Homological invariants relating the super Jordan plane to the Virasoro algebra
aa r X i v : . [ m a t h . K T ] A ug Homological invariants relating the super Jordan plane tothe Virasoro algebra
Sebastián Reca and Andrea Solotar ∗ Abstract
Nichols algebras are an important tool for the classification of Hopf algebras. Within those withfinite GK dimension, we study homological invariants of the super Jordan plane, that is, the Nicholsalgebra A = B( V (− , )) . These invariants are Hochschild homology, the Hochschild cohomologyalgebra, the Lie structure of the first cohomology space - which is a Lie subalgebra of the Virasoroalgebra - and its representations H n ( A , A ) and also the Yoneda algebra. We prove that the algebra A is K . Moreover, we prove that the Yoneda algebra of the bosonization A kZ of A is also finitelygenerated, but not K . Keywords:
Nichols algebra, Hochschild cohomology, Virasoro algebra, Gerstenhaber bracket.
MSC : E , T , W , B . Introduction
Homological methods provide important information about the structure of associative algebras, re-vealing sometimes hidden connections amongst them. In this work, our main task is to compute theHochschild homology and cohomology of the super Jordan plane A , as well as the cup product incohomology, and the Lie structure of H ( A , A ) with the Gerstenhaber bracket, and describe highercohomology spaces as H ( A , A ) -Lie modules. We also describe the Yoneda algebra E ( A ) and theYoneda algebra of the bosonization A kZ , proving that both of them are finitely generated as alge-bras. Roughly speaking, besides the explicit information about cohomology, we obtain three mainresults: Theorem A.
The Gerstenhaber structure endows H ( A , A ) with a Lie algebra structure such that it is iso-morphic to the Cartan subalgebra plus the positive part of the Virasoro algebra. We identify intermediate seriesmodules within the representation H n ( A , A ) . Theorem B.
The Yoneda algebra E ( A ) is generated as an algebra over k by three elements: two in degree oneand one in degree two, subject to relations that we make explicit. As a consequence A is K . Theorem C.
The Yoneda algebra E ( A kZ ) is generated as an algebra over k by three elements: one in degreeone, one in degree two and one in degree three, subject to relations that we make explicit. As a consequence A kZ is not K . As a corollary of this last result, the cohomology algebra H • ( H , k ) , where H is a Hopf algebrawhose associated graded algebra with respect to the coradical fitration is A kZ – see [AAH ]– is alsofinitely generated. Even if this Hopf algebra is not finite dimensional, its Gelfand-Kirillov dimensionis finite, and our result is in line with [GK, MPSW, StVa], all of them related to a conjecture by Etingofand Ostrik [EO]. Notice that in our case the braiding group is infinite and the braiding is not diagonal.We now describe in more detail the objects involved, and the contents of this article. The super Jor-dan plane is the k -algebra generated by two elements x and y , subject to two relations, one monomialquadratic and one cubic: B( V (− , )) ∼ = k h x , y i / ( x , y x − xy − xyx ) . ∗ This work has been supported by the projects UBACYT
BA, PIP-CONICET
CO,MATHAMSUD-REPHOMOL and PICT - . The first named author is a CONICET fellow. The second named author isa research member of CONICET (Argentina) and a Senior Associate of ICTP Associate Scheme. s pointed in [AAH], the second relation is equivalent to the quantum Serre relation ( ad c y ) x − x ( ad c y ) x . Our main motivation to study this algebra comes from the fact that it is a Nichols algebra offinite Gelfand-Kirillov dimension appearing in the classification of pointed Hopf algebras, see [AAH].One of the main tools used is the notion of braided vector space. There are still many open problemswhen the braiding is not of diagonal type. This is the case of the super Jordan plane, where the vectorspace V (− , ) = k x ⊕ k y is a Yetter-Drinfeld module over kZ with braiding c as follows c ( x ⊗ x ) = − x ⊗ x , c ( x ⊗ y ) = x ⊗ x − y ⊗ x , c ( y ⊗ x ) = − x ⊗ y , c ( y ⊗ y ) = x ⊗ y − y ⊗ y .The Gelfand-Kirillov dimension of the super Jordan plane is , the set { x a ( yx ) b y c | a ∈ { , } , b , c ∈ N } is a PBW basis, and clearly it is not a domain, see [AAH]. This algebra is a test case for Nichols algebrascorresponding to non-diagonal braidings: here the described braiding corresponds to a × Jordanblock with eigenvalue − .On the other hand, the Virasoro algebra, that we will denote Vir, is a very important Lie algebraboth in mathematics and in physics, for example in string theory and conformal field theory and itsrepresentations have been extensively - but not completely - studied ([FF], [Z], [MZ]). It is the unique- up to isomorphisms - one dimensional central extension of the Witt algebra, and it is generated by afamily { L n } n ∈ Z together with a central charge C , subject to the relations [ L m , L n ] = ( n − m ) L m + n + δ m , − n m − m12 C , for all n , m ∈ Z .Let us write the triangular decomposition as Vir = Vir + ⊕ h ⊕ Vir − .We compute Hochschild homology and cohomology of A using the minimal resolution obtainedwith methods from [CS]. Our results are explicit enough to afford computations necessary to de-scribe H • ( A , A ) as a graded commutative algebra. The most interesting result is, in our opinion, thecharacterization of the Lie algebra H ( A , A ) - with the Gerstenhaber bracket - as a subalgebra of the Vi-rasoro algebra. Since we also compute the Gerstenhaber brackets h H ( A , A ) , H n ( A , A ) i for all n ≥ ,we obtain induced representations of the Virasoro algebra. We are also very explicit concerning thecomputations of Hochschild homology because it will be useful for a description of the cap product.Next, we compute the Yoneda algebra of A and as a consequence we prove that, even if A is not N -Koszul, it is K [CaSh]. Finally, we compute the Yoneda algebra of A kZ , and we conclude thefinite generation of the aforementioned Hopf algebra H .We do not know yet if there is an intrinsic relation between the Virasoro algebra and the superJordan plane, but we hope to made this clear in subsequent work.The contents of the article are as follows. In Section we construct the minimal projective resolutionof A as bimodule over itself, using an adequate reduction system. This resolution can be nicelyorganized as a total complex of a bicomplex, emphasizing the role played by the monomial relation,which is responsible for the infinity of the global dimension of A .In Section we compute the Hochschild cohomology of A . The very detailed and exhaustivedescription is useful for studying the cup product and the Gerstenhaber bracket. Our method includescomputing the spectral sequence associated to filtering the bicomplex by columns. In Theorem . we exhibit infinite bases for each space H i ( A , A ) , with i > 0 , and prove that the center of A is k .In Section , we use similar methods to describe the Hochschild homology of A , see Theorem . .Section is devoted to the description of the graded commutative algebra H • ( A , A ) , with the cupproduct. This description will also be important for the Gerstenhaber bracket, which is a gradedderivation with respect to the cup product in both variables. We proceed by constructing - sometimespartially - comparison maps between the bar resolution and ours. The results are summarized in Theo-rem . . We prove in particular that this algebra is not finitely generated, not even modulo nilpotents.Moreover, there is a an element u of H ( A , A ) such that the cup product with u is an isomorphismbetween H ( A , A ) and H ( p + ) ( A , A ) , for all p ≥ , and similarly for the odd cohomology spaces.The main part of the article is contained in Sections , , and . In Section we achieve thedescription of H ( A , A ) as Lie algebra - see Theorem . - while Section is dedicated to the detaileddescription of the Lie module structure of H n ( A , A ) , for all n ≥ , and to identify some irreducible ubmodules. The main purpose of Sections and is to describe the Yoneda algebras of A and A kZ ,respectively. In the former we prove that E ( A ) is finitely generated and K , while in the latter westudy E ( A kZ ) . In both cases we find generators and relations to present this algebra.In this article k will be an algebraically closed field of characteristic zero. This last hypothesis isnecessary for our computations, whereas the first one is not, but it comes from the context of Hopfalgebra classification.We would like to thank Mariano Suárez-Álvarez for discussions about this work, in particularabout the last section. We also thank Nicolás Andruskiewitsch for interesting comments about thissubject, and Iván Angiono and Leandro Vendramín for a careful reading of Sebastián Reca’s degreethesis. Projective resolution
From now on A will denote the super Jordan plane, that is, the algebra k h x , y i / (cid:0) x , y x − xy − xyx (cid:1) .It appears as the Nichols algebra B( V (− , )) , where ( V (− , ) , c ) is a braided k -vector space of di-mension not of diagonal type. Here the braiding group is Z and c : V ⊗ V → V ⊗ V is definedby: c ( x ⊗ x ) = − x ⊗ x , c ( x ⊗ y ) = x ⊗ x − y ⊗ x , c ( y ⊗ x ) = − x ⊗ y , c ( y ⊗ y ) = x ⊗ y − y ⊗ y .The super Jordan plane has been studied in [AAH], [AG]. It is graded with deg ( x ) = deg ( y ) = andit has a Poincaré-Birkhoff-Witt basis B = { x a ( yx ) b y c | a ∈ { , } , b , c ∈ N } . It is a noetherian algebraof Gelfand-Kirillov dimension , whose global dimension is infinite. There are lots of open questionsconcerning Nichols algebras appearing in the classification of Hopf algebras.Since our aim is to describe completely the Hochschild cohomology of A as an associative algebrawith the cup product and as a graded Lie algebra with the Gerstenhaber bracket, we will start byconstructing its minimal resolution as bimodule over itself.Our first lemma describes the commutation rules in A . The proof is inductive and, since no diffi-culty arises, we will not include it. It is written in detail in [R]. Lemma . . For all n ∈ N and b ∈ N , the following equalities hold, y x = n X i = n ! i ! x ( yx ) n − i y , ( . ) y + x = n X i = n ! i ! ( yx ) n − i + y , ( . ) y ( yx ) b = n X i = (cid:18) ni (cid:19) ( b + n − i − )!( b − )! ( yx ) b + n − i y , ( . ) y + ( yx ) b = n + X i = (cid:18) n + (cid:19) ( b + n − i )!( b − )! x ( yx ) b + n − i y . ( . )We will use the methods and terminology of [CS] to construct the minimal resolution. For this, wewill consider an order in the set of words in x and y such that y > x , compatible with concatenation,and the induced reduction system R = { ( x , ) , ( y x , xy + xyx ) } . It is clear that for all ( s , f ) ∈ R , f is irreducible. Although we already know that the ambiguities can be solved, since it is proved in[AAH] that B is a PBW basis of A , we will verify it anyway because the process will provide someof the differentials appearing in the resolution as well as the comparison maps that will be neededafterwards. The set of ambiguities is A = { x , y x } . Since x comes from a monomial relation, there s nothing to prove for it. For y x , the possible two paths are the following. ( y x ) x / / ( xy + xyx ) x xy x + xyx / / x ( xy + xyx ) $ $ ❏❏❏❏❏❏❏❏❏❏❏ y x ✇✇✇✇✇✇✇✇✇ ✇✇✇✇✇✇✇✇✇ ●●●●●●●●● ●●●●●●●●● ( x ) / / y ttttttttttttt ttttttttttttt The set of n -ambiguities for n ≥ is A n = { x n + , y x n } . Let A , A denote the sets { x , y } , { x , y x } ,respectively. The associated monomial algebra, taking into account the order in the set of words, is A mon = k h x , y i / (cid:0) x , y x (cid:1) . For a monomial algebra, the minimal resolution as bimodule over itself isBardzell’s resolution [B], that we will denote ( P mon • , δ • ) . Note that the successive modules appearingin this resolution are A mon ⊗ k A monn ⊗ A mon , but A monn = A n for all n , because the n -ambiguitiesdepend on the leading terms of the relations. Moreover, each δ n can be extended to A ⊗ k A n ⊗ A as a map of A -bimodules. Also, the order defined previously induces an order in the monomialsof A , which in turn defines an order in the monomials of A ⊗ k A n ⊗ A for each n . The bimodules A ⊗ k A n ⊗ A are free, hence projective. We consider the following sequence of A e -modules: P • A : · · · d / / A ⊗ k A ⊗ A d / / A ⊗ k A ⊗ A d / / / / A ⊗ k A ⊗ A d / / A ⊗ A / / ( . )where the morphisms between the A e -bimodules are given by d ( ⊗ x ⊗ ) = x ⊗ − ⊗ x , d ( ⊗ y ⊗ ) = y ⊗ − ⊗ y , d ( ⊗ x ⊗ ) = x ⊗ x ⊗ + ⊗ x ⊗ x , d ( ⊗ y x ⊗ ) = y ⊗ x ⊗ + y ⊗ y ⊗ x + ⊗ y ⊗ yx − (cid:16) xy ⊗ y ⊗ + x ⊗ y ⊗ y + ⊗ x ⊗ y (cid:17) − ( xy ⊗ x ⊗ + x ⊗ y ⊗ x + ⊗ x ⊗ yx ) , d n ( ⊗ x n + ⊗ ) = x ⊗ x n ⊗ + (− ) n + ⊗ x n ⊗ x , d n ( ⊗ y x n ⊗ ) = y ⊗ x n ⊗ + (− ) n + ⊗ y x n − ⊗ x − (cid:16) x ⊗ y x n − ⊗ + xy ⊗ x n ⊗ + ⊗ x n ⊗ y + ⊗ x n ⊗ yx (cid:17) for each n ≥ .According to [CS, Thm. . ], to prove that ( P • A , d • ) is a projective resolution of A as A e -bimodulewe only have to verify that • ( P • A , d • ) is a complex such that µd = , where µ : A ⊗ A → A is the multiplication of A . • For all q ∈ k A n , ( d n − δ n )( ⊗ q ⊗ ) is a sum of terms in A ⊗ k A n − ⊗ A that are smaller than ⊗ q ⊗ .Both items can be checked by straightforward computations. Note that the resolution is minimal sincethe image of d n is contained in rad ( A ⊗ k A n − ⊗ A ) , for all n ∈ N .A careful look at this resolution shows that it is the total complex of a double complex X • , • with A ⊗ A at ( , ) , two more columns and infinite rows. The rows correspond to the commutation relation nd the columns correspond to the relation x = , which is responsible of the global dimension beinginfinite. The double complex is A ⊗ k { x } ⊗ A δ (cid:15) (cid:15) (cid:15) (cid:15) ∂ ✤✤✤ A ⊗ k { y x } ⊗ A d o o δ ′ (cid:15) (cid:15) (cid:15) (cid:15) ∂ ′ ✤✤✤ A ⊗ k { x } ⊗ A ∂ (cid:15) (cid:15) A ⊗ k { y x } ⊗ A d o o ∂ ′ (cid:15) (cid:15) A ⊗ k { x } ⊗ A δ (cid:15) (cid:15) A ⊗ k { y x } ⊗ A d o o δ ′ (cid:15) (cid:15) A ⊗ A A ⊗ k { x , y } ⊗ A d o o A ⊗ k { y x } ⊗ A d o o ( . )where, d and d have already been defined, d ( ⊗ y x n ⊗ ) = y ⊗ x n ⊗ − xy ⊗ x n ⊗ − ⊗ x n ⊗ y − ⊗ x n ⊗ yx , n ≥ , ( . ) δ ( ⊗ x n ⊗ ) = x ⊗ x n − ⊗ + ⊗ x n − ⊗ x , ( . ) δ ′ ( ⊗ y x n ⊗ ) = −( x ⊗ y x n − ⊗ + ⊗ y x n − ⊗ x ) , n ≥ , ( . ) ∂ ( ⊗ x n ⊗ ) = x ⊗ x n − ⊗ − ⊗ x n − ⊗ x , ( . ) ∂ ′ ( ⊗ y x n ⊗ ) = −( x ⊗ y x n − ⊗ − ⊗ y x n − ⊗ x ) , n ≥ . ( . )Note that the rows are finite since the cubic relation generates no self ambiguity.The differentials in the first column correspond to the minimal projective resolution of k [ x ] / ( x ) ,while those of the second column differ in a factor y appearing on the left, that can be thought as"indexing" the column, and a factor − , so that we obtain a bicomplex. At this stage, we can alreadydeduce just looking at the resolution that both H • ( A , A ) and H • ( A , A ) will be periodic of period ,starting at H ( A , A ) and H ( A , A ) , respectively. Hochschild cohomology
Applying the functor Hom A e (− , A ) to the double complex ( X • , • ) we obtain the following bicomplexsuch that the homology of its total complex is H • ( A , A ) : A ∂ O O d / / A − ∂ O O A δ O O d / / A − δ O O A ∂ O O d / / A − ∂ O O A d / / A ⊕ A d / / ^ δ O O A − δ O O ith differentials d ( a ) = ([ x , a ] , [ y , a ]) , ^ δ ( a , b ) = xa + ax , d ( a , b ) = [ y , a ] + [ yb + by , x ] − ( xya + ayx ) − xbx , δ ( a ) = xa + ax , d ( a ) = [ y , a ] − ( xya + ayx ) , ∂ ( a ) = [ x , a ] .We will use a spectral sequence argument coming from the filtration by columns to compute thehomology of the total complex. We need to compute the values of δ and ∂ on elements of B . Given z = x a ( yx ) b y c ∈ B , δ ( z ) = x a ( yx ) b y c x + x a + ( yx ) b y c .There are four cases to consider. • a = , c = : δ ( z ) = ( yx ) b y x + x ( yx ) b y = ( yx ) b k X i = k ! i ! x ( yx ) k − i y ! + x ( yx ) b y = k X i = k ! i ! ( yx ) b x ( yx ) k − i y + x ( yx ) b y . – If b = , then δ ( z ) = P ki = ! i ! x ( yx ) k − i y + xy . – If b ≥ , then δ ( z ) = P ki = ! i ! ( yx ) b − ( yx ) x ( yx ) k − i y + x ( yx ) b y = x ( yx ) b y . • a = , c = + : δ ( z ) = ( yx ) b y + x + x ( yx ) b y + = ( yx ) b k X i = k ! i ! ( yx ) k − i + y ! + x ( yx ) b y + = k X i = k ! i ! ( yx ) b + k − i + y + x ( yx ) b y + . • a = , c = : δ ( z ) = x ( yx ) b y x + x ( yx ) b y c = x ( yx ) b y x = x ( yx ) b k X i = k ! i ! x ( yx ) k − i y ! = k X i = k ! i ! x ( yx ) b x ( yx ) k − i y = . • a = , c = + : δ ( x ) = x ( yx ) b y + x = x ( yx ) b k X i = k ! i ! ( yx ) k − i + y ! = k X i = k ! i ! x ( yx ) b + k − i + y .Summarizing, the image of δ is generated by the set (cid:14) k X i = k ! i ! x ( yx ) k − i y + xy , x ( yx ) b + y , k X i = k ! i ! ( yx ) b + k − i + y + x ( yx ) b y + , k X i = k ! i ! x ( yx ) b + k − i + y : b , k ≥ (cid:15) . emma . . The subset (cid:10) x ( yx ) b y , P ki = ! i ! ( yx ) b + k − i + y + x ( yx ) b y + : b , k ≥ (cid:11) is a basis of Im δ .Proof. Let us fix some notation for the generators of Im δ : η k = k X i = k ! i ! x ( yx ) k − i y + xy , θ b , k = x ( yx ) b + y , λ b , k = k X i = k ! i ! ( yx ) b + k − i + y + x ( yx ) b y + , µ b , k = k X i = k ! i ! x ( yx ) b + k − i + y .Since η k = k − X i = k ! i ! x ( yx ) k − i y + = k − X i = k ! i ! θ k − i − , i + and char k = , the element xy belongs to Im δ , for all k ≥ . Also note that µ b , k = k X i = k ! i ! θ b + k − i , i ,so (cid:8) θ b , k , xy , λ b , k : b , k ≥ (cid:9) generates Im δ . Moreover, it is a basis, since the terms appearing ineach generator are different elements of the basis B .Next we compute the values of ∂ on elements of B . Given z = x a ( yx ) b y c ∈ B , ∂ ( z ) = x a + ( yx ) b y c x − x a ( yx ) b y c x .Again, there are four cases to consider. • a = , c = : ∂ ( z ) = x ( yx ) b y − ( yx ) b y x = x ( yx ) b y − P ki = ! i ! ( yx ) b x ( yx ) k − i y . – If b = , ∂ ( z ) = xy − P ki = ! i ! x ( yx ) k − i y = − P k − = ! i ! x ( yx ) k − i y . – If b ≥ , then ∂ ( z ) = x ( yx ) b y . • a = , c = + : ∂ ( z ) = x ( yx ) b y + − ( yx ) b y + x = x ( yx ) b y + − k X i = k ! i ! ( yx ) b + k − i + y . • a = , c = : ∂ ( z ) = x ( yx ) b y − x ( yx ) b y x = . • a = , c = + , ∂ ( z ) = x ( yx ) b y − x ( yx ) b y x = − P ki = ! i ! x ( yx ) b + k − i + y .The following lemma can be proved similarly to the previous one. Lemma . . The set (cid:10) x ( yx ) b + y , P ki = ! i ! ( yx ) b + k − i + y − x ( yx ) b y + : b , k ≥ (cid:11) is a basis of Im ∂ . We are now able to compute the first page E • , • of the spectral sequence. The next propositions willprovide a description of each space E i , j1 . Proposition . . The set X = (cid:14) x ( yx ) b y + − P ki = ! i ! ( yx ) k + b + − i y , x ( yx ) b y | b , k ≥ (cid:15) is a basisof the k -vector space Ker δ .Proof. The elements of X are homogeneous. Given z ∈ Ker δ we may suppose without loss of generalitythat z is homogeneous and consider two cases. deg ( z ) = , z = n X l = α l ( yx ) n − l y + n − X l = β l x ( yx ) n − l − y + .Since z ∈ Ker δ , = δ ( z ) = n X l = α l δ (cid:16) ( yx ) n − l y (cid:17) + n − X l = β l δ (cid:16) x ( yx ) n − l − y + (cid:17) = n X l = (cid:18) α l + α n n ! l ! (cid:19) x ( yx ) n − l y + n − X l = X i = β l l ! i ! x ( yx ) n − i y = n X l = (cid:18) α l + α n n ! l ! (cid:19) x ( yx ) n − l y + n − X i = − X l = i β l l ! i ! x ( yx ) n − i y = n X l = α l + α n n ! l ! + n − X j = l β j j ! l ! x ( yx ) n − l y .All the monomials appearing in the sum are k -linearly independent, so α n = and α l = − n − X j = l β j j ! l ! , for all l , ≤ l < n .We can now write z as a linear combination of elements of X as follows z = − n − X l = n − X j = l β j j ! l ! ( yx ) n − l y + n − X l = β l x ( yx ) n − l − y + = − n − X j = β j j X l = j ! l ! ( yx ) n − l y + x ( yx ) n − j − y + . • deg ( z ) = + , z = n X l = α l ( yx ) n − l y + + n X l = β l x ( yx ) n − l y .Then = δ ( z ) = n X l = α l δ (cid:16) ( yx ) n − l y + (cid:17) + n X l = β l δ (cid:16) x ( yx ) n − l y (cid:17) = n X l = X i = α l l ! i ! ( yx ) n + − i y + n X l = α l x ( yx ) n − l y + .Hence α l = for all l , ≤ l ≤ n and we can write z = P nl = β l x ( yx ) n − l y .We have already proved that X generates Ker δ and it is easy to see that it is linearly independent.By definition there are isomorphisms E , + ∼ = Ker ∂ / Im δ ∼ = E , + for all i ≥ and E , ∼ = Ker δ / Im ∂ ∼ = E , for all i ≥ . Form now on, we will denote by [ z ] the class in E i , j1 of an element z ∈ E i , j0 and we will write ([ a ] , [ b ]) instead of [( a , b )] for ( a , b ) ∈ E , . Proposition . . The set (cid:8) (cid:2) xy (cid:3) | n ≥ (cid:9) is a basis of the vector spaces E , and E , for all i ≥ . roof. The proof follows from Proposition . and Lemma . . Proposition . . The set (cid:10)P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3) : n ≥ (cid:11) is a basis of the vector spaces E , + and E , + for all i ≥ .Proof. Again, given z ∈ Ker ∂ we may suppose that it is homogeneous, and consider two cases. • deg ( z ) = , z = n X l = α l ( yx ) n − l y + n X i = β i x ( yx ) n − i y − .By Lemma . , we know that x ( yx ) n − i y − + P i − = ( i − )! j ! ( yx ) n − j y ∈ Im δ for all i , ≤ i ≤ n .Hence, we may assume that z = P nl = α l ( yx ) n − l y . Since z ∈ Ker ∂ , = n X l = α l ∂ (( yx ) n − l y ) = n − X l = α l ∂ (( yx ) n − l y ) + α n ∂ ( y )= n − X l = α l x ( yx ) n − l y − α n n − X l = n ! l ! x ( yx ) n − l y = n − X l = ( α l − α n n ! l ! ) x ( yx ) n − l y .This implies that α l = α n n ! l ! for all l , ≤ l ≤ n and z = n X l = α l ( yx ) n − l y = α n n X l = n ! l ! ( yx ) n − l y . • deg ( z ) = + , z = n X l = α l ( yx ) n − l y + + n X l = β l x ( yx ) n − l y .By Lemma . , x ( yx ) n − l y ∈ Im δ for all l , ≤ l ≤ n . Therefore, we may assume that z = P nl = α l ( yx ) n − l y + . Since z ∈ Ker ∂ , = n X l = α l ∂ (( yx ) n − l y + ) = n X l = α l − l X i = l ! i ! ( yx ) n − i + y + x ( yx ) n − l y + ! = − n X l = X i = α l l ! i ! ( yx ) n − i + y + n X l = α l x ( yx ) n − l y + ,from which we obtain that α l = for all l , ≤ l ≤ n .So the set in the statement of the lemma generates Ker ∂ / Im δ and it is clear that is linearly indepen-dent.The description of the first page of the spectral sequence ends by observing that E , = A , while , = Ker δ ⊕ A and E , = Ker δ . We summarize this information in the following diagram of E • , • D P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3)E d ( ) / / ✤✤✤ D P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3)E ✤✤✤ (cid:10)(cid:2) xy (cid:3)(cid:11) ✤✤✤ d ( ) / / (cid:10)(cid:2) xy (cid:3)(cid:11) ✤✤✤ D P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3)E d ( ) / / ✤✤✤ D P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3)E ✤✤✤ A d ( ) / / Ker ( δ ) ⊕ A d ( ) / / ✤✤✤ Ker ( δ ) ✤✤✤ where d ( ) , d ( ) and d ( ) are the maps induced by d , d and d respectively.Next we compute the second page of the spectral sequence, which will be the last one since, dueto the shape of our complex, E • , • = E • , • ∞ . For this, we describe in a series of lemmas the images of themaps d ( ) , d ( ) and d ( ) . Lemma . . The map d ( ) : E , + → E , + is null for all i ≥ .Proof. Let [ w ] be an element of the basis of E , + . By Proposition . , we may choose w = n X l = n ! l ! ( yx ) n − l y . • If n = , then d ( w ) = d ( ) = (cid:2) y , (cid:3) − xy − yx = − xy − yx . Thus d ( w ) belongs to Im δ and d ( ) ([ w ]) = . • If n ≥ , we may assume that n = m + , where m ≥ and write w = m + X l = ( m + )! l ! ( yx ) m + − l y = ( m + ) y + x + y + .Applying d , we obtain that d ( w ) = ( m + ) d (cid:0) y + x (cid:1) + d (cid:0) y + (cid:1) . Next we compute eachterm separately d (cid:16) y + x (cid:17) = h y , y + x i − xy + x − y + xyx = y + h y , x i − y + xyx = y + (cid:16) y x − xy − xyx (cid:17) = , d (cid:16) y + (cid:17) = h y , y + i − xy + − y + x = xy + − m + X l = ( m + )! l ! ( yx ) m + − l y .Since xy + − P m + = ( m + )! l ! ( yx ) m + − l y belongs to the image of δ , d ( ) ([ w ]) = . emma . . The map d ( ) : E , → E , is null for all i ≥ .Proof. Let (cid:2) xy (cid:3) be an element of the basis of E , . Since d ( xy ) = h y , xy i − xyxy − xy + x = y xy − xy + − xyxy − n X l = n ! l ! x ( yx ) n − l + y = xy + + xyxy − xy + − xyxy − n X l = n ! l ! x ( yx ) n − l + y = − n X l = n ! l ! x ( yx ) n − l + y ,and x ( yx ) b y c belongs to Im ∂ for all b ≥ and for all c ≥ , we get that d ( ) ( (cid:2) xy (cid:3) ) = for all n ≥ .We compute now the values of d on elements of the basis of E , = A . By definition, d ( z ) =([ x , z ] , [ y , z ]) for all z ∈ A . Since [ x , z ] = ∂ ( z ) , we only need to know the values of [ y , z ] for all z = x a ( yx ) b y c ∈ B . There are three cases to consider: • a = : [ y , z ] = ( yx ) b + y c − x ( yx ) b y c + . • a = , b = : [ y , z ] = . • a = , b ≥ : [ y , z ] = x ( yx ) b − y c + + bx ( yx ) b y c − ( yx ) b y c + .Putting together these computations and the ones we made to obtain the image of ∂ , we get that if z = x a ( yx ) b y c ∈ B , then • a = , c = : – if b = , then d ( ) ([ z ]) = (cid:16) − P k − = ! l ! (cid:2) x ( yx ) k − l y (cid:3) , (cid:17) , – if b ≥ , then d ( ) ([ z ]) = (cid:16)h x ( yx ) b y i , h x ( yx ) b − y + i + b h x ( yx ) b y i − h ( yx ) b y + i(cid:17) . • a = , c = + : – if b = , then d ( ) ([ z ]) = (cid:16) − P kl = ! l ! (cid:2) ( yx ) k + − l y (cid:3) + (cid:2) xy + (cid:3) , (cid:17) , – if b ≥ , then d ( ) ([ z ]) = − k X l = k ! l ! h ( yx ) k + + b − l y i + h x ( yx ) b y + i , h x ( yx ) b − y + i + b h x ( yx ) b y + i − h ( yx ) b y + i ! . • a = , c = , d ( ) ([ z ]) = (cid:0) , (cid:2) ( yx ) b + y (cid:3) − (cid:2) x ( yx ) b y + (cid:3)(cid:1) , • a = , c = + : d ( ) ([ z ]) = − k X l = k ! l ! h x ( yx ) k + + b − l y i , h ( yx ) b + y + i − h x ( yx ) b y + i ! . emma . . The set Y = (cid:8) ρ k , ζ b , k , σ k , τ b , k , ν b , k , ξ b , k | b , k ≥ (cid:9) , where ρ k = k − X l = k ! l ! h x ( yx ) k − l y i , ! , ζ b , k = (cid:16)h x ( yx ) b + y i , h x ( yx ) b y + i + ( b + ) h x ( yx ) b + y i − h ( yx ) b + y + i(cid:17) , σ k = k X l = k ! l ! h ( yx ) k + − l i − h xy + i , ! , τ b , k = − k X l = k ! l ! h ( yx ) k + + b − l y i + h x ( yx ) b + y + i , ( b + ) h x ( yx ) b + y + i ! , ν b , k = (cid:16) , h ( yx ) b + y i − h x ( yx ) b y + i(cid:17) , ξ b , k = − k − X l = k ! l ! h x ( yx ) k + + b − l y i , ( b + ) h x ( yx ) b + y i ! . is a basis of Im d ( ) .Proof. First, notice that d (cid:16) ( yx ) b + y + + x ( yx ) b y ( k + ) (cid:17) = − k X l = k ! l ! ( yx ) k + + b − l y + x ( yx ) b + y + , ( b + ) x ( yx ) b + y + ! and d (cid:16) x ( yx ) b y + + ( yx ) b + y (cid:17) = − k − X l = k ! l ! x ( yx ) k + + b − l y , ( b + ) x ( yx ) b + y ! .This implies that Y generates Im d ( ) . Let us prove that it is linearly independent. For this, supposethat w is a null linear combination of elements of Y . We may suppose, without loss of generality, that w is homogeneous. If deg ( w ) = + , we have = w = αρ k + k − X i = β i ζ k − i − , i + k − X i = γ i ξ k − i − , i = αρ k + k − X i = γ i ξ k − i − , i + k − X i = β i h x ( yx ) k − i y i , h x ( yx ) k − i − y + i + ( k − i ) h x ( yx ) k − i y i − h ( yx ) k − i y + i ! .Since terms of type ( yx ) k − i y + do not appear in ξ ∗ , ∗ , neither in ρ ∗ , we conclude that β i = for all i , ≤ i ≤ k − . Consequently, = αρ k + k − X i = γ i ξ k − i − , i = αρ k + k − X i = γ i − i − X l = i ! l ! h x ( yx ) k − l y i , ( k − i ) h x ( yx ) k − i y i ! .The fact that the second coordinate of ρ k is null implies that γ i = for all i , ≤ i ≤ k − , and thus α = . The case deg ( z ) = is analogous.Now, we want to compute a basis of Im d ( ) . In fact we know that Im d ( ) = Im d ( ) , where d ( ) : Ker δ ⊕ A Im d ( ) → Ker δ is the induced map. Given z ∈ Ker δ ⊕ A , we will thus reduce it modulo m d ( ) . Using the notations and the proof of the previous lemma, we already know that the elements ζ b , k , σ k and τ b , k are in the image of d ( ) for all b , k ≥ , which allows us to suppose that z = (cid:0) xy , (cid:1) or z = (cid:0) , x a ( yx ) b y c (cid:1) . The first possibility is easier, since d ( z ) = h y , xy i − x ( yx ) y − xy + x = − k X i = k ! i ! x ( yx ) k + − i y .For z = (cid:0) , x a ( yx ) b y c (cid:1) , we study several cases. Suppose first that a = , then d ( z ) = h y ( yx ) b y c + ( yx ) b y c + , x i − x ( yx ) b y c x . • If b = , then d ( z ) = (cid:2) y c + , x (cid:3) − xy c x = (cid:0) y c + x − xy c + (cid:1) − xy c x . – If c = , then d ( z ) = (cid:0) y + x − xy + (cid:1) = (cid:16) P ki = ! i ! ( yx ) k + − i y − xy + (cid:17) – If c = + , then d ( z ) = (cid:16) y + x − xy + (cid:17) − xy + x = k X i = ( k + )! i ! ( yx ) k + − i y − k X i = k ! i ! x ( yx ) k + − i y = ( + ) k X i = k ! i ! x ( yx ) k + − i y . • If b ≥ , then d ( z ) = h x ( yx ) b − y c + + bx ( yx ) b y c + ( yx ) b y c + , x i − x ( yx ) b y c = x ( yx ) b − y c + x + bx ( yx ) b y c x + ( yx ) b y c + x − x ( yx ) b y c + − x ( yx ) b y c x = x ( yx ) b − y c + x + ( b − ) x ( yx ) b y c x + ( yx ) b y c + x − x ( yx ) b y c + . – If c = , then d ( z ) = P ki = ! i ! ( yx ) k + b + − i y − x ( yx ) b y + . – If c = + , then d ( z ) = P ki = ! i ! ( k + b ) x ( yx ) k + b + − i y .For a = , d ( z ) = h ( yx ) b + y c x + x ( yx ) b y c + , x i = ( yx ) b + y c x − x ( yx ) b + y c + x ( yx ) b y c + x . • If c = , then d ( z ) = P k − = ! i ! x ( yx ) k + b + − i y . • If c = + , then d ( z ) = P ki = ! i ! ( yx ) k + b + − i y − x ( yx ) b + y + .Since P ki = ! i ! (cid:2) x ( yx ) k + b + − i y (cid:3) and P k − = (cid:2) x ( yx ) k + b + − i y (cid:3) belong to Im d ( ) , the same is truefor their difference. In other words, (cid:2) x ( yx ) b + y (cid:3) belongs to Im d ( ) for all b , i ≥ . The proof of thefollowing lemma is now clear. Lemma . . The set (cid:14) k X i = k ! i ! ( yx ) k + b + − i y − x ( yx ) b y + , x ( yx ) b + y | b , k ≥ (cid:15) is a basis of Im d ( ) . e are now ready to describe the second page of the spectral sequence. We will denote by z theclass in E i , j2 of an element z ∈ E i , j0 and again we will write ( a , b ) instead of ( a , b ) for ( a , b ) ∈ E , . Proposition . . (i) The set (cid:10)P nl = ! l ! ( yx ) n − l y | n ≥ (cid:11) is a basis of the vector spaces E , + and E , + , for all i ≥ .(ii) The set (cid:10) xy | n ≥ (cid:11) is a basis of the vector spaces E , , E , and E , , for all i ≥ .Proof. It is a direct consequence of Lemmas . , . and . .Let us now describe E , . We claim that the set (cid:10) ( , x ) , (cid:16) ( + ) xy , y + (cid:17) | k ≥ (cid:11) is a basis of E , . In order to prove the claim we may suppose, as always, that each coordinate of [ z ] = ([ z ] , [ z ]) ∈ Ker d ( ) is homogeneous, both of the same degree and we can also reduce modulo boundaries to write [ z ] as a linear combination of elements in the set (cid:10) (cid:16)h xy i , (cid:17) , (cid:16) , h x a ( yx ) b y c i(cid:17) | n ≥ , a ∈ { , } , b , c ≥ (cid:11) .In case deg ( z ) = , we write z = (cid:16) , P ki = α i ( yx ) k − i y + P k − = β i x ( yx ) k − i − y + (cid:17) and lookingat the term ( yx ) y in the equality d ( z ) = , we obtain that α k = , and now the equality is: = k − X l = k − X i = l i ! l ! ( α i + β i ) ! ( yx ) k + − l y − k − X i = ( α i + β i ) x ( yx ) k − i y + .We deduce that α i = − β i for all i , ≤ i ≤ k − , so that [ z ] = k − X i = α i (cid:16) , h ( yx ) k − i y i − h x ( yx ) k − i − y + i(cid:17) ,but each term is zero because (cid:2) ( yx ) k − i y (cid:3) − (cid:2) x ( yx ) k − i − y + (cid:3) belongs to Im d ( ) for all i, ≤ i ≤ k − .In case deg ( z ) = + , we write z = k X i = α i (cid:16) , ( yx ) k − i y + (cid:17) + k X i = β i (cid:16) , x ( yx ) k − i y (cid:17) + γ (cid:16) xy , (cid:17) Thus, d ( z ) = k − X i = X l = α i i ! l ! kx ( yx ) k + − l y + k X i = − X l = β i i ! l ! x ( yx ) k + − l y + ( α k ( + ) − γ ) k X i = k ! i ! x ( yx ) k + − i y which is zero since [ z ] ∈ Ker d ( ) . Looking at the term x ( yx ) y we conclude that γ = α k ( + ) .Consequently, = k − X i = X l = α i i ! l ! kx ( yx ) k + − l y + k X i = − X l = β i i ! l ! x ( yx ) k + − l y = k − X l = k − X i = l α i i ! l ! k ! x ( yx ) k + − l y + k − X l = k X i = l + β i i ! l ! ! x ( yx ) k + − l y . rom this, we get P k − = l α i i ! l ! k = − P ki = l + β i i ! l ! = − P k − = l β i + ( i + )! l ! for all l , ≤ l ≤ k − . Let usdenote c l = P k − = l α i i ! k + β i + ( i + ) . Of course c l = , and = c l − c l + = α l l ! k + β l + ( l + )! , so β l + = − α l kl + for all l , ≤ l ≤ k − . Moreover, since = c k − = α k − k ! + β k k ! , we conclude that β l + = − α l kl + for all l , ≤ l ≤ k − . We are now able to write z = k − X i = α i (cid:16) , ( yx ) k − i y + (cid:17) − k X i = α i − ki (cid:16) , x ( yx ) k − i y (cid:17) + α k (cid:16) ( + ) xy , y + (cid:17) + β (cid:16) , x ( yx ) k (cid:17) = k − X i = α i (cid:18) , ( yx ) k − i y + − ki + ( yx ) k − − i y + (cid:19) + α k (cid:16) ( + ) xy , y + (cid:17) + β (cid:16) , x ( yx ) k (cid:17) ,proving that E , is generated by the set (cid:14) (cid:18) , ( yx ) b + y + − b + k + + ( yx ) b y + (cid:19) , (cid:16) , x ( yx ) k (cid:17) , (cid:16) ( + ) xy , y + (cid:17) | k , b ≥ (cid:15) .Using the notation of Lemma . , there are equalities (cid:16) , h x ( yx ) k + i(cid:17) = + , k , (cid:16) , h ( yx ) y + i − h xy + i(cid:17) = − ζ , k + ξ , k + + k + , (cid:18) , h ( yx ) b + y + i − b + k + + h x ( yx ) b y + i(cid:19) = − ζ b , k + ξ b , k − + b − , k + for all b ≥ , k ≥ . Therefore, the set (cid:10) ( , x ) , (cid:16) ( + ) xy , y + (cid:17) | k ≥ (cid:11) generates E , . Toprove that it is linearly independent, pick a linear homogeneous combination of these generators andsuppose that it equals zero. Suppose first that w is a linear combination of elements of degree n ,so that for n = , = w = α ( , x ) + β ( x , y ) for some α , β ∈ k . No monomial of degree appearsas second coordinate of the generators of Im d ( ) , so α = β = . For n = + , we havethat w = (cid:16) ( + ) xy , y + (cid:17) . As before, (cid:2) y + (cid:3) does not appear as a second coordinate of anyelement in Im d ( ) , so that the coefficient must be zero.The only space of the second page of the spectral sequence left to describe is E , . We do it in thenext proposition. Proposition . . The vector space E , is isomorphic to k .Proof. By definition, E , = Ker d ( ) . Given [ z ] ∈ Ker d ( ) , we shall consider two cases, accordingly tothe parity of deg ( z ) . First, suppose that deg ( z ) = and write z = n X l = α l ( yx ) n − l y + n − X l = β l x ( yx ) n − l − y + ∈ Ker d ( ) , hat is, it commutes with y and x . Using the commutation formulas, we obtain: = [ y , z ] = n X l = α l h y , ( yx ) n − l y i + n − X l = β l h y , x ( yx ) n − l − y + i = n − X l = α l (cid:16) x ( yx ) n − l − y + + ( n − l ) x ( yx ) n − l y − ( yx ) n − l y + (cid:17) + n − X l = β l (cid:16) ( yx ) n − l y + − x ( yx ) n − l − y + (cid:17) = n − X l = ( β l − α l ) ( yx ) n − l y + + n − X l =− α l + ( n − l − ) x ( yx ) n − l − y + + n − X l = ( α l − β l ) x ( yx ) n − l − y + .Notice that terms appearing in the first sum will never cancel with terms appearing in the last twosums, so α l = β l for all l , ≤ l ≤ n − , and the equality n − X l =− α l + ( n − l − ) x ( yx ) n − l − y + = .implies that α l = for all l , ≤ l ≤ n − . We deduce that z = α n xy . The equation [ x , z ] = implies that α n = or n = , in which case z belongs to k .Suppose now that deg ( z ) = + , so that z = P nl = α n ( yx ) n − l y + + P nl = β l x ( yx ) n − l y . Theequality [ y , z ] = leads to the following system of equations β = , ( . ) β l − α l − = , for all l , ≤ l ≤ n , ( . ) α n − β = , ( . ) α l − + α l ( n − l ) − β l = , for all l , ≤ l ≤ n − . ( . )From ( . ) and ( . ), we have α l = , for all l , ≤ l ≤ n − , while α = is a consequence of ( . ) and( . ). Finally, ( . ) and ( . ) imply that β l = , for all l , ≤ l ≤ n , so that z = α n y + . The equation [ x , z ] = has just one solution of this type, that is z = .The particular shape of our double complex, provides isomorphismsH ( A , A ) ∼ = E , and H ( A , A ) ∼ = E , .Moreover, for each p ≥ , there is a short exact sequence / / E , p −
12 ι / / H p + ( A , A ) π / / E , p2 / / ✤ / / ( , a ); ( a , b ) ✤ / / b .Extending the basis of ι (cid:16) E , p − (cid:17) by a set of linear independent elements such that their images by π form a basis of E , p2 , we get a basis of H p + ( A , A ) . In this way we have proved the following theorem. Theorem . . There are isomorphisms H ( A , A ) ∼ = k ,H ( A , A ) ∼ = h c , s n : n ≥ i ,H ( A ) ∼ = D t , u : n ≥ E , for all p > 0 ,H + ( A ) ∼ = D v + , w + : n ≥ E , for all p > 0 , here c = ( , x ) , s n = (cid:16) ( + ) xy , y + (cid:17) ; t = (cid:16) , xy (cid:17) , u = n X i = n ! i ! ( yx ) n − i y , − y + ! ; v + = , n X i = n ! i ! ( yx ) n − i y ! , w + = (cid:16) xy , xy + (cid:17) . Remark . . ( ) The element s ∈ H ( A , A ) is the class of the eulerian derivation associated to the gradingsuch that deg ( x ) = = deg ( y ) .( ) Given p , q ∈ N , H ( A , A ) is isomorphic to H ( A , A ) . The isomorphism is induced by an isomorphismbetween the A e -modules P A and P A , namely, P p A / / P q A1 ⊗ x ⊗ ✤ / / ⊗ x ⊗ , ⊗ y x − ⊗ ✤ / / ⊗ y x − ⊗ . The situation for the odd cohomology spaces is similar.
In Section we will describe the multiplicative structure of H • ( A , A ) = ⊕ n ≥ H n ( A , A ) and wewill prove that, as it happens for k [ x ] / (cid:0) x (cid:1) , the above isomorphisms are given by the cup productwith a basis element of H ( A , A ) . Hochschild homology
Now we compute the Hochschild homology of the super Jordan plane. Applying the functor A ⊗ A e − to the resolution . and using the canonical isomorphisms of vector spaces A ⊗ A e ( A ⊗ W ⊗ A ) ∼ = A ⊗ W , we get the following double complex ∂ (cid:15) (cid:15) − ∂ (cid:15) (cid:15) A δ (cid:15) (cid:15) A d o o − δ (cid:15) (cid:15) A ∂ (cid:15) (cid:15) A d o o − ∂ (cid:15) (cid:15) A δ (cid:15) (cid:15) A d o o − δ (cid:15) (cid:15) A A ⊕ A d o o A d o o with differentials d ( a , b ) = [ a , x ] + [ b , y ] , δ ( a ) = ( ax + xa , ) d ( a ) = (cid:16) [ a , y ] − ( axy + yxa ) , [ x , a ] y + y [ x , a ] − xax (cid:17) , δ ( a ) = ax + xa , d ( a ) = [ a , y ] − ( axy + yxa ) , ∂ ( a ) = [ a , x ] .As we have done for cohomology, we shall use the spectral sequence obtained by filtering by columnsin order to compute the homology. We omit the proof of the next proposition, since it is similar towhat we have already proved in the cohomological case. Proposition . . The following sets are bases of Im δ , Im δ and Im ∂ , respectively: (cid:10)P ki = ! i ! ( yx ) b + k − i + y + x ( yx ) b y + : b , k ≥ (cid:11) , • (cid:10) (cid:0) x ( yx ) b y , (cid:1) , (cid:16) P ki = ! i ! ( yx ) b + k − i + y + x ( yx ) b y + , (cid:17) : b , k ≥ (cid:11) , • (cid:10) x ( yx ) b + y , P ki = ! i ! ( yx ) b + k − i + y − x ( yx ) b y + : b , k ≥ (cid:11) . We are now ready to describe E , . We use again brackets to denote the class in E of an elementin E . Proposition . . The vector space E , = A ⊕ A Im δ has a basis (cid:10) h (( yx ) b y c , ) i : b , c ≥ (cid:11) ∪ (cid:10) h ( , x a ( yx ) b y c ) i : a ∈ { , } y b , c ≥ (cid:11) . Proof.
The description of Im δ given in Proposition . tells us that h ( x ( yx ) b y , ) i = and h (− x ( yx ) b y + , ) i = k X i = k ! i ! h ( yx ) b + k − i + y , ) i ,so the given set generates E , , and it is easy to see that it is linearly independent.From now on, we will identify A ⊕ A Im δ with A Im δ ⊕ A and we will write ([ a ] , [ b ]) instead of [( a , b )] for ( a , b ) ∈ E , . Proposition . . The set (cid:8) (cid:2) ( yx ) b y c (cid:3) : b , c ≥ (cid:9) is a basis of E , = A Im ( δ ) .Proof. It is similar to the proof of the previous proposition.We will next describe completely E • , n for n > 0 . There are, as before, two different cases, namely n odd and n even. We will deal with each of them in the following two propositions. Proposition . . The set (cid:8) (cid:2) xy (cid:3) : n ≥ (cid:9) is a basis of E , + ∼ = E , + , for all i ≥ .Proof. Fix i ≥ . The isomorphism E , + ∼ = E , + is a consequence of the following facts: • the kernel and the image of δ : E , + → E , are isomorphic, respectively, to the kernel andthe image of − δ : E , + → E , , • the same happens for ∂ and − ∂ , • Ker δ ∼ = Ker δ .Let us now describe Ker δ . Given z ∈ Ker δ , we suppose without loss of generality that z is homoge-neous. Two cases arise, according to the parity of deg ( z ) . st case: deg ( z ) = . We have z = P nl = α l ( yx ) n − l y + P ni = β i x ( yx ) n − i y − . We get rid of thesecond sum since, by Proposition . , we know that x ( yx ) n − i y − − P i − = ( i − )! j ! ( yx ) n − j y belongsto Im ∂ for i such that ≤ i ≤ n . Now, = δ ( z ) = δ n X l = α l ( yx ) n − l y ! = n X l = α l δ (( yx ) n − l y )= n − X l = α l δ (( yx ) n − l y ) + α n δ ( y )= n − X l = α l x ( yx ) n − l y + α n n X l = n ! l ! x ( yx ) n − l y + α n xy = n − X l = ( α l + α n n ! l ! ) x ( yx ) n − l y + n xy . ooking at the monomial xy , we obtain that α n = and so P n − = α l x ( yx ) n − l y = . But themonomials appearing in the sum are linearly independent, so α l = , for all l , that is z = . nd case: deg ( z ) = + . We have z = P nl = α l ( yx ) n − l y + + P nl = β l x ( yx ) n − l y . Proposition . allows us to suppose that in fact, z = P nl = α l ( yx ) n − l y + + βxy . Now, = δ ( z ) = n X l = α l δ (( yx ) n − l y + ) + βδ ( xy )= n X l = α l l X i = l ! i ! ( yx ) n + − i y + x ( yx ) n − l y + ! = n X l = X i = α l l ! i ! ( yx ) n + − i y + n X l = α l x ( yx ) n − l y + .From the second sum we obtain that α l = for ≤ l ≤ n and z = βxy . The only thing left to proveis that the set { (cid:2) xy (cid:3) : n ≥ } is linearly independent. Suppose that [ w ] ∈ Ker δ Im ∂ is a linear combinationof these elements and that [ w ] = , in other words w ∈ Im ∂ . We may suppose that w is homogeneous,that is w = λxy for some λ ∈ k , n ≥ . The description of Im ∂ implies λ = . Proposition . . For all i ≥ , the vector spaces E , and E , are isomorphic with basis (cid:14) n X l = n ! l ! h ( yx ) n − l y i : n ≥ (cid:15) . Proof.
The proof of the isomorphism E , ∼ = E , is similar to the previous one. As before, given z ∈ Ker ∂ , we may assume that it is homogeneous and split the proof in two cases. st case: deg ( z ) = . Reducing modulo boundaries we may write z = P nl = α l ( yx ) n − l y , thecondition z ∈ Ker ∂ gives = n X l = α l ∂ (( yx ) n − l y ) = n − X l = α l ∂ (( yx ) n − l y ) + α n ∂ ( y )= n − X l = α l (− x ( yx ) n − l y ) + α n n − X l = n ! l ! x ( yx ) n − l y = n − X l = (− α l + α n n ! l ! ) x ( yx ) n − l y ,and we conclude that α l = α n n ! l ! for all l , ≤ l ≤ n so that z = n X l = α l ( yx ) n − l y = α n n X l = n ! l ! ( yx ) n − l y . nd case: deg ( z ) = + . The description of Im δ in Proposition . allows us to suppose that z = P nl = α l ( yx ) n − l y + , and since z ∈ Ker ∂ , we have = n X l = α l ∂ (( yx ) n − l y + ) = n X l = α l l X i = l ! i ! ( yx ) n − i + y − x ( yx ) n − l y + ! = n X l = X i = α l l ! i ! ( yx ) n − i + y − n X l = α l x ( yx ) n − l y + .The first and the last sum should be zero separately, so α l = for all l , ≤ l ≤ n . The set (cid:14) n X l = n ! l ! h ( yx ) n − l y i : n ≥ (cid:15) enerates E , ∼ = E , . Suppose now that [ w ] is a null linear combination of these elements. Thedegree of w is even. If w ∈ Im δ , then w should be a linear combination of elements of type P ki = ! i ! ( yx ) b + k − i + y + x ( yx ) b y + , but this is impossible.Of course E , = A , so the first page of our spectral sequence may be pictured as follows, where d ( ) , d ( ) and d are the morphisms induced by d , d and d . (cid:10)(cid:2) xy (cid:3)(cid:11) ✤✤✤ (cid:10)(cid:2) xy (cid:3)(cid:11) d ( ) o o ✤✤✤ D P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3)E ✤✤✤ D P nl = ! l ! (cid:2) ( yx ) n − l y (cid:3)E d ( ) o o ✤✤✤ (cid:10)(cid:2) xy (cid:3)(cid:11) ✤✤✤ (cid:10)(cid:2) xy (cid:3)(cid:11) d ( ) o o ✤✤✤ A (cid:10)(cid:0)(cid:2) ( yx ) b y c (cid:3) , (cid:1)(cid:11) L A d ( ) o o ✤✤✤ (cid:10)(cid:2) ( yx ) b y c (cid:3)(cid:11) d ( ) o o ✤✤✤ Again, the second page will be the last one. We will next compute it, starting by describing theimages of d ( ) , d ( ) and d . Knowing the images, we will provide bases of E • , n for all n ≥ . Finally,we will describe E , for i = , , . We will write a for the class in E • , • of an element [ a ] ∈ E • , • . Proposition . . For all n ≥ , the map d ( ) : E , n → E , n is zero.Proof. We shall consider separately the cases n odd and n even. st case: n odd. Given a basis element (cid:2) xy (cid:3) of E , + , d (cid:16) xy (cid:17) = h xy , y i − xy xy = h xy , y i − x n X l = n ! l ! x ( yx ) n − l y + = h xy , y i = h x , y i y + x h y , y i = h x , y i y = (cid:16) xy − y x (cid:17) y = (− xyx ) y .This last element belongs to Im ∂ and so d ( ) (cid:0)(cid:2) xy (cid:3)(cid:1) = . nd case: n even. Fix w = P nl = ! l ! ( yx ) n − l y such that [ w ] belongs to the basis of E , . If n = ,then d ( w ) = d ( ) = (cid:2) , y (cid:3) − ( xy + yx ) = − xy − yx and d ( w ) ∈ Im δ . Suppose now that n > 0 and write n = m + with m ≥ . Using the commutation rules we write w = ( m + ) y + x + y + . So, d ( w ) = d (( m + ) y + x + y + ) = ( m + ) d ( y + x ) + d ( y + ) . Computingeach term separately we get d ( y + x ) = − P mj = ( m − j + ) m ! j ! ( yx ) m − j + y and d ( y + ) = P m + = ( m + )! l ! x ( yx ) m − l + y + − yxy + . Therefore, d ( ) ([ w ]) = − m X j = ( m − j + ) ( m + )! j ! h ( yx ) m − j + y i − m + X l = ( m + )! l ! h x ( yx ) m − l + y + i − h yxy + i = − m + X j = ( m − j + ) ( m + )! j ! h ( yx ) m − j + y i − m + X l = ( m + )! l ! h x ( yx ) m − l + y + i .Since P ki = ! i ! ( yx ) b + k − i + y + x ( yx ) b y + ∈ Im δ for all b , k ≥ , we choose k = l and b = m − l + and we obtain that − (cid:2) x ( yx ) m − l + y + (cid:3) equals P lj = ! j ! (cid:2) ( yx ) m − j + y (cid:3) , and using thisequality we conclude that d ( ) ([ w ]) = .Next we apply d ( ) to the elements in the basis of E , . For this, let us denote by f and g thecompositions of d with the first and second canonical projections, respectively, so that d ( a ) = (cid:16)h a , y i − ( axy + yxa ) , [ x , a ] y + y [ x , a ] − xax (cid:17) = ( f ( a ) , g ( a )) , for all a ∈ A .We will also denote by f and g the induced maps in homology. Given b , c ≥ and z = ( yx ) b y c , f ( z ) = h ( yx ) b y c , y i − (cid:16) ( yx ) b y c xy + yx ( yx ) b y c (cid:17) = h ( yx ) b , y i y c − ( yx ) b ( y c x ) y − ( yx ) b + y c = ( yx ) b y c + − y ( yx ) b y c − ( yx ) b ( y c x ) y − ( yx ) b + y c = ( yx ) b y c + − (( yx ) b y + b ( yx ) b + ) y c − ( yx ) b ( y c x ) y − ( yx ) b + y c = −( b + )( yx ) b + y c − ( yx ) b ( y c x ) y .There are two cases: c even and c odd. If c = , then f ([ z ]) = P k − = ( k − l + ) k ! l ! (cid:2) ( yx ) k − l + y (cid:3) if b = , −( b + ) (cid:2) ( yx ) b + y (cid:3) if b ≥ .If c = + , then f ([ z ]) = −( b + ) (cid:2) ( yx ) b + y + (cid:3) − P k − = ! i ! (cid:2) ( yx ) k + b − i + y + (cid:3) .For b , c and z as before, we compute now the image of g , obtaining that g ( z ) = − P k − = ! i ! x ( yx ) k − i y + − P k − = ! i ! ( yx ) k + − i y if c = , b = , x ( yx ) b y + + ( yx ) b + y if c = , b ≥ , − P k − = ! i ! ( yx ) k + − i y + − P ki = ( k + ) k ! i ! x ( yx ) k + − i y if c = + , b ≥ , − P k − = ! i ! ( yx ) b + k + − i y + − P ki = ( k + b + ) k ! i ! x ( yx ) b + k + − i y if c = + , b ≥ . e introduce some notation θ , k = −( b + ) h ( yx ) b + y i , θ , k = h x ( yx ) b y + i + h ( yx ) b + y i , λ , k = ( b + ) h ( yx ) b y + i + k − X i = k ! i ! h ( yx ) k + b − i y + i λ , k = k − X i = k ! i ! h ( yx ) b + k − i y + i + k X i = ( k + b + ) k ! i ! h x ( yx ) b + k − i y i . Proposition . . The set (cid:10) (cid:16) θ , k , θ , k (cid:17) , (cid:16) λ , k , λ , k (cid:17) | k ≥ , b ≥ (cid:11) is a basis of Im d ( ) .Proof. Given k ≥ , consider the elements η = k − X l = ( k − l + ) k ! l ! h ( yx ) k − l + y i , η = − k − X i = k ! i ! h x ( yx ) k − i y + i − k − X i = k ! i ! h ( yx ) k + − i y i .We already know that (cid:10) (cid:16) θ , k , θ , k (cid:17) , (cid:16) λ , k , λ , k (cid:17) , (cid:0) η , η (cid:1) | k ≥ , b ≥ (cid:11) generates Im d ( ) . Noticethat η = = η and that (cid:0) η , η (cid:1) = P k − = (cid:16) θ − l , l , θ − l , l (cid:17) for k ≥ . Moreover, it is easy to see thatthe set (cid:10) (cid:16) θ , k , θ , k (cid:17) , (cid:16) λ , k , λ , k (cid:17) | k ≥ , b ≥ (cid:11) is linearly independent and we obtain the result.Our next step will be to compute a basis of Im d ( ) . We recall that E , is generated by the set (cid:8) (cid:0)(cid:2) ( yx ) b y c (cid:3) , (cid:1) ; (cid:0) , (cid:2) x a ( yx ) b y c (cid:3)(cid:1) : a ∈ { , } , b , c ≥ (cid:9) . For z = (( yx ) b y c , ) ∈ A ⊕ A , d ( z ) = ∂ ( z ) = P k − = ! i ! x ( yx ) k − i y if c = and b = , − x ( yx ) b y if c = and b ≥ , P ki = ! i ! ( yx ) k + b + − i y − x ( yx ) b y + if c = + .while for z = (cid:0) , x a ( yx ) b y c (cid:1) , we know that d ( z ) = d (( , x a ( yx ) b y c )) = x a ( yx ) b y c + − yx a ( yx ) b y c ,consequently d ( z ) = x ( yx ) b y c + − ( yx ) b + y c if a = , if a = = b , ( yx ) b y c + − x ( yx ) b − y c + − bx ( yx ) b y c if a = and b ≥ .In this way we have obtained a set of generators of Im d ( ) , from which we extract a basis. Proposition . . The set (cid:8) (cid:2) x ( yx ) b + y c (cid:3) , (cid:2) ( yx ) b + y c (cid:3) , (cid:2) xy c + − ( yx ) y c (cid:3) | b , c ≥ (cid:9) is a basis of Im d ( ) . roof. Let us write η k = k − X i = k ! i ! x ( yx ) k − i y , θ b , k = x ( yx ) b + y , λ b , k = k X i = k ! i ! ( yx ) k + b + − i y − x ( yx ) b y + , µ b , c = x ( yx ) b y c + − ( yx ) b + y c , ν b , c = ( yx ) b + y c + − x ( yx ) b y c + − ( b + ) x ( yx ) b + y c .Notice that µ b , c + + ν b , c = −( b + ) x ( yx ) b + y c for all b , c ≥ , implying that x ( yx ) b + y c ∈ Im d ( ) and that x ( yx ) b + y c + − µ b + , c , that equals ( yx ) b + y c , also belongs to Im d ( ) .We are now ready to describe the second page of the spectral sequence. The proof of the followingproposition is a direct consequence of Propositions . and . . Proposition . . (i) For all i ≥ , the set (cid:10) xy : n ≥ (cid:11) is a basis of E , + ∼ = E , + .(ii) For all i ≥ , the set (cid:10)P nl = ! l ! ( yx ) n − l y : n ≥ (cid:11) is a basis of E , ∼ = E , .(iii) The set (cid:8) xy n , y n : n ≥ (cid:9) is a basis of E , . We will now complete the description of E • , • by exhibiting bases of E , and E , . Proposition . . The set (cid:10)P nl = ! l ! ( yx ) n − l y : n ≥ (cid:11) is a basis of E , .Proof. Given [ z ] ∈ Ker d ( ) , we may suppose that z is homogeneous of degree r and then treat sepa-rately the cases r even and r odd. Computations similar to those in the proof of Proposition . provethat the set we proposed generates E , and they are clearly linearly independent, taking into accounttheir degrees. Proposition . . The set (cid:14) n X l = n ! l ! (cid:16) ( yx ) n − l y , (cid:17) , (cid:16) −( yx ) , xy + + ( yx ) y (cid:17) , (cid:0) , y n (cid:1) , y + , n X l = n ! l ! n + + − l x ( yx ) n − l y + n X l = n ! l ! − l ( yx ) n − l y + ! , n ≥ (cid:15) . is a basis of E , .Proof. Given ([ z ] , [ w ]) ∈ Ker d ( ) , we can still suppose w and z homogeneous of the same degree.Again, we treat separately the even and odd cases. • Suppose deg ( z ) = deg ( w ) = , ([ z ] , [ w ]) = n X l = α l h ( yx ) n − l y i , n − X l = β l h x ( yx ) n − l − y + i + n X l = γ l h ( yx ) n − l y i! o = d ( ) ([ z ] , [ w ]) = α n n − X l = n ! l ! h x ( yx ) n − l y i − n − X l = α l h x ( yx ) n − l y i + n − X l = β l (cid:16)h x ( yx ) n − l − y + i − h ( yx ) n − l y + i(cid:17) + n − X l = γ l (cid:16)h ( yx ) n − l y + i − h x ( yx ) n − l − y + i − ( n − l ) h x ( yx ) n − l y i(cid:17) = n − X l = ( γ l − β l ) h ( yx ) n − l y + i + n − X l = ( β l − γ l ) h x ( yx ) n − l − y + i + n − X l = (cid:18) α n n ! l ! − α l − ( n − l ) γ l (cid:19) h x ( yx ) n − l y i from which we know that β l = γ l and α l = α n n ! l ! − ( n − l ) γ l for all l , ≤ l ≤ n − . We cannow rewrite ([ z ] , [ w ]) as follows, using that (cid:16) −( n − l ) h ( yx ) n − l y i , h x ( yx ) n − l − y + i + h ( yx ) n − l y i(cid:17) ∈ Im d ( ) for all l , ≤ l ≤ n − , and therefore ([ z ] , [ w ]) = α n n X l = n ! l ! (cid:16)h ( yx ) n − l y i , (cid:17) + γ n (cid:16) , h y i(cid:17) + γ n − (cid:16) − h ( yx ) y − i , h xy − i + h yxy − i(cid:17) . • Suppose deg ( z ) = deg ( w ) = + , ([ z ] , [ w ]) = n X l = α l h ( yx ) n − l y + i , n X l = β l h x ( yx ) n − l y i + n X l = γ l h ( yx ) n − l y + i! ,reducing modulo boundaries = d ( ) ([ z ] , [ w ]) = d ( ) α n h y + i , n X l = β l h x ( yx ) n − l y i + n X l = γ l h ( yx ) n − l y + i! = α n X l = n ! l ! h ( yx ) n + − l y i − α h xy + i + n X l = β l (cid:16)h x ( yx ) n − l y + i − h ( yx ) n − l + y i(cid:17) + n − X l = γ l (cid:16)h ( yx ) n − l y + i − h x ( yx ) n − l − y + i − ( n − l ) h x ( yx ) n − l y + i(cid:17) = ( αn ! − β ) h ( yx ) n + i + n X l = (cid:18) n ! l ! α − β l + γ l − (cid:19) h ( yx ) n + − l y i + ( β − γ n ) [ x ( yx ) n y ] + n − X l = ( β l − γ l − − γ l ( n − l )) h x ( yx ) n − l y + i + ( β n − γ n − − α ) h xy + i . he linear independence of the monomials appearing in the expression gives: β = n ! α , β l = n ! l ! α + γ l − for all l , ≤ l ≤ n , β l = γ l − + γ l ( n − l ) for all l , ≤ n ≤ n − , γ =
1n β , γ n − = β n − α .We deduce that γ l = α n ! l ! − l for all l , ≤ l ≤ n − and that β l = α n ! l ! n + − l + for all l , ≤ l ≤ n .As a consequence, ([ z ] , [ w ]) = α h y + i , n X l = n ! l ! n + − l + h x ( yx ) n − l y i + n X l = n ! l ! − l h ( yx ) n − l y + i ! + γ n (cid:16) , h y + i(cid:17) .So our set generates E , and it is clearly linearly independent.In the homological case we also have a short exact sequence / / E , p ι / / H p + ( A , A ) π / / E , p − / / .from which we can deduce the proof of the next theorem. Notice that all the homology spaces areinfinite dimensional and that starting from degree , they are periodic of period . Theorem . . There are isomorphisms: H ( A , A ) ∼ = (cid:10) xy n , y n | n ≥ (cid:11) ,H ( A , A ) ∼ = * y + , n X i = n ! i ! (cid:18) n + − ( i − ) x ( yx ) n − i y + − i ( yx ) n − i y + (cid:19)! | n ≥ + ⊕ * n X i = n ! i ! ( yx ) n − i y , ! , (cid:0) , y n (cid:1) , (cid:16) − yxy , xy + + yxy (cid:17) | n ≥ + ,H ( A , A ) ∼ = *(cid:16) xy , (cid:17) , , n X i = n ! i ! ( yx ) n − i y ! | n ≥ + ,H + ( A , A ) ∼ = * n X i = n ! i ! ( yx ) n − i y , ! , (cid:16) −( yx ) y , xy (cid:17) | n ≥ + ,H + ( A , A ) ∼ = *(cid:16) xy , (cid:17) , n X i = n ! i ! ( yx ) n − i y + , n X i = n ! i ! ( yx ) n − i y ! | n ≥ + . The ring structure of Hochschild cohomology
We aim to describe the structure of the Hochschild cohomology of the super Jordan plane as associativegraded algebra. By the general theory we already know that it is graded commutative, and sinceH ( A , A ) ∼ = k and H ( A , A ) is infinite dimensional we also know that this algebra cannot be finitelygenerated. he description obtained in this section will be also useful to compute in Section the Gerstenhaberstructure of H • ( A , A ) , using the fact that the Gerstenhaber bracket is a graded derivation with respectto the cup product. Of course, since we have obtained H • ( A , A ) using the minimal resolution of A , we need to do some work before actually computing the cup product. We shall thus constructcomparison maps between the minimal resolution P • A and the bar resolution B • A , that is, morphismsof complexes f • : P • A → B • A and g • : B • A → P • A such that they lift the identity P A ⇆ B A . Asusual, the map f • is easier to describe since P • A is "smaller". Proposition . . Let f • = ( f n ) n ≥ , f n : P n A → B n A be the following sequence of morphisms of A e -modules: • f : A ⊗ A → A ⊗ A , f = id A ⊗ A , • f : A ⊗ k { x , y } ⊗ A → A ⊗ A ⊗ A , f ( ⊗ v ⊗ ) = ⊗ v ⊗ , for v ∈ k { x , y } , • for n ≥ , f n : A ⊗ k (cid:8) x n , y x n − (cid:9) ⊗ A → A ⊗ A ⊗ n ⊗ A , f n ( ⊗ x n ⊗ ) = ⊗ x ⊗ n ⊗ , f n ( ⊗ y x n − ⊗ ) = y ⊗ y ⊗ x ⊗ n − ⊗ + ⊗ y ⊗ yx ⊗ x ⊗ n − ⊗ − x ⊗ y ⊗ y ⊗ x ⊗ n − ⊗ − ⊗ x ⊗ y ⊗ x ⊗ n − ⊗ − x ⊗ y ⊗ x ⊗ n − ⊗ − ⊗ x ⊗ yx ⊗ x ⊗ n − ⊗ + n − X i = (− ) i (cid:16) ⊗ x ⊗ + i ⊗ y ⊗ x ⊗ n − − i ⊗ + ⊗ x ⊗ + i ⊗ yx ⊗ x ⊗ n − − i ⊗ (cid:17) , where we interpret the last sum as when n = .The map f • is a morphism of complexes, lifting the identity.Proof. We only have to prove that for all n ≥ , b n f n + = f n d n , where b • and d • are the differentialsof the bar resolution and of the minimal resolution, respectively. The proof is recursive and it consistsof a straightforward computation that we omit.Notice that the formulas of the components f n of f • come from the terms appearing when applyingthe rewriting rules to the elements of A n , replacing products by tensors.We will not provide the complete description of the maps g n such that g • : B • A → P • A is amorphism of complexes of A e -modules lifting the identity, since we do not need them. We willjust describe in detail those expressions which we need for the cup product and the Gerstenhaberbracket. Notice that since B • A is a free resolution of A as a A e -module, if we define g n on a linearlyindependent subset B n ⊆ A ⊗ A ⊗ n ⊗ A in such a way that d n g n + = g n b n for all z ∈ B n + - anecessary condition is that b n ( B n + ) ⊆ B n -, then it is possible to extend each g n to A ⊗ A ⊗ n ⊗ A asa map of A e -modules in such a way that we get a morphism of complexes. The following propositionprovides the maps we need. Notice that the idea for g n is that we make the product of the intermediatetensors and we look for elements in A n , before and after rewriting. Proposition . . There is a morphism of complexes g • : B • A → P • A such that: • g = id A ⊗ A , • for all a ∈ { , } , b , c ∈ N , g (cid:16) x a ( yx ) b y c (cid:17) = a ⊗ x ⊗ ( yx ) b y c + b − X i = x a ( yx ) i ⊗ y ⊗ x ( yx ) b − − i y c + b − X i = x a ( yx ) i y ⊗ x ⊗ ( yx ) b − − i y c + c − X i = x a ( yx ) b y i ⊗ y ⊗ y c − − i , • for all n ≥ , g n (cid:0) ⊗ y ⊗ x ⊗ n − ⊗ (cid:1) = , – g n (cid:0) ⊗ y ⊗ yx ⊗ x ⊗ n − ⊗ (cid:1) = ⊗ y x n − ⊗ , – g n (cid:0) ⊗ y ⊗ y ⊗ x ⊗ n − ⊗ (cid:1) = , – g n (cid:0) ⊗ x ⊗ n ⊗ (cid:1) = ⊗ x n ⊗ , – g n (cid:0) ⊗ y ⊗ x ⊗ n − ⊗ (cid:1) = ⊗ y x n − ⊗ , – g n (cid:0) ⊗ x ⊗ i ⊗ y ⊗ x n − − i ⊗ (cid:1) = , for all i , ≤ i ≤ n − , – g n (cid:0) ⊗ yx ⊗ x n − ⊗ (cid:1) = y ⊗ x n ⊗ , – g n (cid:0) ⊗ x ⊗ i ⊗ yx ⊗ x ⊗ n − − i ⊗ (cid:1) = , for all i , ≤ i ≤ n − , – g n (cid:0) ⊗ xy ⊗ x ⊗ n − ⊗ (cid:1) = x ⊗ y x n − ⊗ + ⊗ x n ⊗ y + ⊗ x n ⊗ yx , – g n (cid:0) ⊗ x ⊗ i ⊗ xy ⊗ x ⊗ n − − i ⊗ (cid:1) = ⊗ x n ⊗ y + ⊗ x n ⊗ yx , for all i , ≤ i ≤ n − , – g n (cid:0) ⊗ x ⊗ n − ⊗ xy ⊗ (cid:1) = ⊗ x n ⊗ y , – g n (cid:0) ⊗ xyx ⊗ x ⊗ n − ⊗ (cid:1) = xy ⊗ x n ⊗ , – g n (cid:0) ⊗ x ⊗ i ⊗ xyx ⊗ x ⊗ n − − i ⊗ (cid:1) = , for all i , ≤ i ≤ n − , – g n (cid:0) ⊗ x ⊗ n − ⊗ xyx ⊗ (cid:1) = ⊗ x n ⊗ yx .Proof. The proof is, as before, straightforward.The induced morphisms in cohomology, f • and g • , are thus inverse isomorphisms, so the cupproduct may be computed as follows: given ϕ ∈ Hom A e ( P n A , A ) and φ ∈ Hom A e ( P m A , A ) , the cupproduct of their cohomology classes ϕ and φ is: ϕ ⌣ φ = ( ϕg n ⌣ φg m ) f n + m .From now on we omit the bar in the notation of cohomology classes, since we will always work incohomology. The differential in Hom A e ( P • A , A ) will be denoted as usual by d • .Of course, if λ ∈ H ( A , A ) and ϕ ∈ H n ( A , A ) , then ϕ ⌣ λ = λ ⌣ ϕ = λϕ . Suppose nowthat n , m ≥ and that ϕ ∈ H n ( A , A ) and φ ∈ H m ( A , A ) . Their cup product ϕ ⌣ φ belongs toH n + m ( A , A ) , and since the cup product is graded commutative, we know that ϕ ⌣ φ = (− ) nm φ ⌣ ϕ . In order to write these elements in terms of a basis of H n + m ( A , A ) we need to know their value on ⊗ x n + m ⊗ and on ⊗ y x n + m − ⊗ . In each case we will use either ϕ ⌣ φ of φ ⌣ ϕ , accordingto our convenience.Let us start by describing the product of c ∈ H ( A , A ) with all the other generators. The gradedcommutativity implies c = , we will see that in fact all the products involving c are zero.We define a family of cochains that we will use in several of the proofs of the following propositions.For r ≥ let α rm − i , j ∈ Hom A e ( P r A , A ) be defined by α rm − i , j ( ⊗ x r ⊗ ) = , α rm − i , j (cid:16) ⊗ y x r − ⊗ (cid:17) = −( yx ) m − i y . Proposition . . The product of c with any other generator of H • ( A , A ) is zero.Proof. We will denote with the same symbol the cohomology class and its representatives. ) c ⌣ s n : c ⌣ s n (cid:16) ⊗ x ⊗ (cid:17) = ( cg ⌣ s n g ) f (cid:16) ⊗ x ⊗ (cid:17) = ( cg ⌣ s n g ) ( ⊗ x ⊗ ⊗ )= c ( ⊗ x ⊗ ) s n ( ⊗ x ⊗ ) = ( + ) xy = , ⌣ s n (cid:16) ⊗ y x ⊗ (cid:17) = ( cg ⌣ s n g ) f (cid:16) ⊗ y x ⊗ (cid:17) = ( cg ⌣ s n g ) (cid:18) y ⊗ y ⊗ x ⊗ + ⊗ y ⊗ yx ⊗ − x ⊗ y ⊗ y ⊗ − ⊗ x ⊗ y ⊗ − x ⊗ y ⊗ x ⊗ − ⊗ x ⊗ yx ⊗ (cid:19) = yc ( ⊗ y ⊗ ) s n ( ⊗ x ⊗ ) + c ( ⊗ y ⊗ ) s n ( y ⊗ x ⊗ + ⊗ y ⊗ x )− xc ( ⊗ y ⊗ ) s n ( ⊗ y ⊗ ) − c ( ⊗ x ⊗ ) s n ( y ⊗ y ⊗ + ⊗ y ⊗ y )− xc ( ⊗ y ⊗ ) s n ( ⊗ x ⊗ ) − c ( ⊗ x ⊗ ) s n ( y ⊗ x ⊗ + ⊗ y ⊗ x )= yx ( + ) xy + x (cid:16) y ( + ) xy + y + x (cid:17) − x y + − (cid:16) y + + y + (cid:17) − x ( + ) xy − (cid:16) y ( + ) xy + y + x (cid:17) = ( + ) xyxy + n X i = n ! i ! x ( yx ) n + − i y .We will see that reducing modulo coboundaries, this last expression is zero. Defining the -cochains η and ν such that η ( ⊗ x ⊗ ) = − xy , η ( ⊗ y ⊗ ) = , ν ( ⊗ x ⊗ ) = , ν ( ⊗ y ⊗ ) = y + − ( + ) xy ,we obtain that (cid:0) c ⌣ s n − d ( η ) − d ( ν ) (cid:1) (cid:0) ⊗ y x ⊗ (cid:1) = , and the computation for ⊗ x ⊗ re-mains valid, since neither d ( η ) nor d ( ν ) touches ⊗ x ⊗ . ) t ⌣ c : t ⌣ c ( ⊗ x + ⊗ ) = t (cid:16) ⊗ x ⊗ (cid:17) c ( ⊗ x ⊗ ) = while t ⌣ c ( ⊗ y x ⊗ ) = (cid:16) t g ⌣ cg (cid:17) (cid:16) f + (cid:16) ⊗ y x (cid:17)(cid:17) = ,since c ( ⊗ x ⊗ ) = = t (cid:0) ⊗ x ⊗ (cid:1) . ) u ⌣ c : here we compute directly that u m ⌣ c = P mi = ! i ! d (cid:16) α − i , i (cid:17) . ) v + ⌣ c : it is straightforward to verify that it is zero both on ⊗ x + ⊗ and on ⊗ y x + ⊗ . ) w + ⌣ c : w + ⌣ c (cid:16) ⊗ x + ⊗ (cid:17) = w + (cid:16) ⊗ x + ⊗ (cid:17) c ( ⊗ x ⊗ ) = , w + ⌣ c (cid:16) ⊗ y x + ⊗ (cid:17) = w + (cid:16) ⊗ y x ⊗ (cid:17) c ( ⊗ x ⊗ )− w + (cid:16) ⊗ x + ⊗ (cid:17) c ( y ⊗ y ⊗ + ⊗ y ⊗ y )− w + (cid:16) ⊗ x + ⊗ (cid:17) c ( y ⊗ x ⊗ + ⊗ y ⊗ x )= − xy ( yx + xy ) = m X i = m ! i ! x ( yx ) m + − i y .Notice that w + ⌣ c = P mi = ! i ! d + (cid:16) α + + − i , i (cid:17) . ow we are going to prove that the product of two different elements of the infinite family ofgenerators of H ( A , A ) is a scalar multiple of a generator of H ( A , A ) . Proposition . . For all n , m ∈ N , s m ⌣ s n = ( n − m ) t + m + .Proof. Since s m ∈ H ( A , A ) , we know that s m ⌣ s m = . Suppose m = n , s m ⌣ s n (cid:16) ⊗ x ⊗ (cid:17) = ( s m g ⌣ s n g ) f (cid:16) ⊗ x ⊗ (cid:17) = ( s m g ⌣ s n g ) (cid:16) ⊗ x ⊗ ⊗ (cid:17) = s m ( ⊗ x ⊗ ) s n ( ⊗ x ⊗ ) = ( + ) xy ( + ) xy = . ( . )On the other hand, s m ⌣ s n (cid:16) ⊗ y x ⊗ (cid:17) = ( + ) m + X i = ( m + )! i ! x ( yx ) m + − i y ( i + n ) + m + n + X i = ( m + n + )! i ! x ( yx ) m + n + − i y − ( + ) xy ( m + n + ) − ( + )( + ) m + X i = ( m + )! i ! x ( yx ) m + − i y ( i + n ) − ( + ) m + n X i = ( m + n )! i ! x ( yx ) m + n + − i y .Even if this expression is not zero, while trying to write s m ⌣ s n in terms of the basis, we notice thatthe same ν of the previous proof allows us to get rid of terms of type x ( yx ) y in this last expressionby subtracting d ( ν ) to s m ⌣ s n without changing the value on ⊗ x ⊗ . We still need to deal withthe terms of type x ( yx ) b + y , without touching ( . ). Let γ b , i ∈ Hom A e ( A ⊗ k { x , y } ⊗ A , A ) be suchthat γ b , i ( ⊗ x ⊗ ) = and γ b , i ( ⊗ y ⊗ ) = + i ( yx ) b y + − x ( yx ) b y ,so d ( γ b , i ) (cid:0) ⊗ x ⊗ (cid:1) = and d ( γ b , i ) (cid:0) ⊗ y x ⊗ (cid:1) = x ( yx ) b + y . Finally, subtracting d (cid:0) γ b , i (cid:1) for convenient values from b and i to s m ⌣ s n we obtain that s m ⌣ s n = ( n − m ) t + m + .The previous proposition proves that all the generators t of H ( A , A ) , except t and t may beobtained from products of generators of H ( A , A ) . Before continuing with the cup product we provea technical lemma. Lemma . . For all n , m ≥ , P mi = ( n + i )! i ! = ( m + n + )! m !( n + ) .Proof. Dividing by n ! on both sides, we see that the equality is equivalent to P mi = (cid:0) n + ii (cid:1) = (cid:0) m + n + (cid:1) ,which can be proved without difficulty by induction on m .We now compute the products of a generator s n with all generators in degree and + . Proposition . . Given n , m ≥ and p ≥ : (i) t ⌣ s n = , (ii) u ⌣ s n = ( + ) w + + m + + + m + , (iii) v + ⌣ s n = ( + ) t + + m ,(iv) w + ⌣ s n = − + + m + . roof. (i) It is straightforward: t ⌣ s n (cid:16) ⊗ x + ⊗ (cid:17) = t (cid:16) ⊗ x ⊗ (cid:17) s n ( ⊗ x ⊗ ) = , t ⌣ s n (cid:16) ⊗ y x ⊗ (cid:17) = t (cid:16) ⊗ y x − ⊗ (cid:17) s n ( ⊗ x ⊗ )+ t (cid:16) ⊗ x ⊗ (cid:17) s n ( y ⊗ y ⊗ + ⊗ y ⊗ y )+ t (cid:16) ⊗ x ⊗ (cid:17) s n ( y ⊗ x ⊗ + ⊗ y ⊗ x )= xy ( + ) xy = .(ii) The first part is quite direct, since u ⌣ s n (cid:16) ⊗ x + ⊗ (cid:17) = u (cid:16) ⊗ x ⊗ (cid:17) s n ( ⊗ x ⊗ )= m X i = ( yx ) m − i y ( + ) xy = ( + ) m X i = m ! i ! x ( yx ) m − i y ( i + n ) ,and setting η i ∈ Hom A e (cid:0) P A , A (cid:1) for i , ≤ i < m , such that η i (cid:16) ⊗ x ⊗ (cid:17) = ( yx ) m − i y ( i + n ) and η i (cid:16) ⊗ y x − ⊗ (cid:17) = ,it turns out that d ( η i ) (cid:0) ⊗ x + ⊗ (cid:1) = x ( yx ) m − i y ( i + n ) , so u ⌣ s n − m − X i = ( + ) m ! i ! d ( η i ) ! (cid:16) ⊗ x + ⊗ (cid:17) = ( + ) xy ( n + m ) = ( + ) w + + m (cid:16) ⊗ x + ⊗ (cid:17) .Notice that d ( η i ) (cid:16) ⊗ y x ⊗ (cid:17) = ( m − i )( yx ) m − i + y ( i + n ) − i + n X l = ( i + n )! l ! ( yx ) m + n + − l y ,and so this reduction will also have an effect on ⊗ y x ⊗ . Let us now compute the valueon this element. We will write λ l instead of ( yx ) m + n + − l y . u ⌣ s n (cid:16) ⊗ y x ⊗ (cid:17) = u (cid:16) ⊗ y x − ⊗ (cid:17) s n ( ⊗ x ⊗ )+ u (cid:16) ⊗ x ⊗ (cid:17) s n ( y ⊗ y ⊗ + ⊗ y ⊗ y )+ u (cid:16) ⊗ x ⊗ (cid:17) s n ( y ⊗ x ⊗ + ⊗ x ⊗ y )= − y + ( + ) xy + m X i = m ! i ! ( yx ) m − i y y + + m X i = m ! i ! ( yx ) m − i y (cid:16) ( + ) yxy + y + x (cid:17) = m X i = m ! i ! ( yx ) m − i y ( i + n + ) + m − X i = ( + ) m ! i ! ( yx ) m − i y + xy + m X i = m ! i ! ( yx ) m − i y ( i + n )+ x = m + n + X l = n + m !( l − n − )! λ l + m + n − X l = n ( + ) m !( m + n − l )( l − n )! λ l n X l = m ! l ! m X i = ( i + n !) i ! ! λ l + m + n X l = n + m ! l ! m X i = l − n ( i + n !) i ! ! λ l .By Lemma . , we have that n X l = m ! l ! m X i = ( i + n !) i ! ! λ l + m + n X l = n + m ! l ! m X i = l − n ( i + n !) i ! ! λ l = n + m X l = m ! l ! (cid:18) ( n + m + )! m !( n + ) (cid:19) λ l − m + n X l = n + m ! l ! (cid:18) l !( l − n − )!( n + ) (cid:19) λ l ,so u ⌣ s n (cid:16) ⊗ y x ⊗ (cid:17) = m + n + X l = n + m !( l − n − )! λ l + n + m X l = ( n + m + )! l !( n + ) λ l + ( + ) m + n − X l = n m !( l − n )! ( m + n − l ) λ l − m + n X l = n + m !( l − n − )!( n + ) λ l ,and reducing modulo coboundaries, we obtain u ⌣ s n (cid:16) ⊗ y x ⊗ (cid:17) = u ⌣ s n − m − X i = ( + ) m ! i ! d ( η i ) ! (cid:16) ⊗ y x ⊗ (cid:17) = m + n + X l = n + m !( l − n − )! λ l + n + m X l = ( n + m + )! l !( n + ) λ l − m + n X l = n + m !( l − n − )!( n + ) λ l + ( + ) m − X i = m ! i ! i + n X l = ( i + n )! l ! λ l = m + n + X l = n + m !( l − n − )! λ l + n + m X l = ( n + m + )! l !( n + ) λ l − m + n X l = n + m !( l − n − )!( n + ) λ l + ( + ) m X i = + n X l = m ! i ! ( i + n )! l ! λ l − ( + ) m + n X l = ( m + n )! l ! λ l = m + n + X l = n + m !( l − n − )! λ l + ( + ) n + m X l = ( n + m + )! l !( n + ) λ l − ( + ) m + n X l = n + m !( l − n − )!( n + ) λ l − ( + ) m + n X l = ( m + n )! l ! λ l = n + m + + ( + ) n + m X l = ( n + m )! l ! λ l = n + m + X l = ( n + m + )! l ! λ l − n + m X l = ( n + m + )! l ! λ l ! + ( + ) n + m X l = ( n + m )! l ! λ l = n + m + X l = ( n + m + )! l ! λ l − ( + ) n + m X l = ( n + m )! l ! λ l .Choosing β ∈ Hom A e (cid:0) P A , A (cid:1) such that β (cid:16) ⊗ x ⊗ (cid:17) = and β (cid:16) ⊗ y x − ⊗ (cid:17) = − y ( n + m )+ , e have d ( β )( ⊗ x + ⊗ ) = and d ( β )( ⊗ y x ⊗ ) = n + m X l = ( n + m )! l ! λ l + xy ( n + m )+ and using that v + + m (cid:0) ⊗ x + ⊗ (cid:1) = , we obtain u ⌣ s n = u ⌣ s n − m − X i = ( + ) m ! i ! d ( η i ) + d (( + ) β ) = + + m + + ( + ) w + + m .(iii) Clearly v + ⌣ s n (cid:0) ⊗ x + ⊗ (cid:1) = . We compute v + ⌣ s n (cid:16) ⊗ y x + ⊗ (cid:17) = v + (cid:16) ⊗ y x ⊗ (cid:17) s n ( ⊗ x ⊗ )− v + (cid:16) ⊗ x + ⊗ (cid:17) s n ( y ⊗ y ⊗ + ⊗ y ⊗ y )− v + (cid:16) ⊗ x + ⊗ (cid:17) s n ( y ⊗ x ⊗ + ⊗ y ⊗ x )= m X i = m ! i ! ( yx ) m − i y ( + ) xy = ( + ) m X l = m ! l ! x ( yx ) m − l y ( l + n ) ,and we get v + ⌣ s n = v + ⌣ s n − ( + ) m − X l = m ! i ! d + (cid:16) α + − l , l + n (cid:17) = ( + ) t + + m .(iv) The first part is straightforward, since w + ⌣ s n (cid:16) ⊗ x + ⊗ (cid:17) = w + (cid:16) ⊗ x + ⊗ (cid:17) s n ( ⊗ x ⊗ )= xy ( + ) xy = .On the other hand, w + ⌣ s n (cid:16) ⊗ y x + ⊗ (cid:17) = w + (cid:16) ⊗ y x ⊗ (cid:17) s n ( ⊗ x ⊗ )− w + (cid:16) ⊗ x + ⊗ (cid:17) s n ( y ⊗ y ⊗ + ⊗ y ⊗ y )− w + (cid:16) ⊗ x + ⊗ (cid:17) s n ( y ⊗ x ⊗ + ⊗ y ⊗ x )= xy + ( + ) xy − y + − xy (cid:16) ( + ) yxy + y + x (cid:17) = − y + − xy ( n + m )+ x = − ( m + n + ) − n + m X i = ( n + m )! i ! x ( yx ) m + n + − i y .Using again the cochains α + − i , j , we can remove the sum from the last expression. Thus, w + ⌣ s n = − + + m + .We have already proved in Proposition . (i) that the product t ⌣ s n is zero for all m , n ≥ , p ≥ . In fact, the product of t with almost all other generators annihilates, except for one case,where we obtain another generator t . Proposition . . For all m , n ≥ , p , q ≥ we have: i) t ⌣ t = , (ii) u ⌣ t = t ( p + q ) n + m , (iii) t + ⌣ v + = ,(iv) t + ⌣ w + = .Proof. (i) The equality follows from: t ⌣ t (cid:16) ⊗ x + ⊗ (cid:17) = t (cid:16) ⊗ x ⊗ (cid:17) t (cid:16) ⊗ x ⊗ (cid:17) = , t ⌣ t (cid:16) ⊗ y x + − ⊗ (cid:17) = t (cid:16) ⊗ y x − ⊗ (cid:17) t (cid:16) ⊗ x ⊗ (cid:17) + t (cid:16) ⊗ x ⊗ (cid:17) t (cid:16) ⊗ y x − ⊗ (cid:17) + t (cid:16) ⊗ x ⊗ (cid:17) t (cid:16) y ⊗ x ⊗ (cid:17) = .(ii) Clearly u ⌣ t (cid:0) ⊗ x + ⊗ (cid:1) = , while u ⌣ t − m − X l = m ! l ! d ( p + q )− (cid:16) α ( p + q )− − l , l + n (cid:17)! (cid:16) ⊗ y x ( p + q )− ⊗ (cid:17) = xy n + m .(iii) The computations are straightforward: t ⌣ v + (cid:16) ⊗ x + + ⊗ (cid:17) = t (cid:16) ⊗ x ⊗ (cid:17) v + (cid:16) ⊗ x + ⊗ (cid:17) = , t ⌣ v + (cid:16) ⊗ y x + ⊗ (cid:17) = t (cid:16) ⊗ y x − ⊗ (cid:17) v + (cid:16) ⊗ x + ⊗ (cid:17) + t (cid:16) ⊗ x ⊗ (cid:17) v + (cid:16) ⊗ y x ⊗ (cid:17) + t (cid:16) ⊗ x ⊗ (cid:17) v + (cid:16) y ⊗ x + ⊗ (cid:17) = .(iv) Again, it is clear that: t ⌣ w + (cid:16) ⊗ x + + ⊗ (cid:17) = t (cid:16) ⊗ x ⊗ (cid:17) w + (cid:16) ⊗ x + ⊗ (cid:17) = , t ⌣ w + (cid:16) ⊗ y x + ⊗ (cid:17) = t (cid:16) ⊗ y x − ⊗ (cid:17) w + (cid:16) ⊗ x + ⊗ (cid:17) + t (cid:16) ⊗ x ⊗ (cid:17) w + (cid:16) ⊗ y x ⊗ (cid:17) + t (cid:16) ⊗ x ⊗ (cid:17) w + (cid:16) y ⊗ x + ⊗ (cid:17) = xy xy = .Notice that the only case where the product is non zero in the previous proposition is the productwith u , which gives another generator of type t . In fact, the elements of type u act on otherfamilies of generators, as we shall see. Before proving this we need another technical lemma. Lemma . . For all m , n ≥ , there is an equality m X i = m ! i ! ( yx ) m − i y ! n X j = n ! j ! ( yx ) n − j y = m + n X k = ( m + n )! k ! ( yx ) m + n − k y . Proof.
Using the commutation rules the product on the left equals n − X j = X l = m − l X r = (cid:18) n − − j + rr (cid:19)! m ! n ! l ! j ! ( yx ) m + n −( j + l ) y ( l + j ) + m X i = m ! i ! ( yx ) m − i y ( i + n ) . t follows from Lemma . that for all m , n and j , there is an equality P m − lr = (cid:0) n − − j + rr (cid:1) = (cid:0) m + n −( j + l ) n − j (cid:1) ,whence the previous expression is n X j = X l = (cid:18) ml (cid:19)(cid:18) nj (cid:19) ( n + m − ( j + l ))!( yx ) m + n −( j + l ) y ( l + j ) . ( . )The formula we want to prove is symmetric in m and n , so we may suppose that n ≥ m and write m = n + t , with t ≥ . Setting k = j + l and λ k = ( yx ) m + n − k y in ( . ), we get n X k = X j = (cid:18) n + tk − j (cid:19)(cid:18) nj (cid:19) ( + t − k )! λ k + n + t X k = n + X j = (cid:18) n + tk − j (cid:19)(cid:18) nj (cid:19) ( + t − k )! λ k + + t X k = n + t + X j = k −( n + t ) (cid:18) n + tk − j (cid:19)(cid:18) nj (cid:19) ( + t − k )! λ k .Now, the only things left to finish the proof are the equalities: k X j = (cid:18) n + tk − j (cid:19)(cid:18) nj (cid:19) = (cid:18) + tk (cid:19) , for all k , ≤ k ≤ n , n X j = (cid:18) n + tk − j (cid:19)(cid:18) nj (cid:19) = (cid:18) + tk (cid:19) , for all k , n + ≤ k ≤ n + t , n X j = k −( n + t ) (cid:18) n + tk − j (cid:19)(cid:18) nj (cid:19) = (cid:18) + tk (cid:19) , for all k , n + t + ≤ k ≤ + t .The term on the right counts the numbers of subsets of k elements that can be obtained from a set of + t elements. This number can be computed splitting our set in two subsets: one with n elementsand one with n + t elements, and then choosing j from the first subset and k − j from the second onefor all possible values of j .We are now ready to prove our claim. Proposition . . Given m ≥ and p ≥ , the element u ∈ H ( A , A ) acts, via the cup product, on thefamilies of generators (cid:10) u (cid:11) , (cid:10) v + (cid:11) and (cid:10) w + (cid:11) as follows. For all n ≥ and q ≥ , • u ⌣ u = u ( p + q ) m + n , • u ⌣ v + = v ( p + q )+ + n , • u ⌣ w + = w ( p + q )+ + n .Proof. (i) u ⌣ u ( ⊗ x + ⊗ ) = (cid:16) P mi = ! i ! ( yx ) m − i y (cid:17) (cid:16) P nj = ! j ! ( yx ) n − j y (cid:17) . The pre-vious lemma implies that this expression equals P m + nk = ( m + n )! k ! ( yx ) m + n − k y . Also, u ⌣ u ( ⊗ y x + − ⊗ ) = u (cid:16) ⊗ y x − ⊗ (cid:17) u (cid:16) ⊗ x ⊗ (cid:17) + u (cid:16) ⊗ x ⊗ (cid:17) u (cid:16) ⊗ y x − ⊗ (cid:17) + u (cid:16) ⊗ x ⊗ (cid:17) u (cid:16) y ⊗ x ⊗ (cid:17) = − n X l = n ! l ! y + ( yx ) n − l y − m X l = m ! l ! ( yx ) n − l y y + + m X l = m ! l ! ( yx ) m − l y + X i = n ! i ! ( yx ) n − i y = − y ( m + n )+ . e have thus proved that u ⌣ u = u ( p + q ) m + n .(ii) The result follows without difficulty, since v + ⌣ u ( ⊗ x + + ⊗ ) = , while v + ⌣ u ( ⊗ y x + ⊗ ) = v + (cid:16) ⊗ y x ⊗ (cid:17) u (cid:16) ⊗ x ⊗ (cid:17) − v + (cid:16) ⊗ x + ⊗ (cid:17) u (cid:16) ⊗ y x − ⊗ (cid:17) − v + (cid:16) ⊗ x + ⊗ (cid:17) u (cid:16) y ⊗ x ⊗ (cid:17) = n X i = n ! i ! ( yx ) n − i y ! m X j = n ! j ! ( yx ) m − j y = m + n X i = ( m + n )! i ! ( yx ) m + n − i y ,and we know that u ⌣ v + = v + ⌣ u .(iii) Again, it is easy to prove that u ⌣ w + (cid:16) ⊗ x + + ⊗ (cid:17) = m X l = m ! l ! x ( yx ) m − l y ( l + n ) .The computation for u ⌣ w + (cid:0) ⊗ y x + ⊗ (cid:1) is more involved: u ⌣ w + (cid:16) ⊗ y x + ⊗ (cid:17) = m X l = m ! l ! x ( yx ) m − l y ( l + n )+ + m − X i = m !( m − l ) l ! ( yx ) m + − l y ( l + n ) .Reducing modulo coboundaries without modifying the other coordinate, we obtain that u ⌣ w + (cid:16) ⊗ y x + ⊗ (cid:17) = xy ( m + n )+ − m − X l = + n X i = m !( l + n )! l ! i ! ( yx ) m + n + − i y + m − X l = m !( m − l ) l ! ( yx ) m + − l y ( l + n ) .Defining ν l ∈ Hom A e (cid:0) P + A , A (cid:1) , for l such that ≤ l ≤ m − , by the formulas ν l (cid:16) ⊗ x + ⊗ (cid:17) = ( yx ) m − l y ( l + n ) and ν l (cid:16) ⊗ y x + − ⊗ (cid:17) = ,like in Proposition . (ii) we get d + ( ν l ) (cid:16) ⊗ x + + ⊗ (cid:17) = x ( yx ) m − l y ( l + n ) , d + ( ν l ) (cid:16) ⊗ y x + − ⊗ (cid:17) = ( m − l )( yx ) m + − l y ( l + n ) − l + n X i = ( l + n )! i ! ( yx ) m + n + − l y .Finally, u ⌣ w + = u ⌣ w + − m − X l = m ! l ! d + ( ν l ) = w + + + m . he last cup products of generators that we have to describe are those amongst the generators inodd degrees greater than . Most of them are zero except for some cases where we also obtain someof the t generators. We finish the description of the cup product with the following proposition. Proposition . . For all m , n ≥ , p , q ≥ , there are equalities:(i) v + ⌣ v + = ,(ii) w + ⌣ w + = ,(iii) v + ⌣ w + = t ( p + q + ) n + m .Proof. The proof of (i) is straightforward and we omit it.For (ii), w + ⌣ w + (cid:16) ⊗ x ( p + q + ) ⊗ (cid:17) is zero, but the other coordinate needs a reductionmodulo coboundaries. The cochain we will use is, as usual, α ( p + q )+ , k ∈ Hom A e (cid:16) P ( p + q )+ A , A (cid:17) defined by α ( p + q )+ , k (cid:16) ⊗ x ( p + q )+ ⊗ (cid:17) = and α ( p + q )+ , k (cid:16) ⊗ y x ( p + q ) ⊗ (cid:17) = −( yx ) b y .Subtracting d ( p + q )+ (cid:0) α b , k (cid:1) for convenient values of b and k from w + ⌣ w + we obtain thatthe cup product between both generators is zero.To prove (iii) notice that v + ⌣ w + (cid:16) ( ⊗ x + + ⊗ (cid:17) = v + (cid:16) ⊗ x + ⊗ (cid:17) w + (cid:16) ⊗ x + ⊗ (cid:17) = ,while v + ⌣ w + (cid:16) ⊗ y x + + ⊗ (cid:17) = v + (cid:16) ⊗ y x ⊗ (cid:17) w (cid:16) ⊗ x + ⊗ (cid:17) − v + (cid:16) ⊗ x + ⊗ (cid:17) w + (cid:16) ⊗ y x ⊗ (cid:17) − v + (cid:16) ⊗ x + ⊗ (cid:17) w + (cid:16) y ⊗ x + ⊗ (cid:17) = m X i = m ! i ! ( yx ) m − i y xy = m X l = m ! l ! x ( yx ) m − l y ( l + n ) .Reducing modulo coboundaries, we obtain the result.We shall summarize all these propositions in the following table. Theorem . . The cup product endows H • ( A , A ) with an associative graded commutative algebra structure.The next table describes the products amongst generators c s n t u v + w + c 0 0 0 0 0 0s m ( n − m ) t + m + + + m + −( + ) t + + m + + m + +( + ) w + + m t + + n + + m + t + + m u + + n v + + + m w + + + m +( + ) w + + m v + ( + ) t + + m + + + n + + + n w + − + + m + + + + n − t + + + n Corollary . . The graded algebra H • ( A , A ) is generated by the set (cid:8) , c , t , t , s n , u , v : n ≥ (cid:9) . otice that, as we have mentioned before, the cohomology spaces H i ( A , A ) are periodic of period for i ≥ and that this periodicity comes from multiplication by u . We state this fact more preciselyin the next corollary. Corollary . . For all p , q ≥ , there are isomorphisms H ( A , A ) ∼ = H ( q + p ) ( A , A ) and H + ( A , A ) ∼ = H ( q + p )+ ( A , A ) , given by ϕ ∈ H ( A , A ) → ϕ ⌣ u ∈ H ( q + p ) ( A , A ) , ϕ ∈ H + ( A , A ) → ϕ ⌣ u ∈ H ( q + p )+ ( A , A ) . H ( A , A ) as a Lie algebra Given an algebra A , the Gerstenhaber bracket endows H ( A , A ) with a Lie algebra structure. Theaim of this section is to describe it when A is the super Jordan plane. We already have the necessarycomparison maps between the bar resolution and ours. Proposition . . The generator c is central.Proof. We must prove that [ c , s n ] = for all n ∈ N . Straightforward computations using Lemma . lead us to the equalities: c ◦ s n ( ⊗ x ⊗ ) = ( + ) n − X k = n ! l !( n − l ) x ( yx ) n − l y , c ◦ s n ( ⊗ y ⊗ ) = n X i = ( n + )! i !( n − i + ) x ( yx ) n − i y + n − X i = n ! i !( n − i ) ( yx ) n − i y + , s n ◦ c ( ⊗ x ⊗ ) = , s n ◦ c ( ⊗ y ⊗ ) = ( + ) xy ,so that [ c , s n ] ( ⊗ x ⊗ ) = ( + ) n − X k = n ! l !( n − l ) x ( yx ) n − l y , [ c , s n ] ( ⊗ y ⊗ ) = n X l = ( n + )! l !( n − l + ) x ( yx ) n − l y + n − X l = n ! l !( n − l ) ( yx ) n − l y + − ( + ) xy .We will make use of the elements ζ n − l − , l , ξ n − l − , l , ρ n , with ≤ l ≤ n − , of Proposition . . Theybelong to Im d . It is long but not difficult to verify that [ c , s n ] + n − X l = n ! l !( n − l ) ζ n − l − , l − n − X l = ( n + )! l !( n − l + )( n − l ) ξ n − l − , l = ( n + ) ρ n ,which proves that [ c , s n ] is zero in H ( A , A ) .The bracket [ s m , s n ] does not annihilate for n = m . The following computation is easier and weomit it. For details, see [R, p. ] Proposition . . Given n , m ≥∈ N , we have [ s m , s n ] = ( n − m ) s n + m . Recall that the Virasoro algebra, denoted Vir, is the only central extension of the Witt algebra andit is the Lie algebra with basis { C } ∪ { L n : n ∈ Z } , subject to the relations [ L m , C ] = , [ L m , L n ] = ( n − m ) L m + n + δ m , − n m − m12 C . he triangular decomposition of Vir is the following:Vir = Vir + ⊕ h ⊕ Vir − , where Vir + = ∞ M n = k L n , h = k C ⊕ k L , Vir − = ∞ M n = k L − n .The Lie algebra H ( A , A ) can be identified to a Lie subalgebra of Vir. Theorem . . There exists an isomorphism of Lie algebras H ( A , A ) ∼ = Vir + ⊕ h . Proof.
We define the map H ( A , A ) → Vir + ⊕ h : c → C , s m → m + L m , for all n ∈ N .Vir and H ( A , A ) being Lie algebras yields the fact that the map is well-defined. It is clearly anisomorphism of vector spaces, and consequently it is an isomorphism of Lie algebras. Remark . . Any Lie module M over Vir + ⊕ h provides, by induction, a Lie module over Vir . Given n ∈ N , thevector space H n ( A , A ) is a representation of H ( A , A ) . Using the previous theorem, it induces a representationof the Virasoro algebra. The next section will be devoted to the description of these representations. Hochschild cohomology spaces as H ( A , A ) - Lie modules From now on k = C . In this section we will compute the Gerstenhaber brackets h H ( A , A ) , H n ( A , A ) i for n > 1 . These brackets describe the structure of H ( A , A ) -Lie module of each space H n ( A , A ) andthus, their induced structure as Vir-modules.Representation theory of the Virasoro algebra is a very active subject of research since this algebrais important both in mathematics and in physics, in particular in conformal field theory and stringtheory. Recall that a weight-module M is a Vir-module such that the Cartan subalgebra h of Vir actsdiagonally on M . If, additionally, all the weight spaces of a weight module are finite dimensional, then M is called a Harish-Chandra module.Intermediate series modules are modules V a , b with basis { v n } n ∈ Z , a , b ∈ k and action given by L s · v n = ( n + as + b ) v n + s and c · v n = . The Vir-module V a , b is simple if a = , or when b ∈ C \ Z .Notice that according to Mathieu [M], in case V a , b is not simple, the unique simple quotient of V a , b is called an intermediate series module.There are two classical families of simple modules: highest (lowest) weight modules and inter-mediate series modules, and every simple weight Harish-Chandra module belongs to one of thesefamilies [M]. In [CGZ] all V a , b are called intermediate series modules. Even if the algebra consideredis W + := Vir + ⊕ h , the corresponding modules V a , b are called intermediate series modules. We willuse this terminology.We will now start the description of the action of H ( A , A ) on H n ( A , A ) . Proposition . . For all n ≥ , the element c ∈ H ( A , A ) acts trivially on H n ( A , A ) .Proof. We will prove that for all n ≥ and p ≥ , h c , t i = h c , u i = h c , v + i = h c , w + i = .The element c is the class of the derivation in A sending x to a y to x . We will use Suárez-Álvarez’smethod [S] to compute the brackets, so we need liftings of c as in the following diagram. · · · d / / A ⊗ k { x , y x } ⊗ A d / / c (cid:15) (cid:15) A ⊗ k { x , y } ⊗ A d / / c (cid:15) (cid:15) A ⊗ A m A / / c (cid:15) (cid:15) A / / c (cid:15) (cid:15) · · · d / / A ⊗ k { x , y x } ⊗ A d / / / / A ⊗ k { x , y } ⊗ A d / / A ⊗ A m A / / A / / et us briefly recall some facts about this approach: given a derivation δ of an algebra B , δ e = δ ⊗ id + id ⊗ δ is a derivation of B ⊗ B op . For any B -bimodule X, a k -linear map f : X → X is a δ e -operator if f (( a ⊗ b ) · m ) = ( a ⊗ b ) · f ( m ) + δ e ( a ⊗ b ) · m for all a , b ∈ B and x ∈ X . Here, we needthe maps c i to be c e -operators such that the squares are commutative. In this situation, it is enoughto determine the values of each c n in elements of type ⊗ v ⊗ ∈ P n A . It is not difficult to verify thatthe maps { c n } n ≥ defined as c ( ⊗ ) = , c ( ⊗ x ⊗ ) = , c ( ⊗ y ⊗ ) = ⊗ y ⊗ , c n ( ⊗ x n ⊗ ) = , c n ( ⊗ y x n − ⊗ ) = y ⊗ x n ⊗ + (− ) n + ⊗ x n ⊗ y − x ⊗ x n ⊗ ,satisfy all the required conditions.Let us compute (cid:2) c , t (cid:3) . By Proposition . , we know that s ⌣ s n = + and due to thePoisson identity, we get that h c , t + i = [ c , s ⌣ s n ] = ([ c , s ] ⌣ s n + s ⌣ [ c , s n ]) = ,for all n ≥ . It only remains to prove that (cid:2) c , t (cid:3) = (cid:2) c , t (cid:3) = . On the one hand, h c , t i (cid:16) ⊗ x ⊗ (cid:17) = c (cid:16) t (cid:16) ⊗ x ⊗ (cid:17)(cid:17) − t (cid:16) c (cid:16) ⊗ x ⊗ (cid:17)(cid:17) = , h c , t i (cid:16) ⊗ y x ⊗ (cid:17) = c ( x ) − t (cid:16) y ⊗ x ⊗ + (− ) n + ⊗ x ⊗ − x ⊗ x ⊗ (cid:17) = .On the other hand h c , t i (cid:16) ⊗ x ⊗ (cid:17) = c (cid:16) t (cid:16) ⊗ x ⊗ (cid:17)(cid:17) − t (cid:16) c (cid:16) ⊗ x ⊗ (cid:17)(cid:17) = , h c , t i (cid:16) ⊗ y x ⊗ (cid:17) = c (cid:16) xy (cid:17) = xyx .Reducing modulo coboundaries in the same way we did while computing s m ⌣ s n , we obtain that (cid:2) c , t (cid:3) and (cid:2) c , t (cid:3) are zero in H ( A , A ) , and so the bracket (cid:2) c , t (cid:3) is null for all n .We now turn our attention to (cid:2) c , u (cid:3) . Straightforward computations and the usual reductionsmodulo coboundaries show that h c , u i (cid:16) ⊗ x ⊗ (cid:17) = n − X r = n ! r !( n − r ) x ( yx ) n − − r y + + n − X r = n ! r ! n − r X k = ! ( yx ) n − r y , h c , u i (cid:16) ⊗ y x ⊗ (cid:17) = − + n − X r = n !( n − r − ) r !( n − r ) ( yx ) n − r y + .For each r , ≤ r < n , let γ r ∈ Hom A e ( A ⊗ k { x , y } ⊗ A , A ) be defined by γ r ( ⊗ x ⊗ ) = ( yx ) m − − r y + and γ r ( ⊗ y ⊗ ) = .It is not difficult to check the equality P n − = ! r !( n − r ) d ( γ r ) = (cid:2) c , u (cid:3) , so the bracket is zero in coho-mology.Once we have proved that (cid:2) c , t (cid:3) = = (cid:2) c , u (cid:3) for all n ≥ , the identity t ⌣ u = t ( p + q ) n and the fact that [ c , −] : H • ( A , A ) → H • ( A , A ) is a derivation with respect to the cup product allowus to conclude that h c , t i = for all n ≥ , p ≥ . Using a similar argument and the equality u ⌣ u = u ( p + q ) n , we prove that h c , u i = for all n ≥ , p ≥ .The equalities h c , v + i = = h c , w + i can be obtained through analogous methods.Our next step will be to describe the action of H ( A , A ) on the even cohomology spaces. For this,we need to lift the derivations s , s and s as we have done for c . The equalities [ s m , s n ] = ( n − ) s m + n , assure that these liftings will be enough. The liftings of s , s and s will be provided bythe morphisms of complexes α • , β • and γ • , respectively. They are defined by the following formulas: α ( ⊗ ) = β ( ⊗ ) = γ ( ⊗ ) = , α ( ⊗ x ⊗ ) = ⊗ x ⊗ and α ( ⊗ y ⊗ ) = ⊗ y ⊗ , α n ( ⊗ x n ⊗ ) = n ⊗ x n ⊗ , for all n ≥ , α n ( ⊗ y x n − ⊗ ) = ( n + ) ⊗ y x n − ⊗ , for all n ≥ .The formulas for the higher β ’s and γ ’s are more complicated and because of the periodicity of thecohomology we will only need the liftings up to degree : β ( ⊗ ) = , β ( ⊗ x ⊗ ) = ( xy ⊗ y ⊗ + x ⊗ y ⊗ y + ⊗ x ⊗ y ) , β ( ⊗ y ⊗ ) = y ⊗ y ⊗ + y ⊗ y ⊗ y + ⊗ y ⊗ y , β ( ⊗ x ⊗ ) = ⊗ y x ⊗ + ⊗ x ⊗ y + ⊗ x ⊗ yx , β ( ⊗ y x ⊗ ) = ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx − ⊗ y x ⊗ , β ( ⊗ x ⊗ ) = ⊗ y x ⊗ + ⊗ x ⊗ y + ⊗ x ⊗ yx , β ( ⊗ y x ⊗ ) = ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx − ⊗ y x ⊗ , γ ( ⊗ ) = , γ ( ⊗ x ⊗ ) = ( xy ⊗ y ⊗ + xy ⊗ y ⊗ y + xy ⊗ y ⊗ y + x ⊗ y ⊗ y + ⊗ x ⊗ y ) , γ ( ⊗ y ⊗ ) = y ⊗ y ⊗ + y ⊗ y ⊗ y + y ⊗ y ⊗ y + y ⊗ y ⊗ y + ⊗ y ⊗ y , γ ( ⊗ x ⊗ ) = ⊗ x ⊗ y + ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx + ⊗ x ⊗ yxy + ⊗ x ⊗ ( yx ) , γ ( ⊗ y x ⊗ ) = ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx + ⊗ y x ⊗ y + ⊗ y x ⊗ yxy + ⊗ y x ⊗ ( yx ) − ⊗ y x ⊗ − ⊗ y x ⊗ y − ⊗ y x ⊗ yx , γ ( ⊗ x ⊗ ) = ⊗ x ⊗ y + ⊗ y x ⊗ + ⊗ y x ⊗ y ⊗ y x ⊗ yx + ⊗ x ⊗ yxy + ⊗ x ⊗ ( yx ) , γ ( ⊗ y x ⊗ ) = ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx + ⊗ y x ⊗ y + ⊗ y x ⊗ yxy + ⊗ y x ⊗ ( yx ) − ⊗ y x ⊗ − ⊗ y x ⊗ y − ⊗ y x ⊗ yx .We will omit the verifications, since they are direct, long and tedious. Theorem . . For all p ≥ and m , n ≥ (i) h c , t i = = h c , u i ,(ii) h s m , t i = (cid:16) n − ( − ) m − p (cid:17) t + m ,(iii) h s m , u i = (cid:16) ( n − − p ) u + m + pm ( + ) t + m (cid:17) . Before proceeding to the proof of Theorem . we need a technical lemma. We will not prove itsince the computations needed are not complicated but long. emma . . For all n = , s n X l = n ! l ! ( yx ) n − l y ! = n + X l = n ! (cid:0) + n − n + (cid:0) l − l (cid:1)(cid:1) l ! ( yx ) n + − l y + + . Proof of Theorem . . The equalities in (i) have already been proved.(ii) We will prove it by induction on p . Fix p = . Using now induction on m we will show that (cid:2) s m , t (cid:3) = ( n − m − ) t + m . Suppose first that m = : h s , t i (cid:16) ⊗ x ⊗ (cid:17) = s (cid:16) t (cid:16) ⊗ x ⊗ (cid:17)(cid:17) − t (cid:16) α (cid:16) ⊗ x ⊗ (cid:17)(cid:17) = , h s , t i (cid:16) ⊗ y x ⊗ (cid:17) = s (cid:16) xy (cid:17) − t (cid:16) ⊗ y x ⊗ (cid:17) = ( n − ) xy .Thus (cid:2) s , t (cid:3) = ( n − ) t . Supposing now that m = , we have h s , t i (cid:16) ⊗ x ⊗ (cid:17) = − t (cid:16) ⊗ y x ⊗ + ⊗ x ⊗ y + ⊗ x ⊗ yx (cid:17) = , h s , t i (cid:16) ⊗ y x ⊗ (cid:17) = s (cid:16) xy (cid:17) − t (cid:16) ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx − ⊗ y x ⊗ (cid:17) = + + − X i = xy i y y − − i − xy − + − + x + = ( n − ) xy ( n + ) − X j = n ! j ! x ( yx ) n + − j y .Reducing modulo coboundaries we can omit the terms of type x ( yx ) m + − j y from the second co-ordinate, so we obtain that (cid:2) s , t (cid:3) = ( n − ) t + . Finally, suppose m = . Just like before, weget h s , t i (cid:16) ⊗ x ⊗ (cid:17) = ,while h s , t i (cid:16) ⊗ y x ⊗ (cid:17) = s (cid:16) xy (cid:17) − t (cid:16) ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx + ⊗ y x ⊗ y + ⊗ y x ⊗ yxy + ⊗ y x ⊗ ( yx ) − ⊗ y x ⊗ − ⊗ y x ⊗ y − ⊗ y x ⊗ yx (cid:17) = + + + − xy − xy + − xy + x − + − + xy − + xyx + xy + + + + x .Again, reducing modulo coboundaries we conclude that (cid:2) s , t (cid:3) = ( n − ) t + .Given m ≥ , the Jacobi identity reads: h [ s , s m ] , t i = h s , h s m , t ii − h s m , h s , t ii and using the inductive hypothesis and the equality [ s , s m ] = ( m − ) s m + we get the desired result.Assume now that h s m , t i = (cid:16) n − ( − ) m − p (cid:17) t + m . We will compute: h s m , t ( p + ) n i = h s m , u i ⌣ t + u ⌣ h s m , t i = (cid:16) (− − ) u + m ( + ) t (cid:17) ⌣ t + ⌣ ( n − ( − ) m − p ) t = (cid:16) n − ( + ) m − ( p + (cid:17) t ( p + ) n + m . e now turn our attention to the brackets h s m , u i . Again, the equality [ s , s m ] = ( m − ) s m + reduces the computations to those of the brackets h s i , u i with i = , , . We proceed by inductionon p . Set p = , i = and n ≥ : h s , u i (cid:16) ⊗ x ⊗ (cid:17) = s n X l = n ! l ! ( yx ) n − l y ! − u (cid:16) ⊗ x ⊗ (cid:17) = n X l = n ! l ! ( yx ) n − l y − n X l = n ! l ! ( yx ) n − l y = ( n − ) n X l = n ! l ! ( yx ) n − l y .Also, h s , u i (cid:16) ⊗ y x ⊗ (cid:17) = s (cid:16) − y + (cid:17) − u (cid:16) ⊗ y x ⊗ (cid:17) = − ( n − ) y + .Calculating (cid:2) s i , u (cid:3) for i = , is considerably longer. The idea is to compute the values of theseelements of H ( A , A ) on ⊗ x ⊗ and ⊗ y x ⊗ and then reduce modulo coboundaries. We willjust indicate which coboundaries can be used, since the procedure does not differ from what we havealready done in other cases.For i = , h s , u i (cid:16) ⊗ x ⊗ (cid:17) = n X l = n !( − − ) l ! ( yx ) n + − l y + ( n − ) y ( n + ) + + , ( . ) h s , u i (cid:16) ⊗ y x ⊗ (cid:17) = − ( n − ) y + + n X l = ( n + )! i ! x ( yx ) n + − i y . ( . )Taking γ = (cid:0) y + , (cid:1) ∈ Hom A e ( A ⊗ k { x , y } ⊗ A , A ) , gives d ( γ ) (cid:16) ⊗ x ⊗ (cid:17) = xy + + n X l = n ! l ! ( yx ) n + − l y , d ( γ ) (cid:16) ⊗ y x ⊗ (cid:17) = − + − n X l = ( n + )! l ! x ( yx ) n + − l y .With this coboundary we delete the term + from ( . ) and after removing the terms of shape x ( yx ) b + y from the second coordinate, we get that h s , u i = h s , u i − ( γ ) = (cid:16) ( n − ) u + + + (cid:17) .To obtain the value of s ◦ u on ⊗ x ⊗ we use Lemma . . The other term is u (cid:0) γ ( ⊗ x ⊗ ) (cid:1) ,and after some direct computations the result is u (cid:16) γ ( ⊗ x ⊗ ) (cid:17) = − + + n + X l = n ! l ! (cid:16) n + + + l − l (cid:17) ( yx ) n + − l y .Combining both results, h s , u i (cid:16) ⊗ x ⊗ (cid:17) = n + X l = n !( − − − ) l ! ( yx ) n + − l y + ( n − ) y + + + . oreover h s , u i (cid:16) ⊗ y x ⊗ (cid:17) = s (cid:16) − y + (cid:17) − u (cid:16) ⊗ y x ⊗ + ⊗ y x ⊗ y + ⊗ y x ⊗ yx + ⊗ y x ⊗ y + ⊗ y x ⊗ yxy + ⊗ y x ⊗ ( yx ) − ⊗ y x ⊗ − ⊗ y x ⊗ y − ⊗ y x ⊗ yx (cid:17) = −( + ) y + + + + + + + x + + + + xy + + xyx − + − + − + x .The commutation rules and the elimination of the terms x ( yx ) b + y lead to the equality h s , u i (cid:16) ⊗ y x ⊗ (cid:17) = − ( n − ) y + .Setting now γ = (cid:0) y + , (cid:1) ∈ Hom A e ( A ⊗ k { x , y } ⊗ A , A ) , we get h s , u i (cid:16) ⊗ x ⊗ (cid:17) = (cid:16)h s , u i − ( γ ) (cid:17) (cid:16) ⊗ x ⊗ (cid:17) = ( n − ) n + X l = ( n + )! l ! ( yx ) n + − l y ; after deleting again terms of type x ( yx ) b + y of the second coordinate, the result is: h s , u i (cid:16) ⊗ y x ⊗ (cid:17) = (cid:16)h s , u i − ( γ ) (cid:17) (cid:16) ⊗ y x ⊗ (cid:17) = (cid:16) −( n − ) y + + + (cid:17) and the equality (cid:2) s , u (cid:3) = (cid:0) ( n − ) u + + + (cid:1) is thus proved.The formula (cid:2) s m , u (cid:3) = (cid:0) ( n − − ) u + m + m ( + ) t + m (cid:1) is obtained recursively on m ,as follows. Let m ≥ and n ≥ , using the Jacobi identity we get that h [ s , s m ] , u i = h s , h s m , u ii − h s m , h s , u ii = h s , ( n − − ) u + m + m ( + ) t + m i − h s m , ( n − ) u + + + i = ( n − − ) (cid:16) ( n + m − ) u + m + + + m + (cid:17) + ( + )( n + m − ) t + m + − ( n − ) (cid:16) ( n − ) u + m + + m ( + ) t + m + (cid:17) − ( n − m ) t + m + = ( m − )( n − − ) u + m + + ( m − )( m + )( + ) t + m + .On the other hand, we have (cid:2) [ s , s m ] , u (cid:3) = ( m − ) (cid:2) s m + , u (cid:3) , so h s m + , u i = ( n − − ) u + m + + ( m + )( + ) t + m + .Now, using once more the facts that for all ϕ ∈ H ( A , A ) , the bracket [ ϕ , −] is a derivation with respectto the cup product and that u ⌣ u = u ( p + ) n for all p ≥ , we are ready to prove by induction on p that h s m , u i = (cid:16) ( n − − p ) u + m + pm ( + ) t + m (cid:17) . he case p = has already been proved, we suppose now p ≥ : h s m , u i = h s m , u i ⌣ u ( p − ) n + u ⌣ h s m , u ( p − ) n i = (cid:16) (− − ) u + m ( + ) t (cid:17) ⌣ u ( p − ) n + ⌣ (cid:16) ( n − ( p − ) m − ( p − )) u ( p − ) n + m + ( p − ) m ( + ) t ( p − ) n + m (cid:17) = (cid:16) ( n − − p ) u + m + pm ( + ) t + m (cid:17) .The only brackets missing are those corresponding to the action of H ( A , A ) on the odd degreeHochschild cohomology spaces. Proposition . . For all p ≥ and m , n ≥ , H ( A , A ) acts on H + ( A , A ) as follows:(i) h s m , ν + i = (cid:16) n − ( + ) m − ( + p ) (cid:17) v + + m − ( + ) ω + + m − ,(ii) h s m , ω + i = (cid:16) n − − p (cid:17) ω + + m .Proof. For p = , we will not prove the corresponding formulas, since the verifications follow thesame lines of previous computations, while for p > 1 , the periodicity isomorphisms given by the cupproduct with u , and the fact that [ s m , −] is a derivation with respect to the cup product give theresult.We change the bases of the Hochschild cohomology spaces to make the description of the represen-tations clearer. From now on we will use the following notations: L m = s m / m + ; τ = t / n + ; µ = u / n + ; ν + = v + / n + ; ω + = w + / n + .Given p ≥ , let I be the ideal generated by { τ } n ≥ . We recall some facts about intermediateseries modules. As we have already mentioned, it is an important class amongst the simple Harish-Chandra modules of Vir [M]. Given a , b ∈ C , the intermediate series Vir-module V a , b is generated bya family { v n } n ∈ Z and action C · v n = and L s · v n = ( n + as + b ) v n + s .Since W + := Vir + ⊕ h is a Lie subalgebra of Vir, we have that V a , b is also a W + -module containing asubmodule V + a , b generated by { v n } n ≥ . The ideal I is isomorphic, as H ( A , A ) -Lie module, to theintermediate series representation V +−( − ) , − p .We know that if b / ∈ Z or a = , , then V a , b is simple [M]. This is not true for V + a , b because thesubspace generated by { v n } n ≥ k is a submodule of V + a , b , for all k ∈ N . Proposition . . For a , b ∈ N , the W + -module V +− a , − b is indecomposable.Proof. We claim that given v ∈ V + a , b , v = , and k ≥ , there exists m ≥ such that L m · v belongsto h v n i n ≥ k and L m · v = . For this, we write v as a sum of elements in the basis with non zerocoefficients: v = P ki = γ i v ϕ ( i ) , so that L m · v = k X i = γ i L m · v ϕ ( i ) = k X i = γ i ( ϕ ( i ) − am − b ) v ϕ ( i )+ m .The integer m can be chosen big enough to ensure that ϕ ( i ) − am − b = and ϕ ( i ) + m ≥ k for all i ,and our assertion is proved.Let S be a non trivial submodule of V +− a , − b . We have to prove that S is not a direct summandof V +− a , − b . The previous claim shows that it is sufficient to prove that there exists k ∈ N such that v n ∈ S , for all n ≥ k . Choosing a non zero element v of S and writing v = P ki = γ i v ϕ ( i ) as before, e aim to show that there exists j ∈ N such that v j ∈ S . If k = , then we are done. Suppose nowthat k > 1 and that the result holds for elements that are linear combinations of less than k elementsof the basis { v n } n ≥ . Suppose first that there exists m ≥ such that ϕ ( i ) − am − b = for some i , ≤ i ≤ k . In this case L m · v = k X j = γ j L m · v ϕ ( j ) = k X j = , j = i γ j ( ϕ ( j ) − am − b ) v ϕ ( j )+ m and we use the inductive hypothesis for L m · v ∈ S . In case ϕ ( i ) − am − b = for all i , we "shift" v ,acting by L s for some s so as to return to the previous situation: L m · L s · v = k X j = γ j ( ϕ ( j ) − am − b )( ϕ ( j ) + s − am − b ) v ϕ ( j )+ s + m .Since a , b ∈ N , the equation ϕ ( j ) + s = am + b has solutions j , m , s . Finally, since there exists j ∈ N such that v j ∈ S and L m · v j = ( j − am − b ) v j + m , it follows that there exists k ∈ N such that h v n i n ≥ k is a subspace of S . The proposition is thus proved.Our next result concerns isomorphisms between two of such modules. Proposition . . Given a , b , α , β ∈ N , if V +− a , − b is isomorphic to V +− α , − β as W + − modules , then a = α .Proof. We shall denote { v n } and { w n } the generators of V +− a , − b and V +− α , − β , respectively. Let f : V +− a , − b → V +− α , − β be an isomorphism of W + -modules. Since b ∈ N , v b ∈ V +− a , − b and we write f ( v b ) = P kj = γ j w ϕ ( j ) . Notice that L ˙ v b = , which implies = f ( L m · v b ) = L · f ( v b ) = k X j = γ j L · w ϕ ( j ) = k X j = γ j ( ϕ ( j ) − β ) w ϕ ( j )+ m .We conclude from this that there exists j , with ≤ j ≤ k such that ϕ ( j ) = β and γ j = for all j = j , so f ( v b ) = γw β . As a consequence, f ( L m · v b ) = L m · f ( v b ) = − αγmw β + m for all m ≥ . On the other hand, f ( L m · v b ) = f (− amv b + m ) = − amf ( v b + m ) .This means that there exists d ∈ C ∗ such that f ( v b + m ) = dw β + m . If we choose m = , then f ( v b + ) = dw β + . This implies that, for all s ≥ , f ( L s · v b + ) = L s · dw β + = d ( − αs ) w β + + s .Also f ( L s · v b + ) = ( − as ) dw β + + s and we conclude that α = a .We have already mentioned that the H ( A , A ) -module I is isomorphic to V +−( − ) , p . Let us nowintroduce for p ≥ the family of H ( A , A ) -modules I + = D ω + E n ≥ . Note that I + is anideal of H + ( A , A ) and it is isomorphic to V +− , p as H ( A , A ) -modules. The previous propositionallows us to prove that the representations we have obtained are pairwise non isomorphic. Corollary . . Given p , q ≥ , • the H ( A , A ) -modules I and I are isomorphic if and only if p = q , • The H ( A , A ) -modules H ( A , A ) / I ∼ = V +− , p and H ( A , A ) / I ∼ = V +− , q are isomorphic if andonly if p = q , • I is never isomorphic to H ( A , A ) / I ,and similarly for I + . Yoneda algebra
The augmentation map A → k sending x and y to makes k an A -bimodule. We will now compute theHochschild cohomology of A with coefficients in k , together with its cup product. Since Ext • A e ( A , k ) isisomorphic as algebra to Ext • A ( k , k ) [St], we will obtain the Yoneda algebra E ( A ) of A . For this, we aregoing to use again the projective resolution P • A of A as A -bimodule, to which we will apply the functorHom A e (− , k ) . Since P • A is the minimal resolution of the connected graded algebra A , the differentialsobtained in Hom A e ( P n A , k ) are all zero and hence H n ( A , k ) is isomorphic to Hom A e ( P n A , k ) for all n ≥ and we obtain the following description of the cohomology spaces. Proposition . . • The space H ( A , k ) is isomorphic to k with basis { e : ⊗ → } . • H ( A , k ) is -dimensional with basis (cid:8) η , ω (cid:9) defined by η ( ⊗ x ⊗ ) = , η ( ⊗ y ⊗ ) = , ω ( ⊗ x ⊗ ) = , ω ( ⊗ y ⊗ ) = . • For all n ≥ , the space H n ( A , k ) is -dimensional with basis { η n , ω n } defined by: η n ( ⊗ x n ⊗ ) = , η n ( ⊗ y x n − ⊗ ) = , ω n ( ⊗ x n ⊗ ) = , ω n ( ⊗ y x n − ⊗ ) = .The Hilbert series of ⊕ i ≥ H i ( A , k ) is h ( t ) = + P i ≥ t i = + t1 − t . We are interested in the cupproduct of E ( A ) . We will use again the comparison morphisms in order to compute it. Of course e ∈ H ( A , k ) is the unit of the cohomology algebra. We will prove that, denoting η := e , thesubalgebra ⊕ i ≥ k η i is a polynomial algebra in one variable and that the products amongst the ω i ’sare zero. More precisely: Theorem . . Using the notations of the previous proposition, we describe as follows the products amongst thegenerators of H • ( A , k ) : • η p ⌣ η q = η p + q , for all p , q ≥ , • ω p ⌣ ω q = , for all p , q ≥ , • ω ⌣ η q = = η q ⌣ ω , for all q ≥ , • ω p ⌣ η q = ω p + q = (− ) q η q ⌣ ω p , for all p ≥ and q ≥ .As a consequence, the algebra H • ( A , k ) is generated by (cid:8) e , η , ω , ω (cid:9) . Moreover, H • ( A , k ) is isomorphic tothe graded algebra k D η , ω , ω E . (cid:18)(cid:16) ω (cid:17) , (cid:16) ω (cid:17) , ω ω , ω ω , ω η , η ω , ω η + η ω (cid:19) . Proof.
We shall consider four cases and within each case we will compute the product of two genericelements. st case: H ( A , k ) ⌣ H ( A , k ) . Let ϕ , ψ ∈ H ( A , k ) , ϕ ⌣ ψ (cid:16) ⊗ x ⊗ (cid:17) = ϕg ⌣ ψg (cid:16) f (cid:16) ⊗ x ⊗ (cid:17)(cid:17) = ϕg ⌣ ψg (cid:16) ⊗ x ⊗ ⊗ (cid:17) = ϕg ( ⊗ x ⊗ ) ψg ( ⊗ x ⊗ ) = ϕ ( ⊗ x ⊗ ) ψ ( ⊗ x ⊗ ) , ϕ ⌣ ψ (cid:16) ⊗ y x ⊗ (cid:17) = ϕg ⌣ ψg (cid:16) f (cid:16) ⊗ y x ⊗ (cid:17)(cid:17) = ϕg ⌣ ψg (cid:16) y ⊗ y ⊗ x ⊗ + ⊗ y ⊗ yx ⊗ − x ⊗ y ⊗ y ⊗ − ⊗ x ⊗ y ⊗ − x ⊗ y ⊗ x ⊗ − ⊗ x ⊗ yx ⊗ (cid:17) y · ϕ ( ⊗ y ⊗ ) ψ ( ⊗ x ⊗ ) + ϕ ( ⊗ y ⊗ ) ψg ( ⊗ yx ⊗ )− x · ϕ ( ⊗ y ⊗ ) ψ ( ⊗ y ⊗ ) − ϕ ( ⊗ x ⊗ ) ψg ( ⊗ y ⊗ )− x · ϕ ( ⊗ y ⊗ ) ψ ( ⊗ x ⊗ ) − ϕ ( ⊗ x ⊗ ) ψg ( ⊗ yx ⊗ )= ϕ ( ⊗ y ⊗ ) ( y · ψ ( ⊗ x ⊗ ) + ψ ( ⊗ y ⊗ ) · x )− ϕ ( ⊗ x ⊗ ) ( y · ψ ( ⊗ y ⊗ ) + ψ ( ⊗ y ⊗ ) · y )− ϕ ( ⊗ x ⊗ ) ( y · ψ ( ⊗ x ⊗ ) + ψ ( ⊗ y ⊗ ) · x ) .If we look at the generators, we obtain that η ⌣ η = η and all other products are zero. nd case: H ( A , k ) ⌣ H p ( A , k ) , with p ≥ . Suppose that ϕ ∈ H ( A , k ) and ψ ∈ H p ( A , k ) , ϕ ⌣ ψ (cid:16) ⊗ x p + ⊗ (cid:17) = ϕg ⌣ ψg p (cid:16) f p + (cid:16) ⊗ x p + ⊗ (cid:17)(cid:17) = ϕg ⌣ ψg p (cid:16) ⊗ x ⊗ p + ⊗ (cid:17) = ϕg ( ⊗ x ⊗ ) ψg p (cid:0) ⊗ x ⊗ p ⊗ (cid:1) = ϕ ( ⊗ x ⊗ ) ψ ( ⊗ x p ⊗ ) , ϕ ⌣ ψ (cid:16) ⊗ y x p ⊗ (cid:17) = ϕg ⌣ ψg p ⊗ y ⊗ yx ⊗ x ⊗ p − ⊗ − ⊗ x ⊗ y ⊗ x ⊗ p − ⊗ − ⊗ x ⊗ yx ⊗ x ⊗ p − ⊗ + p − X i = (− ) i ⊗ x ⊗ + i ⊗ y ⊗ x ⊗ p − − i ⊗ + p − X i = (− ) i ⊗ x ⊗ + i ⊗ yx ⊗ x ⊗ p − − i ⊗ ! = ϕ ( ⊗ y ⊗ ) ψg p (cid:16) ⊗ yx ⊗ x ⊗ p − ⊗ (cid:17) − ϕ ( ⊗ x ⊗ ) ψg p (cid:16) ⊗ y ⊗ x ⊗ p − ⊗ (cid:17) − ϕ ( ⊗ x ⊗ ) ψg p ( ⊗ yx ⊗ x ⊗ p − ⊗ )+ p − X i = (− ) i ϕ ( ⊗ x ⊗ ) ψg p (cid:16) ⊗ x ⊗ + i ⊗ y ⊗ x ⊗ p − − i ⊗ (cid:17) + p − X i = (− ) i ϕ ( ⊗ x ⊗ ) ψg p (cid:16) ⊗ x ⊗ + i ⊗ yx ⊗ x ⊗ p − − i ⊗ (cid:17) = ϕ ( ⊗ y ⊗ ) ψ ( y ⊗ x p ⊗ ) − ϕ ( ⊗ x ⊗ ) ψ (cid:16) ⊗ y x p − ⊗ (cid:17) − ϕ ( ⊗ x ⊗ ) ψ ( y ⊗ x p ⊗ )+ p − X i = (− ) i ϕ ( ⊗ x ⊗ ) ψ ( ) + p − X i = (− ) i ϕ ( ⊗ x ⊗ ) ψ ( )= − ϕ ( ⊗ x ⊗ ) ψ ( ⊗ y x p − ⊗ ) .We conclude that η ⌣ η p = η p + , η ⌣ ω p = − ω p + and ω ⌣ η p = ω ⌣ ω p = . rd case: Given ψ ∈ H p ( A , k ) with p ≥ and ϕ ∈ H ( A , k ) , ψ ⌣ ϕ (cid:16) ⊗ x p + ⊗ (cid:17) = ψ ( ⊗ x p ⊗ ) ϕ ( ⊗ x ⊗ ) , ψ ⌣ ϕ (cid:16) ⊗ y x p ⊗ (cid:17) = ψ (cid:16) ⊗ y x p − ⊗ (cid:17) φ ( ⊗ x ⊗ )+ (− ) p − ψ ( ⊗ x p ⊗ ) (cid:16) ψg (cid:16) ⊗ y ⊗ (cid:17) + ψg ( ⊗ yx ⊗ ) (cid:17) = ψ (cid:16) ⊗ y x p − ⊗ (cid:17) φ ( ⊗ x ⊗ ) .As a consequence, η p ⌣ η = η p + , ω p ⌣ η = ω p + and η p ⌣ ω = ω p ⌣ ω = . th case: Given ϕ ∈ H q ( A , k ) and ψ ∈ H p ( A , k ) with p , q ≥ , ϕ ⌣ ψ (cid:0) ⊗ x q + p ⊗ (cid:1) = ϕ ( ⊗ x q ⊗ ) ψ ( ⊗ x p ⊗ ) , ϕ ⌣ ψ (cid:16) ⊗ y x q + p − ⊗ (cid:17) = ϕ (cid:16) ⊗ y x q − ⊗ (cid:17) ψ ( ⊗ x p ⊗ )+ (− ) q − ϕ ( ⊗ x p ⊗ ) ψ (cid:16) ⊗ y x q − ⊗ (cid:17) .Thus, we obtain the equalities η q ⌣ η p = η p + q , ω p ⌣ η q = ω p + q = (− ) q η q ⌣ ω p , ω q ⌣ ω p = .Therefore, there is a well defined injective morphism of k -algebras from the algebra in the statementof the theorem to E ( A ) . Since they have the same Hilbert series, the last assertion is now clear.Notice that ( H • ( A , k ) , ⌣ ) is not a graded commutative algebra and in particular, ( H • ( A , k ) , ⌣ ) isnot a subalgebra of ( H • ( A , A ) , ⌣ ) . It is proved in [FS, Theorem . ] that if ( B , ∆ , ε , µ , η ) is a Hopf k -algebra, then H • ( B , k ) is isomorphic to a subalgebra of H • ( B , B ) and as a consequence H • ( B , k ) isgraded commutative. For a Nichols algebra the proof fails since the procedure to obtain a cocycle in C • ( B , B ) from a cocycle in C • ( B , k ) does not provide an actual cocycle if ∆ (cid:16) bb ′ (cid:17) = X b ( ) b ′ ( ) ⊗ b ( ) b ′ ( ) .The algebra A is not N -Koszul. This can be deduced from the minimal projective resolution of k as A -module. There is a generalization of the notion of N -Koszul algebra: the notion of K -algebra, see[CaSh]. Corollary . . The algebra A is K .Proof. The algebra H • ( A , k ) is generated over H ( A , k ) in degrees and . The Yoneda algebra of A kZ In this section we will describe the Yoneda algebra of the bosonization A kZ of the super Jordan plane.We recall that A is a kZ -module algebra, where the action of kZ on A corresponds to the braiding c of V (− , ) . Identifying kZ with the k -algebra of Laurent polynomials k (cid:2) t , t − (cid:3) . The action is definedby t · x = − x ; t · y = − y + x .Our main tool is Grothendieck’s spectral sequence of the derived functors of the composition of twofunctors [G] in Stefan’s version [St], which is multiplicative, see for example [GK]. The spectral se-quence is E p , q2 = H p ( Z , H q ( A , k )) ⇒ H p + q ( A kZ , k ) .The action of kZ on H • ( A , k ) is defined at the complex level as follows, given ϕ ∈ Hom A e ( A ⊗ n , k ) , ( t · ϕ ) ( a ⊗ . . . ⊗ a n ) = tϕ ( t − a ⊗ . . . ⊗ t − a n ) ,where kZ acts on k via the counit map ε : kZ → k .It is well known that the complex / / kZ − t · / / kZ ε / / k / / ( . )provides a projective kZ -resolution of k , and that for any kZ module M H ( Z , M ) ∼ = M Z , H ( Z , M ) ∼ = M / ( m − t · m : m ∈ M ) = M Z , H n ( Z , M ) = , for n ≥ . he filtration we are using to compute the spectral sequence is the second filtration, see [CE, ChapterXV, p. ], so the shape of the second page is H ( Z , H ( A , k )) ❴❴❴✤✤✤ H ( Z , H ( A , k )) ❴❴❴✤✤✤ H ( Z , H ( A , k )) ✤✤✤ ❴❴❴ H ( Z , H ( A , k )) ❴❴❴✤✤✤ H ( Z , H ( A , k )) ❴❴❴✤✤✤ H ( Z , H ( A , k )) ✤✤✤ ❴❴❴ We have thus, in this first quadrant spectral sequence, only two non trivial rows, and the differential d : E p , q2 → E p − , q + can only be non trivial when p = . Moreover, there is a five term exact sequence / / E , / / H ( A kZ , k ) / / E ,
02 d / / E , / / H ( A kZ , k ) . ( . )Due to the shape of the spectral sequence, it will collapse at E • , • .We need to describe H i ( A , k ) Z and H i ( A , k ) Z for all i ≥ . Our computation of H • ( A , k ) hasbeen obtained using the minimal resolution P • A of A as A -bimodule, so we need to use again thecomparison maps f • and g • of Section , more precisely, given ϕ ∈ Hom A e ( P q A , k ) , the action of t on ϕ is t · ϕ = ( t · ϕg q ) f q . It is straightforward to verify that the action is t · e = e , t · η = − η , t · ω = − η − ω , t · η q = (− ) q η q , t · ω q = (− ) q + ω q .As a consequence, we deduce that: E , = h e i , E , = h e i , E , = , E , = , E , = D η E , E , = D η E , for all k ≥ , E , + = D ω + E , E , + = D ω + E , for all k ≥ .Since there are only two non trivial rows, the multiplicative structure of H • ( Z , H • ( A , k )) is inducedby the cup product in E ( A ) . More precisely, the product, that we will denote · , is just the restrictionto the Z -invariants on the -th row, and it is the action of H • ( A , k ) on the coinvariants when one ofthe elements is in the -th row and the other one belongs to the first row, and it is null - by degreearguments - when both of them belong to the first row. We need the products amongst the generatorsof each space. They are as follows: • e · ϕ = ϕ for all ϕ ∈ H • ( Z , H • ( A , k )) , • e · ϕ = ϕ , for all ϕ ∈ E ( A ) Z and e · ϕ = , for all ϕ ∈ E ( A ) Z , • η · η ′ = η ( k + k ′ ) , • η · ω ′ + = ω ( k + k ′ )+ = ω ′ + · η , • ω + · ω ′ + = , • η · η ′ = η ( k + k ′ ) = η · η ′ , • η · ω ′ + = η · ω ′ + = ω ( k + k ′ )+ = ω ′ + · η = ω ′ + · η , • ω + · ω ′ + = = ω + · ω ′ + . ϕ · φ = for all ϕ , φ ∈ E ( A ) Z .The exact sequence ( . ) is / / / / H ( A kZ , k ) / / E ,
02 d / / E , / / H ( A kZ , k ) .Since both E , and E , are one dimensional, d is either zero or an isomorphism, and this willdepend on whether H ( A kZ , k ) is zero or not. We will compute this last space directly, using that itis isomorphic to Ext kZ ( k , k ) , that is, the space of classes of isomorphisms of -extensions of k by k .As vector spaces, every extension will be as follows: / / k i / / k ⊕ k j / / k / / ✤ / / ( , ) The algebra A kZ is graded with generators x , y and t , with x and y in degree and t in degree , so x and y act trivially on ( , ) and ( , ) , while the action of t sends ( , ) → ( , τ ) for some τ and ( , ) → ( , ) . Suppose that s : k → k ⊕ k is a splitting of j , that is s ( ) = ( a , ) , and t · ( a , ) = ( a , aτ + ) . Allthe relations are respected and we have thus nontrivial extensions, indexed by the non-zero elementsof k . As a consequence H ( A kZ , k ) ∼ = k and the map H ( A kZ , k ) → E , is a monomorphism. Wealready know that dim k E , = ; therefore d : E , → E , is zero. The spectral sequence beingmultiplicative and the description of E , j2 , for j ≥ , allow to conclude that d : E , j2 → E , j + is zerofor all j , hence E • , • = E • , • ∞ . Theorem . . . The Hochschild cohomology of A kZ with coefficients in k is as follows: H i ( A kZ , k ) = h e i , if i = , h e i , if i = , (cid:10) η (cid:11) , if i = , D η , ω + E , if i = + , k > 0 D η , ω − E , if i = , k > 1 . . The Yoneda algebra E ( A kZ ) = ⊕ i ≥ H i ( A kZ , k ) is the k -algebra generated by e , η , ω , where deg ( e ) = , deg ( η ) = , deg ( ω ) = . It is graded commutative, summarizing E ( A kZ ) ∼ = k h η i ⊗ Λ ( ω , e ) . In particular, it is finitely generated.Proof.
It follows from the description of the spectral sequence and the fact that it is multiplicative.
Corollary . . The algebra A kZ is not K . Remark . . A kZ is isomorphic as a graded algebra to gr H , where H is the Hopf algebra described in [AAH ]and the filtration considered is the coradical filtration. Adapting the methods of [MPSW, Theorem . ], it ispossible to prove that the Yoneda algebra of H is also finitely generated. References [AAH] Andruskiewitsch, Nicolás; Angiono, Iván; Heckenberger, István.
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Buenos Aires, Argentina [email protected]
A.S.:Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires,Ciudad Universitaria, Pabellón I,
Buenos Aires, Argentina; andIMAS, UBA-CONICET, Consejo Nacional de Investigaciones Científicas y Técnicas, Argentina [email protected]@dm.uba.ar