n−1} maps this to zero. Recall that d_{n−1} is obtained by ’removing brackets’, that is, by substituting f_{i−1}^{n−1} and f_{i}^{n−1} into the relevant terms.

We first substitute f_{i−1}^{n−1}, which occurs in the first two terms of (b). Since i − 1 < (n − 1)/2 we must use the formula from part (a), that is

f_{i−1}^{n−1} = [f_{i−2}^{n−2}](yx)^{k−1}y+(−1)^{n−1}q^{n−i}(yx)^{k−1}y[f_{i−2}^{n−2}]+(−1)^{i−1}x[f_{i−1}^{n−2}]+(−1)^{n−i}q^{i−1}[f_{i−1}^{n−2}]x.

(1) We substitute [f_{i−2}^{n−2}](yx)^{k−1}y into f_{i}^{n}. Since y^{2} = 0, all terms from f_{i}^{n} with a factor [f_{i−1}^{n−1}]y become zero. This leaves only one term, namely

q^{i}(yx)^{k−1}y[f_{i−2}^{n−2}](yx)^{k−1}y.

(2) We substitute (−1)^{n−1}q^{n−i}(yx)^{k−1}y[f_{i−2}^{n−2}]. Similarly almost all terms are zero, and we are left with

(−1)^{n−1}q^{n−i}(yx)^{k−1}y[f_{i−2}^{n−2}](yx)^{k−1}y.

Sine i = n/2 we have n − i = i and also (−1)^{n−1} = (−1) and this term cancels against the term in (1).

(3) Substituting (−1)^{i−1}x[f_{i−1}^{n−2}] + (−1)^{n−i}q^{i−1}[f_{i−1}^{n−2}]x into f_{i}^{n} gives the four terms

(−1)^{i−1}Tr_{xy}(x[f_{i−1}^{n−2}])y + (−1)^{i−1}q^{i}yTr_{xy}(x[f_{i−1}^{n−2}])
+(−1)^{n−i}q^{i−1}(Tr_{xy}(f_{i−1}^{n−2}]x)y + (−1)^{n−i}q^{2i−1}y(Tr_{xy}([f_{i−1}^{n−2}]x).

(4) Now we apply d_{n−1} to the terms of f_{i}^{n} with [f_{i}^{n−1}]. We substitute using the formula from part (c), note that i > (n − 1)/2. This has two terms with [f_{i}^{n−2}]. We substitute these, and exactly as in (1) and (2), most of these are zero and the two terms left cancel out.

(5) We substitute the contribution

[f_{i−1}^{n−2}]y + (−1)^{n−1}q^{n−1−i}y[f_{i−1}^{n−2}]

into the last two terms of f_{i}^{n}. These are four terms, namely

(−1)^{i}xTr_{yx}([f_{i−1}^{n−2}]y) + (−1)^{i}q^{i}(Tr_{yx}[f_{i−1}^{n−2}]y)x
+ (−1)^{i+n−1}q^{n−1−i}x(Tr_{yx}(y[f_{i−1}^{n−2}]) + (−1)^{i+n−1}q^{n−1}(Tr_{yx}(y[f_{i−1}^{n−2}])x.

(6) We compare (3) and (5). The first term of (5) cancels against the first term of (3) since x(Tr_{yx}[f_{i−1}^{n−2}]y) = Tr_{xy}(x[f_{i−1}^{n−2}])y. Similarly the last term of (5) cancels agains the last term of (3).

Now consider the second term (3), which is

(−1)^{i−1}q^{i} ∑_{t=0}^{k−1}(yx)^{t+1}[f_{i−1}^{n−2}](yx)^{k−1−t}.

We add to this the second term of (5). All but two terms cancel and we are left with

(∗)  (−1)^{i−1}q^{i}(yx)^{k}[f_{i−1}^{n−2}] + (−1)^{i}q^{i}[f_{i−1}^{n−2}](yx)^{k}.

Similarly adding the third term of (3) and the third term of (5) has just two terms left,

(∗∗)  (−1)^{n−i}q^{i−1}[f_{i−1}^{n−2}](xy)^{k} + (−1)^{n+i−1}q^{i−1}(xy)^{k}[f_{i−1}^{n−2}].

In (*), we substitute (−q)(yx)^{k} = (xy)^{k}, and adding (*) and (**) gives zero, as required. □

4 Homomorphisms and HH^{n}(A)

We assume from now that q is not a root of unity. The aim is to prove the following.

Theorem 4.1. The dimension of HH^{n}(A) is

k + 1 n = 1
k  n = 2
k  n ≥ 3, char(K) |k
k − 1 n ≥ 3, else.

That is, as a bimodule, A has bounded cohomology. When k = 1, this recovers the result of [4]. On the way we will also see the following.

Theorem 4.2. Let n ≥ 2. If φ : Ω^{n}(A) → A is a homomorphism then the image of φ is contained in the radical of A. Hence the class of φ in HH^{n}(A) is nilpotent.

To prove these, we compute homomorphism.

Proposition 4.3. Let r_{n} := dim Hom_{A^{e}}(Ω^{n}(A), A) for n ≥ 0. We have r_{0} = k + 1 and r_{1} = 4k; and

(i) If n = 2t ≥ 2 then r_{n} = 2kn + k. (ii) If n = 2t + 1 ≥ 3 then

r_{n} = \{ \begin{matrix}2kn + 2k & char(K)|k \\ 2kn + (2k − 1) & else \end{matrix}

4.1 Fix n ≥ 1. We may identify

Hom_{A^{e}}(Ω^{n}(A), A) = {φ : ℙ_{n} → A : φ(Ω^{n+1}(A)) = 0}

Recall that ℙ_{n} is the projective A^{e}-module with generators [f_{j}^{n}] for 0 ≤ j ≤ n. A homomorphism from ℙ_{n} to A is therefore determined by the images

ζ_{j} := φ([f_{j}^{n}]) (0 ≤ j ≤ n).

The submodule Ω^{n+1}(A) of ℙ_{n} is generated by the f_{i}^{n+1} for 0 ≤ i ≤ n + 1 as defined in Proposition 3.1 for n + 1 ≥ 4 and for n = 1, 2 in Notation 2.2 and 3.3. To find the homomorphisms we must determine precisely the ζ_{j} such that φ(f_{i}^{n+1}) = 0 for all i, that is, where we get zero if we substitute the ζ_{j} into the formula for f_{i}^{n+1}. Therefore we must solve the following system of equations for ζ_{j}:

We start with n such that n + 1 ≥ 4, then by Proposition 3.1 the equations are

(E_{0}) xζ_{0} + (−1)^{n+1}ζ_{0}x = 0 and (E_{n+1}) ζ_{n}y + (−1)^{n+1}yζ_{n} = 0

(E_{i}) ζ_{i−1}(yx)^{k−1}y+(−1)^{n+1}q^{n+1−i}(yx)^{k−1}yζ_{i−1} = (−1)^{i+1}(xζ_{i}+(−1)^{n+1}q^{i}ζ_{i}x) (1 ≤ i < (n+1)/2)

(E_{i}) ζ_{i−1}y+(−1)^{n+1}q^{n+1−i}yζ_{i−1} = (−1)^{i+1}[(xy)^{k−1}xζ_{i}+(−1)^{n+1}q^{i}ζ_{i}(xy)^{k−1}x] ((n+1)/2 < i ≤ n)

and if n + 1 is even, also i = (n + 1)/2 and

(E_{(n+1)/2}) Tr_{xy}(ζ_{i−1})y + q^{i}yTr_{xy}(ζ_{i−1}) = (−1)^{i+1}(xTr_{yx}(ζ_{i} + q^{i}(Tr_{yx}(ζ_{i})x))

The form of these identities shows that we must understand the spaces

X_{α} := {η ∈ A : xη + αηx = 0}, and Y_{α} := {η ∈ A : ηy + αyη = 0}

for 0 ≠ α ∈ K. One checks that

X_{α} = \{ \begin{matrix}Sp\{1, x\} + X̂_{α} & α = −1 \\ Sp\{x, (yx)^{k−1}y\} + X̂_{α} & α = q \\ Sp\{x\} + X̂_{α} & else \end{matrix}

Y_{α} = \{ \begin{matrix}Sp\{1, y\} + X̂_{α} & α = −1 \\ Sp\{y, (xy)^{k−1}x\} + Ŷ_{α} & α = q \\ Sp\{y\} + Ŷ_{α} & else \end{matrix}

where

X̂_{α} := Sp{(xy)^{i} − α(yx)^{i} : 1 ≤ i ≤ k − 1} ∪ {(xy)^{i}x, 1 ≤ i ≤ k − 1} ∪ {(xy)^{k}}
Ŷ_{α} := Sp{(xy)^{i} − α(yx)^{i} : 1 ≤ i ≤ k − 1} ∪ {(yx)^{i}y, 1 ≤ i ≤ k − 1} ∪ {(xy)^{k}}

and X̂_{α} and Ŷ_{α} each has dimension 2k − 1.

4.1 The proof of 4.3 for n + 1 ≥ 4

Assume n + 1 ≥ 4, where n = 2t or n = 2t + 1. We will prove the following

(1) ζ_{0} = b_{0}x + ζ̂_{0} with ζ̂_{0} ∈ X̂_{(−1)^{n+1}}. and b_{0} ∈ K. Moreover b_{0} = 0 if (−1)^{n} = 1.

(2) For 1 ≤ i ≤ t we have ζ_{i} = b_{1}x+c_{i}(yx)^{k−1}y+ζ̂_{i} where ζ̂_{1} ∈ X̂_{(−1)^{n+1}q^{i}} and b_{i}, c_{i} ∈ K and c_{i} is a function of b_{i−1}, except that if (−1)^{n} = 1 then c_{1} is arbitrary.

(3) ζ_{n} = b_{n}y + ζ̂_{n} with ζ̂_{n} ∈ Ŷ_{(−1)^{n+1}} and b_{n} ∈ K. If (−1)^{n} = 1 then b_{n} = 0.

(4) For t + 1 ≤ i ≤ n − 1, we have ζ_{i} = b_{i}y + c_{i}(xy)^{k−1}x + ζ̂_{i} with ζ̂_{i} ∈ Ŷ_{(−1)^{n+1}q^{i}} and b_{i}, c_{i} ∈ K and where c_{i} is a function of b_{i+1}; except if (−1)^{n} = 1 then c_{n−1} is arbitrary.

When n = 2t there is an additional condition on ζ_{t}, and when n = 2t + 1 we may have an extra condition relating ζ_{t} and ζ_{t+1}:

(5) Suppose n = 2t. Then ζ_{t} = c_{t}(yx)^{k−1}y + c′_{t}(xy)^{k−1}x + ζ̃_{t} with ζ̂_{t} ∈ X̂_{−q^{t}} ∩ Ŷ_{−q^{t}} and c_{t}, c′_{t} ∈ K where c_{t} is a function of b_{t−1} and c′_{t} is a function of b_{t+1}.

(6) Suppose n = 2t + 1. If char(K) does not divide k then b_{t} = (−1)^{t}b_{t+1}. Otherwise b_{t}, b_{t+1} are arbitrary.

Proof One checks that the elements listed satify the identities. We show that these are all.

(1) For identity (E_{0}) we require ζ_{0} ∈ X_{(−1)^{n+1}}, and hence ζ_{0} = a_{0} + b_{0}x + ζ̂_{0} with a_{0}, b_{0} ∈ K and ζ̂_{0} ∈ X̂_{−1}. We substitute into (E_{1}) and get

a_{0}(1 − q^{n})(yx)^{k−1}y + b_{0}(1 + q^{n−1})(xy)^{k} = xζ_{1} + (−1)^{n+1}qζ_{1}x.

The element (yx)^{k−1}y is not of the form xη + qηx for η ∈ A and hence a_{0} = 0, and the claim holds for ζ_{0}. Moreover, the element (xy)^{k} is not of the form xη + qηx for η ∈ A. So if (−1)^{n+1} = 1 then also b_{0} = 0. If (−1)^{n+1} = 1 then ζ_{1} ∈ X_{(−1)^{n+1}q} and we may write ζ_{1} = b_{1}x + c_{1}(yx)^{k−1}Y ζ̂_{1} with b_{1}, c_{1} ∈ K arbitrary, and ζ̂_{1} ∈ X̂_{(−1)^{n+1}q}. Suppose (−1)^{n+1} ≠ 1, then the identity (E_{1}) is satisfied for ζ_{1} = c_{1}(yx)^{k−1}y + ζ′_{1} where c_{1} is a function of b_{0}, and where xζ′_{1} + (−1)^{n+1}qζ′_{1}x = 0. That is ζ′_{1} = b_{1}x + ζ̂_{1} and ζ̂_{1} ∈ X̂_{(−1)^{n+1}q}. This proves the claim for ζ_{1}.

(2) The case i = 1 is done, and we continue by induction: Assume true for i − 1, and i < t. Then we substitute ζ_{i−1} into (E_{i}) and get

b_{i−1}(1 + (−1)^{n}q^{n−i})(xy)^{k} = (−1)^{i+1}(xζ_{i} + (−1)^{n+1}q^{i}ζ_{i}x).

This is satisfied with ζ_{i} = c_{i}(yx)^{k−1}y + ζ′_{i} where c_{i} ∈ K is a function of b_{i−1} and where xζ′_{i} + (−1)^{n+1}q^{i}ζ′_{i}x = 0. Then

ζ′_{i} = b_{i}x + ζ̂_{i}

and b_{i} ∈ K and ζ̂_{i} ∈ X̂_{(−1)^{n+1}q^{i}}.

(3) For (E_{n+1}) we require ζ_{n} ∈ Y_{(−1)^{n+1}} and hence ζ_{n} = a_{n} + b_{n}y + ζ̂_{(−1)^{n+1}}. We substitute into identity (E_{n}) and get

ζ_{n−1}y+(−1)^{n+1}qζ_{n−1}y = (−1)^{n+1}[a_{n}(1+(−1)^{n+1}q^{n})(xy)^{k−1}x+b_{n}(1+(−1)^{n}q^{n−1})(xy)^{k}].

As in (1) we deduce a_{n} = 0. If (−1)^{n} = 1 then ζ_{n} ∈ Y_{(−1)^{n+1}q} and ζ_{n−1} = b_{n−1}x + c_{n−1}(yx)^{k−1}y + ζ̂_{n−1} with b_{n−1} and c_{n−1} ∈ K and ζ̂_{n−1} ∈ Ŷ_{(−1)^{n+1}q}. Otherwise we get the same description but now c_{n−1} is a function of b_{n−1}.

(4) By downwards induction, one obtains the stated expression for ζ_{i} from (E_{i+1}) for t + 1 ≤ i ≤ n.

(5) Assume n = 2t. Then we have an expression for ζ_{t} from the last step of (4). We also have an expression for ζ_{t} from the last step of (3). These must be the same, and we deduce that it has no terms with x, y and that it is of the stated form.

(6) Assume n = 2t + 1. Then by (1) to (4) we have expressions for all ζ_{i}, and all identities (E_{j}) for j ≠ t + 1 are satisfied. We substitute ζ_{t} and ζ_{t+1} into (E_{t+1}). This gives

b_{t}[Tr_{xy}(x)y + q^{t+1}yTr_{xy}(x)] = (−1)^{t+2}b_{t+1}[xTr_{yx}(y) + q^{t+1}Tr_{yx}(y)x].

That is

kb_{t}(1 − q^{t})(xy)^{k} = (−1)^{t}kb_{t+1}(1 − q^{t})(xy)^{k}.

If char(K) does not divide k, we need b_{t} = (−1)^{t}b_{t+1}. Otherwise b_{t}, b_{t+1} are arbitrary.

We compute now the dimension to complete the proof of Propositon 4.3. We have n = 2t or 2t + 1 and n ≥ 4.

(1) Assume n is even. For each i ≠ t with 0 ≤ i ≤ n we have the summand ζ̂_{i} which can be arbitrary in a space of dimension 2k−1. We have also ζ̃_{t} in a space of dimension k, the intersection of X̂_{−q^{t}} ∩ Ŷ_{−q^{t}}. Moreover we have n independent scalar parameters (in all cases). This gives in total

r_{n} = n(2k − 1) + k + n = 2kn + k,

as stated.

(2) Now assume n is odd. Then for each i with 0 ≤ i ≤ n we have the summand ζ̂_{i} which can be arbitrary in a space of dimension 2k − 1. Moreover we have n + 1 independent scalars if char(K) divides k and n independent scalars otherwise. So the dimension is either r_{n} = (n + 1)(2k − 1) + (n + 1) = (n + 1)2k or is (n + 1)(2k) − 1.

4.2 Small cases

4.2.0 Let n = 0. The dimension r_{0} = dim Z(A) is equal to the dimension of X_{−1} ∩Y_{−1} which is equal to k + 1, see 4.1.

4.2.1 We find r_{1}. Let φ : ℙ_{1} → A be a homomorphism such that φ(Ω^{2}(A)) = 0. Then φ is determined by elements φ([f_{i}^{1}]) = ζ_{i} in A, for i = 0, 1. We have φ(Ω^{2}(A)) = 0 if and only if the follwoing identities (E_{i}) hold, obtained from Notation 2.2:

(E_{0}) xζ_{0} + ζ_{0}x = 0 and (E_{2}) ζ_{1}y + yζ_{1} = 0
(E_{1}) Tr_{xy}(ζ_{0})y + qyTr_{xy}(ζ_{0}) = −(xTr_{yx}(ζ_{1}) + qTr(ζ_{1})x).

We will show that these are satisfied if and only if

(1) ζ_{0} = b_{0}x + ζ̂_{0} with b_{0} ∈ K and ζ̂_{0} ∈ X̂_{1} and

(2) ζ_{1} = b_{1}y + ζ̂_{1} with b_{1} ∈ K and ζ̂_{1} ∈ Ŷ_{1}. These satify the (E_{i}) and we show that there are no others. By (E_{0}) we may write ζ_{0} = a_{0} + b_{0}x + ζ̂_{0} where a_{0}, b_{0} ∈ K and where ζ̂_{0} ∈ X̂_{1}. Similarly by (E_{2}) we have ζ_{1} = a_{1} + b_{1}y + ζ̂_{1} with a_{1}, b_{1} ∈ K and where ζ̂_{1} ∈ Ŷ_{1}. We substitute ζ_{0} and ζ_{1} into (E_{1}), the terms with ζ̂_{i} become zero. The terms with b_{0} and b_{1} satisfy

kb_{0}((xy)^{k} + q(yx)^{k}) = −[kb_{1}((xy)^{k} + q(yx)^{k})],

namely both sides are zero because of the relation. The terms with with a_{0} and a_{1} must satisfy

a_{0}(1 + q)(yx)^{k−1}y = −[a_{1}(1 + q)(xy)^{k−1}y],

and hence a_{0} = 0 = a_{1}. With (1) and (2) we have

r_{1} = dim X̂_{1} + dim Ŷ_{1} + 2 = 2(2k − 1) + 2 = 4k.

4.2.2 We find r_{2}. Consider a homomorphism φ : ℙ_{2} → A which maps Ω^{3}(A) to zero. The homomorphism φ is determined by φ[f_{i}^{2}] = ζ_{i} ∈ A for 0 ≤ i ≤ 2. Then φ(Ω^{3}(A)) = 0 if and only if the following identities (E_{i}) hold, obtained from 3.3:

(E_{0}) xζ_{0} − ζ_{0}x = 0 and (E_{3}) ζ_{2}y − yζ_{2} = 0;
(E_{1}) ζ_{0}(yx)^{k−1}y − q^{2}(yx)^{k−1}yζ_{0} = xζ_{1} − qζ_{1}x;
(E_{2}) ζ_{1}y − qyζ_{1} = (xy)^{k−1}xζ_{2} − q^{2}ζ_{2}(xy)^{k−1}x.

Precisely as in the proof for the general case, one shows that these hold if and only if (1) ζ_{0} = b_{0}x + ζ̂_{0} with b_{0} ∈ K and ζ̂_{0} ∈ X̂_{−1}, and b_{0} = 0 if char(K) = 2.

(2) ζ_{2} = b_{2}y + ζ̂_{2} with b_{2} ∈ K and ζ̂_{2} ∈ Ŷ_{−1}, and b_{2} = 0 if char(K) = 2.

(3) ζ_{1} = c_{1}(yx)^{k−1}y + c′_{1}(xy)^{k−1}x + ζ̃_{1} with ζ̃_{1} ∈ X̂_{−q} ∩ Ŷ_{−q}, and where c_{1}, c′_{1} ∈ K. If char(K) ≠ 2 then c_{1} is a function of b_{0} and c′_{1} is a function of b_{2}. If char(K) = 2 then c_{1}, c′_{1} are arbitrary.

With this, we note that the number of independent scalar parameters is the same whether or not the characteristic is 2, and we compute that

r_{2} = dim X̂_{−1} + dim Ŷ_{−1} + 2 + dim(X̂_{−q} ∩ Ŷ_{−q} = 4kn + k.

We prove Theorem 4.1, using the exact sequence

0 → Hom_{A^{e}}(Ω^{n−1}(A), A) → Hom_{A^{e}}(ℙ_{n−1}, A) → Hom_{A^{e}}(Ω^{n}(A), A) → HH^{n}(A) → 0.

The dimension of Hom_{A^{e}}(ℙ_{n−1}, A) is 4kn and hence

dim HH^{n}(A) = r_{n} + r_{n−1} − 4kn

and we get the stated answer.

We prove Theorem 4.2. The description of the most general homomorphism Ω^{n}(A) → A in 4.3 and in the small cases shows that its image is contained in the radical of A. Hence by Proposition 2.3 every element of positive degree is in HH^{∗}(A) is nilpotent.

References

[1] D.J. Benson, Representation rings of finite groups, Representations of algebras (Durham, 1985), 181–199, London Math. Soc. Lecture Note Ser. 116, Cambridge Univ. Press, Cambridge, 1986.

[2] D. J. Benson, Modular representation theory: new trends and methods. Lecture Notes in Mathematics, 1081 Springer-Verlag, Berlin, 1984.

[3] D. J. Benson, Representations and cohomology I, Cambridge studies in advanced mathematics 30, 1991

[4] R. Buchweitz, E.L. Green, D. Madsen and Ø. Solberg, Finite Hochschild cohomology without finite global dimension. Math. Res. Lett. 12 (2005), no. 5-6, 805–816.

[5] K. Erdmann, Blocks of tame representation type and related algebras, Springer Lecture Notes in Mathematics 1428 (1990).

[6] K. Erdmann, M. Holloway, N. Snashall, Ø. Solberg, R. Taillefer, Support varieties for selfinjective algebras, K-Theory, 33 (2004), 67–87.

[7] Erdmann, K., Solberg, Ø., Radical cube zero weakly symmetric algebras and support varieties, J. Pure Appl. Algebra 215(2011), 185-200.

[8] K. Erdmann, Algebras with non-periodic bounded modules, to appear in J. Algebra.

[9] D. Happel, Hochschild cohomology of finite-dimensional algebras, Springer Lecture Notes in Mathematics 1404 (1989), 108-126.

[10] T. Holm, Hochschild cohomology of tame blocks. J. Algebra 271 (2004)(2), 798-826.

[11] C. M. Ringel, The indecomposable representations of the dihedral 2groups. Math. Ann. 214 (1975), 19-34.

[12] N. Snashall, Ø. Solberg, Support varieties and Hochschild cohomology rings. Proc. London Math. Soc. (3) 88 (2004), no. 3, 705–732.

[13] Ø. Solberg, Support varieties for modules and complexes. Trends in representation theory of algebras and related topics, 239–270, Contemp. Math., 406, Amer. Math. Soc., Providence, RI, 2006.

Karin Erdmann, Mathematical Institute, University of Oxford, ROQ Woodstock Road, Oxford OX2 6GG, UK

email: erdmann@maths.ox.ac.uk