Cyclic homology of algebras of global dimension at most two
aa r X i v : . [ m a t h . K T ] N ov Cyclic homology of algebras of globaldimension at most two
Seminar on Cyclic Homology, November 7, 2017, held atSTOCKHOLM UNIVERSITYbyClas L¨ofwall
Abstract
We study graded connected algebras over a field of characteristiczero and give an explicit formula for the cyclic homology of a tensoralgebra. By means of a slightly new definition of David Anick’s notion”strongly free” we are able to prove that cyclic homology of an algebraof global dimension two is zero in homological degree greater than oneand is zero also in homological degree equal to one in case the relationsare monomials. We give also explicit formulas for the cyclic homologyof a tensor algebra modulo one symmetric quadratic form.
We will study N × Z -graded connected algebras, A = k ⊕ I where k is a fieldof characteristic zero, I = ⊕ q ≥ ,ǫ ∈ Z I q,ǫ , | I q,ǫ | < ∞ , I q,ǫ · I q ′ ,ǫ ′ ⊆ I q + q ′ ,ǫ + ǫ ′ . Weuse | a | as notation for the Z -degree which is called the sign degree , the firstdegree is called the weight . The series for I is defined as I ( z, y ) = X q ≥ ,ǫ ∈ Z | I q,ǫ | ∈ Z [[ z ]][ y ] / ( y − Z [[ z ]][ y ] / ( y −
1) is isomorphic to the ring of Z + × Z -graded locallyfinite dimensional vector spaces with ⊕ and ⊗ as operations and with formaladditive inverses adjoined.We use the same notation for a space and its series. Example T ( V ) · V + 1 = T ( V ) implies that T ( V ) = 1 / (1 − V )We have HC ( I ) = HC ( A ) = I/ [ I, I ] where [
I, I ] is the subspace of I (not ideal!) generated by all commutators ab − ( − | a || b | ba . Taking modulo[ I, I ] is the same as allowing circular permutation of monomials, e.g., abcd = − dabc = cdab = − bcda in I/ [ I, I ] if a, b, c, d are odd elements in I .The Hochschild homology HH( A ) is obtained by tensoring the two-sidedbar resolution ( A ⊗ I ⊗ n ⊗ A, b ′ ) of A as an A -bimodule with the bimodule A .Thus HH( A ) may be seen as the homology of the complex ( n ≥ I ⊗ = k )( I ⊗ n ⊕ I ⊗ n +1 , (cid:18) − b ′ − t b (cid:19) )This is the mapping cone in positive degrees of the map of complexes (1 − t ) :( I ⊗ n , b ′ ) → ( I ⊗ n , b ) where n ≥ . . . −→ I ⊗ b ′ −→ I ⊗ b ′ −→ I ⊗ b ′ −→ I −→ k (1 − t ) ↓ (1 − t ) ↓ (1 − t ) ↓ ↓ . . . −→ I ⊗ b −→ I ⊗ b −→ I ⊗ b −→ I Taking homology first vertically and then horizontally gives a long exactsequence by spectral sequence theory. There is only one differential d . . . . → H n +2 (coker(1 − t )) d −−→ H n (ker(1 − t )) → HH n ( A ) → H n +1 (coker(1 − t )) d −−→ H n − (ker(1 − t )) → HH n − ( A ) → H n (coker(1 − t )) → . . . The homology of the complex C t ( A ) = (coker(1 − t ) , ¯ b ) is by definition HC( A ),with a degree shift, so HC n ( A ) = H n +1 ((coker(1 − t ) , ¯ b )). The norm mapdefines an isomorphism P t i : (coker(1 − t ) , ¯ b ) → (ker(1 − t ) , b ′ ) and henceHC( A ) may also be seen as the homology of (ker(1 − t ) , b ′ ).In our case the differential d = 0, so the long exact sequence is short:0 −→ HC n − ( A ) −→ HH n ( A ) −→ HC n ( A ) −→ n > n = 0 we have HH ( A ) = k ⊕ HC ( A ) = k ⊕ I/ [ I, I ].The series for HH( A ) and HC( A ) have three variables x for the homo-logical degree, z for the weight and y for the sign degree. The series belongto the ring Z [[ x, z ]][ y ] / ( y − n − ( A ) → HH n ( A ) has bothhomological degree and sign degree equal to one, hence we have the followingrelation HH( A )( x, z, y ) = 1 + (1 + xy ) HC( A )( x, z, y ) (1)where HH( A )( x, z, y ) = X q ≥ n ≥ ,ǫ =0 , | HH n,q,ǫ ( A ) | x n z q y ǫ HC( A )( x, z, y ) = X q − ≥ n ≥ ,ǫ =0 , | HC n,q,ǫ ( A ) | x n z q y ǫ . For commutative algebras
A, B we haveHH( A ⊗ B ) = HH( A ) ⊗ HH( B ) (2) There is a subcomplex of the two-sided bar resolution of the form A ⊗ ( A ! ) ∗ ⊗ A , where A ! is the Koszul dual of A . The differential in this subcomplex hasonly two terms with no middle terms. In fact, ( A ! ) ∗ in homological degree n is exactly the subspace of I ⊗ n consisting of the elements for which all middleterms of the differential are zero. Thus( A ! ) ∗ = n − \ i =0 I ⊗ i ⊗ ker( µ ) ⊗ I ⊗ ( n − − i )1 and where µ is multiplication I ⊗ I → I .If A is Koszul, then this complex is exact (the opposite is also true) andwe may use this to compute the Hochschild homology as the homology of A ⊗ ( A ! ) ∗ where the differential again has two terms but now one of them iscirculating. Taking the vector space dual of the complex we get the complex A ∗ ⊗ A ! with isomorphic homology and since ( A ! ) ! = A , this is the complexwe get if we compute the Hochschild homology of A ! with the same method.Hence we get that HH( A ) and HH( A ! ) are in a sense dual. One has to becareful with the gradings, the homological degree of A r ⊗ (( A ! ) ∗ ) p is p andthe weight is p + r , while in ( A ∗ ) r ⊗ ( A ! ) p the homological degree is r and the3eight is p + r . The sign degree is preserved in passing from one complex tothe other. We thus get the following relationHH p,q,ǫ ( A ! ) ∼ = HH q − p,q,ǫ ( A ) (3)yielding the following formula of Hilbert seriesHH( A ! )( x, z, y ) = HH( A )(1 /x, xz, y ) . (4)We may obtain a similar formula for HC using (1),HC( A ! )( x, z, y ) = yx HC( A )(1 /x, xz, y ) , (5)which gives the followingHC p,q,ǫ ( A ! ) ∼ = HC q − p − ,q,ǫ +1 ( A ) . (6)As an application, we get an easy proof of the Hochschild-Kostant-Rosenbergtheorem in the graded case. Let A = k [ x , . . . , x n ] be a polynomial ring in n even variables of weight one. Then it is easy to see that the differential inthe two-sided Koszul complex A ⊗ ( A ! ) ∗ is zero giving that, as a vector space,HH( A ) = k [ x , . . . , x n ] ⊗ k [¯ x , . . . , ¯ x n ]where ¯ x i are odd anti-commutative variables of homological degree and weightone. This gives the following series in three variables,HH( A )( x, z, y ) = (1 + yxz ) n / (1 − z ) n . (7)Using (4) we get HH( A ! )( x, z, y ) = (1 + yz ) n / (1 − xz ) n which (as expected) is the series for the exterior algebra on n odd variablesof weight one tensor with the polynomial algebra on n even variables ofhomological degree and weight one.Putting p = 0 in (3) we get ⊕ q HH qq ( A ) ∼ = A ! / [ A ! , A ! ]. This is in fact truein general and may be compared with the result that ⊕ q Ext qq ( A ) = A ! . Proposition 1.1
Let A be a k -algebra. Then ⊕ q HH qq ( A ) ∼ = ( A ! / [ A ! , A ! ]) ∗ and if A is commutative, the isomorphism is an isomorphism as algebras,where the right hand side is an algebra since [ A ! , A ! ] is a co-ideal in the Hopfalgebra A ! . ✷ .3 Free models A free model of a k -algebra A is a differential graded algebra ( T ( V ) , d ) to-gether with a surjective map ( T ( V ) , d ) → A which induces an isomorphismin homology. Here V has homological degree ≥ ≥ d is a derivation of homological degree − A has differential zero and is concentrated inhomological degree zero. There is always a free model for A (even a minimalone, but that is not important for us). Example A = k h x , x i / ( x x ). Then a free model is (as follows from Propo-sition 3.3) k h x , x , S ; dx = dx = 0 , dS = x x i The map to A maps S to 0.There is also a notion of relative model for a map f : A → B . This isa differential graded algebra which is a free extension of A , ( A ∗ T ( V ) , d ),together with a surjective map ( A ∗ T ( V ) , d ) → B inducing an isomorphismin homology and which equals f when it is restricted to A . We will use thenotation A h V i for A ∗ T ( V ).The relative cyclic homology, HC( A → B ), of a map f : A → B isdefined as the homology of the mapping cone for the map of complexesC t ( f ) : C t ( A ) → C t ( B ). The exact sequence of complexes gives rise to a longexact sequence (where s denotes shift of sign degree), . . . → HC n ( A ) → HC n ( B ) → HC n ( A → B ) → s HC n − ( A ) → s HC n − ( B ) → s HC n − ( A → B ) → HC n − ( A ) → . . . (8)Free models may be used to compute cyclic homology (and also Tor A ( k, k )and Ext A ( k, k )). Suppose ( T ( V ) , d ) is a model for A and that ( A h V i , d )is a relative model for a map A → B , then we have the following result,which may be proven by a spectral sequence argument, using the fact thatHC n ( T ( V )) = 0 for n > Proposition 1.2
HC( A ) = H( T ( V ) / ([ T ( V ) , T ( V )] + k ) , ¯ d )HC( A → B ) = H( A h V i / ([ A h V i , A h V i ] + A ) , ¯ d ) where ¯ d is the induced differential ( d maps a commutator to a linear combi-nation of commutators). A )and HC( A ! ) when A is Koszul. Indeed, we have that the cobar constructionfor A , ( T ( I ∗ ) , d ), where d is the derivation which extends the dual of themultiplication map I ⊗ I → I , is a (minimal) model for A ! when A is Koszul.From this the relation (6) may be deduced.The following result by Feigin and Tsygan, see [2], is fundamental for us. Proposition 1.3
Let A och B be graded k -algebras. Then HC n ( A ∗ B ) = HC n ( A ) ⊕ HC n ( B ) for n > ( A ∗ B ) = HC ( A ) ⊕ HC ( B ) ⊕ HC ( T ( A + ⊗ B + )) Proof.
Let R A and R B be models for A and B . Then R A ∗ R B is a modelfor A ∗ B . This follows from K¨unneth’s formula and since R A ∗ R B is free asan algebra. We have( R A ∗ R B ) + = R + A ⊕ R + B ⊕ ( R + A ⊗ R + B ) ⊕ ( R + B ⊗ R + A ) ⊕ ( R + A ⊗ R + B ⊗ R + A ) ⊕ . . . Modulo commutators we get( R A ∗ R B ) + / [ , ] = R + A / [ , ] ⊕ R + B / [ , ] ⊕ ( R + A ⊗ R + B ) ⊕ (( R + A ⊗ R + B ) ⊗ ( R + A ⊗ R + B )) t ⊕ . . . where t indicates that the two blocks ( R + A ⊗ R + B ) may change place. Theclaim now follows taking homology and using K¨unneth’s formula and thatthe functor coker(1 − t ) : ( · ) ⊗ n → ( · ) ⊗ n ) is exact, since it is right exact andisomorphic to the left exact functor ker(1 − t ) : ( · ) ⊗ n → ( · ) ⊗ n ). ✷ In this section we will get a formula for the series ofHC ( T ( V )) = T ( V ) + / [ T ( V ) , T ( V )]for a Z + × Z -graded locally finite dimensional vector space V . On the waywe also get a formula for the series of a Lie algebra given the series for itsenveloping algebra.Let P denote the additive group of formal power series X n ≥ a n z n + y X n ≥ b n z n ∈ Z [[ z ]][ y ] / ( y − U denote the multiplicative group 1+ P. The inverse of 1 − P n ≥ a n z n − y P n ≥ b n z n is1 + ( X n ≥ a n z n + y X n ≥ b n z n ) + ( X n ≥ a n z n + y X n ≥ b n z n ) + . . . (observe that any series in P may be inserted in any formal power series inone variable to get a new well-defined series).A topology on P and U is defined as follows. We say that f n → f if,for any r, the coefficients in f n and f are the same up to z r and yz r for bigenough n .Instead of assuming that the coefficients in the formal power series areintegers, we will also consider the case when the coefficients are rationalnumbers. In this case we write P Q and U Q for the corresponding additiveand multiplicative groups. Definition 2.1
A continuous group isomorphism P → U (or P Q → U Q ) iscalled an “exponential” and a continuous isomorphism U → P (or U Q → P Q )is called a “logarithm”. Proposition 2.2 exp( X ) = X k ≥ X k k ! , X ∈ P Q , is an exponential P Q → U Q log( X ) = − X k ≥ (1 − X ) k k , X ∈ U Q , is a logarithm U Q → P Q S ( X ) = ∞ Y k =1 (1 + yz k ) b k (1 − z k ) a k , X = ∞ X k =1 a k z k + y ∞ X k =1 b k z k is an exponential P → U Proof.
The usual proof of e x + y = e x · e y using formal power series depends onlyon the binomial theorem and that x and y are commuting symbols. Hencealso exp( X + Y ) = exp( X ) · exp( Y ) , since P Q is commutative. The identitiesexp(log( X )) = X and log(exp( X )) = X hold, since they hold for smallenough convergent series with complex coefficients. Hence exp and log arebijective and log is also a homomorphism and clearly they are continuous.The series S ( X ) is the series for the symmetric algebra of a graded andweighted vector space X = ⊕ i ≥ X i . (By means of the Taylor expansion of(1 + z ) α the operator S may be extended to a transformation P Q → U Q . )7his gives indeed an exponential P → U . It is evident that S is ahomomorphism and that it is continuous. To prove bijectivity, suppose f ∈ U and suppose inductively that there are uniquely determined a , b , . . . , a k − , b k − ∈ Z such that k − Y j =1 (1 + yz j ) b j (1 − z j ) a j ≡ f ( z, y ) mod z k Hence k − Y j =1 (1 − z j ) a j (1 + yz j ) b j f ( z, y ) = 1 + a k z k + b k yz k + higher termsand hence(1 − z k ) a (1 + yz k ) b k − Y j =1 (1 − z j ) a j (1 + yz j ) b j f ( z, y ) ≡ z k +1 ⇔ a = a k and b = b k ✷ By means of “log” we get a lot of other logarithms:
Example
Let c = { c k } ∞ k =1 be any sequence of rational numbers with c = 0 . ThenLog c ( X ) = ∞ X k =1 c k log( X ( z k , ( − k +1 y k )) , X ∈ U Q is a logarithm U Q → P Q . The substitution ( − k +1 y k for y in the formula will be useful below. Toget a logarithm we could also have used just y instead. The proof of thehomomorphism law is evident. Definition 2.3
The inverse of S is a logarithm, Lie( X ) , which gives theseries for the Lie algebra whose enveloping algebra has series X. In particular Lie( − X ) gives the series for the free Lie algebra on a gradedand weighted vector space with series X. Proposition 2.4
Lie( X ) = ∞ X k =1 µ ( k ) k log( X ( z k , ( − k +1 y k )) for all X ∈ U. roof. We will determine c, such that Lie = Log c . In fact, it is enough todetermine c such that Lie( X ) = Log c ( X ) for X = 1 − z and X = 1 + yz. This follows from the fact that S is surjective and both Lie and Log c arelogarithms which commute with the substitution z → z k . Now Lie( − z ) = z and Lie(1 + yz ) = yz and − Log c (1 − z ) = − ∞ X k =1 c k log(1 − z k ) = ∞ X j,k =1 c k z kj j = ∞ X n =1 ( X k | n kc k ) z n n (9)Hence, Lie(1 − z ) = Log c (1 − z ) if c = 1 and P k | n kc k = 0 for n > . M¨obiusinversion formula gives that the solution is given by kc k = µ ( k ) . By definitionof Log c we haveLog c (1 + yz ) = ∞ X k =1 c k log(1 + ( − k +1 y k z k )= ∞ X k =1 c k log(1 − ( − yz ) k ) = Log c (1 − z ) (cid:12)(cid:12) z = − yz (10)With c chosen as above, we hence get that Log c (1 + yz ) = Lie(1 − z ) (cid:12)(cid:12) z = − yz = yz = Lie(1 + yz ) and the result follows. ✷ From the proof above it follows that any logarithm L is of the form Log c for a unique c if the following conditions hold: • L commutes with the substitution z → z k for any k • L (1 + yz ) = L (1 − z ) (cid:12)(cid:12) z = − yz If we apply the formula for Lie on 1 − dz we get Serre’s formula for theseries of the free Lie algebra on d even generators: ∞ X n =1 ( 1 n X j | n µ ( nj ) d j ) z n . We now turn to the series for HC ( T ( V )). Let V ( z, y ) ∈ P be the seriesfor V and let hc( V ) denote the series for HC ( T ( V )). Here we use the samename for the series and the corresponding vector space.By Proposition 1.3 we have the following rule for hc, where we assumethat the coefficients in the series are natural numbershc( V + V ) = hc( V ) + hc( V ) + hc( V · V / (1 − V ) / (1 − V )) . (11)9ow define the transformation hcfree : P → P ashcfree( V )( z, y ) = − ∞ X k =1 φ ( k ) k log(1 − V ( z k , ( − k +1 y k )) , where φ is Euler’s φ -function. Observe that this is similar to the formula forLie(1 − V ) above, the difference is only that the M¨obius function has beenreplaced by Euler’s φ -function. The transformation hcfree also satifies (11),which follows from the algebraic identity1 − V − V = (1 − V )(1 − V )(1 − V · V / (1 − V ) / (1 − V )) . We will now first prove that hc( V ) = hcfree( V ) when V is one-dimensional.Since both transformations commute with z → z k , it is enough to considerthe two cases V = z and V = yz . For a free algebra on one even generatorHC is one-dimensional in each degree ≥
1, hence hc( z ) = z/ (1 − z ). Usingthe computation in (9), we see that the series hcfree( z ) ishcfree( z ) = ∞ X n =1 z n n X k | n φ ( k ) . Since P k | n φ ( k ) = n for all n , we get that hc( z ) = hcfree( z ). For a freealgebra on one odd generator HC is zero in each even degree and one-dimensional in each odd degree ≥
1, hence hc( yz ) = yz/ (1 − z ). Since(1 + yz )(1 − yz ) = 1 − z and hcfree( V ) is logarithmic in 1 − V , we havehcfree( yz ) = hcfree( z ) − hcfree( − yz ) = hcfree( z ) (cid:12)(cid:12) z = z − hcfree( − yz ) . Using (10), we get that hcfree( − yz ) = hcfree( z ) (cid:12)(cid:12) z = − yz = − yz/ (1 + yz ) andhence hcfree( yz ) = z / (1 − z ) + yz/ (1 + yz ) = yz/ (1 − z ) = hc( yz ) . The proof that hc( V ) = hcfree( V ) for all V ∈ P with natural numbers ascoefficients is now completed by a double induction, using that both transfor-mations satisfy (11). Indeed, fixing a degree k we claim that the coefficientsfor z k and yz k agree in hc( V ) and hcfree( V ). For V = P ∞ r =1 a r z r + b r yz r defineord( V ) = min { r ; a r + b r = 0 } (ord(0) = ∞ ). We have that ord(hc( V )) ≥ ord( V ) and ord(hcfree( V )) ≥ ord( V ). The claim is true for all V withord( V ) > k . Suppose the claim is true for all V such that ord( V ) > r andlet V = a r z r + b r yz r + W where ord( W ) > r . We now prove the claim for V by induction over a r + b r . Suppose the claim is true if a r + b r < N andsuppose a r + b r = N ≥
1. Let V = z r if a r > V = yz r otherwise.Then the claim is true for V by the above and for V − V by induction. Sinceord( V ( V − V )) > r the claim is also true for V ( V − V ) / (1 − V ) / (1 − V + V )and hence, by (11), the claim is true for V .10 Free algebras modulo one relation
Definition 3.1
Suppose A is a differential graded k -algebra (DGA for short)and ω ∈ A is a nonzero homogeneous cycle. A new DGA, A ω , is defined inthe following way. Let s ω A denote the graded vector space which is obtainedfrom A by raising all degrees (sign degree, weight and homological degree) withthe corresponding degrees for ω . The elements in s ω A are written s ω a for ahomogeneous element a ∈ A (even a = 1 is possible) and s ω is considered asa linear operator A → A ω . Then A ω = k ⊕ s ω A as a vector space d ( s ω a ) = ( − ω s ω d ( a )( s ω a ) · ( s ω a ) = s ω ( a ωa ) . Now suppose A is just a k -algebra without differential. Choose a homoge-neous k -section f of the natural map A → A/AωA (with f (1) = 1) anddefine an algebra homomorphism F : T ( s ω ( A/AωA )) −→ A ω induced by s ω a s ω f ( a ) . We have F ( s ω a ⊗ · · · ⊗ s ω a n ) = s ω ( f ( a ) ωf ( a ) ω . . . ωf ( a n )) . Since A is generated as an algebra by ω and the image of f (proof by inductionover the weight), we get that F is a surjective homomorphism of algebras.The notion of strongly free was introduced by D. Anick in [1]. The slightlydifferent definition given below is however more suitable for our purposes. Definition 3.2
A homogeneous nonzero element ω in a k -algebra A is called strongly free if the map F above is an isomorphism. A sequence ω =( ω , ω , . . . ) of homogeneous elements in A is called a strongly free set in A if,for all i , ω i is strongly free in A modulo the ideal generated by { ω , . . . , ω i − } . The definition is independent of the choice of the section f , since a sur-jective graded linear map between two isomorphic graded vector spaces is anisomorphism. As always, we use the same notation of a vector space and itsseries (in two variables). Since F is always surjective, we get the inequality1 / (1 − ωB ) ≥ A ω = 1 + ωA B ≥ A/ (1 + ωA ) (12)may be deduced. Observe that the inequality of series C > C is positive and that A ≥ B , C ≥ AC ≥ BC .We have that equality holds in (12) iff ω is strongly free in A . The sameformula (and inequality) holds also in the case when ω is a strongly free setin A . Indeed, suppose ω = { ω , ω } is a strongly free set in A . Then theseries for A/AωA is A/ (1 + ω A ) / (1 + ω A/ (1 + ω A )) = A/ (1 + ω A + ω A ) . It follows from the inequality of series that any reordering of a strongly freeset ω is still a strongly free set and hence the use of the word set instead of sequence is appropriate.In particular, if ω is any set of homogeneous elements in T ( V ), then T ( V ) / h ω i ≥ / (1 − V + ω )with equality iff ω is a strongly free set in T ( V ). We will now in detail study a free extension with one variable, A → A h S ; dS = ω i , in particular the case when A is concentrated in homological degree zeroand hence S has homological degree one. The reason for introducing A ω inthe previous subsection is the following proposition, which also gives anothercharacterization of the property strongly free. Proposition 3.3
Let i : A → A h S ; dS = w i be a free extension of analgebra A , concentrated in homological degree zero, with a variable S killingthe homogeneous nonzero element ω ∈ A . Then, for n, q ≥ , ǫ ∈ Z H n ( A h S i ) = Tor A ω n +1 ( k, k )H n +1 ,q,ǫ (( A h S i / ([ A h S i , A h S i ] + A ) , d )) = HC n,q,ǫ +1 ( A ω ) . Proof.
The elements in A h S i can uniquely be represented as a Sa S . . . Sa n ,where a , . . . , a n ∈ A . The elements in ( A h S i / [ , ] + A, d ) can uniquely berepresented as a Sa S . . . a n S , where a , . . . , a n ∈ A . Consider the maps φ : A h S i → B( A ω ) of homological degree +1 (and weight equal to the weight of12 and sign degree | ω | + 1) and φ t : ( A h S i / [ , ] + A, d ) → C t ( A ω ) of homologicaldegree −
1, weight zero and sign degree one defined by φ ( a Sa S . . . Sa n ) = ( − n | w | ( s ω a | . . . | s ω a n ) φ t ( a Sa S . . . a n S ) = ( − n | w | ( s ω a ⊗ . . . ⊗ s ω a n ) / (1 − t ) . It is easy to verify that φ och φ t are isomorphisms for complexes. ✷ Remark 3.4
The proposition is true also in the case when A is a DGA (thesign for φ and φ t in the proof is changed to ( − p | w | where p is the homologicaldegree of the argument on the left hand side). Then the fact H( A ω ) = H( A ) ω may be used to prove that a set { ω i } is strongly free in A iff A h S i ; dS i = ω i i is acyclic. This could be compared with the result in commutative algebra thata sequence is regular iff the Koszul complex is acyclic. Since a k -algebra B is free iff Tor Bn ( k, k ) = 0 for n >
1, it follows fromProposition 3.3 that A h S ; dS = w i is acyclic iff ω is strongly free in A andin this case A h S ; dS = w i is a relative model for A → A/AωA . Proposition 3.5
Suppose A is a graded algebra and ω is strongly free in A .Put B = A/AωA and let s denote shift of sign degree. Then a ) HC n ( B ) ≃ HC n ( A ) for n ≥ b ) there is an exact sequence → s HC ( A ) −→ s HC ( B ) −→ HC ( A ω ) −→ HC ( A ) −→ HC ( B ) → c ) if HC ( A ) = 0 then s HC ( B ) ∼ = ( ωA ∩ [ A, A ]) / [ ωA, ωA ] Proof.
By the argument above and by Proposition 1.2 and Proposition 3.3,HC n ( A → B ) ∼ = s HC n − ( A ω ) and then the long exact sequence for cyclichomology of a map (8) gives . . . → HC n ( A ) → HC n ( B ) → s HC n − ( A ω ) → s HC n − ( A ) → s HC n − ( B ) → HC n − ( A ω ) → HC n − ( A ) → HC n − ( B ) → . . . . Since A ω is free, a) and b) follows. To prove c), we use the following exactsequence which holds for any homogeneous element ω ∈ A and is proved quiteelementary. 0 → ( ωA ∩ [ A, A ]) / [ ωA, ωA ] → ωA/ [ ωA, ωA ] → A + / [ A, A ] → A + / ( ωA + [ A, A ]) → ω is strongly free in A , it follows that ω is a non-zerodivisor and hencethe map s ω A → ωA given by s ω a ωa is an isomorphism of vector spaces.Under this map [ A ω , A ω ] is mapped to [ ωA, ωA ], since ( s ω a )( s ω b ) = s ω ( aωb )is mapped to ωaωb .Hence, in the sequence above, ωA/ [ ωA, ωA ] may be replaced by s ω A/ [ A ω , A ω ] = HC ( A ω ). Comparing the above sequence and the sequencein b) and using HC ( A ) = 0 we getHC ( B ) ∼ = s (( ωA ∩ [ A, A ]) / [ ωA, ωA ]) . ✷ Proposition 3.6
Suppose ω is a strongly free set in T ( V ) and put B = T ( V ) / ( ω ) . Then a ) HC n ( B ) = 0 for n ≥ b ) HC ( B ) = y HC ( B ) + hcfree( V − ω ) c ) if ω consists of monomials, then HC ( B ) = 0 Proof.
The proof is by induction over the size of ω , so suppose the claim istrue for A = T ( V ) / h α i and that ω is a strongly free element in A and put B = A/AωA . Then a) is true for B by Proposition 3.5. To prove b) we usethe same proposition to getHC ( B ) − y HC ( B ) = HC ( A ) − y HC ( A ) − HC ( A ω ) . Since A ω is free on s ω B and B has series 1 / (1 − V + α + ω ) by the above,we get by inductionHC ( B ) − y HC ( B ) = hcfree( V − α ) − hcfree( ω/ (1 − V + α + ω ))We now use that hcfree( V ) is logarithmic in 1 − V and the fact that(1 − V + α ) / (1 − ω/ (1 − V + α + ω )) = 1 − V + α + ω to get HC ( B ) − y HC ( B ) = hcfree( V − α − ω )which proofs the induction step for b). The induction start, ω = 0, for a),b) and c) is true, since HC n ( T ( V )) = 0 for n > ( T ( V )) =hcfree( V ) by definition of hcfree. 14o prove the induction step for c) we will use Proposition 3.5 and provethat ωA ∩ [ A, A ] ⊆ [ ωA, ωA ].Since A is a free algebra modulo monomials it has a fine grading wherethe graded components consist of the nonzero monomials in A . Consider thecoarser grading where the components (called cyclic components ) consist ofthe direct sum of the cyclic permutations of a given monomial. Then [ A, A ]respects this grading and hence a ∈ [ A, A ] iff each cyclic component of a isin [ A, A ].Suppose x is a nonzero element in ωA ∩ [ A, A ] lying in a cyclic component C and let ωa be a nonzero monomial term in x . We claim that all cyclicpermutations of ωa are nonzero, which implies that C has dimension equalto the length of ωa and hence the cyclic operator t circulates a basis for C .Here t is defined on a monomial x x . . . x n as t ( x x . . . x n ) = ( − | x | ( | x | + ···| x n | ) x x . . . x n x Since C ∩ [ A, A ] = im(1 − t ), the claim also implies that x = n − X i =0 c i t i ( ωa ) where n − X i =0 c i = 0 . Indeed, if the claim is not true, then there is an overlap of ω with a mono-mial in the set α which defines A or ω is a submonomial of some monomial in α or ω overlaps itself. In each case it is easy to construct a nonzero elementin ker( F ); e.g., if α α ∈ α , ω = α ω and α is nonzero as an element in A/AωA then F ( s ω ( ¯ α ) s ω (1)) = 0.Moreover, since ω is strongly free in A , the occurences of ω in ωa do notoverlap, so we may write ωa = ωa ωa . . . ωa r where for all i , a i does not contain ω as a submonomial. Since each monomialterm in x belongs to ωA we have ( n = 0): x = r − X i =0 d i t n i ( ωa ) where r − X i =0 d i = 0 and n i = length( ωa ωa . . . ωa i )(13)If bc is a monomial and length( b ) = m , then it is easy to prove by inductionover m that t m ( bc ) = ( − | b || c | cb i = 0 to r −
1, we have t n i ( ωa ) = ( − | b i || c i | c i b i where b i = ωa ωa . . . ωa i and c i = ωa i +1 a i +2 . . . ωa r .Since one may compute | b i || c i | + | b i +1 || c i +1 | to be | ωa i +1 | ( | b i | + | c i +1 | ) itfollows that t n i ( ωa ) − t n i +1 ( ωa ) = ( − | b i || c i | ( ωa i +1 c i +1 b i − ( − | ωa i +1 | ( | b i | + | c i +1 | ) c i +1 b i ωa i +1 )= ( − | b i || c i | [ ωa i +1 , c i +1 b i ] ∈ [ ωA, ωA ]By (13) we have x = r − X i =0 ( i X j =0 c j )( t n i ( ωa ) − t n i +1 ( ωa ))and it follows that x ∈ [ ωA, ωA ]. ✷ Observe that D. Anick has proved that a strongly free set of monomials isjust a set of monomials such that no two monomials have a common beginningand ending and no monomial is contained in another monomial. Moreover,if there is an order of the variables such that the leading monomials of a set ω of homogeneous polynomials yield a strongly free set of monomials, then ω is strongly free. As an application of the results in the previous section, we will compute thecyclic homology of a free algebra modulo some generic quadratic forms. Wecall some forms generic if the N coefficients are algebraically independent, orthat the coefficients defines a point in k N which lies in a non-empty Zariski-open subset of K N . We will assume that the variables are odd (in which casethe Koszul dual has even variables).Assume ω is a generic quadratic form in T ( V ) = k h T , . . . T n i with n ≥ ω is strongly free in T ( V ), since the Hilbert series is minimal for ageneric relation and there is an example of a strongly free form, namely ω = T T . By Proposition 3.6 we getHC ( T ( V ) / ( ω )) ≥ hcfree( nzy − z )But since HC ( T ( V ) / ( ω )) = T ( V ) / [ T ( V ) , T ( V )] + ( ω ) it follows that theseries for HC ( T ( V ) / ( ω )) is the minimal possible in the generic case. Sinceby Proposition 3.6, HC ( T ( V ) / ( T T ) = hcfree( nzy − z ), we get16 roposition 4.1 If ω is a generic quadratic form in T ( V ) = k h T , . . . T n i with n ≥ , then HC ( T ( V ) / ( ω )) = hcfree( nzy − z )HC n ( T ( V ) / ( ω )) = 0 for n > ✷ See also Proposition 4.5, which is a generalization.In the symmetric case we have the following
Proposition 4.2
Suppose ω is a generic symmetric quadratic form in n oddvariables of weight one, a , . . . , a n , where n ≥ , and let A = k h a , . . . , a n i and B = A/ ( ω ) . Then HC ( B ) = z + hcfree( nzy − z )HC ( B ) = yz HC n ( B ) = 0 for n > Remark 4.3
In fact, it is enough to assume that the symmetric form hasrank at least three in the proposition above.
Proof.
As above it follows that ω is strongly free since [ a , a ] is stronglyfree. Since ω ∈ [ A, A ], it follows that ω is a non-zero element in ωA ∩ [ A, A ] / [ ωA, ωA ] and hence by Proposition 3.5 HC ( B ) ≥ yz and by Propo-sition 3.6, HC ( B ) ≥ z + hcfree( nzy − z )Hence, it is enough to give an example of a B which attains the lower bound.We will consider B = k h a, b, c i / ([ a, b ]+ c ). We have that [ a, b ]+ c is stronglyfree in k h a, b, c i since ab is a strongly free leading monomial. It is enough toprove that HC ( B ) = z + hcfree(3 zy − z ), since then by Proposition 1.3,HC ( B ∗ k h a , . . . a n i ) = z + hcfree(3 zy − z ) + hcfree(( n − zy )+hcfree( B + ( n − zy/ (1 − ( n − zy ))But B = 1 / (1 − zy + z ) and1 − (3 zy − z )( n − zy/ (1 − ( n − zy ) / (1 − zy + z ) =(1 − nzy + z ) / (1 − zy + z ) / (1 − ( n − zy )and then, since hcfree( V ) is logarithmic in 1 − V ,HC ( B ∗ k h a , . . . a n i ) = z + hcfree( nzy − z )17 basis for HC ( k h a, b i /ab ) is easily seen to be { b j , a k ; j and k odd ≥ } .By Proposition 1.3 a basis for HC ( k h a, b, c i /ab ) is { b j , a k , c l ; j, k, l odd ≥ } ∪ { ( b j a k c l ) . . . ( b j r a k r c l r ); j i + k i , l i > } where the last set is taken modulo cyclic permutation of groups. The seriesfor the basis is hcfree(3 zy − z ). Considering the relation [ a, b ] + c as there-write rule ab → − ba − c one can see that the elements in the set abovetogether with { b j a k } generate HC ( B ). We will now prove that for any j > , k > , j + k > b j a k or not b j − a k − c . From this the result follows since then the series for HC ( B ) isat most z + hcfree(3 zy − z ) which is also the minimal possible value.The following computation is done modulo c . We have, b j a k = ( − j + k − ab j a k − = ( − k − b j a k b j a k = ( − j + k − b j − a k b = ( − j − b j a k Hence, if j or k is even, then b j a k may be expressed in terms of the generatorscontaining c . Suppose now j and k are odd. We have, b j a k = − ab j a k − = b j a k + j X i =1 ( − i b i − c b j − i a k − and b i − c b j − i a k − = ( − ( i − j − i + k − b j − i a k − b i − c But modulo c we have b j − i a k − b i − c = ( − ( i − k − b j − a k − c whence( − i b i − c b j − i a k − = ( − i +( i − i − b j − a k − c = − b j − a k − c and hence − jb j − a k − c belongs to the ideal generated by c in k h a, b, c i which implies that b j − a k − c is not needed as a generator (char( k ) = 0). ✷ Remark 4.4
The anti-symmetric case, i.e., when the variables are even,may be treated similarly (and more easily) by studying ω = [ a, b ] + [ c, d ] in k h a, b, c, d i . There might however be some more exceptional cases in threevariables. Proposition 4.5
Suppose there is a strongly free set of monomials in T ( V ) with series ω ( z, y ) . Then for any set g of generic polynomials with series ω ( z, y ) we have, HC ( T ( V ) / ( g )) = hcfree( V − ω )HC n ( T ( V ) / ( g )) = 0 for n > Proof.
Put B = T ( V ) / ( g ). We have B ≥ / (1 − V + ω ). By assumption,the minimal value is attained and hence B = 1 / (1 − V + ω ) since g is generic.It follows that g is a strongly free set. Also HC ( B ) is minimal in the genericcase and by Proposition 3.6, HC ( B ) ≥ hcfree( V − ω ). Again, this minimalvalue is attained by assumption and Proposition 3.6, hence it is also the seriesin the general case. Moreover, Proposition 3.6 also gives that HC n ( B ) = 0for n > ✷ Proposition 4.6
Suppose ω , . . . , ω r are r generic symmetric quadratic formsin n odd variables of weight one, where r ≤ jk and j + k + jk ≤ n for somepositive integers j, k and let A = k h a , . . . , a n i and B = A/ ( ω . . . , ω r ) . Then HC ( B ) = rz + hcfree( nzy − rz )HC ( B ) = ryz HC n ( B ) = 0 for n > Proof.
Consider ω m,n = [ a m , b n ] + c m,n for m = 1 , . . . , j and n = 1 , . . . , k in k h a , . . . , a j , b , . . . , b k , c , , . . . , c j,k i . ✷ Taking Koszul dual, we obtain a result for quotients of a polynomial ring bysufficiently many generic quadratic relations.
Proposition 4.7
Suppose f , . . . , f t are generic quadratic relations in A = k [ x , . . . , x n ] , where x i are even variables of weight one and t = (cid:0) n +12 (cid:1) − r and where r ≤ jk , j + k + jk ≤ n for some positive integers j, k . Let R = A/ ( f , . . . , f t ) and hcfree( nzy − rz ) = P i ≥ a i z i + y P i ≥ b i z i . Then HC( R ) = rz + rxyz + X i ≥ b i x i − z i + y X i ≥ a i x i − z i xample Let R = k [ x , . . . , x ] / ( f , . . . , f ) where f , . . . , f are genericquadratic polynomials in the even variables x , . . . , x . Thenhcfree(7 yz − z ) = 18 z + 465 z + . . . + y (7 z + 98 z + 2401 z + . . . )and HC( R ) = 7 z + 3 z + 21 yxz + 98 x z + 465 yx z + 2401 x z + . . . Exceptional cases : A = k h T , . . . , T n i / ( T ) T i even A = k h T , . . . , T n i / ( T ) T i odd B = k h T , . . . , T n i / ([ T , T ]) T i even B = k h T , . . . , T n i / ([ T , T ]) T i oddHC( A )( x, y, z ) = z/ (1 − x z ) + hcfree(( n − z + ( n − z )HC( A )( x, y, z ) = yz/ (1 − xz ) + hcfree(( n − yz + ( n − z )HC( B )( x, y, z ) = z / (1 − z ) + yxz / (1 − z ) + hcfree( nz − z )HC( B )( x, y, z ) = z / (1 − z ) + 2 yz/ (1 − z ) + yxz / (1 − z ) ++ hcfree( nyz − z ) Proof.
For each case we start to assume n = 1 for A , A and n = 2 for B , B . Except for B we first compute the series for the Koszul dual. Wehave that A !0 is free on one odd generator hence, HC( A !0 ) = yz/ (1 − z ) andby (5), HC( A ) = z/ (1 − x z ).Now, by Proposition 1.3 we get for general n ,HC( k h T i / ( T ) ∗ k h T , . . . , T n i ) = z/ (1 − x z ) + hcfree(( n − z )+hcfree( z ( n − z/ (1 − ( n − z ))but(1 − ( n − z )(1 − ( n − z / (1 − ( n − z )) = 1 − ( n − z − ( n − z whence the result, since hcfree( V ) is logarithmic in 1 − V .The proof of A is similar. The algebra B for n = 2 is a commutativepolynomial ring in two even variables, hence by (7), HH( B ) = (1+ xyz ) / (1 − z ) . Now by (1),HC( B ) = ((1 + xyz ) / (1 − z ) − / (1 + xy ) = (2 z − z + xyz ) / (1 − z ) . xy = y ( x + y ) so in order to divide a polynomial p ( x ) by1 + xy one may instead perform the division of p ( x ) by x + y (and checkdivisibility by putting x = − y ).We have for a general n ,HC( k h T , T i / ([ T , T ]) ∗ k h T , . . . , T n i ) = (2 z − z + xyz ) / (1 − z ) +hcfree(( n − z ) + hcfree((1 / (1 − z ) − n − z/ (1 − ( n − z ))but(1 − ( n − z )(1 − (2 z − z )( n − z/ (1 − z ) / (1 − ( n − z )) = (1 − nz + z ) / (1 − z ) whence,hcfree(( n − z ) + hcfree((1 / (1 − z ) − n − z/ (1 − ( n − z )) =hcfree( nz − z ) − z )and the result for B follows since, hcfree( z ) = z/ (1 − z ).For B and n = 2 the Koszul dual is B !1 = k [ a ] / ( a ) ⊗ k [ b ] / ( b ) where a, b are even. Hence by (2), HH( B !1 ) = (HH( k [ a ] / ( a ))) and by the result for A ,HH( k [ a ] / ( a )) = 1 + (1 + xy ) HC( k [ a ] / ( a )) = 1 + (1 + xy ) z/ (1 − x z )Hence,HH( B !1 ) = 1 + (1 + xy ) z / (1 − x z ) + 2(1 + xy ) z/ (1 − x z )and then by (1)HC( B !1 ) = (1 + xy ) z / (1 − x z ) + 2 z/ (1 − x z )and finally by (5)HC( B ) = y/x ((1 + y/x ) x z / (1 − z ) + 2 xz/ (1 − z )) == yxz / (1 − z ) + z / (1 − z ) + 2 yz/ (1 − z )A similar computation as for B now gives the result for B for general n . ✷ eferences [1] D. Anick, Non-Commutative Graded Algebras and Their Hilbert Series,J. of Algebra,78