Hochschild cohomology of some quantum complete intersections
aa r X i v : . [ m a t h . K T ] M a r HOCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETEINTERSECTIONS
KARIN ERDMANN, MAGNUS HELLSTRØM-FINNSEN
Abstract.
We compute the Hochschild cohomology ring of the algebras A = k h X, Y i / ( X a , XY − qY X, Y a ) over a field k where a ≥ q ∈ k is a primitive a -th root of unity. We find the thedimension of HH n ( A ) and show that it is independent of a . We compute explicitly the ring structure ofthe even part of the Hochschild cohomology modulo homogeneous nilpotent elements. Introduction
Let k be a field, and let 0 = q ∈ k . Quantum complete intersections originate from work of Manin[Man87]. Here we focus on the algebras A q = k h X, Y i / ( X a , XY − qY X, Y a ) . Such algebras have provided several examples giving answers to homological conjectures and questions.Perhaps most spectacular amongst these is Happel’s question. In [Hap89] Happel asked whether analgebra whose Hochschild cohomology is finite-dimensional, must have finite global dimension. The mainresult of [BGMS05] gave a negative answer: It shows that the Hochschild cohomology of the quantumcomplete intersection A q as above, when a = 2 and q not a root of unity, is finite-dimensional. Howeverthe algebra A q is selfinjective, hence has infinite global dimension. Already earlier, R. Schulz discoveredunusual properties for these algebras A q , see [Sch86] and [Sch94].Furthermore, there is a theory of support varieties in terms of Hochschild cohomology provided thealgebra satisfies suitable finite generation properties, known as condition (Fg) (see [EHSST04] and [Sol06]).For A q , this condition is satisfied precisely when q is a root of unity. The general theory of these supportvarieties has now been well established in several papers. However, in order to actually compute thevarieties over a given algebra, one needs to determine the ring structure of the Hochschild cohomology,or at least modulo homogeneous nilpotent elements.The results in this paper will be a contribution towards this goal. We determine the ring structureof the even part of HH ∗ ( A ) modulo the ideal of homogeneous nilpotent elements for A q when q is aprimitive a -th root of unitity. The proofs are quite technical, but this illustrates the typical difficultiesand computations one is faced with when trying to compute Hochschild cohomology.First we present an unpublished result by P. Bergh and K. Erdmann which determines the dimensionsof the Hochschild cohomology groups; this is done via exploiting Hochschild homology. Surprisingly, theanswer is independent of a (see Theorem 3.1 and Corollary 3.2). This suggests that perhaps also the ringstructure might not depend too much on the parameter a . We determine explicit bases of the even partHH ∗ ( A ) (see Section 5.2).Furthermore, we compute the algebra structure of HH ∗ ( A ) modulo the largest homogeneous nilpotentideal. We show that it is Z -graded, with degree zero part isomorphic to the polynomial ring in twovariables, generated in degree 2. The explicit description is given in 4.2 when a = 2, and in 5.4 when a ≥ Date : September 24, 2018.2010
Mathematics Subject Classification.
Primary 16E40; Secondary 16U80; 16S80; 81R50.
Key words and phrases.
Hochschild Cohomology; Quantum complete intersections.
An explicit description when a = 2 was also given in [BGMS05, Section 3.4]. We include this case (inSection 4), as it shows that it is part of the general pattern.2. Preliminaries
More generally, let A be any finite-dimensional algebra over a field k , and let A e = A ⊗ k A op de-note the enveloping algebra . We view bimodules over A as left modules over A e . In this setting, the Hochschild cohomology of A can be taken as HH n ( A ) = Ext nA e ( A, A ), the n -th cohomology of the complexHom A e ( P A , A ), i.e. Ext nA e ( A, A ) = ker d ∗ n +1 / im d ∗ n , (2.1)where d ∗ n = Hom A e ( d n , A ) and where d n are the maps in a minimal projective resolution: P : · · · → P d −→ P d −→ P µ −→ A → . (2.2)Then the Hochschild cohomology HH ∗ ( A ) = Ext ∗ A e ( A, A )(2.3)is a k -algebra which is graded-commutative. There are various equivalent ways to define the product;here we will work with the Yoneda product.We specialize now to the quantum complete intersections. Let a be an integer such that a ≥
2. Wealso let q ∈ k be a primitive a th root of unity, and A is the k -algebra defined by A = k h X, Y i / ( X a , XY − qY X, Y a ) . (2.4)We write x and y for the residue classes of X and Y , respectively.In [BE08], for arbitrary parameter q = 0, an explicit minimal projective bimodule resolution P as in(2.2) was constructed. The n th bimodule in P is P n = n M i =0 A e f ni , (2.5)the free A e -module of rank n + 1 having generators { f n , f n , ..., f nn } . For each s ≥ A e : τ ( s ) = q s (1 ⊗ x ) − ( x ⊗ τ ( s ) = (1 ⊗ y ) − q s ( y ⊗ γ ( s ) = a − X j =0 q js ( x a − − j ⊗ x j )(2.8) γ ( s ) = a − X j =0 q js ( y j ⊗ y a − − j )(2.9)The maps d n : P n → P n − in P are given by d t : f ti ( γ (cid:0) ai (cid:1) f t − i + γ (cid:0) at − ai (cid:1) f t − i − for i even − τ (cid:0) ai − a +22 (cid:1) f t − i + τ (cid:0) at − ai − a +22 (cid:1) f t − i − for i odd(2.10) d t +1 : f t +1 i ( τ (cid:0) ai (cid:1) f ti + γ (cid:0) at − ai +22 (cid:1) f ti − for i even − γ (cid:0) ai − a +22 (cid:1) f ti + τ (cid:0) at − ai + a (cid:1) f ti − for i odd(2.11)where the convention f n − = f nn +1 = 0 has been used. So far, q is arbitrary. Later in our setting we willsimplify these expressions.We will wish to identify nilpotent elements of Hochschild cohomology. This can be done by exploitingthe following result of N. Snashall and Ø. Solberg, see Proposition 4.4 in [SS04]. OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 3
Proposition 2.1.
Assume k is a field and A is a finite-dimensional k -algebra. Suppose η is a map into A representing an element of HH n ( A ) . If im( η ) is in the radical of A then η is nilpotent in HH ∗ ( A ) . Dimensions of Hochschild cohomology groups
We recall an unpublished result by Petter A. Bergh and Karin Erdmann which determines the dimen-sions.By viewing A as a left A e -module, it follows from [CE99, p. VI.5.3] that D (HH ∗ ( A, A )) is isomorphicto Tor A e ∗ ( D ( A ) , A ) as a vector space, where D denotes the usual k -dual i.e. D ( − ) := Hom k ( − , k ). Inparticular, we see that dim HH n ( A ) = dim Tor A e n ( D ( A ) , A ) for all n ≥
0. Moreover, it follows from[BE08] that A is a Frobenius algebra with Nakayama automorphism ν : A → A defined by ν : ( x q − a xy q a − y. (3.1)The bimodules D ( A ) and ν A are isomorphic; here the left action on ν A is taken as a · m := ν ( a ) m .Consequently the dimensions of the Hochschild cohomology of A are given bydim HH n ( A ) = dim Tor A e n ( ν A , A )(3.2)for all n ≥ A e n ( ν A , A ), we tensor the deleted projective bimodule resolution P A with the right A e -module ν A . We then obtain an isomorphism · · · ν A ⊗ A e P n +1 ν A ⊗ A e P n ν A ⊗ A e P n − · · ·· · · ⊕ n +1 i =0 ( ν A ) e n +1 i ⊕ n +1 i =0 ( ν A ) e ni ⊕ n +1 i =0 ( ν A ) e n − i · · · ⊗ d n +1 ⊗ d n δ n +1 δ n ∼ = ∼ = ∼ = of complexes, where { e n , e n , . . . , e nn } is the standard generating set of n + 1 copies of ν A . Now given anelement α ∈ k and a positive integer t , define an element K t ( α ) ∈ k by K t ( α ) := t − X j =0 α j . (3.3)The map δ n is then given by δ t : y u x v e ti ( qK a ( q v +1 ) y u + a − x v e t − i + K a ( q u +1 ) y u x v + a − e t − i − for i even[ q v +1 − q a − ] y u +1 x v e t − i + [ q u +2 − y u x v +1 e t − i − for i odd(3.4) δ t +1 : y u x v e t +1 i ( [ q a − − q v ] y u +1 x v e ti + K a ( q u +2 ) y u x v + a − e ti − for i even qK a ( q v +2 ) y u + a − x v e ti + [ q u +1 − y u x v +1 e ti − for i odd(3.5)where we use the convention e n − = e nn +1 = 0. This was proved in [BE08] in a more general setting, and byspecializing q and using that x, y have the same nilpotency index, we obtain precisely the above formulae.For the following result we use this complex to compute the Hochschild cohomology of our algebra A , inthe case when q is a primitive a th root of unity. The result shows that the dimensions of the cohomologygroups do not depend on the characteristic of the field. Theorem 3.1. If q is a primitive a th root of unity, then dim k HH n ( A ) = 2 n + 2 for all n ≥ . KARIN ERDMANN, MAGNUS HELLSTRØM-FINNSEN
Proof.
Since HH ( A ) is isomorphic to the centre of A , we see immediately that HH ( A ) is 2-dimensional.To find the dimension of HH n ( A ) for n >
0, we compute ker δ t for t ≥ δ t +1 for t ≥ δ t for t ≥
1. Since k contains a primitive q -th root of unity, the characteristicof k does not divide a . The equalities 0 = 1 − ( q m ) a = (1 − q m ) K a ( q m ), valid for any integer m , implythat K a ( q v + 1) = 0 if and only if 0 ≥ v ≥ a −
2, whereas K a ( q u + 1) = 0 if and only if 0 ≥ u ≥ a − δ t ( y u x v e ti ) ⇔ u ∈ { , . . . , a − } , v ∈ { , . . . , a − } , i even u = a − v ∈ { , . . . , a − } , i even u ∈ { , . . . , a − } , v = a − , i even u = 0 , v = a − , i = 2 tu = a − , v = 0 , t = 0 u = a − , v = a − , i odd u = a − , v = a − , i odd(3.6)and there are a t + a such elements. As for linear combinations in ker δ t of at least two basis vectorsfrom ⊕ ti =0 ( ν A ) e ti , they appear in pairs. Namely, they are linear combinations of the pairs x a − e ti + Cy a − x a − e ti +1 i = 0 , , . . . , t − y a − e ti + C ′ y a − x a − e ti − i = 2 , , . . . , t (3.8)where C and C ′ are scalars whose values depend on all the parameters involved. There are 2 t suchelements, hence dim k ker δ t = ( a + 2) t + a .Next we compute ker δ t +1 for t ≥
0, recall that the characteristic of k does not divide a . We see that δ t +1 ( y u x v e t +1 i ) = 0 ⇔ ( u = a − v ∈ { , . . . , a − } , i arbitrary u ∈ { , . . . , a − } , v = a − , i arbitrary(3.9)and there are (2 a − t + 2) such elements. As for linear combinations in ker δ t +1 of at least two basisvectors from ⊕ t +1 i =0 ( ν A ) e t +1 i , also here they appear in pairs. Namely, they are linear combinations ofthe pairs y u x v e t +1 i + Cy u +1 x v − e t +1 i +1 (3.10)for 0 ≤ u ≤ a −
3, 1 ≤ v ≤ a − i = 0 , , . . . , t , and where C is a scalar. There are ( a − ( t + 1)such pairs (note that, if a = 2, then the requirements on u and v are empty, but this causes no problemsince by the formula there are no pairs in this situation), and so dim k ker δ t +1 = ( a + 2) t + a + 2.We have now computed ker δ t for t ≥ δ t +1 for t ≥ . Using the equalitiesdim k im δ n + dim k ker δ n = dim k ⊕ ni =0 ( ν A ) e ni ) = ( n + 1) a , (3.11)we see that dim k im δ t +1 = dim k im δ t +2 = ( a − t + 1). Consequentlydim k HH t +1 ( A ) = dim k ker δ t +1 − dim k im δ t +2 (3.12) = 4 t + 4(3.13) dim k HH t +2 ( A ) = dim k ker δ t +2 − dim k im δ t +3 (3.14) = 4 t + 6(3.15)for t ≥
0, and the proof is complete. (cid:3)
This result implies immediately the following:
Corollary 3.2.
The dimension of the cohomology groups HH n ( A ) is independent of a . OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 5 Hochschild cohomology when a = 2In this section we let a = 2 and q = − k ) = 2), so we have that A = k h X, Y i / ( X , XY + Y X, Y ) . (4.1)We write x, y again for the images of X, Y in A . We will rewrite the differentials for the minimal projectiveresolution before studying the even cohomology ring HH ∗ ( A ) for this case.4.1. Minimal projective resolution when a = 2 . We introduce the following notation: β y = (1 ⊗ y ) + ( y ⊗ β x = (1 ⊗ x ) + ( x ⊗ α y = (1 ⊗ y ) − ( y ⊗ α x = (1 ⊗ x ) − ( x ⊗ P in Equation 2.10 and 2.11;we get: d n ( f ni ) = ( ( − i ( β y f n − i + β x f n − i − ) when n is even( − i ( α y f n − i − α x f n − i − ) when n is odd . (4.4)4.2. Description of cohomology groups.
In Section 3 we have seen that dim HH n ( A ) = 2 n + 2.Knowing this, we will determine a basis for HH n ( A ) for arbitrary even degrees n . We write δ ir as usualfor the Kronecker symbol. Lemma 4.1.
Let n = 2 t . For r = 0 , , . . . , t define maps ξ r , η r : P t → A as follows. ξ r ( f ti ) = δ ir · A , η r ( f ti ) = δ ir · xy. (4.5) (a) The classes of these maps form a basis for HH t ( A ) .(b) The classes of the η r give nilpotent elements in HH ∗ ( A ) .Proof. Part (b) will follow from Proposition 2.1. We prove now part (a). Note that these are 2 n + 2elements, so we only have to show that the maps are in the kernel of d ∗ t +1 , and that they are linearlyindependent modulo the image of d ∗ t .(1) We apply ξ r to d t +1 ( f t +1 i ), this gives ξ r [( − i ( α y f ti − α x f ti − )] = ( − i [ α y [ δ ir · A ] − α x [ δ i − ,r · A ]] = 0(4.6) (we view A as a left A e module, and α y · A = 0 = α x · A ). Similarly we apply η r to d t +1 ( f t +1 i )and get η r [( − i ( α y f ti − α x f ti − )] = ( − i [ α y [ δ ir · xy ] − α x [ δ i − ,r · xy ]] = 0(4.7) (since xy is in the socle of A we see that α y · xy = 0 and α x · xy = 0).(2) Let c r , d r ∈ K and ρ : P t − → A such that t X r =0 c r ξ r + d r η r = ρ ◦ d t ∈ im( d ∗ t ) . (4.8) We must show that c r = 0 = d r for all r . Write ρ ( f t − i ) = p i = a i + b i x + c i y + d i xy ∈ A . Thenwe have ρ ◦ d t ( f ti ) = ( − i [ β y p i + β x p i − ] = ( − i [2 a i y + 2 a i − x ](4.9) which are elements in A . On the other hand if we apply the map given by the sum to f ti thenwe get c i + d i xy (4.10) also elements in A . We assume these are equal, and it follows that all scalars are zero. (cid:3) KARIN ERDMANN, MAGNUS HELLSTRØM-FINNSEN
Products in even degrees of HH ∗ ( A ) . Recall that the even part HH ∗ ( A ) is a subring of theHochschild cohomology, and it is commutative. The aim of this section is to prove the following: Theorem 4.2.
Let k be a field with char ( k ) = 2 , and let A = k h X, Y i / ( X , XY + Y X, Y ) . R be thesubalgebra R := k h ξ ti : t ≥ , ≤ i ≤ t i of the even Hochschild cohomology ring of A . This is Z -graded,with R := k h ξ ti : i even i , R := k h ξ ti : i odd i . (4.11) Then R is isomorphic to the polynomial ring k [ z , z ] where we identify ξ with z and ξ with z .Moreover, the odd part is equal to R = R ξ and ξ · ξ = ξ . Corollary 4.3.
Let N be the largest homogeneous nilpotent ideal of HH ∗ ( A ) . Then HH ∗ ( A ) / N ∼ = R. (4.12)We fix a degree 2 t , and we will compute the product of a general element ξ of degree 2 t with an element χ of degree 2 m and we let 2 m vary. We take representatives ξ : P t → A and χ : P m → A which are k -linear combinations of the basis. Let ξ ( f ti ) = p i ∈ A with 0 ≤ i ≤ t (4.13) χ ( f mi ) = p i ∈ A with 0 ≤ i ≤ m. (4.14)By (4.5), the elements p i and ¯ p i are then in the centre of A , we will use this freely. Definition 4.4.
The Yoneda product ξ • χ is the residue class of χ ◦ h m where the family ( h s ) with h s : P t + s → P s is a lifting of ξ . That is, we have the following diagram: P t +2 m P t +2 m − · · · P t + s P t + s − · · · P t +1 P t P m P m − · · · P s P s − · · · P P AA d t +2 m d t + s d t +1 d m d s d µh m h m − h s h s − h h ξχ where ξ = µ ◦ h and where all squares commute. We define maps h s (0 ≤ s ≤ m ), and will show thatthey are a lifting. h s ( f t + si ) = (P sj =0 p i − j f sj when s even( − i (cid:16)P sj =0 ( − j p i − j f sj (cid:17) when s odd(4.15) Proposition 4.5.
The maps h s for ≤ s ≤ s make the lifting diagram commutative, i.e. d s ◦ h s = h s − ◦ d t + s .Proof. When 2 t is fixed the proof of this result is an examination when s is even and when s odd, andthe result follows from explicit calculations. OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 7
Case s even : We have( d s ◦ h s )( f t + si ) = d s ◦ s X j =0 p i − j f sj = s X j =0 p i − j d s ( f sj )(4.16) = s X j =0 p i − j ( − j (cid:0) β y f s − j + β x f s − j − (cid:1) ( h s − ◦ d t + s )( f ti ) = h s − ◦ (cid:0) ( − i (cid:0) β y f t + s − i + β x f t + s − i − (cid:1)(cid:1) (4.17) = β y h s − ( f t + s − i ) + β x h s − ( f t + s − i − )= s − X j =0 ( − j β y p i − j f s − j + ( − i − s X j =0 ( − j β x p i − ( j +1) f s − j = s X j =0 ( − j p i − j ( β y f s − j + β x f s − j − )We observe that the expressions are equal, hence d s ◦ h s = h s − ◦ d t + s and we conclude that h s is alifting map when s is even. Case s odd : We calculate( d s ◦ h s )( f t + si ) = d s ◦ ( − i s X j =0 ( − j p i − j f sj = s X j =0 ( − j p i − j d s f sj (4.18) = ( − i s X j =0 ( − j p i − j ( − j ( α y f s − j − α x f s − j − )= ( − i s X j =0 p i − j ( α y f s − j − α x f s − j − )( h s − ◦ d t + s )( f ti ) = h s − ◦ (cid:0) ( − i (cid:0) α y f t + s − i − α x f t + s − i − (cid:1)(cid:1) (4.19) = ( − i s X j =0 ( − j p i − j ( α y f s − j − α x f s − j − )The expressions are equal, which proves the odd case as well, and that h s is a lifting map after all. (cid:3) Description of Yoneda products.
In Section 4.2 we have described a basis for HH t +2 m ( A ). Nowwe compute the Yoneda product of ξ ∈ HH t ( A ) and χ ∈ HH m ( A ) as linear combination of κ r and η r for 0 ≤ r ≤ t + 2 m : χ • ξ = t +2 m X r =0 u r κ r + v r η r (4.20)where κ r ( f t +2 mi ) = δ ir · A η r ( f t + si ) = δ ir · xy (4.21)hence we get ( ξ • χ )( f t +2 mi ) = u i · A + v i · xy. (4.22)By letting u i and v i denote maps u i : f t +2 mi x and v i : f t + si y (4.23)we observe that this is isomorphic to the polynomial algebra with two generators. KARIN ERDMANN, MAGNUS HELLSTRØM-FINNSEN
Corollary 4.6.
Let ξ ( f tr ) = p r ∈ A and χ ( f mr ) = ¯ p r ∈ A . Then χ ◦ h m ( f t +2 mi ) = X ≤ j ≤ m and ≤ i − j ≤ t p i − j ¯ p j . (4.24) In particular if we let ξ mi and ξ tj denote the basis elements of Lemma 4.1 then we have ξ mi · ξ tj = ξ t +2 mi + j . (4.25) Proof.
We apply the lifting formula and obtain the first part directly. If we take χ = ξ mi and ξ = ξ tj then p i = 1 and p r = 0 for r = i and similarly ¯ p j = 1 and ¯ p r = 0 otherwise. So we get that the image of f t +2 mν is 1 if ν = i + j and is zero otherwise. The last part follows. (cid:3) Completing the proof of Theorem 4.2.
To see that R is as stated, we let the ξ t correspondto the powers of z , and we let the ξ t t correspond to the powers of z and in general ξ ti for i evencorresponds to the product z t − i z i . The formula in the Corollary may be used to see that this gives analgebra isomorphism. The rest follows from the previous results.The Corollary is a direct consequence: The intersection of R with N is zero, and as we have observed,any element in the span of maps η r is in N . Cohomology for a ≥ a ≥
3. Still let q be an a th root of unity and assume the algebrais A = k h X, Y i / ( X a , XY − qY X, Y a ) . (5.1)We write again x, y for the images of X, Y in A .5.1. Differentials.
We assume a ≥
3, then we can simplify the differentials defined in 2.10 and 2.11. Weobserve that the elements in A introduced in 2.6 to 2.9 depend only on the parity of s modulo a , and thearguments in 2.10 and 2.11 make only use of the cases where s ≡ s ≡ a . Using this thedifferentials take the following form which we will use from now: d t : f ti ( γ y (0) f t − i + γ x (0) f t − i − for i even − τ y (1) f t − i + τ x (1) f t − i − for i odd(5.2) d t +1 : f t +1 i ( τ y (0) f ti + γ x (1) f ti − for i even − γ y (1) f ti + τ x (0) f ti − for i odd(5.3)where we have replaced τ = τ x τ = τ y γ = γ x γ = γ x (5.4)5.2. A basis for HH t ( A ) for a ≥ . As observed the dimension of the degree 2 t part is always 4 t + 2which is independent of a . We therefore expect that there should be a basis when a ≥ a = 2. Definition 5.1.
Let ζ j : P t → A be the map ζ j ( f ti ) = ( i = j j be even, then define η + j ( f ti ) = ( x a − y a − i = j OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 9
Now let j be odd, then define η − j ( f ti ) = ( xy i = j Lemma 5.2.
We fix a degree t .(a) The classes of the elements ζ i and η ± j as defined above form a basis of HH t ( A ) .(b) The classes of the elements η ± j give nilpotent elements of HH ∗ ( A ) .Proof. Part (b) will follow again from Proposition 2.1. We prove now part (a). These are in total 4 t + 2maps, so we only have to show that the maps lie in the kernel of d ∗ t +1 , and that they are linearlyindependent modulo the image of d ∗ t .(1) Let ξ be one these maps. We write ξ ( f ti ) = p i ∈ A , so that p i is either 0 or 1 or one of x a − y a − or xy depending on the parity of i . We need to check that ξ ( d t +1 ( f ti )) = 0.(a) Assume i is even, then this is equal to ξ ( τ y (0) f ti + γ x (1) f ti − ) = τ y (0) p i + γ x (1) p i − . (5.8) This has to be calculated in A which is viewed as an A e left module. We have τ y (0) p i = p i y − yp i (5.9) This is zero if p i = 1. Otherwise since i is even we only need to consider p i = x a − y a − and then p i y = 0 and yp i = 0. Next, if p i − = 1 then γ x (1) p i − = a − X j =0 q j x a − − j · · x j = ( a − X j =0 q j ) x a − (5.10) and this is zero, note that 1 + q + . . . + q a − = 0 since q is an a -th root of 1. Otherwise p i − = xy and then γ x (1) p i − = a − X j =0 q j ( x a − − j xyx j )(5.11) and this is a scalar multiple of x a y and hence is zero.(b) Let i be odd, we get ξ ( − γ y (1) f ti + τ x (0) f ti − ) = − γ y (1) p i + τ x (0) p i − . (5.12) By calculations similar to part (a) we see that this is zero in all cases to be considered.(2) We consider a linear combination of the above elements and assume that it lies in the image of d ∗ t .Explicitly let t X j =0 c j ζ j + X j even s + j η + j + X j odd s − j η − j = ξ ◦ d t (5.13)where ξ : P t − → A , with c j and s ± j in k . We must show that this is only possible, as ξ varies, with all c j and s ± j equal to zero.(a) Apply the LHS to f ti with i even, this gives c i + s + i ( x a − y a − ) . (5.14) On the other hand, ξ ◦ d t ( f ti ) = γ y (0) ξ ( f t − i ) + γ x (0) ξ ( f t − i − ) . (5.15) This is an element in A viewed as an A e left module. For any element z ∈ A , γ x (0) z or γ y (0) z cannever have a constant term since γ x (0) and γ y (0) are in the radical of A e . Hence the Equation(5.15) does never have a non-zero constant term and c i = 0.We claim that we also cannot get a term which is a multiple of x a − y a − . Namely if so thiscould only come from either γ y (0) x a − or from γ x (0) y a − . Now, γ y (0) x a − = a − X j =0 y j x a − y a − − j = a − X j =0 ( q − ) j ( a − x a − y a − = 0(5.16) since P a − j =1 q j = 0. Hence s + i = 0.(b) Apply the LHS to f ti with i odd, this gives c i + s − i ( xy ) . (5.17) On the other hand, ξ ◦ d t ( f ti ) = − τ y (1) ξ ( f t − i ) + τ x (1) ξ ( f t − i − ) . (5.18) As before, since τ y (1) and τ x (1) are in the radical of A e , this cannot have non-zero constant terms.Hence c i = 0.We must check that we cannot get xy . If xy should occur in τ y (1) this can only come from τ y (1) x but this is equal to xy − qyx = 0. Similarly τ x (1) y = 0 and we do not get xy . Hence s − i = 0.We have proved that the 4 t + 2 maps are linearly independent modulo the image of d ∗ t . By dimensions,they are a basis of HH t ( A ). (cid:3) The aim of this section is to prove the following.
Theorem 5.3.
Let k be a field, a ≥ an integer, q ∈ k a primitive a th root of unity, and A the quantumcomplete intersection k h X, Y i / ( X a , XY − qY X, Y a ) . Assume R = k − Sp { ζ ti : t ≥ and ≤ i ≤ t } . Then R is a subalgebra of HH ∗ ( A ) . It is Z -graded with R = h ζ ti : i even i and R = h ζ ti : i odd i .Moreover ζ ml · ζ tr = (cid:26) l, r odd ζ m +2 tl + r otherwise (5.19)As for the case a = 2 we can see: Corollary 5.4.
The even part R of R is isomorphic to the polynomial ring in two variables. Corollary 5.5.
Assume A is as in the Theorem, and let N be the largest homogeneous nilpotent ideal of HH ∗ ( A ) . Then HH ∗ ( A ) / N is isomorphic to R . Lifting.
We compute the Yoneda product χ • ξ where χ, ξ are k -linear combinations of maps ζ j asin Definition 5.1.For ξ in the span of the ζ j , the values of ξ are scalars and therefore they commute with elements of A e . Luckily, we are only interested in the even Hochschild cohomology modulo homogeneous nilpotentelements.Similar as for the case where a = 2 we use liftings along the minimal projective resolution to definethe Yoneda products in the cohomology ring. Let ξ : P t → A where ξ ( f ti ) := p i (5.20) OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 11 and we assume p i is a scalar multiple of 1, for all i . We restricted to the subring of this kind. Consequentlythe values p i commute with all elements in A e . As usual we set p i = 0 if i > t or if i < h : P t → P is defined by h ( f ti ) := p i f (0 ≤ i ≤ t ) . (5.21)Moreover, we search explicit formulae for maps h s : P t + s → P s . (5.22)For s ≥ h s − ◦ d t + s = d s ◦ h s . (5.23)5.3.1. Some formulae in A e . In order to define these lifting maps h s for s > A e . Let c i = 1 + q + . . . + q i for 0 ≤ i ≤ a − . (5.24) Definition 5.6.
For an integer s we define β x ( s ) = a − X i =0 c i q si ( x a − − i ⊗ x i )(5.25) β y ( s ) = a − X i =0 c i q si ( y i ⊗ y a − − i )(5.26)Recall now the elements in A e which occur in the definition of the differentials: γ y ( s ) = a − X j =0 q js ( y j ⊗ y a − − j )(5.27) γ x ( s ) = a − X j =0 q js ( x a − − j ⊗ x j )(5.28)At the end we only need s = 0 and s = 1 τ y (1) = (1 ⊗ y ) − q ( y ⊗ τ x (1) = q (1 ⊗ x ) − ( x ⊗ τ y (0) = (1 ⊗ y ) − ( y ⊗ τ x (0) = (1 ⊗ x ) − ( x ⊗ . (5.32)With this notation, we will define maps h s : P t + s → P s , defined on the generators f t + si of the free A e module P t + s , and we will show below that they lift ξ : Definition 5.7.
Assume s is even. For an integer i we define the following elements in the algebra, ω + ( j ) = ( β x ( − β y (1) j odd1 j even ω − ( j ) = 1(5.33)We will show that a lifting formula is given by h s ( f t + si ) := (P sj =0 p i − j ω + ( j ) f sj i even P sj =0 p i − j ω − ( j ) f sj i odd.(5.34)Now assume s is odd. Here we need two parameters in A e , one for x and one for y . We set ε x ( j ) = ( − β x (1) j odd1 j even ε y ( j ) = ( j odd − β y (0) j even.(5.35) We will show that a lifting formula is given by h s ( f t + si ) := (P sj =0 p i − j ε x ( j ) f sj i even P sj =0 p i − j ε y ( j ) f sj i odd.(5.36) Lemma 5.8.
We have that the following relations hold: β y (1) τ y (1) = γ y (2)(a) β x ( − γ y (2) = γ y (0) β x (0)(b) β x ( − β y (1) = β y ( − β x (1)(c) β x (1) τ x (1) = − γ x (2)(d) β y ( − γ x (2) = γ x (0) β y (0)(e) β y (0) τ y (0) = γ y (1)(f) β x (0) τ x (0) = − γ x (1)(g) τ y (1) β y (0) = γ y (0)(h) τ x (0) β x ( −
1) = − γ x ( − γ x ( − β y (1) = β y (0) γ x (1)(j) τ y (0) β y ( −
1) = γ y ( − τ x (1) β x (0) = − γ x (0)(l) β x (0) γ y (1) = γ y ( − β x (1)(m) Proof.
We prove (a) and (b), and the other relations follows from the same kind of reasoning. Start with(a), we have β y (1) τ y (1) = a − X i =0 c i q i ( y i ⊗ y a − − i ) ! ((1 ⊗ y ) − q ( y ⊗ a − X i =0 (cid:0) c i q i ( y i ⊗ y a − − i ) − c i q i +1 ( y i +1 ⊗ y a − − i ) (cid:1) (5.38) = c (1 ⊗ y a − ) + c q ( y ⊗ y a − ) + · · · + c a − q a − ( y a − ⊗ y )(5.39) − c q ( y ⊗ y a − ) − · · · − c a − q a − ( y a − ⊗ y ) − c a − q a − ( y a − ⊗ c (1 ⊗ y a − ) + q ( c − c )( y ⊗ y a − ) + q ( c − c )( y ⊗ y a − )+(5.41) · · · + q a − ( c a − − c a − )( y a − ⊗ y ) − q a − c a − ( y a − ⊗ c = 1, c − c = 1 + q − qc i +1 − c i = (1 + q + · · · + q i +1 ) − (1 + q + · · · + q i ) = q i +1 (5.43)We also observe c a − = 1 + q + · · · + q a − = − q a − (5.44)since a is a root of unity and hence 1 + q + · · · + q a − + q a − = 0. Then we have, β y (1) τ y (1) = (1 ⊗ y a − ) + q ( y ⊗ y a − ) + · · · + q a − ( y a − ⊗
1) = γ y (2)(5.45)For the relation (b) we inspect a typical element in this sum: c i q − i ( x a − − i ⊗ x i ) q j ( y j ⊗ y a − − j ) = c i q − i q j ( x a − − i y j ⊗ x i ∗ y a − − j )(5.46) OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 13 (where ∗ denotes the multiplication in A op ). Now we recall that xy = qyx (and x ∗ y = q − y ∗ x ) hence x a − − i y j = q j ( a − − i ) y j x a − − i and x i ∗ y a − − j = q − i ( a − − j ) y a − − j ∗ x i . We get c i q − i q j q j ( a − − i ) q − i ( a − − j ) ( y j x a − − i ⊗ y a − − j ∗ x i ) = ( y j ⊗ y a − − j ) c j ( x a − − i ⊗ x i )(5.47)which is the most typical element in the sum γ y (0) β x (0). (cid:3) The relations (a) to (m) in Lemma 5.8 can be used to prove that the maps h s are liftings for the givenmap ξ : Proposition 5.9.
The lifting formulas make the suggested squares commutative, that is h s − ◦ d t + s = d s ◦ h s when s ≥ and ξ = µ ◦ h .Proof. We give details when s and i are even, the other cases are similar. The strategy is to apply bothsides to f t + si and express the answer in terms of the basis { f s − j } , with coefficients in A e and then showthat the coefficients of the f s − j in the two expressions are equal.We have ( d s ◦ h s )( f t + si ) = d s ◦ ( s X j =0 p i − j ω + ( j ) f sj )(5.48) = X j even, 0 ≤ j ≤ s p i − j ω + ( j )[ γ y (0) f s − j + γ x (0) f s − j − ](5.49) + X j odd, 0 ≤ j ≤ s p i − j ω + ( j )[ − τ y (1) f s − j + τ x (1) f s − j − ](5.50)We split each of the two sums, and when the index is j − l = j − j = l + 1 and noting that l has opposite parity as j . As well we set ω + ( j ) = 1 for j even. Then thisbecomes = X j even, 0 ≤ j ≤ s p i − j γ y (0) f s − j + X l odd , − ≤ l ≤ s − p i − l − γ x (0) f s − l (5.51) + X j odd , 0 ≤ j ≤ s − p i − j ω + ( j ) τ y (1) f s − j + X l even, − ≤ l ≤ s − p i − l − ω + ( l + 1) τ x (1) f s − l (5.52)The range of summation can be unified since f sj = 0 for j = − j = s . We write this now as acombination in the A e -basis f s − j for 0 ≤ j ≤ s −
1, (writing j for l ) and we get= X j even, 0 ≤ j ≤ s − [ p i − j γ y (0) + p i − j − ω + ( j + 1) τ x (1)] f s − j + X j odd, 0 ≤ j ≤ s − [ p i − j − γ x (0) − p i − j ω + ( j ) τ y (1)] f s − j (5.53)On the other hand( h s − ◦ d t + s )( f t + si ) = h s − ◦ ( γ y (0) f t + s − i + γ x (0) f t + s − i − )(5.54) = γ y (0)[ s − X j =0 p i − j ε x ( j ) f s − j ] + γ x (0)[ s − X j =0 p i − − j ε y ( j ) f s − j ](5.55) = s − X j =0 [ p i − j γ y (0) ε x ( j ) + p i − − j γ x (0) ε y ( j )] f s − j (5.56)We must show that for each j the coefficients of f s − j in (5.53) and in (5.56) are equal. (a) Assume first j is even. We require p i − j γ y (0) + p i − j − ω + ( j + 1) τ x (1) = p i − γ y (0) ε x ( j ) + p i − j − γ x (0) ε y ( j )(5.57) For j even, ε x ( j ) = 1 and the first terms agree. The second terms agree provided ω + ( j + 1) τ x (1) = γ x (0) ε y ( j )(5.58) Consider the LHS, by identities (c), (d) and (e) it is equal to β y ( − β x (1) τ x (1) = − β y ( − γ x (2) = − γ x (0) β y (0) = γ x (0) ε y ( j )(5.59) from the definition of ε y ( j ) in this case. Hence the second terms agree as well.(b) Now assume j is odd. We require p i − j − γ x (0) − p i − j ω + ( j ) τ y (1) = p i − j γ y (0) ε x ( j ) + p i − − j γ x (0) ε y ( j )(5.60) For j odd, ε y ( j ) = 1 and the terms with p i − j − agree. For the other two terms to agree we need γ y (0) ε x ( j ) = − ω + ( j ) τ y (1)(5.61) We have using the definition and identities (a) and (b) that − ω + ( j ) τ y (1) = − β x ( − β y (1) τ y (1) = − β x ( − γ y (2) = − γ y (0) β x (0) = γ y (0) ε x ( j )(5.62) as required. (cid:3) Similar as for the case a = 2 we define the Yoneda product of the residue classes represented by ξ and χ to be the residue class represented by the composition χ • ξ = χ ◦ h s . (5.63)5.4. Description of Yoneda products of basis elements when a ≥ . In the definition 5.7 of thelifting maps, we have the term ω + ( j ) = β x ( − β y (1) ∈ A e (for j odd). When this is evaluated in A , itbecomes ω + ( j ) · A . We claim that this is always zero, in fact β y (1) · A = 0.Namely, we must view A as an A e bimodule and then β y (1) · A = a − X i =0 c i q i y a − = ( a − X i =0 c i q i ) y a − (5.64)The following shows that this is zero: Lemma 5.10.
Let q be a primitive a -th root of unity for a ≥ . Let c i = 1 + q + . . . + q i for i ≥ , then a − X i =0 c i q i = 0(5.65) Proof.
Set also c − := 0. Then we have for i ≥ c i − c i − = q i . We get a − X i =0 c i q i = X i c i ( c i − c i − )(5.66) OCHSCHILD COHOMOLOGY OF SOME QUANTUM COMPLETE INTERSECTIONS 15
Therefore (all summations from i = 0 to a − q )( X i c i q i ) = X i c i q i + X i c i q i +1 = X i c i ( c i − c i − ) + X i c i ( c i +1 − c i )= X i ( c i c i +1 − c i c i − )= c a − c a − − c c − =0since c a − = 1 + q + . . . + q a − = 0 and c − = 0. But q = −
1, so we can cancel by (1 + q ) and get theclaim. (cid:3) We analyse now the products, and this will complete the proof of Theorem 5.3. Define R = Sp { ζ ti : t ≥ , ≤ i ≤ t } . (5.67)We compute products of elements in R .Let χ be of degree 2 m and ξ of degree 2 t , both in R . Let ξ ( f ti ) = p i ∈ K for 0 ≤ i ≤ t and χ ( f mj ) = ¯ p j ∈ K for 0 ≤ j ≤ m . As before we set p i = 0 for 0 < i of i > t , and similarly we define ¯ p j for any j ∈ Z .Then χ • ξ is the class of χ ◦ h m where ( h s ) is a lifting of ξ , where we use the formula computed above.Note that we only need the case when s = 2 m is even. We have χ ◦ h s ( f t + si ) = χ ( s X j =0 p i − j ω + ( j ) f sj ) = (cid:26) P sj =0 p i − j ω + ( j )¯ p j i even P sj =0 p i − j ¯ p j i odd(5.68)where we have already used that ω − ( j ) = 1.Now assume χ = ζ l for some 0 ≤ l ≤ s , so ¯ p l = 1 and p j = 0 otherwise. Then the above simplifies to f t + si (cid:26) p i − l ω + ( l ) · i even p i − l · i odd(5.69)Now take ξ = ζ r for some 0 ≤ r ≤ t . Then p i − l = 1 if i − l = r , and = 0 otherwise.Note that ω + ( l ) · A = 0 for l odd and = 1 otherwise. The zero occurs precisely when l is odd and i = l + r is even, i.e. if both l, r are odd. So we get ζ ml · ζ tr = (cid:26) ζ m +2 tl + r l, r not both odd0 l, r odd . (5.70)As for the case a = 2 we see that R is isomorphic to the polynomial ring in two variables.Furthermore, we see that elements in R are also nilpotent. The subalgebra R intersects the largesthomogeneous nilpotent ideal N trivially, and the span of the η ± is contained in N . Acknowledgements.
Both authors thank Petter A. Bergh for his joint notes with Karin Erdmann. Thesecond author (Magnus Hellstrøm-Finnsen) would like to thank the first author (Karin Erdmann) a lot forthe invitation to Oxford and is very grateful for spending these months on this wonderful place. Thanksfor valuable discussions on this project and on various topics in general, and for including me into themathematical community in Oxford. Thanks also to Petter for arranging and help with applications. TheResearch Council of Norway supports the PhD work of the second author through the research project
Triangulated categories in algebra (NFR 221893). The Research Council of Norway has also supportedthe second author with oversea grant in the occasion of the stay in Oxford.
References [BE08] P. A. Bergh and K. Erdmann,
Homology and cohomology of quantum complete intersections ,Algebra Number Theory vol. (2008), no. 5, 501–522.[BGMS05] R.-O. Buchweitz, E. L. Green, D. Madsen, and Ø. Solberg, Finite Hochschild cohomologywithout finite global dimension , Math. Res. Lett. vol. (2005), no. 5-6, 805–816.[CE99] H. Cartan and S. Eilenberg, Homological algebra , Princeton Landmarks in Mathematics,With an appendix by David A. Buchsbaum, Reprint of the 1956 original, Princeton Univer-sity Press, Princeton, NJ (1999), xvi+390.[EHSST04] K. Erdmann, M. Holloway, N. Snashall, Ø. Solberg, and R. Taillefer,
Support varieties forselfinjective algebras , K -Theory vol. (2004), no. 1, 67–87.[Hap89] D. Happel, Hochschild cohomology of finite-dimensional algebras , S´eminaire d’Alg`ebre PaulDubreil et Marie-Paul Malliavin, 39`eme Ann´ee (Paris, 1987/1988), vol. , Lecture Notesin Math. Springer, Berlin (1989), 108–126.[Man87] Y. I. Manin,
Some remarks on Koszul algebras and quantum groups , Ann. Inst. Fourier(Grenoble) vol. (1987), no. 4, 191–205.[Sch86] R. Schulz, Boundedness and periodicity of modules over QF rings , J. Algebra vol. (1986),no. 2, 450–469.[Sch94] R. Schulz,
A nonprojective module without self-extensions , Arch. Math. (Basel) vol. (1994), no. 6, 497–500.[Sol06] Ø. Solberg, Support varieties for modules and complexes , Trends in representation theory ofalgebras and related topics, vol. , Contemp. Math. Amer. Math. Soc., Providence, RI(2006), 239–270.[SS04] N. Snashall and Ø. Solberg,
Support varieties and Hochschild cohomology rings , Proc. LondonMath. Soc. (3) vol. (2004), no. 3, 705–732. Karin Erdmann, Mathematical Institute, University of Oxford, OX2 6GG Oxford, United Kingdom
E-mail address : [email protected] Magnus Hellstrøm-Finnsen, Institutt for matematiske fag, Norges teknisk-naturvitenskapelige universitet,N-7491 Trondheim, Norway
E-mail address ::