Hochschild cohomology of some quantum complete intersections

HOCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS

KARIN ERDMANN, MAGNUS HELLSTRØM-FINNSEN

Abstract. We compute the Hochschild cohomology ring of the algebras A = khX, Yi/(X^a, XY − qY X, Y^a) over a fieldkwherea≥2 and whereq∈kis a primitivea-th root of unity. We find the the dimension of HHⁿ(A) and show that it is independent ofa. We compute explicitly the ring structure of the even part of the Hochschild cohomology modulo homogeneous nilpotent elements.

1. Introduction

Letkbe a field, and let 06=q∈k. Quantum complete intersections originate from work of Manin [8].

Here we focus on the algebras

Aq =khX, Yi/(X^a, XY −qY X, Y^a).

Such algebras have provided several examples giving answers to homological conjectures and questions.

Perhaps most spectacular amongst these is Happel’s question. In [6] Happel asked whether an algebra whose Hochschild cohomology is finite-dimensional, must have finite global dimension. The main result of [3] gave a negative answer: It shows that the Hochschild cohomology of the quantum complete intersection A_q as above, when a = 2 and q not a root of unity, is finite-dimensional. However the algebra A_q is selfinjective, hence has infinite global dimension. Already earlier, R. Schulz discovered unusual properties for these algebrasAq, see [11] and [10].

Furthermore, there is a theory of support varieties in terms of Hochschild cohomology provided the algebra satisfies suitable finite generation properties, known as condition (Fg) (see [5] and [13]). ForA_q, this condition is satisfied precisely whenqis a root of unity. The general theory of these support varieties has now been well established in several papers. However, in order to actually compute the varieties over a given algebra, one needs to determine the ring structure of the Hochschild cohomology, or at least modulo homogeneous nilpotent elements.

The results in this paper will be a contribution towards this goal. We determine the ring structure of the even part of HH^∗(Aq) (which will be denoted HH^2∗(Aq)) modulo the ideal of homogeneous nilpotent elements forAq whenqis a primitivea-th root of unity. The proofs are quite technical, but this illustrates the typical difficulties and computations one is faced with when trying to compute Hochschild cohomology.

First we present an unpublished result by P. Bergh and K. Erdmann which determines the dimensions of the Hochschild cohomology groups; this is done via exploiting Hochschild homology. Surprisingly, the answer is independent ofa(see Theorem 3.1 and Corollary 3.2). This suggests that perhaps also the ring structure might not depend too much on the parameter a. We determine explicit bases of HH^2∗(A) (see Section 5.2).

Furthermore, we compute the algebra structure of HH^2∗(A) modulo the largest homogeneous nilpotent ideal. We show that it is Z2-graded, with degree zero part isomorphic to the polynomial ring in two variables, generated in degree 2. The explicit description is given in 4.2 when a = 2, and in 5.4 when a≥3.

Date: November 1, 2017.

2010Mathematics Subject Classification. Primary 16E40; Secondary 16U80; 16S80; 81R50.

Key words and phrases. Hochschild Cohomology; Quantum complete intersections.

1

An explicit description whena= 2 was also given in [3, Section 3.4]. We include this case (in Section 4), as it shows that it is part of the general pattern.

S. Oppermann gave also a description of the Hochschild cohomology and homology of more general quantum complete intersections in [9]. The products are, however, not computed completely explicitly, though it discusses a more general setting. However, in this paper we calculate products explicitly by liftings along a minimal projective resolution (which will be discussed in Section 2). This illustrates techniques that might be of independent interest.

In a larger context, there is even more structure in some classes of Hochschild cohomology than the well known Gerstenhaber algebra structure. In [7] T. Lambre, G. Zhou and A. Zimmermann prove that the Hochschild cohomology ring of quantum complete intersections is a so called Batalin–Vilkovisky algebra (Corollary 5.8). Roughly speaking a Batalin–Vilkovisky algebra is a Gerstenhaber algebra with an additional operation ∆ : HHⁿ → HHⁿ⁻¹ which squares to zero and which, together with the cup product, can express the Lie Bracket.

2. Preliminaries

More generally, let A be any finite-dimensional algebra over a field k, and let A^e = A⊗kA^op de- note the enveloping algebra. We view bimodules over A as left modules over A^e. In this setting, the Hochschild cohomology ofAcan be taken as HHⁿ(A) = Extⁿ_Ae(A, A), then-th cohomology of the complex HomA^e(P, A), i.e.

Extⁿ_Ae(A, A) = kerd^∗_n+1/imd^∗_n, (2.1)

whered^∗_n= Hom_Ae(d_n, A) and whered_n are the maps in a minimal projective resolution:

P:· · · →P₂−→^d2 P₁−→^d1 P₀−→^µ A→0.

(2.2)

Then theHochschild cohomology

HH^∗(A) = Ext^∗_Ae(A, A) (2.3)

is a k-algebra which is graded-commutative. There are various equivalent ways to define the product;

here we will work with the Yoneda product.

We specialize now to the quantum complete intersections. Let abe an integer such that a≥2. We also letq∈kbe a primitivea-th root of unity, andAis thek-algebra defined by

A=khX, Yi/(X^a, XY −qY X, Y^a).

(2.4)

We writexandyfor the residue classes ofX andY, respectively.

In [2], for arbitrary parameter q6= 0, an explicit minimal projective bimodule resolution Pas in (2.2) was constructed. Thenth bimodule inPis

Pn =

n

M

i=0

A^ef_iⁿ, (2.5)

the freeA^e-module of rankn+ 1 having generators{f_n⁰, f_n¹, ..., f_nⁿ}. For eachs≥0 define the following four elements ofA^e:

τ1(s) =q^s(1⊗x)−(x⊗1) (2.6)

τ2(s) = (1⊗y)−q^s(y⊗1) (2.7)

γ₁(s) =

a−1

X

j=0

q^js(x^a−1−j⊗x^j) (2.8)

γ2(s) =

a−1

X

j=0

q^js(y^j⊗y^a−1−j) (2.9)

The mapsd_n:P_n→P_n−1in Pare given by

d2t:f_i^2t7→

(γ₂ ^ai₂

f_i^2t−1+γ₁ ^2at−ai₂

f_i−1^2t−1 forieven

−τ2 ai−a+2 2

f_i^2t−1+τ1 2at−ai−a+2 2

f_i−1^2t−1 foriodd (2.10)

d2t+1:f_i^2t+17→

(τ2 ai 2

f_i^2t+γ1 2at−ai+2 2

f_i−1^2t forieven

−γ₂ ^ai−a+2₂

f_i^2t+τ₁ ^2at−ai+a₂

f_i−1^2t foriodd (2.11)

where the convention f₋₁ⁿ =f_n+1ⁿ = 0 has been used. So far, qis arbitrary. Later in our setting we will simplify these expressions.

We will wish to identify nilpotent elements of Hochschild cohomology. This can be done by exploiting the following result of N. Snashall and Ø. Solberg, see Proposition 4.4 in [12].

Proposition 2.1. Assume k is a field andA is a finite-dimensionalk-algebra. Supposeη is a map into A representing an element ofHHⁿ(A). Ifim(η)is in the radical ofA thenη is nilpotent inHH^∗(A).

3. Dimensions of Hochschild cohomology groups

We recall an unpublished result by Petter A. Bergh and Karin Erdmann which determines the dimen- sions.

By viewing A as a left A^e-module, it follows from [4, p. VI.5.3] that D(HH^∗(A, A)) is isomorphic to Tor^A_∗^e(D(A), A) as a vector space, where D denotes the usual k-dual i.e. D(−) := Homk(−, k). In particular, we see that dim HHⁿ(A) = dim Tor^A_n^e(D(A), A) for all n ≥0. Moreover, it follows from [2]

thatA is a Frobenius algebra with Nakayama automorphismν:A→Adefined by

ν :

(x7→q^1−ax y7→q^a−1y.

(3.1)

The bimodules D(A) and_νA₁ are isomorphic; here the left action on _νA₁ is taken as a·m := ν(a)m.

Consequently the dimensions of the Hochschild cohomology ofAare given by dim HHⁿ(A) = dim Tor^A_n^e(νA1, A)

(3.2)

for alln≥0.

To compute Tor^A_n^e(νA1, A), we tensor the deleted projective bimodule resolution P with the right A^e-moduleνA1. We then obtain an isomorphism

· · · νA1⊗A^ePn+1 νA1⊗A^ePn νA1⊗A^eP_n−1 · · ·

· · · ⊕ⁿ⁺¹_i=0(νA1)eⁿ⁺¹_i ⊕ⁿ_i=0(νA1)eⁿ_i ⊕ⁿ⁻¹_i=0(νA1)eⁿ⁻¹_i · · ·

1⊗dn+1 1⊗d_n

δ_n+1 δ_n

∼= ∼= ∼=

of complexes, where{eⁿ₀, eⁿ₁, . . . , eⁿ_n} is the standard generating set ofn+ 1 copies ofνA1. Now given an elementα∈kand a positive integer t, define an elementK_t(α)∈kby

Kt(α) :=

t−1

X

j=0

α^j. (3.3)

The mapδ_n is then given by δ2t:y^ux^ve^2t_i 7→

(qK_a(q^v+1)y^u+a−1x^ve^2t−1_i +K_a(q^u+1)y^ux^v+a−1e^2t−1_i−1 forieven [q^v+1−q^a−1]y^u+1x^ve^2t−1_i + [q^u+2−1]y^ux^v+1e^2t−1_i−1 foriodd (3.4)

δ2t+1:y^ux^ve^2t+1_i 7→

([q^a−1−q^v]y^u+1x^ve^2t_i +K_a(q^u+2)y^ux^v+a−1e^2t_i−1 forieven

−qKa(q^v+2)y^u+a−1x^ve^2t_i + [q^u+1−1]y^ux^v+1e^2t_i−1 foriodd (3.5)

where we use the conventioneⁿ₋₁ =eⁿ_n+1 = 0. This was proved in [2] in a more general setting, and by specializingqand using thatx, yhave the same nilpotency index, we obtain the above formulae (correcting an unimportant sign error in [2]).

For the following result we use this complex to compute the Hochschild cohomology of our algebra A, in the case when q is a primitive a-th root of unity. The result shows that the dimensions of the cohomology groups do not depend on the characteristic of the field, except that the characteristic of k does not divideasincekcontains a primitivea-th root of unity.

Theorem 3.1. If qis a primitivea-th root of unity, then dimkHHⁿ(A) = 2n+ 2 for alln≥0.

Proof. Since HH⁰(A) is isomorphic to the centre ofA, we see immediately that HH⁰(A) is 2-dimensional.

To find the dimension of HHⁿ(A) forn >0, we compute kerδ_2tfort≥1 and kerδ_2t+1 fort≥0.

Sincekcontains a primitiveq-th root of unity, the characteristic ofkdoes not dividea. The equalities 0 = 1−(q^m)^a= (1−q^m)K_a(q^m), valid for any integerm, show thatK_a(q^m) = 0 if and only ifmis not divisible bya. We will use this fact throughout.

We first compute kerδ2t fort≥1. By the previous observation,Ka(q^v+1) = 0 if and only if 0≤v≤ a−2, whereasKa(q^u+1) = 0 if and only if 0≤u≤a−2. Therefore

δ_2t(y^ux^ve^2t_i ) = 0⇔



























u∈ {1,2, . . . , a−1}, v∈ {1,2, . . . , a−1}, ieven, ,0≤i≤2t u∈ {0,1, . . . , a−2}, v= 0, ieven,0≤i≤2t

u= 0, v∈ {0,1, . . . , a−2}, ieven, ,0≤i≤2t u= 0, v=a−1, i= 2t

u=a−1, v= 0, i= 0

u=a−2, v=a−2, iodd,1≤i≤2t−1 u=a−1, v=a−1, iodd,1≤i≤2t−1 (3.6)

and there area²t+a² such elements.

LetB:={y^ux^ve^2t_i : 0≤u, v ≤a−1,0≤i≤2t}, a basis forνA1⊗A^eP2t. We split this basis into three parts. LetXbe the set of basis vectors which are in the kernel ofδ2t, so thatXis given by the above list.

Next, let

Y:={y^ux^ve^2t_i : iodd, ,0≤u, v < a−2}

One checks directly that ker(δ2t)∩Sp(Y) ={0}. LetZ:=B\(X∪Y). We find thatZis equal to {x^a−1e^2t_2j : 0≤j < t} ∪ {y^a−1e^2t_2j: 0< j≤t}

∪ {y^a−2x^a−1e^2t_2j+1: 0≤j < t} ∪ {y^a−1x^a−2e^2t_2j+1: 0≤j < t}.

The map δ2t takes each member ofZ to a non-zero scalar multiple ofy^a−1x^a−1e^2t−1_i , and each of these occurs, for 0≤i≤2t−1. Hence the image ofδ_2trestricted to the span ofZ has dimension 2t, and the size of Zis 4t. This means that the restriction ofδ2tto the span of Zhas kernel of dimension 2t, by the rank-nullity formula. Hence we get 2tfurther linearly independent elements of the kernel ofδ2t. In total, this shows that dimkkerδ2t= (a²+ 2)t+a².

Next we compute kerδ_2t+1fort≥0, recall that the characteristic ofkdoes not dividea. We see that δ_2t+1(y^ux^ve^2t+1_i ) = 0⇔

(u=a−1, v∈ {0, . . . , a−1}, iarbitrary u∈ {0, . . . , a−2}, v=a−1, iarbitrary (3.7)

and there are (2a−1)(2t+ 2) such elements.

LetB:={y^ux^ve^2t+1_i : 0≤u, v≤a−1,0≤i≤2t+ 1}, a basis for νA1⊗A^eP2t+1. We split this basis into three parts. LetXbe the set of basis vectors which are in the kernel ofδ_2t+1, that is Xis given by the above list. Next, consider

Y= {y^a−2x^ve^2t+1_i :ieven,0≤v≤a−2} ∪ {y^ue^2t+1_i :ieven,0≤u≤a−3}

∪ {y^ux^a−2e^2t+1_i :iodd,0≤u≤a−2} ∪ {x^ve^2t+1_i :iodd,0≤v≤a−3}.

One checks that Sp(Y)∩Ker(δ_2t+1) ={0}. Now let Z:=B\(X∪Y). This is the disjoint union of two sets,Z=Ze∪Zowhere

Ze:={y^ux^ve^2t+1_i :ieven, 0≤u≤a−3,1≤v≤a−2}

Zo:={y^ux^ve^2t+1_i : iodd , 1≤u≤a−2, 0≤v≤a−3}

both of size (a−2)²(t+ 1). Thenδ2t+1(khZei) =khZ˜iand δ2t+1(khZoi) =khZ˜iwhere Z˜ :={y^ux^ve^2t_j :j even,1≤u≤a−2,1≤v≤a−2}

which also has size (a−2)²(t+ 1). By the rank-nullity formula, the kernel of δ_2t+1 restricted toZhas dimension (a−2)²(t+ 1). (Note that, if a = 2, then Z = ∅). In total we get that dimkkerδ2t+1 = (a²+ 2)(t+ 1).

We have now computed kerδ2tfort≥1 and kerδ2t+1 fort≥0. Using the equalities dimkimδn+ dimkkerδn= dimk⊕ⁿ_i=0(νA1)eⁿ_i = (n+ 1)a², (3.8)

we see that dimkimδ2t+1= dimkimδ2t+2= (a²−2)(t+ 1). Consequently dimkHH^2t+1(A) = dimkkerδ2t+1−dimkimδ2t+2

(3.9)

= 4t+ 4 (3.10)

dim_kHH^2t+2(A) = dim_kkerδ_2t+2−dim_kimδ_2t+3 (3.11)

= 4t+ 6 (3.12)

fort≥0, and the proof is complete.

This result implies immediately the following:

Corollary 3.2. The dimension of the cohomology groupsHHⁿ(A) is independent ofa.

4. Hochschild cohomology whena= 2 In this section we leta= 2 and q=−1 (and char(k)6= 2), so we have that

A=khX, Yi/(X², XY +Y X, Y²).

(4.1)

We write x, y again for the images of X, Y in A. We also mention related work by P. A. Bergh in [1]

where the main objective is to compute the homology and cohomology ofAwith coefficients in the twisted bimodule1Λφ for anyk-linear automorphismφof the algebra Λ.

We will simplify the differentials of the minimal projective resolution, before studying the even part of cohomology ring for this case.

4.1. Minimal projective resolution whena= 2. We introduce the following notation:

βy= (1⊗y) + (y⊗1) βx= (1⊗x) + (x⊗1) (4.2)

α_y= (1⊗y)−(y⊗1) α_x= (1⊗x)−(x⊗1) (4.3)

Now we can rewrite the differentials for the minimal projective resolutionP in Equation 2.10 and 2.11;

we get:

d_n(f_iⁿ) =

((−1)ⁱ(βyf_iⁿ⁻¹+βxf_i−1ⁿ⁻¹) whennis even (−1)ⁱ(αyf_iⁿ⁻¹−αxf_i−1ⁿ⁻¹) whennis odd. (4.4)

4.2. Description of cohomology groups. In Section 3 we have seen that dim HHⁿ(A) = 2n+ 2.

Knowing this, we will determine a basis for HHⁿ(A) for arbitrary even degreesn. We writeδ_ir as usual for the Kronecker symbol.

Lemma 4.1. Let n= 2t. Forr= 0,1, . . . ,2t define maps ξr, ηr:P2t→A as follows.

ξr(f_i^2t) =δir·1A, ηr(f_i^2t) =δir·xy.

(4.5)

(a) The classes of these maps form a basis forHH^2t(A).

(b) The classes of the η_r give nilpotent elements inHH^∗(A).

Proof. Part (b) will follow from Proposition 2.1. We prove now part (a). Note that these are 2n+ 2 elements, so we only have to show that the maps are in the kernel of d^∗_2t+1, and that they are linearly independent modulo the image ofd^∗_2t.

(1) We apply ξr tod2t+1(f_i^2t+1), this gives

ξ_r[(−1)ⁱ(α_yf_i^2t−α_xf_i−1^2t )] = (−1)ⁱ[α_y[δ_ir·1_A]−α_x[δ_i−1,r·1_A]] = 0 (4.6)

(we viewAas a leftA^emodule, andαy·1A= 0 =αx·1A). Similarly we applyηrtod2t+1(f_i^2t+1) and get

ηr[(−1)ⁱ(αyf_i^2t−αxf_i−1^2t )] = (−1)ⁱ[αy[δir·xy]−αx[δi−1,r·xy]] = 0 (4.7)

(sincexy is in the socle ofAwe see thatα_y·xy= 0 andα_x·xy= 0).

(2) Let cr, dr∈K andρ:P2t−1→A such that

2t

X

r=0

c_rξ_r+d_rη_r=ρ◦d_2t∈im(d^∗_2t).

(4.8)

We must show thatcr= 0 =dr for allr. Writeρ(f_i^2t−1) =pi=ai+bix+ciy+dixy∈A. Then we have

ρ◦d_2t(f_i^2t) = (−1)ⁱ[β_yp_i+β_xp_i−1] = (−1)ⁱ[2a_iy+ 2a_i−1x]

(4.9)

which are elements inA. On the other hand if we apply the map given by the sum to f_i^2t then we get

c_i + d_ixy, (4.10)

also elements inA. We assume these are equal, and it follows that all scalars are zero.

4.3. Products in even degrees of HH^∗(A). Recall that the even part HH^2∗(A) is a subring of the Hochschild cohomology, and it is commutative. The aim of this section is to prove the following:

Theorem 4.2. Let k be a field with char(k)6= 2, and letA=khX, Yi/(X², XY +Y X, Y²). Assume R= Sp{ξ_i^2t:t≥0, and0≤i≤2t}.

(4.11)

ThenR is a subalgebra ofHH^2∗(A). It is Z2-graded, with

R₀:=khξ_i^2t:i eveni, and R₁:=khξ_i^2t:iodd i.

(4.12)

We have ξ_l^2mξ^2t_r =ξ_l+r^2m+2t. The subalgebra R₀ is isomorphic to the polynomial ring k[z₀, z₁] where we identifyξ²₀ withz₀ andξ²₂ with z₁. Moreover, R₁=R₀ξ²₁ and ξ₁²·ξ²₁=ξ⁴₂.

Corollary 4.3. Let Nbe the largest homogeneous nilpotent ideal ofHH^2∗(A). Then HH^2∗(A)/N∼=R.

(4.13)

We fix a degree 2t, and we will compute the product of a general elementξof degree 2twith an element χ of degree 2m, and we let 2m vary. We take representativesξ :P_2t→A and χ: P_2m →A which are k-linear combinations of the basis. Let

ξ(f_i^2t) =pi∈A with 0≤i≤2t (4.14)

χ(f_i^2m) =p_i∈A with 0≤i≤2m.

(4.15)

By (4.5), the elementsp_i and ¯p_i are then in the centre ofA, we will use this freely.

Definition 4.4. The Yoneda product χ•ξ is the residue class of χ◦h2m where the family (hs) with hs:P2t+s→Psis a lifting ofξ. That is, we have the following diagram:

P_2t+2m P_2t+2m−1 · · · P_2t+s P_2t+s−1 · · · P_2t+1 P_2t

P2m P2m−1 · · · Ps Ps−1 · · · P1 P0 A

A

d_2t+2m d_2t+s d_2t+1

d2m ds d1 µ

h2m h2m−1 hs hs−1 h1 h0

ξ

χ

where ξ=µ◦h₀ and where all squares commute. We define mapsh_s (0≤s≤2m), and will show that they are a lifting.

h_s(f_i^2t+s) = (Ps

j=0pi−jf_j^s whenseven

(−1)ⁱ Ps

j=0(−1)^jp_i−jf_j^s

whensodd (4.16)

Proposition 4.5. The maps hs for0≤s≤2mmake the lifting diagram commutative, that isds◦hs= h_s−1◦d_2t+s.

Proof. When 2tis fixed the proof of this result is an examination when sis even and when sodd, and the result follows from explicit calculations.

Case s even: We have

(ds◦hs)(f_i^2t+s) =ds





s

X

j=0

pi−jf_j^s



=

s

X

j=0

pi−jds(f_j^s) (4.17)

=

s

X

j=0

pi−j(−1)^j βyf_j^s−1+βxf_j−1^s−1 .

(h_s−1◦d_2t+s)(f_i^2t) =h_s−1 (−1)ⁱ β_yf_i^2t+s−1+β_xf_i−1^2t+s−1 (4.18)

=βyh_s−1(f_i^2t+s−1) +βxh_s−1(f_i−1^2t+s−1)

=

s−1

X

j=0

(−1)^jβ_yp_i−jf_j^s−1+ (−1)ⁱ⁻¹

s

X

j=0

(−1)^jβ_xp_i−(j+1)f_j^s−1

=

s

X

j=0

(−1)^jpi−j(βyf_j^s−1+βxf_j−1^s−1).

We observe that the expressions are equal, henceds◦hs=h_s−1◦d2t+s. Case s odd: We calculate

(ds◦hs)(f_i^2t+s) =ds



(−1)ⁱ

s

X

j=0

(−1)^jp_i−jf_j^s



=

s

X

j=0

(−1)^jp_i−jdsf_j^s (4.19)

= (−1)ⁱ

s

X

j=0

(−1)^jp_i−j(−1)^j(αyf_j^s−1−αxf_j−1^s−1)

= (−1)ⁱ

s

X

j=0

p_i−j(α_yf_j^s−1−α_xf_j−1^s−1).

(h_s−1◦d2t+s)(f_i^2t) =h_s−1 (−1)ⁱ αyf_i^2t+s−1−αxf_i−1^2t+s−1 (4.20)

= (−1)ⁱ

s

X

j=0

(−1)^jp_i−j(αyf_j^s−1−αxf_j−1^s−1).

The expressions are equal, which deals with the odd case. In total, these show that (h_s)_s≥0 defines a

lifting map.

4.4. Description of Yoneda products. In Section 4.2 we have described a basis for HH^2t+2m(A). Now we compute the Yoneda product ofξ∈HH^2t(A) andχ∈HH^2m(A).

Corollary 4.6. Let ξ(f_r^2t) =pr∈A andχ(f_r^2m) = ¯pr∈A. Then χ◦h2m(f_i^2t+2m) = X

0≤j≤2mand0≤i−j≤2t

pi−jp¯j. (4.21)

In particular if we let ξ_u^2mandξ_v^2tdenote the basis elements of Lemma 4.1 then we have ξ_u^2m·ξ_v^2t=ξ^2t+2m_u+v

(4.22)

showing thatR is closed under multiplication.

Proof. We apply the lifting formula and obtain the first part directly. If we take χ =ξ_u^2m and ξ=ξ_v^2t thenpv = 1 andpr= 0 forr6=vand similarly ¯pu= 1 and ¯pr= 0 otherwise. So we get that the image of f_ν^2t+2mis 1 ifν =u+v and is zero otherwise. The last part follows.

4.5. Completing the proof of Theorem 4.2. We are left to show that R₀ is isomorphic to the polynomial ringk[z₀, z₁], the rest follows from (4.22). We definez₀7→ξ²₀ andz₁7→ξ₂²; this extends to an algebra map (recall thatR0 is commutative). This takesz₀^rz₁^s to (ξ₀²)^r(ξ²₂)^s=ξ₀^2rξ^2s_2s=ξ^2r+2s_2s . The map is bijective: namely a general basis vectorξ_2r^2t, where we must haver≤t, factorizes uniquely as

ξ_2r^2t=ξ₀^2(t−r)ξ_2r^2r.

Corollary 4.3 is a direct consequence: The intersection ofR withNis zero, and as we have observed, any element in the span of mapsηr is inN. 2

5. Cohomology for a≥3

Now we study the cohomology whena≥3. Still letqbe ana-th root of unity and assume the algebra is

A=khX, Yi/(X^a, XY −qY X, Y^a).

(5.1)

We write againx, yfor the images ofX, Y inA.

5.1. Differentials. We assumea≥3, then we can simplify the differentials defined in 2.10 and 2.11. We observe that the elements inAintroduced in 2.6 to 2.9 depend only on the parity ofsmoduloa, and the arguments in 2.10 and 2.11 make only use of the cases wheres ≡0 or s≡1 moduloa. Using this the differentials take the following form which we will use from now:

d2t:f_i^2t7→

(γy(0)f_i^2t−1+γx(0)f_i−1^2t−1 forieven

−τy(1)f_i^2t−1+τ_x(1)f_i−1^2t−1 foriodd (5.2)

d_2t+1:f_i^2t+17→

(τy(0)f_i^2t+γx(1)f_i−1^2t forieven

−γy(1)f_i^2t+τx(0)f_i−1^2t foriodd (5.3)

where we have replaced

τ₁=τ_x τ₂=τ_y γ₁=γ_x γ₂=γ_y (5.4)

5.2. A basis for HH^2t(A) for a≥3. As observed the dimension of the degree 2t part is always 4t+ 2 which is independent ofa. We therefore expect that there should be a basis when a≥3 which is not so different from the one we had fora= 2.

Definition 5.1. Letζ_j:P_2t→Abe the map ζj(f_i^2t) =

(1 i=j 0 else.

(5.5)

Letj be even, then define

η⁺_j(f_i^2t) =

(x^a−1y^a−1 i=j

0 else.

(5.6)

Now letj be odd, then define

η⁻_j (f_i^2t) =

(xy i=j 0 else.

(5.7)

Lemma 5.2. We fix a degree2t.

(a) The classes of the elementsζi andη_j^± as defined above form a basis of HH^2t(A).

(b) The classes of the elementsη^±_j give nilpotent elements ofHH^∗(A).

Proof. Part (b) will follow again from Proposition 2.1. We prove now part (a). These are in total 4t+ 2 maps, so we only have to show that the maps lie in the kernel of d^∗_2t+1, and that they are linearly independent modulo the image ofd^∗_2t.

(1) Letξbe one these maps. We writeξ(f_i^2t) =p_i∈A, so thatp_i is either 0 or 1 or one ofx^a−1y^a−1 or xydepending on the parity of i. We need to check thatξ(d_2t+1(f_i^2t+1)) = 0.

(a) Assumeiis even, then this is equal to

ξ(d2t+1(f_i^2t+1)) =ξ(τy(0)f_i^2t+γx(1)f_i−1^2t ) =τy(0)pi+γx(1)p_i−1. (5.8)

This has to be calculated inA which is viewed as anA^eleft module. We have τ_y(0)p_i=p_iy−yp_i

(5.9)

This is zero ifpi= 1. Otherwise sinceiis even we only need to considerpi=x^a−1y^a−1and then p_iy= 0 and yp_i= 0. Next, if p_i−1= 1 then

γx(1)pi−1=

a−1

X

j=0

q^jx^a−1−j·1·x^j= (

a−1

X

j=0

q^j)x^a−1 (5.10)

and this is zero, note that 1 +q+. . .+q^a−1= 0 sinceqis ana-th root of 1. Otherwisepi−1=xy and then

γ_x(1)p_i−1=

a−1

X

j=0

q^j(x^a−1−jxyx^j) (5.11)

and this is a scalar multiple ofx^ay and hence is zero.

(b) Letibe odd, we get

ξ(−γ_y(1)f_i^2t+τ_x(0)f_i−1^2t ) =−γ_y(1)p_i+τ_x(0)p_i−1. (5.12)

By calculations similar to part (a) we see that this is zero in all cases to be considered.

(2) We consider a linear combination of the above elements and assume that it lies in the image ofd^∗_2t. Explicitly let

2t

X

j=0

c_jζ_j+ X

jeven

s⁺_jη_j⁺+ X

jodd

s⁻_jη_j⁻=ξ◦d_2t (5.13)

whereξ:P_2t−1→A, withcj ands^±_j ink. We must show that this is only possible, asξ varies, with all cj ands^±_j equal to zero.

(a) Apply the LHS to f_i^2twithieven, this gives

c_i+s⁺_i (x^a−1y^a−1).

(5.14)

On the other hand,

ξ◦d2t(f_i^2t) =γy(0)ξ(f_i^2t−1) +γx(0)ξ(f_i−1^2t−1).

(5.15)

This is an element inA viewed as an A^e left module. For any element z ∈A, γx(0)z or γy(0)z can never have a non-zero constant term sinceγx(0) andγy(0) are in the radical of A^e. Hence the Equation (5.15) does never have a non-zero constant term andc_i= 0.

We claim that we also cannot get a term which is a multiple ofx^a−1y^a−1. Namely if so this could only come from eitherγy(0)x^a−1 or fromγx(0)y^a−1. Now,

γ_y(0)x^a−1=

a−1

X

j=0

y^jx^a−1y^a−1−j=

a−1

X

j=0

(q⁻¹)^j(a−1)x^a−1y^a−1= 0 (5.16)

sincePa−1

j=0q^j= 0. Hence s⁺_i = 0. Similarly one sees thatγ_x(0)y^a−1= 0.

(b) Apply the LHS tof_i^2twithiodd, this gives c_i+s⁻_i (xy).

(5.17)

On the other hand,

ξ◦d2t(f_i^2t) =−τy(1)ξ(f_i^2t−1) +τx(1)ξ(f_i−1^2t−1).

(5.18)

As before, sinceτy(1) andτx(1) are in the radical ofA^e, this cannot have non-zero constant terms.

Hencec_i= 0.

We must check that we cannot getxy. If xy should occur in τy(1) this can only come from τy(1)x but this is equal to xy−qyx = 0. Similarly τx(1)y = 0 and we do not get xy. Hence s⁻_i = 0.

We have proved that the 4t+ 2 maps are linearly independent modulo the image ofd^∗_2t. By dimensions,

they are a basis of HH^2t(A).

The aim of this section is to prove the following.

Theorem 5.3. Letkbe a field,a≥3an integer,q∈ka primitivea-th root of unity, andAthe quantum complete intersection khX, Yi/(X^a, XY −qY X, Y^a). Assume

R:= Sp{ζ_i^2t: t≥0 and 0≤i≤2t}.

(5.19)

ThenRis a subalgebra of HH^2∗(A). It isZ2-graded withR0:=khζ_i^2t:ieveniandR1:=khζ_i^2t:iodd i.

Moreover

ζ_l^2m·ζ_r^2t=

0 l, r odd ζ_l+r^2m+2t otherwise.

(5.20)

As for the casea= 2 we can see:

Corollary 5.4. The even partR₀ of R is isomorphic to the polynomial ring in two variables.

Corollary 5.5. AssumeA is as in the Theorem, and letNbe the largest homogeneous nilpotent ideal of HH^2∗(A). Then HH^2∗(A)/Nis isomorphic to R0.

5.3. Lifting. We compute the Yoneda productχ•ξ whereχ, ξare k-linear combinations of mapsζj as in Definition 5.1.

Forξ in the span of theζ_j, the values ofξ are scalars and therefore they commute with elements of A^e. Luckily, we are only interested in the even Hochschild cohomology modulo homogeneous nilpotent elements.

Similar as for the case where a= 2 we use liftings along the minimal projective resolution to define the Yoneda products in the cohomology ring. Letξ:P_2t→Awhere

ξ(f_i^2t) :=p_i (5.21)

and we assumep_iis a scalar multiple of 1, for alli. Consequently the valuesp_icommute with all elements inA^e. As usual we setpi= 0 ifi >2tor ifi <0.

The maph0:P2t→P0 is defined by

h0(f_i^2t) :=pif₀⁰ (0≤i≤2t).

(5.22)

Moreover, we search explicit formulae for maps

h_s:P_2t+s→P_s. (5.23)

Fors≥1 we require

h_s−1◦d2t+s=ds◦hs. (5.24)

If so, then (hs)_s≥0 liftsξ along the minimal projective resolution.

5.3.1. Some formulae in A^e. In order to define such lifting mapsh_sfors >0 we establish some formulae inA^e. Let

ci= 1 +q+. . .+qⁱ for 0≤i≤a−2.

(5.25)

Definition 5.6. For an integerswe define βx(s) =

a−2

X

i=0

ciq^si(x^a−2−i⊗xⁱ) (5.26)

β_y(s) =

a−2

X

i=0

c_iq^si(yⁱ⊗y^a−2−i) (5.27)

Recall now the elements inA^ewhich occur in the definition of the differentials:

γy(s) =

a−1

X

j=0

q^js(y^j⊗y^a−1−j) (5.28)

γ_x(s) =

a−1

X

j=0

q^js(x^a−1−j⊗x^j) (5.29)

At the end we will only needs= 0 ands= 1. Recall also τy(1) = (1⊗y)−q(y⊗1) (5.30)

τ_x(1) =q(1⊗x)−(x⊗1) (5.31)

τy(0) = (1⊗y)−(y⊗1) (5.32)

τx(0) = (1⊗x)−(x⊗1).

(5.33)

With this notation, we will define mapsh_s: P_2t+s→P_s, defined on the generators f_i^2t+s of the free A^e moduleP2t+s, and we will show below that they liftξ:

Definition 5.7.

Assumesis even. For an integeriwe define the following elements in the algebra, ω(j) =

(βx(−1)βy(1) j odd

1 j even

(5.34)

With this, we define forseven

hs(f_i^2t+s) :=

(Ps

j=0p_i−jω(j)f_j^s ieven Ps

j=0p_i−jf_j^s iodd.

(5.35)

Now assumesis odd. Here we need two parameters inA^e, one forxand one fory. We set ε_x(j) =

(−β_x(0) j odd

1 j even ε_y(j) =

(1 j odd

−βy(0) j even.

(5.36)

With these, we define forsodd,

h_s(f_i^2t+s) :=

(Ps

j=0p_i−jεx(j)f_j^s ieven Ps

j=0pi−jεy(j)f_j^s iodd.

(5.37)

We will show that (hs)_s≥0 is a lifting forξ. For this, we need some formulae.

Lemma 5.8. We have that the following relations hold:

β_y(1)τ_y(1) =γ_y(2) (a)

β_x(−1)γ_y(2) =γ_y(0)β_x(0) (b)

βx(−1)βy(1) =βy(−1)βx(1) (c)

βx(1)τx(1) =−γx(2) (d)

β_y(−1)γ_x(2) =γ_x(0)β_y(0) (e)

βy(0)τy(0) =γy(1) (f)

βx(0)τx(0) =−γx(1) (g)

τ_y(1)β_y(0) =γ_y(0) (h)

τx(0)βx(−1) =−γx(−1) (i)

γx(−1)βy(1) =βy(0)γx(1) (j)

τ_y(0)β_y(−1) =γ_y(−1) (k)

τx(1)βx(0) =−γx(0) (l)

βx(0)γy(1) =γy(−1)βx(1) (m)

Proof. We prove (a) and (b), and the other relations follows from the same kind of reasoning. Start with (a), we have

β_y(1)τ_y(1) =

a−2

X

i=0

c_iqⁱ(yⁱ⊗y^a−2−i)

!

((1⊗y)−q(y⊗1)) (5.38)

=

a−2

X

i=0

ciqⁱ(yⁱ⊗y^a−1−i)−ciqⁱ⁺¹(yⁱ⁺¹⊗y^a−2−i) (5.39)

=c₀(1⊗y^a−1) +c₁q(y⊗y^a−2) +· · ·+c_a−2q^a−2(y^a−2⊗y) (5.40)

−c0q(y⊗y^a−2)− · · · −ca−3q^a−2(y^a−2⊗y)−ca−2q^a−1(y^a−1⊗1) (5.41)

=c0(1⊗y^a−1) +q(c1−c0)(y⊗y^a−2) +q²(c2−c1)(y²⊗y^a−3)+

(5.42)

· · ·+q^a−2(c_a−2−c_a−3)(y^a−2⊗y)−q^a−1c_a−2(y^a−1⊗1) (5.43)

where we have thatc₀= 1, c₁−c₀= 1 +q−1 =q, . . .

c_i+1−c_i= (1 +q+· · ·+qⁱ⁺¹)−(1 +q+· · ·+qⁱ) =qⁱ⁺¹ (5.44)

We also observe

c_a−2= 1 +q+· · ·+q^a−2=−q^a−1 (5.45)

sinceais a root of unity and hence 1 +q+· · ·+q^a−2+q^a−1= 0. Then we have, β_y(1)τ_y(1) = (1⊗y^a−1) +q²(y⊗y^a−2) +· · ·+q^2(a−1)(y^a−1⊗1) =γ_y(2) (5.46)

For the relation (b) we inspect a typical element in this sum:

ciq⁻ⁱ(x^a−2−i⊗xⁱ)q^2j(y^j⊗y^a−1−j) =ciq⁻ⁱq^2j(x^a−2−iy^j⊗xⁱ∗y^a−1−j) (5.47)

(where∗ denotes the multiplication inA^op). Now we recall thatxy=qyx(andx∗y =q⁻¹y∗x) hence x^a−2−iy^j=q^j(a−2−i)y^jx^a−2−i andxⁱ∗y^a−1−j =q^{−i(a−1−j)}y^a−1−j∗xⁱ. We get

ciq⁻ⁱq^2jq^j(a−2−i)q^{−i(a−1−j)}(y^jx^a−2−i⊗y^a−1−j∗xⁱ) = (y^j⊗y^a−2−j)ci(x^a−2−i⊗xⁱ) (5.48)

which is the most typical element in the sumγy(0)βx(0).

The relations (a) to (m) in Lemma 5.8 can be used to prove that the mapsh_sare liftings for the given mapξ:

Proposition 5.9. The lifting formulas make the suggested squares commutative, that is h_s−1◦d2t+s= ds◦hs whens≥1 andξ=µ◦h0.

Proof. We give details whens andiare even, the other cases are similar. The strategy is to apply both sides tof_i^2t+sand express the answer in terms of the basis{f_j^s−1}, with coefficients inA^eand then show that the coefficients of thef_j^s−1 in the two expressions are equal.

We have

(d_s◦h_s)(f_i^2t+s) =d_s





s

X

j=0

p_i−jω(j)f_j^s

 (5.49) 

= X

jeven, 0≤j≤s

pi−jω(j)

γy(0)f_j^s−1+γx(0)f_j−1^s−1 (5.50)

+ X

jodd, 0≤j≤s

p_i−jω(j)

−τy(1)f_j^s−1+τx(1)f_j−1^s−1 (5.51)

We split each of the two sums, and when the index is j−1 we change variables, setting l=j−1 so thatj=l+ 1 and noting thatl has opposite parity asj. As well we recallω(j) = 1 forjeven. Then this becomes

= X

jeven, 0≤j≤s

p_i−jγy(0)f_j^s−1+ X

lodd ,−1≤l≤s−1

p_i−l−1γx(0)f_l^s−1 (5.52)

+ X

jodd , 0≤j≤s

−pi−jω(j)τy(1)f_j^s−1+ X

leven,−1≤l≤s−1

pi−l−1ω(l+ 1)τx(1)f_l^s−1 (5.53)

The range of summation can be unified since f_j^s = 0 for j = −1 or j = s. We write this now as a combination in theA^e-basisf_j^s−1 for 0≤j ≤s−1, (writingj forl) and we get

= X

jeven, 0≤j≤s−1

[pi−jγy(0) +pi−j−1ω(j+ 1)τx(1)]f_j^s−1

+ X

jodd, 0≤j≤s−1

[p_i−j−1γx(0)−p_i−jω(j)τy(1)]f_j^s−1 (5.54)

On the other hand

(hs−1◦d2t+s)(f_i^2t+s) =hs−1(γy(0)f_i^2t+s−1+γx(0)f_i−1^2t+s−1) (5.55)

=γy(0)[

s−1

X

j=0

p_i−jεx(j)f_j^s−1] +γx(0)[

s−1

X

j=0

p_i−1−jεy(j)f_j^s−1] (5.56)

=

s−1

X

j=0

[p_i−jγ_y(0)ε_x(j) +p_i−1−jγ_x(0)ε_y(j)]f_j^s−1. (5.57)

We must show that for eachj the coefficients off_j^s−1 in (5.54) and in (5.57) are equal.

(a) Assume firstj is even. We require

p_i−jγy(0) +p_i−j−1ω(j+ 1)τx(1) =p_i−jγy(0)εx(j) +p_i−j−1γx(0)εy(j) (5.58)

Forj even,ε_x(j) = 1 and the first terms agree. The second terms agree provided ω(j+ 1)τx(1) =γx(0)εy(j)

(5.59)

Consider the LHS, by identities (c), (d) and (e) it is equal to

βy(−1)βx(1)τx(1) =−βy(−1)γx(2) =−γx(0)βy(0) =γx(0)εy(j) (5.60)

from the definition ofεy(j) in this case. Hence the second terms agree as well.

(b) Now assumej is odd. We require

p_i−j−1γx(0)−p_i−jω(j)τy(1) =p_i−jγy(0)εx(j) +p_i−1−jγx(0)εy(j) (5.61)

Forj odd,ε_y(j) = 1 and the terms withp_i−j−1 agree. For the other two terms to agree we need γ_y(0)ε_x(j) =−ω(j)τ_y(1)

(5.62)

We have using the definition and identities (a) and (b) that

−ω(j)τy(1) =−βx(−1)βy(1)τy(1) =−βx(−1)γy(2) =−γy(0)βx(0) =γy(0)εx(j) (5.63)

as required.

Similar as for the casea= 2 we define the Yoneda product of the residue classes represented byξ of degree 2tandχ of degree 2mto be the residue class represented by the composition

ξ•χ=χ◦h_2m. (5.64)

5.4. Description of Yoneda products of basis elements when a≥3. In the definition 5.7 of the lifting maps, we have the term ω(j) = β_x(−1)β_y(1) ∈ A^e (for j odd). When this is evaluated inA, it becomesω(j)·1A. We claim that this is always zero, in factβy(1)·1A= 0.

Namely, we must viewAas anA^ebimodule and then β_y(1)·1_A=

a−2

X

i=0

c_iqⁱy^a−2=

a−2

X

i=0

c_iqⁱ

! y^a−2 (5.65)

The following shows that this is zero:

Lemma 5.10. Let qbe a primitivea-th root of unity for a≥3. Let ci= 1 +q+. . .+qⁱ fori≥0, then

a−2

X

i=0

c_iqⁱ= 0 (5.66)

Proof. Set alsoc−1:= 0. Then we have fori≥0 thatci−ci−1=qⁱ. We get

a−2

X

i=0

c_iqⁱ=X

i

c_i(c_i−c_i−1) (5.67)

Therefore (all summations are fromi= 0 toa−2) (1 +q)(X

i

ciqⁱ) =X

i

ciqⁱ+X

i

ciqⁱ⁺¹

=X

i

ci(ci−c_i−1) +X

i

ci(ci+1−ci)

=X

i

(cici+1−cic_i−1)

=c_a−2c_a−1−c0c₋₁

=0

sincec_a−1 = 1 +q+. . .+q^a−1 = 0 andc₋₁= 0. But q6=−1, so we can cancel by (1 +q) and get the

claim.