HOMOLOGY AND COHOMOLOGY OF QUANTUM COMPLETE INTERSECTIONS

PETTER ANDREAS BERGH & KARIN ERDMANN

Dedicated to Luchezar Avramov on the occasion of his sixtieth birthday

Abstract. We construct a minimal projective bimodule resolution for every finite dimensional quantum complete intersection of codimension two. Then we use this resolution to compute both the Hochschild cohomology and homology for such an algebra. In particular, we show that the cohomology vanishes in high degrees, while the homology is always nonzero.

1. 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], Benson, Erdmann and Holloway defined and studied a new rank variety theory for modules over finite dimensional quantum complete intersections. For this theory to work, it is essential that the commutators defining the quantum complete intersection be roots of unity, so that a linear combination of the generators behave itself as a generator. In this setting, at least for quantum complete intersections of codimension two, the Hochschild cohomology ring is infinite dimensional, and a priori there might be connections between rank varieties and the support varieties defined by Snashall and Solberg (cf. [SnS], [EHSST]).

Whether or not the higher Hochschild cohomology groups of a finite dimensional algebra of infinite global dimension can vanish, known as “Happel’s question”, was unknown until the appearance of [BGMS]. In that paper, the authors constructed a four dimensional selfinjective algebra whose total Hochschild cohomology is five dimensional, thus giving a negative answer to Happel’s question. The algebra they constructed is the smallest possible noncommutative quantum complete intersection.

In this paper we study finite dimensional quantum complete intersections of codimension two. For such an algebra, we construct a minimal projective bimodule resolution, and use this to compute the Hochschild homology and cohomology. In particular, we show that the higher Hochschild cohomology groups vanish if and only if the commutator element is not a root of unity, whereas the Hochschild homology groups never vanish. Thus we obtain a large class of algebras having the same homological properties as the algebra used in [BGMS].

2000 Mathematics Subject Classification. 16E40, 16S80, 16U80, 81R50.

Key words and phrases. Hochschild (co)homology, quantum complete intersections.

2. The minimal projective resolution

Throughout this paper, let k be a field and q ∈ k a nonzero element. In the main results, this element is assumed not to be a root of unity, implying indirectly that k is an infinite field. We fix two integers a, b ≥ 2, and denote by A the k-algebra

A = k⟨X, Y ⟩/(X^{a}, XY − qY X, Y^{b}).

This is a finite dimensional algebra of dimension ab, and it is justifiably a quantum complete intersection of codimension 2; it is the quotient of the quantum symmetric algebra k⟨X, Y ⟩/(XY −qY X) by the quantum regular sequence X^{a}, Y^{b} (as defined in [AGP, Section 2]). We denote the generators of A by x and y, and use the set

{y^{i}x^{j}}_{0≤i<b, 0≤j<a}

as a k-basis. The opposite algebra of A is denoted by A^{op}, and the enveloping algebra A ⊗_{k} A^{op} by A^{e}.

We now construct explicitly a minimal projective bimodule resolution

ℙ: ⋯ → P_{2} ⟶^{d_{2}} P_{1} ⟶^{d_{1}} P_{0} ⟶^{μ} A → 0,

in which P_{n} is free of rank n + 1, viewing the bimodules as left A^{e}-modules. The generators 1 ⊗ 1 of P_{n} are labeled ϵ(i, j) for i, j ≥ 0, such that

P_{n} = ⨁_{i+j=n} A^{e}ϵ(i, j).

For each s ≥ 0, define the following four elements of A^{e}:

τ_{1}(s) = q^{s}(1 ⊗ x) − (x ⊗ 1)
τ_{2}(s) = (1 ⊗ y) − q^{s}(y ⊗ 1)
γ_{1}(s) = ∑_{j=0}^{a−1} q^{js}(x^{a−1−j} ⊗ x^{j})

γ_{2}(s) = ∑_{j=0}^{b−1} q^{js}(y^{j} ⊗ y^{b−1−j}).

Let P_{0} ⟶^{μ} A be the multiplication map w ⊗ z ↦ wz. The kernel of this map is generated by τ_{1}(0) and τ_{2}(0). Now let R_{1} and R_{2} be the commutative subalgebras of A generated by x and y, respectively. The annihilator of τ_{i}(0), viewed as an element of R_{i}^{e}, is γ_{i}(0), and the complex

⋯ → R_{i}^{e} ⟶^{τ_{i}(0)} R_{i}^{e} ⟶^{γ_{i}(0)} R_{i}^{e} ⟶^{τ_{i}(0)} R_{i}^{e} ⟶^{μ} R_{i} → 0

is a minimal projective bimodule resolution of R_{i} (cf. [Hol]).

In general, given any algebra Γ and an automorphism Γ ⟶^{ψ} Γ, we may endow every Γ-module X with a new module structure by restricting scalars via ψ. In this way, we obtain a new module _{ψ}X, whose underlying set is the same as that of X, but where scalar multiplication is given by

γ · x = ψ(γ)x

for γ ∈ Γ and x ∈ X. This new module is the twist of X with respect to ψ. A homomorphism X → Y of Γ-modules induces a homomorphism _{ψ}X → _{ψ}Y of twisted modules.

Now for i = 1, 2, define an algebra automorphism R_{i}^{e} ⟶^{σ_{i}} R_{i}^{e} by

σ_{1} : \begin{matrix}x ⊗ 1 ↦ x ⊗ 1 \\ 1 ⊗ x ↦ q(1 ⊗ x) \end{matrix} σ_{2} : \begin{matrix}y ⊗ 1 ↦ q(y ⊗ 1) \\ 1 ⊗ y ↦ 1 ⊗ y. \end{matrix}

When we twist the above resolution of R_{i} by the automorphism σ_{i}^{s} for some s ≥ 0, then multiplication by τ_{i}(0) and γ_{i}(0) become multiplication by τ_{i}(s) and γ_{i}(s), respectively. We denote this twisted resolution by R_{i}(s).

We now define a double complex

\xymatrix{⋮ \ar[d] & ⋮ \ar[d] & ⋮ \ar[d] \\ A^{e}ϵ(0, 2) \ar[d] & A^{e}ϵ(1, 2) \ar[d] \ar[l] & A^{e}ϵ(2, 2) \ar[d] \ar[l] & ⋯ \ar[l] \\ A^{e}ϵ(0, 1) \ar[d] & A^{e}ϵ(1, 1) \ar[d] \ar[l] & A^{e}ϵ(2, 1) \ar[d] \ar[l] & ⋯ \ar[l] \\ A^{e}ϵ(0, 0) & A^{e}ϵ(1, 0) \ar[l] & A^{e}ϵ(2, 0) \ar[l] & ⋯ \ar[l]}

whose total complex ℙ turns out to be the projective bimodule resolution we are seeking. Along row 2s we use the resolution R_{1}(bs), and along row 2s + 1 we use the resolution R_{1}(bs + 1). Explicitly, the row maps are given by

ϵ(2r, 2s) ↦ γ_{1}(bs)ϵ(2r − 1, 2s)
ϵ(2r + 1, 2s) ↦ τ_{1}(bs)ϵ(2r, 2s)
ϵ(2r, 2s + 1) ↦ γ_{1}(bs + 1)ϵ(2r − 1, 2s + 1)
ϵ(2r + 1, 2s + 1) ↦ τ_{1}(bs + 1)ϵ(2r, 2s + 1).

Similarly, along column 2r we use the resolution R_{2}(ar), and along column 2r + 1 we use the resolution R_{2}(ar + 1), introducing a sign in the odd columns. The column maps are therefore given by

ϵ(2r, 2s) ↦ γ_{2}(ar)ϵ(2r, 2s − 1)
ϵ(2r, 2s + 1) ↦ τ_{2}(ar)ϵ(2r, 2s)
ϵ(2r + 1, 2s) ↦ −γ_{2}(ar + 1)ϵ(2r + 1, 2s − 1)
ϵ(2r + 1, 2s + 1) ↦ −τ_{2}(ar + 1)ϵ(2r + 1, 2s).

It is straightforward to verify that these maps indeed define a double complex; all the four different types of squares commute. The transpose of the matrices defining the maps in the resulting double complex are given by

( \begin{matrix}γ_{1}(0) & −τ_{2}(as+1) & 0 & 0 & 0 & 0 & ··· & 0 \\ 0 & τ_{1}(1) & γ_{2}(as) & 0 & 0 & 0 & ··· & 0 \\ 0 & 0 & γ_{1}(b) & −τ_{2}(a[s−1]+1) & 0 & 0 & ··· & 0 \\ 0 & 0 & 0 & τ_{1}(b+1) & γ_{2}(a[s−1]) & 0 & ··· & 0 \\ ⋮ & ⋮ & & ... & ... & ... & ... & ⋮ \\ 0 & 0 & ··· & & 0 & γ_{1}(bs) & −τ_{2}(1) & 0 \\ 0 & 0 & ··· & & 0 & 0 & τ_{1}(bs+1) & γ_{2}(0) \end{matrix} )

for the map at stage 2(s + 1), and

( \begin{matrix}τ_{1}(0) & τ_{2}(as) & 0 & 0 & 0 & 0 & ··· & 0 \\ 0 & γ_{1}(1) & −γ_{2}(a[s−1]+1) & 0 & 0 & 0 & ··· & 0 \\ 0 & 0 & τ_{1}(b) & τ_{2}(a[s−1]) & 0 & 0 & ··· & 0 \\ 0 & 0 & 0 & γ_{1}(b+1) & −γ_{2}(a[s−2]+1) & 0 & ··· & 0 \\ ⋮ & ⋮ & & ... & ... & ... & ... & ⋮ \\ 0 & 0 & ··· & & 0 & γ_{1}(b[s−1]+1) & −γ_{2}(1) & 0 \\ 0 & 0 & ··· & & 0 & 0 & τ_{1}(bs) & τ_{2}(0) \end{matrix} )

for the map at stage 2s + 1.

Now, for each n ≥ 0, denote the generator ϵ(i, n − i) by f_{i}^{n}, so that the nth bimodule in the total complex ℙ is

P_{n} = ⨁_{i=0}^{n} A^{e}f_{i}^{n},

the free A^{e}-module of rank n + 1 having generators {f_{0}^{n}, f_{1}^{n}, …, f_{n}^{n}}. Then the maps P_{n} ⟶^{d_{n}} P_{n−1} in ℙ are given by

d_{2t}: f_{i}^{2t} ↦ \{ \begin{matrix}γ_{2}(\frac{ai}{2})f_{i}^{2t−1} & + & γ_{1}(\frac{2bt−bi}{2})f_{i−1}^{2t−1}, & for i even \\ −τ_{2}(\frac{ai−a+2}{2})f_{i}^{2t−1} & + & τ_{1}(\frac{2bt−bi−b+2}{2})f_{i−1}^{2t−1}, & for i odd \end{matrix}

d_{2t+1} : f_{i}^{2t+1} ↦ \{ \begin{matrix}τ_{2}(\frac{ai}{2})f_{i}^{2t} & + & γ_{1}(\frac{2bt−bi+2}{2})f_{i−1}^{2t}, & for i even \\ −γ_{2}(\frac{ai−a+2}{2})f_{i}^{2t} & + & τ_{1}(\frac{2bt−bi+b}{2})f_{i−1}^{2t}, & for i odd, \end{matrix}

where we use the convention f_{−1}^{n} = f_{n+1}^{n} = 0. The following result shows that the complex is exact.

Proposition 2.1. The complex ℙ is exact, and is therefore a minimal projective resolution

ℙ: ⋯ → P_{2} ⟶^{d_{2}} P_{1} ⟶^{d_{1}} P_{0} ⟶^{μ} A → 0

of the left A^{e}-module A.

Proof. We will show that the complex ℙ ⊗_{A} k is exact, and a minimal projective resolution of the A-module k. Then the arguments in [GrS] show that the complex ℙ is exact.

When applying − ⊗_{A} k to A^{e} = A ⊗_{k} A^{op}, the elements x and y in A^{op} become zero, and so the elements τ_{i}(s) ⊗ 1 and γ_{i}(s) ⊗ 1 are just given by

τ_{1}(s) ⊗ 1 = −(x ⊗ 1)
τ_{2}(s) ⊗ 1 = −q^{s}(y ⊗ 1)
γ_{1}(s) ⊗ 1 = (x^{a−1} ⊗ 1)
γ_{2}(s) ⊗ 1 = q^{(b−1)s}(y^{b−1} ⊗ 1).

We shall identify these elements with −x, −q^{s}y, x^{a−1} and q^{(b−1)s}y^{b−1}, respectively. Moreover, whenever the commutator element q is involved, its precise power does not affect the dimensions of the vector spaces we are considering, so we shall write q^{∗} for simplicity.

Fix a number n ≥ 0. The free bimodule P_{n} has generators ϵ(i, j), with n = i + j and i, j ≥ 0. When the degree is not ambiguous, we shall denote the element

ϵ(i, j) ⊗ 1 ∈ P_{n} ⊗_{A} k by e_{j}, and we shall denote the map P_{n} ⊗_{A} k ⟶^{d_{n}⊗1} P_{n−1} ⊗_{A} k

by d̂_{n}. Moreover, we denote by U_{i} the left A-submodule of P_{n−1} ⊗_{A} k generated by d̂_{n}(e_{j}), so that

Im d̂_{n} = U_{0} + ⋯ + U_{n} ⊆ P_{n−1} ⊗_{A} k.

We now compute the dimensions of these modules U_{i}. Assume first that n is even. Then

U_{0} = Ax^{a−1}e_{0}
U_{i} = A [(q^{∗}y)e_{i−1} + (q^{∗}x)e_{i}] , for odd 0 < i < n
U_{i} = A [(q^{∗}y^{b−1})e_{i−1} + (q^{∗}x^{a−1})e_{i}] , for even 0 < i < n
U_{n} = Ay^{b−1}e_{n−1},

and so we see that dim U_{0} = b, dim U_{n} = a, and otherwise dim U_{i} = ab − 1 and dim U_{j} = a + b + 1 for i odd and j even. When n is odd, then

U_{0} = Axe_{0}
U_{i} = A [(−q^{∗}y)e_{i−1} + (q^{∗}x^{a−1})e_{i}] , for odd 0 < i < n
U_{i} = A [(q^{∗}y^{b−1})e_{i−1} + (q^{∗}x)e_{i}] , for even 0 < i < n
U_{n} = Aye_{n−1},

and so in this case we see that dim U_{0} = b(a − 1), dim U_{n} = a(b − 1), and otherwise dim U_{i} = a(b − 1) + 1 and dim U_{j} = b(a − 1) + 1 for i odd and j even.

Our aim is to compute the dimensions of various intersections and sums obtained from the modules U_{i}. In order to do this, we need the following fact: for any elements z_{1}, z_{2} ∈ A, the implication

(†)  z_{1}x^{s} = z_{2}y^{t} ⟹ z_{1} = v_{1}y^{t} + w_{1}x^{a−s} and z_{2} = v_{2}x^{s} + w_{2}y^{b−t}

holds, where v_{i} and w_{i} are some elements in A depending on z_{1} and z_{2}. To see this, write z_{1} = g_{0} + g_{1}y + ⋯ + g_{b−1}y^{b−1} and z_{2} = h_{0} + h_{1}y + ⋯ + h_{b−1}y^{b−1}, where the g_{i} and h_{i} are polynomials in x. Then

∑_{i} h_{i}y^{t+i} = z_{2}y^{t} = z_{1}x^{s} = ∑_{j}(q^{−js}g_{j}x^{s})y^{j},

and comparing the coefficients of y^{j}, we find that g_{j}x^{s} = 0 for j < t. Therefore, for these values of j, the polynomial g_{j} must be a multiple of x^{a−s}. Then we can write ∑_{j<t} g_{j}y^{j} = w_{1}x^{a−s} for some w_{1} ∈ A, giving

z_{1} = ∑_{j<t} g_{j}y^{j} + ∑_{j≥t} g_{j}y^{j} = w_{1}x^{a−s} + v_{1}y^{t},

where v_{1} = ∑_{j≥t} g_{j}y^{j−t}. This proves the statement for z_{1}, and the proof for z_{2} is similar.

We now compute the intersections of pairs of the modules U_{i}. Suppose n is even, and fix an even integer 0 ≤ j ≤ n. If u belongs to U_{j} ∩U_{j+1}, then there are elements z_{1}, z_{2} ∈ A such that

u = z_{1} [(q^{∗}y^{b−1})e_{j−1} + (q^{∗}x^{a−1})e_{j}] = z_{2} [(q^{∗}y)e_{j} + (q^{∗}x)e_{j+1}] .

The coefficients of e_{j−1} and e_{j+1} must be zero, whereas those of e_{j} must be equal, giving (z_{1}q^{∗})x^{a−1} = (z_{2}q^{∗})y. By (†), there are elements v_{1}, v_{2}, w_{1}, w_{2} ∈ A such that

z_{1}q^{∗} = v_{1}y + w_{1}x, z_{2}q^{∗} = v_{2}x^{a−1} + w_{2}y^{b−1},

hence u ∈ Ayx^{a−1}e_{j}. Conversely, any element in Ayx^{a−1}e_{j} belongs to U_{j} ∩ U_{j+1}, showing U_{j} ∩ U_{j+1} = Ayx^{a−1}e_{j}, and that the dimension of this intersection is b − 1.

Similarly, we compute three other types of intersections using the same method, and record everything in the following table:

n  j  intersection

dimension

even even U_{j} ∩ U_{j+1} = Ayx^{a−1}e_{j}  b − 1
even odd U_{j} ∩ U_{j+1} = Ay^{b−1}xe_{j}  a − 1
odd even U_{j} ∩ U_{j+1} = Ayxe_{j}  (a − 1)(b − 1)
odd odd U_{j} ∩ U_{j+1} = Ay^{b−1}x^{a−1}e_{j} 1

Next we show that the equality

(U_{0} + U_{1} + ⋯ + U_{s}) ∩ U_{s+1} = U_{s} ∩ U_{s+1}

holds for any s ≥ 1. Suppose first that both n and s are even. The inclusion U_{s} ∩ U_{s+1} ⊆ (U_{0} + U_{1} + ⋯ + U_{s}) ∩ U_{s+1} obviously holds, so suppose u is an element belonging to (U_{0} + U_{1} + ⋯ + U_{s}) ∩ U_{s+1}. Then u can be written as

u = z_{0}x^{a−1}e_{0} + z_{1} [(q^{∗}y)e_{0} + (q^{∗}x)e_{1}] + ⋯ + z_{s} [(q^{∗}y^{b−1})e_{s−1} + (q^{∗}x^{a−1})e_{s}]
= z_{s+1} [(q^{∗}y)e_{s} + (q^{∗}x)e_{s+1}] ,

in which the coefficient of e_{s+1} must be zero. Moreover, the coefficients of e_{s} must be equal, i.e. (z_{s+1}q^{∗})y = (z_{s}q^{∗})x^{a−1}, and so from (†) we see that there exist elements v, w ∈ A such that z_{s+1} = vx^{a−1} + wy^{b−1}. This gives

u = (vx^{a−1} + wy^{b−1})q^{∗}ye_{s} = vq^{∗}x^{a−1}ye_{s},

and we see directly that u belongs to U_{s} ∩U_{s+1}. The equality (U_{0} +U_{1} +⋯+U_{s})∩ U_{s+1} = U_{s} ∩ U_{s+1} therefore holds when n and s are even, and the same arguments show that the equality holds regardless of the parity of n and s.

Using what we just showed, an induction argument gives the equality

dim(U_{0} + ⋯ + U_{s}) = ∑_{i=0}^{s} dim U_{i} − ∑_{i=0}^{s−1} dim(U_{i} ∩ U_{i+1}).

Then by counting dimensions, we see that the dimension of Im d̂_{n} is given by

dim Im d̂_{n} = \{ \begin{matrix}tab + 1, & when n = 2t \\ (t + 1)ab − 1, & when n = 2t + 1. \end{matrix}

The exactness of the complex ℙ ⊗_{A} k now follows easily; the image of d̂_{n+1} is contained in the kernel of d̂_{n}, and the dimension of P_{n} ⊗_{A} k is ab(n + 1). It follows that Im d̂_{n+1} and Ker d̂_{n} are of the same dimension.

As for minimality, it suffices to show that Im d̂_{n} does not have a projective summand. This follows from the description of this module as the sum of the U_{i}. Namely, we see directly that the element y^{b−1}x^{a−1} ∈ A annihilates each U_{i}, and therefore also Im d̂_{n}. □

3. Hochschild (co)homology

Having obtained the bimodule resolution of A = k⟨X, Y ⟩/(X^{a}, XY − qY X, Y^{b}), we turn now to its Hochschild homology and cohomology groups. Let B be a bimodule, and recall that the Hochschild homology of A with coefficients in B, denoted HH_{∗}(A, B), is the k-vector space

HH_{∗}(A, B) = Tor_{∗}^{A^{e}}(B, A),

where B is viewed as a right A^{e}-module. Dually, the Hochschild cohomology of A with coefficients in B, denoted HH^{∗}(A, B), is the k-vector space

HH^{∗}(A, B) = Ext_{A^{e}}^{∗}(A, B),

where B is viewed as a left A^{e}-module. Of particular interest is the case B = A, namely the Hochschild homology and cohomology of A, denoted HH_{∗}(A) and HH^{∗}(A), respectively. Now by viewing A and B as left A^{e}-modules, it follows from [CaE, VI.5.3] that D (HH^{∗}(A, B)) is isomorphic as a vector space to Tor_{∗}^{A^{e}} (D(B), A), where D denotes the usual k-dual Hom_{k}(−, k). In particular, by taking B = A we see that dim_{k} HH^{n}(A) = dim_{k} Tor_{n}^{A^{e}} (D(A), A) for all n ≥ 0.

Our algebra A is Frobenius; it is easy to check that the map A ⟶^{ϕ} D(A) of left A-modules defined by

ϕ(1): ∑_{\begin{matrix}0≤j≤b−1 \\ 0≤i≤a−1 \end{matrix}} c_{j,i}y^{j}x^{i} ↦ c_{b−1,a−1}

is an isomorphism. To such a Frobenius isomorphism one can always associate a k-algebra automorphism A ⟶^{ν} A, a Nakayama automorphism, with the (defining) property that w ·ϕ(1) = ϕ(1)·ν(w) for all elements w ∈ A. In our case the elements x and y generate A, and since x · ϕ(1) = ϕ(1) · q^{1−b}x and y · ϕ(1) = ϕ(1) · q^{a−1}y, we see that the automorphism defined by

ν: \begin{matrix}x ↦ q^{1−b}x \\ y ↦ q^{a−1}y \end{matrix}

is a Nakayama automorphism. The composite map ϕ ∘ ν^{−1} is then a bimodule isomorphism between the right A^{e}-modules _{ν}A_{1} and D(A), where the scalar action on _{ν}A_{1} is given by u · (w_{1} ⊗ w_{2}) = ν(w_{2})uw_{1}. Consequently, we see that

dim_{k} HH^{n}(A) = dim_{k} Tor_{n}^{A^{e}}(_{ν}A_{1}, A)

for all n ≥ 0.

Now let α, β ∈ k be nonzero scalars, and let A ⟶^{ψ} A be the automorphism defined by x ↦ αx and y ↦ βy. Tensoring the deleted projective bimodule resolution ℙ_{A} with the right A^{e}-module _{ψ}A_{1}, we obtain an isomorphism

\xymatrix{⋯ \ar[r] & _{ψ}A_{1} ⊗_{A^{e}} P_{n+1} \ar[d]^{≀} \ar[r]^{1⊗d_{n+1}} & _{ψ}A_{1} ⊗_{A^{e}} P_{n} \ar[d]^{≀} \ar[r]^{1⊗d_{n}} & _{ψ}A_{1} ⊗_{A^{e}} P_{n−1} \ar[d]^{≀} \ar[r] & ⋯ \\ ⋯ \ar[r] & ⊕_{i=0}^{n+1}(_{ψ}A_{1})e_{i}^{n+1} \ar[r]^{δ_{n+1}^{ψ}} & ⊕_{i=0}^{n}(_{ψ}A_{1})e_{i}^{n} \ar[r]^{δ_{n}^{ψ}} & ⊕_{i=0}^{n−1}(_{ψ}A_{1})e_{i}^{n−1} \ar[r] & ⋯}

of complexes, where {e_{0}^{n}, e_{1}^{n}, …, e_{n}^{n}} is the standard generating set of n + 1 copies of _{ψ}A_{1}. The map δ_{n}^{ψ} is then given by

δ_{2t}^{ψ}: y^{u}x^{v}e_{i}^{2t} ↦

K_{1}^{ψ}(t, i, u, v)y^{u+b−1}x^{v}e_{i}^{2t−1} + K_{2}^{ψ}(t, i, u, v)y^{u}x^{v+a−1}e_{i−1}^{2t−1},  for i even
[q^{\frac{ai−a+2+2v}{2}} − β] y^{u+1}x^{v}e_{i}^{2t−1} + [αq^{\frac{2bt−bi−b+2+2u}{2}} − 1] y^{u}x^{v+1}e_{i−1}^{2t−1}, for i odd

δ_{2t+1}^{ψ}: y^{u}x^{v}e_{i}^{2t+1} ↦

[β − q^{\frac{ai+2v}{2}} ] y^{u+1}x^{v}e_{i}^{2t} + K_{3}^{ψ}(t, i, u, v)y^{u}x^{v+a−1}e_{i−1}^{2t},  for i even

K_{4}^{ψ}(t, i, u, v)y^{u+b−1}x^{v}e_{i}^{2t} + [αq^{\frac{2bt−bi+b+2u}{2}} − 1] y^{u}x^{v+1}e_{i−1}^{2t}, for i odd,

where we use the convention e_{−1}^{n} = e_{n+1}^{n} = 0. Here the elements K_{j}^{ψ}(t, i, u, v), which are scalars whose values depend on the parameters ψ, t, i, u and v, are defined as follows:

K_{1}^{ψ}(t, i, u, v) = ∑_{j=0}^{b−1} q^{j( \frac{ai+2v}{2} )}β^{b−1−j}

K_{2}^{ψ}(t, i, u, v) = ∑_{j=0}^{a−1} q^{j( \frac{2bt−bi+2u}{2} )}α^{j}

K_{3}^{ψ}(t, i, u, v) = ∑_{j=0}^{a−1} q^{j( \frac{2bt−bi+2+2u}{2} )}α^{j}

K_{4}^{ψ}(t, i, u, v) = ∑_{j=0}^{b−1} q^{j( \frac{ai−a+2+2v}{2} )}β^{b−1−j}

Note that when q is not a root of unity and the characteristic of k does not divide a or b, then these scalars are all nonzero when the automorphism ψ is either the identity or the Nakayama automorphism. For in this case the elements are of the form q^{s}(1 + q^{m} + q^{2m} + ⋯ + q^{rm}) for some m, s ∈ ℤ and r = a − 1 or r = b − 1. When m = 0 this element is nonzero since the characteristic of k does not divide a or b, and if it was zero for some m ≠ 0 then q would be a root of unity because of the equality (1 + q^{m} + q^{2m} + ⋯ + q^{rm})(1 − q^{m}) = 1 − q^{(r+1)m}.

In the following result we use this complex to compute the Hochschild homology of our algebra A.

Theorem 3.1. When q is not a root of unity, the Hochschild homology of A is given by

dim_{k} HH_{n}(A) = \{ \begin{matrix}a + b − 1, & when n = 0 \\ a + b, & when n ≥ 1 and char k divides both a and b \\ a + b − 1, & when n ≥ 1 and char k divides one of a and b \\ a + b − 2, & when n ≥ 1 and char k does not divide a or b \end{matrix}

Proof. We need to compute the homology groups of the above complex in the case when ψ is the identity automorphism on A, i.e. when α = 1 = β. We do this by computing Ker δ_{2t}^{1} for t ≥ 1 and Ker δ_{2t+1}^{1} for t ≥ 0, and we treat these two cases separately.

Ker δ_{2t}^{1}:

The image under the map δ_{2t}^{1} of a basis vector y^{u}x^{v}e_{i}^{2t} ∈ ⊕_{i=0}^{2t}Ae_{i}^{2t} is given by

K_{1}^{1}(t, i, u, v)y^{u+b−1}x^{v}e_{i}^{2t−1} + K_{2}^{1}(t, i, u, v)y^{u}x^{v+a−1}e_{i−1}^{2t−1},  for i even
[q^{\frac{ai−a+2+2v}{2}} − 1] y^{u+1}x^{v}e_{i}^{2t−1} + [q^{\frac{2bt−bi−b+2+2u}{2}} − 1] y^{u}x^{v+1}e_{i−1}^{2t−1}, for i odd.

From the definition of the scalars K_{1}^{1} and K_{2}^{1} we see that

K_{1}^{1}(t, i, u, v) = 0 ⇔ i = 0, v = 0, char k|b
K_{2}^{1}(t, i, u, v) = 0 ⇔ i = 2t, u = 0, char k|a,

and therefore we first compute the dimension of Ker δ_{2t}^{1} under the assumption that the characteristic of k does not divide a or b.

First we count the number of single basis vectors in ⊕_{i=0}^{2t}Ae_{i}^{2t} belonging to Ker δ_{2t}^{1}. For even i, we have

δ_{2t}^{1}(y^{u}x^{v}e_{i}^{2t}) = 0 for all even i ⇔ u + b − 1 ≥ b and v + a − 1 ≥ a
⇔ 1 ≤ u ≤ b − 1 and 1 ≤ v ≤ a − 1,

from which we obtain (b − 1)(a − 1)(t + 1) vectors (there are t + 1 even numbers in the set {0, 1, …, 2t}). For odd i, we have

δ_{2t}^{1}(y^{u}x^{v}e_{i}^{2t}) = 0 for all odd i ⇔ u + 1 ≥ b and v + 1 ≥ a
⇔ u = b − 1 and v = a − 1,

giving t vectors (there are t odd numbers in the set {0, 1, …, 2t}). Next we count the other single basis vectors which are mapped to zero, starting with those for which i is even. The element e_{2t}^{2t−1} is zero by definition, hence when i = 2t and v +a−1 ≥ a, that is when 1 ≤ v ≤ a−1, we see that y^{u}x^{v}e_{i}^{2t} maps to zero. But the vectors for which u is nonzero were counted above, hence the new vectors are x^{v}e_{2t}^{2t} for 1 ≤ v ≤ a − 1. Similarly, the element e_{−1}^{2t−1} is zero by definition, hence when i = 0 and u+b−1 ≥ b, that is when 1 ≤ u ≤ b−1, we see that y^{u}x^{v}e_{i}^{2t} maps to zero. But here the vectors for which v is nonzero were counted above, and so the new vectors are y^{u}e_{0}^{2t} for 1 ≤ u ≤ b − 1. It is easy to see that except for these a + b − 2 new vectors, there is no other single basis vector y^{u}x^{v}e_{i}^{2t} in Ker δ_{2t}^{1} for which i is even, since both K_{1}^{1}(t, i, u, v) and K_{2}^{1}(t, i, u, v) are always nonzero. Moreover, when i is odd, neither e_{i}^{2t−1} nor e_{i−1}^{2t−1} are zero, and the coefficients [q^{\frac{ai−a+2+2v}{2}} − 1] and [q^{\frac{2bt−bi−b+2+2u}{2}} − 1] are both nonzero. Hence in this case there are no new basis vectors mapped to zero.

Now we count the number of nontrivial linear combinations of two or more basis vectors in ⊕_{i=0}^{2t}Ae_{i}^{2t} belonging to Ker δ_{2t}^{1}. Let i be even. If the first term of δ_{2t}^{1}(y^{u}x^{v}e_{i}^{2t}) is nonzero, then the only way to “kill” it is to involve the second term of δ_{2t}^{1}(y^{u+b−1}x^{v−1}e_{i+1}^{2t}). Thus to get a nontrivial linear combination we see that u, v and i must satisfy u = 0, 1 ≤ v ≤ a − 1 and i = 0, 2, …, 2t − 2. For these parameter values the second term of δ_{2t}^{1}(y^{u}x^{v}e_{i}^{2t}) vanishes, as does the first term of δ_{2t}^{1}(y^{u+b−1}x^{v−1}e_{i+1}^{2t}). Therefore, for a suitable nonzero scalar C(a, b, i, u, v), the linear combination

x^{v}e_{i}^{2t} + C(a, b, i, u, v)y^{b−1}x^{v−1}e_{i+1}^{2t}

is mapped to zero for 1 ≤ v ≤ a − 1 and i = 0, 2, …, 2t − 2, and there are (a − 1)t such elements. If the second term of δ_{2t}^{1}(y^{u}x^{v}e_{i}^{2t}) is nonzero, then the only way to “kill” it is to involve the first term of δ_{2t}^{1}(y^{u−1}x^{v+a−1}e_{i−1}^{2t}). To get a nontrivial linear combination, the parameters u, v, and i must satisfy 1 ≤ u ≤ b − 1, v = 0 and i = 2, 4, …, 2t, and for these values the first term of δ_{2t}^{1}(y^{u}x^{v}e_{i}^{2t}) and the second term of δ_{2t}^{1}(y^{u−1}x^{v+a−1}e_{i−1}^{2t}) vanish. Thus, for a suitable nonzero scalar C′(a, b, i, u, v), the linear combination

y^{u}e_{i}^{2t} + C′(a, b, i, u, v)y^{u−1}x^{a−1}e_{i−1}^{2t}

is mapped to zero for 1 ≤ u ≤ b − 1 and i = 2, 4, …, 2t, and there are (b − 1)t such elements.

We have now accounted for all the elements of Ker δ_{2t}^{1} when the characteristic of k does not divide a or b. If the characteristic of k divides a, then we must add to our list the element e_{2t}^{2t}. Similarly, if the characteristic of k divides b, then we must add to our list the element e_{0}^{2t}. Finally, if the characteristic of k divides both a and b, then we must add both these two elements to our list (and they are different elements since t ≥ 1). Summing up, we see that the total dimension of Ker δ_{2t}^{1} is given by

dim_{k} Ker δ_{2t}^{1} = \{ \begin{matrix}abt + ab − 1, & when char k does not divide a or b \\ abt + ab + 1, & when char k divides both a and b \\ abt + ab, & otherwise. \end{matrix}

Ker δ_{2t+1}^{1}:

The image under the map δ_{2t+1}^{1} of a basis vector y^{u}x^{v}e_{i}^{2t+1} ∈ ⊕_{i=0}^{2t+1}Ae_{i}^{2t+1} is given by

[1 − q^{\frac{ai+2v}{2}} ] y^{u+1}x^{v}e_{i}^{2t} + K_{3}^{1}(t, i, u, v)y^{u}x^{v+a−1}e_{i−1}^{2t},  for i even

K_{4}^{1}(t, i, u, v)y^{u+b−1}x^{v}e_{i}^{2t} + [q^{\frac{2bt−bi+b+2u}{2}} − 1] y^{u}x^{v+1}e_{i−1}^{2t}, for i odd.

From the definition of the elements K_{3}^{1} and K_{4}^{1} we see that they are always nonzero, contrary to the case above where there were parameters for which K_{1}^{1} and K_{2}^{1} vanished. Therefore the characteristic of k does not matter when we compute the dimension of Ker δ_{2t+1}^{1}.

We follow the same procedure as we did for Ker δ_{2t}^{1}; first we count the number of single basis vectors in ⊕_{i=0}^{2t+1}Ae_{i}^{2t+1} belonging to Ker δ_{2t+1}^{1}. For even i, we have

δ_{2t+1}^{1}(y^{u}x^{v}e_{i}^{2t+1}) = 0 for all even i ⇔ u + 1 ≥ b and v + a − 1 ≥ a
⇔ u = b − 1 and 1 ≤ v ≤ a − 1,

resulting in (a − 1)(t + 1) vectors (there are (t + 1) even numbers in the set {0, 1, …, 2t + 1}). When i is odd, we have

δ_{2t+1}^{1}(y^{u}x^{v}e_{i}^{2t+1}) = 0 for all odd i ⇔ u + b − 1 ≥ b and v + 1 ≥ a
⇔ 1 ≤ u ≤ b − 1 and v = a − 1,

giving (b−1)(t+1) vectors (there are (t+1) odd numbers in the set {0, 1, …, 2t+1}). Next we count the other single basis vectors in ⊕_{i=0}^{2t+1}Ae_{i}^{2t+1} belonging to Ker δ_{2t+1}^{1}, starting with those for which i is even. The element e_{−1}^{2t} is zero, hence for i = 0 the second term in δ_{2t+1}^{1}(y^{u}x^{v}e_{i}^{2t+1}) vanishes. If now v = 0, then the coefficient [1−q^{\frac{ai+2v}{2}} ] vanishes, and therefore the vector y^{u}e_{0}^{2t+1} maps to zero for 0 ≤ u ≤ b−1. There are b such vectors, and none of them was counted above. Moreover, it is not hard to see that there is no other vector y^{u}x^{v}e_{i}^{2t+1} in Ker δ_{2t+1}^{1} for which i is even. As for the case when i is odd, the element e_{2t+1}^{2t} is zero by definition, and the coefficient [q^{\frac{2bt−bi+b+2u}{2}} − 1] vanishes for i = 2t + 1 and u = 0. Therefore the vector x^{v}e_{2t+1}^{2t+1} maps to zero for 0 ≤ v ≤ a − 1. These a vectors have not been counted before, and Ker δ_{2t+1}^{1} does not contain more vectors y^{u}x^{v}e_{i}^{2t+1} for which i is odd.

At last we count the number of nontrivial linear combinations of two or more basis vectors in ⊕_{i=0}^{2t+1}Ae_{i}^{2t+1} belonging to Ker δ_{2t+1}^{1}. Let i be even, and suppose the first term of δ_{2t+1}^{1}(y^{u}x^{v}e_{i}^{2t+1}) is nonzero. The only way to cancel this term is to involve the second term of δ_{2t+1}^{1}(y^{u+1}x^{v−1}e_{i+1}^{2t+1}). Now the first term in the latter vanishes, as does the second term of δ_{2t+1}^{1}(y^{u}x^{v}e_{i}^{2t+1}) since v must be nonzero. Thus for a suitable nonzero scalar C′′(a, b, i, u, v), the element

y^{u}x^{v}e_{i}^{2t+1} + C′′(a, b, i, u, v)y^{u+1}x^{v−1}e_{i+1}^{2t+1}

belongs to Ker δ_{2t+1}^{1} when the parameters satisfy 0 ≤ u ≤ b − 2, 1 ≤ v ≤ a − 1 and i = 0, 2, …, 2t. There are (a − 1)(b − 1)(t + 1) such elements. Finally, suppose the second term of δ_{2t+1}^{1}(y^{u}x^{v}e_{i}^{2t+1}) is nonzero. To cancel it, we must involve the first term in δ_{2t+1}^{1}(y^{u−b+1}x^{v+a−1}e_{i−1}^{2t+1}), and so we see that the only possibility for u and v is u = b − 1 and v = 0. Therefore, for a suitable nonzero scalar C′′′(a, b, i, u, v), the element

y^{b−1}e_{i}^{2t+1} + C′′′(a, b, i, u, v)x^{a−1}e_{i−1}^{2t+1}

for i = 2, 4, …, 2t. There are t such linear combinations.

All the elements of Ker δ_{2t+1}^{1} are now accounted for, and so when summing up we obtain the dimension of this vector space:

dim_{k} Ker δ_{2t+1}^{1} = abt + ab + a + b − 1.

Using the identities dim_{k} Ker δ_{n}^{1} + dim_{k} Im δ_{n}^{1} = dim_{k} A^{n+1} = (n + 1)ab, we can now calculate the Hochschild homology of A. The dimension formula gives dim_{k} Im δ_{2t+1}^{1} = abt + ab − a − b + 1, in particular dim_{k} Im δ_{1}^{1} = 2ab − a − b + 1, giving

dim_{k} HH_{0}(A) = dim_{k} A − dim_{k} Im δ_{1}^{1} = a + b − 1.

Applying the formula to the results we obtained when computing Ker δ_{2t}^{1}, we get

dim_{k} Im δ_{2t+2}^{1} = \{ \begin{matrix}abt + ab + 1, & when char k does not divide a or b \\ abt + ab − 1, & when char k divides both a and b \\ abt + ab, & otherwise, \end{matrix}

and so by calculating dim_{k} HH_{n}(A) = dim_{k} Ker δ_{n}^{1} − dim_{k} Im δ_{n+1}^{1} for n ≥ 1 we get

dim_{k} HH_{n}(A) = \{ \begin{matrix}a + b − 2, & when char k does not divide a or b \\ a + b, & when char k divides both a and b \\ a + b − 1, & otherwise. \end{matrix}

This completes the proof. □

In particular, since a and b are both at least 2, the Hochschild homology of A does not vanish in high degrees (or in any degree). In [Han] this was conjectured by Han to hold for all finite dimensional algebras of infinite global dimension, and in the same paper it was proved that this conjecture holds for monomial algebras.

The converse of this conjecture always holds when the algebra modulo its radical is separable over the ground field. Namely, in this situation, if the global dimension of the algebra is finite, then the algebra has finite projective dimension as a bimodule, and hence its Hochschild homology vanishes in high degrees. The same holds of course for Hochschild cohomology, and in [Hap], following this easy observation, Happel remarked that “the converse seems to be not known”. Thus the cohomology version of Han’s conjecture came to be known as “Happel’s question”. However, this cohomology version is false in general; it was proved in [BGMS] that there do exist finite dimensional algebras of infinite global dimension for which Hochschild cohomology vanishes in high degrees. The counterexample used in the paper was precisely our algebra A with a = 2 = b, and the following result shows that the same holds for arbitrary a and b. Contrary to the homology case, the dimensions of the cohomology groups do not depend on the characteristic of k.

Theorem 3.2. When q is not a root of unity, the Hochschild cohomology of A is given by

dim_{k} HH^{n}(A) = \{ \begin{matrix}2, & for n = 0 \\ 2, & for n = 1 \\ 1, & for n = 2 \\ 0, & for n ≥ 3. \end{matrix}

In particular, the Hochschild cohomology of A vanishes in high degrees.

Proof. It is well known and easy to see that in general HH^{0}(A) is isomorphic to the center of A, that is, the subalgebra {w ∈ A | wz = zw for all z ∈ A}. The center of our algebra A is the vector space spanned by the “first” and the “last” elements in its basis, namely the elements 1 and y^{b−1}x^{a−1}. Hence HH^{0}(A) is 2-dimensional.

To compute the Hochschild cohomology groups of positive degree, we compute the homology of the complex obtained prior to Theorem 3.1, in the case when ψ is the Nakayama automorphism ν. In this case the scalars α and β are given by α = q^{1−b} and β = q^{a−1}. We apply the same method as we did when computing homology; we compute Ker δ_{2t}^{ν} for t ≥ 1 and Ker δ_{2t+1}^{ν} for t ≥ 0, treating the two cases separately.

Ker δ_{2t}^{ν}:

The result when applying the map δ_{2t}^{ν} to a basis vector y^{u}x^{v}e_{i}^{2t} ∈ ⊕_{i=0}^{2t} (_{ν}A_{1}) e_{i}^{2t} is given by

K_{1}^{ν}(t, i, u, v)y^{u+b−1}x^{v}e_{i}^{2t−1} + K_{2}^{ν}(t, i, u, v)y^{u}x^{v+a−1}e_{i−1}^{2t−1},  for i even
[q^{\frac{ai−a+2+2v}{2}} − q^{a−1}] y^{u+1}x^{v}e_{i}^{2t−1} + [q^{\frac{2bt−bi−3b+4+2u}{2}} − 1] y^{u}x^{v+1}e_{i−1}^{2t−1}, for i odd.

From the definition of the elements K_{1}^{ν} and K_{2}^{ν} we see that

K_{1}^{ν}(t, i, u, v) = 0 ⇔ i = 0, v = a − 1, char k|b
K_{2}^{ν}(t, i, u, v) = 0 ⇔ i = 2t, u = b − 1, char k|a,

and so we first compute the dimension of Ker δ_{2t}^{ν} in the case when the characteristic of k does not divide a or b.

First we count the number of single basis vectors in ⊕_{i=0}^{2t}(_{ν}A_{1})e_{i}^{2t} belonging to Ker δ_{2t}^{ν}. As in the homology case, we have

δ_{2t}^{ν}(y^{u}x^{v}e_{i}^{2t}) = 0 for all even i ⇔ u + b − 1 ≥ b and v + a − 1 ≥ a
⇔ 1 ≤ u ≤ b − 1 and 1 ≤ v ≤ a − 1,
δ_{2t}^{ν}(y^{u}x^{v}e_{i}^{2t}) = 0 for all odd i ⇔ u + 1 ≥ b and v + 1 ≥ a
⇔ u = b − 1 and v = a − 1,

from which we obtain (b − 1)(a − 1)(t + 1) + t vectors. Next we count the other single basis vectors in ⊕_{i=0}^{2t}(_{ν}A_{1})e_{i}^{2t} belonging to Ker δ_{2t}^{ν}. Since K_{1}^{ν} and K_{2}^{ν} are always nonzero, the number of such vectors for which i is even is the same as in the homology case, namely a+b−2. As for the vectors for which i is odd, it is no longer true that the coefficients are always nonzero. The coefficient [q^{\frac{ai−a+2+2v}{2}} − q^{a−1}] vanishes when i = 1 and v = a − 2, whereas [q^{\frac{2bt−bi−3b+4+2u}{2}} − 1] vanishes when i = 2t − 1 and u = b − 2. Both these cases will occur, since t is at least 1 when we compute Ker δ_{2t}^{ν}. However, these coefficients need to vanish simultaneously for the basis vector to belong to Ker δ_{2t}^{ν}, and this only happens when t = 1, since then 2t − 1 = 1. Thus, when t = 1 the vector y^{b−2}x^{a−2}e_{1}^{2} maps to zero, whereas when t ≥ 2 there are no new basis vectors in Ker δ_{2t}^{ν} for which i is odd.

Now we count the number of nontrivial linear combinations of two or more basis vectors in ⊕_{i=0}^{2t}(_{ν}A_{1})e_{i}^{2t} belonging to Ker δ_{2t}^{ν}. These elements are precisely the same as in the homology case, and we do not encounter problems because of the “new” basis vector in Ker δ_{2}^{ν} we obtained above. Therefore the number of such linear combinations is (a − 1)t + (b − 1)t.

We now look at what happens when the characteristic of k divides a or b. If char k divides a, then we must add the vector x^{a−1}e_{0}^{2t} to the list of single basis vectors mapped to zero. However, this vector already appears in one of the nontrivial linear combinations, hence it does not contribute to the total dimension. Similarly, when char k divides b, then the new vector y^{b−1}e_{2t}^{2t} belongs to the list of single basis vectors mapped to zero. But again this vector already appears in one of the nontrivial linear combinations, and it will therefore not contribute to the total dimension. This argument is still valid if char k divides both a and b. This shows that the dimension of Ker δ_{2t}^{ν} is independent of the characteristic of k.

In total we see that the dimension of Ker δ_{2t}^{ν} is almost the same as it was in the homology case when the characteristic of k did not divide a or b; we need one additional vector when t = 1. Therefore the dimension is given by

dim_{k} Ker δ_{2t}^{ν} = \{ \begin{matrix}2ab, & when t = 1 \\ abt + ab − 1, & when t ≥ 2. \end{matrix}

Ker δ_{2t+1}^{ν}:

The image under the map δ_{2t+1}^{ν} of a basis vector y^{u}x^{v}e_{i}^{2t+1} ∈ ⊕_{i=0}^{2t+1} (_{ν}A_{1}) e_{i}^{2t+1} is given by

[q^{a−1} − q^{\frac{ai+2v}{2}} ] y^{u+1}x^{v}e_{i}^{2t} + K_{3}^{ν}(t, i, u, v)y^{u}x^{v+a−1}e_{i−1}^{2t},  for i even

K_{4}^{ν}(t, i, u, v)y^{u+b−1}x^{v}e_{i}^{2t} + [q^{\frac{2bt−bi−b+2+2u}{2}} − 1] y^{u}x^{v+1}e_{i−1}^{2t}, for i odd.

Now from the definition of the scalars K_{3}^{ν} and K_{4}^{ν} we see that K_{3}^{ν} is always nonzero, while we have

K_{4}^{ν}(t, i, u, v) = 0 ⇔ i = 1, v = a − 2, char k|b.

Therefore we first compute the dimension of Ker δ_{2t+1}^{ν} under the assumption that the characteristic of k does not divide b.

First we count the number of single basis vectors in ⊕_{i=0}^{2t+1}(_{ν}A_{1})e_{i}^{2t+1} belonging to Ker δ_{2t+1}^{ν}. As in the homology case, we have

δ_{2t+1}^{ν}(y^{u}x^{v}e_{i}^{2t+1}) = 0 for all even i ⇔ u + 1 ≥ b and v + a − 1 ≥ a
⇔ u = b − 1 and 1 ≤ v ≤ a − 1,
δ_{2t+1}^{ν}(y^{u}x^{v}e_{i}^{2t+1}) = 0 for all odd i ⇔ u + b − 1 ≥ b and v + 1 ≥ a
⇔ 1 ≤ u ≤ b − 1 and v = a − 1,

from which we obtain (a−1)(t+1)+(b−1)(t+1) vectors. Next we count the other single basis vectors in ⊕_{i=0}^{2t+1}(_{ν}A_{1})e_{i}^{2t+1} belonging to Ker δ_{2t+1}^{ν}, treating first the ones for which i is even. When i = 0 the second term of δ_{2t+1}^{ν}(y^{u}x^{v}e_{i}^{2t+1}) vanishes, and the first term then vanishes if u = b − 1 or v = a − 1. Some of these vectors are among the ones counted above, the new ones are y^{b−1}e_{0}^{2t+1} and y^{u}x^{a−1}e_{0}^{2t+1} for 0 ≤ u ≤ b − 2. Except for these b elements, there are no other single basis elements in Ker δ_{2t+1}^{ν} for which i is even. As for those for which i is odd, we see that the first term of δ_{2t+1}^{ν}(y^{u}x^{v}e_{i}^{2t+1}) vanishes when i = 2t + 1. In this case the second term vanishes if u = b−1 or v = a−1, and of these vectors the ones which have not been counted before are the a elements x^{a−1}e_{2t+1}^{2t+1} and y^{b−1}x^{v}e_{2t+1}^{2t+1} for 0 ≤ v ≤ a − 2. It is not hard to see that Ker δ_{2t+1}^{ν} does not contain any other element y^{u}x^{v}e_{i}^{2t+1} for which i is odd.

Finally we count the number of nontrivial linear combinations of two or more basis elements in ⊕_{i=0}^{2t+1}(_{ν}A_{1})e_{i}^{2t+1} belonging to Ker δ_{2t+1}^{ν}. In the homology case, these were

y^{u}x^{v}e_{i}^{2t+1} + C′′(a, b, i, u, v)y^{u+1}x^{v−1}e_{i+1}^{2t+1}

for 0 ≤ u ≤ b − 2, 1 ≤ v ≤ a − 1 and i = 0, 2, …, 2t, and

y^{b−1}e_{i}^{2t+1} + C′′′(a, b, i, u, v)x^{a−1}e_{i−1}^{2t+1}

for i = 2, 4, …, 2t, where C′′ and C′′′ are suitable scalars. The t latter elements also belong to Ker δ_{2t+1}^{ν}, but among the (a − 1)(b − 1)(t + 1) first elements there are some combinations that are not mapped to zero. Namely, we must discard the b − 1 elements for which i = 0 and v = a − 1, since we showed above that y^{u}x^{a−1}e_{0}^{2t+1} maps to zero for 0 ≤ u ≤ b − 2. Similarly, we must discard the a − 1 combinations for which i = 2t and u = b − 2, since y^{b−1}x^{v}e_{2t+1}^{2t+1} maps to zero for 0 ≤ v ≤ a − 2. However, when t = 0 then the situations i = 0 and i = 2t are the same, and the element y^{b−2}x^{a−1}e_{0}^{1} + C′′(a, b, i, u, v)y^{b−1}x^{a−2}e_{1}^{1} has been discarded twice. Thus the total number of nontrivial linear combinations is (a − 1)(b − 1)(t + 1) + t − (a − 1) − (b − 1) when t ≥ 1, and one more when t = 0.

What happens when char k divides b? The element y^{u}x^{a−2}e_{1}^{2t+1} is not mapped to zero for any u, and it does not “interfere” with one of the nontrivial linear combinations. Hence the dimension of Ker δ_{2t+1}^{ν} is also independent of the characteristic of k.

In total we see that the dimension of Ker δ_{2t+1}^{ν} differs from that in the homology case since we need to subtract (a − 1) + (b − 1) when t ≥ 1 and (a − 1) + (b − 1) − 1 when t = 0. Thus the dimension is given by

dim_{k} Ker δ_{2t+1}^{ν} = \{ \begin{matrix}ab + 2, & when t = 0 \\ abt + ab + 1, & when t ≥ 1. \end{matrix}

We can now calculate the positive degree cohomology groups. We have dim_{k} Ker δ_{1}^{ν} = ab + 2, and since dim_{k} Ker δ_{2}^{ν} = 2ab we must have dim_{k} Im δ_{2}^{ν} = ab, giving

dim_{k} HH^{1}(A) = dim_{k} Ker δ_{1}^{ν} − dim_{k} Im δ_{2}^{ν} = 2.

Furthermore, since dim_{k} Ker δ_{3}^{ν} = 2ab + 1, we must have dim_{k} Im δ_{3}^{ν} = 2ab − 1, giving

dim_{k} HH^{2}(A) = dim_{k} Ker δ_{2}^{ν} − dim_{k} Im δ_{3}^{ν} = 1.

Similarly, direct computations show that the cohomology groups HH^{n}(A) vanish when n ≥ 3, thereby completing the proof. □

When the commutator element q is a root of unity, then it is not hard to see that the dimensions of infinitely many of the kernels in the complex we used to compute (co)homology will increase. Therefore the Hochschild homology of A is still nonzero in all degrees, while it is no longer true that all the higher Hochschild cohomology groups vanish. We record this fact in the final result, which also gives the multiplicative structure of the Hochschild cohomology ring when q is not a root of unity.

Theorem 3.3. The Hochschild cohomology ring HH^{∗}(A) is finite dimensional if and only if q is not a root of unity. When this is the case, the algebra is isomorphic to the (five dimensional graded) fibre product

k[U]/(U^{2}) ×_{k} k⟨V, W ⟩/(V^{2}, V W + W V, W^{2}),

where U is in degree zero and V and W are in degree one.

Proof. Suppose q is not a root of unity. Recall first the initial part

P_{2} ⟶^{d_{2}} P_{1} ⟶^{d_{1}} P_{0} ⟶^{μ} A → 0

of the projective bimodule resolution of A, where μ is the multiplication map. The maps d_{1} and d_{2} are defined on generators as follows:

d_{1} : f_{0}^{1} ↦ [(1 ⊗ y) − (y ⊗ 1)] f_{0}^{0}
f_{1}^{1} ↦ [(1 ⊗ x) − (x ⊗ 1)] f_{0}^{0}
d_{2} : f_{0}^{2} ↦ [(1 ⊗ y^{b−1}) + (y ⊗ y^{b−2}) + ⋯ + (y^{b−1} ⊗ 1)] f_{0}^{1}
f_{1}^{2} ↦ [q(1 ⊗ x) − (x ⊗ 1)] f_{0}^{1} + [(1 ⊗ y) − q(y ⊗ 1)] f_{1}^{1}
f_{2}^{2} ↦ [(1 ⊗ x^{a−1}) + (x ⊗ x^{a−2}) + ⋯ + (x^{a−1} ⊗ 1)] f_{1}^{1}.

Define two bimodule maps g: P_{1} → A and h: P_{1} → A by

g: f_{0}^{1} ↦ y
f_{1}^{1} ↦ 0
h: f_{0}^{1} ↦ 0
f_{1}^{1} ↦ x.

One checks directly that g ∘ d_{2} = 0 = h ∘ d_{2}, and that neither of the two maps are liftable through d_{1}. Consequently they represent the two basis elements of HH^{1}(A) = Ext_{A^{e}}^{1}(A, A).

We may identify the degree zero part of HH^{∗}(A) with the center of A, the two dimensional vector space spanned by the elements 1 and y^{b−1}x^{a−1}. The latter element annihilates both g and h, hence HH^{∗}(A) is isomorphic to the k-fibre product of the algebra generated by y^{b−1}x^{a−1} with the algebra generated by g and h. Since the Hochschild cohomology ring of a finite dimensional algebra is always graded commutative (cf. [SnS, Corollary 1.2]), both g and h square to zero. Therefore, as HH^{2}(A) is one dimensional, we are done if we can show that the product hg ∈ HH^{2}(A) is nonzero.

Define a bimodule map g_{0} : P_{1} → P_{0} by

\begin{aligned}g_{0} : & f_{0}^{1} & ↦ & (y ⊗ 1)f_{0}^{0} \\ & f_{1}^{1} & ↦ & 0. \end{aligned}

It is not hard to see that there exists an element w ∈ A^{e} such that the map g_{1} : P_{2} → P_{1} defined by

g_{1} : f_{0}^{2} ↦ wf_{0}^{1}
f_{1}^{2} ↦ q(y ⊗ 1)f_{1}^{1}
f_{2}^{2} ↦ 0

gives a commutative diagram

\xymatrix{P_{2} \ar[d]^{g_{1}} \ar[r]^{d_{2}} & P_{1} \ar[d]^{g_{0}} \ar[dr]^{g} \\ P_{1} \ar[r]^{d_{1}} & P_{0} \ar[r]^{μ} & A}

The product hg ∈ HH^{2}(A) is then represented by the composite map h ∘ g_{1}, under which the images of the generators in P_{2} are given by

h ∘ g_{1} : f_{0}^{2} ↦ 0
f_{1}^{2} ↦ qyx
f_{2}^{2} ↦ 0.

This map is not liftable through d_{2}, and therefore it represents a nonzero element of HH^{2}(A). Consequently the product hg is nonzero. □

References

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

[CaE] H. Cartan, S. Eilenberg, Homological Algebra, Princeton University Press, 1956.

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

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[Han] Y. Han, Hochschild (co)homology dimension, J. London Math. Soc. (2) 73 (2006), no. 3, 657-668.

[Hap] D. Happel, Hochschild cohomology of finite-dimensional algebras, in Séminaire d’Algèbre Paul Dubreil et Marie-Paul Malliavin, 39ème Année (Paris, 1987/1988), Lecture Notes in Mathematics 1404 (1989), 108-126.

[Hol] T. Holm, Hochschild cohomology rings of algebras k[X]/(f), Beiträge Algebra Geom. 41 (2000), no. 1, 291-301.

[Man] I. Manin, Some remarks on Koszul algebras and quantum groups, Ann. Inst. Fourier (Grenoble) 37 (1987), 191-205.

[SnS] N. Snashall, Ø. Solberg, Support varieties and Hochschild cohomology rings, Proc. London Math. Soc. 88 (2004), 705-732.

Petter Andreas Bergh Mathematical Institute, 24-29 St. Giles, Oxford, OX1 3LB, United Kingdom Present address: Institutt for matematiske fag, NTNU, N-7491 Trondheim, Norway

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

Karin Erdmann Mathematical Institute, 24-29 St. Giles, Oxford, OX1 3LB, United Kingdom

   E-mail address: erdmann@maths.ox.ac.uk