# On the Homology Theory of Operator Algebras

Authors: Alaa Hassan Noreldeen
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We investigate the cyclic homology and free resolution effect of a commutative unital Banach algebra. Using the free resolution operator, we define the relative cyclic homology of commutative Banach algebras. Lemmas and theorems of this investigation are studied and proved. Finally, the relation between cyclic homology and relative cyclic homology of Banach algebra is deduced.


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## Body

## 1. Introduction

Many years ago, cyclic homology has been introduced by Connes and Tsygan and defined on suitable categories of algebras, as the homology of a natural chain complex and the target of a natural Chern character from topological (or algebraic) K-Theory.In order to extend the classical theory of the Chern character to the noncommutative setting, Connes [1] and Tsygan [2] have developed the cyclic homology of associative algebras. Recently, there has been increasing interest in general algebraic structures than associative algebras, characterized by the presence of several algebraic operations. Such structures appear, for example, in homotopy theory [3, 4] and topological field theory [5].Brylinski and Nistor [6] have extended Conne’s computation of the cyclic cohomology  groups  of smooth algebras arising from foliations with separated graphs and explained some results of Atiyah and Segal on orbifold  Euler characteristic in the setting of cyclic homology. Kazhdan [7] studied Hochschild and cyclic homology of finite type algebras using abelian stratifications of their primitive ideal spectrum.Victor Nistor [8] has studied associative p-summable quasi homomorphism’s and p-summable extensions elements in a bivariant cyclic cohomology group defined by Connes, and showed that this generalizes his character on K-homology; furthermore, he studied the properties of this character and showed that it is compatible with analytic index.Results of Connes [1] have led much research interest into the computation of cyclic (co)homology groups in recent years (see, [4, 6, 9–13]).A promising approach to the calculation of cyclic cohomology groups is to break it down by making use of extensions of Banach algebras; this is a standard device in the study of various properties of Banach algebras.The Banach cyclic (co)homology of Banach algebra has been studied by Christensen and Sinclair [3], Helemskii [4, 9], among others. The dihedral cohomology in Banach category and its relation with the cyclic cohomology, the triviality, and nontriviality of dihedral cohomology groups of some classes of operator algebras have been studied [14].Suppose thatA and B be commutative unital Banach algebra with involution (in short B*-algebra). And let Cn(A) denote the (n+1) fold projective tensor power of A. The elements of this Banach space n-dimensional will be called chains. Let tn:Cn(A)→Cn(A), n=1,2,… denote the operator uniquely defined by
(1.1)tn(a0⊗⋯⊗an)=(-1)nan⊗a0⊗⋯⊗an-1.
And let CCn(A) denote the quotient space of Cn(A) modulo the closure of the linear span of elements of the form  x-tnx(n=1,2,…). Note that Im(idcn(A)-tn) is closed in Cn(A) and so  CCn(A)=Cn(A)/Im(idCn(A)-tn) and also  CC0(A)=C0(A)=A.Define the complexC(A)=(C*(A),δ*),  where   Cn(A)=A⊗⋯⊗A and,      b*:Cn(A)→Cn-1(A)  is the boundary operator;
(1.2)bn(a0⊗⋯⊗an)=∑i=0n-1(-1)ia0⊗⋯⊗aiai-1⊗⋯⊗an-1.We can easily verify thatbn+1bn=0 and hence Kerbn⊃Imbn-1. The group Hn(A)=H(C(A))=(Kerbn)/(Imbn-1)  is  called  the simplicial  (Hochschild) homology of  Banach algebra  A.Note that:Kerbn  is always closed, but, in general  Imbn-1  is not closed.Considering  a unital Banach algebraA, one acts on the complex C(A), by the cyclic group of order  (n+1)  by  means of  the operator  tn:Cn(A)→Cn(A)  which we denote.The quotient complexCCn(A)=Cn(A)/Im(1-tn)  is a subcomplex of the complex  Cn(A).  Following [15],  the cyclic homology of a Banach algebra  A  is the homology of the complex CCn(A).Given commutative unital Banach algebrasA and B, let f:A→B be  algebras homomorphism. We define a free resolution  of a Banach algebra  B   over the homomorphism f:A→iR→πB, where  i is an inclusion and  π  is a quasi-isomorphism, and use this fact to define the relative cyclic homology
(1.3)HC*(A→fB)=H*(R(A+[R,R]+(1-tn)),
where [R,R]  is the commutant of  Banach algebra R.Definition 1.1.
A graded Banach algebra is a Banach algebra that has a graded normed algebra as a dense subalgebra.
We discuss the existence of the free Banach algebra resolution. LetV=∑n=0∞Vn be a graded vector space over ring   F(F=RorC).   Suppose  that   R  is a differential graded F-algebra and let  R〈V〉=R*Tk(V)  be the free product of Banach algebras, where  Tk(V)=∑j≥0V⊗j  is the tensor Banach algebra over   F. The product in  R〈V〉  is given by
(1.4)(r1e1,…,rnenrn+1)·(r^1e^1,…,r^ke^kr^k+1)=(r1e1,…,rnen(rn+1r^1)e^1,…,r^ke^kr^k+1),ri,r^j∈R,eie^j∈Tk(V).Definition 1.2.
Letf:R1→R2  be a homomorphism of differential graded  F-algebras. An algebra   R2    is a free Banach algebra over the homomorphismf  if there exists an isomorphism  α:R1〈V〉≈R2, where  E   is a differential graded vector space with the following commutative diagram:
(1.5)
where iis an inclusion map.Lemma 1.3.
Letf:A→B  be a homomorphism of F-algebras. Then there exists a differential graded Banach algebra R=∑i=0Ri  with the following properties.(i)
π is surjection, and the  following diagram is commutative:
(1.6)where i  is the inclusion map.
Clearly, there is an isomorphismj:R→A  such that  j∘i=1A.(ii)
π   is quasi-isomorphism,  that is,   π*:H*(A)→H*(B)=B,    where B  is a  differential graded Banach  algebra,
(1.7)(B)i={B,i=00,i>0,andthedifferential∂B=0.(iii)
The differential  graded  Banach  algebraR   is free over  the  homomorphism  i:A→R.Proof.
We proof this theorem by two steps.(1)
We construct a commutative diagram of  a Banach algebras(1.8)where R(0)  is free  over the homomorphism i0:A→R(0), π0  an surjection.       Define  A〈(ti)〉=E(ti), where E(ti)  is an involutive vector space generated by {ti}, or generated by the family {titi*}. The automorphism *:E(ti)→E(ti)  is given as follows  (ti)*=(ti*)(ti*)*=ti.
We choose a system{ℜi(0)} of generators in a Banach algebraB.This family is assumed to be closed under an involutive on B.
Now, letR(0)=A〈ti(0)〉, where  ti(0) is equivalent to the generator {ℜ(0)} in a Banach algebra B, and suppose that βi(0)=ti(0)  or (ti(0))*. We define  π0  using the universal property of  R(0). Let π0 be the unique involutive Banach algebras R(0)→B homomorphism, which restricts f on  A  and sends  ti(0) to  ℜi(0).
Sincei0:A→A〈ti(0)〉 is an inclusion map,    i0(a)=a, i0  is a Banach algebras homomorphism, and  π0i0(a)=π0(a)=f(a). Hence, diagram (1.8) is commutative and   π0  is surjective.
Letj0:R(0)→A be the unique algebras homomorphism restricting to the identity on  A  and mapping ti(0)  to zero.R(0)  is a differential graded K-Banach algebra (see[13]);
(1.9)(R(0))i={R(0),i=00,i>0,andthedifferential∂R(0)Bi(0)=0.
The algebra   R(0)   is free over the homomorphism  i0:A→R(0)  since  R(0)=A〈ti(0)〉.(2)
We construct the second commutative diagram(1.10)where R(1) is free over the homomorphism  i1:A→R(1)  and π1  surjection. Choose a system {ℜj(1)} of generators of  Kerπ0, which is closed under involution. Let ti(1)  be indeterminate which are bijection with the ℜj(1). Define R(1)=A〈ti(0),tj(1)〉, where ti(0) is as defined above. Suppose that  βj(1)  denotes  tj(1)  or  (tj(1))*.  The homomorphism  π1  is defined as to be the unique algebras        R(1)→B  restricting to π0 on  R(0)  and sending tj(1) to zero. As can be seen, the homomorphism  π1 can be defined as  π0 and that  π1 is surjective since  π1(βi(0))=ℜi,   π1(βj(1))=0i. The homomorphism   i1:A→A〈ti(0),tj(1)〉  is inclusion. The diagram  (1.10)  is commutative since (π1i1)(a)=π1(a)=f(a).
The homomorphismj1   is  defined  to  be the unique homomorphism: R(1)→A of involutive Banach algebras restricting to identity on A   and mapping  ti(1)  to zero.  The algebra  R(1)=A〈ti(0),tj(1)〉 is free over   i.
Finally, we have a differential graded Banach algebra(1.11)R(1)=(R(1))0⊕(R(1))1⊕,…,degβi(1)=0,degβj(1)=1.
The differential  ∂iR(1) is the unique derivation on  R(1)  satisfying the graded Leibntiz rule  and commuting with the involution which restricts to zero on  R(1) and sends ti(1)  to ℜj(1).  So    ∂0R(1)βi(0)=0,  ∂1R(1)βj(0)=ℜj(1)∈Kerπ0,i>1.
Similarly, we can consider the commutative diagram(1.12)
where R(2)=A〈ti(0),tj(1),tk(2)〉  is a differential graded  Banach algebra
(1.13)R(2)=(R(2))0⊕(R(2))1⊕(R(2))2⊕,…,degβi(0)=0,degβj(0)=1,degβj(0)=2.
The differential Banach algebra  R(2) is also defined by using a universal property and, hence,
(1.14)∂0R(2)βi(0)=0,∂1R(2)βj(0)=ℜj(1),∂1R(2)βj(2)=ℑk(2)=0,i>2.
Consequently, we can construct an involutive  Banach algebra    R(i),       i≥0    with the following commutative diagram:
(1.15)
where πi  is surjection, i≥0, in=Pn-1∘⋯∘P0∘io  is an inclusion map from  A   to  R(n),     Pi    is also an inclusion map from
(1.16)Pi:A〈tmo(0),tm1(1),…,tmi(i)〉toA〈tmo(0),tm1(1),…,tmi(i),tmi+1(i+1)〉.
Define in=qn-1∘⋯∘q0∘jo,    where qn  is the projection of    Pi.
The diagram (1.15) is commutative since      in+1(βi(n))=πn(βi(n))=0,n≥o.
DefineR=limRn,  π=limπn, i=limin, j=limjn. Then the differential  Banach graded algebra  R  satisfies the items of Lemma 1.3  since:(1)
π=limπn   is surjection, the diagram
(1.17)is commutative sincei(a)=a,   π(a)=f(a)(2)
π is quasi-isomorphism of differential graded algebras
(1.18)where∂iR=lim∂iR, (R)0=ker(π)0=B, Im∂R=ker∂0R, that  is,
(1.19)H0(R)=Hi(R)=0.(3)
the differential graded Banach algebrasR  is free over the homomorphism i:A→R,  since R=E,    E  is a vector space generated by the system:
(1.20){ti0(0),ti1(1),…,tin(n)}.Definition 1.4.
The  differential  graded  Banach algebra  which  satisfies the conditions  (i), (ii), and  (iii) of  Lemma1.3  is called a free resolution of  Banach algebra  B over f.


## 2. The Relative  Cyclic Homology of  Banach Algebra

In this part, we define the relative cyclic homology of commutative unital Banach algebra and study its properties. Letf  be a homomorphism of Banach algebras  A  and  B over a field  K (K is a real or a complex number set). Let RfB be a free resolution of  Banach algebra  B  over  f   and,  for    r1,r2∈RfB,  let  [r1,r2]=r1,r2-(-1)|r1||r2|r2r1  where |ri|=degri, i=1,2.LetC=[RfB,RfB] be the linear space generated by    [r1,r2],    r1,r2∈RfB.We construct the complexC=[RfB,RfB]. Clearly, from the definition of RfB, that    Im(1-tn)  is a subcomplex of RfB.  We have
(2.1)∂[r1r2]=r1r2-(-1)|lr1||r2|r2r1=∂r1r2+(-1)|r2|r1∂r2-(-1)|lr1||r2|(∂r2r1-(-1)|r2|r2∂r1)=∂r1r2-(-1)|r2|(|r1|+1)r2∂r1+(-1)|r1|(r1∂r2-(-1)|r1|(|r2|+1)∂r2r1)=[∂r1r2]+(-1)|r1|[r1,∂r2],|∂ri|=|r1|-1,i=1,2.
Then  [RfB,RfB] is subcomplex in  RfB. Therefore, the chain complex of  K-module  [RfB,RfB]  is a subcomplex of  RfB.Definition 2.1.
Letf:A→B  be F-algebras (char F=0) homomorphism, and RfB  be a free resolution of a Banach algebra B over  f. Then the relative cyclic homology is defined as follows:
(2.2)HC*(A→fB)=H*(RfB(A+[R,R]+Im(1-tn)).Definition 2.2.
TheF-algebra A〈t〉 generated by the elements a0ta1t,…,tan, n≥0  can be considered as differential graded Banach algebras by requiring that the  morphism  A→A〈t〉 is a morphism of differential graded algebras (A is viewed as a differential graded Banach algebra concentrated in degree 0) and the degt=1, ∂t=0  and  t*=t.Lemma 2.3.
A Banach algebraA〈t〉 is splitable.  One has a free algebra resolution of Banach algebra  B=0  over the homomorphism  A→0.Proof.
Using [6] from the chain complex
(2.3)A⟵∂AtA⟵∂AtAt⟵∂⋯⟵∂At⋯tA⟵∂⋯,
where  At,…,tA (n-times)  is  a K-module and the boundary operator   ∂  is given by
(2.4)∂(a0ta1t,…,tan-1tan)=∑i=0n-1(-1)ia0ta1t,…,tai(∂t)ai+1t,…,tan=∑i=0n-1(-1)ia0ta1t,…,t(aiai+1)t,…,tan.
Note that the differential   ∂ in A〈t〉 is equivalent to the operator  δn′:Cn(A)→Cn-1(A), defined by
(2.5)δn′(a0⊗⋯⊗an)=∑i=0n-1(-1)ia0⊗⋯⊗aiai+1⊗⋯⊗an.
Following [11], the complex (Cn(A),δn′)  is splitable and so is the complex  A〈t〉, that is, H*(A〈t〉)=0.  Therefore,  Banach algebra  A〈t〉 is free resolution of the Banach algebra  B=0   over the homomorphism A→0.Lemma 2.4.
The complexA〈t〉/[A,A〈t〉] is a standard simplicial (Hochschild) complex.Proof.
Consider the factor complexA〈t〉/[A,A〈t〉].    It is generated by the elements  a0ta1t,…,tan-1t, since
(2.6)a0ta1t,…,tan-1tan=ana0×ta1t,…,tan-1t(mod[A,A〈t〉]).
The action of the differential    ∂ on the complex     A〈t〉/[A,A〈t〉]    is given by
(2.7)∂(a0ta1t,…,tan-1tan)=∑i=0n-1(-1)ia0ta1t,…,t(aiaai+1)t,…,tan+(-1)nana0ta1t,…,an-1t.
Consider the complex(2.8)A⟵idA⟵δA⨂2⟵δ⋯⟵δA⨂n⟵δ⋯,
where  δ   is the differential  in the standard Hochschild complex (see [7, 16]). Since the space    (A〈t〉/[A,A〈t〉])n+1    identifies with the space
(2.9)A⨂n+1:a0ta1,…,tant→a0⊗·a1⊗⋯⊗an
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 complex.Theorem 2.5.
LetA be Banach algebra with unity and involution.  Then  HCi(A→B)=HCi-1(A), where HCi(A)  is the cyclic homology of  F-algebras (char F = 0).Proof.
Consider the factor complex:A〈t〉/[A〈t〉,A〈t〉]+Im(1-rn),    such that
(2.10)a0ta1t,…,tan-1t=(-1)nanta0t…an-1t,
where       dega0ta1t,…,tan-1t=n,      dega0ta1t,…,tant=n+1.
The cyclic homology ofA〈t〉 is the  homology of the complex  A〈t〉/[A〈t〉,A〈t〉]+Im(1-tn).   By factoring  A〈t〉,   first by the subcomplex    A←0←0←⋯  and second  by the subcomplex  [A〈t〉,A〈t〉]+Im(1-tn)],  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).Theorem 2.6.
Letf:A→B be a homomorphism of a Banach algebras over a field K(charK=0). Then the relative cyclic homology  HCi(A→fB)  does not depend on the choice of the resolution.Proof.
The homomorphismf  induces homomorphism of chain complexes
(2.11)f*:CC*(A)→CC*(B),
where   CC*(A)   is a cyclic complex. Consider the diagram
(2.12)
where RfB  is defined above, and   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 (see [8, 17]), to the complex     RfB/[RfB,RfB]+Im(1-tn),       Since
(2.13)Hi(RfB)={B,i=00,i>0.
Then the isomorphism  π*:CC*(RfB)→CC*(B)  induces an isomorphism of the homology of these complexes. Since     i*:CC*(A)→CC*(RfB)  is an inclusion, then
(2.14)HCi(A→fB)→HCi(A→g∘fC)→HCi(A→gC)→HCi-1(A→fB)→⋯M(i*)≈CC*(RfB)/CC*(A),    where  M(i*) is a cone of  i (see [12, 15, 18]).
Note that, the symbol≈ denotes a quasi-isomorphism. It is clear, from the above discussion, that the following diagram is commutative
(2.15)
and hence     M(f*)≈CC*(RfB)/CC*(A).
Following [11], we have  CC*(RfB)/CC*(A)≈RfB/A+[RfB,RfB],  where  CC*  is the Connes cyclic complex, and by using the spectral sequence  Eij2=H*ε(Z/2,H*(RfB))=HεCi+j(RfB),    we have
(2.16)CC*(RfB)CC*(A)≈RfB/A+[RfB,RfB]+Im(1-tn).
So,  M(f*)≈RfB/A+[RfB,RfB]+Im(1-tn). Then      HCi(A→fB) does not depend on the choice of RfB.Theorem 2.7.
LetA, B, and D be involutive Banach algebra. Then the following sequence A→fB→gD  induces the long exact sequence of relative cyclic homology;
(2.17)HCi(A→fB)→HCi(A→g∘fD)→HCi(B→gD)→HCi-1(A→fB)→⋯.Proof.
In Theorem2.6,   it  has  been  proved  that any homomorphism f:A→B  of involutive algebra in an arbitrary category is equivalent to an inclusion  i:A→RfB.  Then, for a sequence  A→fB→gC  of  involutive Banach algebra, we have the following complex:
(2.18)
Consider the following sequence of mapping cones(2.19)o→M(i*)→M(i*′)→M(i*∘i*′)→o.
In general, the sequence (2.17) is not exact. The composition of two  morphisms will be zero. However, the cone over the morphism M(i*)→M(i*′)  is canonically homotopy equivalent to  M(i*∘i*′). So, we get the following exact sequence of the relative cyclic homology
(2.20)HCi(A→fB)→HCi(A→g∘fC)→HCi(B→gC)→HCi-1(A→fB)→⋯.
In the following we give an example of the cyclic homology of tensor algebra by using the free resolution fact.  Let a Banach A  be F-algebra, (charF=0) and  M  be  A-bimodule.  For a chain complex  V• of modules,  consider the complex  Sn(A,V•)=A⊗A⊗AopV•⊗(k+1). If we act on  Sn(A,V•) by the  cyclic group zn+1 of order (n+1) by means of automorphisms, we get
(2.21)tn(v0⊗⋯⊗vn)=(-1)μvn⊗v0⊗⋯⊗vn-1,
where  μ=(degpn)(∑i=0n-1degpi).
IfV•   is a free resolution of A-bimodule M, then the complex Sn(A,V•) can be considered as complex  Sn(A,M).Example 2.8.
LetM  be  A-bimodule, where  A is K-Banach algebra, TA(M)  be  a tensor Banach algebra and  ToriA(M,M)=0,i>0, then
(2.22)HCi(TA(M))=HCi(A)⊕(⊕n=0∞Hi(Zn+1;Sn(A,M))).Proof.
SupposeV•  is a free resolution of  A-bimodule M. Then according to the condition ToriA(M,M)=0,i>0, the space TA(V•) is a free resolution of  algebra TA(M) over inclusion i:A→TA(M). Using Theorem 2.7 the long exact sequence of relative cyclic homology of the following sequence,  A→iTA(M)→0, we get
(2.23)⋯→HCi(A→iTA(M))→HCi(A→00)→HCi(TA(M)→0)→HCi-1(A→iTA(M))→0⋯.
Since  A is a direct sum of  TA(M),  we have
(2.24)0→HCi(A→iTA(M))→HCi(A)→HCi(TA(M))→0
and hence
(2.25)HCi(TA(M))=HCi(A)⊕HCi(A→iTA(M)).
To prove the theorem we show that
(2.26)HCi(A→iTA(M))=⊕n=0∞Hi(Zn+1;Sn(A,M)).
Clearly,
(2.27)TA(V•)(A+[TA(V•),TA(V•)]+Im(1-tn))=⊕n=0∞V•⊗(n+1)([TA(V•),TA(V•)]+Im(1-tn)).
Then we have the following isomorphism:
(2.28)⊕n=0∞P•⊗(n+1)([TA(V•),TA(V•)]+Im(1-tn))≈⊕n=0∞A⊗A⊗AopP•⊗(n+1)Im(1-tn).
The homology of the chain complex:
(2.29)⊕n=0∞A⊗A⊗AopV•⊗(n+1)Im(1-tn)
is equivalent to
(2.30)⊕n=0∞Hi(Zn+1;Sn(A,M)).
From (2.25) and (2.26), the proof  is completed.



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*Source: 368527-2012-12-25.xml*