On the Homology Theory of Operator Algebras

: The homology (cyclic and dihedral) of unital and involuted Banach algebra are studied. The extension of results in [6] for dihedral case is studied. Also the Dihedral homolog of commutative involutive unital Banach algebra, Laurent polynomial algebra and involutive tensor Banach algebra are is studied. The main results are theorems (2,2), (3.1) (3.2), (4.1) and (4,2).


Introduction.
Given Banach algebrasA, B, let f : A → B be algebras homomorphism.We define a free resolution of algebra B over the homomorphism f : A i − → R π − → B, where i is an inclusion and π is a quasi-isomorphism, and use this fact to define the relative cyclic homology where[R, R] is the commutant of algebra R, and study its main properties.First, we recall some definitions and facts from [2, 4, 6].Let A be a unital Banach algebra over a commutative ring K (K = R or C).

The group H n (A) = H(C(A)) = Kerbn
Im b n−1 is called the simplicial (hochschild) homology of algebra A.
Note that Ker b n is always closed, but Im b n−1 , in general , is not closed.Considering a unital Banach algebra A, one acts on the complex C(A), by the cyclic group of order n + 1 by means of the operator t n : C n (A) → C n (A) such that : t n (a 0 ⊗ ... ⊗ a n ) = (−1) n a n ⊗ a 0 ⊗ ... ⊗ a n−1 (3)   The quotient complex CC n (A) = C n (A)/ Im(1 − t n ) is a subcomplex of a complex C n (A).Following [Hel, wood 4] the cyclic homology of algebraA is the homology of the complex CC * (A).

Free resolution. [[1]].
In this part, we discuss the existenc of the free algebra resolution.Let V = ∞ n−0 V n be a graded vector space over ring F (F = R or C).Suppose that R is a differential graded F − algebra and let R V = R * T K (V ) be the free product of algebras, where T K (V ) = j≥0 V ⊗j is the tensor algebra over F. The product in R V is given by (r 1 e 1 ...r n e n r n+1 ).( An algebra R 2 is a free algebra over the homomorphism f if there exists an isomorphismα : R 1 V ≈ R 2 , where E is a differential graded vector space with the following commutative diagram: where i is inclusion map.
Then there exists a differential graded algebra R = i=0 R i with the following properties: (i) πis surjection and the following diagram is commutative where i is the inclusion map.Clearly there is an isomorphism j : R → A such thatj (iii) The differential graded algebra R is free over the homomorphism i : Definition 2 1.3.The differential graded algebra which satisfies the conditions (i), (ii) and (iii) of lemma 1.2 is called a free resolution of algebra B overf.
Proof of lemma 1.2.First step.We construct a commutative diagram of algebras where R (0) is free over the homomorphism i 0 : A → R (0) , π 0 an surjection.Define A (t i ) = A (t i ) , where E(t i )is an involutive vector space generated by {t i }, or generated by the family {t i t * i } .The automorphism * : of generators in algebra B. This family is assumed to be closed under an involutive onB.Now, let in algebraB, and suppose that β We define π 0 using the universal property of R (0) .Let π 0 be the unique involutive algebras R (0) → B homomorphism, which restrictsf onA and sends t is an inclusion map, i 0 (a) = a, i 0 is an algebras homomorphism and π 0 i 0 (a) = π 0 (a) = f (a).Hence, diagram (8) is commutative and π 0 is surjective.Letj 0 : R (0) → A be the unique algebras homomorphism restricting to the identity on A and mapping t (0) The algebra R (0) is free over the homomorphism i 0 : where R (1) is free over the homomorphism I 1 = A → R (1) and π 1 surjection.Choose a system i ) * .The homomorphism π 1 is defined to be the unique algebras R (1) → B restricting to π 0 on R (0) and sending t (1) i to zero.We can see, from the above discussion, that the homomorphism π 1 can be defined as π 0 and that π 1 is surjective since π 1 ( β The homomrphism j 1 is defined to be the unique homomorphism: R (1) → A of involutive algebras restricting to identity on A and mapping t (1)   i to zero.The algebra R (1) (1) j is free over i.Finally, we have a differential graded algebra The differential ∂ R (1) i is the unique derivation on R (1) satisfying the graded Leibntiz rule and commuting with the involution which restricts to zero onR (1) and sends t In the same manner, we can consider the commutative diagram The differential algebra R (2) is also defined by using a universal property and, hence, Consequently, we can construct an involutive algebra R (i) , i ≥ 0 with the following commutative diagram: Where π i is surjection, i ≥ 0, i n = P n−1 o...oP 0 oi o is an inclusion map from A to R (n) , P i is also is an inclusion map from m1 , ..., t (i) Define i n = q n−1 o...oq 0 oj o , where q n is the projection of P i .The diagram 15 is commutative since i n+1 (β Then the differential graded algebra R satisfies the items of lemma 1.2 since: (1) π = lim π n is surjection, the diagram where (3) The differential graded algebrasR is free over the homomorphismi : A → R,since R = E, E is a vector space generated by the system: 3 The cyclic homology.
In this part, we define the relative cyclic homology and study its properties.Let f be a homomorphism of Banach algebras A and B over a field K ( K is real or complex number set ).Let R B f be a free resolutionof algebra B over f and, for r = ∂r 1 r 2 + (−1) f be a free resolution of algebra B over f.Then the relative cyclic homology is defined as follows: The main properties of the relative cyclic homology are submitted in Theorem 2.2, 2.6, 2.7.
Theorem 2.2.Let A be an algebra.Then HC i (A → B) = HC i−1 (A) , where HC i−1 (A) is the cyclic homology of F-algebras (charF = 0).Proof.To do this, we need the following definition and lemmas.Definition 4 2.3.The F -algebra A t generated by the elements a 0 ta 1 t...ta n , n ≥ 0,can be considered as differential graded algebras by requiring that the morphism A → A t is a morphism of differential graded algebras (A is viewed as a differential graded algebra concentrated indegree 0) and the degt = 1, ∂t = 0 and t * = t.Lemma 2 2.4.The algebra A t is splitable.It is free algebra resolution of the algebraB = 0over the homomorphism A → 0.

Proof. Define the following chain complex
where At...tA (n-times) is a K-module and the boundary operator ∂ is given by Note that the differential ∂ in A t is equivalent to the operator δ n : Following [ [4]], the complex (C n (A), δ n ) is splitable and so the complex A t is also splitable, that is, H * (A t ) = 0. Therefore, algebra A t is free resolution of the algebra B = 0 over the homomorphism A → 0.

Lemma 3 2.5. The complex A t / [A,
A t ] is standard simplicial (Hochschild) complex.

Proof. Consider the factor complex
The action of the differential ∂ on the complex A t / [A, A t ] is given by; Consider the complex where δ is the differential in the standard Hochschild complex ( [4]).Since the space (A t / [A, A t ]) n+1 identifies with the space and the differential in A t / [A, A t ] identifies with the differential in the standard Hochschild complex, then the complex A t / [A, A t ] is the Hochschild (simplicial) complex .Now ,we prove Theorem 2.2.Consider the factor complex : where deg , deg a 0 ta 1 t...ta n−1 t = 0 deg a 0 ta 1 t...ta n t = n + 1.The cyclic homology of A t is the homology of the complex A t / [A t , A t ]+ Im(1 − t n ).By factoring A t , first by the subcomplex A ←− 0 ←− 0 ←− ... and second by the subcomplex [A t , A t ] + Im(1 − t n ) we get a homomorphism CC * (A → 0) → CC * −1 (A) , which induces an isomorphism of the cyclic one homology groups HC * (A → 0) → HC * −1 (A).Proof.The homomorphism f induces homomorphism of chain complexes Where CC * (A) is a cyclic complex.Consider the diagram Where R B f is defined above, i is an inclusion map.The idea of proof is to show that the cone of the map i is quasi-isomorphic to an arbitrary category ( [5]), to the complex Then the isomorphismπ * CC * (R B f ) → CC * (B) induces an isomorphism of the homology of these complexes.Since , where M(i * ) is a cone of i (see [5]).Note that, the symbol ≈ denotes a quasi-isomorphism.It is clear, from the above discussion, that the following diagram is commutative: where CC * is the Connes cyclic complex, and by using the spectral sequence

the long exact sequence of relative cyclic homology HC
Proof.In Theorem 2.6, it has been proved that any homomorphism f : A → B of involutive algebra in an arbitrary category is equivalent to an inclusion of involutive algebra, we have the following complex Consider the following sequence of mapping cones In general, the sequence (37) is not exact.The composition of two morphism will be zero.However , the cone over the morphism M(i * ) → M(i * ) is canonically homotopy equivalent to M(i * oi * ).So we get the following exact sequence of the relative cyclicl homology (38) In the following we give an example of the cyclic homology of tensor algebra by using the free resolution fact.LetA be F − algebra, (charF = 0) and M is A-bimodule.For a chain complex V • of modules, consider the complexS n (A, . If we act on S n (A, V • ) by the cyclic group Z n+1 of order (n + 1) by means of automorphisms where μ = (deg p n )( n−1 i=0 deg p i ).If V • is a free resolution of A − bimodule M, then the complex S n (A, V • ) can be considered by as complex S n (A, M).Example 1 2.8.Let, M be A − bimodule, where A is K − algebra (charK = 0), is a tensor algebra and T or A i (M, M) = 0, i 0 , then Proof.Suppose V • is a free resolution of A−bimodule M, then.According the condition T or A i (M, M) = 0, i 0 , the space T A (A) is a free resolution of algebra T A (M) over inclusion i : A → T A (M).Using theorem 2.7. the long exact sequence of relative cyclic homology of the following sequence A
Ker π 0 , which is closed under involution.Let t(1) i be indeterminate which are bijection with the(1)

Theorem 4 2
.6:f : A → B be a homomorphism of algebras over a fieldK(charK = 0).Then the relative cyclic homologyHC i (A f −→ B) does not depends on the choice of the resolution.