Cup Product on Hochschild Cohomology of a Family of Quiver Algebras

CUP PRODUCT ON HOCHSCHILD COHOMOLOGY OF A FAMILY OF QUIVER ALGEBRAS

Tolulope Oke October 29, 2019

Abstract Let k be a ﬁeld, q ∈ k. We derive a cup product formula on the ∗ Hochschild cohomology HH (Λq) of a family Λq of quiver algebras. Using ∗ this formula, we determine a subalgebra of k[x, y] isomorphic to HH (Λq)/N , where N is the ideal generated by homogeneous nilpotent elements. We explicitly construct non-nilpotent Hochschild cocycles which cannot be generated by lower homological degree elements, thus disproving the Snashall- Solberg ﬁnite generation conjecture.

1 Introduction

The theory of support varieties has been well developed for finite groups using group cohomology. Several efforts were made to develop similar theories for finitely generated modules over finite dimensional algebras using Hochschild cohomology. Hochschild cohomology HH∗(Λ) is graded commutative. If the characteristic char(k) 6= 2, then every homogeneous element of odd degree is nilpotent. Let N be the set of nilpotent elements of HH∗(Λ), Hochschild cohomology modulo nilpotents HH∗(Λ)/N is therefore a commutative k-algebra. For some finite dimensional

arXiv:2004.00780v1 [math.RA] 2 Apr 2020 algebras, it is well known that the Hochschild cohomology ring modulo nilpotents is finitely generated as an algebra. N. Snashall described many classes of such algebras in section 3 of [6]. Before the expository paper [6], it was conjectured in [7] that Hocschild cohomology modulo nilpotents is always finitely generated as an algebra for finite dimensional algebras. The first counterexample to this conjecture appeared in [9] where F. Xu used certain techniques in category theory to construct a seven-dimensional category algebra whose Hochschild cohomology ring modulo nilpotents is not finitely generated. There have been since then several constructions of different counter examples to this conjecture[10]. While it is of great use to produce a counterexample, it is equally of importance to discuss the techniques used to produce such examples. Snashall gave a different presentation of the Xu counterexample which we will summarize briefly. A quiver is a directed graph where loops and multiple arrows (also called paths) between two vertices are possible. For a field k, the path algebra kQ, is the k- vector space generated by all paths in the quiver Q. A vertex is a path of length 0. By taking multiplication of two paths x and y to be the concatenation xy if the terminal vertex t(x) of x and the origin vertex o(y) of y are equal, and otherwise 0, kQ becomes an associative ring. Let I be an ideal of kQ. The quotient Λ = kQ/I is called a quiver algebra.

Let Q be the quiver; 1 c 2

b and let kQ Λ = Λ = ,I = ha2, b2, ab − qba, aci, q ∈ k (1.1) q I be a family of quiver algebras. We note the following about Λq for each q. Remarks

• Λ is finitely generated since Q is a finite quiver with finite vertices and arrows.

• Λ is a Koszul graded quiver algebra.

! • Let Λ = ⊕i≥0Λi be a grading for Λ, the Koszul dual Λ of Λ is connected to the Yoneda algebra of Λ by the following;

∗ ! ∼ opp ⊥ E(Λ) = ExtΛ(Λ0, Λ0) = Λ = kQ /I (1.2)

where Qopp is the quiver with opposite arrow, I⊥ = haob0 +q−1b0a0, b0c0i and any v ∈ KQ, v0 is the correponding arrow in opposite quiver algebra kQopp. Note also that Λ! is generated in degrees 0 and 1.

• The case where q = ±1, I belongs to a class of (anti-)commutative ideals studied by Gawell and Xantcha. There is an associated generator graph (of the orthogonal ideal I⊥ of I) which has no directed cycles. This means that the ideal I is admissible [3].

2 For the case q = 1 of (1.1), the graded center Zgr(E(Λq)) is given by the following ( k ⊕ k[a, b]b, if char(k) = 2 Zgr(E(Λ1)) = k ⊕ k[a2, b2]b2, if char(k) 6= 2 where the degree of b is 1, and that of ab is 2. We now present a theorem of Snashall’s with respect to the ﬁnite generation conjecture.

Theorem 1.3. Let k be a ﬁeld and Λ1 be a member of the class of quiver algebras ∗ given in (1.1), and N be the set of nilpotent elements of HH (Λ1), then ( ∗ ∼ k ⊕ k[a, b]b, if char(k) = 2 HH (Λ1)/N = Zgr(E(Λ1)) = k ⊕ k[a2, b2]b2, if char(k) 6= 2 where the degree of b is 1, and that of ab is 2.

Our Result: In this paper, we study the Hochschild cohomology ring of the family Λq of quiver algebras of equation (1.1). We give a formula for the dimension of the ∗ ∗ space of Hochschild cocycles Ker(d ) where d : HomΛe (K∗, Λ) → HomΛe (K∗, Λ). We show that this number increases as the homological dimension grows. However, with a generalized cup product formula, we show that many of these cocycles are nilpotent with respect to the cup product and all cocycles of odd homological degrees are nilpotent. While some authors have presented the result of Theorem 1.3 using the graded center of the Yoneda algebra E(Λ), we give the same result using a diﬀerent approach. We present our result in Theorem 3.10 without using the isomorphism ∗ ∼ HH (Λ1)/N = Zgr(E(Λ1)) of Theorem (1.3) ( or the map φM of equation (2.3) deﬁned in the next section). This approach involves a direct computation and the use of the generalized cup product formula on elements of Hochschild cohomology in Propostion (3.5). Furthermore, we give explicit presentation of the elements a2n−rbr in k[a, b] of Theorem 1.3 as cocycles in Hochschild cohomology. We noted that these elements cannot be generated from other elements of lower homological degrees. Our main results are the following;

Proposition 1.4. Let φ : Km → Λ, and µ : Kn → Λ, be two Hochschild cocycles, m tm m m and {εr }r=0 are basis elements of Km such that for each 0 ≤ r ≤ tm, φ(εr ) = φr . Then the following gives a formula for the cup product on Hochschild cohomology.

3 (−1)mnφmµn, when k = 0  0 0 (−1)mnT m+n when 0 < k < m + n (φ µ)(εm+n) = (φµ)m+n = k ` k k (−1)mnφmµn, when k = m + n  m n  mn m n (−1) φ0 µn+1, when k = m + n + 1

min{m,k} m+n X j(n−k+j) m n Tk = (−q) φj µk−j, 0 < k < m + n. j=max{0,k−n}

kQ Theorem 1.5. Let k (char(k) 6= 2) be a ﬁeld and Λq = I be the family of quiver ∗ algebras of (1.1), and N the set of nilpotent elements of HH (Λq), then ( HH0(Λ )/N ∼= Z(Λ )∗ ∼= k, if q 6= ±1 HH∗(Λ )/N = q q q ∗ 2 2 2 ∼ 2 2 2 Z(Λq) ⊕ k[x , y ]y = k ⊕ k[x , y ]y , if q = ±1 where the degree of y is 1, and that of xy is 2.

2 Preliminary

A k-algebra Λ, is a graded quiver algebra if and only if there exists a finite quiver ∼ Q and a homogeneous admissible ideal I ⊆ kQ for which Λ = kQ/I [5]. Let (kQ)n be the vector subspace of kQ containing paths and linear combination of paths of ∼ length of n. Let I ⊂ (kQ)2, then Λ = kQ/I is a quadratic quiver algebra. We define the quadratic dual Λ! of Λ to be Λ! ∼= kQopp/I⊥, where Qopp is the quiver Q with opposite arrows. Since kQopp is also a k-vector space, it has a dual basis with opp respect to kQ. We define for each basis element vi ∈ (kQ )2, a corresponding ( 1, if i = j xi ∈ (kQ)2 such that the bilinear form hvi, xji = 0 if i 6= j.

⊥ opp Let I = {vi ∈ (kQ )2|hvi, xji = 0, ∀ xj ∈ I ⊂ (kQ)2}. ∼ L L If Λ is a graded quiver algebra, then Λ = i=0 Λi. We denote by r = i>0 Λi, the Jacobson radical of Λ and Λ0 = Λ/r is isomorphic to a direct sum of a ﬁ- nite number of copies of k. The Yoneda algebra E(Λ) of Λ is given by E(Λ) = ∗ ExtΛ(Λ0, Λ0). For Koszul algebras Λ = kQ/I,(kQ)2 ⊇ I is quadratic and generated by minimal uniform relations of homogeneous elements of degree 2, E(Λ) ∼= Λ! and E(Λ) is generated in degrees 0 and 1 [5]. For any two left Λ-modules M and N, the Hochschild cohomology ring HH∗(Λ) ∗ ∗ ∗ acts on ExtΛ(M,N) in such a way that ExtΛ(M,N) is an HH (Λ)-module. That is, there is a map

∗ ∗ ∗ HH (Λ) × ExtΛ(M,N) −→ ExtΛ(M,N) (2.1)

4 ∗ ∗ that deﬁnes a right (also left) module action of HH (Λ) on ExtΛ(M,N) [8]. A consequence of the module action of equation (2.1) is the following Proposition given in [7].

Proposition 2.2. Let M be a Λ−module. The map

∗ ∗ φM : HH (Λ) −→ ExtΛ(M,M) (2.3) deﬁned at the chain level by φM (f) = f ⊗1M is a ring homomorphism whose image ∗ is contained in the graded center Zgr(ExtΛ(M,M)).

For Koszul algebras, the image of the map φM was shown to be equal to the ∗ ∼ graded center Zgr(ExtΛ(M,M)), where M = Λ/r [2]. The graded center is ∗ |z||α| the set of all homogeneous z ∈ ExtΛ(M,M) for which zα = (−1) αz, for all homogeneous α. For a k-algebra Λ, denote by Λe := Λ ⊗ Λop its enveloping algebra having the tensor product algebra structure. Λop is Λ with the opposite multiplication. We have then that a left Λe-module is a Λ-bimodule and vice versa. In [2], the following ideas were used to define a minimal projective resolution of Λ as a Λe-module. For a finite quiver Q, let R = kQ, and I ⊂ kQ an ideal of R. It was shown in n tn [4] that there exist integers tn and uniform elements {fi }i=0 in R such that for n ≥ 0, there is a filtration of right ideals

tn tn−1 t1 t0 M n M n−1 M 1 M 0 · · · ⊆ fi R ⊆ fi R ⊆ · · · ⊆ fi R ⊆ fi R = R i=0 i=0 i=0 i=0 in R. This ﬁltration was then employed to construct a minimal projective resolution K• of Λ which will be used in deﬁning Hochschild cohomology. For the family of quiver algebras in (1.1), take Λ0 to be the subalgebra of Λ generated by the vertices. We immediately view R as the tensor algebra TΛ0 (Λ1) where Λ1 is n the Λ0-bimodule generated by the arrows in Q. Since Λ is Koszul, each fi can be viewed as a linear combination of paths of length n. Hence it can be viewed as ⊗Λ0 n an element in Λ1 for all n. This makes it possible to embed them into the bar 0 t0 1 t1 resolution. We choose {fi }i=0 to be the set of vertices in Q, {fi }i=0 to be the set 2 t2 of arrows in Q while {fi }i=0 to be a minimal set of homogeneous generators of degree two for I. We use the notation Γn to denote the set containing all linear ⊗Λ0 n combinations of homogeneous elements of degree n, viewed as elements of Λ1 . In summary, by taking ei (i = 0, 1) to be the arrow of length 0 (also idempotents)

5 associated with the vertex i, we have

0 0 0 0 Γ = {e1, e2} = {f0 , f1 }, t0 = |Γ |, 1 1 1 1 1 Γ = {a, b, c} = {f0 , f1 , f2 }, t1 = |Γ |, Γ2 = {a ⊗ a, a ⊗ b − qb ⊗ a, b ⊗ b, a ⊗ c} 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 = {f0 ⊗ f0 , f0 ⊗ f1 − qf1 ⊗ f0 , f1 ⊗ f1 , f0 ⊗ f2 } = {f0 , f1 , f2 , f3 }, t2 = |Γ |. and for n ≥ 2

a⊗n, when i = 0  f n−1 ⊗ b + (−q)sf n−1 ⊗ a, when (0 < i < n) Γn = f n = s−1 s (2.4) i ⊗n b , when i = n  a⊗(n−1) ⊗ c, when i = n + 1

n e and set tn = |Γ |. The minimal projective resolution (K, d) of Λ over Λ -modules is given by tn M n n Kn = Λo(fi ) ⊗k t(fi )Λ (2.5) i=0

To describe the differentials d, we need to define the basis elements of Kn for each n. n n n n n Let εi = (0,..., 0, o(fi ) ⊗k t(fi ), 0,..., 0) ∈ Kn where the element o(fi ) ⊗k t(fi ) is in the i-th position counting from 0. We now define the differentials on the resolution dn+1 dn dn−1 d2 d1 π ··· → Kn → Kn−1 → · · · → K1 → K0 → Λ → 0 (2.6) as follows

1 0 0 d1(ε2) = cε1 − ε0c n n−1 n−i i n−1 dn(εi ) = (1 − ∂n,i)[aεi + (−1) q εi a] n−i n−1 n n−1 + (1 − ∂i,0)[(−q) bεi−1 + (−1) εi−1 b], for i ≤ n n n−1 n n−1 dn(εn+1) = aεn + (−1) ε0 c, when n ≥ 2 ( 1 if i = j where ∂i,j = . Let B = B•(Λ) denote the bar resolution of Λ 0, otherwise given by;

⊗(n+2) δn ⊗(n+1) δn−1 δ2 ⊗3 δ1 ⊗2 π B• := · · · → Λ → Λ → · · · → Λ → Λ → Λ → 0 (2.7)

The diﬀerentials δi are given by n X i δn(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = (−1) a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1 (2.8) i=0

6 for each element ai ∈ Λ (0 ≤ i ≤ n + 1) and π, the multplication map. The Hochschild cohomology of Λ is deﬁned to be

∗ ∗ M n HH (Λ) = ExtΛe (Λ, Λ) = H (HomΛe (B•(Λ), Λ)) n≥0 There is a natural embedding of K into B. That is, there is a map

ι• : K• −→ B• n n lifting the identity on Λ such that for each n, ιn(εi ) = 1 ⊗ fi ⊗ 1. We recall that ⊗(n+2) n Bn = Λ , and so, the following diagram is commutative that is ιn−1dn(εi ) = n δnιn(εi ) (see Proposition 2.1 of [2] for a proof).

d2 d1 π K• : ··· / K2 / K1 / K0 / Λ / 0 (2.9)

ι2 ι1 ι0 1Λ

δ2 δ1 π B• : ··· / B2 / B1 / B0 / Λ / 0

e We apply the functor HomΛe (−, Λ) to the projective resolution K of Λ - modules. ˆ Deﬁning Ki = HomΛe (Ki, Λ), we obtain the following complex

d∗ d∗ d∗ d∗ d∗ d∗ ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 0 −→ K0 −→ K1 −→ K2 −→ K3 −→ K4 −→ K5 −→ K6 −→ · · ·

We note that t0 = 1, therefore, ˆ 0 0 0 0 K0 = HomΛe (K0, Λ) = HomΛe (Λo(f0 ) ⊗k t(f0 )Λ ⊕ Λo(f1 ) ⊗k t(f1 )Λ, Λ) 0 0 0 0 = HomΛe (Λo(f0 ) ⊗k t(f0 )Λ, Λ) ⊕ HomΛe (Λo(f1 ) ⊗k t(f1 )Λ, Λ) 0 0 0 0 = o(f0 )Λt(f0 ) ⊕ o(f1 )Λt(f1 ) n n n For a ﬁxed n and i, let φ ∈ HomΛe (Λo(fi )⊗kt(fi )Λ, Λ), suppose that φ(0, ··· , 0, o(fi )⊗k n n n t(fi ), 0, ··· , 0) = φ(εi ) = λi ∈ Λ, then n n n n n n n o(fi )λi t(fi ) = o(fi )φ(o(fi ) ⊗k t(fi ))t(fi ) n 2 n 2 n n n = φ(o(fi ) ⊗k t(fi ) ) = φ(o(fi ) ⊗k t(fi )) = λi n n where each o(fi ), t(fi ) is any of the idempotents e1 or e2. We have the following e n n n n isomorphisms of Λ -modules: HomΛe (Λo(fi ) ⊗k t(fi )Λ, Λ) ' o(fi )Λ t(fi ). So

tn ˆ M n n Kn = HomΛe (Kn, Λ) = o(fi )Λ t(fi ) (2.10) i=0 ˆ n n In general when φ ∈ Kn such that φ(εi ) = λi for 0 ≤ i ≤ tn, we simply write n n n φ = (λ0 , ··· , λi , ··· , λtn ).

7 3 Proof of Main Results

In this section, we give a sequence of propositions leading to the main results. We start by ﬁnding elements of Hochschild 0-cocycles, which in theory corresponds to elements in the center of the algebra.

∗ 0 ker d1 The 0th Hochschild cohomology (HH (Λ) = Im(0) ). ∗ ˆ 0 0 0 0 Let φ ∈ ker d1 ⊆ K0 = HomΛe (K0, Λ), such that φ = (λ0, λ1), for some λ1, λ1 ∈ 0 ∗ 1 Λ. We solve for the λi (i = 0, 1) for which d1φ(εi ) = 0 as follows

∗ 1 1 0 1 0 0 d1φ(ε0) = φd1(ε0) = φ(a(ε0) + (−1) q (ε0)a) 0 0 = aλ0 − λ0a = 0 ∗ 1 1 0 0 0 d1φ(ε1) = φd1(ε1) = φ((−q) b(ε0) − (ε0)b) 0 0 = bλ0 − λ0b = 0 ∗ 1 1 0 0 d1φ(ε2) = φd1(ε2) = φ(c(ε1) − (ε0)c) 0 0 = cλ1 − λ0c = 0 If q = 1, then ab − ba = 0, we get the following set of solutions: φ = (a, 0), 0 0 (ab, 0),(0, a), (0, b), (e1, e2) or (0, e1). By identifying each solution (λ0, λ1) with 0 0 0 0 0 0 0 0 0 0 (o(f0 )λ0t(f0 ), o(f1 )λ1t(f1 )) = (e1λ0e1, e2λ1e2), we need to have o(λ0) = t(λ0) = e1 0 0 and o(λ1) = t(λ1) = e2, we eliminate some solutions to have the following unique set of solutions φ1 = (a, 0), φ2 = (ab, 0) and φ3 = (e1, e2)

If q = −1, then ab + ba = 0, we get the same unique set of solutions: φ1 = (a, 0) (if char(k) = 2), φ2 = (ab, 0) and φ3 = (e1, e2).

If q 6= 1, then ab − qba = 0, we get φ2 = (ab, 0) and φ3 = (e1, e2). Therefore, e ∗ the Λ -module homomorphisms φ1, φ2, φ3 form a basis for the kernel of d1 as a k-vector space. That is,

∗ ker d1 = spank{φ1, φ2, φ3}. In summary we obtain for any q ∈ k that,  spank{(a, 0), (ab, 0), (e1, e2)}, if q = 1  ker d∗ span {(ab, 0), (e , e )}, if q = −1 HH0(Λ) = 1 = k 1 2 Im(0) span {(a, 0), (ab, 0), (e , e )}, if q = −1 and char(k) = 2  k 1 2  spank{(ab, 0), (e1, e2)}, for every other q 6= ±1 where each ordered pairs are in e1Λe1 ⊕ e2Λe2.

8 Remark 3.1. We note that each Hochschild 0-cocycle of the set {(a, 0), (ab, 0)} corresponds to an element in the set {a, ab} of the center of the algebra Λq. As we will see later in Remark 3.6, these elements are nilpotent with respect to the cup product but the 0-cocycle φ3 = (e1, e2) is not, since e1, e2 are idempotent elements. ∗ We then identify spank{(e1, e2)} to be the subalgebra of HH (Λq) isomorphic to k because e1 +e2 = 1Λq . This brings us to make the following deduction for any q ∈ k ker d∗ HH0(Λ)/N = 1 = span {(e , e )} ∼= k. (3.2) Im(0) k 1 2

Higher Hochschild cocycles We now give the following counting proposition about the dimension of the ∗ ˆ ˆ kernels of the diﬀerentials dn+1 : Kn → Kn+1. kQ Proposition 3.3. Let k be a ﬁeld and let Λq = I where Q is the quiver of ∗ n kerdn+1 equation (1.1). Hochschild cohomology group is given by HH (Λ) = ∗ , and Im dn for n 6= 0 ( ∗ 2(n + 2), when n is odd dim(ker dn+1) = 5n q = 1 2 + 4, when n is even ( 5n + 4, when n is odd dim(ker d∗ ) = 2 q = −1 n+1 2(n + 2), when n is even ∗ dim(ker dn+1) = n + 2, for any integer n q 6= ±1 as a k-vector space. ∗ n n n n n Proof. Let φ ∈ ker dn+1, with φ = (φ0 , φ1 , ··· , φn, φn+1). The elements φi = n φ(εi ), i = 0, ··· , n + 1 are obtained by solving the following set of equations For any n or q ∗ n+1 n n+1 n dn+1φ(ε0 ) = aφ(ε0 ) + (−1) φ(ε0 )a n n = aφ0 ± φ0 a = 0 and ∗ n+1 n n+1 n dn+1φ(εn+2) = aφ(εn+1) + (−1) φ(ε0 )c n n = aφn+1 ± φ0 c = 0 n for these set of equations to be zero, we should have φ0 ∈ spank{a, c, ab, bc} and n n n φn+1 ∈ spank{a, c, ab, bc}. But we recall that φ0 ∈ e1Λe1, and φn+1 ∈ e1Λe2, we n n thus obtain the following φ0 ∈ spank{a, ab} and φn+1 ∈ spank{c, bc}. The rest of this proof involves solving the general set of equations; ∗ n+1 n n+1−r r n n+1−r n n+1 n dn+1φ(εr ) = aφ(εr ) + (−1) q φ(εr )a + (−q) bφ(εr−1) + (−1) φ(εr−1)b n n+1−r r n n+1−r n n+1 n = aφr + (−1) q φr a + (−q) bφr−1 + (−1) φr−1b = 0 ∗ n+1 n n−r r+1 n n−r n n+1 n dn+1φ(εr+1 ) = aφr+1 + (−1) q φr+1a + (−q) bφr + (−1) φr b = 0

9 for diﬀerent values of n, r and q. We recall that q = ±1 implies ab ∓ ab = 0 When n is even, r is even, q = 1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving

∗ n+1 n n dn+1φ(εr ) = aφr − φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = bφr − φr b = 0.

n We can only obtain both equations equal to 0 if φr ∈ spank{a, b, ab, bc, e1}. Again n n we recall that φr ∈ e1Λe1, so φr ∈ spank{a, b, ab, e1}. When n is even, r is odd, q = 1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving

∗ n+1 n n dn+1φ(εr ) = aφr + φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = −bφr − φr b = 0

n n We can only obtain both equations equal to 0 if φr ∈ spank{ab, bc}. Again φr ∈ n e1Λe1, so φr ∈ spank{ab}. When n is odd, r is even, q = 1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving

∗ n+1 n n dn+1φ(εr ) = aφr + φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = −bφr + φr b = 0

n We can only obtain both equations equal to 0 if φr ∈ spank{a, ab, bc} and ﬁnally n we get φr ∈ spank{a, ab}. When n is odd, r is odd, q = 1, n n n we obtain φr by setting φr−1 = φr+1 = 0 solving

∗ n+1 n n dn+1φ(εr ) = aφr − φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = bφr + φr b = 0

n we can only obtain both equations equal to 0 if φr ∈ spank{ab, bc, b}. Like before n we obtain φr ∈ spank{b, ab}. When n is even, r is even, q = −1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving

∗ n+1 n n dn+1φ(εr ) = aφr − φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = bφr − φr b = 0

n We have the solution φr ∈ spank{ab, e1}.

10 When n is even, r is odd, q = −1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving ∗ n+1 n n dn+1φ(εr ) = aφr + φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = bφr − φr b = 0 n We can only obtain both equations equal to 0 if φr ∈ spank{ab, b}. When n is odd, r is even, q = −1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving ∗ n+1 n n dn+1φ(εr ) = aφr + φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = bφr + φr b = 0 n We can only obtain both equations equal to 0 if φr ∈ spank{a, b, ab}. When n is odd, r is odd, q = −1 n n n we obtain φr by setting φr−1 = φr+1 = 0 solving ∗ n+1 n n dn+1φ(εr ) = aφr − φr a = 0 and ∗ n+1 n n dn+1φ(εr+1 ) = bφr + φr b = 0 n we obtain φr ∈ spank{a, ab}. For any other q 6= ±1 If n is even and r is even, we obtain ∗ n+1 n r n dn+1φ(εr ) = aφr − q φr a = 0 and ∗ n+1 n−r n n dn+1φ(εr+1 ) = q bφr − φr b = 0 n we obtain the trivial solutions φr ∈ spank{ab}. If n is even and r is odd, we obtain ∗ n+1 n r n dn+1φ(εr ) = aφr + q φr a = 0 and ∗ n+1 n−r n n dn+1φ(εr+1 ) = −q bφr − φr b = 0 n with the trivial solutions φr ∈ spank{ab}. In similar ways we obtain the trivial solutions for n odd and r even or odd. But n n we must have φr ∈ e1Λe1, (0 ≤ r ≤ n) and φn+1 ∈ e1Λe2 , we come to conclude n that φr ∈ spank{ab}, whenever q 6= ±1.

n The following table is a summary of solutions stating generators for each φr ; q = 1 n is even n is odd r is even r is odd r is even r is odd n φ0 a, ab a, ab n φr a, b, ab, e1 ab a, ab b, ab n φn+1 c, bc c, bc

11 q = −1 n is even n is odd r is even r is odd r is even r is odd n φ0 a, ab a, ab n φr ab, e1 b, ab a, b, ab a, ab n φn+1 c, bc c, bc q 6= ±1 n is even n is odd r is even r is odd r is even r is odd n φ0 a, ab a, ab n φr ab ab ab ab n φn+1 c, bc c, bc From these tables, we make the following deductions;

(odd−positions) (even−positions) n n n (n is even and q = +1) : dim(Kerd∗ ) = 2+( × 1 + × 4 )+2 = 5( )+4 n+1 2 2 2

(odd−positions) (even−positions) n n (n is odd and q = +1) : dim(Kerd∗ ) = 2+( × 2 + × 2 )+2 = 2(n+2) n+1 2 2 (odd−positions) (even−positions) n n (n is even and q = −1) : dim(Kerd∗ ) = 2+( × 2 + × 2 )+2 = 2(n+2) n+1 2 2 (odd−positions) (even−positions) n n n (n is odd and q = −1) : dim(Kerd∗ ) = 2+( × 2 + × 3 )+2 = 5( )+4 n+1 2 2 2 (odd−positions) (even−positions) n n ( for any n, q 6= ±1) : dim(Kerd∗ ) = 2+( × 1 + × 1 )+2 = n+4 n+1 2 2

Remark 3.4. We note that these dimensions grow linearly as the homological dimension n grows. We also observe from the tables in the proof of Proposition 3.3 that if q = ±1, there are Hochschild n-cocycles of the form φ = (0, ··· , 0, e1, 0, ··· , 0) n i.e φi = e1, where both n and i are even. We will later see that these are the only n non-nilpotent elements. Whenever n is odd, there is no φ for which φi = e1. This is equivalent to saying that all elements of odd homological degrees are nilpotent with respect to the cup product.

A Cup product formula on Hochschild cohomology We will now deﬁne a cup product formula at the chain level for cocycles in Hochschild cohomology. It was shown in [4] that the minimal projective resolution

12 K of Λ posseses a “comultiplicative structure”. That is, there is a comultiplication map ∆K : K → K ⊗Λ K such that (∆ ⊗ 1)∆ = (1 ⊗ ∆)∆. Recall that K embeds nicely into the bar complex B via ι : K → B. There is a map ι⊗ι : K⊗K → B⊗B such that (ι ⊗ ι)(K ⊗ K) = ι(K) ⊗ ι(K) ⊆ B ⊗ B. The following diagram is commutative.

∆K K −→ K ⊗Λ K ι ↓ ↓ ι ⊗ ι

∆B B −→ B ⊗Λ B ∼ ⊗m that is (ι⊗ι)◦∆K = ∆B ◦ι. We recall that if α ∈ HomΛe (Bm, Λ) = Homk(Λ , Λ), ⊗n and β ∈ Homk(Λ , Λ) are two cocycles, then one way to deﬁne the cup product is the composition of the following maps;

∆ α⊗β π α ` β : B −→ B ⊗Λ B −→ Λ ⊗Λ Λ ' Λ where (α ⊗ β)(x ⊗ y) = (−1)|β||x|α(x) ⊗ β(y), and we take |β| = n since it is an n-cocycle. This convention matches the traditional deﬁnition of the cup product on the bar resolution given as

mn (α ` β)(a1⊗· · ·⊗am+n) = (−1) α(a1⊗· · ·⊗am)·β(am+1⊗· · ·⊗am+n), ∀ ai ∈ Λ This deﬁnition extends to the resolution K using a similar composition map but replacing the ∆B, that is

φ⊗µ π ∆K φ ` µ : K −→ K ⊗Λ K −→ Λ ⊗Λ Λ ' Λ so that φ ` µ = π(φ ⊗ µ)∆K. Let φ : Km → Λ, and µ : Kn → Λ be two cocycles of homological degrees m and n respectively, we use the follow- m m m m n n n n n ing notation φ ` µ = (φ0 , φ1 , ··· , φm, φm+1) ` (µ0 , µ1 , µ2 , ··· , µn, µn+1) = m+n m+n m+n m+n m+n ((φµ)0 , (φµ)1 , (φµ)2 , ··· , (φµ)m+n, (φµ)m+n+1) for their cup product.

Proposition 3.5. Let φ : Km → Λ, and µ : Kn → Λ, be two Hochschild cocycles. Then the following gives a formula for their cup product.

(−1)mnφmµn, when k = 0  0 0 (−1)mnT m+n when 0 < k < m + n (φ µ)(εm+n) = (φµ)m+n = k ` k k (−1)mnφmµn, when k = m + n  m n  mn m n (−1) φ0 µn+1, when k = m + n + 1

min{m,k} m+n X j(n−k+j) m n Tk = (−q) φj µk−j, 0 < k < m + n. j=max{0,k−n}

13 1 1 1 1 1 1 Proof. Suppose m = n = 1, take φ = (φ0, φ1, φ2) = (a, b, c) = (f0 , f1 , f2 ) and µ = 1 1 1 1 1 1 1 1 1 (µ0, µ1, µ2) = (a, b, c). We then realize that φ ` µ = (φ0, φ1, φ2) ` (µ0, µ1, µ2) = 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 (φ0µ0, φ0µ1 − qφ1µ0, φ1µ1, φ0µ2) = (a , ab − qba, b , ac) = (f0 , f1 , f2 , f3 ) after applying π(φ⊗µ)∆K. This consists of paths and linear combination of paths of length 2, that is, elements of Γ2 given in (2.4). Similarly, if we take m = 1, n = 2, that is, 1 1 1 2 2 2 2 2 2 take φ = (φ0, φ1, φ2) = (a, b, c) and µ = (µ0, µ1, µ2, µ3) = (a , ab − qba, b , ac). We 1 2 1 2 2 1 2 1 2 obtain after applying π(φ ⊗ µ)∆K, φ ` µ = (φ0µ0, φ0µ1 + (−q) φ1µ0, φ0µ2 + 1 1 2 1 2 1 2 3 2 2 2 2 2 2 3 2 (−q) φ1µ1, φ1µ2, φ0µ3) = (a , a b − qaba + q ba , ab − qbab + q b a, b , a c) = 3 3 3 3 3 3 (f0 , f1 , f2 , f3 , f4 ) which are elements of Γ in (2.4). It is enough to ﬁnd a general m+n m+n m+n m+n formula for each element fr ∈ Γ , so that ι(εr ) = 1 ⊗ fr ⊗ 1. We will m+n then ﬁnd the image of εr under ∆K. Since φ ` µ : Km+n → Λ, we simply evaluate m+n m+n (φ ` µ)(εr ) = π(φ ⊗ µ)∆K(εr ), for 0 ≤ r ≤ m + n + 1. Using the fact that m+n m+n (ι ⊗ ι)∆K(ε0 ) = ∆Bι(ε0 ) m+n times m+n 1 1 1 = ∆B(1 ⊗ f0 ⊗ 1) =∆B(1 ⊗ f0 ⊗ f0 ⊗ · · · ⊗ f0 ⊗ 1) m+n X r m+n−r = (1 ⊗ f0 ⊗ 1) ⊗ (1 ⊗ f0 ⊗ 1) r=0 m+n X r m+n−r = (ι ⊗ ι)( ε0 ⊗ ε0 ), so that r=0 m+n m+n X r m+n−r ∆K(ε0 ) = ( ε0 ⊗ ε0 ), similarly r=0

m+n m+n X r m+n−r ∆K(εm+n) = εr ⊗ εm+n−r and r=0 m+n m+n X r m+n−r ∆K(εm+n+1) = ε0 ⊗ εm+n−r+1 r=0 We therefore obtain for r = 0, m + n, m + n + 1; m+n m n mn m n mn m n π(φ ⊗ µ)∆K(ε0 ) = π(φ ⊗ µ)(ε0 ⊗ ε0 ) = (−1) π(φ0 ⊗ µ0 ) = (−1) φ0 µ0 . m+n m n mn m n mn m n π(φ ⊗ µ)∆K(εm+n) = π(φ ⊗ µ)(εm ⊗ εn) = (−1) π(φm ⊗ µn) = (−1) φmµn. m+n m n mn m n mn m n π(φ⊗µ)∆K(εm+n+1) = π(φ⊗µ)(ε0 ⊗εn+1) = (−1) π(φ0 ⊗µn+1) = (−1) φ0 µn+1. It was shown in [1] that for r = 1, 2, ··· , n − 1,

min{t,r} n X j(n−r+j−t) t n−t fr = (−q) fj ⊗ fr−j , j=max{0,r+t−n}

14 therefore

min{t,r} m+n h X j(m+n−r+j−t) t m+n−ti ι(εr ) = 1 ⊗ (−q) fj ⊗ fr−j ⊗ 1 j=max{0,r+t−m−n} letting t = m min{m,r} X j(n−r+j) m n = (−q) 1 ⊗ fj ⊗ fr−j ⊗ 1 j=max{0,r−n} applying the diagonal map ∆B and retaining the part that is nonzero when you apply φ ⊗ µ

min{m,r} m+n X j(n−r+j) m n (∆Bι)(εr ) = ··· + (−q) (1 ⊗ fj ⊗ 1) ⊗ (1 ⊗ fr−j ⊗ 1) + ··· j=max{0,r−n} min{m,r} X j(n−r+j) m n = ··· + (−q) (ι ⊗ ι)(εj ⊗ εr−j) + ··· j=max{0,r−n}

using the relation that (ι ⊗ ι) ◦ ∆K = ∆B ◦ ι min{m,r} m+n X j(n−r+j) m n (ι ⊗ ι)∆K(εr ) = (ι ⊗ ι) ··· + (−q) (εj ⊗ εr−j) + ··· j=max{0,r−n}

min{m,r} m+n X j(n−r+j) m n ∆K(εr ) = ··· + (−q) εj ⊗ εr−j + ··· j=max{0,r−n} therefore we obtain after applying φ ⊗ µ and multiplication π min{m,r} m+n mn X j(n−r+j) m n (φ ` µ)(εr ) = (−1) (−q) φ(εj )µ(εr−j) j=max{0,r−n} min{m,r} mn X j(n−r+j) m n = (−1) (−q) φj µr−j j=max{0,r−n} mn m+n = (−1) Tr

Remark 3.6. We give the following as a support to our previous Remark 3.4. ∗ m From Proposition 3.3, we observed that kerdn+1 is generated by φ such that φ(εi ) =

15 m m φi ∈ spank{a, b, ab, c, bc, e1}. But any φ having any of its φi to be any of a, b, ab, c, bc is nilpotent. This is because

min{m,r} m+n mn X j(n−r+j) m n (φ ` φ)(εr ) = (−1) (−q) φj φr−j (3.7) j=max{0,r−n}

m n where φj φr−j is a product of any two elements in the set {a, b, ab, c, bc} which is equal to 0 in the algebra. If it is not zero, we simply take a triple cup product using the following;

n+n+n m+n (φ ` φ ` φ)(εr ) = (µ ` φ)(εr )(take µ = φ ` φ, m = n + n) min{m,r} mn X j(n−r+j) m n = (−1) (−q) µ(εj )φ(εr−j) j=max{0,r−n} min{m,r} mn X j(n−r+j) n+n n = (−1) (−q) [φ ` φ(εj )]φ(εr−j) j=max{0,r−n} min{m,r} min{n,l} mn X j(n−r+j)h n2 X i(n−l+i) n n i n = (−1) (−q) (−1) (−q) φ(εi )φ(εl−i) φ(εr−j) j=max{0,r−n} i=max{0,l−n} min{m,r} min{n,l} 3n2 X X ij(n−r+j)(n−l+i) n n n = (−1) (−q) φ(εi )φ(εl−i)φ(εr−j) j=max{0,r−n} i=max{0,l−n}

n n n n n n n The product φ(εi )φ(εl−i)φ(εr−j) = φi φl−iφr−j is always 0 in Λq except each φi = m m m e1. Therefore a cocycle φ ∈ HH (Λ) is non-nilpotent if and only if φi = φl−i = m φr−j = e1 for some i, j, l, r. According to Proposition 3.3 this is the case only when q = ±1, n is even and i is even.

The following Proposition summarizes Remarks 3.4 and 3.6.

Proposition 3.8. Let φ : Kn → Λq, be a cocycle. Then φ is non-nilpotent if, and only if q = ±1 and ( e , if n, r are even φ(εn) = φn = 1 r r 0 otherwise.

n e Let C (Λ, Λ) = HomΛe (Kn, Λ) be the Λ -module generated by all n-Hochschild ∗ L cochains. Denote by C (Λ, Λ) = n≥0 HomΛe (Kn, Λ) the algebra of Hochschild ∗ cochains with coeﬃcients in Λ. The n-cocycles of HH (Λ±1) given in Proposi- tion 3.8 are therefore given by Z∗(Λ, Λ) = C∗(Λ, Λ)/N .

16 Before we state the next proposition, we will show that no two cocycles of Proposition 3.8 differ by a coboundary. This is important as we will later define a 1-1 module homomorphism from Z∗(Λ, Λ) to the polynomial ring k[x, y]. 2n 2n 2n For a fixed n, let φ, β be two cocycles such that φ(εr ) = φr = e1, β(εs ) = 2n ∗ βs = e1, where r < s are both even and there is α such that d (α) = φ − β = (0, ··· , 0, e1, 0, ··· , 0, e1, 0, ··· , 0), the idempotents e1 is in the r and s position. We have the following when i = 0,

∗ 2n 2n d (α)(εi ) = αd(εi ), 2n−1 n 2n−1 2n−1 2 2n−1 0 = α(aε0 + (−1) ε0 a) = aα(ε0 ) + (−1) nα(ε0 )a 2n−1 hence, α(ε0 ) = 0 and in general, we must have

∗ 2n 2n−1 2n−r r 2n−1 2n−r 2n−1 2n 2n−1 e1 = d (α)(εr ) = aα(εr ) + (−1) q α(εr )a + (−q) bα(εr−1 ) + (−1) α(εr−1 )b and ∗ 2n 2n−1 2n−s s 2n−1 2n−s 2n−1 2n 2n−1 e1 = d (α)(εs ) = aα(εs ) + (−1) q α(εs )a + (−q) bα(εs−1 ) + (−1) α(εs−1 )b.

2n−1 2n−1 There is no way we can deﬁne α(εr ) and α(εs ) so that the equation above is true i.e., the right hand side equals e1. Hence there is no such α. Therefore each cocycle is distinct and do not diﬀer by a coboundary. We now make the following Proposition.

Proposition 3.9. Z∗(Λ, Λ) is graded with respect to the cup product and can be expressed as a subalgebra of k[x, y], that is

Z∗(Λ, Λ) ∼= k[x2, y2]y2 where the degree of y is 1, and xy is 2.

Proof. We ﬁrst show that Z∗(Λ, Λ) can be expressed as ( n e , if r is even o L n 1 spank φ : K2n → Λ±1 φr = . It is straightforward to n>0 0 otherwise see that if φ ∈ Z∗(Λ, Λ), there are pairs of positive even integers m, i such that m φ : Km → Λ±1 and φ(εi ) = e1 and 0 at other positions not equal to i. Hence ( n e , if i is even o ∗ m 1 Z (Λ, Λ) ⊆ spank φ : Km → Λ±1 φi = 0 otherwise ( n e , if r is even o M n 1 ⊆ spank φ : K2n → Λ±1 φr = n>0 0 otherwise

17 The grading comes from the fact that its elements are Hochschild cocycles and L all odd degree elements vanish. We will now show that spank φ : K2n → n>0 n Λ±1 φr = e1, for some r is graded with respect to the cup product, hence contained in Z∗(Λ, Λ). ( 2m e1, if r is even Let φ : K2m → Λq, be given by φr = , and 0 otherwise ( 2n e1, if s is even µ : K2n → Λq, be given by µs = . 0 otherwise 2(m+n) 2m 2n First note that (φ ` µ)(ε0 ) = φ0 µ0 = 0. Since 2m + 2n is even, whenever 2(m+n) 2m 2n r = 2m and s = 2n, we get (φ ` µ)(ε2(m+n)) = φ2mµ2s = e1 · e1 = e1. Also 2(m+n) 2m 2n (φ ` µ)(ε2(m+n)+1) = φ0 µ2n+1 = 0. Whenever 0 ≤ t ≤ 2(m + n)

min{2m,t} 2m+2n X j(2n−t+j) 2m 2n (φ ` µ)(εt ) = (−q) φj µt−j = ±e1 j=max{0,t−2n} ( e , if t, j is even whenever t, j are even since φ2mµ2n = 1 j t−j 0 otherwise. ( n e , if r is even o n 1 This shows that (φ µ) ∈ spank φ : K2n → Λq φr = . We ` 0 otherwise deﬁne a map from Z∗(Λ, Λ) → k[x, y] by 2(n−1) 2 2(n−2) 4 (0, 0, e1, 0, ··· , 0) 7→ x y , (0, 0, 0, 0, e1, 0, ··· , 0) 7→ x y , 2n−r r 2n ··· , (0, ··· , 0, e1, 0, ··· , 0) 7→ x y , ··· , (0, 0, ··· , 0, e1, 0) 7→ y . Under this | {z } r−th position map, the image of Z∗(Λ, Λ) is the subalgebra k[x2, y2]y2 which is not ﬁnitely generated as an algebra. Also note how the cup product corresponds with multiplication in k[x, y], that is

= 2n−r r 2m−s s (0, ··· , 0, e1 , 0, ··· , 0) ` (0, ··· , 0, e1 , 0, ··· , 0) / (x y ) · (x y ) |{z} |{z} r s = = = 2(n+m)−(r+s) r+s (0, ··· , 0, e1 , 0, ··· , 0) / x y |{z} r+s

2(n−1) 2 For each n, the element x y which we identify with (0, 0, e1, 0, ··· , 0) cannot be generated by any element of lower homological degree. This brings us to conclude that the graded copies of cocycles of Proposition 3.8 can be compactly

18 given as

L span φ : → Λ φn = e , for some r = k[x2, y2]y2. n>0 k K2n ±1 r 1

Before we give the ﬁnal Theorem, we illustrate with an example. Example To show that

2 2 2 x y · y = (0, 0, e1, 0, 0, 0) ` (0, 0, e1, 0) 2 4 = (0, 0, 0, 0, e1, 0, 0, 0) = x · y

2 2 4 4 4 4 4 4 2 2 2 2 2 Take φ = x y = (φ0, φ1, φ2, φ3, φ4, φ5) = (0, 0, e1, 0, 0, 0) and µ = y = (φ0, φ1, φ2, φ3) = (0, 0, e1, 0)

6 4 2 (φ ` µ)(ε0) = φ0µ0 = 0 1 6 X j(1+j) 4 2 4 2 4 2 (φ ` µ)(ε1) = (−1) φj µ1−j = φ0µ1 + φ1µ0 = 0 j=0 2 6 X j2 4 2 4 2 4 2 4 2 (φ ` µ)(ε2) = (−1) φj µ2−j = φ0µ2 − φ1µ1 + φ2µ0 = 0 j=0 3 6 X j(−1+j) 4 2 4 2 4 2 4 2 (φ ` µ)(ε3) = (−1) φj µ3−j = φ1µ2 + φ2µ1 + φ3µ0 = 0 j=1 4 6 X j(−2+j) 4 2 4 2 4 2 4 2 (φ ` µ)(ε4) = (−1) φj µ4−j = φ2µ2 − φ3µ1 + φ4φ0 = e1 j=2 4 6 X j(−3+j) 4 2 4 2 4 2 (φ ` µ)(ε5) = (−1) φj µ5−j = φ3µ2 + φ4µ1 = 0 j=3 6 4 2 (φ ` µ)(ε6) = φ4µ4 = 0 6 4 2 (φ ` µ)(ε7) = φ0µ3 = 0

kQ Theorem 3.10. Let k (char(k) 6= 2) be a ﬁeld and Λq = I be the family of quiver ∗ algebras of (1.1), and N the set of nilpotents elements of HH (Λq), then ( HH0(Λ )/N ∼= Z(Λ )∗ ∼= k, if q 6= ±1 HH∗(Λ )/N = q q q ∗ 2 2 2 ∼ 2 2 2 Z(Λq) ⊕ k[x , y ]y = k ⊕ k[x , y ]y , if q = ±1 where the degrees of y is 1, and that of xy is 2.

19 Proof. If q 6= ±1, then all φ : Kn → Λq are nilpotent elements by Remark3.6 and Proposition 3.8. From Remark 3.1, we have then that

∗ 0 ∼ ∗ ∼ HH (Λq)/N = HH (Λq)/N = Z(Λq) = k

If q = ±1, then the only non-nilpotent elements are those of Proposition 3.8. From Remarks 3.1 and Proposition 3.9 we have that Hochschild cohomology ring modulo nilpotent elements of the family of quiver algebras in (1.1) is spanned by graded copies of cocycles given by Proposition 3.8. That means that

∗ 0 ∗ HH (Λq)/N = HH (Λ±1)/N ⊕ Z (Λ, Λ) ( ∼ M n n e1, if n, r is even o = k ⊕ spank φ : K2n → Λq φr = n>0 0 otherwise = k ⊕ k[x2, y2]y2.

Acknowledgment: I want to thank Dr. Sarah Witherspoon for reading through the manuscript and making invaluable suggestions.

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