HOCHSCHILD COHOMOLOGY OF ALGEBRAS OF QUATERNION … · additive structure of the algebra HH∗(R) was described in [10] for group blocks having tame representation type and one or

Algebra i analiz St. Petersburg Math. J.Tom. 18 (2006), 1 Vol. 18 (2007), No. 1, Pages 37–76

S 1061-0022(06)00942-3Article electronically published on November 27, 2006

HOCHSCHILD COHOMOLOGY OF ALGEBRAS OF QUATERNIONTYPE, I: GENERALIZED QUATERNION GROUPS

A. I. GENERALOV

Abstract. In terms of generators and defining relations, a description is given ofthe Hochschild cohomology algebra for one of the series of local algebras of quater-nion type. As a corollary, the Hochschild cohomology algebra is described for thegroup algebras of generalized quaternion groups over algebraically closed fields ofcharacteristic 2.

Introduction

Let R be a finite-dimensional algebra over a field K, let Λ = R e = R ⊗K Rop be itsenveloping algebra, and let HHn(R) = Extn

Λ(R, R) be the nth Hochschild cohomologygroup of the algebra R (with coefficients in the R-bimodule R). On the vector space

HH∗(R) =⊕n≥0

HHn(R) =⊕n≥0

ExtnΛ(R, R),

we can introduce the -product with respect to which it becomes an associative K-algebra (see [1, §5], [2, Chapter XI], [3]); this algebra is called the Hochschild cohomologyalgebra. The algebra HH∗(R) is a graded commutative algebra [3]; moreover, the -product coincides with the Yoneda product on the Ext-algebra

⊕n≥0 Extn

Λ(R, R) of theΛ-module R [4, p. 120].

In recent years, progress has been made in the investigation of the multiplicativestructure of the Hochschild cohomology algebra for finite-dimensional algebras. In [5, 6]it was proved that HH∗(K[G]) H∗(G)⊗K K[G] if G is a commutative finite group. In[7], a description was obtained of the Hochschild cohomology algebra for the symmetricgroup S3 over the field F3 and for the alternating group A4 and the dihedral 2-groupsover the field F2. In [8], the algebra HH∗(R) was described in the case where R is aself-injective Nakayama algebra, and in [9], the subalgebra HHr∗(R) generated by thehomogeneous elements of degrees divisible by r was identified, where R is the so-calledMobius algebra (here r is a parameter related to the algebra R). We note also that theadditive structure of the algebra HH∗(R) was described in [10] for group blocks havingtame representation type and one or three simple modules.

In the recent paper [11] by the author, a description of the Hochschild cohomologyalgebra was given for algebras of dihedral type in the family D(3K) over an algebraicallyclosed ground field of characteristic two. We recall that algebras of dihedral, semidihedral,and quaternion types appeared in the work of Erdmann on classification of group blocksof tame representation type (see [12]). By using the results of [13], our descriptionof the algebra HH∗(R) was extended additionally to algebras of three families in theclassification of Erdmann, namely, to those of the families D(3A)1, D(3B)1, and D(3D)1

2000 Mathematics Subject Classification. Primary 13D03.Key words and phrases. Algebras of quaternion type, Hochschild cohomology, generalized quaternion

groups.

c©2006 American Mathematical Society

37

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38 A. I. GENERALOV

(in the notation of [12]). In particular, this allowed us to give a description of theHochschild cohomology algebra for all group blocks with dihedral defect group and threesimple modules.

In the present paper, we use the technique of [11] to compute the Hochschild coho-mology algebra for a family of local algebras of quaternion type, namely, for the familythat contains the group algebras of generalized quaternion groups over an algebraicallyclosed field of characteristic 2 (see [12]). The corresponding approach was used earlierin the computation of Yoneda algebras for algebras of dihedral or semidihedral type (see[14]–[19]). Its specific feature is that, on the basis of some empirical observations, aconjecture is stated concerning the structure of the minimal projective resolution of themodules under consideration and then, upon verifying this conjecture, the resolutions areused in the calculation of the corresponding cohomology algebras. For local algebras ofquaternion type to be treated here, in the same way we build the minimal Λ-projectiveresolution of the module R (see §2). With the help of this resolution, we pick a (finite)set of generators for the algebra HH∗(R) and find relations satisfied by these generators(see §§3 and 4). We see that the minimal Λ-projective resolution of the bimodule R hasperiod 4 and consequently, the Hochschild cohomology of the local algebras mentionedabove is also periodic. It should be noted that the initial part (containing three arrows)of the bimodule resolution was described in [20] for an arbitrary algebra. However, inthe context of the family of algebras under consideration, the proof of the 4-periodicityof this resolution requires additional effort (cf. Proposition 2.5). So, it seems that ourdirect construction of bimodule resolutions is more efficient. We also note that a similarperiodicity of bimodule resolutions is valid for the self-injective algebras of the tree classAn; see [8, 9].

For completeness, in the Appendix we briefly describe the results of the calculation ofthe Yoneda algebra for the local algebras under consideration.

§1. Statement of the main result

Let K be an algebraically closed field of arbitrary characteristic. For any k ∈ N withk ≥ 2, we introduce the K-algebra Rk = K〈X, Y 〉/I, where I is an ideal of the freealgebra K〈X, Y 〉 generated by the elements

(1.1) X2 − Y (XY )k−1, Y 2 − X(Y X)k−1, (XY )k − (Y X)k, X(Y X)k.

The images of the elements X and Y under the canonical homomorphism from K〈X, Y 〉to Rk will be denoted by x and y, respectively. The algebra Rk is a symmetric localalgebra of tame representation type [12, III.1]; moreover, in terms of [12, Chapter VII],Rk is an algebra of quaternion type. If G is the generalized quaternion group of order2n (n ≥ 3) and char K = 2, then the group algebra KG is isomorphic to the algebra Rk

with k = 2n−2.To describe the Hochschild cohomology algebra HH∗(Rk) for the algebras Rk (k ≥ 2),

we construct several graded algebras. Set

(1.2) X1 = p1, p2, p′2, p3, u1, u

′1, u2, v1, v2, v

′2, w, z.

The grading on the algebra K[X1] will be such that

(1.3)

deg p1 = deg p2 = deg p′2 = deg p3 = 0, deg u1 = deg u′

1 = deg u2 = 1,

deg v1 = deg v2 = deg v′2 = 2, deg w = 3, deg z = 4.


HOCHSCHILD COHOMOLOGY OF ALGEBRAS OF QUATERNION TYPE. I 39

Then we introduce the graded K-algebra A1 = K[X1]/I1, where the ideal I1 is gener-ated by the following elements:of degree 0:

pk1 , p2

2, (p′2)2, p1p2, p1p

′2, p2p

′2,

p23, p1p3, p2p3, p′2p3;

(1.4)

of degree 1:

p2u1 − p′2u′1, p′2u1 − p1u

′1, p1u1 − p2u

′1;(1.5)

pk−11 u2, p2u2, p′2u2, p3u2, p2u1 − pk−2

1 u2;(1.6)

of degree 2: ⎧⎪⎪⎨⎪⎪⎩pk−11 v1, p2v2, p′2v

′2, p3v1, p3v2, p3v

′2,

p2v1 − p1v′2, p2v1 − p′2v2, p2v1 − p3u

21,

p′2v1 − p1v2, p′2v1 − p2v′2, p′2v1 − p3(u′

1)2,u1u

′1 − kpk−2

1 v1;

(1.7)

u22, u1u2 − kp3(u′

1)2, u′

1u2 − kp3u21;(1.8)

of degree 3:

u′1v2 − u1v

′2, u′

1v1 − u1v2, u1v1 − u′1v

′2, u3

1 − (u′1)

3;(1.9)

p2w, p′2w, p3w, u′1v2 − pk−2

1 w;(1.10)

u2v2, u2v′2, u2v1 − p1w;(1.11)

of degree 4:

v22 , (v′2)

2, v1v2, v1v′2, v2v

′2, v2

1 − p21z;(1.12)

u1w, u′1w, u2w;(1.13)

of degree 5:

(1.14) v2w, v′2w, v1w + p1u2z;

of degree 6:

(1.15) w2.

Moreover, we induce a grading on the algebra A1 from the grading on K[X1].Next, we consider the algebra A2 = K[X2]/I2, where

(1.16) X2 = X1 \ u2, w

with X1 as in (1.2); the grading on this algebra is introduced by restricting the grading ofK[X1] (see (1.3)), and the ideal I2 is spanned by the generators of the ideal I1 occurringin (1.4) with k = 2, in (1.5) and (1.7) with k = 2, in (1.9), and in (1.12). Since the idealI2 is homogeneous, the grading on the algebra K〈X2〉 induces a grading on A2.

Next, consider the set

(1.17) X3 = p1, p2, p′2, u2, u3, v0, v

′0, v1, w, z.

We introduce a grading on the algebra K〈X3〉 so that

(1.18)

deg p1 = deg p2 = deg p′2 = 0, deg u2 = deg u3 = 1,

deg v0 = deg v′0 = deg v1 = 2, deg w = 3, deg z = 4.


40 A. I. GENERALOV

Then we define a graded K-algebra A3 = K〈X3〉/I3, where the ideal I3 is generated bythe following elements:of degree 0:

pk+11 , p2

2, (p′2)2, p1p2, p1p

′2, p2p

′2;(1.19)

of degree 1:

pk−11 u2, p2u2, p′2u2;(1.20)

p1u3;(1.21)

of degree 2:

p1v0, p1v′0, p2v0, p′2v

′0, p2v1, p′2v1, pk−1

1 v1;(1.22)

p2v′0 + p′2v0;(1.23)

of degree 3:

u2v0, u2v′0;(1.24)

p2w, p′2w, u2v1 − p1w;(1.25)

u3v1, p′2u3v0 − pk−11 w;(1.26)

of degree 4:

v0v′0, v0v1, v′0v1, v2

0 − 2p′2z, (v′0)2 − 2p2z, v2

1 + p21z;(1.27)

u2w, pk1z;(1.28)

u3w;(1.29)

of degree 5:

(1.30) v0w, v′0w, v1w + p1u2z;

and, additionally, by the elements of the form

(1.31) ab − (−1)deg a deg bba with a, b ∈ X3.

Since the ideal I3 is homogeneous, the grading on the algebra K〈X3〉 induces a gradingon A3.

Next, put

(1.32) X4 =(X3 \ u2, u3, w

)∪ u1, u

′1, w.

A grading on the algebra K〈X4〉 is introduced by the requirement that relations (1.18)be satisfied for the elements of the set X3 \ u2, u3, w, and, moreover,

deg u1 = deg u ′1 = 1, deg w = 3.

Then we define a graded K-algebra A4 = K〈X4〉/I4, where the ideal I4 is generated bythe generators of I3 listed in (1.19), (1.22), and (1.27), by the elements

pk1 u1, pk

1 u ′1, p2u1, p′2u

′1;

u1u′1, p′2v0 + p2v

′0 − pk−1

1 ;

u ′1v0, u1v

′0, p2w, p′2w, u1v1 − u ′

1v1, p1w − 2u1v1;

u1w, u ′1w, v0w, v′0w,

and by the elements of the form

ab − (−1)deg a deg bba with a, b ∈ X4.




Now, we consider the set

(1.33) X5 = p1, p2, p′2, u2, u4, u

′4, v0, v

′0, v1, w1, w

′1, w, z.

We introduce the grading on the algebra K〈X3〉 so that

(1.34)

⎧⎪⎨⎪⎩deg p1 = deg p2 = deg p′2 = 0, deg u2 = deg u4 = deg u′

4 = 1,

deg v0 = deg v′0 = deg v1 = 2, deg w1 = deg w′1 = deg w = 3,

deg z = 4.

Then we define a graded K-algebra A5 = K〈X5〉/I5, where the ideal I4 is generated bythe generators of I3 of the form (1.19), (1.20), (1.22), (1.24), (1.25), (1.27), (1.28), and(1.30), by the elements⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

p1u4, p1u′4, p2u4, p2u

′4, p′2u4, p′2u

′4, p2v

′0, p′2v0,

u2u4, u2u′4, u4u

′4, p2w1, p′2w

′1, p1w1, p1w

′1,

u4v′0, u′

4v0, u4v1, u′1v1,

p′2w1 − p2w′1, p′2w1 − u4v0, p′2w1 − u′

4v′0,

u2w1, u2w′1, u4w1, u4w

′1, u′

4w1, u′4w

′1,

v0w′1, v′0w1, v1w1, v1w

′1, v0w1 − 2u4z, v′0w

′1 − 2u′

4z, w1w′1,

(1.35)

p′2w1 − pk−11 w, u4w, u′

4w, w1w, w′1w,




Next, we consider the set

(1.36) X6 =(X5 \ w

)∪ w

and introduce a grading on the algebra K〈X6〉 by the requirement that relations (1.34)be satisfied for the elements of the set X5 \ w, and, moreover,

deg w = 3.

Then we define a graded K-algebra A6 = K〈X6〉/I6, where the ideal I6 is generated bythe generators of I3 occurring in (1.19), (1.20), (1.22), (1.24), (1.27), and (1.35), by theelements

p2w, p′2w, p2w1 − 12pk

1w, u2v1 − 12p2

1w,u2w, u4w, u′

4w, ww1, ww′1,



The algebra A6 inherits the natural grading from the algebra K〈X6〉.Finally, we consider the set

(1.37) X7 =(X5 \ u2, u4, u

′4, w1, w

′1, w

)∪ u3

and introduce a grading on the algebra K〈X7〉 so that relations (1.34) be satisfied for theelements of the set X7 \ u3, and, moreover,

deg u3 = 1.

Then we define a graded K-algebra A7 = K〈X7〉/I7; here the ideal I7 is generated bythe generators of I3 occurring in (1.19), (1.22), and (1.27) by the elements

pk1 u3, p2v

′0 − p′2v0, pk

1z,


42 A. I. GENERALOV



The algebra A7 inherits the natural grading from the algebra K〈X7〉.

Theorem 1.1. Let R = Rk where k ∈ N \ 1, and let p = char K.1) Assume that p = 2.

1 a) If k ≥ 3, then, as a graded K-algebra, the Hochschild cohomology algebra HH∗(R)is isomorphic to A1.

1 b) If k = 2, then HH∗(R) A2 as graded K-algebras.2) Assume that p = 3.

2 a) If 3 does not divide k, then HH∗(R) A3 as graded K-algebras.2 b) If 3 divides k, then HH∗(R) A4 as graded K-algebras.

3) Assume that p ∈ 2, 3.3 a) If p divides neither 3 − 2k, nor k, then HH∗(R) A5 as graded K-algebras.3 b) If p does not divide 3−2k but divides k, then HH∗(R) A6 as graded K-algebras.3 c) If p divides 3 − 2k, then HH∗(R) A7 as graded K-algebras.

This description of the Hochschild cohomology algebra immediately implies the fol-lowing statement.

Corollary 1.2. Let G be a generalized quaternion group of order 2n (n ≥ 3), and let Kbe an algebraically closed field of characteristic 2. Then, with k = 2n−2, we have

HH∗(KG) A1 if n > 3,

A2 if n = 3.

In the course of the proof of Theorem 1.1 we also compute the dimensions of thegroups HHn(R). This is of independent interest; we collect the results in the followingstatement.

Proposition 1.3. Let R = Rk.1) If p = 2, then

dimK HHn(R) =

k + 3 if n ≡ 0, 3 (mod 4),k + 5 if n ≡ 1, 2 (mod 4).

2) Assume that p = 3.2a) If 3 does not divide k, then

dimK HHn(R) =

k + 3 if n = 0,

k + 2 if n > 0.

2b) If 3 divides k, then dimK HHn(R) = k + 3 for any n ≥ 0.3) Assume that p ∈ 2, 3.

3a) If p divides neither 3 − 2k, nor k, then

dimK HHn(R) =

⎧⎪⎨⎪⎩k + 3 if n = 0,

k + 1 if n ≡ 1, 2 (mod 4),k + 2 if n > 0 and n ≡ 0, 3 (mod 4).

3b) If p does not divide 3 − 2k, but divides k, then

dimK HHn(R) =

k + 3 if n ≡ 0, 3 (mod 4),k + 1 if n ≡ 1, 2 (mod 4).



3c) If p divides 3 − 2k, then

dimK HHn(R) =

k + 3 if n = 0,

k + 2 if n > 0.

Remark 1.4. In the case where char K = 2, Part 1 of Proposition 1.3 extends the cor-responding result of [10] for group algebras of the generalized quaternion groups to theentire family of the algebras Rk.

§2. Resolution

Put R = Rk (k ≥ 2). The algebra R admits the following set as a K-basis:

Bst = (xy)i | 0 ≤ i ≤ k − 1∪ (yx)i | 1 ≤ i ≤ k ∪ x(yx)i, y(xy)i | 0 ≤ i ≤ k − 1.

This set consists of all nonzero paths of the quiver of R (this quiver has one vertex andtwo loops x and y). The set Bst is called the standard basis of R. In its turn, theenveloping algebra Λ of the algebra R has a K-basis consisting of elements of the form

(2.1) u ⊗ v with u, v ∈ Bst.

This basis of Λ is also said to be standard. We introduce the following grading on Λ: foran element of the form (2.1), we define its degree as the sum of the lengths of the pathsu and v, and then we put Λ =

∑n≥0 Λn, where Λn is the K-vector space spanned by the

elements (2.1) with degree n.Multiplication from the right by an element λ ∈ Λ induces an endomorphism λ∗ of the

left Λ-module Λ; for simplicity, we often keep denoting this endomorphism by λ. Some-times, we consider an endomorphism of the right Λ-module Λ induced by multiplicationfrom the left by an element λ. This endomorphism will be denoted by ∗λ.

We build the following 4-periodic sequence in the category of Λ-modules:

(2.2) Q0d0←− Q1

d1←− Q2d2←− Q3

d3←− · · · ,

where Q0 = Q3 = Λ, Q1 = Q2 = Λ2,

d0 =(x ⊗ 1 − 1 ⊗ x, y ⊗ 1 − 1 ⊗ y

),

(2.3)

d1 =

⎛⎜⎜⎝x ⊗ 1 + 1 ⊗ x −k−2∑i=0

(yx)iy ⊗ y(xy)k−2−i, −k−1∑i=0

(xy)i ⊗ (yx)k−1−i

−k−1∑i=0

(yx)i ⊗ (xy)k−1−i, y ⊗ 1 + 1 ⊗ y −k−2∑i=0

(xy)ix ⊗ x(yx)k−2−i

⎞⎟⎟⎠ ,

d2 =(

x ⊗ 1 − 1 ⊗ xy ⊗ 1 − 1 ⊗ y

),

(2.4) d3 = g∗ with g = (x ⊗ 1 + 1 ⊗ x) ·k−1∑i=0

(yx)i ⊗ (xy)k−1−i · (y ⊗ 1 + 1 ⊗ y),

and, for 0 ≤ j ≤ 3 and l ∈ N,

Q4l+j = Qj , d4l+j = dj .

Also, we consider the homomorphism µ : Q0 = Λ → R induced by the product in R:µ(a ⊗ b) = ab.


44 A. I. GENERALOV

Theorem 2.1. The sequence (2.2) together with the augmentation map µ is the minimalΛ-projective resolution of the module R; in particular, the Λ-module R is Ω-periodic withperiod 4, i.e., Ω4(ΛR) ΛR.

Remark 2.2. With the help of Happel’s lemma [22] (see also a refinement in [23]), thedescription of the modules Qj in the resolution (2.2) could be deduced from the con-struction of the minimal projective resolution of a unique simple R-module that is givenin the Appendix below. However, we obtain this description directly in the course of theproof of Theorem 2.1.

Remark 2.3. The fact that (2.2) is a differential sequence is established by direct calcu-lations. In order to verify the formula d2d3 = 0 = d3d4, we may use the relation

g = (y ⊗ 1 + 1 ⊗ y) ·k−1∑i=0

(xy)i ⊗ (yx)k−1−i · (x ⊗ 1 + 1 ⊗ x).

The exactness at the member Q0, i.e., the relation Kerµ = Im d0, is a well-known fact(see, e.g., [21, Proposition 2.1]).

We establish the exactness of the sequence at Q1 by proving the following lemma.

Lemma 2.4. Ker d0 ⊂ Im d1.

Proof. Let s = (q, q) ∈ Ker d0 ⊂ Q1 = Λ2, i.e.,

(2.5) q · (x ⊗ 1 − 1 ⊗ x) + q · (y ⊗ 1 − 1 ⊗ y) = 0.

Let q(t) and q (t) be the homogeneous components of degree t (0 ≤ t ≤ 4k) of q andq, respectively. From (2.5) it easily follows that q(0) = 0 = q (0). We fix t such thatthe homogeneous components of q and q of degree i with i < t are equal to zero. Weshall modify s successively, by adding elements of Im d1, so as to get an element allhomogeneous components of which of degrees i ≤ t are zero. After finitely many steps,we replace the initial s ∈ Ker d0 by the zero element; this will yield the inclusion Ker d0 ⊂Im d1.

We introduce additional notation, letting h1 (respectively, h2) denote the first (respec-tively, second) column of the matrix of the differential d1, which is viewed as an elementin Im d1 ⊂ Q1.

Step 1. First, we consider the case where t = 2 + 1 with 0 ≤ ≤ k − 2. We representthe homogeneous components q(t) and q (t) as linear combinations of the standard basiselements in Λ (with coefficients αi, . . . , fi, αi, . . . , fi in K):

q(t) =∑

i=0

αi(xy)i ⊗ x(yx)−i +∑

i=0

βi(xy)i ⊗ y(xy)−i

+∑

i=1

γi(yx)i ⊗ x(yx)−i +∑

i=1

ϕi(yx)i ⊗ y(xy)−i

+∑

i=0

aix(yx)i ⊗ (xy)−i +−1∑i=0

bix(yx)i ⊗ (yx)−i

+∑

i=0

ciy(xy)i ⊗ (xy)−i +−1∑i=1

fiy(xy)i ⊗ (yx)−i,

(2.6)



q (t) =∑

i=0

αi(yx)i ⊗ y(xy)−i +∑

i=0

βi(yx)i ⊗ x(yx)−i

+∑

i=1

γi(xy)i ⊗ y(xy)−i +∑

i=1

ϕi(xy)i ⊗ x(yx)−i

+∑

i=0

aiy(xy)i ⊗ (yx)−i +−1∑i=0

biy(xy)i ⊗ (xy)−i

+∑

i=0

cix(yx)i ⊗ (yx)−i +−1∑i=1

fix(yx)i ⊗ (xy)−i.

(2.7)

It should be noted that the form of q (t) corresponds to the form of q(t) via the obvioussymmetry x → y, y → x. Substituting (2.6) and (2.7) in (2.5), we obtain the followingequations for the scalar coefficients in (2.6) and (2.7) (for the reader’s convenience, inbrackets we indicate the elements of the standard basis of R at which the correspondingscalars are compared):

αi = bi for 0 ≤ i ≤ − 1((xy)ix ⊗ (xy)−ix

);(2.8)

α = a

((xy)x ⊗ x

);(2.9)

βi = fi for 0 ≤ i ≤ − 1((xy)ix ⊗ (yx)−iy

);(2.10)

βi = fi−1 for 1 ≤ i ≤ ((xy)i ⊗ (xy)+1−i

);(2.11)

ϕi = ci−1 for 1 ≤ i ≤ ((yx)i ⊗ (xy)l+1−i

);(2.12)

c = β

((yx)y ⊗ x

);

c = 0((yx)+1 ⊗ 1

);

β0 = 0(1 ⊗ (xy)+1

).(2.13)

By the symmetry mentioned above, we also obtain the following equations (here we donot indicate the corresponding basis vectors):

αi = bi (0 ≤ i ≤ − 1); α = a; βi = fi (0 ≤ i ≤ − 1);βi = fi−1 (1 ≤ i ≤ ); ϕi = ci−1 (1 ≤ i ≤ ); c = β = 0; β0 = 0.

Relations (2.10), (2.11), and (2.13) imply that

0 = β0 = f0 = β1 = f1 = β2 = · · · = f−1 = β.

By (2.8), for 0 ≤ i ≤ − 1 we have

(2.14) αi(xy)i ⊗ x(yx)−i + bix(yx)i ⊗ (yx)−i = αi(xy)i ⊗ (yx)−i · (1 ⊗ x + x ⊗ 1).

Hence, replacing s with

(2.15) s′ = (q′, q ′) := s −−1∑i=0

αi(xy)i ⊗ (yx)−i · h1,


46 A. I. GENERALOV

we get an element for which the corresponding homogeneous component q′(t) in q′ ofdegree t involves no nonzero summands of the form

−1∑i=0

α′i(xy)i ⊗ x(yx)−i +

−1∑i=0

b′ix(yx)i ⊗ (yx)−i

with α′i, b

′i ∈ K. Moreover, under such a replacement we can change only homogeneous

components of q and q with degrees exceeding t. Hence, we may assume that, alreadyfor the initial element s, we have αi = 0 = bi with 0 ≤ i ≤ − 1.

Next, we have

ϕi(yx)i ⊗ y(xy)−i + ci−1y(xy)i−1 ⊗ (xy)+1−i

= ϕiy(xy)i−1 ⊗ y(xy)−i · (x ⊗ 1 + 1 ⊗ x),(2.16)

γi(yx)i ⊗ x(yx)−i

= γi(yx)i ⊗ (yx)−i · (x ⊗ 1 + 1 ⊗ x)(2.17)

for 1 ≤ i ≤ ,

(2.18) aix(yx)i ⊗ (xy)−i = ai(xy)i ⊗ (xy)−i · (x ⊗ 1 + 1 ⊗ x)

for 0 ≤ i ≤ − 1, and

(2.19) α(xy) ⊗ x + ax(yx) ⊗ 1 = α(xy) ⊗ 1 · (x ⊗ 1 + 1 ⊗ x).

As before (cf. (2.15)), we modify s with the help of a suitable multiple of the column h1,obtaining a new element such that all coefficients in the decomposition of its componentof degree t (this decomposition is similar to (2.6)) are zero. Hence, we may assume thatfor the initial s we have q(t) = 0. By symmetry, we may also assume that q (t) = 0 (toestablish this, we can argue as above, using the column h2 in place of h1).

Step 2. Now we put t = 2 with 1 ≤ ≤ k − 1. Then

q(t) =∑

i=0

αi(xy)i ⊗ (xy)−i +−1∑i=0

βi(xy)i ⊗ (yx)−i

+∑

i=1

γi(yx)i ⊗ (xy)−i +−1∑i=1

ϕi(yx)i ⊗ (yx)−i

+−1∑i=0

aix(yx)i ⊗ x(yx)−1−i +−1∑i=0

bix(yx)i ⊗ y(xy)−1−i

+−1∑i=0

ciy(xy)i ⊗ x(yx)−1−i +−1∑i=0

fiy(xy)i ⊗ y(xy)−1−i,

(2.20)

q (t) =∑

i=0

αi(yx)i ⊗ (yx)−i +−1∑i=0

βi(yx)i ⊗ (xy)−i

+∑

i=1

γi(xy)i ⊗ (yx)−i +−1∑i=1

ϕi(xy)i ⊗ (xy)−i

+−1∑i=0

aiy(xy)i ⊗ y(xy)−1−i +−1∑i=0

biy(xy)i ⊗ x(yx)−1−i

+−1∑i=0

cix(yx)i ⊗ y(xy)−1−i +−1∑i=0

fix(yx)i ⊗ x(yx)−1−i,

(2.21)



where αi, . . . , fi, αi, . . . , fi ∈ K. As at Step 1, the notation in (2.20) and (2.21) employsthe symmetry mentioned above. Substituting (2.20) and (2.21) in (2.5), we obtain thefollowing equations (again we indicate the corresponding basis elements):

αi = bi for 0 ≤ i ≤ − 1((xy)ix ⊗ (xy)−i

);

α = f−1

((xy) ⊗ x

);

βi = fi for 0 ≤ i ≤ − 1((xy)ix ⊗ (yx)−i

);(2.22)

βi = fi−1 for 1 ≤ i ≤ − 1((xy)i ⊗ (xy)−ix

);(2.23)

ϕi = ci−1 for 1 ≤ i ≤ − 1((yx)i ⊗ (xy)−ix

);

γ = c−1

((yx) ⊗ x

);

β0 = 0(1 ⊗ (xy)x

);(2.24)

α = 0((xy)x ⊗ 1

).

Relations (2.22)–(2.24) imply

0 = β0 = f0 = β1 = f1 = · · · = f−2 = β−1 = f−1.

Since

αi(xy)i ⊗ (yx)−i + bix(yx)i ⊗ y(xy)−1−i = αi(xy)i ⊗ y(xy)−1−i · (1 ⊗ x + x ⊗ 1)

for 0 ≤ i ≤ − 1, we have

ϕi(yx)i ⊗ (yx)−i + ci−1y(xy)i−1 ⊗ x(yx)−i = ϕiy(xy)i−1 ⊗ (yx)−i · (x ⊗ 1 + 1 ⊗ x),γi(yx)i ⊗ (xy)−i = γi(yx)i−1y ⊗ (xy)−i · (x ⊗ 1 + 1 ⊗ x)

for 1 ≤ i ≤ − 1, and

γ(yx) ⊗ 1 + c−1y(xy)−1 ⊗ x = γy(xy)−1 ⊗ 1 · (x ⊗ 1 + 1 ⊗ x),

aix(yx)i ⊗ x(yx)−1−i = ai(xy)i ⊗ x(yx)−1−i · (x ⊗ 1 + 1 ⊗ x)

for 0 ≤ i ≤ − 2. Moreover, for > 1 we have

a−1x(yx)−1 ⊗ x = a−1(xy)−1 ⊗ 1 · (x ⊗ 1 + 1 ⊗ x),

and if = 1, then

a0x ⊗ x = a0x ⊗ 1 · (x ⊗ 1 + 1 ⊗ x − y(xy)k−1 ⊗ 1).

Now, changing s with the help of a suitable multiple of the column h1 (cf. Step 1),we may assume that q(t) = 0 for the initial s; by symmetry, we may also assume thatq (t) = 0.

Step 3. Let t = 2k − 1. Again we represent the homogeneous components q(t) andq (t) in the form (2.6) and (2.7), respectively; here we put = k − 1. Substituting this in(2.5) and arguing as at Step 1, we obtain equations (2.8)–(2.12) (again, = k − 1) andthe equations

ck−1 + ck−1 = 0;

ck−1 = βk−1;(2.25)

β0 + β0 = 0.(2.26)


48 A. I. GENERALOV

Relations (2.10) and (2.11) imply

β := β0 = f0 = β1 = f1 = · · · = fk−2 = βk−1.

Using symmetry and also (2.25) and (2.26), we obtain

β0 = f0 = β1 = f1 = · · · = fk−2 = βk−1 = ck−1 = −β.

Arguing as at Step 1 (see (2.14) and (2.16)–(2.19)), we may assume that for q(t),

αi = 0 (0 ≤ i ≤ k − 1); bi = 0 (0 ≤ i ≤ k − 2); γi = 0 (1 ≤ i ≤ k − 1);

ai = 0 (0 ≤ i ≤ k − 1); ϕi = 0 (1 ≤ i ≤ k − 1); ci = 0 (0 ≤ i ≤ k − 2).

Symmetry allows us to assume that for the homogeneous component q (t) the correspond-ing scalars are also zero (i.e., in the above formulas we can insert the tilde above thescalars). Thus,

u :=(

q(t)

q (t)

)= β ·

( ∑k−1i=0 (xy)i ⊗ y(xy)k−1−i −

∑k−1i=0 y(xy)i ⊗ (yx)k−1−i

−∑k−1

i=0 (yx)i ⊗ x(yx)k−1−i +∑k−1

i=0 x(yx)i ⊗ (xy)k−1−i

).

A direct calculation shows that

u = β · (1 ⊗ x − x ⊗ 1) · h1 ∈ Im d1.

Hence, replacing s by s − u, we may assume that q(t) = 0 = q (t) for the initial s.Step 4. Let t = 2 with k ≤ ≤ 2k − 1, and put = − k. The decomposition of the

homogeneous components q(t) and q (t) in the standard basis can be written as follows:

q(t) =k∑

i=

αi(xy)i ⊗ (xy)−i +k∑

i=+1

βi(xy)i ⊗ (yx)−i

+k−1∑i=

γi(yx)i ⊗ (xy)−i +k−1∑

i=+1

ϕi(yx)i ⊗ (yx)−i

+k−1∑i=

aix(yx)i ⊗ x(yx)−1−i +k−1∑i=

bix(yx)i ⊗ y(xy)−1−i

+k−1∑i=

ciy(xy)i ⊗ x(yx)−1−i +k−1∑i=

fiy(xy)i ⊗ y(xy)−1−i,

(2.27)

q (t) =k∑

i=

αi(yx)i ⊗ (yx)−i +k∑

i=+1

βi(yx)i ⊗ (xy)−i

+k−1∑i=

γi(xy)i ⊗ (yx)−i +k−1∑

i=+1

ϕi(xy)i ⊗ (xy)−i

+k−1∑i=

aiy(xy)i ⊗ y(xy)−1−i +k−1∑i=

biy(xy)i ⊗ x(yx)−1−i

+k−1∑i=

cix(yx)i ⊗ y(xy)−1−i +k−1∑i=

fix(yx)i ⊗ x(yx)−1−i,

(2.28)

where αi, . . . , fi, αi, . . . , fi ∈ K. In fact, the case where = k differs a little from theother cases treated at this step. We include this case in the arguments below, assumingthat, for = k, in (2.27) and (2.28) we have αk = 0 = αk.



Substituting (2.27) and (2.28) in (2.5), we obtain

αi = bi and ϕi = ci−1 for + 1 ≤ i ≤ k − 1;(2.29)

βi = fi and βi = fi−1 for + 1 ≤ i ≤ k − 1;(2.30)

βk − ck−1 − fk−1 = 0;(2.31)

α − b − f = 0.(2.32)

As at Step 2, relations (2.29) allow us to assume that, for the initial element s ∈ Ker d0,we have

αi = bi = ϕi = 0 if + 1 ≤ i ≤ k − 1;

ci = 0 if ≤ i ≤ k − 2;

ai = 0 if ≤ i ≤ k − 1.

By symmetry, we may assume that for the homogeneous component q (t) the correspond-ing coefficients are also zero (i.e., in the above formulas we can insert the tilde above thecoefficients). Moreover, relation (2.30) implies

β := f = β+1 = f+1 = · · · = βk−1 = fk−1.

By symmetry, we also have

β := f = β+1 = f+1 = · · · = βk−1 = fk−1.

Next, replacing s by the element

s −(ck−1y(xy)k−1 ⊗ (yx) + b(xy) ⊗ y(xy)k−1

)· h1,

we may assume that ck−1 = 0 = b for the initial s. Then relations (2.31) and (2.32)imply that α = β = βk. Similarly, we may assume that α = β = βk. Thus,

(q(t)

q (t)

)=

(β ·

∑ki=(xy)i ⊗ (yx)−i + β ·

∑k−1i=

y(xy)i ⊗ y(xy)−1−i

β ·∑k

i=(yx)i ⊗ (xy)−i + β ·∑k−1

i=x(yx)i ⊗ x(yx)−1−i

).

If > k, then, replacing s with the element

s + β(x ⊗ x(yx) − y(xy)k−1 ⊗ (yx)

)· h1,

we may assume that β = 0. For = k, the replacement of s with

s + β(x ⊗ x − x2 ⊗ 1 − 1 ⊗ x2

)· h1

again allows us to assume that β = 0. Arguing by symmetry (with the use of the columnh2) we may assume that for the initial s we also have β = 0.


50 A. I. GENERALOV

Step 5. Consider the case where t = 2 + 1 with k ≤ ≤ 4k − 2. The homogeneouscomponent q(t) is represented in the form

q(t) =k∑

i=+1

αi(xy)i ⊗ x(yx)−i +k∑

i=+1

βi(xy)i ⊗ y(xy)−i

+k−1∑

i=+1

γi(yx)i ⊗ x(yx)−i +k−1∑

i=+1

ϕi(yx)i ⊗ y(xy)−i

+k−1∑i=

aix(yx)i ⊗ (xy)−i +k−1∑

i=+1

bix(yx)i ⊗ (yx)−i

+k−1∑i=

ciy(xy)i ⊗ (xy)−i +k−1∑

i=+1

fiy(xy)i ⊗ (yx)−i.

The homogeneous component q (t) has a similar representation, obtained by using thesymmetry mentioned above (the corresponding coefficients can be supplied with thetilde; cf. (2.6) and (2.7)). Then equation (2.5) implies the relations

αi = bi for + 1 ≤ i ≤ k − 1; ϕi = ci−1 for + 2 ≤ i ≤ k − 1;(2.33)

βi = fi for + 1 ≤ i ≤ k − 1; βi = fi−1 for + 2 ≤ i ≤ k − 1;(2.34)

βk − ck−1 − fk−1 = 0;(2.35)

β+1 + ϕ+1 − c = 0;

β+1 + ϕ+1 − c = 0.(2.36)

From (2.34) it follows that

β := β+1 = f+1 = · · · = fk−2 = βk−1 = fk−1.

By symmetry, we obtain

β := β+1 = f+1 = · · · = fk−2 = βk−1 = fk−1.

As at Step 1, relations (2.33) allow us to assume that for s we have

αi = bi = γi = 0 for + 1 ≤ i ≤ k − 1; ai = 0 for ≤ i ≤ k − 1;

ϕi = 0 for + 2 ≤ i ≤ k − 1; ci = 0 for + 1 ≤ i ≤ k − 2.

Moreover, replacing s with

(2.37) s − αk(xy)k ⊗ (yx) · h1,

we assume also that αk = 0. Observe that we may assume additionally that ck−1 = 0.Indeed, the replacement of s = (q, q) with the element

s − ck−1y(xy)k−1 ⊗ y(xy) · h1

affects only βk (besides ck−1) and several homogeneous components of q and q of degreegreater than t. Similarly, replacing s with

s − ϕ+1y(xy) ⊗ y(xy)k−1 · h1,

we may assume that ϕ+1 = 0. Consequently, using (2.35) and (2.36), we obtain addi-tionally that βk = β and c = β; by symmetry, we have βk = β and c = β. Similar



arguments, by symmetry, apply to the homogeneous component q (t). Thus,(q(t)

q (t)

)=

(β ·

∑ki=+1(xy)i ⊗ y(xy)−i + β ·

∑k−1i=

y(xy)i ⊗ (yx)−i

β ·∑k

i=+1(yx)i ⊗ x(yx)−i + β ·∑k−1

i=x(yx)i ⊗ (xy)−i

).

Replacing s with the element

s + β(x ⊗ (xy)+1 − y(xy)k−1 ⊗ y(xy)

)· h1,

we may assume that for the initial s we have β = 0. By symmetry, we may also assumethat β = 0.

Step 6. Consider the case where t = 4k−1. We represent the homogeneous componentsq(t) and q (t) in the form

q(t) = α(xy)k ⊗ x(yx)k−1 + β(xy)k ⊗ y(xy)k−1

+ ax(yx)k−1 ⊗ (xy)k + cy(xy)k−1 ⊗ (xy)k,

q (t) = α(yx)k ⊗ y(xy)k−1 + β(yx)k ⊗ x(yx)k−1

+ ay(xy)k−1 ⊗ (yx)k + cx(yx)k−1 ⊗ (yx)k

with α, β, a, c, α, β, a, c ∈ K. Equation (2.5) implies a unique relation

β − c + β − c = 0.

As at Step 1, using an analog of (2.18) we may assume that a = 0. Moreover, as at Step5, replacing s with the element of the form (2.37), where = k − 1, we may assume thatα = 0. By symmetry, a = 0 = α. Next, replacing s by s − c · y(xy)k−1 ⊗ y(x)k−1 · h1,we may assume additionally that c = 0 and, by symmetry, also c = 0. Then β + β = 0;replacing s with s − β(xy)k ⊗ x · h1, we may also assume that β = 0 = β.

Step 7. Finally, if t = 4k, then(q(t)

q (t)

)=

(α(xy)k ⊗ (xy)k

α(xy)k ⊗ (xy)k

)= α ·(xy)k⊗y(xy)k−1 ·h1+α ·(xy)k⊗x(yx)k−1 ·h2 ∈ Im d1,

where α, α ∈ K. A suitable replacement allows us to assume that for the initial s wehave q(t) = q (t) = 0.

We return to the proof of Theorem 2.1. Consider the following sequence in the categoryof right Λ-modules:

(2.38) Λ d0←− Λ2 d1←− Λ2

with

(2.39) d0 =(∗(x ⊗ 1 − 1 ⊗ x), ∗(y ⊗ 1 − 1 ⊗ y)

),

where the matrix of the homomorphism d1 is the transpose of the matrix of d1 (see(2.3)); the entries of the matrix of d1 determine the corresponding homomorphisms ofmultiplication by these elements from the left (cf. the notation in (2.39)). Using theright-left-symmetric version of Lemma 2.4, we see that Ker d0 = Im d1. To the sequence(2.38), we apply the duality functor

D = HomK(−, K) : Mod -Λ −→ Λ- Mod .

Clearly, D(d1) = d1 and D(d0) = d2, whence, by the exactness of the functor, we obtainKer d1 = Im d2.

Proposition 2.5. Im d3 = Λg R.


52 A. I. GENERALOV

Proof. First, we prove that dimK(Λg) = 4k. Since (x ⊗ 1)g = (1 ⊗ x)g and (y ⊗ 1)g =(1 ⊗ y)g, the vector space Λg is spanned by the elements(

(xy)i ⊗ 1)g,

((yx)i ⊗ 1

)g (1 ≤ i ≤ k − 1),(2.40) (

x(yx)i ⊗ 1)g,

(y(xy)i ⊗ 1

)g (0 ≤ i ≤ k − 1),(2.41)

together with g and((xy)k ⊗ 1

)g. Clearly, all these elements are nonzero. Suppose

that this set is linearly dependent. This means that there is a nonzero element r ∈ Rsuch that (r ⊗ 1)g = 0. Using the grading on Λ introduced above and the fact thatg is a homogeneous element (of degree 2k), we may assume that r is homogeneous. Ifdeg r ∈ 0, 2k, we obtain a contradiction immediately. If deg r = 2i with 1 ≤ i ≤ k − 1,then the corresponding two elements in (2.40) are linearly dependent. However, thedecompositions of these elements in the standard basis of Λ, namely,(

(xy)i ⊗ 1)g =

k∑t=i

(xy)t ⊗ (xy)k+i−t +k−1∑t=i

(xy)tx ⊗ y(xy)k+i−t−1,

((yx)i ⊗ 1

)g =

k∑t=i

(yx)t ⊗ (xy)k+i−t +k−1∑t=i

(yx)ty ⊗ x(yx)k+i−t−1,

show that they are linearly independent. Similarly, using the decompositions of elementsin (2.41) (with fixed i), we obtain a contradiction again.

Now we consider a Λ-homomorphism ψ : Λ → Λg such that ψ(1 ⊗ 1) = g. Since

Kerψ ⊃ Λ〈x ⊗ 1 − 1 ⊗ x, y ⊗ 1 − 1 ⊗ y〉 = Kerµ,

there exists an epimorphism ψ′ : R → Λg with ψ′ µ = ψ. Since dimK Λg = 4k, ψ′ is anisomorphism.

To complete the proof of Theorem 2.1, it remains to show that the complex (2.2) isexact at the member Q3, i.e., that dimK Ker d2 = dimK Im d3. We have already provedthe exactness of the sequence

0 ←− Rµ←− Q0

d0←− Q1d1←− Q2

d2←− Q3 ←− Ker d2 ←− 0;

it follows that

dimK R + dimK Q1 + dimK Q3 = dimK Q0 + dimK Q2 + dimK Ker d2.

Finally, we obtaindimK Ker d2 = dimK R = dimK Im d3.

§3. Additive structure of the cohomology algebra

Let Q•µ−→ R be the minimal Λ-projective resolution described in §2. Then we have

HHn(R) = ExtnΛ(R, R) = Hn(HomΛ(Q•, R)),

where HomΛ(Q•, R) is the complex

(3.1)(

HomΛ(Qn, R), δn = HomΛ(dn, R))

n≥0.

Let Zn(R) (respectively, Bn(R)) denote the vector space of n-cocycles (respectively,n-coboundaries), i.e., Zn(R) = Ker δn, Bn(R) = Im δn−1.

Since for any n ≥ 0 the module Qn is isomorphic either to Λ or to Λ2, any element inHomΛ(Qn, R) can be identified either with an element of R, or with a pair of elements of



R, respectively. Upon this identification, the differential δ0 : R → R 2 can be describedas follows: for any r ∈ R,

(3.2) δ0(r) =((x ⊗ 1 − 1 ⊗ x)r, (y ⊗ 1 − 1 ⊗ y)r

)= (xr − rx, yr − ry).

Proposition 3.0.1. dimK HH0(R) = k + 3, dimB1(R) = 3(k − 1).

Proof. By [12, III.14], the center Z(R) = Z0(R) of the algebra R admits the followingbasis:

(3.3) 1, xy + yx, (xy)2 + (yx)2, . . . , (xy)k−1 + (yx)k−1, x(yx)k−1, y(xy)k−1, (xy)k.

Therefore, dimK HH0(R) = k+3 and dim B1(R) = dimK R−dimK Z0(R) = 3(k−1).

Remark 3.0.2. In the sequel, for some of the elements in (3.3) we use an abbreviatednotation:

(3.4) p1 := xy + yx, p2 := x(yx)k−1, p′2 := y(xy)k−1, p3 := (xy)k.

Observe that (xy)i + (yx)i = pi1 with 1 ≤ i ≤ k − 1. Moreover, if char K = 2, then

(xy)k = 12pk

1 .

Remark 3.0.3. Letting r in (3.2) run over the standard basis of the algebra R, we easilysee that B1(R) admits the following basis:

((xy)i − (yx)i, 0), (0, (yx)i − (xy)i), (x(yx)i,−y(xy)i), where 1 ≤ i ≤ k − 1.

3.1. The differential δ1. To study the differential δ1 : R2 → R2, we need to imposeadditional assumptions on the parameter k (see (1.1)) and on p = char K.

Case 1: k ≥ 3. In this case the condition (r1, r2) ∈ Z1(R) is equivalent to the followingsystem of equations (over R):

(3.5)

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩xr1 + r1x −

k−2∑i=0

(yx)iy · r1 · y(xy)k−2−i −k−1∑i=0

(yx)i · r2 · (xy)k−1−i = 0,

−k−1∑i=0

(xy)i · r1 · (yx)k−1−i + yr2 + r2y −k−2∑i=0

(xy)ix · r2 · x(yx)k−2−i = 0.

Consider the decompositions of r1 and r2 in the standard basis of the algebra R:

(3.6) r1 =∑b∈B

λbb, r2 =∑b∈B

µbb;

here B := Bst and λb, µb ∈ K. Substituting this in the first equation in (3.5), we obtainthe relations

2λ1 = 0, µ1 = λy(xy)k−2 ;(3.7)λy(xy)i−1 = 0 for 1 ≤ i ≤ k − 2;

λ(yx)i + λ(xy)i = 0 for 1 ≤ i ≤ k − 1;

(3 − k)λx = kµy;(3.8)

2λy(xy)k−1 = µxy + µyx.(3.9)


54 A. I. GENERALOV

By symmetry, the second equation in (3.5) implies

2µ1 = 0, λ1 = µx(yx)k−2 ;(3.10)µx(yx)i−1 = 0 for 1 ≤ i ≤ k − 2;

µ(xy)i + µ(yx)i = 0 for 1 ≤ i ≤ k − 1;(3.11)

(3 − k)µy = kλx;(3.12)2µx(yx)k−1 = λyx + λxy.

It should be noted that the scalars λx(yx)i and µy(xy)i with 1 ≤ i ≤ k − 1 do not occurin these two groups of relations, so that the elements in HomΛ(Q1, R) = R2 of the form

(3.13) (x(yx)i, 0), (0, y(xy)i) for 1 ≤ i ≤ k − 1

lie in Z1(R).Next, (3.9) and (3.11) yield

(3.14) 2λy(xy)k−1 = 0.

By symmetry,

(3.15) 2µx(yx)k−1 = 0.

Moreover, from (3.8) and (3.12) it follows that the scalars λx and µy satisfy a homoge-neous system of linear equations (over K) with the matrix

C =(

3 − k −k−k 3 − k

).

Case 1.1: p = 2. We continue to study system (3.5) under the additional assumptionchar K = 2.

Proposition 3.1.1. If p = 2 and k ≥ 3, then a K-basis of the vector space HH1(R) isformed by the cohomology classes of the following elements in HomΛ(Q1, R) = R2:

(1, x(yx)k−2), (y(xy)k−2, 1);(3.16)

(x(yx)i, 0) for 1 ≤ i ≤ k − 1;(3.17)

(y(xy)k−1, 0), (0, x(yx)k−1);(3.18)

((xy)k, 0), (0, (xy)k).(3.19)

Proof. Now we have detC = 1, whence λx = 0 = µy. Arguing as above, we easily showthat as a K-basis of the vector space Z1(R) we can take the set formed by the elementslisted in (3.13), (3.16), (3.18), and (3.19), and also by the elements

(3.20) ((xy)i − (yx)i, 0), (0, (yx)i − (xy)i) for 1 ≤ i ≤ k − 1.

Combining this and Remark 3.0.3, we arrive at the desired statement.

Corollary 3.1.2. If p = 2 and k ≥ 3, then

dimK Z1(R) = 4k + 2, dimK HH1(R) = k + 5, dimK B2(R) = 4k − 2.



Remark 3.1.3. It is verified directly that the vector space B2(R) is generated by thefollowing elements:

((xy)i + (yx)i, 0), (0, (yx)i + (xy)i) for 1 ≤ i ≤ k − 1,(3.21)

(x(yx)i, 0), (0, y(xy)i) for 2 ≤ i ≤ k − 1,

δ1(x, 0) =((3 − k)y(xy)k−1,−KB(yx)k−1

),(3.22)

δ1(0, y) =(− ky(xy)k−1, (3 − k)x(yx)k−1

),(3.23)

δ1(xy, 0) = (xyx,−(xy)k), δ1(0, yx) = (−(xy)k, yxy).

Since the number of these elements is equal to 4k − 2, they form a basis of B2(R).Furthermore, detC = 0, so that we can replace δ1(x, 0) and δ1(0, y) in this basis by theelements (

y(xy)k−1, 0)

and(0, x(yx)k−1

).

Remark 3.1.4. In the sequel, if f ∈ Zn(R) is an n-cocycle, its cohomology class is oftendenoted still by f .

Case 1.2: p = 3. Relations (3.7), (3.10), (3.14), and (3.15) imply that

(3.24) λ1 = 0 = µ1, λy(xy)k−2 = 0 = µx(yx)k−2 , λy(xy)k−1 = 0 = µx(yx)k−1 .

Now we consider two subcases.Case 1.2a: Assume additionally that 3 does not divide k.

Proposition 3.1.5. If p = 3, k ≥ 3, and 3 does not divide k, then a K-basis of thevector space HH1(R) is formed by the cohomology classes of the elements listed in (3.17)and (3.19), together with the cohomology class of the element (x,−y).

Proof. In this case we have rk C = 1; hence, the homogeneous system of linear equationswith the matrix C reduces to a single equation, namely, to λx + µy = 0, whence we seethat (x,−y) ∈ Z1(R). The above arguments show that the elements occurring in (3.13),(3.19), and (3.20), together with (x,−y), form a basis of Z1(R). Now, Remark 3.0.3completes the proof of Proposition 3.1.5. Corollary 3.1.6. If p = 3, k ≥ 3, and 3 does not divide k, then

dimK Z1(R) = 4k − 1, dimK HH1(R) = k + 2, dimK B2(R) = 4k + 1.

Remark 3.1.7. It is easily seen that, in the case under consideration (i.e., p = 3 does notdivide k and k ≥ 3), the vector space B2(R) is generated by the set consisting of theelements occurring in (3.21) and the elements

δ1(1, 0) = (2x,−(yx)k−1 − (xy)k−1), δ1(0, 1) = (−(xy)k−1 − (yx)k−1, 2y),

−1k

δ1(x, 0) = (y(xy)k−1, x(yx)k−1),(3.25)

12δ1(y(xy)k−1, 0) = ((xy)k, 0),

12δ1(0, x(yx)k−1) = (0, (xy)k).(3.26)

This set is a basis of B2(R), because its cardinality is equal to 4k + 1. Moreover, since(0, (yx)k−1+(xy)k−1

),((xy)k−1+(yx)k−1, 0

)∈ B2(R), the elements δ1(1, 0) and δ1(0, 1)

can be replaced in this basis with (x, 0) and (0, y).

Case 1.2b: 3 divides k.

Proposition 3.1.8. If p = 3 and 3 divides k, then a K-basis of the vector space HH1(R)is formed by the cohomology classes of the elements occurring in (3.17) and (3.19), to-gether with the cohomology classes of the elements (x, 0) and (0, y).


56 A. I. GENERALOV

Proof. Now we have rk C = 0; hence, to get a basis of the vector space Z1(R), we needto adjoin the elements (x, 0) and (0, y) to the elements listed in (3.13), (3.19), and (3.20).Then the proof is completed in the same way as that of Proposition 3.1.5.

Corollary 3.1.9. If p = 3 and 3 divides k, then

dimK Z1(R) = 4k, dimK HH1(R) = k + 3, dimK B2(R) = 4k.

Remark 3.1.10. Under the assumptions of Proposition 3.1.8, a basis of B2(R) is obtainedby deleting the element

(y(xy)k−1, x(yx)k−1

)from the set indicated in Remark 3.1.7

(because now we have δ1(x, 0) = 0; see (3.22)).

Case 1.3: p ∈ 2, 3. Again we consider two subcases.Case 1.3a: Assume additionally that 3 − 2k = 0 (in the field K), i.e., rkC = 2.

Proposition 3.1.11. If p ∈ 2, 3, k ≥ 3, and p does not divide 3 − 2k, then a K-basisof the vector space HH1(R) is formed by the cohomology classes of the elements occurringin (3.17) and (3.19).

Proof. First, relations (3.24) are satisfied since p = 2 (cf. case 1.2). Moreover, we haverkC = 2, whence λx = 0 = µy. Consequently, the vector space Z1(R) admits a basisconsisting of the elements of the form (3.13), (3.19), and (3.20). Using Remark 3.0.3, weobtain the desired statement.

Corollary 3.1.12. If p ∈ 2, 3, k ≥ 3, and p does not divide 3 − 2k, then


Remark 3.1.13. Under the assumptions of Proposition 3.1.11, the vector space B2(R) isgenerated by the set of elements indicated in (3.21) and (3.26), together with the elementsδ1(x, 0), δ1(0, y) (see (3.22) and (3.23)). Since the cardinality of this set is equal to 4k+2,this is a basis of B2(R). Moreover, since rk C = 2, we see that the elements δ1(x, 0) andδ1(0, y) can be replaced in this basis with the elements (y(xy)k−1, 0) and (0, x(yx)k−1).

Case 1.3b: Assume that 3 − 2k = 0 in K. Clearly, in this case p does not divide kand rkC = 1.

Proposition 3.1.14. If p ∈ 2, 3 and p divides 3 − 2k, then a K-basis of the vectorspace HH1(R) is formed by the cohomology classes of the elements occurring in (3.17)and (3.19), together with the cohomology class of the element (x, y).

Proof. Since rk C = 1, the homogeneous system of linear equations with the matrix Creduces to a single equation, namely, to λx −µy = 0, whence we see that (x, y) ∈ Z2(R).Now the proof can be completed like that of Proposition 3.1.5.

Corollary 3.1.15. If p ∈ 2, 3 and p divides 3 − 2k, then


Remark 3.1.16. Under the assumptions of Proposition 3.1.14, the vector space B2(R) isgenerated by the set indicated in Remark 3.1.7. To see this, we need a minor modificationin the corresponding argument; in fact, now we have (cf. (3.25))

−1k

δ1(x, 0) = (−y(xy)k−1, x(yx)k−1).



Case 2: k = 2. If k = 2, then the condition δ1(r1, r2) = 0 implies the followingrelations for coefficients in the decompositions of the elements r1 and r2 (see (3.6)):

2λ1 = 0 = µ1, λy = µ1, µx = λ1,

λ1 = λxy + λyx, µ1 = µxy + µyx;λx − 2µy = 0,

2λx − µy = 0;(3.27)

λy + µxy + µyx = 2λyxy,

µx + λxy + λyx = 2µxyx.(3.28)

Case 2.1: p = 2. We continue the analysis of the above relations under the additionalassumption that p = 2.

Proposition 3.1.17. If p = 2 and k = 2, then a K-basis of the vector space HH1(R) isformed by the cohomology classes of the elements

(3.29) (1 + xy, x), (y, 1 + yx), (yxy, 0), (0, xyx), (xyx, 0), ((xy)2, 0), (0, (xy)2).

Proof. If p = 2, then (3.27) implies λx = 0 = µy, while (3.28) is a consequence of theremaining relations and can be omitted. It follows that, as a basis for Z1(R), we cantake the set of elements listed in (3.29) together with

(xy − yx, 0), (0, yx − xy), (0, yxy).

Using Remark 3.0.3, we arrive at the desired statement.

Remark 3.1.18. The proof of Proposition 3.1.17 shows that the formulas for the dimen-sions of the vector spaces Z1(R), HH1(R), and B2(R), as indicated in Corollary 3.1.2,remain valid in the case in question. Moreover, we can argue as in Remark 3.1.3 to provethat, as a basis for B2(R), we can take the set of the following elements:

(xy + yx, yxy), (xyx, yx + xy), (yxy, 0), (0, xyx), (xyx, (xy)2), ((xy)2, yxy).

Case 2.2: p = 2. If p = 2 (and k = 2), then it is easily seen that the arguments ofCase 1.2 (for p = 3) and Case 1.3 (for p ∈ 2, 3) apply without change.

3.2. The differential δ2. In this subsection we study the differential δ2 : R2 → R,where

δ2(r1, r2) = xr1 − r1x + yr2 − r2y,

and also we describe HH2(R).

Proposition 3.2.1. a) The vector space Z2(R) admits a K-basis formed by the followingelements:

(y(xy)i, x(yx)i) for 0 ≤ i ≤ k − 2;(3.30)

((xy)i + (yx)i, 0), (0, (yx)i + (xy)i), (x(yx)i, 0), (0, y(xy)i) for 1 ≤ i ≤ k − 1;(3.31)

(1, 0), (0, 1), (x, 0), (0, y);(3.32)

(y(xy)k−1, 0), (0, x(yx)k−1);(3.33)

((xy)k, 0), (0, (xy)k).(3.34)

b) The vector space B3(R) admits a K-basis formed by the elements

(3.35) (xy)i − (yx)i, x(yx)i, y(xy)i for 1 ≤ i ≤ k − 1.


58 A. I. GENERALOV

Proof. Looking at decompositions of the form (3.6) in the standard basis of R, we seethat the relation δ2(r1, r2) = 0 is equivalent to the following equations for the coefficientsλb, µb (b ∈ Bst) of these decompositions:

λy(xy)i = µx(yx)i for 0 ≤ i ≤ k − 2;λ(yx)i = λ(xy)i , µ(xy)i = µ(yx)i for 1 ≤ i ≤ k − 1.

This implies directly that the set of elements listed in (3.30)–(3.34) is a basis of Z2(R).In particular, we have dimK Z2(R) = 5k + 3, dimK B3(R) = 3k − 3.

Next, computing all elements of the form δ2(b, 0) and δ2(0, b) with b ∈ Bst, we seethat the elements in (3.35) generate B3(R). Since their number is equal to dimK B3(R),they form a basis of this vector space.

Corollary 3.2.2. a) dimK Z2(R) = 5k + 3, dimK B3(R) = 3k − 3.b)

dimK HH2(R) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

k + 5 if p = 2,

k + 3 if p = 3 and 3 divides k,

k + 2 if p = 3 and 3 does not divide k,

or p ∈ 2, 3 and p divides 3 − 2k,

k + 1 if p ∈ 2, 3 and p does not divide 3 − 2k.

Proof. Statement a) follows from Proposition 3.2.1. Statement b) follows from Corollaries3.1.2, 3.1.9, 3.1.12, and 3.1.15, and from Remark 3.1.18.

Proposition 3.2.3. a) If p = 2, then a K-basis of the vector space HH2(R) is formedby the cohomology classes of the elements occurring in (3.30), (3.32), and (3.34).

b) Let p = 3. If 3 does not divide k, then a K-basis of the vector space HH2(R) isformed by the cohomology classes of the elements (3.30) and the cohomology classes ofthe elements

(3.36) (1, 0), (0, 1), (y(xy)k−1, 0).

If 3 divides k, we should add the element (0, x(yx)k−1) to the preceding set in order toget a basis of HH2(R).

c) Assume that p ∈ 2, 3. If p divides 3 − 2k, then a K-basis for the vector spaceHH2(R) is formed by the set of the cohomology classes of the elements listed in (3.30)and (3.36). If p does not divide 3− 2k, then we should remove the element (y(xy)k−1, 0)from the preceding set in order to get a basis of HH2(R).

Proof. In all cases, the statements follow (with the help of Corollary 3.2.2) immediatelyfrom Proposition 3.2.1 and Remarks 3.1.3, 3.1.7, 3.1.10, 3.1.13, 3.1.16, and 3.1.18, inwhich the corresponding bases for B2(R) were presented.

3.3. The differential δ3. Since the resolution (2.2) is 4-periodic, it remains to studythe differential δ3 : R → R. Since δ3 = ∗g, we have

δ3(r) =k∑

i=1

(yx)i · r · (yx)k−i +k−1∑i=0

y(xy)i · r · x(yx)k−1−i(3.37)

+k−1∑i=0

x(yx)i · r · y(xy)k−1−i +k−1∑i=0

(xy)i · r · (xy)k−i.

Proposition 3.3.1. a) If p does not divide 2k, then the set Bst\1 is a basis for Z3(R).b) If p divides 2k, then δ3 = 0, whence Z3(R) = R.



Proof. Since g is a homogeneous element of degree 2k (see (2.4)), for any b ∈ Bst \1 wehave δ3(b) = 0. Moreover, relation (3.37) implies δ3(1) = 4k(xy)k. Now, both statementsfollow immediately. Proposition 3.3.2. a) If p divides 2k, then a K-basis of the vector space HH3(R) isformed by the cohomology classes of the following elements:

(3.38)

1, x, y,(xy)i with 1 ≤ i ≤ k.

If p does not divide 2k, then, in order to get a basis of HH3(R), we should remove theelement 1 from (3.38).

b) If p divides 2k, thenHH4(R) HH0(R).

If p does not divide 2k, then a K-basis for the vector space HH4(R) is formed by thecohomology classes of the elements

(xy)i + (yx)i for 1 ≤ i ≤ k − 1,

together with the cohomology classes of the elements 1, x(yx)k−1, y(xy)k−1.

Proof. The first statement follows from Propositions 3.3.1,a) and 3.2.1,b); the second isa consequence of the proofs of Propositions 3.3.1 and 3.0.1. Corollary 3.3.3.

dimK HH3(R) = dimK HH4(R) =

k + 3 if p divides 2k,

k + 2 if p does not divide 2k.

§4. Generators and relations

We briefly recall an interpretation of the Yoneda product in

HH∗(R) = Ext∗Λ(R, R) =⊕m≥0

ExtmΛ (R, R),

which was used in [11].Let µ : Q• → R be the minimal Λ-projective resolution (see §2). Let HomΛ(Q•, R) =

(HomΛ(Qn, R), δn) be the complex (3.1). Any n-cocycle f ∈ Zn(R) = Ker δn is lifted(uniquely up to homotopy) to a chain map of complexes ϕi : Qn+i → Qii≥0. Thehomomorphism ϕi is called the ith translate of the cocycle f and will be denoted byTi(f). For cocycles f ∈ Ker∆n and g ∈ Ker ∆t, we have cl g · cl f = cl(µ T0(g) Tt(f)).Consequently, for us it suffices to know the composition of the translates T0(g) andTt(f). In the sequel, we shall use direct decompositions of the modules Qi to describe thetranslates of cocycles and their products with the help of the matrices that correspond tosuch decompositions. Observe that, by the graded-commutativity of the algebra HH∗(R),to compute the product of a pair of homogeneous elements in HH∗(R) it suffices to know,for an element of degree i, its translates of order not exceeding i.

4.1. char K = 2. In this subsection, we assume that the ground field K has characteristictwo, if not stipulated otherwise.

Case 1: k ≥ 3. Consider the following homogeneous elements in HH∗(R):

of degree 1: u1 := (1, x(yx)k−2), u′1 := (y(xy)k−2, 1), u2 := (xyx, 0);

of degree 2: v1 := (y, x), v2 := (x, 0), v′2 := (0, y);of degree 3: w := xy;of degree 4: z := 1.


60 A. I. GENERALOV

Recall that earlier we have distinguished (in the case of a field K with arbitrarycharacteristic) certain elements in HH0(R): p1, p2, p

′2, p3 (see (3.4)).

Proposition 4.1.1. Assume that p = 2 and k ≥ 3. Then, in the algebra HH∗(R), theelements of the set

(4.1) Y1 = p1, p2, p′2, p3, u1, u

′1, u2, v1, v2, v

′2, w, z

satisfy the following relations:pk1 = p2

2 = (p′2)2 = p1p2 = p1p′2 = p2p

′2 = 0,

p23 = p1p3 = p2p3 = p′2p3 = 0;(4.2)

pk−11 u2 = p2u2 = p′2u2 = p3u2 = 0,

p2u1 = p′2u′1 = pk−2

1 u2;(4.3)

p′2u1 = p1u′1, p1u1 = p2u

′1,

pk−11 v1 = p2v2 = p′2v

′2 = p3v1 = p3v2 = p3v

′2 = 0;

(4.4)

p2w = p′2w = p3w = 0;(4.5)p2v1 = p1v

′2 = p′2v2 = p3u

21, p′2v1 = p1v2 = p2v

′2 = p3(u′

1)2;(4.6)

u1u′1 = kpk−2

1 v1;(4.7)u1u2 = kp3(u′

1)2, u′

1u2 = kp3u21, u2

2 = 0;(4.8)

u2v2 = u2v′2 = 0, u2v1 = p1w, u′

1v2 = u1v′2 = pk−2

1 w;u1v1 = u′

1v′2, u′

1v1 = u1v2, u31 = (u′

1)3;(4.9)

v22 = (v′2)

2 = v1v2 = v1v′2 = v2v

′2 = 0;(4.10)

v21 = p2

1z, u1w = u′1w = u2w = v2w = v′2w = 0;

v1w = p1u2z, w2 = 0.

Proof. Relations (4.2)–(4.5) are verified directly. To prove the remaining relations, weneed to compute the translates of suitable order for the elements in Y1 that have apositive degree. By the 4-periodicity of the minimal Λ-projective resolution (2.2), for theelement z the identity maps of the corresponding modules can be taken as the translatesTi(z).

Lemma 4.1.2. For the role of those translates of the elements in Y1 \ z that have apositive degree, we can take the homomorphisms determined by the following matrices:

T0(u1) =(1 ⊗ 1, x(yx)k−2 ⊗ 1

),

T1(u1) =

⎛⎜⎜⎝1 ⊗ 1 + (xy)k−2 ⊗ (xy)k−1 + (yx)k−2 ⊗ (yx)k−1k−2∑i=0

y(xy)i ⊗ (yx)k−2−i

k−2∑i=0

(yx)i ⊗ y(xy)k−2−i

⎞⎟⎟⎠with

T1(u1)2,2 =k−2∑i=1

(xy)i ⊗ x(yx)k−2−i + x(yx)k−2 ⊗ 1 +k−2∑i=0

(yx)i ⊗ x(yx)k−2−i



at the position (2.2);

T0(u′1) =

(y(xy)k−2 ⊗ 1, 1 ⊗ 1

),

T1(u′1) =

⎛⎜⎜⎝ k−2∑i=0

(xy)i ⊗ x(yx)k−2−i

k−2∑i=0

x(yx)i ⊗ (xy)k−2−i 1 ⊗ 1 + (yx)k−2 ⊗ (yx)k−1 + (xy)k−2 ⊗ (xy)k−1

⎞⎟⎟⎠with

T1(u′1)1,1 =

k−2∑i=1

(yx)i ⊗ y(xy)k−2−i + y(xy)k−2 ⊗ 1 +

k−2∑i=0

(xy)i ⊗ y(xy)k−2−i;

T0(u2) =(xyx ⊗ 1, 0

),

T1(u2) =

⎛⎜⎜⎝−xyx ⊗ 1 +k−1∑i=2

(i − 1)y(xy)i ⊗ y(xy)k−1−ik∑

i=2

(i − 1)(xy)i ⊗ (yx)k−i

k∑i=2

(i − 1)(yx)i ⊗ (xy)k−ik−1∑i=1

ix(yx)i ⊗ x(yx)k−1−i

⎞⎟⎟⎠ ;

T0(v1) =(1 ⊗ y, x ⊗ 1

), T1(v1) =

(y ⊗ 11 ⊗ x

),

T2(v1) =

(−x ⊗ xy + y(xy)k−1 ⊗ y + x(yx)k−1 ⊗ (yx)k−1

−y ⊗ yx + x(yx)k−1 ⊗ x + y(xy)k−1 ⊗ (xy)k−1

);

T0(v2) =(x ⊗ 1, 0

), T1(v2) =

(x ⊗ 1

0

),

T2(v2) =

⎛⎝ k∑i=1

(yx)i ⊗ (yx)k−i +k−1∑i=0

(xy)i ⊗ (xy)k−i

y ⊗ y(xy)k−1 + x(yx)k−1 ⊗ (xy)k−1 + (xy)k ⊗ x(yx)k−2 + (xy)k−1 ⊗ x(yx)k−1

⎞⎠ ;

T0(v′2) = (0, y ⊗ 1), T1(v′

2) =

(0

y ⊗ 1

),

T2(v′2) =

⎛⎝x ⊗ x(yx)k−1 + y(xy)k−1 ⊗ (yx)k−1 + (yx)k ⊗ y(xy)k−2 + (yx)k−1 ⊗ y(xy)k−1

k∑i=1

(xy)i ⊗ (xy)k−i +k−1∑i=0

(yx)i ⊗ (yx)k−i

⎞⎠ ;

T0(w) = (x ⊗ y)∗,

T1(w) =

⎛⎜⎜⎝y ⊗ (yx)k −k−1∑i=1

i(xy)i+1 ⊗ y(xy)k−1−i −k−2∑i=1

iy(xy)i+1 ⊗ (yx)k−1−i

−k−1∑i=1

i(yx)i+1 ⊗ x(yx)k−1−i −k−1∑i=1

ix(yx)i ⊗ (xy)k−i

⎞⎟⎟⎠ ,

T2(w) =

(−x ⊗ xyx + y(xy)k−1 ⊗ yx, (xy)k−1 ⊗ (yx)k + (k − 1)(xy)k ⊗ (yx)k−1

(k − 1)(yx)k ⊗ (xy)k−1 0

),

T3(w) =

(, −

k∑i=2

(i − 1)(xy)i ⊗ x(yx)k−i −k−1∑i=1

ix(yx)i ⊗ (yx)k−i

)with

T3(w)1,1 =k−2∑i=0

(k − 1 − i)y(xy)i ⊗ (xy)k−i +k−2∑i=1

(k − 1 − i)(yx)i ⊗ y(xy)k−i

+ y(xy)k−1 ⊗ yx + (k − 1)x(yx)k−1 ⊗ x(yx)k−1.

The proof of the lemma is a direct verification of the relations µ T0(b) = b, di−1 Ti(b) =Ti−1(b)di+deg b (0 < i ≤ deg b), where b ∈ Y1 \ z with deg b > 0, and we leave it to thereader.


62 A. I. GENERALOV

Remark 4.1.3. For what follows, it should be noted that the formulas for the translatesof the elements u2, v1, and w remain valid if the field K has an arbitrary characteristic.

Now the proof of Proposition 4.1.1 is completed with the help of calculations withthe matrices presented in Lemma 4.1.2. We illustrate the corresponding techniques byproving the relation u1u2 = kp3(u′

1)2 (see (4.8)). The remaining relations are establishedsimilarly.

With the help of Lemma 4.1.2, we obtain T0(u′1) T1(u′

1) = (A1, A2), where

A1 = y(xy)k−1 ⊗ y(xy)k−3 + y(xy)k−2 ⊗ y(xy)k−2 +k−2∑i=0

x(yx)i ⊗ (xy)k−2−i,

A2 = y(xy)k−2 ⊗ x(yx)k−2 + 1 ⊗ 1 + (yx)k−2 ⊗ (yx)k−1 + (xy)k−2 ⊗ (xy)k−1.

Consequently,

µ · T0(u′1) T1(u′

1) =

(x(yx)k−2, 1) if k > 3,

(xyx, 1 + (xy)3) if k = 3.

In both cases the 2-cocycle µ T0(u′1) T1(u′

1) is cohomological to the 2-cocycle (0,1) (seeRemark 3.1.3). On the other hand, using Lemma 4.1.2 again, we obtain

T0(u1) T1(u2) =(

xyx ⊗ 1 +k−1∑i=2

(i − 1)y(xy)i ⊗ y(xy)k−1−i,k∑

i=2

(i − 1)(xy)i ⊗ (yx)k−i

).

Thus, µ T0(u1) T1(u2) = (xyx, (k − 1)(xy)k). This 2-cocycle is cohomological to thecocycle (0, k(xy)k). Consequently, in HH∗(R) we have u1u2 = kp3(u′

1)2.

Proposition 4.1.4. Assume that p = 2 and k ≥ 3. The set Y1 defined in (4.1) generatesHH∗(R) as a K-algebra.

Proof. Let H denote the K-subalgebra of HH∗(R) generated by the set Y1 ∪ 1 (here 1denotes the unity of the algebra HH∗(R)). Since multiplication by z gives an isomorphismHHi(R) → HHi+4(R) (i ≥ 0), it suffices to prove that HHi(R) ⊂ H for i ≤ 3.

By Remark 3.0.2, we have HH0(R) ⊂ H. The following relations are satisfied inHH∗(R):

(x(yx)i, 0) = pi−11 u2 with 2 ≤ i ≤ k − 2;

(y(xy)k−1, 0) = p′2u1, (0, x(yx)k−1) = p2u′1,

((xy)k, 0) = p3u1, (0, (xy)k) = p3u′1.

Now, Proposition 3.1.1 implies HH1(R) ⊂ H.Next, in the proof of Proposition 4.1.1 we saw that (0, 1) = (u′

1)2; by symmetrywe also have (1, 0) = u2

1. Moreover, in HH∗(R) we have (y(xy)i, x(yx)i) = pi1v1 with

1 ≤ i ≤ k − 2. Using Proposition 3.2.3,a), we see that HH2(R) ⊂ H.As in the proof of Proposition 4.1.1, it is established that for the elements 1, x, y ∈

Z3(R), the following relations are satisfied in the algebra HH∗(R):

1 = u31, x = u′

1v1, y = u1v1.

Since in HH∗(R) we also have (xy)i = pi−11 w for 1 ≤ i ≤ k − 1, Proposition 3.3.2 implies

the inclusion HH3(R) ⊂ H.

Let A1 = K[X1]/I1 be the graded K-algebra defined in §1, with X1 as in (1.2), I1 beingthe corresponding ideal of relations (see (1.4)–(1.15)). The (nonzero) images of monomi-als in K[X1] under the canonical epimorphism K[X1] → A1 are also called monomials.Any element a ∈ A1 is represented as a linear combination of monomials (with coefficients



in K). Propositions 4.1.1 and 4.1.4 imply that there exists a surjective homomorphismϕ : A1 → HH∗(R) of graded K-algebras that takes the generators belonging to the setX1 to the corresponding generators in Y1 (see (4.1)). Note that no ambiguity is causedby employing the same letter to denote elements of both sets that correspond to eachother. Let A1 =

⊕m≥0 Am

1 be the direct decomposition of the algebra A1 into homoge-neous direct summands. Now, part 1 a) of Theorem 1.1 is a consequence of the followingstatement.

Proposition 4.1.5. For any m ≥ 0, we have

dimK Am1 = dimK HHm(R).

Before proving this proposition, we state the following auxiliary fact.

Lemma 4.1.6. The following relations are satisfied in the algebra A1:

p21u1 = p2

1u′1 = 0,

p1u21 = p2u

21 = p′2u

21 = p1(u′

1)2 = p2(u′

1)2 = p′2(u

′1)

2 = 0,u2

1u′1 = u1(u′

1)2 = u21v1 = (u′

1)2v1 = u21v2 = (u′

1)2v2 = u41 = (u′

1)4 = 0,

p1u1v1 = p1u′1v1 = p′2u1v1 = 0, p3u

31 = pk−1

1 w.

Proof. To prove the lemma, it suffices to deduce the above relations directly from thedefining relations of the algebra A1, and we leave this to the reader.

Proof of Proposition 4.1.5. On the polynomial ring K[X1], where X1 is as in (1.2), weintroduce a lexicographic order such that

v′2 > v2 > u2 > u′1 > v1 > u1 > w > z > p1 > p2 > p′2 > p3.

Let

(4.11) f = (v′2)α′

vα2 uβ

2 (u′1)

m′vγ1 um

1 wεznpt1p

ζ2(p

′2)

ζ′pη3

be a nonzero monomial in A1. The defining relations of the algebra A1 imply (see, inparticular, Lemma 4.1.6) that in (4.11) we necessarily have α, α′, β, ε, ζ, ζ ′, η ∈ 0, 1,m ≤ 3, m′ ≤ 3 and t ≤ k− 1. There are several additional restrictions on the occurrenceof generators in (4.11) (e.g., we have at most one nonzero number among t, ζ, ζ ′, η);we shall analyze such restrictions below.

By definition, the reduction of a monomial f in A1 is the process of replacement ofsome submonomials of f by other elements of A1 in accordance with the following rules(a → b means the replacement of every entrance of the monomial a by the element b):

(4.12)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v1w → p1u2z, v21 → p2

1z,

u1v′2 → u′

1v2 → pk−21 w, p3u

31 → pk−1

1 w,u2v1 → p1w, u1v2 → u′

1v1,

u′1v

′2 → u1v1, u1u

′1 → pk−2

1 v1 (if k odd),p1v

′2 → p′2v2 → p2v1 → p3u

21, p2v

′2 → p1v2 → p3(u′

1)2 → p′2v1,

u1u2 → p3(u′1)2 (if k odd), u′

1u2 → p3u21 (if k odd),

(u′1)

3 → u31, pk−2

1 u2 → p′2u′1 → p2u1,

p1u′1 → p′2u1, p2u

′1 → p1u1.

Any replacement on the above list is called an elementary step of reduction. Undersuch an elementary step, any nonzero monomial turns into a strictly smaller monomialwith respect to the lexicographic order. Hence, after finitely many steps, we obtain amonomial to which we cannot apply any elementary step of reduction. We say that apresentation of an element a ∈ A1 as a linear combination of monomials has a normalform if reduction cannot be applied to any of these monomials. Since these elementary


64 A. I. GENERALOV

steps correspond to some relations satisfied in the algebra A1 (see (1.4)–(1.15) and Lemma4.1.6), any element a ∈ A1 admits at least one presentation in the normal form.

Put qi = dimK Ai1. We denote the number of monomials in Ai

1 presented in the normalform by qi. It is clear that qi ≥ qi. Since there is an epimorphism Ai

1 → HHi(R), we seethat qi ≥ dimK HHi(R); consequently, it suffices to show that

(4.13) qi = dimK HHi(R).

We prove that all (nonzero) monomials having normal form are contained in thefollowing list:

Degree Normal forms Number

4nznpi

1 (0 ≤ i ≤ k − 1),znp2, znp′2, znp3

k + 3

4n + 1 u2znpi

1 (0 ≤ i ≤ k − 3),u1z

n, u1znp1, u1z

np2, u1znp′2, u1z

np3, u′1z

n, u′1z

np3

k + 5

4n + 2 v1znpi

1 (0 ≤ i ≤ k − 2),v1z

np′2, v2zn, v′2z

n, u21z

n, u21z

np3, (u′1)

2k + 5

4n + 3 wznpi1 (0 ≤ i ≤ k − 1),

v1u1zn, v1u

′1z

n, u31z

nk + 3

It is easily seen that all monomials in this list have normal form.Let f be a nonzero monomial in (4.11) represented in the normal form. It is clear

that γ ≤ 1 (by the reduction v21 → p2

1z) and m′ ≤ 2 (by (u′1)3 → u3

1). Next, we considerseveral cases successively.

1) Assume that the monomial f contains a factor w. Some of the monomial relationssatisfied in the algebra A1 (see (1.4)–(1.15) and Lemma 4.1.6) imply that f does notcontain v1, v2, v

′2, u1, u

′1, u2, p2, p

′2, p3. Consequently, we have

f = wznpi1 for 0 ≤ i ≤ k − 1.

2) Assume that f does not involve w, but involves u1. Then f cannot involve any oneof the following elements: u′

1 (because we have the reduction u1u′1 → pk−1

1 v1 if k is odd,and u1u

′1 = 0 if k is even), u2 (because we have the reduction u1u2 → p3(u′

1)2 if k isodd, and u1u2 = 0 if k is even), v2 (because we have u1v2 → u′

1v1), v′2 (because we haveu1v

′2 → pk−2

1 w). Consequently,

f = vγ1 um

1 znpt1p

ζ2(p

′2)

ζ′pη3 .

Here t ≤ 1 (because p21u1 = 0; see Lemma 4.1.6). Moreover, if m > 1, then t = ζ = ζ ′ = 0

by Lemma 4.1.6; if m = 3, then also η = 0; and if γ = 1, then, by Lemma 4.1.6, m = 1and also t = ζ = ζ ′ = η = 0. Consequently, f coincides with one of the followingmonomials:

u1zn, u1z

np1, u1znp2, u1z

np′2, u1znp3, u2

1zn, u2

1znp3, u3

1zn, v1z

nu1.

3) Assume that f involves neither w nor u1, and that v1 occurs in f . Arguing asabove, we see that f does not involve v2, v

′2, u2, p2, p3. Consequently,

f = (u′1)

m′v1z

npt1(p

′2)

ζ′;

here t ≤ k−2 and m′ ≤ 1. If additionally we have m′ = 1, then t = ζ ′ = 0. Consequently,f coincides with one of the following monomials:

v1znpi

1 with 0 ≤ i ≤ k − 2; v1znp′2, u′

1v1zn.



4) Assume that w, u1, v1 do not occur and u′1 occurs in f . Then it is easily seen that f

does not involve v2, v′2, u2, p1, p2, p

′2. Consequently, f = (u′

1)m′

znpη3 , and if m′ > 1, then

η = 0. Hence, f is one of the monomials

u′1z

n, u′1z

np3, (u′1)

2zn.

5) Assume that w, u1, v1, u′1 do not occur and u2 occurs in f . Then f does not involve

v2, v′2, p2, p

′2, p3. Consequently, f = u2p

i1 for i ≤ k − 3, because we have the reduction

pk−21 u2 → p2u1.

6) Assume that w, u1, v1, u′1, u2 do not occur and v2 occurs in f . Then f does not

involve v′2, p1, p2, p′2, p3. Consequently, f = v2z

n.7) Assume that w, u1, v1, u

′1, u2, v2 do not occur and v′2 occurs in f . Then f does not

involve p1, p2, p′2, p3. Consequently, f = v′2z

n.8) Finally, assume that w, u1, v1, u

′1, u2, v2, v

′2 do not occur in f . In this case, clearly,

f coincides with one of the monomials

znpi1 for 0 ≤ i ≤ k − 1; znp2, znp′2, znp3.

Consequently, we have shown that any monomial having the normal form is on thelist above, and this completes the proof of (4.13).

Case 2: k = 2. Consider the following homogeneous elements of the algebra HH∗(R):

of degree 1: u1 := (1 + xy, x), u′1 := (y, 1 + yx);

of degree 2: v1 := (y, x), v2 := (x, 0), v′2 := (0, y);of degree 4: z := 1.

Proposition 4.1.7. Assume that p = 2 and k = 2. Then, in the algebra HH∗(R), theelements of the set

(4.14) Y2 = p1, p2, p′2, p3, u1, u

′1, v1, v2, v

′2, z

satisfy relations (4.2), (4.4), (4.6), (4.7), (4.9), and (4.10), and also the relations

p2u1 = p′2u′1, u′

1v2 = u1v′2, v2

1 = 0.

Proof. As in the proof of Proposition 4.1.1, we need to know the translates for theelements of the set Y2 that have positive degree. It is easily seen that the formulas forthe translates of v1, v2, v

′2 presented in Lemma 4.1.2 remain valid for k = 2. Hence, we

must compute additionally the corresponding translates for the elements u1 and u′1.

Lemma 4.1.8. For the role of the translates (up to first order inclusive) of the elementsu1 and u′

1 we can take the following maps:

T0(u1) =((1 + xy) ⊗ 1, x ⊗ 1

),

T1(u1) =(

1 ⊗ (1 + xy + yx) (y + yxy) ⊗ 11 ⊗ y + x ⊗ x + x2 ⊗ 1 x ⊗ 1 + 1 ⊗ x + x ⊗ yx

);

T0(u′1) =

(y ⊗ 1, (1 + yx) ⊗ 1

),

T1(u′1) =

(y ⊗ 1 + 1 ⊗ y + y ⊗ xy 1 ⊗ x + y ⊗ y + y2 ⊗ 1

(x + xyx) ⊗ 1 1 ⊗ (1 + yx + xy)

).

This lemma is proved by direct inspection (cf. Lemma 4.1.2).Now, we can finish the proof of Proposition 4.1.7 with the help of the same technique

as in the proof of Proposition 4.1.1. Proposition 4.1.9. Assume that p = 2 and k = 2. The set Y2 defined in (4.14)generates HH∗(R) as a K-algebra.


66 A. I. GENERALOV

Proof. The proof is similar to that of Proposition 4.1.4. Additionally, it should be ob-served that the element w := xy ∈ Z3(R) satisfies w = u′

1v2.

Let A2 = K[X2]/I2 be the graded K-algebra defined in §1 with X2 as in (1.16), I2

being the corresponding ideal of relations. Propositions 4.1.7 and 4.1.9 imply that thereexists a surjective homomorphism ϕ : A2 → HH∗(R) of graded K-algebras that takes thegenerators belonging to X2 to the corresponding generators in Y2. Let A2 =

⊕m≥0 Am

2

be the direct decomposition into homogeneous direct summands. Now, part 1 b) ofTheorem 1.1 is a consequence of the following statement.


dimK Am2 = dimK HHm(R).

Proof. The proof of this proposition repeats the proof of Proposition 4.1.5 with minorchanges. We only remark that on K[X2] we introduce the lexicographic order such that

v′2 > v2 > u′1 > v1 > u1 > z > p1 > p2 > p′2 > p3.

Moreover, from the list (4.12) we remove the elementary steps of reduction in which u2

or w occurs.

4.2. char K = 3. In this subsection, we assume that the ground field K has characteristicthree, if not stipulated otherwise.

Case 1: 3 does not divide k. Consider the following homogeneous elements in HH∗(R):

of degree 0: p1, p2, p′2 (see (3.4));

of degree 1: u2 := (xyx, 0), u3 := (x,−y);

of degree 2: v0 := (1, 0), v′0 := (0, 1), v1 := (y, x);(4.15)of degree 3: w := xy;of degree 4: z := 1.

Proposition 4.2.1. Assume that p = 3 and 3 does not divide k. Then, in the algebraHH∗(R), the elements of the set

(4.16) Y3 = p1, p2, p′2, u2, u3, v0, v

′0, v1, w, z

satisfy the following relations:

pk+11 = p2

2 = (p′2)2 = p1p2 = p1p

′2 = p2p

′2 = 0;(4.17)

p1v0 = p1v′0 = p2v0 = p′2v

′0 = p2v1 = p′2v1 = pk−1

1 v1 = 0;(4.18)

p2u2 = p′2u2 = pk−11 u2 = 0;(4.19)

p2v′0 + p′2v0 = 0, p1u3 = u2u3 = u3v1 = 0;

p2w = p′2w = u2v0 = u2v′0 = u2w = pk

1z = 0, u2v1 = p1w;(4.20)

p′2u3v0 = pk−11 w, u3w = 0;

v0v′0 = v0v1 = v′0v1 = 0, v2

0 = 2p′2z, (v′0)2 = 2p2z, v2

1 = −p21z;(4.21)

v0w = v′0w = 0, v1w = −p1u2z.(4.22)

Proof. To prove these relations, we need to know the translates (of suitable orders) forthe elements of Y3 that have positive degree. By Remark 4.1.3, we must calculate onlythe translates for u3, v0, v

′0.



Lemma 4.2.2. For the role of the translates of the elements u3, v0, and v′0, we can takethe following maps:

T0(u3) =(x ⊗ 1, −y ⊗ 1

),

T1(u3) =

(−x ⊗ 1 −

∑k−2i=0 y(xy)i ⊗ y(xy)k−2−i 0

0 y ⊗ 1 +∑k−2

i=0 x(yx)i ⊗ x(yx)k−2−i

);

T0(v0) = (1 ⊗ 1, 0), T1(v0) =(

1 ⊗ 10

),

T2(v0) =(∑k−1

i=0 y(xy)i ⊗ (yx)k−1−i +∑k−1

i=0 (xy)i ⊗ y(xy)k−1−i

(yx)k−1 ⊗ (xy)k−1

);

T0(v′0) = (0, 1 ⊗ 1), T1(v′0) =(

01 ⊗ 1

),

T2(v′0) =(

(xy)k−1 ⊗ (yx)k−1∑k−1i=0 x(yx)i ⊗ (xy)k−1−i +

∑k−1i=0 (yx)i ⊗ x(yx)k−1−i

).

This lemma is proved by direct calculation (cf. Lemma 4.1.2).

Remark 4.2.3. It should be noted that the formulas for the translates of the elements v0

and v′0 remain valid in the cases where k is divisible by p = 3 or p ∈ 2, 3.

Now, the proof of Proposition 4.2.1 can be completed as was done in the proof ofProposition 4.1.1.

Proposition 4.2.4. Assume that p = 3 and 3 does not divide k. The set Y3 defined in(4.16) generates HH∗(R) as a K-algebra.

Proof. The proof is similar to that of Proposition 4.1.4 and employs, in particular, Propo-sitions 3.1.5, 3.2.3, b) and 3.3.2. Besides the relations obtained in the proof of Proposition4.1.4, the following relations should be used:

in HH1(R) : ((xy)k, 0) = p′2u3, (0, (xy)k) = −p2u3;

in HH2(R) : (y(xy)k−1, 0) = p′2v0;

in HH3(R) : x = u3v0, y = −u3v′0;

in HH4(R) : (xy)i + (yx)i = pi1z (1 ≤ i ≤ k − 1), x(yx)k−1 = p2z, y(xy)k−1 = p′

1z.

Let A3 = K〈X3〉/I3 be the graded K-algebra defined in §1, with X3 as in (1.17),I3 being the corresponding ideal of relations (see (1.19)–(1.31)). Propositions 4.2.1 and4.2.4 imply that there exists a surjective homomorphism ϕ : A3 → HH∗(R) of graded K-algebras that takes the generators belonging to X3 to the corresponding generators in Y3

(see (4.16)). Let A3 =⊕

m≥0 Am3 be the direct decomposition into homogeneous direct

summands. Now, part 2 a) of Theorem 1.1 is a consequence of the following statement.


(4.23) dimK Am3 = dimK HHm(R).

Proof. On the polynomial ring K[X3], we introduce the lexicographic order such that

v′0 > v0 > v1 > u2 > u3 > w > z > p1 > p2 > p′2.

Since the algebra A3 is graded-commutative, any nonzero monomial f ∈ A3 is repre-sented, up to a scalar factor, in the form

f = (v′0)α′

vα0 vβ

1 uγ2uε

3wζznpt

1pθ2(p

′2)

θ′.


68 A. I. GENERALOV

Such representations of elements of the algebra A3 can be viewed as elements of K[X3].Consider the following list of elementary steps of reduction:

(4.24)

⎧⎪⎪⎨⎪⎪⎩v1w → p1u2z, u2v1 → p1w,p2v

′0u3 → p′2v0u3 → pk−1

1 w, v21 → p2

1z,v20 → p′2z, (v′0)2 → p2z,

p2v′0 → p′2v0.

As in the proof of Proposition 4.1.5, we can define the normal form of an element a ∈ A3

(with respect to the steps of reduction (4.24)) and then verify that any element a ∈ A3

admits at least one normal form. Finally, studying several cases successively, we provethat all (nonzero) monomials having normal form are on the following list:


0 pi1(1 ≤ i ≤ k),

1, p2, p′2

k + 3

4n, n > 0 znpi1(0 ≤ i ≤ k − 1),znp2, z

np′2k + 2

4n + 1 u2znpi

1(0 ≤ i ≤ k − 2),u3z

n, u3znp2, u3z

np′2k + 2

4n + 2 v1znpi

1(0 ≤ i ≤ k − 2),v0z

n, v0znp′2, v

′0z

n k + 2

4n + 3wznpi

1(0 ≤ i ≤ k − 1),v0u3z

n, v′0u3zn k + 2

As in the proof of Proposition 4.1.5, this implies (4.23). The details are left to thereader.

Case 2: 3 divides k. Consider the following homogeneous elements of HH∗(R):

of degree 0: p1, p2, p′2 (see (3.4));

of degree 1: u1 := (x, 0), u ′1 := (0, y);

of degree 2: v0, v′0, v1 (see (4.15));

of degree 3: w := 1;of degree 4: z := 1.

Proposition 4.2.6. Assume that p = 3 and 3 divides k. Then, in the algebra HH∗(R),the elements of the set

(4.25) Y4 = p1, p2, p′2, u1, u

′1, v0, v

′0, v1, w, z

satisfy relations (4.17), (4.18), and (4.21), and the relations

(4.26)

⎧⎪⎪⎨⎪⎪⎩p2u1 = p′2u

′1 = pk

1 u1 = pk1 u ′

1 = 0, p1u1 = p1u′1;

u1u′1 = 0, pk−1

1 v1 = p′2v0 + p2v′0;

p2w = p′2w = u ′1v0 = u1v

′0 = 0, u ′

1v1 = u1v1, p1w = 2u1v1;u1w = u ′

1w = v0w = v′0w = 0, v1w = −2u2z.

Proof. By Remark 4.2.3, to prove (4.26) it remains to calculate the translates of theelements u1, u

′1, and w.



Lemma 4.2.7. For the role of the translates of u1, u′1, and w, we can take the following

maps:

T0(u1) =(x ⊗ 1, 0

),

T1(u1) =

(−x ⊗ 1 +

∑k−2i=1 iy(xy)i ⊗ y(xy)k−2−i

∑k−1i=1 i(xy)i ⊗ (yx)k−1−i∑k−1

i=1 i(yx)i ⊗ (xy)k−1−i∑k−2

i=0 (i + 1)x(yx)i ⊗ x(yx)k−2−i

);

T0(u ′1) =

(0, y ⊗ 1

),

T1(u ′1) =

(∑k−2i=0 (i + 1)y(xy)i ⊗ y(xy)k−2−i

∑k−1i=1 i(xy)i ⊗ (yx)k−1−i∑k−1

i=1 i(yx)i ⊗ (xy)k−1−i −y ⊗ 1 +∑k−2

i=1 ix(yx)i ⊗ x(yx)k−2−i

);

T0(w) = (1 ⊗ 1)∗,

T1(w) =

(−

∑k−1i=0 (2i + 1)y(xy)i ⊗ (yx)k−1−i −

∑k−1i=0 (2i + 1)(xy)i ⊗ y(xy)k−1−i

−∑k−1

i=1 2i(yx)i ⊗ x(yx)k−1−i −∑k−1

i=0 (2i + 2)x(yx)i ⊗ (xy)k−1−i

),

T2(w) =(−2x ⊗ x + x2 ⊗ 1 + 1 ⊗ x2 − (xy)k−1 ⊗ (yx)k−1

−(yx)k−1 ⊗ (xy)k−1 0

),

T3(w) =(−

k−1∑i=0

(2i + 1)(yx)i ⊗ (xy)k−1−i · (y ⊗ 1 + 1 ⊗ y), )

with

T3(w)1,2 = −2k−2∑i=0

(i + 1)x(yx)i ⊗ (yx)k−1−i −k−1∑i=0

2i(xy)i ⊗ x(yx)k−1−i.

This lemma is proved by direct calculation (cf. Lemma 4.1.2).

Remark 4.2.8. Observe that the formulas for the translates of w remain valid in the casewhere p ∈ 2, 3 and p divides k.

Now, the proof of Proposition 4.2.6 can be completed as was done in the proof ofProposition 4.1.1.

Proposition 4.2.9. Assume that p = 3 and 3 divides k. The set Y4 defined in (4.25)generates HH∗(R) as a K-algebra.

Proof. The proof is similar to that of Proposition 4.1.4 and employs, in particular, Propo-sitions 3.1.8, 3.2.3 b), and 3.3.2.

Let A4 = K〈X4〉/I4 be the graded K-algebra defined in §1, with X4 as in (1.32), I4

being the corresponding ideal of relations. Propositions 4.2.6 and 4.2.9 imply that thereexists a surjective homomorphism ϕ : A4 → HH∗(R) of graded K-algebras that takesthe generators belonging to X4 to the corresponding generators in Y4 (see (4.25)). LetA4 =

⊕m≥0 Am

4 be the direct decomposition into homogeneous direct summands. Now,part 2 b) of Theorem 1.1 is a consequence of the following statement.



Proof. The proof repeats that of Proposition 4.2.5. On the polynomial ring K[X4], weintroduce the lexicographic order such that

v1 > v′0 > v0 > u ′1 > u1 > w > z > p1 > p2 > p′2.


70 A. I. GENERALOV

Then we consider the following list of elementary steps of reduction:

u ′1v1 → u1v1 → p1w, p2u

′1v

′0 → p′2u1v0 → pk

1w,

pk−11 v1 → p′2v0 + p2v

′0, v2

1 → p21z,

v20 → p′2z, (v′0)2 → p2z,

p1u′1 → p1u1.

This list includes the steps p2u′1v

′0 → p′2u1v0 → pk

1w, because the defining relations ofthe algebra A4 imply that pk

1w = 2p2u′1v

′0 = 2p′2u1v0. Finally, it can be checked that all

(nonzero) monomials having normal form are contained in the following list:


4nznpi

1(0 ≤ i ≤ k),znp2, z

np′2k + 3

4n + 1 u1znpi

1(0 ≤ i ≤ k − 1),u1z

np′2, u′1z

n, u ′1z

np2k + 3

4n + 2 v1znpi

1(0 ≤ i ≤ k − 2),v0z

n, v0znp′2, v

′0z

n, v′0znp2

k + 3

4n + 3 wznpi1(0 ≤ i ≤ k),

v0u1, v′0u

′1

k + 3

This leads to (4.27).

4.3. char K ∈ 2, 3. In this subsection, we assume that the ground field K has charac-teristic different from 2 and 3.

Case 1: p divides neither 3− 2k nor k. Consider the following homogeneous elementsin HH∗(R):

of degree 0: p1, p2, p′2 (see (3.4));

of degree 1: u2 := (xyx, 0), u4 := ((xy)k, 0), u′4 := (0, (xy)k);(4.28)

of degree 2: v0, v′0, v1 (see (4.15));

of degree 3: w1 := x, w′1 := y, w := xy;

of degree 4: z := 1.

Proposition 4.3.1. Assume that p divides neither 3 − 2k nor k. Then, in the algebraHH∗(R), the elements of the set

(4.29) Y5 = p1, p2, p′2, u2, u4, u

′4, v0, v

′0, v1, w1, w

′1, w, z

satisfy relations (4.17), (4.18), (4.19), (4.20), (4.21), and (4.22), and also the relations⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

p1u4 = p1u′4 = p2u4 = p2u

′4 = p′2u4 = p′2u

′4 = 0;

u2u4 = u2u′4 = u4u

′4 = 0;

p1w1 = p1w′1 = p2w1 = p′2w

′1 = 0;

p2w′1 = p′2w1 = u4v0 = u′

4v′0;

u4v′0 = u′

4v0 = u1v1 = u′1v1 = 0;

u2w1 = u2w′1 = u4w1 = u4w

′1 = u′

4w1 = u′4w

′1 = 0;

v0w′1 = v′0w1 = v1w1 = v1w

′1 = 0, v0w1 = 2u4z, v′0w

′1 = 2u′

4z;w1w

′1 = 0;

(4.30)

u′4v

′0 = pk−1

1 w, u4w = u′4w = w1w = w′

1w = 0.

Proof. By Remarks 4.1.3 and 4.2.3, to prove the relations mentioned above it remains tocalculate the translates (of suitable orders) of the elements u4, u

′4, w1, and w′

1.



Lemma 4.3.2. For the role of the translates of u4, u′4, w1, and w′

1, we can take thefollowing maps:

T0(u4) =((xy)k ⊗ 1, 0

), T1(u4) =

(−(xy)k ⊗ 1 0

0 (xy)k ⊗ x(yx)k−2

);

T0(u′4) =

(0, (xy)k ⊗ 1

), T1(u′

4) =(

(xy)k ⊗ y(xy)k−2 00 −(xy)k ⊗ 1

);

T0(w1) = (x ⊗ 1)∗,

T1(w1) =

⎛⎝k−1∑i=1

y(xy)i−1 ⊗ x(yx)k−i +k−1∑i=0

(xy)i ⊗ (xy)k−i − (xy)k ⊗ 1

(yx)k−1 ⊗ y(xy)k−1

⎞⎠ ,

T2(w1) =(

(xy)k ⊗ 1 + 1 ⊗ (xy)k y(xy)k−1 ⊗ y + (xy)k−1 ⊗ x(yx)k−1

x(yx)k−1 ⊗ (xy)k−1 0

),

T3(w1) =( k−1∑

i=0

y(xy)i ⊗ x(yx)k−1−i +k−1∑i=0

(yx)i ⊗ (yx)k−i, −2y(xy)k−1 ⊗ (yx)k−1

);

T0(w′1) = (x ⊗ 1)∗,

T1(w′1) =

⎛⎝ (xy)k−1 ⊗ x(yx)k−1

k−1∑i=1

x(yx)i−1 ⊗ y(xy)k−i +k−1∑i=0

(yx)i ⊗ (yx)k−i − (xy)k ⊗ 1

⎞⎠ ,

T2(w′1) =

(0 y(xy)k−1 ⊗ (yx)k−1

x(yx)k−1 ⊗ x + (yx)k−1 ⊗ y(xy)k−1 (xy)k ⊗ 1 + 1 ⊗ (xy)k

),

T3(w′1) =

(− 2y(xy)k−1 ⊗ (yx)k−1,

k−1∑i=0

x(yx)i ⊗ y(xy)k−1−i +k−1∑i=0

(xy)i ⊗ (xy)k−i

).

Using Lemma 4.3.2, we can complete the proof of Proposition 4.3.1 as was done inthe proof of Proposition 4.1.1.

Proposition 4.3.3. Assume that p divides neither 3 − 2k nor k. The set Y5 defined in(4.29) generates HH∗(R) as a K-algebra.

The proof is similar to that of Proposition 4.1.4. Let A5 = K〈X5〉/I5 be the graded K-algebra defined in §1, with X5 as in (1.33), I5

being the corresponding ideal of relations. Propositions 4.3.1 and 4.3.3 imply that thereexists a surjective homomorphism ϕ : A5 → HH∗(R) of graded K-algebras that takesthe generators belonging to X5 to the corresponding generators in Y5 (see (4.29)). LetA5 =

⊕m≥0 Am

5 be the direct decomposition into homogeneous direct summands. Now,part 3 a) of Theorem 1.1 is a consequence of the following statement.



Proof. On K[X5], we introduce the lexicographic order such that

v′0 > v0 > v1 > u2 > u′4 > u4 > w > w′

1 > w1 > z > p1 > p2 > p′2.


72 A. I. GENERALOV

Consider the following list of elementary steps of reduction:

(4.32)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩v0w1 → u4z, v′0w

′1 → u′

4z,v20 → p′2z, (v′0)

2 → p2z,v21 → p2

1z, u2v1 → p1w,

u′4v

′0 → u4v0 → p′2w1, pk−1

1 w → p2w′1 → p′2w1,

v1w → p1u2z.

As in the proof of Proposition 4.1.5, we can define the normal form of an element a ∈ A5

and then verify that any element a ∈ A5 admits at least one normal form (see also theproof of Proposition 4.2.5). After that, we check that all (nonzero) monomials havingnormal form are on the following list:


0 pi1(1 ≤ i ≤ k),

1, p2, p′2

k + 3

4n, n > 0 znpi1(0 ≤ i ≤ k − 1),znp2, z

np′2k + 2

4n + 1 u2znpi

1(0 ≤ i ≤ k − 2),u4z

n, u′4z

n k + 1

4n + 2v1z

npi1(0 ≤ i ≤ k − 2),v0z

n, v′0zn k + 1

4n + 3wznpi

1(0 ≤ i ≤ k − 2),w1z

n, w1znp′2, w

′1z

n k + 2

This leads to (4.31).

Case 2: p does not divide 3 − 2k and divides k. Consider the following homogeneouselements in HH∗(R):

of degree 0: p1, p2, p′2 (see (3.4));

of degree 1: u2, u4, u′4 (see (4.28));

of degree 2: v0, v′0, v1 (see (4.15));

of degree 3: w := 1, w1 := x, w′1 := y;

of degree 4: z := 1.

Proposition 4.3.5. Assume that p does not divide 3 − 2k and divides k. Then, in thealgebra HH∗(R), the elements of the set

Y6 = p1, p2, p′2, u2, u4, u

′4, v0, v

′0, v1, w, w1, w

′1, z

satisfy relations (4.17), (4.18), (4.19), and (4.30), and the relations

u4v0 = 12pk

1w, u2v1 = 12p2

1w, p2w = p′2w = 0;u2w = u4w = u′

4w = v0w = v′0w = 0, v1w = −2u2z;w1w = w′

1w = 0.

Proof. The arguments are similar to those in the proof of Proposition 4.1.1. We observeadditionally that the required translates of the elements of Y6 have been calculated earlier,and a large part of the relations mentioned above have already been established.

Now, the proof of part 3 b) of Theorem 1.1 can be completed as in the previous cases.In particular, on the ring K[X6] (with X6 as in (1.36)), we use the lexicographic ordersuch that

v′0 > v0 > v1 > u2 > u′4 > u4 > w > w′

1 > w1 > z > p1 > p2 > p′2.



Then we introduce a list of elementary steps of reduction, obtained from the list (4.32)by deleting the steps that involve w and adding the following steps to this list:

u2v1 → p21w, v1w → u2z, pkw → p2w

′1.

After that, we show that the monomials having a normal form (with respect to the stepsof reduction mentioned above) are as follows:


4nznpi

1(0 ≤ i ≤ k),znp2, z

np′2k + 3

4n + 1 u2znpi

1(0 ≤ i ≤ k − 2),u4z

n, u′4z

n k + 1

4n + 2 v1znpi

1(0 ≤ i ≤ k − 2),v0z

n, v′0zn k + 1

4n + 3 wznpi1(0 ≤ i ≤ k − 1),

w1zn, w1z

np′2, w′1z

n k + 3

Case 3: p divides 3 − 2k; it is clear that p does not divide k. Consider the followinghomogeneous elements in HH∗(R):

of degree 0: p1, p2, p′2 (see (3.4));

of degree 1: u3 := (x, y);

of degree 2: v0, v′0, v1 (see (4.15));of degree 4: z := 1.

Proposition 4.3.6. Assume that p divides 3 − 2k. Then, in the algebra HH∗(R), theelements of the set

Y7 = p1, p2, p′2, u3, v0, v

′0, v1, z

satisfy relations (4.17), (4.18), and (4.21), and also the relations

p2v′0 = p′2v0, pk

1 u3 = pk1z = 0.

The proof is similar to that of Proposition 4.1.1. We need to use the translates of theelement u3 that are presented in the following lemma.

Lemma 4.3.7. For the role of the translates (up to first order inclusive) of the elementu3, we can take the following maps:

T0(u3) = (x ⊗ 1, y ⊗ 1),

T1(u3) =⎛⎜⎜⎝−x ⊗ 1 +k−2∑i=1

(2i + 1)y(xy)i ⊗ y(xy)k−2−ik−1∑i=1

2i(xy)i ⊗ (yx)k−1−i

k−1∑i=1

2i(yx)i ⊗ (xy)k−1−i −y ⊗ 1 +k−2∑i=0

(2i + 1)x(yx)i ⊗ x(yx)k−2−i

⎞⎟⎟⎠ .

Now, the proof of part 3 c) of Theorem 1.1 can be completed as in the preceding cases.On the ring K[X7] (with X7 as in (1.37)), we use the lexicographic order such that

v′0 > v0 > v1 > u3 > z > p1 > p2 > p′2.


74 A. I. GENERALOV

Then we introduce the following list of elementary steps of reduction:

v20 → p′2z, (v′0)

2 → p2z,v21 → p2

1z, p2v′0 → p′2v0,

and show that all monomials having a normal form (with respect to the steps of reductionmentioned above) are on the following list:


0 pi1(1 ≤ i ≤ k),

1, p2, p′2

k + 3

4n, n > 0 znpi1(0 ≤ i ≤ k − 1),znp2, z

np′2k + 2

4n + 1u3z

npi1(0 ≤ i ≤ k − 1),

u3znp2, u3z

np′2k + 2

4n + 2v1z

npi1(0 ≤ i ≤ k − 2),

v0zn, v0z

np′2, v′0z

n k + 2

4n + 3 v1u3znpi

1(0 ≤ i ≤ k − 2),v0u3z

n, v0u3znp′2, v

′0u3z

n k + 2

This completes the proof of Theorem 1.1.

§5. Appendix. The Yoneda algebra

Here we give (without a detailed proof) the description of the Yoneda algebra for thealgebras Rk (k ≥ 2).

Let R = Rk, and let S be a unique simple left R-module. The Yoneda algebra Y(R) ofthe algebra R is a direct sum

∑m≥0 Extm

R (S, S) with multiplication given by the Yonedaproduct.

Proposition 5.1. The module S is Ω-periodic with period 4, and its minimal projectiveresolution is of the form

0 ←− S ←− R(x,y)←− R2

(x −(xy)k−1

−(yx)k−1 y

)←−−−−−−−−−−−−−−−−− R2

(xy

)←−−−− R

(xy)k

←− R(x,y)←− R2 ←− . . .

The proof consists in a direct verification of the fact that the sequence presented aboveis exact. Recall that the entries of the matrices of the differentials are endomorphisms ofthe left R-module R that are induced by multiplication on the right by the correspondingelements of the algebra R.

Now, using the techniques of the papers [24, 25], we can prove the following statement.

Theorem 5.2. Let R = Rk with k ≥ 2. Then, as a graded algebra, the Yoneda algebraY(R) is isomorphic to the algebra

K[ξ, ζ, η]/(ξζ, ξ3 − ζ3, ξ4, ζ4),

where deg ξ = deg ζ = 1, deg η = 4.

Remark 5.3. Let G be the generalized quaternion group, and let K = F2. The corre-sponding cohomology ring H∗(G, K) = Y(KG) was calculated earlier in [26].



References

[1] S. Eilenberg and S. MacLane, Cohomology theory in abstract groups. I, Ann. of Math. (2) 48 (1947),51–78. MR0019092 (8:367f)

[2] H. Cartan and S. Eilenberg, Homological algebra, Princeton Univ. Press, Princeton, NJ, 1956.MR0077480 (17:1040e)

[3] M. Gerstenhaber, The cohomology structure of an associative ring, Ann. of Math. (2) 78 (1963),267–288. MR0161898 (28:5102)

[4] M. Gerstenhaber and S. D. Schack, Algebraic cohomology and deformation theory, DeformationTheory of Algebras and Structures and Applications (Il Ciocco, 1986) (M. Hazewinkel, M. Ger-stenhaber, eds.), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., vol. 247, Kluwer Acad. Publ.,Dordrecht, 1988, pp. 11–264. MR0981619 (90c:16016)

[5] Th. Holm, The Hochschild cohomology ring of a modular group algebra: The commutative case,Comm. Algebra 24 (1996), 1957–1969. MR1386022 (97c:13012)

[6] C. Cibils and A. Solotar, Hochschild cohomology algebra of abelian groups, Arch. Math. (Basel) 68(1997), 17–21. MR1421841 (97k:13018)

[7] S. F. Siegel and S. J. Witherspoon, The Hochschild cohomology ring of a group algebra, Proc.London Math. Soc. (3) 79 (1999), 131–157. MR1687539 (2000b:16016)

[8] K. Erdmann and Th. Holm, Twisted bimodules and Hochschild cohomology for self-injective algebrasof class An, Forum Math. 11 (1999), 177–201. MR1680594 (2001c:16018)

[9] K. Erdmann, Th. Holm, and N. Snashall, Twisted bimodules and Hochschild cohomology for self-injective algebras of class An. II, Algebr. Represent. Theory 5 (2002), 457–482. MR1935856

(2004a:16013)[10] Th. Holm, Hochschild cohomology of tame blocks, J. Algebra 271 (2004), 798–826. MR2025551

(2005c:20019)[11] A. I. Generalov, Hochschild cohomology of algebras of dihedral type. I. The family D(3K) in char-

acteristic 2, Algebra i Analiz 16 (2004), no. 6, 53–122; English transl., St. Petersburg Math. J. 16(2005), no. 6, 961–1012. MR2117449 (2005i:16013)

[12] K. Erdmann, Blocks of tame representation type and related algebras, Lecture Notes in Math.,vol. 1428, Springer-Verlag, Berlin, 1990. MR1064107 (91c:20016)

[13] Th. Holm, Derived equivalence classification of algebras of dihedral, semidihedral, and quaterniontype, J. Algebra 211 (1999), 159–205. MR1656577 (2000a:16019)

[14] A. I. Generalov, Cohomology of algebras of semidihedral type. I, Algebra i Analiz 13 (2001), no. 4,54–85; English transl., St. Petersburg Math. J. 13 (2002), no. 4, 549–573. MR1865495 (2002h:16020)

[15] A. I. Generalov and E. A. Osiyuk, Cohomology of algebras of dihedral type, III. The D(2A) series,Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 289 (2002), 113–133; Englishtransl., J. Math. Sci. (N.Y.) 124 (2004), no. 1, 4741–4753. MR1949737 (2003k:16015)

[16] A. I. Generalov, Cohomology of algebras of dihedral type, IV: D(2B) series, Zap. Nauchn. Sem.S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 289 (2002), 76–89; English transl., J. Math. Sci.(N.Y.) 124 (2004), no. 1, 4719–4726. MR1949735 (2003k:16016)

[17] M. A. Antipov and A. I. Generalov, Cohomology of algebras of semidihedral type. II, Zap. Nauchn.Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 289 (2002), 9–36; English transl., J. Math.Sci. (N.Y.) 124 (2004), no. 1, 4681–4697. MR1949731 (2003k:16022)

[18] A. I. Generalov, Cohomology of algebras of semidihedral type, III. The series SD(3K), Zap. Nauchn.Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 305 (2003), 84–100; English transl., J. Math.

Sci. (N.Y.) 130 (2005), no. 3, 4689–4698. MR2033615 (2004m:16016)[19] , Cohomology of algebras of semidihedral type. IV, Zap. Nauchn. Sem. S.-Peterburg. Otdel.

Mat. Inst. Steklov. (POMI) 319 (2004), 81–116; English transl., J. Math. Sci. (N.Y.) 134 (2006),no. 6, 2489–2510. MR2117855 (2005k:16019)

[20] E. L. Green and N. Snashall, Projective bimodule resolutions of an algebra and vanishing of thesecond Hochschild cohomology group, Forum Math. 16 (2004), 17–36. MR2034541 (2005c:16011)

[21] M. J. Bardzell, The alternating syzygy behavior of monomial algebras, J. Algebra 188 (1997), 69–89.MR1432347 (98a:16009)

[22] D. Happel, Hochschild cohomology of finite-dimensional algebras, Seminaire d’Algebre Paul Dubreilet Marie-Paul Malliavin, 39eme Annee (Paris, 1987/1988), Lecture Notes in Math., vol. 1404,Springer, Berlin, 1989, pp. 108–126. MR1035222 (91b:16012)

[23] A. I. Generalov and M. A. Kachalova, Bimodule resolution of the Mobius algebra, Zap. Nauchn.Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 321 (2005), 36–66. (Russian) MR2138411(2006a:16014)


http://www.ams.org/mathscinet-getitem?mr=0019092














































76 A. I. GENERALOV

[24] A. I. Generalov, Cohomology of algebras of dihedral type. I, Zap. Nauchn. Sem. S.-Peterburg. Otdel.Mat. Inst. Steklov. (POMI) 265 (1999), 139–162 (2000); English transl., J. Math. Sci. (N.Y.) 112(2002), no. 3, 4318–4331. MR1757820 (2001b:20090)

[25] O. I. Balashov and A. I. Generalov, The Yoneda algebras for some class of dihedral algebras, VestnikS.-Peterburg. Univ. Ser. 1 1999, vyp. 3, 3–10; English transl., Vestnik St. Petersburg Univ. Math.32 (1999), no. 3, 1–8 (2000). MR1794988 (2001m:16020)

[26] J. Martino and S. Priddy, Classification of BG for groups with dihedral or quaternion Sylow 2-subgroups, J. Pure Appl. Algebra 73 (1991), 13–21. MR1121628 (92f:55022)

Department of Mathematics and Mechanics, St. Petersburg State University, Universi-

tetskiı Pr. 28, Staryı Peterhof, St. Petersburg 198904, Russia

Received 19/SEP/2005

Translated by THE AUTHOR








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