Secondary Hochschild and Cyclic (Co)Homologies

SECONDARY HOCHSCHILD AND CYCLIC (CO)HOMOLOGIES

Jacob Laubacher

A Dissertation

Submitted to the Graduate College of Bowling Green State University in partial fulﬁllment of the requirements for the degree of

DOCTOR OF PHILOSOPHY

May 2017

Committee:

Mihai D. Staic, Advisor

John Laird, Graduate Faculty Representative

Xiangdong Xie

Mihai D. Staic, Advisor

Hochschild cohomology was originally introduced in 1945. Much more recently in 2013 a gen- eralization of this theory, the secondary Hochschild cohomology, was brought to light. In this dissertation we provide the details behind the simplicial structure for the chain complexes associated to the (secondary) Hochschild (co)homology. For this we introduce the notion of simplicial algebras and simplicial modules. The key results are two lemmas (3.4.1 and 3.4.2) that can be thought of as analogues of the Tor and Ext functors in the context of simplicial modules. It was a pleasant surprise that the higher order Hochschild homology over the 2-sphere can also be described using simplicial structures. We study some other related concepts like the secondary Hochschild and cyclic homologies associated to the triple (A, B, ε), as well as some of their properties. iv ACKNOWLEDGMENTS

I would ﬁrst like to thank my parents, KC and Laura, for distilling in me the notion to do what I love. Without them I would be lost and miserable, albeit a bit richer. I would like to thank my siblings Rachael, Gretchen, and Seth for their patience and tolerance for a brother who was inter- minably in school, as well as Kathy for putting up with me during this ﬁnal stretch. My colleagues also deserve a heartfelt thanks for their support and collaboration: Dave Walmsley, Mark Medwid, Sam Carolus, John Haman, Todd Romutis, and Rob Kelvey. I would also like to thank my committee members Xiangdong Xie, Juan Bes,` and John Laird for their helpful comments, feedback, and suggestions. Finally, a special thanks to my advisor Mihai Staic. The task of mathematical research seemed daunting, yet with his careful guidance and instruction, he navigated me to a dissertation worthy of the title that I have pursued for so long. v

TABLE OF CONTENTS Page

CHAPTER 1 INTRODUCTION ...... 1

CHAPTER 2 PRELIMINARIES ...... 4 2.1 Notations and Conventions ...... 4 2.2 Basic Homological Algebra ...... 4 2.3 The Simplicial Category ...... 7 2.4 Bicomplexes ...... 9 2.5 Hochschild (Co)homology ...... 13 2.6 Cyclic (Co)homology ...... 17 2.7 Secondary Hochschild Cohomology ...... 18 2.8 Higher Order Hochschild (Co)homology ...... 21

CHAPTER 3 SIMPLICIAL STRUCTURES ...... 23 3.1 Simplicial Algebras ...... 23 3.1.1 Definition ...... 23 3.1.2 Examples of Simplicial Algebras ...... 24 3.2 Simplicial Modules ...... 29 3.2.1 Definition ...... 29 3.2.2 Examples of Simplicial Modules ...... 29 3.3 Co-simplicial Modules ...... 54 3.3.1 Definition ...... 54 3.3.2 Examples of Co-simplicial Modules ...... 55 3.4 The Hom and Tensor Lemmas ...... 62 vi CHAPTER 4 APPLICATIONS OF THE HOM AND TENSOR LEMMAS ...... 68 4.1 Hochschild (Co)homology Examples ...... 68 4.2 Higher Order Hochschild Homology Examples ...... 75 4.2.1 Higher Order Hochschild Homology Over the Circle ...... 76 4.2.2 Higher Order Hochschild Homology Over the Sphere ...... 79 4.2.3 A Special Case Over the Sphere ...... 113

CHAPTER 5 FOUNDATIONS OF THE SECONDARY CYCLIC HOMOLOGY . . . . . 117 5.1 Deﬁnition and Construction ...... 117 5.1.1 The Secondary Connes’ Complex ...... 122 5.1.2 The Secondary Cyclic Bicomplex ...... 129 5.1.3 The Secondary Diagonal Bicomplex ...... 132 5.2 Proofs of the Equivalences ...... 137 5.3 The Connes’ Long Exact Sequence for the Secondary Case ...... 142 5.3.1 Proof 1 ...... 142 5.3.2 Proof 2 ...... 145 5.3.3 Proof 3 ...... 147

CHAPTER 6 PROBLEMS FOR FUTURE RESEARCH ...... 149

BIBLIOGRAPHY ...... 150 1

CHAPTER 1 INTRODUCTION

Hochschild introduced the cohomology bearing his name in 1945 in [21]. His main motivation was to study extensions of associative algebras over a ﬁeld k. Employing the bar resolution B(A), one can show that the Hochschild cohomology of an associative k-algebra A with coefﬁcients in the A-bimodule M is

n n n H (A, M) := H (HomA⊗Aop (B(A),M)) = ExtA⊗Aop (A, M).

In 1964, Gerstenhaber made a connection between Hochschild cohomology and deformation theory. The main result from [16] (Theorem 2.5.7) determines the obstruction for the multiplication law

2 3 mt(a ⊗ b) = ab + c1(a ⊗ b)t + c2(a ⊗ b)t + c3(a ⊗ b)t + ··· to be associative. The Hochschild homology can be deﬁned in a similar manner. More precisely,

A⊗Aop Hn(A, M) := Hn(M ⊗A⊗Aop B(A)) = Torn (A, M).

Notice again the use of the bar resolution. A special case is obtained by taking M = A, and this is denoted HHn(A) := Hn(A, A). The Hochschild homology associated to A has many nice properties, and under certain conditions it is a good substitute for the module of differential forms

n ΩA|k. In 1962 this fact was made clear through the celebrated Hochschild-Kostant-Rosenberg Theorem (see [22]). Connes ﬁrst introduced cyclic homology in 1981 in [8] (for detailed construction and further reading we refer to [24], [25], [28], and [42]). Cyclic homology has close ties to the de Rham cohomology and differential forms. For more on this, one can see [1], [28], and [32]. Most 2 importantly, the Hochschild and cyclic homologies were united in Connes’ long exact sequence:

B∗ I∗ S∗ B∗ I∗ ... −−→ HHn(A) −−→ HCn(A) −−→ HCn−2(A) −−→ HHn−1(A) −−→ ...

There is an analogous long exact sequence for the cohomology, and both have proved useful for computations. In 2013 Staic introduced the secondary Hochschild cohomology of the triple (A, B, ε) with coefﬁcients in M in [39]. Here A is an associative k-algebra, B a commutative k-algebra, and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A) (i.e., A is a B-algebra). We now let M be an A-bimodule which is B-symmetric (that is, mε(α) = ε(α)m). We deﬁne

n ⊗n ⊗ n(n−1) C ((A, B, ε); M) = Homk(A ⊗ B 2 ,M), and in [39] it was shown that there exists a

• ε n complex (C ((A, B, ε); M), δn). Its cohomology is denoted H ((A, B, ε); M) and is called the secondary Hochschild cohomology of the triple (A, B, ε) with coefficients in M. The secondary Hochschild cohomology is used to study deformations A[[t]] for a B-algebra A. More properties of this new cohomology were seen in [11] and [40], including relations to the higher order Hochschild cohomology, cup and bracket products (establishing a Gerstenhaber algebra structure), as well as a Hodge-type decomposition (for characteristic zero). In [27] we introduce the notion of simplicial algebras and simplicial modules. In particular, we showed how they can be used to describe the secondary Hochschild cohomology. Using simplicial modules allows one to vary the module of coefficients, which was the main impediment for describing Hn((A, B, ε); M) as an Ext functor. The main ingredient in our construction is the so-called secondary bar resolution (i.e., the simplicial left module B(A, B, ε)). On top of its original use for the secondary cohomology, B(A, B, ε) was employed to define a secondary homology, as well as cyclic counterparts. The working force behind these descriptions are Lemma 3.4.1 and Lemma 3.4.2. In the case where B = k, our results are essentially equivalent to the classical description of the Hochschild cohomology. 3 Here is a short summary of the results in this dissertation. Chapter 2 is dedicated to introducing the necessary material so as to keep this dissertation as self-contained as possible. Chapter 3 focuses on the results in [27]. Whereas in a paper the details can be lacking, here we give all sufficient reasoning and justification. In [27] we studied the simplicial structure of the complex C•((A, B, ε); M), which is associated to the secondary Hochschild cohomology of the triple (A, B, ε) with coefficients in M. We were able to define a secondary cyclic (co)homology as well as study some of its properties. In Chapter 4 we present a few more applications to the above-mentioned lemmas. Besides the Hochschild and secondary Hochschild examples that were presented in [27] (Section 4.1), we also give a presentation of the higher order Hochschild homology (Section 4.2). Chapter 5 is devoted to establishing the secondary cyclic homology associated to the triple (A, B, ε). This was introduced in [27], where we also proved the existence of a Connes-like long exact sequence concerning the secondary Hochschild and cyclic homology:

B∗ I∗ S∗ B∗ I∗ ... −−→ HHn(A, B, ε) −→ HCn(A, B, ε) −−→ HCn−2(A, B, ε) −−→ HHn−1(A, B, ε) −→ ...

Here we prove this in three different ways, one of which is done without using bicomplexes (under

the condition that k contains Q). Chapter 6 lists a few problems and possible research directions that give avenues for future work. 4

CHAPTER 2 PRELIMINARIES

In this chapter we deﬁne and present all material deemed necessary for the remaining chapters. The aim is to keep this dissertation as self-contained as possible.

2.1 Notations and Conventions

For this dissertation we ﬁx k to be a ﬁeld. Furthermore, all tensor products are over k unless

otherwise stated (that is, ⊗ = ⊗k). We ﬁx the following: A is an associative k-algebra and M is an A-bimodule. We ﬁx B to be a commutative k-algebra and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A). Here ε induces a B-symmetric structure on M. Furthermore, we let all k-algebras have multiplicative unit. For module theory, we commonly refer the reader to [12]. For homological algebra, we regu- larly use [28] and [42]. Many of the concepts regarding the secondary Hochschild (co)homology can be found in [27], [39], or [40].

2.2 Basic Homological Algebra

For this section we ﬁx R to be a ring with unit. We refer to [6], [12], [14], [28], [30], and [42].

Deﬁnition 2.2.1. ([12]) A left module over R is a set M consisting of an abelian group (M, +) along with the operation · : R × M −→ M with (r, m) 7−→ r · m such that for all r, s ∈ R and for all m, n ∈ M we have that

(i) r · (m + n) = r · m + r · n,

(ii) (r + s) · m = r · m + s · m,

(iii) r · (s · m) = (rs) · m, and

(iv) 1R · m = m.

One can deﬁne a right module and bimodule in a similar fashion. 5 Deﬁnition 2.2.2. ([12],[42]) A chain of modules

dn+3 dn+2 dn+1 dn dn−1 dn−2 ... −−−→ Cn+2 −−−→ Cn+1 −−−→ Cn −−→ Cn−1 −−−→ Cn−2 −−−→ ...

is a complex if Im(dn+1) ⊆ Ker(dn) for every n. Equivalently, dn ◦ dn+1 = 0. We denote this

complex C•.

th Deﬁnition 2.2.3. ([6],[28],[30],[42]) The n homology group of a complex C• is

Ker(dn) Hn(C•) = . Im(dn+1)

Deﬁnition 2.2.4. ([6],[28],[30],[42]) A chain complex C• is exact at Cn if Im(dn+1) = Ker(dn).

In particular, Hn(C•) = 0 if and only if C• is exact at Cn.

Deﬁnition 2.2.5. ([6],[30],[42]) A sequence of modules

f g 0 −→ M −→ N −→ Q −→ 0 is short exact if it is exact at M, N, and Q. In particular, f is injective, g is surjective, and Ker(g) = Im(f).

Lemma 2.2.6. ([28],[42]) For a chain complex C•, if there is a collection of contracting maps

{sn} where sn : Cn −→ Cn+1 such that

idCn = dn+1 ◦ sn + sn−1 ◦ dn,

then Hn(C•) = 0 for all n.

Deﬁnition 2.2.7. ([6],[28],[30],[42]) Given two chain complexes C• and D•, we say that f :

C• −→ D• is a chain map if fn : Cn −→ Dn such that fn−1 ◦ cn = dn ◦ fn for every n. That is, 6 the following diagram commutes.

cn+2 cn+1 cn cn−1 cn−2 ... Cn+1 Cn Cn−1 Cn−2 ...

fn+1 fn fn−1 fn−2

dn+2 dn+1 dn dn−1 dn−2 ... Dn+1 Dn Dn−1 Dn−2 ...

Remark 2.2.8. ([42]) Using the diagram above, we note that fn sends elements of Ker(cn) (called cycles) to elements of Ker(dn). Likewise, fn sends elements of Im(cn+1) (called boundaries) to elements of Im(dn+1). Thus, the chain map f induces a map f∗ such that

f∗ :Hn(C•) −→ Hn(D•) for every n.

Deﬁnition 2.2.9. ([6],[30],[42]) A chain map f : C• −→ D• is called a quasi-isomorphism if the induced maps

f∗ :Hn(C•) −→ Hn(D•) are isomorphisms for all n.

Remark 2.2.10. ([42]) A sequence of chain complexes

f g 0 −→ M• −→ N• −→ Q• −→ 0 is short exact whenever

fn gn 0 −→ Mn −−→ Nn −−→ Qn −→ 0 is short exact for every n. 7 Theorem 2.2.11 (Fundamental Theorem of Homological Algebra). ([6],[28],[30],[42]) Let

f g 0 −→ M• −→ N• −→ Q• −→ 0

be a short exact sequence of complexes. Then there are natural maps ∂ :Hn(Q•) −→ Hn−1(M•) for all n such that

g∗ ∂ f∗ g∗ ∂ f∗ ... −−→ Hn+1(Q•) −→ Hn(M•) −−→ Hn(N•) −−→ Hn(Q•) −→ Hn−1(M•) −−→ ...

is a long exact sequence.

2.3 The Simplicial Category

The results from this section can be found in [28], [30], or [42]. Let ∆ be the category whose objects are ordered sets n = {0, 1, . . . , n} for any integer n ≥ 0. The morphisms in ∆ are nondecreasing maps. We note that any morphism in ∆ can be written as the composition of face maps di : n −→ n + 1 and degeneracy maps si : n −→ n − 1 which are given by    u if u < i di(u) = ,  u + 1 if u ≥ i

and    u if u ≤ i si(u) = .  u − 1 if u > i

It can be shown that the above satisfy the following identities:

djdi = didj−1 if i < j,

sjsi = sisj+1 if i ≤ j,

sjdi = disj−1 if i < j, (2.1)

sjdi = id if i = j or i = j + 1, 8 sjdi = di−1sj if i > j + 1.

Deﬁnition 2.3.1. ([28],[30],[42]) Let C be a category. A simplicial object in C is a functor X : ∆op −→ C. Here ∆op signiﬁes the opposite category of ∆. In other words, the simplicial object X is a contravariant functor from ∆ to C. The image of any morphism ϕ in ∆ is denoted X(ϕ). Due to being contravariant, we have that for any ϕ, ψ ∈ ∆, X(ϕ ◦ ψ) = X(ψ) ◦ X(ϕ).

One can see that a simplicial object is a collection of objects X0,X1,X2,... in C such that for any morphism ϕ ∈ ∆ with ϕ : n −→ m we have that X(ϕ): Xm −→ Xn for all n and m.

i i Setting the notation δi = X(d ) and σi = X(s ) for 0 ≤ i ≤ n, one can apply X to (2.1) and see that we have the following identities:

δiδj = δj−1δi if i < j, (2.2)

and

(i) σiσj = σj+1σi if i ≤ j,

(ii) δiσj = σj−1δi if i < j, (2.3)

(iii) δiσj = id if i = j or i = j + 1,

(iv) δiσj = σjδi−1 if i > j + 1.

Dually, we have the following deﬁnition:

Deﬁnition 2.3.2. ([28],[30],[42]) A co-simplicial object in a category C is a functor Y : ∆ −→ C. In contrast to simplicial objects, these co-simplicial objects are given by a covariant functor. That is, for any ϕ, ψ ∈ ∆, we have that Y (ϕ ◦ ψ) = Y (ϕ) ◦ Y (ψ).

Again setting the notation δi = Y (di) and σi = Y (si), one can apply Y to (2.1) and see that we have the following identities:

δjδi = δiδj−1 if i < j, (2.4) 9 and (i) σjσi = σiσj+1 if i ≤ j,

(ii) σjδi = δiσj−1 if i < j, (2.5) (iii) σjδi = id if i = j or i = j + 1,

(iv) σjδi = δi−1σj if i > j + 1.

2.4 Bicomplexes

The results from this section can be found in most homological algebra texts. (See [14], [28], [30], or [42] for more reading.)

Deﬁnition 2.4.1. ([28],[42]) A bicomplex C•• is a collection of modules Cp,q (indexed by integers p and q) together with horizontal differentials

h dp,q : Cp,q −→ Cp−1,q and vertical differentials

v dp,q : Cp,q −→ Cp,q−1 such that

h h (i) dp−1,q ◦ dp,q = 0,

v v (ii) dp,q−1 ◦ dp,q = 0, and

v h h v (iii) dp−1,q ◦ dp,q + dp,q−1 ◦ dp,q = 0.

Notice that (i) and (ii) say that each row and each column are themselves chain complexes, 10 while (iii) says that each square anticommutes. A picture for the bicomplex C•• may help:

......

h h dp,q+1 dp+1,q+1 ··· Cp−1,q+1 Cp,q+1 Cp+1,q+1 ···

v v v dp−1,q+1 dp,q+1 dp+1,q+1

h h dp,q dp+1,q ··· Cp−1,q Cp,q Cp+1,q ···

v v v dp−1,q dp,q dp+1,q

h h dp,q−1 dp+1,q−1 ··· Cp−1,q−1 Cp,q−1 Cp+1,q−1 ···

......

Next we suppose that we have a bicomplex C•• which is restricted to the ﬁrst quadrant. That

is, Cp,q = 0 if p < 0 or q < 0. We can then deﬁne the k-module

M (Tot C••)n := Cp,q. p+q=n

This is a ﬁnite sum, and the induced chain is endowed with the differentials

dn : (Tot C••)n −→ (Tot C••)n−1

given by

h v dn = d + d .

One can check that dn−1 ◦ dn = 0, and thus Tot C•• is a complex. This is called the total complex 11 of the bicomplex C•• and the homology groups

Hn(Tot C••)

are called the homology groups of the bicomplex C••. Again, a diagram may help.

......

h h d1,2 d2,2 C0,2 C1,2 C2,2 ···

v v v d0,2 d1,2 d2,2

h h d1,1 d2,1 C0,1 C1,1 C2,1 ···

v v v d0,1 d1,1 d2,1

h h d1,0 d2,0 C0,0 C1,0 C2,0 ···

The total complex Tot C•• would have the form:

d4 d3 d2 d1 ... −−→ C0,3 ⊕ C1,2 ⊕ C2,1 ⊕ C3,0 −−→ C0,2 ⊕ C1,1 ⊕ C2,0 −−→ C0,1 ⊕ C1,0 −−→ C0,0 −→ 0.

For example, with a ∈ C0,2, b ∈ C1,1, and c ∈ C2,0, we have that

v h v h d1 ◦ d2(a, b, c) = d1(d0,2(a) + d1,1(b), d1,1(b) + d2,0(c))

v v v h h v h h = d0,1d0,2(a) + d0,1d1,1(b) + d1,0d1,1(b) + d1,0d2,0(c)

= 0.

Deﬁnition 2.4.2. ([28],[42]) A morphism of bicomplexes f : CC•• −→ DD•• is a collection of 12 maps

fp,q : Cp,q −→ Dp,q such that

v v (i) dp,q ◦ fp,q = fp,q−1 ◦ cp,q, and

h h (ii) dp,q ◦ fp,q = fp−1,q ◦ cp,q

for all p, q ∈ Z. In other words, the following diagram commutes:

h dp,q Dp−1,q Dp,q

f p−1,q v dp,q fp,q

Cp−1,q Cp,q Dp,q−1 h cp,q

v cp,q fp,q−1

Cp,q−1

As discussed in [28], one can take the homology of a bicomplex C•• in other ways. For instance, the theory of spectral sequences can be used. We do not discuss spectral sequences in this dissertation, but instead we will make use out of bicomplexes in the form of cyclic homology, which will be discussed in Chapter 5. The following proposition is essential.

Theorem 2.4.3. ([28]) Let f : C•• −→ D•• be a morphism of bicomplexes which is a quasi- isomorphism when restricted to each column. Then the induced map on the total complexes is a quasi-isomorphism.

In particular, suppose that for all q the horizontal homology groups Hp(C•,q) are zero for

v p > 0, and put Kn = H0(C•,n). Then Hn(Tot C••) = Hn(K•, d ). In other words, under the 13 above hypothesis, the homology of the bicomplex C•• is the homology of the cokernel of the ﬁrst two columns.

2.5 Hochschild (Co)homology

We refer the reader to [21], [24], [28], [31], and [42].

Deﬁnition 2.5.1. ([12]) We say that an associative ring A is a k-algebra if there exists a morphism of rings ϕ : k −→ A such that ϕ(k) ⊆ Z(A).

In this section we ﬁx A to be an associative k-algebra and M an A-bimodule. Deﬁne Cn(A, M) =

⊗n n n n+1 Homk(A ,M) and d : C (A, M) −→ C (A, M) determined by

n d (f)(a0 ⊗ a1 ⊗ · · · ⊗ an) = a0f(a1 ⊗ a2 ⊗ · · · ⊗ an) n X i n+1 + (−1) f(a0 ⊗ · · · ⊗ ai−1ai ⊗ · · · ⊗ an) + (−1) f(a0 ⊗ a1 ⊗ · · · ⊗ an−1)an. i=1

One can show that dn+1 ◦ dn = 0. We call this complex C•(A, M).

Deﬁnition 2.5.2. ([21],[28],[42]) The cohomology of C•(A, M) is denoted by Hn(A, M) and is called the Hochschild cohomology of A with coefﬁcients in M.

∗ n n ∗ ∗ When we take M = Homk(A, k) = A , we denote HH (A) := H (A, A ). Here the dual A has the A-bimodule structure given by (a · f · b)(x) = f(bxa).

Example 2.5.3. We can compute H0(A, M). Notice that

d0(f)(a) = af(1) − f(1)a,

where f : k −→ M is determined by f(1). Therefore we can say that the kernel of d0 consists of 14 all elements of M such that they commute with all elements of A. Formally, we have

H0(A, M) = Ker(d0) = {m ∈ M | am = ma for all a ∈ A}.

In notation, we write H0(A, M) = [M,A].

Deﬁnition 2.5.4. ([28],[42]) A derivation of A with values in M is a k-linear map D : A −→ M such that D(ab) = aD(b) + D(a)b

for all a, b ∈ A. The module of all k-linear derivations of A in M is denoted by Derk(A, M).

Deﬁnition 2.5.5. ([28],[42]) The set of all k-linear derivations of A in M of the form fm(a) = am − ma is called the module of inner derivations of A in M. This is denoted Innk(A, M).

Proposition 2.5.6. ([28],[42]) We have that

Der (A, M) H1(A, M) = k Innk(A, M)

and in particular, for A commutative

1 ∗ 1 ∼ ∗ H (A, A ) = HH (A) = Derk(A, A ).

Hochschild introduced the cohomology Hn(A, M) in 1945 in [21] in order to study extensions of associative algebras over ﬁelds. Almost twenty years later, Gerstenhaber made a connection between Hochschild cohomology and deformation theory. His result is below.

Theorem 2.5.7. ([16]) Let A be a k-algebra, and suppose that for each i ≥ 1 we have a k-linear

map ci : A ⊗ A −→ A. We consider the multiplication law

2 3 mt(a ⊗ b) = ab + c1(a ⊗ b)t + c2(a ⊗ b)t + c3(a ⊗ b)t + ··· 15 Then

2 (i) The multiplication law mt is associative mod t if and only if c1 is a 2-cocycle. That is,

2 c1 ∈ Z (A, A).

n n+1 (ii) Suppose that mt is associative mod t . Then mt can be extended to be associative mod t if and only if

3 c1 ◦ cn−1 + c2 ◦ cn−2 + ··· + cn−1 ◦ c1 = 0 ∈ H (A, A).

2 (iii) The class of c1 ∈ H (A, A) is determined by the isomorphism class of mt.

Remark 2.5.8. We note that in Theorem 2.5.7 the multiplication law mt is such that

mt : A[[t]] ⊗ A[[t]] −→ A[[t]].

Furthermore, mt is k[[t]]-bilinear and therefore it is sufﬁcient to only deﬁne mt on elements of A ⊗ A.

⊗n Next we recall the dual notion of homology. We deﬁne Cn(A, M) = M ⊗ A and dn :

Cn(A, M) −→ Cn−1(A, M) determined by

dn(m ⊗ a1 ⊗ · · · ⊗ an) = ma1 ⊗ a2 ⊗ · · · ⊗ an n−1 X i n + (−1) m ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an + (−1) anm ⊗ a1 ⊗ · · · ⊗ an−1. i=1

One can show that dn ◦ dn+1 = 0. We denote this complex by C•(A, M).

Deﬁnition 2.5.9. ([21],[28],[42]) The homology of C•(A, M) is denoted by Hn(A, M) and is called the Hochschild homology of A with coefﬁcients in M.

When we take M = A we denote HHn(A) := Hn(A, A) and call this the Hochschild homology associated to A. 16

Example 2.5.10. We can compute H0(A, M). Notice that Ker(d0) = M, and Im(d1) = {ma − am : m ∈ M, a ∈ A}. Thus,

M H (A, M) = . 0 {ma − am : m ∈ M, a ∈ A}

M In notation we write H (A, M) = . 0 [M,A]

Recall the following:

1 Deﬁnition 2.5.11. ([28],[42]) For a commutative k-algebra A, let ΩA|k be the left A-module of Kahler¨ differentials. It is generated by the symbols d(a) for a ∈ A with the relations

(i) d(λa + µb) = λd(a) + µd(b), and

(ii) d(ab) = ad(b) + bd(a) for all a, b ∈ A

for all λ, µ ∈ k and a, b ∈ A

Remark 2.5.12. Observe that d(1) = 0, and therefore d(λ) = 0 for all λ ∈ k.

Proposition 2.5.13. ([28],[42]) For A commutative and M an A-bimodule which is A-symmetric, we have that ∼ 1 H1(A, M) = M ⊗A ΩA|k.

In particular, ∼ 1 H1(A, A) = HH1(A) = ΩA|k.

Under certain conditions, derivations and differentials can be good substitutes for the Hochschild (co)homology groups. This is displayed in a theorem by Hochschild, Kostant, and Rosenberg given below. See [42] for the deﬁnition of a smooth algebra.

Theorem 2.5.14 (HKR). ([22]) For any smooth (commutative) algebra A over k, the antisym- metrization map

• ε∗ :ΩA|k −→ H•(A, A) 17 is an isomorphism of graded algebras. Additionally, there is an isomorphism ϕ∗ between Hochschild cohomology and the wedge product of derivations:

∗ ^ • ϕ : Derk(A, A) −→ H (A, A).

2.6 Cyclic (Co)homology

The results from this section can be found in [8], [24], [25], [28], [33], and [42]. Again letting A

⊗n+1 ⊗n+1 be a k-algebra, we consider the cyclic group action of Cn+1 = hλni on Homk(A , k) and A

⊗n+1 in order to define cyclic (co)homology. For cohomology, we first define λn : Homk(A , k) −→

⊗n+1 Homk(A , k) by

n λn(f)(a0 ⊗ · · · ⊗ an) = (−1) f(an ⊗ a0 ⊗ · · · ⊗ an−1).

n Following Connes, we then set Cλ (A) := Ker(1 − λn). One can see in [8] that the maps for Cn(A, A) (traditionally labeled bn in this case) are well-deﬁned, and the subsequent chain complex

• Cλ(A) is commonly referred to as Connes’ complex (see below).

b0 b1 b2 b3 0 −→ Homk(A, k) −−→ Ker(1 − λ1) −−→ Ker(1 − λ2) −−→ Ker(1 − λ3) −−→ ...

• n Deﬁnition 2.6.1. ([8],[28],[42]) The cohomology of Cλ(A) is denoted HC (A) and is called the cyclic cohomology of A.

⊗n+1 ⊗n+1 For cyclic homology we deﬁne λn : A −→ A by

n λn(a0 ⊗ · · · ⊗ an) = (−1) (an ⊗ a0 ⊗ · · · ⊗ an−1).

λ With Cn (A) := Cn(A, A)/(1−λn), Connes again showed in [8] that the maps for Cn(A, A) induce 18 λ λ maps on Cn (A). Connes’ complex C• (A) takes the form

b3 ⊗3 b2 ⊗2 b1 ... −−→ A /(1 − λ2) −−→ A /(1 − λ1) −−→ A −→ 0.

λ Deﬁnition 2.6.2. ([8],[28],[42]) The homology of C• (A) is denoted HCn(A) and is called the cyclic homology of A.

0 0 Example 2.6.3. By observation, HC (A) = HH (A) and HC0(A) = HH0(A).

Proposition 2.6.4. ([28],[42]) For A commutative, we have that

Ω1 HC (A) ∼= A|k . 1 d(A)

More generally, as consequence of the HKR Theorem, one can show:

Theorem 2.6.5. ([28],[42]) Let k contain Q. For any smooth (commutative) algebra A over k we have that ∼ n n−1 n−2 n−4 HCn(A) = ΩA|k/dΩA|k ⊕ HDR (A) ⊕ HDR (A) + ··· ,

n where HDR(A) is the de Rham cohomology of A in dimension n.

2.7 Secondary Hochschild Cohomology

As we saw, Hochschild cohomology describes deformations that admit a k-algebra structure. In 2013 Staic introduced the secondary Hochschild cohomology in [39] which describes deformations that admit a B-algebra structure. We give here a short overview. Let A be an associative k-algebra, B a commutative k-algebra, ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A) (so in fact, A is a B-algebra), and M an A-bimodule which is B-symmetric (that is, ε(α)m = mε(α) for all α ∈ B and all m ∈ M). Recall the following construction from [39] and [40]: Deﬁne

n ⊗n ⊗ n(n−1) C ((A, B, ε); M) := Homk(A ⊗ B 2 ,M) 19 and

ε n n+1 δn : C ((A, B, ε); M) −→ C ((A, B, ε); M)

determined by

   a0 b0,1 b0,2 ··· b0,n−2 b0,n−1 b0,n         1 a1 b1,2 ··· b1,n−2 b1,n−1 b1,n          1 1 a ··· b b b    2 2,n−2 2,n−1 2,n     ε   ......  δn(f) ⊗  ......          1 1 1 ··· an−2 bn−2,n−1 bn−2,n         1 1 1 ··· 1 a b    n−1 n−1,n    1 1 1 ··· 1 1 an

   a1 b1,2 ··· b1,n−2 b1,n−1 b1,n         1 a ··· b b b    2 2,n−2 2,n−1 2,n       ......    ......  = a0ε(b0,1b0,2 ··· b0,n−2b0,n−1b0,n)f ⊗        1 1 ··· an−2 bn−2,n−1 bn−2,n         1 1 ··· 1 a b    n−1 n−1,n    1 1 ··· 1 1 an

   a0 b0,1 ··· b0,i−1b0,i ··· b0,n−1 b0,n         1 a1 ··· b1,i−1b1,i ··· b1,n−1 b1,n       ......    ......  n    X    + (−1)if ⊗     1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,n−1bi,n−1 bi−1,nbi,n    i=1   ......    ......          1 1 ··· 1 ··· a b    n−1 n−1,n     1 1 ··· 1 ··· 1 an 20    a0 b0,1 b0,2 ··· b0,n−2 b0,n−1         1 a b ··· b b    1 1,2 1,n−2 1,n−1       1 1 a ··· b b  n+1   2 2,n−2 2,n−1  +(−1) f ⊗   anε(bn−1,nbn−2,n ··· b2,nb1,nb0,n).   ......    ......          1 1 1 ··· a b    n−2 n−2,n−1    1 1 1 ··· 1 an−1

ε ε • It was shown in [39] that δn+1 ◦ δn = 0. We denote the resulting chain complex C ((A, B, ε); M).

Deﬁnition 2.7.1. ([39]) The cohomology of C•((A, B, ε); M) is denoted by Hn((A, B, ε); M) and is called the secondary Hochschild cohomology of the triple (A, B, ε) with coefﬁcients in M.

Example 2.7.2. ([39]) Notice that when we take B = k, Hn((A, k, ε); M) = Hn(A, M).

Proposition 2.7.3. ([40]) In low dimension we have

H0((A, B, ε); M) = H0(A, M) = [M,A], and Der (A, M) H1((A, B, ε); M) = B . Innk(A, M)

Theorem 2.7.4. ([39]) Let A be a k-algebra, B a commutative k-algebra, and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A). Suppose that for each i ≥ 1 we have a k-linear

map ci : A ⊗ A ⊗ B −→ A. We consider a family of products M = {mα,t}α∈B where

2 3 mα,t(a ⊗ b) = ε(α)ab + c1(a ⊗ b ⊗ α)t + c2(a ⊗ b ⊗ α)t + c3(a ⊗ b ⊗ α)t + ···

Then

2 (i) The family M is associative mod t if and only if c1 is a 2-cocycle.

(ii) Suppose that M is associative mod tn. Then M can be extended to be associative mod tn+1 21 if and only if

3 c1 ◦ cn−1 + c2 ◦ cn−2 + ··· + cn−1 ◦ c1 = 0 ∈ H ((A, B, ε); A).

2 (iii) The class of c1 ∈ H ((A, B, ε); A) is determined by the isomorphism class of M.

Remark 2.7.5. We note that in Theorem 2.7.4 each mα,t is such that

mα,t : A[[t]] ⊗ A[[t]] −→ A[[t]].

Note that mα,t is k[[t]]-bilinear and therefore it is sufﬁcient to only deﬁne mα,t on elements of A ⊗ A. By associativity of M we mean

mβγ,t(mα,t ⊗ id) = mαβ,t(id ⊗mγ,t) for all α, β, γ ∈ B.

2.8 Higher Order Hochschild (Co)homology

For this section we refer to [2], [9], [10], [11], [18], [19], and [35]. The explicit construction involving a simplicial set was discovered in [35]. We recall it here. Let A be a commutative k-algebra and M an A-bimodule which is A-symmetric. Let V be a

finite pointed set. We can identify V with v+ = {0, 1, 2, . . . , v} where |V | = v + 1. We let L(A, M) be a functor from the category of finite pointed sets to the category of k-vector spaces. Here we define

⊗v L(A, M)(V ) = L(A, M)(v+) = M ⊗ A ,

and for ϕ : V = v+ −→ W = w+ we have

L(A, M)(ϕ): L(A, M)(v+) −→ L(A, M)(w+) 22 which is determined as follows:

L(A, M)(ϕ)(m0 ⊗ a1 ⊗ · · · ⊗ av) = m0b0 ⊗ b1 ⊗ · · · ⊗ bw where Y bi = aj.

{j∈v+ : j6=0, ϕ(j)=i}

Take X• to be a pointed simplicial set such that |Xn| = sn + 1. We identify Xn with (sn)+ =

{0, 1, 2, . . . , sn}. Deﬁne

X• ⊗sn Cn = L(A, M)(Xn) = M ⊗ A .

∗ X• X• For 0 ≤ i ≤ n and di : Xn −→ Xn−1 we deﬁne di := L(A, M)(di) and take ∂n : Cn −→ Cn−1 as n X i ∗ ∂n := (−1) di . i=0

X• The homology of this complex is denoted Hn (A, M) and is called the higher order Hochschild

homology group of A with values in M over the simplicial set X•. The cohomology groups are deﬁned in a similar manner, yet we will not use them in this dissertation.

Remark 2.8.1. The higher order Hochschild (co)homology groups depend only on the homotopy

type of the simplicial set X•.

The two concepts of higher order Hochschild cohomology and secondary Hochschild cohomology were united in [11]. In this dissertation we will use the theory of higher order Hochschild homology in Section 4.2. The simplicial sets we employ will be the circle S1 and the sphere S2. 23

CHAPTER 3 SIMPLICIAL STRUCTURES

The contents of this chapter were introduced in [27]. Unlike in the paper, however, we have supplemented the examples and results with details.

3.1 Simplicial Algebras

In this section we deﬁne simplicial k-algebras. We will use these simplicial k-algeras for the remaining sections of Chapter 3, as well as Chapter 4.

3.1.1 Deﬁnition

A simplicial k-algebra is simply a simplicial object in the category of k-algebras. There is a detailed deﬁnition below.

Deﬁnition 3.1.1. ([27]) A simplicial k-algebra is a collection of k-algebras An together with morphisms of k-algebras

A δi : An −→ An−1 and

A σi : An −→ An+1 for 0 ≤ i ≤ n such that (2.2) and (2.3) are satisﬁed.

Remark 3.1.2. Notice that the maps δi and σi do in fact rely on n. Typically it is understood how to compose these maps, but when it is unclear, we will take the time to put the superscripts and denote it as

n δi : An −→ An−1 and

n σi : An −→ An+1, for 0 ≤ i ≤ n. Here the A is omitted since it is understood the operation is on the simplicial k-algebra A. 24 3.1.2 Examples of Simplicial Algebras

Here we present the main examples on which we will rely. They can also be found in [27].

Example 3.1.3. ([27]) Let A be a k-algebra. We deﬁne the simplicial k-algebra A(A ⊗ Aop) by

op A A setting An = A ⊗ A for all n, δi = idA⊗Aop , and σi = idA⊗Aop .

Proof. It is trivial to check that (2.2) and (2.3) are satisﬁed.

Example 3.1.4. ([27]) Let A be a k-algebra, B a commutative k-algebra, and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A). We deﬁne the simplicial k-algebra A(A, B, ε)

⊗2n+1 op A by setting An = A ⊗ B ⊗ A , and δi : An −→ An−1 determined by

A δ0 (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= aε(α1) ⊗ α2 ⊗ · · · ⊗ αn ⊗ γβ1 ⊗ β2 ⊗ · · · ⊗ βn ⊗ b,

A δi (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · βn ⊗ b for 1 ≤ i ≤ n − 1, and

A δn (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αn−1 ⊗ αnγ ⊗ β1 ⊗ · · · ⊗ βn−1 ⊗ ε(βn)b.

A Moreover, let σi : An −→ An+1 be determined by

A σ0 (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ 1 ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ 1 ⊗ β1 ⊗ · · · ⊗ βn ⊗ b, 25

A σi (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · βn ⊗ b for 1 ≤ i ≤ n − 1, and

A σn (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αn ⊗ 1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ 1 ⊗ b.

Proof. We ﬁrst verify equation (2.2).

δiδi+1(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi(a ⊗ α1 ⊗ · · · ⊗ αi+1αi+2 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi+1βi+2 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αiαi+1αi+2 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1βi+2 ⊗ · · · ⊗ βn ⊗ b

= δi(a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)

= δiδi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Suppose that i < j. Then we have

δiδj(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi(a ⊗ α1 ⊗ · · · ⊗ αjαj+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βjβj+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αjαj+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βjβj+1 ⊗ · · · ⊗ βn ⊗ b

= δj−1(a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)

= δj−1δi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b). 26 Next we show (2.3) (i).

σiσi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= σi(a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b

= σi+1(a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

= σi+1σi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Suppose that i ≤ j. Then we have

σiσj(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= σi(a ⊗ α1 ⊗ · · · ⊗ αj ⊗ 1 ⊗ αj+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βj ⊗ 1 ⊗ βj+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αj ⊗ 1 ⊗ αj+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βj ⊗ 1 ⊗ βj+1 ⊗ · · · ⊗ βn ⊗ b

= σj+1(a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

= σj+1σi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Now we show (2.3) (ii).

δiσi+1(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi(a ⊗ α1 ⊗ · · · ⊗ αi+1 ⊗ 1 ⊗ αi+2 ⊗ · · · ⊗ αn ⊗ γ 27

⊗ β1 ⊗ · · · ⊗ βi+1 ⊗ 1 ⊗ βi+2 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ 1 ⊗ αi+2 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ 1 ⊗ βi+2 ⊗ · · · ⊗ βn ⊗ b

= σi(a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)

= σiδi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Suppose that i < j. Then we have

δiσj(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi(a ⊗ α1 ⊗ · · · ⊗ αj ⊗ 1 ⊗ αj+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βj ⊗ 1 ⊗ βj+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αj ⊗ 1 ⊗ αj+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βj ⊗ 1 ⊗ βj+1 ⊗ · · · ⊗ βn ⊗ b

= σj−1(a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)

= σj−1δi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Next we show (2.3) (iii).

δiσi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi(a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αi · 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi · 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b

= a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b.

δiσi−1(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) 28

= δi(a ⊗ α1 ⊗ · · · ⊗ αi−1 ⊗ 1 ⊗ αi ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi−1 ⊗ 1 ⊗ βi ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αi−1 ⊗ 1 · αi ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi−1 ⊗ 1 · βi ⊗ · · · ⊗ βn ⊗ b

= a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b.

Finally we show (2.3) (iv).

δi+2σi(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi+2(a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1αi+2 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1βi+2 ⊗ · · · ⊗ βn ⊗ b

= σi(a ⊗ α1 ⊗ · · · ⊗ αi+1αi+2 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi+1βi+2 ⊗ · · · ⊗ βn ⊗ b)

= σiδi+1(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Suppose that i > j + 1. Then we have

δiσj(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)

= δi(a ⊗ α1 ⊗ · · · ⊗ αj ⊗ 1 ⊗ αj+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βj ⊗ 1 ⊗ βj+1 ⊗ · · · ⊗ βn ⊗ b)

= a ⊗ α1 ⊗ · · · ⊗ αj ⊗ 1 ⊗ αj+1 ⊗ · · · ⊗ αi−1αi ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βj ⊗ 1 ⊗ βj+1 ⊗ · · · ⊗ βi−1βi ⊗ · · · ⊗ βn ⊗ b

= σj(a ⊗ α1 ⊗ · · · ⊗ αi−1αi ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi−1βi ⊗ · · · ⊗ βn ⊗ b)

= σjδi−1(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Thus, by Deﬁnition 3.1.1, A(A, B, ε) is a simplicial k-algebra. 29 Remark 3.1.5. Notice that if one takes B = k for Example 3.1.4, one ends up with Example 3.1.3.

3.2 Simplicial Modules

Here we introduce simplicial modules over simplicial algebras. This construction is vital to- wards working with the secondary Hochschild (co)homologies.

3.2.1 Deﬁnition

For this subsection we ﬁx A to be a simplicial k-algebra.

Deﬁnition 3.2.1. ([27]) We say that M is a simplicial left module over the simplicial k-algebra

A if M = (Mn)n≥0 is a simplicial k-vector space (satisﬁes (2.2) and (2.3)) together with a left

An-module structure on Mn for all n ≥ 0 such that we have the following natural compatibility conditions: M A M δi (anmn) = δi (an)δi (mn), (3.1) M A M σi (anmn) = σi (an)σi (mn)

for all an ∈ An, for all mn ∈ Mn, and for 0 ≤ i ≤ n.

With a slight modiﬁcation, we have an analogous deﬁnition for simplicial right modules.

Deﬁnition 3.2.2. ([27]) We say that M is a simplicial right module over the simplicial k-algebra

A if M = (Mn)n≥0 is a simplicial k-vector space (satisﬁes (2.2) and (2.3)) together with a right

An-module structure on Mn for all n ≥ 0 such that we have the following natural compatibility conditions: M M A δi (mnan) = δi (mn)δi (an), (3.2) M M A σi (mnan) = σi (mn)σi (an)

for all an ∈ An, for all mn ∈ Mn, and for 0 ≤ i ≤ n.

3.2.2 Examples of Simplicial Modules

In this subsection we present some examples of simplicial modules that will be used in the remaining sections of Chapter 3 and Section 4.1. 30 Example 3.2.3. ([27]) Let A be a k-algebra and M an A-bimodule. We deﬁne the simplicial right module M(M) over the simplicial k-algebra A(A ⊗ Aop) (from Example 3.1.3) by setting

M M Mn = M for all n, δi = idM , and σi = idM .

op For an ⊗ bn ∈ An = A ⊗ A and mn ∈ Mn = M, the product mn · (an ⊗ bn) is given by

bnmnan.

M M Proof. We begin by noticing that since δi = idM and σi = idM , the equations from (2.2) and (2.3) are immediately satisﬁed. It now sufﬁces to check the compatibility conditions from (3.2). Notice that

M M M A δi (mn · (an ⊗ bn)) = δi (bnmnan) = bnmnan = mn · (an ⊗ bn) = δi (mn)δi (an ⊗ bn) and

M M M A σi (mn · (an ⊗ bn)) = σi (bnmnan) = bnmnan = mn · (an ⊗ bn) = σi (mn)σi (an ⊗ bn).

Hence, M(M) is a simplicial right module over A(A ⊗ Aop).

Example 3.2.4 (Bar Resolution). ([27]) Let A be a k-algebra. We deﬁne the simplicial left module

op ⊗n+2 B(A) over the simplicial k-algebra A(A ⊗ A ) (from Example 3.1.3) by setting Bn = A for all n ≥ 0. The left module structure is given by

(a ⊗ b) · (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1) = aa0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1b.

Furthermore, for 0 ≤ i ≤ n we take

B δi (a0 ⊗ · · · ⊗ an+1) = a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1, and

B σi (a0 ⊗ · · · ⊗ an+1) = a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1. 31 Proof. We ﬁrst verify that equations (2.2) and (2.3) are satisﬁed. We start with (2.2).

δiδi+1(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ ai+1ai+2 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ aiai+1ai+2 ⊗ · · · ⊗ an+1

= δi(a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1)

= δiδi(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Suppose that i < j. Then we have

δiδj(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ an+1

= δj−1(a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1)

= δj−1δi(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Next we show (2.3) (i).

σiσi(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = σi(a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1

= σi+1(a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1)

= σi+1σi(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Suppose that i ≤ j. Then we have

σiσj(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = σi(a0 ⊗ a1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ an+1

= σj+1(a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1)

= σj+1σi(a0 ⊗ a1 ⊗ · · · ⊗ an+1). 32 Now we show (2.3) (ii).

δiσi+1(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ ai+1 ⊗ 1 ⊗ ai+2 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ 1 ⊗ ai+2 ⊗ · · · ⊗ an+1

= σi(a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1)

= σiδi(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Suppose that i < j. Then we have

δiσj(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ an+1

= σj−1(a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1)

= σj−1δi(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Next we show (2.3) (iii).

δiσi(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ ai · 1 ⊗ ai+1 ⊗ · · · ⊗ an+1

= a0 ⊗ a1 ⊗ · · · ⊗ an+1.

δiσi−1(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ ai−1 ⊗ 1 ⊗ ai ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ ai−1 ⊗ 1 · ai ⊗ · · · ⊗ an+1

= a0 ⊗ a1 ⊗ · · · ⊗ an+1. 33 Finally we show (2.3) (iv).

δi+2σi(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi+2(a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1ai+2 ⊗ · · · ⊗ an+1

= σi(a0 ⊗ a1 ⊗ · · · ⊗ ai+1ai+2 ⊗ · · · ⊗ an+1)

= σiδi+1(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Suppose that i > j + 1. Then we have

δiσj(a0 ⊗ a1 ⊗ · · · ⊗ an+1) = δi(a0 ⊗ a1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ an+1)

= a0 ⊗ a1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ ai−1ai ⊗ · · · ⊗ an+1

= σj(a0 ⊗ a1 ⊗ · · · ⊗ ai−1ai ⊗ · · · ⊗ an+1)

= σjδi−1(a0 ⊗ a1 ⊗ · · · ⊗ an+1).

Next we check the compatibility conditions (3.1). Notice that we have

B δi ((a ⊗ b) · (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1))

B = δi (aa0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1b)

= aa0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an ⊗ an+1b

= (a ⊗ b) · (a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an ⊗ an+1)

A B = δi (a ⊗ b)δi (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1),

and

B σi ((a ⊗ b) · (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1))

B = σi (aa0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1b)

= aa0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an ⊗ an+1b 34

= (a ⊗ b) · (a0 ⊗ a1 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an ⊗ an+1)

A B = σi (a ⊗ b)σi (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1), which gives us the simplicial left module structure. Thus, by Deﬁnition 3.2.1, B(A) is a simplicial left module over A(A ⊗ Aop).

The next three examples are lengthy, but will be useful with the construction of the secondary

Hochschild homologies Hn((A, B, ε); M) and HHn(A, B, ε).

Example 3.2.5. ([27]) Let A be a k-algebra, B a commutative k-algebra, ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A), and M an A-bimodule which is B-symmetric. We deﬁne the simplicial right module S(M) over the simplicial k-algebra A(A, B, ε) (from Example 3.1.4)

S S by setting Sn = M for all n, δi = idM , and σi = idM . The product is determined by the bimodule structure of M over A, which is given by

mn · (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) = bmnaε(α1 ··· αnγβ1 ··· βn).

S S Proof. Just as in Example 3.1.3 and Example 3.2.3, we again notice that since δi = idM and σi =

idM , the equations from (2.2) and (2.3) are immediately satisﬁed. Next we check the compatibility conditions (3.2). We have that

S δi (mn · (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b))

S = δi (bmnaε(α1 ··· αnγβ1 ··· βn))

= bmnaε(α1 ··· αnγβ1 ··· βn)

= mn · (a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)

S A = δi (mn)δi (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b), 35 and

S σi (mn · (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b))

S = σi (bmnaε(α1 ··· αnγβ1 ··· βn))

= bmnaε(α1 ··· αnγβ1 ··· βn)

= mn · (a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

S A = σi (mn)σi (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b).

Thus by Deﬁnition 3.2.2, S(M) is a simplicial right module over A(A, B, ε).

The following example is important because it is one of the key examples of the paper [27].

Example 3.2.6 (Secondary Bar Resolution). ([27]) Let A be a k-algebra, B a commutative k- algebra, and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A). We deﬁne the simplicial left module B(A, B, ε) over the simplicial k-algebra A(A, B, ε) (from Example 3.1.4)

⊗n+2 ⊗ (n+1)(n+2) by setting Bn = A ⊗ B 2 for all n ≥ 0. The left module structure is given by

   a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1      . . . . .  (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) · ⊗  ......             1 1 ··· an bn,n+1    1 1 ··· 1 an+1

  aa0 α1b0,1 ··· αnb0,n γb0,n+1      1 a1 ··· b1,n b1,n+1β1     . . . . .  = ⊗  ......  .        1 1 ··· an bn,n+1βn   1 1 ··· 1 an+1b 36 Furthermore, for 0 ≤ i ≤ n we take

   a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    B   . . . . .  δ ⊗  ......  i            1 1 ··· an bn,n+1    1 1 ··· 1 an+1

  a0 b0,1 ··· b0,ib0,i+1 ··· b0,n b0,n+1      1 a1 ··· b1,ib1,i+1 ··· b1,n b1,n+1     ......   ......   . . . . .    = ⊗   ,  1 1 ··· aiε(bi,i+1)ai+1 ··· bi,nbi+1,n bi,n+1bi+1,n+1    ......   ......       1 1 ··· 1 ··· a b   n n,n+1    1 1 ··· 1 ··· 1 an+1 and    a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    B   . . . . .  σ ⊗  ......  i            1 1 ··· an bn,n+1    1 1 ··· 1 an+1 37   a0 b0,1 ··· b0,i 1 b0,i+1 ··· b0,n b0,n+1      1 a ··· b 1 b ··· b b   1 1,i 1,i+1 1,n 1,n+1     ......   ......       1 1 ··· ai 1 bi,i+1 ··· bi,n bi,n+1      = ⊗  1 1 ··· 1 1 1 ··· 1 1  .        1 1 ··· 1 1 ai+1 ··· bi+1,n bi+1,n+1    ......   ......       1 1 ··· 1 1 1 ··· a b   n n,n+1    1 1 ··· 1 1 1 ··· 1 an+1

Proof. We begin by checking the compatibility conditions (3.1). We have that

   a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    ! B   . . . . .  δ (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) · ⊗  ......  i            1 1 ··· an bn,n+1    1 1 ··· 1 an+1    aa0 α1b0,1 ··· αnb0,n γb0,n+1         1 a1 ··· b1,n b1,n+1β1     B   . . . . .  = δ ⊗  ......  i            1 1 ··· an bn,n+1βn    1 1 ··· 1 an+1b 38   aa0 α1b0,1 ··· αib0,iαi+1b0,i+1 ··· αnb0,n γb0,n+1      1 a1 ··· b1,ib1,i+1 ··· b1,n b1,n+1β1     ......   ......   . . . . .    = ⊗    1 1 ··· aiε(bi,i+1)ai+1 ··· bi,nbi+1,n bi,n+1βibi+1,n+1βi+1    ......   ......       1 1 ··· 1 ··· a b β   n n,n+1 n    1 1 ··· 1 ··· 1 an+1b   aa0 α1b0,1 ··· αiαi+1b0,ib0,i+1 ··· αnb0,n γb0,n+1      1 a1 ··· b1,ib1,i+1 ··· b1,n b1,n+1β1     ......   ......   . . . . .    = ⊗    1 1 ··· aiε(bi,i+1)ai+1 ··· bi,nbi+1,n bi,n+1bi+1,n+1βiβi+1    ......   ......       1 1 ··· 1 ··· a b β   n n,n+1 n    1 1 ··· 1 ··· 1 an+1b

= (a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)    a0 b0,1 ··· b0,ib0,i+1 ··· b0,n b0,n+1         1 a1 ··· b1,ib1,i+1 ··· b1,n b1,n+1       . . . . .    ......    . . . . .     · ⊗     1 1 ··· aiε(bi,i+1)ai+1 ··· bi,nbi+1,n bi,n+1bi+1,n+1      ......    ......          1 1 ··· 1 ··· a b    n n,n+1     1 1 ··· 1 ··· 1 an+1 39    a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    A B   . . . . .  = δ (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)δ ⊗  ......  , i i            1 1 ··· an bn,n+1    1 1 ··· 1 an+1 and

   a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    ! B   . . . . .  σ (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) · ⊗  ......  i            1 1 ··· an bn,n+1    1 1 ··· 1 an+1    aa0 α1b0,1 ··· αnb0,n γb0,n+1         1 a1 ··· b1,n b1,n+1β1     B   . . . . .  = σ ⊗  ......  i            1 1 ··· an bn,n+1βn    1 1 ··· 1 an+1b   aa0 α1b0,1 ··· αib0,i 1 αi+1b0,i+1 ··· αnb0,n γb0,n+1      1 a ··· b 1 b ··· b b β   1 1,i 1,i+1 1,n 1,n+1 1     ......   ......       1 1 ··· ai 1 bi,i+1 ··· bi,n bi,n+1βi      = ⊗  1 1 ··· 1 1 1 ··· 1 1         1 1 ··· 1 1 ai+1 ··· bi+1,n bi+1,n+1βi+1    ......   ......       1 1 ··· 1 1 1 ··· a b β   n n,n+1 n    1 1 ··· 1 1 1 ··· 1 an+1b 40

= (a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)    a0 b0,1 ··· b0,i 1 b0,i+1 ··· b0,n b0,n+1         1 a ··· b 1 b ··· b b    1 1,i 1,i+1 1,n 1,n+1       ......    ......          1 1 ··· ai 1 bi,i+1 ··· bi,n bi,n+1        · ⊗  1 1 ··· 1 1 1 ··· 1 1             1 1 ··· 1 1 ai+1 ··· bi+1,n bi+1,n+1      ......    ......          1 1 ··· 1 1 1 ··· a b    n n,n+1     1 1 ··· 1 1 1 ··· 1 an+1    a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    A B   . . . . .  = σ (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b)σ ⊗  ......  i i            1 1 ··· an bn,n+1    1 1 ··· 1 an+1 establish the simplicial left module structure over A(A, B, ε). Next we check that the equations (2.2) and (2.3) are satisﬁed. First we check (2.2).

   a0 ··· b0,n+1      . . .  δiδi+1 ⊗  . .. .        1 ··· an+1    a0 ··· b0,i+1b0,i+2 ··· b0,n+1      ......    . . . . .        = δi ⊗  1 ··· a ε(b )a ··· b b    i+1 i+1,i+2 i+2 i+1,n+1 i+2,n+1      ......    . . . . .     1 ··· 1 ··· an+1 41   a0 ··· b0,ib0,i+1b0,i+2 ··· b0,n+1    ......   . . . . .      = ⊗  1 ··· a ε(b )a ε(b )a ··· b b b   i i,i+1 i+1 i+1,i+2 i+2 i,n+1 i+1,n+1 i+2,n+1    ......   . . . . .    1 ··· 1 ··· an+1    a0 ··· b0,ib0,i+1 ··· b0,n+1      ......    . . . . .        = δi ⊗  1 ··· a ε(b )a ··· b b    i i,i+1 i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1    a0 ··· b0,n+1      . . .  = δiδi ⊗  . .. .  .       1 ··· an+1

Suppose that i < j. Then we have

   a0 ··· b0,n+1      . . .  δiδj ⊗  . .. .        1 ··· an+1    a0 ··· b0,jb0,j+1 ··· b0,n+1      ......    . . . . .        = δi ⊗  1 ··· a ε(b )a ··· b b    j j,j+1 j+1 j,n+1 j+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1 42   a0 ··· b0,ib0,i+1 ··· b0,jb0,j+1 ··· b0,n+1    ......   ......       1 ··· a ε(b )a ··· b b b b ··· b b   i i,i+1 i+1 i,j i,j+1 i+1,j i+1,j+1 i,n+1 i+1,n+1     ......  = ⊗  ......       1 ··· 1 ··· ajε(bj,j+1)aj+1 ··· bj,n+1bj+1,n+1    . . . .   ......   . . . .    1 ··· 1 ··· 1 ··· an+1    a0 ··· b0,ib0,i+1 ··· b0,n+1      ......    . . . . .        = δj−1 ⊗  1 ··· a ε(b )a ··· b b    i i,i+1 i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1    a0 ··· b0,n+1      . . .  = δj−1δi ⊗  . .. .  .       1 ··· an+1

Next we show (2.3) (i).

   a0 ··· b0,i 1 b0,i+1 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   i i,i+1 i,n+1          . .. .     σiσi ⊗  . . .  = σi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 ai+1 ··· bi+1,n+1      ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1 43   a0 ··· b0,i 1 1 b0,i+1 ··· b0,n+1    ......   ......       1 ··· a 1 1 b ··· b   i i,i+1 i,n+1       1 ··· 1 1 1 1 ··· 1  = ⊗      1 ··· 1 1 1 1 ··· 1       1 ··· 1 1 1 a ··· b   i+1 i+1,n+1    ......   ......    1 ··· 1 1 1 1 ··· an+1    a0 ··· b0,i 1 b0,i+1 ··· b0,n+1      ......    ......          1 ··· a 1 b ··· b    i i,i+1 i,n+1        = σi+1 ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+1 ··· bi+1,n+1      . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1    a0 ··· b0,n+1      . . .  = σi+1σi ⊗  . .. .  .       1 ··· an+1 44 Suppose that i ≤ j. Then we have

   a0 ··· b0,j 1 b0,j+1 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   j j,j+1 j,n+1          . .. .     σiσj ⊗  . . .  = σi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 aj+1 ··· bj+1,n+1      . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1   a0 ··· b0,i 1 b0,i+1 ··· b0,j 1 b0,j+1 ··· b0,n+1    ......   ......       1 ··· a 1 b ··· b 1 b ··· b   i i,i+1 i,j i,j+1 i,n+1       1 ··· 1 1 1 ··· 1 1 1 ··· 1       1 ··· 1 1 ai+1 ··· bi+1,j 1 bi+1,j+1 ··· bi+1,n+1    ......  = ⊗  ......         1 ··· 1 1 1 ··· aj 1 bj,j+1 ··· bj,n+1       1 ··· 1 1 1 ··· 1 1 1 ··· 1       1 ··· 1 1 1 ··· 1 1 a ··· b   j+1 j+1,n+1    ......   ......    1 ··· 1 1 1 ··· 1 1 1 ··· an+1    a0 ··· b0,i 1 b0,i+1 ··· b0,n+1      ......    ......          1 ··· a 1 b ··· b    i i,i+1 i,n+1        = σj+1 ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+1 ··· bi+1,n+1      . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1 45    a0 ··· b0,n+1      . . .  = σj+1σi ⊗  . .. .  .       1 ··· an+1

Now we check (2.3) (ii).

   a0 ··· b0,i+1 1 b0,i+2 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   i+1 i+1,i+2 i+1,n+1         . .. .     δiσi+1 ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 ai+2 ··· bi+2,n+1      ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1   a0 ··· b0,ib0,i+1 1 b0,i+2 ··· b0,n+1    ......   ......       1 ··· a ε(b )a 1 b b ··· b b   i i,i+1 i+1 i,i+2 i+1,i+2 i,n+1 i+1,n+1     = ⊗  1 ··· 1 1 1 ··· 1       1 ··· 1 1 ai+2 ··· bi+2,n+1     ......   ......   . . . . .    1 ··· 1 1 1 ··· an+1    a0 ··· b0,ib0,i+1 ··· b0,n+1      ......    . . . . .        = σi ⊗  1 ··· a ε(b )a ··· b b    i i,i+1 i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1 46    a0 ··· b0,n+1      . . .  = σiδi ⊗  . .. .  .       1 ··· an+1

Suppose i < j. Then we have

   a0 ··· b0,j 1 b0,j+1 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   j j,j+1 j,n+1          . .. .     δiσj ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 aj+1 ··· bj+1,n+1      . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1   a0 ··· b0,ib0,i+1 ··· b0,j 1 b0,j+1 ··· b0,n+1    ......   ......         1 ··· aiε(bi,i+1)ai+1 ··· bi,jbi+1,j 1 bi,j+1bi+1,j+1 ··· bi,n+1bi+1,n+1    ......   ......      = ⊗  1 ··· 1 ··· a 1 b ··· b   j j,j+1 j,n+1       1 ··· 1 ··· 1 1 1 ··· 1       1 ··· 1 ··· 1 1 aj+1 ··· bj+1,n+1     ......   ......      1 ··· 1 ··· 1 1 1 ··· an+1    a0 ··· b0,ib0,i+1 ··· b0,n+1      ......    . . . . .        = σj−1 ⊗  1 ··· a ε(b )a ··· b b    i i,i+1 i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1 47    a0 ··· b0,n+1      . . .  = σj−1δi ⊗  . .. .  .       1 ··· an+1

Next we show (2.3) (iii).

   a0 ··· b0,i 1 b0,i+1 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   i i,i+1 i,n+1          . .. .     δiσi ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 ai+1 ··· bi+1,n+1      ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1   a0 ··· b0,i · 1 b0,i+1 ··· b0,n+1    ......   ......         1 ··· aiε(1)1 bi,i+1 · 1 ··· bi,n+1 · 1 = ⊗      1 ··· 1 ai+1 ··· bi+1,n+1     ......   ......      1 ··· 1 1 ··· an+1   a0 ··· b0,n+1    . . .  = ⊗  . .. .  .     1 ··· an+1 48    a0 ··· b0,i−1 1 b0,i ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   i−1 i−1,i i−1,n+1         . .. .     δiσi−1 ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 ai ··· bi,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1   a0 ··· b0,i−1 · 1 b0,i ··· b0,n+1    ......   ......         1 ··· ai−1ε(1)1 bi−1,i · 1 ··· bi−1,n+1 · 1 = ⊗      1 ··· 1 ai ··· bi,n+1     ......   ......      1 ··· 1 1 ··· an+1   a0 ··· b0,n+1    . . .  = ⊗  . .. .  .     1 ··· an+1

Finally we verify (2.3) (iv).

   a0 ··· b0,i 1 b0,i+1 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   i i,i+1 i,n+1          . .. .     δi+2σi ⊗  . . .  = δi+2 ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 ai+1 ··· bi+1,n+1      ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1 49   a0 ··· b0,i 1 b0,i+1b0,i+2 ··· b0,n+1    ......   ......       1 ··· a 1 b b ··· b   i i,i+1 i,i+2 i,n+1      = ⊗  1 ··· 1 1 1 ··· 1       1 ··· 1 1 ai+1ε(bi+1,i+2)ai+2 ··· bi+1,n+1bi+2,n+1    ......   ......   . . . . .    1 ··· 1 1 1 ··· an+1    a0 ··· b0,i+1b0,i+2 ··· b0,n+1      ......    . . . . .        = σi ⊗  1 ··· a ε(b )a ··· b b    i+1 i+1,i+2 i+2 i+1,n+1 i+2,n+1      ......    . . . . .     1 ··· 1 ··· an+1    a0 ··· b0,n+1      . . .  = σiδi+1 ⊗  . .. .  .       1 ··· an+1

Suppose i > j + 1. Then we have

   a0 ··· b0,j 1 b0,j+1 ··· b0,n+1      ......    ......           a ··· b   1 ··· a 1 b ··· b  0 0,n+1   j j,j+1 j,n+1          . .. .     δiσj ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1   1 ··· 1 1 aj+1 ··· bj+1,n+1      ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1 50   a0 ··· b0,j 1 b0,j+1 ··· b0,i−1b0,i ··· b0,n+1    ......   ......         1 ··· aj 1 bj,j+1 ··· bj,i−1bj,i ··· bj,n+1       1 ··· 1 1 1 ··· 1 ··· 1      = ⊗  1 ··· 1 1 a ··· b b ··· b   j+1 j+1,i−1 j+1,i j+1,n+1     ......   ......       1 ··· 1 1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,n+1bi,n+1    ......   ......      1 ··· 1 1 1 ··· 1 ··· an+1    a0 ··· b0,i−1b0,i ··· b0,n+1      ......    . . . . .        = σj ⊗  1 ··· a ε(b )a ··· b b    i−1 i−1,i i i−1,n+1 i,n+1      ......    . . . . .     1 ··· 1 ··· an+1    a0 ··· b0,n+1      . . .  = σjδi−1 ⊗  . .. .  .       1 ··· an+1

This is what we wanted. One can now observe that Deﬁnition 3.2.1 is satisﬁed, and therefore B(A, B, ε) is a simplicial left module over A(A, B, ε).

Example 3.2.7. ([27]) Let A be a k-algebra, B a commutative k-algebra, and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A). We deﬁne the simplicial right module L(A, B, ε)

⊗n over the simplicial k-algebra A(A, B, ε) (from Example 3.1.4) by setting Ln = A ⊗ B for all n ≥ 0. The right module structure is given by

(a0 ⊗ b1 ⊗· · ·⊗bn)·(a⊗α1 ⊗· · ·⊗αn ⊗γ ⊗β1 ⊗· · ·⊗βn ⊗b) = ba0aε(γ)⊗α1b1β1 ⊗· · · αnbnβn. 51 L Furthermore, we take δi : Ln −→ Ln−1 determined by

L δ0 (a0 ⊗ b1 ⊗ · · · ⊗ bn) = a0ε(b1) ⊗ b2 ⊗ · · · ⊗ bn,

L δi (a0 ⊗ b1 ⊗ · · · ⊗ bn) = a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn for 1 ≤ i ≤ n − 1, and

L δn (a0 ⊗ b1 ⊗ · · · ⊗ bn) = a0ε(bn) ⊗ b1 ⊗ · · · ⊗ bn−1,

L and σi : Ln −→ Ln+1 for 0 ≤ i ≤ n determined by

L σi (a0 ⊗ b1 ⊗ · · · ⊗ bn) = a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn.

L L Proof. We start by showing that δi and σi satisfy equations (2.2) and (2.3). We ﬁrst show that (2.2) is satisﬁed.

δiδi+1(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bi+1bi+2 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bibi+1bi+2 ⊗ · · · ⊗ bn+1

= δi(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

= δiδi(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Suppose i < j. Then we have

δiδj(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bjbj+1 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bjbj+1 ⊗ · · · ⊗ bn+1

= δj−1(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

= δj−1δi(a0 ⊗ b1 ⊗ · · · ⊗ bn+1). 52 Next we check (2.3) (i).

σiσi(a0 ⊗ b1 ⊗ · · · ⊗ bn−1) = σi(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

= a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1

= σi+1(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

= σi+1σi(a0 ⊗ b1 ⊗ · · · ⊗ bn−1).

Suppose i ≤ j. Then we have

σiσj(a0 ⊗ b1 ⊗ · · · ⊗ bn−1) = σi(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn−1)

= a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn−1

= σj+1(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

= σj+1σi(a0 ⊗ b1 ⊗ · · · ⊗ bn−1).

Now we show (2.3) (ii).

δiσi+1(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bi+1 ⊗ 1 ⊗ bi+2 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ 1 ⊗ bi+2 ⊗ · · · ⊗ bn+1

= σi(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

= σiδi(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Suppose i < j. Then we have

δiσj(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn+1

= σj−1(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

= σj−1δi(a0 ⊗ b1 ⊗ · · · ⊗ bn+1). 53 Next we verify (2.3) (iii).

δiσi(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bi · 1 ⊗ bi+1 ⊗ · · · ⊗ bn+1

= a0 ⊗ b1 ⊗ · · · ⊗ bn+1.

δiσi−1(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bi−1 ⊗ 1 ⊗ bi ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bi−1 ⊗ 1 · bi ⊗ · · · ⊗ bn+1

= a0 ⊗ b1 ⊗ · · · ⊗ bn+1.

Finally we show (2.3) (iv).

δi+2σi(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi+2(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1bi+2 ⊗ · · · ⊗ bn+1

= σi(a0 ⊗ b1 ⊗ · · · ⊗ bi+1bi+2 ⊗ · · · ⊗ bn+1)

= σiδi+1(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Suppose i > j + 1. Then we have

δiσj(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δi(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn+1)

= a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bi−1bi ⊗ · · · ⊗ bn+1

= σj(a0 ⊗ b1 ⊗ · · · ⊗ bi−1bi ⊗ · · · ⊗ bn+1)

= σjδi−1(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Thus the simplicial identities are satisﬁed. Next we must verify that L(A, B, ε) fulﬁlls the com- 54 patibility conditions (3.2). We have

L δi ((a0 ⊗ b1 ⊗ · · · ⊗ bn) · (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b))

L = δi (ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αnbnβn)

= ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αibiβiαi+1bi+1βi+1 ⊗ · · · ⊗ αnbnβn

= ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αiαi+1bibi+1βiβi+1 ⊗ · · · ⊗ αnbnβn

= (a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn)

· (a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn ⊗ b)

L A = δi (a0 ⊗ b1 ⊗ · · · ⊗ bn) · δi (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b), and

L σi ((a0 ⊗ b1 ⊗ · · · ⊗ bn) · (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b))

L = σi (ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αnbnβn)

= ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αibiβi ⊗ 1 ⊗ αi+1bi+1βi+1 ⊗ · · · ⊗ αnbnβn

= (a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn) · (a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn

⊗ γ ⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn ⊗ b)

L A = σi (a0 ⊗ b1 ⊗ · · · ⊗ bn) · σi (a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b),

which establishes the simplicial right module structure over A(A, B, ε). Hence L(A, B, ε) is a simplicial right module over A(A, B, ε).

3.3 Co-simplicial Modules

Here we introduce co-simplicial modules, which can be seen as the dual to simplicial modules.

3.3.1 Deﬁnition

In this subsection we ﬁx A to be a simplicial k-algebra. 55 Deﬁnition 3.3.1. ([27]) We say that M is a co-simplicial left module over the simplicial k-algebra

n A if M = (M )n≥0 is a co-simplicial k-vector space (satisﬁes (2.4) and (2.5)) together with a left

n An-module structure on M for all n ≥ 0 such that we have the following natural compatibility conditions: i A i δM(δi (an+1) · mn) = an+1 · δM(mn) (3.3) i A i σM(σi (an−1) · mn) = an−1 · σM(mn) for all an−1 ∈ An−1, for all an+1 ∈ An+1, for all mn ∈ Mn, and for 0 ≤ i ≤ n.

Remark 3.3.2. It is easy to lose track of where each of the maps are going, so we note the following: for 0 ≤ i ≤ n we have

i n n+1 δM : M −→ M ,

i n n−1 σM : M −→ M ,

A δi : An+1 −→ An, and

A σi : An−1 −→ An.

Therefore, the compatibility conditions (3.3) are consistent. We will see in the next section that these co-simplicial modules will be used for the Hom groups in a simplicial setting.

3.3.2 Examples of Co-simplicial Modules

In this subsection we present some examples of co-simplicial modules that will be used in the remaining sections of Chapter 3 and Section 4.1.

Example 3.3.3. ([27]) Let A be a k-algebra and M an A-bimodule. We deﬁne the co-simplicial left module N (M) over the simplicial k-algebra A(A ⊗ Aop) (from Example 3.1.3) by setting

i i Nn = M for all n ≥ 0, δN = idM , and σN = idM .

op For an ⊗ bn ∈ An = A ⊗ A and mn ∈ Nn = M, the product (an ⊗ bn) · mn is given by anmnbn.

i i Proof. We begin by noticing that since δN = idM and σN = idM , the equations from (2.4) and 56 (2.5) are immediately satisﬁed. It now sufﬁces to check the compatibility conditions from (3.3). Notice that

i A i i δN (δi (an+1 ⊗ bn+1) · mn) = δN ((an+1 ⊗ bn+1) · mn) = δN (an+1mnbn+1)

i = an+1mnbn+1 = (an+1 ⊗ bn+1) · δN (mn) and

i A i i σN (σi (an−1 ⊗ bn−1) · mn) = σN ((an−1 ⊗ bn−1) · mn) = σN (an−1mnbn−1)

i = an−1mnbn−1 = (an−1 ⊗ bn−1) · σN (mn).

Hence, N (M) is a co-simplicial left module over A(A ⊗ Aop).

Example 3.3.4. Let A be a k-algebra, B a commutative k-algebra, ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A), and M an A-bimodule which is B-symmetric. We deﬁne the co-simplicial left module C(M) over the simplicial k-algebra A(A, B, ε) (from Example 3.1.4) by

n i i setting C = M for all n, δC = idM , and σC = idM . The product is determined by the bimodule structure of M over A, which is given by

(a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) · mn = amnbε(α1 ··· αnγβ1 ··· βn).

i i Proof. Since δC = idM and σC = idM , the equations from (2.4) and (2.5) are immediately satisﬁed. Next we check the compatibility conditions (3.3). We have that:

i A δC(δi (a ⊗ α1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn+1 ⊗ b) · mn)

i = δC((a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn+1 ⊗ b) · mn)

i = δC(amnbε(α1 ··· αiαi+1 ··· αn+1γβ1 ··· βiβi+1 ··· βn+1)) 57

= amnbε(α1 ··· αn+1γβ1 ··· βn+1)

= (a ⊗ α1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn+1 ⊗ b) · mn

i = (a ⊗ α1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn+1 ⊗ b) · δC(mn), and

i A σC(σi (a ⊗ α1 ⊗ · · · ⊗ αn−1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn−1 ⊗ b) · mn)

i = σC((a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn−1

⊗ γ ⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn−1 ⊗ b) · mn)

i = σC(amnbε(α1 ··· αi · 1 · αi+1 ··· αn−1γβ1 ··· βi · 1 · βi+1 ··· βn−1))

= amnbε(α1 ··· αn−1γβ1 ··· βn−1)

= (a ⊗ α1 ⊗ · · · ⊗ αn−1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn−1 ⊗ b) · mn

i = (a ⊗ α1 ⊗ · · · ⊗ αn−1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn−1 ⊗ b) · σC(mn), which show the co-simplicial left module structure over A(A, B, ε). Thus, C(M) is a co-simplicial left module over A(A, B, ε).

Example 3.3.5. ([27]) Let A be a k-algebra, B a commutative k-algebra, and ε : B −→ A a morphism of k-algebras such that ε(B) ⊆ Z(A). We deﬁne the co-simplicial left module H(A, B, ε)

n over the simplicial k-algebra A(A, B, ε) (from Example 3.1.4) by setting H = Homk(A ⊗

⊗n B , k). The left An-module structure is given by

((a ⊗ α1 ⊗ · · · ⊗ αn ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn ⊗ b) · f)(a0 ⊗ b1 ⊗ · · · ⊗ bn)

= f(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αnbnβn). 58 i n n+1 Furthermore, we take δH : H −→ H by

0 δH(f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = f(a0ε(b1) ⊗ b2 ⊗ · · · ⊗ bn+1),

i δH(f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = f(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1) for 1 ≤ i ≤ n, and

n+1 δH (f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = f(a0ε(bn+1) ⊗ b1 ⊗ · · · ⊗ bn),

i n n−1 and σH : H −→ H for 0 ≤ i ≤ n − 1 by

i σH(f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1) = f(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1).

i i Proof. We start by showing that δH and σH satisfy equations (2.4) and (2.5). We ﬁrst check that (2.4) is satisﬁed.

i+1 i i δ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi+1bi+2 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1bi+2 ⊗ · · · ⊗ bn+1)

i = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

i i = δ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Suppose i < j. Then we have

j i i δ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bjbj+1 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bjbj+1 ⊗ · · · ⊗ bn+1)

j−1 = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

i j−1 = δ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1). 59 Next we check (2.5) (i).

i i i σ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1) = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

i+1 = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

i i+1 = σ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1).

Suppose i ≤ j. Then we have

j i i σ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1) = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn−1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn−1)

j+1 = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

i j+1 = σ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1).

Now we show (2.5) (ii).

i+1 i i σ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi+1 ⊗ 1 ⊗ bi+2 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ 1 ⊗ bi+2 ⊗ · · · ⊗ bn+1)

i = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

i i = δ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Suppose i < j. Then we have

j i i σ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn+1)

j−1 = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

i j−1 = δ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1). 60 Next we verify (2.5) (iii).

i i i σ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bi · 1 ⊗ bi+1 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

i−1 i i σ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi−1 ⊗ 1 ⊗ bi ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bi−1 ⊗ 1 · bi ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Finally we show (2.5) (iv).

i i+2 i+2 σ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1bi+2 ⊗ · · · ⊗ bn+1)

i = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi+1bi+2 ⊗ · · · ⊗ bn+1)

i+1 i = δ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Suppose i > j + 1. Then we have

j i i σ (δ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1) = δ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bn+1)

= f(a0 ⊗ b1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bi−1bi ⊗ · · · ⊗ bn+1)

j = σ (f)(a0 ⊗ b1 ⊗ · · · ⊗ bi−1bi ⊗ · · · ⊗ bn+1)

i−1 j = δ (σ f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1).

Thus the co-simplicial identities are satisﬁed. Next we must verify that H(A, B, ε) fulﬁlls the 61 compatibility conditions (3.3). We note that

i A δH(δi (a ⊗ α1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn+1 ⊗ b) · f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1)

i = δH((a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn+1 ⊗ b)

· f)(a0 ⊗ b1 ⊗ · · · ⊗ bn+1)

= ((a ⊗ α1 ⊗ · · · ⊗ αiαi+1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βiβi+1 ⊗ · · · ⊗ βn+1 ⊗ b) · f)

(a0 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn+1)

= f(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αiαi+1bibi+1βiβi+1 ⊗ · · · ⊗ αn+1bn+1βn+1)

= f(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αibiβiαi+1bi+1βi+1 ⊗ · · · ⊗ αn+1bn+1βn+1)

i = δH(f)(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αn+1bn+1βn+1)

i = ((a ⊗ α1 ⊗ · · · ⊗ αn+1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn+1 ⊗ b) · δH(f))(a0 ⊗ b1 ⊗ · · · ⊗ bn+1), and likewise

i A σH(σi (a ⊗ α1 ⊗ · · · ⊗ αn−1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn−1 ⊗ b) · f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1)

i = σH((a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn−1 ⊗ γ

⊗ β1 ⊗ · · · ⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn−1 ⊗ b) · f)(a0 ⊗ b1 ⊗ · · · ⊗ bn−1)

= ((a ⊗ α1 ⊗ · · · ⊗ αi ⊗ 1 ⊗ αi+1 ⊗ · · · ⊗ αn−1 ⊗ γ ⊗ β1 ⊗ · · ·

⊗ βi ⊗ 1 ⊗ βi+1 ⊗ · · · ⊗ βn−1 ⊗ b) · f)(a0 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn−1)

= f(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αibiβi ⊗ 1 · 1 · 1 ⊗ αi+1bi+1βi+1 ⊗ · · · ⊗ αn−1bn−1βn−1)

= f(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αibiβi ⊗ 1 ⊗ αi+1bi+1βi+1 ⊗ · · · ⊗ αn−1bn−1βn−1)

i = σH(f)(ba0aε(γ) ⊗ α1b1β1 ⊗ · · · ⊗ αn−1bn−1βn−1)

i = ((a ⊗ α1 ⊗ · · · ⊗ αn−1 ⊗ γ ⊗ β1 ⊗ · · · ⊗ βn−1 ⊗ b) · σH(f))(a0 ⊗ b1 ⊗ · · · ⊗ bn−1).

Hence by Deﬁnition 3.3.1, H(A, B, ε) is a co-simplicial left module over A(A, B, ε). 62 3.4 The Hom and Tensor Lemmas

This section is the culmination of the previous sections. Again, the statements of these lemmas are in [27], but here we present the proofs in detail. The Tensor Lemma unites simplicial modules in the form of the tensor product. We ﬁx A to be a simplicial k-algebra.

X X Lemma 3.4.1 (Tensor Lemma). ([27]) Suppose that (X , δi , σi ) is a simplicial right module over

Y Y the simplicial k-algebra A (see Deﬁnition 3.2.2), and (Y, δi , σi ) is a simplicial left module over

the simplicial k-algebra A (see Deﬁnition 3.2.1). Then M = (X ⊗A Y,Di,Si) is a simplicial

k-module where Mn = Xn ⊗An Yn for all n ≥ 0, and we take

Di : Mn = Xn ⊗An Yn −→ Xn−1 ⊗An−1 Yn−1 = Mn−1

determined by

X Y Di(xn ⊗An yn) = δi (xn) ⊗An−1 δi (yn).

Similarly, take

Si : Mn = Xn ⊗An Yn −→ Xn+1 ⊗An+1 Yn+1 = Mn+1

determined by

X Y Si(xn ⊗An yn) = σi (xn) ⊗An+1 σi (yn).

Proof. We show that both Di and Si are well-deﬁned. Notice that

X Y Di(xn · an ⊗An yn) = δi (xn · an) ⊗An−1 δi (yn)

X A Y = δi (xn)δi (an) ⊗An−1 δi (yn)

X A Y = δi (xn) ⊗An−1 δi (an)δi (yn)

X Y = δi (xn) ⊗An−1 δi (an · yn)

= Di(xn ⊗An an · yn). 63 Likewise,

X Y Si(xn · an ⊗An yn) = σi (xn · an) ⊗An+1 σi (yn)

X A Y = σi (xn)σi (an) ⊗An+1 σi (yn)

X A Y = σi (xn) ⊗An+1 σi (an)σi (yn)

X Y = σi (xn) ⊗An+1 σi (an · yn)

= Si(xn ⊗An an · yn).

Next we will show that Di and Si satisfy the simplicial identities (2.2) and (2.3). This will rely upon both X and Y satisfying equations (2.2) and (2.3). We start with (2.2), where i < j.

X Y X Y X X Y Y DiDj = (δi ⊗ δi )(δj ⊗ δj ) = δi δj ⊗ δi δj

X X Y Y X Y X Y = δj−1δi ⊗ δj−1δi = (δj−1 ⊗ δj−1)(δi ⊗ δi ) = Dj−1Di.

Next, we show (2.3) (i). Let i ≤ j. Then we have

X Y X Y X X Y Y SiSj = (σi ⊗ σi )(σj ⊗ σj ) = σi σj ⊗ σi σj

X X Y Y X Y X Y = σj+1σi ⊗ σj+1σi = (σj+1 ⊗ σj+1)(σi ⊗ σi ) = Sj+1Si.

Now we show (2.3) (ii), where i < j.

X Y X Y X X Y Y DiSj = (δi ⊗ δi )(σj ⊗ σj ) = δi σj ⊗ δi σj

X X Y Y X Y X Y = σj−1δi ⊗ σj−1δi = (σj−1 ⊗ σj−1)(δi ⊗ δi ) = Sj−1Di.

Next we check (2.3) (iii).

X Y X Y X X Y Y DiSi = (δi ⊗ δi )(σi ⊗ σi ) = δi σi ⊗ δi σi = id ⊗ id = id . 64

X Y X Y X X Y Y DiSi−1 = (δi ⊗ δi )(σi−1 ⊗ σi−1) = δi σi−1 ⊗ δi σi−1 = id ⊗ id = id .

Finally we verify (2.3) (iv). Suppose i > j + 1. Then we have

X Y X Y X X Y Y DiSj = (δi ⊗ δi )(σj ⊗ σj ) = δi σj ⊗ δi σj

X X Y Y X Y X Y = σj δi−1 ⊗ σj δi−1 = (σj ⊗ σj )(δi−1 ⊗ δi−1) = SjDi−1.

Thus, X ⊗A Y is a simplicial k-module.

Again employing the simplicial structures, we give an analogous representation of the Hom functor.

X X Lemma 3.4.2 (Hom Lemma). ([27]) Suppose that (X , δi , σi ) is a simplicial left module over

i i the simplicial k-algebra A (see Deﬁnition 3.2.1), and (Y, δY , σY ) is a co-simplicial left module

i i over the simplicial k-algebra A (see Deﬁnition 3.3.1). Then M = (HomA(X , Y),D ,S ) is a

n n co-simplicial k-module where M = HomAn (Xn,Y ) for all n ≥ 0, and where for any f ∈

n HomAn (Xn,Y ) we take

i n n n+1 n+1 D : M = HomAn (Xn,Y ) −→ HomAn+1 (Xn+1,Y ) = M

determined by

i i X D (f) = δY fδi ,

and

i n n n−1 n−1 S : M = HomAn (Xn,Y ) −→ HomAn−1 (Xn−1,Y ) = M

determined by

i i X S (f) = σY fσi . 65 Proof. We use the following diagram to help keep us organized,.

X X δi σi Xn+1 Xn Xn−1

Di(f) f Si(f)

i i δY σY Y n+1 Y n Y n−1

First observe that these maps are landing in their appropriate places. Notice that

i i X D (f)(an+1 · xn+1) = δY fδi (an+1 · xn+1)

i A X = δY f(δi (an+1) · δi (xn+1))

i A X = δY (δi (an+1) · f(δi (xn+1)))

X = an+1 · δY fδi (xn+1)

i = an+1 · D (f)(xn+1).

Moreover,

i 0 i X 0 D (f)(xn+1 + xn+1) = δY fδi (xn+1 + xn+1)

i X X 0 = δY f(δi (xn+1) + δi (xn+1))

i X X 0 = δY (f(δi (xn+1)) + f(δi (xn+1)))

i X i X 0 = δY fδi (xn+1) + δY fδi (xn+1)

i i 0 = D (f)(xn+1) + D (f)(xn+1).

n i n+1 So indeed for f ∈ HomAn (Xn,Y ), we have that D (f) ∈ HomAn+1 (Xn+1,Y ). Likewise, we 66 i n−1 conclude S (f) is in fact in HomAn−1 (Xn−1,Y ) because

i i X S (f)(an−1 · xn−1) = σY fσi (an−1 · xn−1)

i A X = σY f(σi (an−1) · σi (xn−1))

i A X = σY (σi (an−1) · f(σi (xn−1)))

X = an−1 · δY fσi (xn−1)

i = an−1 · S (f)(xn−1),

and

i 0 i X 0 S (f)(xn−1 + xn−1) = σY fσi (xn−1 + xn−1)

i X X 0 = σY f(σi (xn−1) + σi (xn−1))

i X X 0 = σY (f(σi (xn−1)) + f(σi (xn−1)))

i X i X 0 = σY fσi (xn−1) + σY fσi (xn−1)

i i 0 = S (f)(xn−1) + S (f)(xn−1).

Next we verify that Di and Si satisfy (2.4) and (2.5). These will be easy to check, mostly

X X i i because (X , δi , σi ) satisﬁes (2.2) and (2.3), while (Y, δY , σY ) satisﬁes (2.4) and (2.5). We start with (2.4), where i < j.

j i j i X j i X X i j−1 X X i j−1 X i j−1 D (D f) = δY (D f)δj = δY δY fδi δj = δY δY fδj−1δi = δY (D f)δi = D (D f).

Likewise for (2.5) (i). Suppose that i ≤ j. We have that

j i j i X j i X X i j+1 X X i j+1 X i j+1 S (S f) = σY (S f)σj = σY σY fσi σj = σY σY fσj+1σi = σY (S f)σi = S (S f). 67 Next we check (2.5) (ii). Suppose i < j. We have

j i j i X j i X X i j−1 X X i j−1 X i j−1 S (D f) = σY (D f)σj = σY δY fδi σj = δY σY fσj−1δi = δY (S f)δi = D (S f).

Now we verify (2.5) (iii).

i i i i X i i X X S (D f) = σY (D f)σi = σY δY fδi σi = id f id = f.

i−1 i i−1 i X i−1 i X X S (D f) = σY (D f)σi−1 = σY δY fδi σi−1 = id f id = f.

Finally we check (2.5) (iv). Suppose that i > j + 1. Then we have that

j i j i X j i X X i−1 j X X i−1 j X i−1 j S (D f) = σY (D f)σj = σY δY fδi σj = δY σY fσj δi−1 = δY (S f)δi−1 = D (S f).

Thus, HomA(X , Y) is a co-simplicial k-module.

Remark 3.4.3. Notice that X ⊗A Y and HomA(X , Y) are not simplicial and co-simplicial modules over A, respectively, but over the ﬁeld k. In a certain way they will be replacements for the Tor and Ext functors, respectively. This will be made more precise in the next Chapter. 68

CHAPTER 4 APPLICATIONS OF THE HOM AND TENSOR LEMMAS

In this chapter we give applications of the Hom and Tensor Lemmas. The results from Section 4.1 come directly from [27], but with added details where appropriate. Section 4.2, however, houses results involving the higher order Hochschild homology (see Section 2.8).

4.1 Hochschild (Co)homology Examples

In this section we show that (secondary) Hochschild (co)homology can be obtained from the Hom and Tensor Lemmas.

Proposition 4.1.1. ([27]) M(M) ⊗A(A⊗Aop) B(A) is a simplicial k-module. Moreover we have,

∼ Hn(M(M) ⊗A(A⊗Aop) B(A)) = Hn(A, M).

Proof. We know from Example 3.2.3 that M(M) is a simplicial right module over A(A ⊗ Aop), and B(A) is a simplicial left module over A(A ⊗ Aop) from Example 3.2.4.

Therefore, by Lemma 3.4.1 we get that M(M) ⊗A(A⊗Aop) B(A) is a simplicial k-module. Moreover, we have that

⊗n+2 op Mn ⊗An Bn = M ⊗A⊗A A in dimension n. Following Lemma 3.4.1 for 0 ≤ i ≤ n, the maps

⊗n+2 ⊗n+1 Di : M ⊗A⊗Aop A −→ M ⊗A⊗Aop A are

Di(m ⊗A⊗Aop (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1))

M B = δi (m) ⊗A⊗Aop δi (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1) 69

= m ⊗A⊗Aop (a0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an ⊗ an+1).

⊗n+2 ⊗n Identifying M ⊗A⊗Aop A with M ⊗ A under the isomorphisms

⊗n+2 ⊗n ϕn : M ⊗A⊗Aop A −→ M ⊗ A given by

ϕn(m ⊗A⊗Aop (a0 ⊗ a1 ⊗ · · · ⊗ an ⊗ an+1)) = an+1ma0 ⊗ a1 ⊗ · · · ⊗ an, and

−1 ⊗n ⊗n+2 ϕn : M ⊗ A −→ M ⊗A⊗Aop A given by

−1 ϕn (m ⊗ a1 ⊗ · · · ⊗ an) = m ⊗A⊗Aop (1 ⊗ a1 ⊗ · · · ⊗ an ⊗ 1),

we get that

−1 (ϕn−1 ◦ D0 ◦ ϕn )(m ⊗ a1 ⊗ · · · ⊗ an) = ma1 ⊗ a2 ⊗ · · · ⊗ an,

−1 (ϕn−1 ◦ Di ◦ ϕn )(m ⊗ a1 ⊗ · · · ⊗ an) = m ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an

for 1 ≤ i ≤ n − 1, and

−1 (ϕn−1 ◦ Dn ◦ ϕn )(m ⊗ a1 ⊗ · · · ⊗ an) = anm ⊗ a1 ⊗ · · · ⊗ an−1.

⊗n ⊗n−1 Pn i −1 Taking dn : M ⊗ A −→ M ⊗ A by setting dn = i=0(−1) (ϕn−1 ◦ Di ◦ ϕn ), we get

exactly the chain complex C•(A, M) from Deﬁnition 2.5.9. The homology of C•(A, M) is the

Hochschild homology of A with coefﬁcients in M, denoted Hn(A, M).

Proposition 4.1.2. ([27]) HomA(A⊗Aop)(B(A), N (M)) is a co-simplicial k-module. Moreover we 70 have, n ∼ n H (HomA(A⊗Aop)(B(A), N (M))) = H (A, M).

Proof. We know from Example 3.3.3 that N (M) is a co-simplicial left module over A(A ⊗ Aop), and B(A) is a simplicial left module over A(A ⊗ Aop) from Example 3.2.4.

Therefore, by Lemma 3.4.2 we get that HomA(A⊗Aop)(B(A), N (M)) is a co-simplicial k- module. n ∼ n n The fact that H (HomA(A⊗Aop)(B(A), N (M))) = H (A, M), where H (A, M) is the Hochschild cohomology of A with coefﬁcients in M, follows similarly to the argument in the proof of Propo- sition 4.1.1, so we omit it here.

Proposition 4.1.3. ([27]) HomA(A,B,ε)(B(A, B, ε), C(M)) is a co-simplicial k-module. Moreover we have, n ∼ n H (HomA(A,B,ε)(B(A, B, ε), C(M))) = H ((A, B, ε); M).

Proof. We know from Example 3.3.4 that C(M) is a co-simplicial left module over A(A, B, ε), and B(A, B, ε) is a simplicial left module over A(A, B, ε) from Example 3.2.6.

Therefore, by Lemma 3.4.2 we get that HomA(A,B,ε)(B(A, B, ε), C(M)) is a co-simplicial k- module. Moreover, we have that

n ⊗n+2 ⊗ (n+1)(n+2) ⊗2n+1 op 2 HomAn (Bn,C ) = HomA⊗B ⊗A (A ⊗ B ,M)

in dimension n. Following Lemma 3.4.2 for 0 ≤ i ≤ n + 1, the maps

i n n+1 D : HomAn (Bn,M ) −→ HomAn+1 (Bn+1,M ) 71 are

   a b0 b1 ··· bn−1 bn d        1 a0 b0,1 ··· b0,n−1 b0,n c0         1 1 a ··· b b c    1 1,n−1 1,n 1     i   ......  D (f) ⊗  ......         1 1 1 ··· an−1 bn−1,n cn−1        1 1 1 ··· 1 a c    n n     1 1 1 ··· 1 1 b    a b0 b1 ··· bn−1 bn d        1 a0 b0,1 ··· b0,n−1 b0,n c0         1 1 a ··· b b c    1 1,n−1 1,n 1     i B   ......  = δCfδi ⊗  ......         1 1 1 ··· an−1 bn−1,n cn−1        1 1 1 ··· 1 a c    n n     1 1 1 ··· 1 1 b    a b0 ··· bi−1bi ··· bn d        1 a0 ··· b0,i−1b0,i ··· b0,n c0       ......    ......    . . . . .     = δi f ⊗   C  1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,nbi,n ci−1ci      ......    ......         1 1 ··· 1 ··· a c    n n     1 1 ··· 1 ··· 1 b 72    a b0 ··· bi−1bi ··· bn d        1 a0 ··· b0,i−1b0,i ··· b0,n c0       ......    ......    . . . . .     = f ⊗   .  1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,nbi,n ci−1ci      ......    ......         1 1 ··· 1 ··· a c    n n     1 1 ··· 1 ··· 1 b

⊗n+2 ⊗ (n+1)(n+2) ⊗n ⊗ n(n−1) Identifying HomA⊗B⊗2n+1⊗Aop (A ⊗ B 2 ,M) with Homk(A ⊗ B 2 ,M) under the isomorphisms

⊗n+2 ⊗ (n+1)(n+2) ⊗n ⊗ n(n−1) ϕn : HomA⊗B⊗2n+1⊗Aop (A ⊗ B 2 ,M) −→ Homk(A ⊗ B 2 ,M)

given by

   1 1 1 ··· 1 1 1          a1 b1,2 ··· b1,n−1 b1,n  1 a1 b1,2 ··· b1,n−1 b1,n 1            1 a ··· b b   1 1 a ··· b b 1   2 2,n−1 2,n    2 2,n−1 2,n          ......   ......  ϕn(g) ⊗  . . . . .  = g ⊗ ......  ,               1 1 ··· an−1 bn−1,n  1 1 1 ··· an−1 bn−1,n 1             1 1 ··· 1 a  1 1 1 ··· 1 a 1 n   n     1 1 1 ··· 1 1 1

and

n(n−1) (n+1)(n+2) −1 ⊗n ⊗ 2 ⊗n+2 ⊗ 2 ϕn : Homk(A ⊗ B ,M) −→ HomA⊗B⊗2n+1⊗Aop (A ⊗ B ,M) 73 given by    a0 b0,1 ··· b0,n b0,n+1         1 a1 ··· b1,n b1,n+1    −1   . . . . .  ϕ (f) ⊗  ......  n            1 1 ··· an bn,n+1    1 1 ··· 1 an+1    a1 ··· b1,n      . . .  = a0f ⊗  . .. .  an+1ε(b0,1 ··· b0,n+1 ··· bn,n+1),       1 ··· an we get that    a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n     0 −1   . . . . .  (ϕn+1 ◦ D ◦ ϕ )(f) ⊗  ......  n            1 1 ··· an−1 bn−1,n    1 1 ··· 1 an

   a1 ··· b1,n−1 b1,n      . . . .    . .. . .     = a0ε(b0,1 ··· b0,n−1b0,n)f ⊗   ,   1 ··· a b    n−1 n−1,n    1 ··· 1 an

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n     i −1   . . . . .  (ϕn+1 ◦ D ◦ ϕ )(f) ⊗  ......  n            1 1 ··· an−1 bn−1,n    1 1 ··· 1 an 74    a0 ··· b0,i−1b0,i ··· b0,n      ......    . . . . .        = f ⊗  1 ··· a ε(b )a ··· b b    i−1 i−1,i i i−1,n i,n      ......    . . . . .     1 ··· 1 ··· an for 1 ≤ i ≤ n, and

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n     n+1 −1   . . . . .  (ϕn+1 ◦ D ◦ ϕ )(f) ⊗  ......  n            1 1 ··· an−1 bn−1,n    1 1 ··· 1 an

   a0 b0,1 ··· b0,n−1         1 a1 ··· b1,n−1 = f ⊗   a ε(b ··· b b ).   . . .  n n−1,n 1,n 0,n   ......    . . .     1 1 ··· an−1

ε ⊗n ⊗ n(n−1) ⊗n+1 ⊗ n(n+1) ε Taking δn : Homk(A ⊗ B 2 ,M) −→ Homk(A ⊗ B 2 ,M) by setting δn = Pn+1 i i −1 • i=0 (−1) (ϕn+1 ◦D ◦ϕn ), we get exactly the chain complex C ((A, B, ε); M) from Deﬁnition 2.7.1. The homology of C•((A, B, ε); M) is the secondary Hochschild cohomology of the triple (A, B, ε) with coefﬁcients in M, denoted Hn((A, B, ε); M).

The next three examples, once again using the Hom and Tensor Lemmas, are actually deﬁni- tions. The ﬁrst one was studied in depth in [27], and so we refer the reader there for details. For more information about Proposition 4.1.5, we refer the reader to Section 5.1.

Proposition 4.1.4. ([27]) HomA(A,B,ε)(B(A, B, ε), H(A, B, ε)) is a co-simplicial k-module. We denote the associated cohomology groups by HHn(A, B, ε), and we call them the secondary 75 Hochschild cohomology groups associated to the triple (A, B, ε).

Proof. We know from Example 3.3.5 that H(A, B, ε) is a co-simplicial left module over A(A, B, ε), and B(A, B, ε) is a simplicial left module over A(A, B, ε) from Example 3.2.6.

Therefore, by Lemma 3.4.2, we get that HomA(A,B,ε)(B(A, B, ε), H(A, B, ε)) is a co-simplicial k-module.

Proposition 4.1.5. ([27]) L(A, B, ε) ⊗A(A,B,ε) B(A, B, ε) is a simplicial k-module. We denote

the associated homology groups by HHn(A, B, ε), and we call them the secondary Hochschild homology groups associated to the triple (A, B, ε).

Proof. We know from Example 3.2.7 that L(A, B, ε) is a simplicial right module over A(A, B, ε), and B(A, B, ε) is a simplicial left module over A(A, B, ε) from Example 3.2.6.

Therefore, by Lemma 3.4.1 we get that L(A, B, ε) ⊗A(A,B,ε) B(A, B, ε) is a simplicial k- module.

Proposition 4.1.6. ([27]) S(M) ⊗A(A,B,ε) B(A, B, ε) is a simplicial k-module. We denote the as-

sociated hohomology groups by HHn((A, B, ε); M), and we call them the secondary Hochschild homology groups of the triple (A, B, ε) with coefﬁcients in M.

Proof. We know from Example 3.3.5 that S(M) is a simplicial right module over A(A, B, ε), and B(A, B, ε) is a simplicial left module over A(A, B, ε) from Example 3.2.6.

Hence by Lemma 3.4.1 we get that S(M) ⊗A(A,B,ε) B(A, B, ε) is a simplicial k-module.

4.2 Higher Order Hochschild Homology Examples

S1 S2 In this section we give a representation for Hn (A, M) and Hn (A, M), which is the higher order Hochschild homology of A with values in M over the simplicial set S1 and S2, respectively. For this section we ﬁx A to be a commutative k-algebra and M an A-bimodule which is A- symmetric. 76 4.2.1 Higher Order Hochschild Homology Over the Circle

Let’s take the simplicial set S1 with the simplicial structure given in the following way: let the circle S1 be obtained from the 1-simplex Γ = [01] by identifying the endpoints. Calling the complex Y•, we have that Y0 = {∗0}, Y1 = {∗1} ∪ {[01]}, and Y2 = {∗2} ∪ {[001], [011]}.

a Continuing in this way, we get that Yn = {∗n} ∪ {Γb : a, b ∈ N, a + b = n − 1} for n ≥ 1, where

a Γb ∈ Yn has a copies of the vertex [0] and b copies of the vertex [1]. Notice that |Yn| = n + 1. For 0 ≤ i ≤ n we have the following:

 ∗ if a = 0 and i = 0,  n−1   Γa−1 if a 6= 0 and i ≤ a, a  b di(Γb ) = (4.1)  ∗n−1 if b = 0 and i = n = a + 1,    a Γb−1 if b 6= 0 and a < i ≤ a + b + 1 = n.

Y• ⊗n Following Section 2.8, we see that Cn = L(A, M)(Yn) = M ⊗ A . Moreover, we identify

a Yn with y+ = {0, 1, 2, . . . , n} in the following way: we identify Γb to the position (a + 1) of an element in M ⊗ A⊗n, which is represented as

m0 ⊗ a1 ⊗ a2 ⊗ · · · ⊗ an. (4.2)

We add (0) to correspond to ∗n. Making this identiﬁcation, (4.1) becomes:

  (0) if a = 0 and i = 0,    (a) if a 6= 0 and i ≤ a, di((a + 1)) =  (0) if b = 0 and i = n = a + 1,    (a + 1) if b 6= 0 and a < i ≤ a + b + 1 = n. 77 ∗ We then have that di : Yn −→ Yn−1 induces di := L(A, M)(di), and so

∗ di (m0 ⊗ a1 ⊗ · · · ⊗ an) = m0b0 ⊗ b1 ⊗ · · · ⊗ bn−1

where Y bi = aj.

{j∈v+ : j6=0, di(j)=i}

1 0 Remark 4.2.1. Let’s consider dimension 2. Recall Y2 consists of the elements ∗2, Γ0, and Γ1. Making the identiﬁcations, we have

1 0 d0((2)) = d0(Γ0) = Γ0 = (1),

1 0 d1((2)) = d1(Γ0) = Γ0 = (1),

1 d2((2)) = d2(Γ0) = ∗1 = (0),

and

0 d0((1)) = d0(Γ1) = ∗1 = (0),

0 0 d1((1)) = d1(Γ1) = Γ0 = (1),

0 0 d2((1)) = d2(Γ1) = Γ0 = (1).

Therefore, we see that

∗ d0(m0 ⊗ a1 ⊗ a2) = m0b0 ⊗ b1 = m0a1 ⊗ a2

since Y Y b0 = aj = aj = a1

{j∈y+ : j6=0, d0(j)=0} {j=1,2 : d0(j)=0} 78 and Y Y b1 = aj = aj = a2.

{j∈y+ : j6=0, d0(j)=1} {j=1,2 : d0(j)=1} Likewise

∗ d1(m0 ⊗ a1 ⊗ a2) = m0b0 ⊗ b1 = m0 ⊗ a1a2 since Y Y b0 = aj = aj = id

{j∈y+ : j6=0, d1(j)=0} {j=1,2 : d1(j)=0} and Y Y b1 = aj = aj = a1a2.

{j∈y+ : j6=0, d1(j)=1} {j=1,2 : d0(j)=1} Finally,

∗ d2(m0 ⊗ a1 ⊗ a2) = m0b0 ⊗ b1 = m0a2 ⊗ a1 since Y Y b0 = aj = aj = a2

{j∈y+ : j6=0, d2(j)=0} {j=1,2 : d2(j)=0} and Y Y b1 = aj = aj = a1.

{j∈y+ : j6=0, d2(j)=1} {j=1,2 : d2(j)=1}

Thus, ∂2 : M ⊗ A ⊗ A −→ M ⊗ A is given by

∗ ∗ ∗ ∂2(m0 ⊗ a1 ⊗ a2) = (d0 − d1 + d2)(m0 ⊗ a1 ⊗ a2) = m0a1 ⊗ a2 − m0 ⊗ a1a2 + m0a2 ⊗ a1.

In general we have that

∗ d0(m0 ⊗ a1 ⊗ · · · ⊗ an) = m0a1 ⊗ a2 ⊗ · · · ⊗ an,

∗ di (m0 ⊗ a1 ⊗ · · · ⊗ an) = m0 ⊗ a1 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an 79 for 1 ≤ i ≤ n − 1, and

∗ dn(m0 ⊗ a1 ⊗ · · · ⊗ an) = m0an ⊗ a1 ⊗ · · · ⊗ an−1.

⊗n ⊗n−1 Taking ∂n : M ⊗ A −→ M ⊗ A by

n X i ∗ ∂n := (−1) di , i=0

1 Y• S we now have defined the chain complex C• (A, M), the homology of which is H• (A, M). This is the higher order Hochschild homology of A with values in M over the simplicial set S1. Comparing to Definition 2.5.9 of the Hochschild homology of A with coefficients in M, we get the following classical result:

Proposition 4.2.2. We have that

∼ S1 Hn(A, M) = Hn (A, M).

Proof. Follows from the above discussion.

4.2.2 Higher Order Hochschild Homology Over the Sphere

The results of this subsection is not is [27]. The description of the chain complex associated to the 2-sphere is new, where we use techniques developed in Chapter 3.

S2 2 Here we give an explicit description for Hn (A, M). Let the sphere S be obtained from the

2-simplex ∆ = [012] by identifying the boundary to a single point. Calling the complex X•, we have that X0 = {∗0}, X1 = {∗1}, X2 = {∗2} ∪ {[012]}, and X3 = {∗3} ∪ {[0012], [0112], [0122]}.

a b Continuing in this way, we get that Xn = {∗n} ∪ { ∆c : a, b, c ∈ N, a + b + c = n − 2} for

a b n ≥ 2, where ∆c ∈ Xn has a copies of the vertex [0], b copies of the vertex [1], and c copies of n(n−1) the vertex [2]. Notice that |Xn| = 2 + 1. 80 For 0 ≤ i ≤ n we have the following:

  ∗n−1 if a = 0 and i = 0,   a−1 b  ∆c if a 6= 0 and i ≤ a,    ∗n−1 if b = 0 and i = a + 1, a b  di( ∆c) = (4.3) a b−1  ∆c if b 6= 0 and a < i ≤ a + b + 1,    ∗n−1 if c = 0 and i = n = a + b + 2,   a b  ∆c−1 if c 6= 0 and i ≥ a + b + 2.

n(n−1) X• ⊗ Following Section 2.8, we see that Cn = L(A, M)(Xn) = M ⊗ A 2 . Moreover, we

identify Xn with

x+ = {(0, 0), (1, 1), (1, 2),..., (1, n − 1), (2, 2),..., (2, n − 1), (3, 3),..., (n − 1, n − 1)}

a b in the following way: we identify ∆c to the entry position (a + 1, a + b + 1) of an element in

⊗ n(n−1) M ⊗ A 2 , which is represented as a matrix

  a1,1 a1,2 a1,3 ··· a1,n−3 a1,n−2 a1,n−1      1 a2,2 a2,3 ··· a2,n−3 a2,n−2 a2,n−1       1 1 a ··· a a a   3,3 3,n−3 3,n−2 3,n−1     ......  m0,0 ⊗  ......  . (4.4)      1 1 1 ··· an−3,n−3 an−3,n−2 an−3,n−1      1 1 1 ··· 1 a a   n−2,n−2 n−2,n−1   1 1 1 ··· 1 1 an−1,n−1 81

Like [11] add (0, 0) to correspond to ∗n. Making this identiﬁcation, (4.3) becomes:

  (0, 0) if a = 0 and i = 0,    (a, a + b) if a 6= 0 and i ≤ a,    (0, 0) if b = 0 and i = a + 1, di((a + 1, a + b + 1)) =  (a + 1, a + b) if b 6= 0 and a < i ≤ a + b + 1,    (0, 0) if c = 0 and i = n = a + b + 2,    (a + 1, a + b + 1) if c 6= 0 and i ≥ a + b + 2.

∗ We then have that di : Xn −→ Xn−1 induces di := L(A, M)(di), and so

   a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     ∗   . . . . .  d m0,0 ⊗  ......  i            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1

  b1,1 b1,2 ··· b1,n−3 b1,n−2      1 b2,2 ··· b2,n−2 b2,n−2     . . . . .  = m0,0b0,0 ⊗  ......         1 1 ··· bn−3,n−3 bn−3,n−2   1 1 ··· 1 bn−2,n−2 where Y bi,j = ak,l.

{(k,l)∈x+ :(k,l)6=(0,0), di((k,l))=(i,j)}

1 0 0 1 Remark 4.2.3. Let’s consider dimension 3. Recall X3 consists of the elements ∗3, ∆0, ∆0, and 82 0 0 ∆1. Making the identiﬁcations, we have

1 0 0 0 d0((2, 2)) = d0( ∆0) = ∆0 = (1, 1),

1 0 0 0 d1((2, 2)) = d1( ∆0) = ∆0 = (1, 1),

1 0 d2((2, 2)) = d2( ∆0) = ∗2 = (0, 0),

1 0 d3((2, 2)) = d3( ∆0) = ∗2 = (0, 0),

and

0 1 d0((1, 2)) = d0( ∆0) = ∗2 = (0, 0),

0 1 0 0 d1((1, 2)) = d1( ∆0) = ∆0 = (1, 1),

0 1 0 0 d2((1, 2)) = d2( ∆0) = ∆0 = (1, 1),

0 1 d3((1, 2)) = d3( ∆0) = ∗2 = (0, 0),

and

0 0 d0((1, 1)) = d0( ∆1) = ∗2 = (0, 0),

0 0 d1((1, 1)) = d1( ∆1) = ∗2 = (0, 0),

0 0 0 0 d2((1, 1)) = d2( ∆1) = ∆0 = (1, 1),

0 0 0 0 d3((1, 1)) = d3( ∆1) = ∆0 = (1, 1).

Therefore, we see that

   a1,1 a1,2 d∗ m ⊗   = m b ⊗ b = m a a ⊗ a 0  0,0   0,0 0,0 1,1 0,0 1,1 1,2 2,2 1 a2,2 83 since

Y Y b0,0 = ak,l = ak,l = a1,1a1,2

{(k,l)∈x+ :(k,l)6=(0,0), d0(k,l)=(0,0)} {(k,l)=(1,1),(1,2),(2,2) : d0(k,l)=(0,0)} and

Y Y b1,1 = ak,l = ak,l = a2,2.

{(k,l)∈x+ :(k,l)6=(0,0), d0(k,l)=(1,1)} {(k,l)=(1,1),(1,2),(2,2) : d0(k,l)=(1,1)}

Likewise    a1,1 a1,2 d∗ m ⊗   = m b ⊗ b = m a ⊗ a a 1  0,0   0,0 0,0 1,1 0,0 1,1 1,2 2,2 1 a2,2 since

Y Y b0,0 = ak,l = ak,l = a1,1

{(k,l)∈x+ :(k,l)6=(0,0), d1(k,l)=(0,0)} {(k,l)=(1,1),(1,2),(2,2) : d1(k,l)=(0,0)} and

Y Y b1,1 = ak,l = ak,l = a1,2a2,2.

{(k,l)∈x+ :(k,l)6=(0,0), d1(k,l)=(1,1)} {(k,l)=(1,1),(1,2),(2,2) : d1(k,l)=(1,1)}

Next    a1,1 a1,2 d∗ m ⊗   = m b ⊗ b = m a ⊗ a a 2  0,0   0,0 0,0 1,1 0,0 2,2 1,1 1,2 1 a2,2 since

Y Y b0,0 = ak,l = ak,l = a2,2

{(k,l)∈x+ :(k,l)6=(0,0), d2(k,l)=(0,0)} {(k,l)=(1,1),(1,2),(2,2) : d2(k,l)=(0,0)} 84 and

Y Y b1,1 = ak,l = ak,l = a1,1a1,2.

{(k,l)∈x+ :(k,l)6=(0,0), d2(k,l)=(1,1)} {(k,l)=(1,1),(1,2),(2,2) : d2(k,l)=(1,1)}

Finally,    a1,1 a1,2 d∗ m ⊗   = m b ⊗ b = m a a ⊗ a 3  0,0   0,0 0,0 1,1 0,0 1,2 2,2 1,1 1 a2,2 since

Y Y b0,0 = ak,l = ak,l = a1,2a2,2

{(k,l)∈x+ :(k,l)6=(0,0), d3(k,l)=(0,0)} {(k,l)=(1,1),(1,2),(2,2) : d3(k,l)=(0,0)} and

Y Y b1,1 = ak,l = ak,l = a1,1.

{(k,l)∈x+ :(k,l)6=(0,0), d3(k,l)=(1,1)} {(k,l)=(1,1),(1,2),(2,2) : d3(k,l)=(1,1)}

Thus, ∂3 : M ⊗ A ⊗ A ⊗ A −→ M ⊗ A is given by

      a1,1 a1,2 a1,1 a1,2 ∂ m ⊗   = (d∗ − d∗ + d∗ − d∗) m ⊗   3  0,0   0 1 2 3  0,0   1 a2,2 1 a2,2

= m0,0a1,1a1,2 ⊗ a2,2 − m0,0a1,1 ⊗ a1,2a2,2 + m0,0a2,2 ⊗ a1,1a1,2 − m0,0a1,2a2,2 ⊗ a1,1.

In general we have that

   a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     ∗   . . . . .  d m0,0 ⊗  ......  0            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1 85   a2,2 ··· a2,n−2 a2,n−1    . . . .   . .. . .    = m0,0a1,1a1,2 ··· a1,n−1 ⊗   ,  1 ··· a a   n−2,n−2 n−2,n−1   1 ··· 1 an−1,n−1

   a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     ∗   . . . . .  d m0,0 ⊗  ......  = i            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1

  a1,1 ··· a1,i−2 a1,i−1a1,i a1,i+1 a1,i+2 ··· a1,n−1    ......   ......       1 ··· a a a a a ··· a   i−2,i−2 i−2,i−1 i−2,i i−2,i+1 i−2,i+2 i−2,n−1       1 ··· 1 ai−1,i−1ai−1,i ai−1,i+1 ai−1,i+2 ··· ai−1,n−1  m0,0ai,i ⊗      1 ··· 1 1 ai,i+1ai+1,i+1 ai,i+2ai+1,i+2 ··· ai,n−1ai+1,n−1      1 ··· 1 1 1 a ··· a   i+2,i+2 i+2,n−1     ......   ......    1 ··· 1 1 1 1 ··· an−1,n−1 for 1 ≤ i ≤ n − 1, and

   a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     ∗   . . . . .  d m0,0 ⊗  ......  n            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1 86   a1,1 a1,2 ··· a1,n−2      1 a2,2 ··· a2,n−2  = m a a ··· a ⊗   . 0,0 1,n−1 2,n−1 n−1,n−1  . . . .   . . .. .   . . .    1 1 ··· an−2,n−2

⊗ n(n−1) ⊗ (n−1)(n−2) Taking ∂n : M ⊗ A 2 −→ M ⊗ A 2 by

n X i ∗ ∂n := (−1) di , i=0

2 X• S we now have deﬁned the chain complex C• (A, M), the homology of which is H• (A, M). This is the higher order Hochschild homology of A with values in M over the simplicial set S2.

S2 Our goal is to represent H• (A, M) as the homology of a complex corresponding to simplicial structures. To do this, we will need to deﬁne an appropriate simplicial k-algebra along with simplicial modules (as seen in Chapter 3).

2 ⊗3n+3 Example 4.2.4. Deﬁne the simplicial k-algebra A (A) by setting An = A and

A δ0 (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) =

a0a1b1 ⊗ a2 ⊗ · · · ⊗ an+1 ⊗ b2 ⊗ · · · ⊗ bn ⊗ dc1 ⊗ c2 ⊗ · · · ⊗ cn,

A δi (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) =

a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cn for 1 ≤ i ≤ n − 1, and

A δn (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) =

a0 ⊗ · · · ⊗ an−2 ⊗ anan+1cn ⊗ b1 ⊗ · · · ⊗ bn−1 ⊗ bnd ⊗ c1 ⊗ · · · ⊗ cn−1. 87 Furthermore, deﬁne

A σ0 (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) =

a0 ⊗ 1 ⊗ a1 ⊗ · · · ⊗ an+1 ⊗ 1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ 1 ⊗ c1 ⊗ · · · ⊗ cn,

A σi (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) =

a0⊗· · ·⊗ai⊗1⊗ai+1⊗· · ·⊗an+1⊗b1⊗· · ·⊗bi⊗1⊗bi+1⊗· · ·⊗bn⊗d⊗c1⊗· · ·⊗ci⊗1⊗ci+1⊗· · ·⊗cn for 1 ≤ i ≤ n − 1, and

A σn (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) =

a0 ⊗ · · · ⊗ an ⊗ 1 ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ 1 ⊗ d ⊗ c1 ⊗ · · · ⊗ cn ⊗ 1.

⊗3n+3 Notice it may be helpful to view elements of An = A as a matrix:

  a0 b1 b2 ··· bn−1 bn d      a1 c1       a c   2 2     .. .  ⊗  . .  .      an−1 cn−1       a c   n n    an+1

Proof. In order to satisfy Deﬁnition 3.1.1, we need to check (2.2) and (2.3). We ﬁrst consider (2.2).

δiδi+1(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ ai+1ai+2 ⊗ · · · ⊗ bj+1bj+2 ⊗ · · · ⊗ cj+1cj+2 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ aiai+1ai+2 ⊗ · · · ⊗ bibi+1bi+2 ⊗ · · · ⊗ cici+1ci+2 ⊗ · · · ⊗ cn 88

= δi(a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cn)

= δiδi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Suppose i < j. Then we have

δiδj(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ bjbj+1 ⊗ · · · ⊗ cjcj+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bjbj+1

⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cjcj+1 ⊗ · · · ⊗ cn

= δj−1(a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cn)

= δj−1δi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Next we look at (2.3) (i).

σiσi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= σi(a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn

= σi+1(a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn)

= σi+1σi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Suppose i ≤ j. Then we have

σiσj(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= σi(a0 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ cj ⊗ 1 ⊗ cj+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1

⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cj ⊗ 1 ⊗ cj+1 ⊗ · · · ⊗ cn 89

= σj+1(a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn)

= σj+1σi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Now we check (2.3) (ii).

δiσi+1(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ ai+1 ⊗ 1 ⊗ ai+2 ⊗ · · · ⊗ bi+1 ⊗ 1 ⊗ bi+2 ⊗ · · · ⊗ ci+1 ⊗ 1 ⊗ ci+2 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ aiai+1 ⊗ 1 ⊗ ai+2 ⊗ · · · ⊗ bibi+1 ⊗ 1 ⊗ bi+2 ⊗ · · · ⊗ cici+1 ⊗ 1 ⊗ ci+2 ⊗ · · · ⊗ cn

= σi(a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cn)

= σiδi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Suppose i < j. Then we have

δiσj(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ cj ⊗ 1 ⊗ cj+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1

⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cj ⊗ 1 ⊗ cj+1 ⊗ · · · ⊗ cn

= σj−1(a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cn)

= σj−1δi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Next we show (2.3) (iii).

δiσi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ ai · 1 ⊗ ai+1 ⊗ · · · ⊗ bi · 1 ⊗ bi+1 ⊗ · · · ⊗ ci · 1 ⊗ ci+1 ⊗ · · · ⊗ cn

= a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn. 90

δiσi−1(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ ai−1 ⊗ 1 ⊗ ai ⊗ · · · ⊗ bi−1 ⊗ 1 ⊗ bi ⊗ · · · ⊗ ci−1 ⊗ 1 ⊗ ci ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ ai−1 ⊗ 1 · ai ⊗ · · · ⊗ bi−1 ⊗ 1 · bi ⊗ · · · ⊗ ci−1 ⊗ 1 · ci ⊗ · · · ⊗ cn

= a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn.

Finally we verify (2.3) (iv).

δi+2σi(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi+2(a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1ai+2 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1bi+2 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1ci+2 ⊗ · · · ⊗ cn

= σi(a0 ⊗ · · · ⊗ ai+1ai+2 ⊗ · · · ⊗ bi+1bi+2 ⊗ · · · ⊗ ci+1ci+2 ⊗ · · · ⊗ cn)

= σiδi+1(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Suppose i > j + 1. Then we have

δiσj(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn)

= δi(a0 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ cj ⊗ 1 ⊗ cj+1 ⊗ · · · ⊗ cn)

= a0 ⊗ · · · ⊗ aj ⊗ 1 ⊗ aj+1 ⊗ · · · ⊗ ai−1ai ⊗ · · · ⊗ bj ⊗ 1 ⊗ bj+1 ⊗ · · · ⊗ bi−1bi

⊗ · · · ⊗ cj ⊗ 1 ⊗ cj+1 ⊗ · · · ⊗ ci−1ci ⊗ · · · ⊗ cn

= σj(a0 ⊗ · · · ⊗ ai−1ai ⊗ · · · ⊗ bi−1bi ⊗ · · · ⊗ ci−1ci ⊗ · · · ⊗ cn)

= σjδi−1(a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Thus, A2(A) is a simplicial k-algebra.

Example 4.2.5. We deﬁne the simplicial left module B2(A) over the simplicial k-algebra A2(A) 91 ⊗ (n+2)(n+3) (see Example 4.2.4) by setting Bn = A 2 . The multiplication on Bn is given by

   a0 b1 b2 ··· bn−1 bn d         a1 c1          a c    2 2       .. .  ⊗  . .          an−1 cn−1          a c    n n     an+1

   a0,0 a0,1 a0,2 ··· a0,n−1 a0,n a0,n+1         1 a1,1 a1,2 ··· a1,n−1 a1,n a1,n+1          1 1 a ··· a a a    2,2 2,n−1 2,n 2,n+1       ......  · ⊗  ......          1 1 1 ··· an−1,n−1 an−1,n an−1,n+1         1 1 1 ··· 1 a a    n,n n,n+1     1 1 1 ··· 1 1 an+1,n+1

  a0a0,0 b1a0,1 b2a0,2 ··· bn−1a0,n−1 bna0,n da0,n+1      1 a1a1,1 a1,2 ··· a1,n−1 a1,n c1a1,n+1       1 1 a a ··· a a c a   2 2,2 2,n−1 2,n 2 2,n+1     ......  = ⊗  ......  .      1 1 1 ··· an−1an−1,n−1 an−1,n cn−1bn−1,n+1       1 1 1 ··· 1 a a c a   n n,n n n,n+1    1 1 1 ··· 1 1 an+1an+1,n+1 92 Next, for 0 ≤ i ≤ n we deﬁne

   a0,0 a0,1 ··· a0,n a0,n+1         1 a1,1 ··· a1,n a1,n+1     B   . . . . .  δ ⊗  ......  i            1 1 ··· an,n an,n+1     1 1 ··· 1 an+1,n+1

  a0,0 a0,1 ··· a0,ia0,i+1 ··· a0,n a0,n+1      1 a1,1 ··· a1,ia1,i+1 ··· a1,n a1,n+1     ......   ......   . . . . .    = ⊗   ,  1 1 ··· ai,iai,i+1ai+1,i+1 ··· ai,nai+1,n ai,n+1ai+1,n+1    ......   ......       1 1 ··· 1 ··· a a   n,n n,n+1    1 1 ··· 1 ··· 1 an+1,n+1

and    a0,0 a0,1 ··· a0,n a0,n+1         1 a1,1 ··· a1,n a1,n+1     B   . . . . .  σ ⊗  ......  i            1 1 ··· an,n an,n+1     1 1 ··· 1 an+1,n+1 93   a0,0 a0,1 ··· a0,i 1 a0,i+1 ··· a0,n a0,n+1      1 a ··· a 1 a ··· a a   1,1 1,i 1,i+1 1,n 1,n+1     ......   ......       1 1 ··· ai,i 1 ai,i+1 ··· ai,n ai,n+1      = ⊗  1 1 ··· 1 1 1 ··· 1 1  .        1 1 ··· 1 1 ai+1,i+1 ··· ai+1,n ai+1,n+1     ......   ......       1 1 ··· 1 1 1 ··· a a   n,n n,n+1    1 1 ··· 1 1 1 ··· 1 an+1,n+1

Proof. First we check that the equations (2.2) and (2.3) are satisﬁed. We start with (2.2).

   a0,0 ··· a0,n+1      . . .  δiδi+1 ⊗  . .. .        1 ··· an+1,n+1    a0,0 ··· a0,i+1a0,i+2 ··· a0,n+1      ......    . . . . .        = δi ⊗  1 ··· a a a ··· a a    i+1,i+1 i+1,i+2 i+2,i+2 i+1,n+1 i+2,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1   a0,0 ··· a0,ia0,i+1a0,i+2 ··· a0,n+1    ......   . . . . .      = ⊗  1 ··· a a a a a ··· a a a   i,i i,i+1 i+1,i+1 i+1,i+2 i+2,i+2 i,n+1 i+1,n+1 i+2,n+1    ......   . . . . .    1 ··· 1 ··· an+1,n+1 94    a0,0 ··· a0,ia0,i+1 ··· a0,n+1      ......    . . . . .        = δi ⊗  1 ··· a a a ··· a a    i,i i,i+1 i+1,i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = δiδi ⊗  . .. .  .       1 ··· an+1,n+1

Suppose i < j. Then we have

   a0,0 ··· a0,n+1      . . .  δiδj ⊗  . .. .        1 ··· an+1,n+1    a0,0 ··· a0,ja0,j+1 ··· a0,n+1      ......    . . . . .        = δi ⊗  1 ··· a a a ··· a a    j,j j,j+1 j+1,j+1 j,n+1 j+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1   a0,0 ··· a0,ia0,i+1 ··· a0,ja0,j+1 ··· a0,n+1    ......   ......       1 ··· a a a ··· a a a a ··· a a   i,i i,i+1 i+1,i+1 i,j i,j+1 i+1,j i+1,j+1 i,n+1 i+1,n+1     ......  = ⊗  ......       1 ··· 1 ··· aj,jaj,j+1aj+1 ··· aj,n+1aj+1,n+1    . . . .   ......   . . . .    1 ··· 1 ··· 1 ··· an+1,n+1 95    a0,0 ··· a0,ia0,i+1 ··· a0,n+1      ......    . . . . .        = δj−1 ⊗  1 ··· a a a ··· a a    i,i i,i+1 i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = δj−1δi ⊗  . .. .  .       1 ··· an+1,n+1

Next we show (2.3) (i).

   a0,0 ··· a0,i 1 a0,i+1 ··· a0,n+1      ......    ......           a ··· a   1 ··· a 1 a ··· a  0,0 0,n+1   i,i i,i+1 i,n+1          . .. .     σiσi ⊗  . . .  = σi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1,n+1   1 ··· 1 1 ai+1,i+1 ··· ai+1,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,i 1 1 a0,i+1 ··· a0,n+1    ......   ......       1 ··· a 1 1 a ··· a   i,i i,i+1 i,n+1       1 ··· 1 1 1 1 ··· 1  = ⊗      1 ··· 1 1 1 1 ··· 1       1 ··· 1 1 1 a ··· a   i+1,i+1 i+1,n+1     ......   ......    1 ··· 1 1 1 1 ··· an+1,n+1 96    a0,0 ··· a0,i 1 a0,i+1 ··· a0,n+1      ......    ......          1 ··· a 1 a ··· a    i,i i,i+1 i,n+1        = σi+1 ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+1,i+1 ··· ai+1,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = σi+1σi ⊗  . .. .  .       1 ··· an+1,n+1

Suppose i ≤ j. Then we have

   a0,0 ··· a0,j 1 a0,j+1 ··· a0,n+1      ......    ......           a ··· a   1 ··· a 1 a ··· a  0,0 0,n+1   j,j j,j+1 j,n+1          . .. .     σiσj ⊗  . . .  = σi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1,n+1   1 ··· 1 1 aj+1,j+1 ··· aj+1,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1 97   a0,0 ··· a0,i 1 a0,i+1 ··· a0,j 1 a0,j+1 ··· a0,n+1    ......   ......       1 ··· a 1 a ··· a 1 a ··· a   i,i i,i+1 i,j i,j+1 i,n+1       1 ··· 1 1 1 ··· 1 1 1 ··· 1       1 ··· 1 1 ai+1,i+1 ··· ai+1,j 1 ai+1,j+1 ··· ai+1,n+1     ......  = ⊗  ......         1 ··· 1 1 1 ··· aj,j 1 aj,j+1 ··· aj,n+1       1 ··· 1 1 1 ··· 1 1 1 ··· 1       1 ··· 1 1 1 ··· 1 1 a ··· a   j+1,j+1 j+1,n+1     ......   ......    1 ··· 1 1 1 ··· 1 1 1 ··· an+1,n+1    a0,0 ··· a0,i 1 a0,i+1 ··· a0,n+1      ......    ......          1 ··· a 1 a ··· a    i,i i,i+1 i,n+1        = σj+1 ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+1,i+1 ··· ai+1,n+1       . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = σj+1σi ⊗  . .. .  .       1 ··· an+1,n+1

Now we check (2.3) (ii).

   a0,0 ··· a0,n+1      . . .  δiσi+1 ⊗  . .. .        1 ··· an+1,n+1 98    a0,0 ··· a0,i+1 1 a0,i+2 ··· a0,n+1      ......    ......          1 ··· a 1 a ··· a    i+1,i+1 i+1,i+2 i+1,n+1        = δi ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+2,i+2 ··· ai+2,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,ia0,i+1 1 a0,i+2 ··· a0,n+1    ......   ......       1 ··· a a a 1 a a ··· a a   i,i i,i+1 i+1,i+1 i,i+2 i+1,i+2 i,n+1 i+1,n+1     = ⊗  1 ··· 1 1 1 ··· 1       1 ··· 1 1 ai+2,i+2 ··· ai+2,n+1     ......   ......   . . . . .    1 ··· 1 1 1 ··· an+1,n+1    a0,0 ··· a0,ib0,i+1 ··· a0,n+1      ......    . . . . .        = σi ⊗  1 ··· a a a ··· a a    i,i i,i+1 i+1,i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = σiδi ⊗  . .. .  .       1 ··· an+1,n+1 99 Suppose i < j. Then we have

   a0,0 ··· a0,j 1 a0,j+1 ··· a0,n+1      ......    ......           a ··· a   1 ··· a 1 a ··· a  0,0 0,n+1   j,j j,j+1 j,n+1          . .. .     δiσj ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1,n+1   1 ··· 1 1 aj+1,j+1 ··· aj+1,n+1       . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,ia0,i+1 ··· a0,j 1 a0,j+1 ··· a0,n+1    ......   ......         1 ··· ai,iai,i+1ai+1,i+1 ··· ai,jai+1,j 1 ai,j+1ai+1,j+1 ··· ai,n+1ai+1,n+1    ......   ......      = ⊗  1 ··· 1 ··· a 1 a ··· a   j,j j,j+1 j,n+1       1 ··· 1 ··· 1 1 1 ··· 1       1 ··· 1 ··· 1 1 aj+1,j+1 ··· aj+1,n+1     ......   ......      1 ··· 1 ··· 1 1 1 ··· an+1,n+1    a0,0 ··· a0,ia0,i+1 ··· a0,n+1      ......    . . . . .        = σj−1 ⊗  1 ··· a a a ··· a a    i,i i,i+1 i+1,i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = σj−1δi ⊗  . .. .  .       1 ··· an+1,n+1 100 Next we show (2.3) (iii).

   a0,0 ··· a0,i 1 a0,i+1 ··· a0,n+1      ......    ......           a ··· a   1 ··· a 1 a ··· a  0,0 0,n+1   i,i i,i+1 i,n+1          . .. .     δiσi ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1,n+1   1 ··· 1 1 ai+1,i+1 ··· ai+1,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,i · 1 a0,i+1 ··· a0,n+1    ......   ......         1 ··· ai,i · 1 · 1 ai,i+1 · 1 ··· ai,n+1 · 1 = ⊗      1 ··· 1 ai+1,i+1 ··· ai+1,n+1     ......   ......      1 ··· 1 1 ··· an+1,n+1   a0,0 ··· a0,n+1    . . .  = ⊗  . .. .  .     1 ··· an+1,n+1

   a0,0 ··· a0,n+1      . . .  δiσi−1 ⊗  . .. .        1 ··· an+1,n+1 101    a0,0 ··· a0,i−1 1 a0,i ··· a0,n+1      ......    ......          1 ··· a 1 a ··· a    i−1,i−1 i−1,i i−1,n+1        = δi ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai,i ··· ai,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,i−1 · 1 a0,i ··· a0,n+1    ......   ......         1 ··· ai−1,i−1 · 1 · 1 ai−1,i · 1 ··· ai−1,n+1 · 1 = ⊗      1 ··· 1 ai,i ··· ai,n+1     ......   ......      1 ··· 1 1 ··· an+1,n+1   a0,0 ··· a0,n+1    . . .  = ⊗  . .. .  .     1 ··· an+1,n+1

Finally we verify (2.3) (iv).

   a0,0 ··· a0,n+1      . . .  δi+2σi ⊗  . .. .        1 ··· an+1,n+1 102    a0,0 ··· a0,i 1 a0,i+1 ··· a0,n+1      ......    ......          1 ··· a 1 a ··· a    i,i i,i+1 i,n+1        = δi+2 ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+1,i+1 ··· ai+1,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,i 1 a0,i+1a0,i+2 ··· a0,n+1    ......   ......       1 ··· a 1 a a ··· a   i,i i,i+1 i,i+2 i,n+1      = ⊗  1 ··· 1 1 1 ··· 1       1 ··· 1 1 ai+1,i+1ai+1,i+2ai+2,i+2 ··· ai+1,n+1ai+2,n+1    ......   ......   . . . . .    1 ··· 1 1 1 ··· an+1,n+1    a0,0 ··· a0,i+1a0,i+2 ··· a0,n+1      ......    . . . . .        = σi ⊗  1 ··· a a a ··· a a    i+1,i+1 i+1,i+2 i+2,i+2 i+1,n+1 i+2,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = σiδi+1 ⊗  . .. .  .       1 ··· an+1,n+1 103 Suppose i > j + 1. Then we have

   a0,0 ··· a0,j 1 a0,j+1 ··· a0,n+1      ......    ......           a ··· a   1 ··· a 1 a ··· a  0,0 0,n+1   j,j j,j+1 j,n+1          . .. .     δiσj ⊗  . . .  = δi ⊗  1 ··· 1 1 1 ··· 1              1 ··· an+1,n+1   1 ··· 1 1 aj+1,j+1 ··· aj+1,n+1       . . . . .    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1   a0,0 ··· a0,j 1 a0,j+1 ··· a0,i−1a0,i ··· a0,n+1    ......   ......         1 ··· aj,j 1 aj,j+1 ··· aj,i−1aj,i ··· aj,n+1       1 ··· 1 1 1 ··· 1 ··· 1      = ⊗  1 ··· 1 1 a ··· a a ··· a   j+1,j+1 j+1,i−1 j+1,i j+1,n+1     ......   ......       1 ··· 1 1 1 ··· ai−1,i−1ai−1,iai,i ··· ai−1,n+1ai,n+1    ......   ......      1 ··· 1 1 1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,i−1a0,i ··· a0,n+1      ......    . . . . .        = σj ⊗  1 ··· a a a ··· a a    i−1,i−1 i−1,i i,i i−1,n+1 i,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1    a0,0 ··· a0,n+1      . . .  = σjδi−1 ⊗  . .. .  .       1 ··· an+1,n+1 104 Next we show that the compatibility conditions (3.1) are satisﬁed. We have that

      a0 b1 ··· bn d a0,0 a0,1 ··· a0,n a0,n+1               a1 c1    1 a1,1 ··· a1,n a1,n+1        B   . .    . . . . .  δ ⊗  .. .  · ⊗  ......  i                     an cn    1 1 ··· an,n an,n+1        an+1 1 1 ··· 1 an+1,n+1    a0a0,0 b1a0,1 b2a0,2 ··· bn−1a0,n−1 bna0,n da0,n+1         1 a1a1,1 a1,2 ··· a1,n−1 a1,n c1a1,n+1          1 1 a a ··· a a c a    2 2,2 2,n−1 2,n 2 2,n+1     B   ......  = δi ⊗  ......          1 1 1 ··· an−1an−1,n−1 an−1,n cn−1bn−1,n+1          1 1 1 ··· 1 a a c a    n n,n n n,n+1     1 1 1 ··· 1 1 an+1an+1,n+1   a0a0,0 b1a0,1 ··· bia0,ibi+1a0,i+1 ··· bna0,n da0,n+1      1 a1a1,1 ··· a1,ia1,i+1 ··· a1,n c1a1,n+1     ......   ......   . . . . .    = ⊗    1 1 ··· aiai,iai+1ai+1,i+1ai,i+1 ··· ai,nai+1,n ciai,n+1ci+1ai+1,n+1    ......   ......       1 1 ··· 1 ··· a a c a   n n,n n n,n+1    1 1 ··· 1 ··· 1 an+1an+1,n+1    a0 ··· bibi+1 ··· d      .. .    . .        = ⊗  a a c c    i i+1 i i+1      .. .    . .     an+1 105    a0,0 ··· a0,ia0,i+1 ··· a0,n+1      ......    . . . . .        · ⊗  1 ··· a a a ··· a a    i,i i,i+1 i+1,i+1 i,n+1 i+1,n+1      ......    . . . . .     1 ··· 1 ··· an+1,n+1       a0 b1 ··· bn d a0,0 a0,1 ··· a0,n a0,n+1               a1 c1    1 a1,1 ··· a1,n a1,n+1        A   . .  B   . . . . .  = δ ⊗  .. .  δ ⊗  ......  , i    i                  an cn    1 1 ··· an,n an,n+1        an+1 1 1 ··· 1 an+1,n+1 and

      a0 b1 ··· bn d a0,0 a0,1 ··· a0,n a0,n+1               a1 c1    1 a1,1 ··· a1,n a1,n+1        B   . .    . . . . .  σ ⊗  .. .  · ⊗  ......  i                     an cn    1 1 ··· an,n an,n+1        an+1 1 1 ··· 1 an+1,n+1    a0a0,0 b1a0,1 b2a0,2 ··· bn−1a0,n−1 bna0,n da0,n+1         1 a1a1,1 a1,2 ··· a1,n−1 a1,n c1a1,n+1          1 1 a a ··· a a c a    2 2,2 2,n−1 2,n 2 2,n+1     B   ......  = σi ⊗  ......          1 1 1 ··· an−1an−1,n−1 an−1,n cn−1bn−1,n+1          1 1 1 ··· 1 a a c a    n n,n n n,n+1     1 1 1 ··· 1 1 an+1an+1,n+1 106   a0a0,0 b1a0,1 ··· bia0,i 1 bi+1a0,i+1 ··· bna0,n da0,n+1      1 a a ··· a 1 a ··· a c a   1 1,1 1,i 1,i+1 1,n 1 1,n+1     ......   ......       1 1 ··· ai 1 ai,i+1 ··· ai,n ciai,n+1      = ⊗  1 1 ··· 1 1 1 ··· 1 1         1 1 ··· 1 1 ai+1 ··· ai+1,n ci+1ai+1,n+1     ......   ......       1 1 ··· 1 1 1 ··· a a c a   n n,n n n,n+1    1 1 ··· 1 1 1 ··· 1 an+1an+1,n+1    a0 ··· bi 1 bi+1 ··· d      . .    .. .          a c    i i        = ⊗  1 1          ai+1 ci+1      . .    .. .    .     an    a0,0 ··· a0,i 1 a0,i+1 ··· a0,n+1      ......    ......          1 ··· a 1 a ··· a    i,i i,i+1 i,n+1        · ⊗  1 ··· 1 1 1 ··· 1          1 ··· 1 1 ai+1,i+1 ··· ai+1,n+1       ......    ......    . . . . .     1 ··· 1 1 1 ··· an+1,n+1 107       a0 b1 ··· bn d a0,0 a0,1 ··· a0,n a0,n+1               a1 c1    1 a1,1 ··· a1,n a1,n+1        A   . .  B   . . . . .  = σ ⊗  .. .  σ ⊗  ......  . i    i                  an cn    1 1 ··· an,n an,n+1        an+1 1 1 ··· 1 an+1,n+1

Thus, B2(A) is a simplicial left module over the simplicial k-algebra A2(A).

Example 4.2.6. Deﬁne the simplicial right module M2(M) over the simplicial k-algebra A2(A)

M M (see Example 4.2.4) by setting Mn = M, δi = idM , and σi = idM for all n. The multiplication

on Mn is given by

m · (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn) = ma0 ··· an+1b1 ··· bndc1 ··· cn.

Proof. The simplicial identities (2.2) and (2.3) are immediately satisﬁed. Now we just need to check the compatibility conditions (3.2). We have that

M δi (m · (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn))

M = δi (ma0 ··· an+1b1 ··· bndc1 ··· cn)

= ma0 ··· an+1b1 ··· bndc1 ··· cn

= m · (a0 ⊗ · · · ⊗ aiai+1 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bibi+1 ⊗ · · · ⊗ bn

⊗ d ⊗ c1 ⊗ · · · ⊗ cici+1 ⊗ · · · ⊗ cn)

M A = δi (m)δi (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn), and

M σi (m · (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn))

M = σi (ma0 ··· an+1b1 ··· bndc1 ··· cn) 108

= ma0 ··· an+1b1 ··· bndc1 ··· cn

= m · (a0 ⊗ · · · ⊗ ai ⊗ 1 ⊗ ai+1 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bi ⊗ 1 ⊗ bi+1 ⊗ · · · ⊗ bn

⊗ d ⊗ c1 ⊗ · · · ⊗ ci ⊗ 1 ⊗ ci+1 ⊗ · · · ⊗ cn)

M A = σi (m)σi (a0 ⊗ · · · ⊗ an+1 ⊗ b1 ⊗ · · · ⊗ bn ⊗ d ⊗ c1 ⊗ · · · ⊗ cn).

Thus, M2(M) is a simplicial right module over the simplicial k-algebra A2(A).

2 2 From Lemma 3.4.1 we note that M (M) ⊗A2(A) B (A) is a simplicial k-module. In dimension n, we have

⊗ (n+2)(n+3) ⊗3n+3 2 Mn ⊗An Bn = M ⊗A A .

Following Lemma 3.4.1 for 0 ≤ i ≤ n, the maps

⊗ (n+2)(n+3) ⊗ (n+1)(n+2) Di : M ⊗A3n+3 A 2 −→ M ⊗A⊗3n A 2 are

   a0,0 a0,1 ··· a0,n a0,n+1         1 a1,1 ··· a1,n a1,n+1       . . . . .  Di m ⊗A⊗3n+3  ......             1 1 ··· an,n an,n+1     1 1 ··· 1 an+1,n+1    a0,0 a0,1 ··· a0,n a0,n+1         1 a1,1 ··· a1,n a1,n+1     M B   . . . . .  = δ (m) ⊗A⊗3n δ ⊗  ......  i i            1 1 ··· an,n an,n+1     1 1 ··· 1 an+1,n+1 109   a0,0 a0,1 ··· a0,ia0,i+1 ··· a0,n a0,n+1      1 a1,1 ··· a1,ia1,i+1 ··· a1,n a1,n+1     ......   ......   . . . . .    = m ⊗ ⊗3n ⊗   . A  1 1 ··· ai,iai,i+1ai+1,i+1 ··· ai,nai+1,n ai,n+1ai+1,n+1    ......   ......       1 1 ··· 1 ··· a a   n,n n,n+1    1 1 ··· 1 ··· 1 an+1,n+1

⊗ (n+2)(n+3) ⊗ n(n−1) Identifying M ⊗A⊗3n+3 A 2 with M ⊗ A 2 under the isomorphisms

⊗ (n+2)(n+3) ⊗ n(n−1) ϕn : M ⊗A⊗3n+3 A 2 −→ M ⊗ A 2

given by    a0,0 a0,1 ··· a0,n a0,n+1         1 a1,1 ··· a1,n a1,n+1       . . . . .  ϕn m ⊗A⊗3n+3 ⊗  ......             1 1 ··· an,n an,n+1     1 1 ··· 1 an+1,n+1

  a1,2 a1,3 ··· a1,n−1 a1,n      1 a2,3 ··· a2,n−1 a2,n     . . . . .  = ma0,0a0,1 ··· a0,n+1a1,n+1 ··· an+1,n+1an,n ··· a1,1 ⊗  ......         1 1 ··· an−2,n−1 an−2,n   1 1 ··· 1 an−1,n and

n(n−1) (n+2)(n+3) −1 ⊗ 2 ⊗ 2 ϕn : M ⊗ A −→ M ⊗A⊗3n+3 A 110 given by    a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     −1   . . . . .  ϕ m ⊗  ......  n            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1

  1 1 1 1 ··· 1 1 1     1 1 a1,1 a1,2 ··· a1,n−2 a1,n−1 1     1 1 1 a ··· a a 1  2,2 2,n−2 2,n−1    ......  ......  m ⊗A⊗3n+3 ⊗   ,  ..  1 1 1 1 . an−2,n−2 an−2,n−1 1     1 1 1 1 ··· 1 a 1  n−1,n−1      1 1 1 1 ··· 1 1 1   1 1 1 1 ··· 1 1 1 we get that

   a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     −1   . . . . .  (ϕn−1 ◦ D0 ◦ ϕ ) m ⊗  ......  n            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1

  a2,2 ··· a2,n−2 a2,n−1    . . . .   . .. . .    = ma1,1a1,2 ··· a1,n−1 ⊗   ,  1 ··· a a   n−2,n−2 n−2,n−1   1 ··· 1 an−1,n−1 111    a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     −1   . . . . .  (ϕn−1 ◦ Di ◦ ϕ ) m ⊗  ......  = n            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1

  a1,1 ··· a1,i−2 a1,i−1a1,i a1,i+1 a1,i+2 ··· a1,n−1    ......   ......       1 ··· a a a a a ··· a   i−2,i−2 i−2,i−1 i−2,i i−2,i+1 i−2,i+2 i−2,n−1       1 ··· 1 ai−1,i−1ai−1,i ai−1,i+1 ai−1,i+2 ··· ai−1,n−1  mai,i ⊗      1 ··· 1 1 ai,i+1ai+1,i+1 ai,i+2ai+1,i+2 ··· ai,n−1ai+1,n−1      1 ··· 1 1 1 a ··· a   i+2,i+2 i+2,n−1     ......   ......    1 ··· 1 1 1 1 ··· an−1,n−1 for 1 ≤ i ≤ n − 1, and

   a1,1 a1,2 ··· a1,n−2 a1,n−1         1 a2,2 ··· a2,n−2 a2,n−1     −1   . . . . .  (ϕn−1 ◦ Dn ◦ ϕ ) m ⊗  ......  n            1 1 ··· an−2,n−2 an−2,n−1    1 1 ··· 1 an−1,n−1

  a1,1 a1,2 ··· a1,n−2      1 a2,2 ··· a2,n−2  = ma a ··· a ⊗   . 1,n−1 2,n−1 n−1,n−1  . . . .   . . .. .   . . .    1 1 ··· an−2,n−2

n(n−1) (n−1)(n−2) n ⊗ 2 ⊗ 2 P i −1 Taking ∂n : M ⊗ A −→ M ⊗ A by setting ∂n = i=0(−1) (ϕn−1 ◦ Di ◦ ϕn ), 112 we produce a chain complex of the form

n(n−1) ∂n+1 ⊗ ∂n ∂4 ⊗3 ∂3 ∂2 ∂1 ... −−−→ M ⊗ A 2 −−→ ... −−→ M ⊗ A −−→ M ⊗ A −−→ M −−→ M −→ 0.

Let’s actually go through the details and construct ∂3. We have that

      3 a b X a b ∂ m ⊗   := ϕ ◦ (−1)iD ◦ ϕ−1 m ⊗   3    2 i 3    1 c i=0 1 c    1 1 1 1 1        1 1 a b 1       = ϕ2 ◦ (D0 − D1 + D2 − D3) m ⊗A⊗12 1 1 1 c 1           1 1 1 1 1    1 1 1 1 1      1 a b 1 1 1 1 1            1 1 c 1 1 a bc 1      = ϕ2 m ⊗A⊗9   − m ⊗A⊗9    1 1 1 1 1 1 1 1           1 1 1 1 1 1 1 1      1 1 1 1 1 1 1 1            1 1 ab 1 1 1 a b      + ϕ2 m ⊗A⊗9   − m ⊗A⊗9    1 1 c 1 1 1 1 c           1 1 1 1 1 1 1 1

= mab ⊗ c − ma ⊗ bc + mc ⊗ ab − mbc ⊗ a.

One can do similar constructions for n = 1, n = 2, and n = 4 and get

∂1(m) = m − m = 0,

∂2(m ⊗ a) = ma − ma + ma = ma, 113 and    a b c           d e bd ce ∂4 m ⊗ 1 d e = mabc ⊗   − ma ⊗             1 f 1 f 1 1 f

      ab c a bc a b       +md ⊗   − mf ⊗   + mcef ⊗   . 1 ef 1 de 1 d

S2 Comparing this to the construction of Hn (A, M), we now have the following:

Proposition 4.2.7. We have that

2 2 ∼ S2 Hn(M (M) ⊗A2(A) B (A)) = Hn (A, M).

Proof. Follows from the above discussion.

4.2.3 A Special Case Over the Sphere

We next deﬁne a special case involving the bar complex B2(A) for S2. This will permit us to identify the resulting chain complex with the secondary Hochschild homology of the triple (A, A, id) with coefﬁcients in M. Again A is commutative and M is a symmetric A-bimodule.

Example 4.2.8. Deﬁne the simplicial right module S2(A) over the simplicial k-algebra A2(A) by

⊗2n+1 setting Sn = A . The multiplication on Sn is given by

      a0 b1 b2 ··· bn−1 bn d α β β ··· β β    0 1 2 n−1 n         a1 c1          α1          a c       2 2    α2          .. .  ⊗   · ⊗  . .    ...                an−1 cn−1          αn−1          a c       n n  αn    an+1 114   α0a0an+1d β1b1c1 β2b2c2 ··· βn−1bn−1cn−1 βnbncn      α a   1 1       α2a2  = ⊗   .  ..   .       α a   n−1 n−1    αnan

Next we deﬁne

   α0 β1 β2 ··· βn−1 βn         α0α1β1 β2 ··· βn−1 βn   α      1        α2    α    S   2   .  δ ⊗   = ⊗  ..  , 0   ..      .             αn−1    α      n−1     αn αn

     α0 β1 β2 ··· βn−1 βn α0 β1 ··· βiβi+1 ··· βn             α   α    1   1      .    α   ..  S   2    δ ⊗   = ⊗   i   ..      .   αiαi+1           .    α   ..    n−1         αn αn 115 for 1 ≤ i ≤ n − 1, and

   α0 β1 β2 ··· βn−1 βn         α0αnβn β1 β2 ··· βn−1   α      1        α1    α    S   2    δ ⊗   = ⊗  α  n   ..   2    .        ..      .    α      n−1     αn−1. αn

Moreover,

  α0 β1 ··· βi 1 βi+1 ··· βn        α0 β1 β2 ··· βn−1 βn  α1           .    α   ..    1           α   α  S   2   i  σ ⊗   = ⊗   i   ..      .   1              α   α    n−1   i+1        ..  αn  .    αn for 0 ≤ i ≤ n.

Proof. One checks these details.

2 2 Remark 4.2.9. From Lemma 3.4.1 we note that S (A) ⊗A2(A) B (A) is a simplicial k-module. Making an identiﬁcation and following a process similar to that done in Subsection 4.2.2, we have that

⊗2n+1 ⊗ (n+2)(n+3) ⊗ (n+1)(n+2) ⊗3n+3 2 2 Sn ⊗An Bn = A ⊗A A = A 116 in dimension n. This produces a chain complex of the form

(n+1)(n+2) bn+1 ⊗ bn b4 ⊗10 b3 ⊗6 b2 ⊗3 b1 ... −−→ A 2 −−→ ... −−→ A −−→ A −−→ A −−→ A −→ 0 with maps for n = 1, n = 2, and n = 3 given as

   a b    b1 ⊗   = abc − abc = 0, 1 c

   a b c             abd ce a bc acf be b2 ⊗ 1 d e = ⊗   − ⊗   + ⊗   ,             1 f 1 def 1 d 1 1 f and    a b c d        1 e f g    b3 ⊗   =  1 1 h i       1 1 1 j

        abe cf dg a bc d a b cd adj bg ci                 ⊗  1 h i  − ⊗ 1 efh gi + ⊗ 1 e fg  − ⊗  1 e f  .                 1 1 j 1 1 j 1 1 hij 1 1 h

2 2 Remark 4.2.10. We can consider Hn(S (A)⊗A2(A)B (A)) as the “HHn”-equivalent of Proposition 4.2.7. Furthermore, we note that

2 2 ∼ Hn(S (A) ⊗A2(A) B (A)) = HHn(A, A, id), which is the secondary Hochschild homology associated to the triple (A, A, id), as we will see in detail in Chapter 5. 117

CHAPTER 5 FOUNDATIONS OF THE SECONDARY CYCLIC HOMOLOGY

This chapter is dedicated to the construction of the secondary cyclic homology associated to the triple (A, B, ε). The construction of HCn(A, B, ε) appears in [27], but lacks some detail. Here

we present three analogous deﬁnitions of HCn(A, B, ε), along with three proofs of the secondary analogue of Connes’ long exact sequence.

5.1 Deﬁnition and Construction

The following construction appeared in [27]. It is similar to that seen in [8] or [28]. We ﬁrst recall the secondary Hochschild homology associated to the triple (A, B, ε).

Deﬁnition 5.1.1. ([27]) The homology of the complex which is associated to the simplicial k-

module L(A, B, ε)⊗A(A,B,ε)B(A, B, ε) (see Proposition 4.1.5) is called the secondary Hochschild homology associated to the triple (A, B, ε). The groups are denoted by

HHn(A, B, ε).

As done in [27], we note that in dimension n we have

Cn(A, B, ε) := Ln ⊗A⊗B⊗2n+1⊗Aop Bn

⊗n ⊗n+2 ⊗ (n+1)(n+2) = A ⊗ B ⊗A⊗B⊗2n+1⊗Aop ⊗A ⊗ B 2

⊗n ⊗n ⊗ n(n−1) = A ⊗ B ⊗ A ⊗ B 2

⊗n+1 ⊗ n(n+1) = A ⊗ B 2 .

We can consider the induced chain

n(n+1) ∂n+1 ⊗n+1 ⊗ ∂n ∂3 ⊗3 ⊗3 ∂2 ⊗2 ∂1 ... −−−→ A ⊗ B 2 −−→ ... −−→ A ⊗ B −−→ A ⊗ B −−→ A −→ 0. 118 The morphisms

∂n : Cn(A, B, ε) −→ Cn−1(A, B, ε) are given by    a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  ∂n ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an

   a0 b0,1 ··· b0,ib0,i+1 ··· b0,n−1 b0,n         1 a1 ··· b1,ib1,i+1 ··· b1,n−1 b1,n       ......    ......  n−1    X    = (−1)i ⊗     1 1 ··· aiε(bi,i+1)ai+1 ··· bi,n−1bi+1,n−1 bi,nbi+1,n    i=0   ......    ......          1 1 ··· 1 ··· a b    n−1 n−1,n     1 1 ··· 1 ··· 1 an

   anε(b0,n)a0 b1,nb0,1 ··· bn−2,nb0,n−2 bn−1,nb0,n−1         1 a1 ··· b1,n−2 b1,n−1     n   . . . . .  +(−1) ⊗  ......  .            1 1 ··· an−2 bn−2,n−1     1 1 ··· 1 an−1

Remark 5.1.2. For low dimensions, we have

   a α    ∂1 ⊗   = aε(α)b − bε(α)a. 1 b 119 and

   a α β             aε(α)b βγ a αβ cε(β)a αγ ∂2 ⊗ 1 b γ = ⊗   − ⊗   + ⊗   .             1 c 1 bε(γ)c 1 b 1 1 c

It follows from Lemma 3.4.1 that any two consecutive morphisms will yield zero, and therefore

the above is a complex which we will denote by C•(A, B, ε).

∼ Example 5.1.3. ([27]) For B = k, we get that HHn(A, k, ε) = HHn(A) for all n.

A Remark 5.1.4. ([27]) Observe that HH (A, B, ε) ∼= HH (A) = . 0 0 [A, A]

th i Notice that ∂n has n + 1 terms. We deﬁne δi to be the i term in the sum without the (−1) , and can therefore write n X i ∂n = (−1) δi. i=0

0 We deﬁne ∂n to be the ﬁrst n terms of ∂n. As above,

n−1 0 X i ∂n = (−1) δi. i=0

0 Proposition 5.1.5. The chain complex C•(A, B, ε) with the maps ∂ is acyclic. We denote this by

0 0 (C•(A, B, ε), ∂ ), and in particular, Hn((C•(A, B, ε), ∂ )) = 0 for all n ≥ 0.

Proof. First we deﬁne maps

sn : Cn(A, B, ε) −→ Cn+1(A, B, ε) 120 by

     1 1 1 ··· 1 1   a0 b0,1 ··· b0,n−1 b0,n      1 a b ··· b b      0 0,1 0,n−1 0,n    1 a1 ··· b1,n−1 b1,n           . . . . .  1 1 a1 ··· b1,n−1 b1,n  sn ⊗  ......  = ⊗      ......     ......         1 1 ··· an−1 bn−1,n      1 1 1 ··· a b   n−1 n−1,n 1 1 ··· 1 an   1 1 1 ··· 1 an

for n ≥ 0, where s−1 = 0. We notice that

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n     0   . . . . .  ∂ ◦ sn ⊗  ......  n+1            1 1 ··· an−1 bn−1,n    1 1 ··· 1 an    a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n     0   . . . . .  + sn−1 ◦ ∂ ⊗  ......  n            1 1 ··· an−1 bn−1,n    1 1 ··· 1 an    1 1 1 ··· 1 1        1 a b ··· b b    0 0,1 0,n−1 0,n      1 1 a ··· b b  0   1 1,n−1 1,n  = ∂ ⊗   n+1  ......   ......         1 1 1 ··· a b    n−1 n−1,n    1 1 1 ··· 1 an 121     a0 ··· b0,ib0,i+1 ··· b0,n        ......     . . . . .  n−1    X i    + sn−1  (−1) ⊗  1 ··· a ε(b )a ··· b b     i i,i+1 i+1 i,n i+1,n  i=0       ......     . . . . .      1 ··· 1 ··· an   a0 b0,1 ··· b0,n−1 b0,n      1 a1 ··· b1,n−1 b1,n     . . . . .  = ⊗  ......         1 1 ··· an−1 bn−1,n   1 1 ··· 1 an    1 1 1 1 1 1        1 a ··· b b ··· b    0 0,i 0,i+1 0,n   ......  n−1  ......  X i+1    + (−1) ⊗      i=0  1 1 ··· aiε(bi,i+1)ai+1 ··· bi,nbi+1,n     ......   ......        1 1 ··· 1 ··· an    1 1 1 1 1 1        1 a ··· b b ··· b    0 0,i 0,i+1 0,n   ......  n−1  ......  X i    + (−1) ⊗      i=0  1 1 ··· aiε(bi,i+1)ai+1 ··· bi,nbi+1,n     ......   ......        1 1 ··· 1 ··· an 122   a0 b0,1 ··· b0,n−1 b0,n      1 a1 ··· b1,n−1 b1,n     . . . . .  = ⊗  ......  .        1 1 ··· an−1 bn−1,n   1 1 ··· 1 an

Thus, we have that

0 0 id n(n+1) = ∂n+1 ◦ sn + sn−1 ◦ ∂n. A⊗n+1⊗B⊗ 2

We then apply Lemma 2.2.6 to get our desired result.

5.1.1 The Secondary Connes’ Complex

We now continue with a construction that was done in [27]. Again, this is similar to what is done is [28]. When B = k it reduces to the construction in [28].

We consider the permutation λn = (0, 1, 2, . . . , n) and the cyclic group Cn+1 = hλni, which has a left action on Cn(A, B, ε) deﬁned by

   a0 b0,1 b0,2 ··· b0,n−2 b0,n−1 b0,n         1 a1 b1,2 ··· b1,n−2 b1,n−1 b1,n          1 1 a ··· b b b    2 2,n−2 2,n−1 2,n       ......  λn ⊗  ......          1 1 1 ··· an−2 bn−2,n−1 bn−2,n         1 1 1 ··· 1 a b    n−1 n−1,n    1 1 1 ··· 1 1 an 123    an b0,n b1,n ··· bn−3,n bn−2,n bn−1,n         1 a0 b0,1 ··· b0,n−3 b0,n−2 b0,n−1          1 1 a ··· b b b    1 1,n−3 1,n−2 1,n−1     n   ......  = (−1) ⊗  ......  .         1 1 1 ··· an−3 bn−3,n−2 bn−3,n−1         1 1 1 ··· 1 a b    n−2 n−2,n−1    1 1 1 ··· 1 1 an−1

n+1 So λn is a cyclic operator from Cn(A, B, ε) to itself and is a well-deﬁned action since λn = 1.

Remark 5.1.6. For λ1 : C1(A, B, ε) −→ C1(A, B, ε), we have

      a α b α       λ1 ⊗   = − ⊗   . 1 b 1 a

Likewise, for λ2 : C2(A, B, ε) −→ C2(A, B, ε), we have

      a α β c β γ             λ2 ⊗ 1 b γ = ⊗ 1 a α .             1 1 c 1 1 b

n X i Lemma 5.1.7. With ∂n = (−1) δi, we have that i=0

(i) δiλn = −λn−1δi−1 for 0 < i ≤ n, and

i (ii) δ0λn = (−1) δn. 124 Proof. For (i) we start by checking the case where 0 < i < n. Observe we have

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  δiλn ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an     an b0,n ··· bn−2,n bn−1,n            1 a0 ··· b0,n−2 b0,n−1       n   . . . . .  = δi (−1) ⊗  ......                 1 1 ··· an−2 bn−2,n−1     1 1 ··· 1 an−1    an b0,n ··· bi−1,nbi,n ··· bn−2,n bn−1,n         1 a0 ··· b0,i−1b0,i ··· b0,n−2 b0,n−1       ......    ......    . . . . .     = (−1)n ⊗     1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,n−2bi,n−2 bi−1,n−1bi,n−1      ......    ......          1 1 ··· 1 ··· a b    n−2 n−2,n−1     1 1 ··· 1 ··· 1 an−1 for the left hand side, and

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  −λn−1δi−1 ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an 125    a0 b0,1 ··· b0,i−1b0,i ··· b0,n−1 b0,n         1 a1 ··· b1,i−1b1,i ··· b1,n−1 b1,n       ......    ......    . . . . .     = −λ ⊗   n−1   1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,n−1bi,n−1 bi−1,nbi,n      ......    ......          1 1 ··· 1 ··· a b    n−1 n−1,n     1 1 ··· 1 ··· 1 an    an b0,n ··· bi−1,nbi,n ··· bn−2,n bn−1,n         1 a0 ··· b0,i−1b0,i ··· b0,n−2 b0,n−1       . . . . .    ......    . . . . .     = (−1)n ⊗     1 1 ··· ai−1ε(bi−1,i)ai ··· bi−1,n−2bi,n−2 bi−1,n−1bi,n−1      ......    ......          1 1 ··· 1 ··· a b    n−2 n−2,n−1     1 1 ··· 1 ··· 1 an−1 for the right hand side. For i = n we get that

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  δiλn ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an     an b0,n ··· bn−2,n bn−1,n            1 a0 ··· b0,n−2 b0,n−1       n   . . . . .  = δi (−1) ⊗  ......                 1 1 ··· an−2 bn−2,n−1     1 1 ··· 1 an−1 126    an−1ε(bn−1,n)an b0,n−1b0,n ··· bn−2,n−1bn−2,n         1 a0 ··· b0,n−2  = (−1)n ⊗     . . . .    . . .. .    . . .     1 1 ··· an−2 for the left, and

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  −λn−1δn−1 ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an    a0 b0,1 ··· b0,n−1b0,n         1 a1 ··· b1,n−1b1,n  = −λ ⊗   n−1   . . . .    . . .. .    . . .     1 1 ··· an−1ε(bn−1,n)an    an−1ε(bn−1,n)an b0,n−1b0,n ··· bn−2,n−1bn−2,n         1 a0 ··· b0,n−2  = (−1)n ⊗     . . .    ......    . . .     1 1 ··· an−2 for the right. This establishes (i). 127 For (ii), we verify

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  δ0λn ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an     an b0,n ··· bn−2,n bn−1,n            1 a0 ··· b0,n−2 b0,n−1       n   . . . . .  = δ0 (−1) ⊗  ......                 1 1 ··· an−2 bn−2,n−1     1 1 ··· 1 an−1    anε(b0,n)a0 ··· bn−2,nb0,n−2 bn−1,nb0,n−1      . . . .    . .. . .  n    = (−1) ⊗     1 ··· a b    n−2 n−2,n−1     1 ··· 1 an−1    a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n     n   . . . . .  = (−1) δn ⊗  ......  ,            1 1 ··· an−1 bn−1,n    1 1 ··· 1 an which was what we wanted.

0 Lemma 5.1.8. For all n ≥ 0 we have that ∂n(1 − λn) = (1 − λn−1)∂n.

Proof. We will start with the identity found above and then build on that.

δiλn = −λn−1δi−1 by Lemma 5.1.7 (i) 128

i i (−1) δiλn = −(−1) λn−1δi−1

i i−1 (−1) δiλn = λn−1(−1) δi−1 n n X i X i−1 (−1) δiλn = λn−1 (−1) δi−1 i=1 i=1

0 (∂n − δ0)λn = λn−1∂n

0 ∂nλn − δ0λn = λn−1∂n

n 0 ∂nλn − (−1) δn = λn−1∂n by Lemma 5.1.7 (ii)

0 0 ∂nλn − (∂n − ∂n) = λn−1∂n

0 0 ∂nλn − ∂n = λn−1∂n − ∂n

0 ∂n(λn − 1) = (λn−1 − 1)∂n

0 ∂n(1 − λn) = (1 − λn−1)∂n.

This proves the lemma.

Deﬁne

λ Cn(A, B, ε) Cn (A, B, ε) := . Im(1 − λn)

As consequence of Lemma 5.1.8, ∂n sends the image of 1−λn to the image of 1−λn−1. Therefore,

λ λ ∂n : Cn (A, B, ε) −→ Cn−1(A, B, ε)

λ is well-deﬁned. We can then consider the chain complex C• (A, B, ε):

∂n+1 λ ∂n ∂3 λ ∂2 λ ∂1 λ ... −−−→ Cn (A, B, ε) −−→ ... −−→ C2 (A, B, ε) −−→ C1 (A, B, ε) −−→ C0 (A, B, ε) −→ 0.

Deﬁnition 5.1.9. ([27]) The secondary cyclic homology groups HCn(A, B, ε) are the homology

λ groups of the complex C• (A, B, ε). That is,

λ HCn(A, B, ε) = Hn(C• (A, B, ε)). 129 λ • Remark 5.1.10. Observe that C• (A, B, ε) is the secondary analogue of Connes’ complex Cλ(A), which can be seen in Section 2.6 and extensively in [28].

5.1.2 The Secondary Cyclic Bicomplex

The goal of this subsection is to explore an equivalent way to deﬁne cyclic homology. In Section 5.2, we will actually show that the two deﬁnitions are the same. This method is adapted from [28].

We ﬁrst deﬁne the map Nn. Let

Nn : Cn(A, B, ε) −→ Cn(A, B, ε)

be given by

2 3 n Nn = 1 + λn + λn + λn + ··· + λn.

Lemma 5.1.11. For all n ≥ 0 we have that (1 − λn)Nn = 0 = Nn(1 − λn).

Proof. Observe that

2 3 n (1 − λn)Nn = (1 − λn)(1 + λn + λn + λn + ··· + λn)

n+1 = 1 − λn

= 0

n+1 = 1 − λn

2 3 n = (1 + λn + λn + λn + ··· + λn)(1 − λn)

= Nn(1 − λn).

This completes our lemma.

n X i Lemma 5.1.12. With ∂n = (−1) δi, we have that i=0

j j j (i) δiλn = (−1) λn−1δi−j whenever i ≥ j, and 130 j n−j+1 j−1 (ii) δiλn = (−1) λn−1δn+1+i−j whenever i < j.

Proof. For i ≥ j, we notice that

j j−1 δiλn = (δiλn)λn

j−1 = −λn−1δi−1λn by Lemma 5.1.7 (i)

j−2 = −λn−1(δi−1λn)λn

2 2 j−2 = (−1) λn−1δi−2λn again by Lemma 5.1.7 (i)

j j = ··· = (−1) λn−1δi−j.

This establishes (i). Whenever i < j we have that

j i j−i δiλn = (δiλn)λn

i i j−i = (−1) λn−1δ0λn by (i) above

i i j−i−1 = (−1) λn−1(δ0λn)λn

i i n j−i−1 = (−1) λn−1(−1) δnλn by Lemma 5.1.7 (ii)

n+i i j−i−1 = (−1) λn−1(δnλn )

n+i i j−i−1 j−i−1 = (−1) λn−1(−1) λn−1 δn−j+i+1 by Lemma 5.1.7 (i)

n−j+1 j−1 = (−1) λn−1δn+1+i−j,

which gives us (ii).

0 Lemma 5.1.13. For all n ≥ 0 we have that ∂nNn = Nn−1∂n.

Proof. We have

n−1 ! n ! 0 X i X j ∂nNn = (−1) δi λn i=0 j=0 131

X i j = (−1) δiλn i,j

X i j X i j = (−1) δiλn + (−1) δiλn 0≤j≤i≤n−1 0≤i

X i j j X i n−j+1 j−1 = (−1) (−1) λn−1δi−j + (−1) (−1) λn−1δn+1+i−j 0≤j≤i≤n−1 0≤i

by Lemma 5.1.12 (i) and (ii), respectively

X i−j j X n+1+i−j j−1 = (−1) λn−1δi−j + (−1) λn−1δn+1+i−j. 0≤j≤i≤n−1 0≤i

p Following [28], in the above summation, for 0 ≤ p ≤ n, the coefﬁcient of (−1) δp is

X j X j−1 λn−1 + λn−1 = Nn−1. 0≤j≤n−1−p n−p≤j−1≤n−1

0 X p Thus, ∂nNn = Nn−1 (−1) δp = Nn−1∂n. 0≤p≤n We now construct a ﬁrst quadrant bicomplex as follows:

......

∂3 −∂3 ∂3

1 − λ2 N2 1 − λ2 A⊗3 ⊗ B⊗3 A⊗3 ⊗ B⊗3 A⊗3 ⊗ B⊗3 ···

0 ∂2 −∂2 ∂2

1 − λ1 N1 1 − λ1 A⊗2 ⊗ B A⊗2 ⊗ B A⊗2 ⊗ B ···

0 ∂1 −∂1 ∂1

1 − λ0 N0 1 − λ0 A A A ··· 132 Notice that the rows are chain complexes because of Lemma 5.1.11, and all the columns are chain

complexes because those are the original maps from the complex C•(A, B, ε). Every square anticommutes due to Lemma 5.1.8 and Lemma 5.1.13. Therefore, by Deﬁnition 2.4.1, the above is

indeed a bicomplex which we will denote CC••(A, B, ε). For this secondary cyclic bicomplex, we have

⊗q ⊗ q(q−1) CCp,q(A, B, ε) = Cq(A, B, ε) = A ⊗ B 2 .

We can consider the homology of the total complex associated to CC••(A, B, ε), denoted

Tot CC••(A, B, ε). That is,

Hn(Tot CC••(A, B, ε)).

5.1.3 The Secondary Diagonal Bicomplex

This subsection explores yet another equivalent construction of the secondary cyclic homology, which will employ a diagonal bicomplex. Again, see Section 5.2 for the proofs that all constructions yield the same homology groups. A similar construction appears in [28] for the cyclic homology.

0 For every column in the secondary cyclic bicomplex CC••(A, B, ε) with −∂ as the morphisms, we note by Proposition 5.1.5 that it is split exact. Therefore it is endowed with a contracting homotopy sn. In essence, this allows us to bypass those columns and write new morphisms connecting the remaining columns.

1 − λn+1 ⊗n+2 ⊗ (n+1)(n+2) ⊗n+2 ⊗ (n+1)(n+2) A ⊗ B 2 A ⊗ B 2

Nn ⊗n+1 ⊗ n(n+1) ⊗n+1 ⊗ n(n+1) A ⊗ B 2 A ⊗ B 2 133 We deﬁne

⊗n+1 ⊗ n(n+1) ⊗n+2 ⊗ (n+1)(n+2) Bn : A ⊗ B 2 −→ A ⊗ B 2 by

Bn = (1 − λn+1)snNn.

We can then rearrange to make it have the following appearance:

......

B2 B1 B0 A⊗4 ⊗ B⊗6 A⊗3 ⊗ B⊗3 A⊗2 ⊗ B A

∂3 ∂2 ∂1

B1 B0 A⊗3 ⊗ B⊗3 A⊗2 ⊗ B A

∂2 ∂1

B0 A⊗2 ⊗ B A

∂1

Remark 5.1.14. We have that

   1 1    B0(a) = (1 − λ1)s0N0(a) = (1 − λ1)s0(a) = (1 − λ1) ⊗   1 a     1 1 a 1     = ⊗   + ⊗   . 1 a 1 1 134 Similarly, we have that

           a α a α a α b α            B1 ⊗   = (1 − λ2)s1N1 ⊗   = (1 − λ2)s1 ⊗   − ⊗   1 b 1 b 1 b 1 a      1 1 1 1 1 1           = (1 − λ2) ⊗ 1 a α − ⊗ 1 b α           1 1 b 1 1 a         1 1 1 1 1 1 a 1 α b 1 α                 = ⊗ 1 a α − ⊗ 1 b α + ⊗ 1 1 1 − ⊗ 1 1 1 .                 1 1 b 1 1 a 1 1 b 1 1 a

Likewise,

      a α β a α β             B2 ⊗ 1 b γ = (1 − λ3)s2Nn ⊗ 1 b γ             1 1 c 1 1 c        a α β c β γ b γ α               = (1 − λ3)s2 ⊗ 1 b γ + ⊗ 1 a α + ⊗ 1 c β               1 1 c 1 1 b 1 1 a        1 1 1 1 1 1 1 1 1 1 1 1                1 a α β 1 c β γ 1 b γ α        = (1 − λ3) ⊗   + ⊗   + ⊗    1 1 b γ 1 1 a α 1 1 c β               1 1 1 c 1 1 1 b 1 1 1 a       1 1 1 1 1 1 1 1 1 1 1 1             1 a α β 1 b γ α 1 c β γ       = ⊗   + ⊗   + ⊗   1 1 b γ 1 1 c β 1 1 a α             1 1 1 c 1 1 1 a 1 1 1 b 135       a 1 α β b 1 γ α c 1 β γ             1 1 1 1 1 1 1 1 1 1 1 1       + ⊗   + ⊗   + ⊗   . 1 1 b γ 1 1 c β 1 1 a α             1 1 1 c 1 1 1 a 1 1 1 b

In general, we have that

   a0 b0,1 ··· b0,n−1 b0,n         1 a1 ··· b1,n−1 b1,n       . . . . .  Bn ⊗  ......             1 1 ··· an−1 bn−1,n    1 1 ··· 1 an

   1 1 1 ··· 1 1 ··· 1 1        1 a b ··· b b ··· b b    i i,i+1 i,n 0,i i−2,i i−1,i         1 1 ai+1 ··· bi+1,n b0,i+1 ··· bi−2,i+1 bi−1,i+1     ......   ......  n    X ni    = (−1) ⊗ 1 1 1 ··· a b ··· b b    n 0,n i−2,n i−1,n  i=0        1 1 1 ··· 1 a0 ··· b0,i−2 b0,i−1      ......   ......         1 1 1 ··· 1 1 ··· a b    i−2 i−2,i−1    1 1 1 ··· 1 1 ··· 1 ai−1 136    ai 1 bi,i+1 ··· bi,n b0,i ··· bi−2,i bi−1,i         1 1 1 ··· 1 1 ··· 1 1             1 1 ai+1 ··· bi+1,n b0,i+1 ··· bi−2,i+1 bi−1,i+1      ......    ......  n    X ni    + (−1) ⊗  1 1 1 ··· a b ··· b b  .   n 0,n i−2,n i−1,n  i=0         1 1 1 ··· 1 a0 ··· b0,i−2 b0,i−1       ......    ......          1 1 1 ··· 1 1 ··· a b    i−2 i−2,i−1    1 1 1 ··· 1 1 ··· 1 ai−1

Lemma 5.1.15. For all n > 0 we have that BnBn−1 = 0.

Proof. Notice that

BnBn−1 = (1 − λn+1)snNn(1 − λn)sn−1Nn−1 = 0 by Lemma 5.1.11.

Lemma 5.1.16. For all n ≥ 0 we have that ∂n+1Bn + Bn−1∂n = 0.

Proof. For n = 0, we need to show that ∂1B0 = 0. Notice by Remark 5.1.14 that

     1 1 a 1      ∂1B0(a) = ∂1 ⊗   + ⊗   = 1ε(1)a − aε(1)1 + aε(1)1 − 1ε(1)a = 0. 1 a 1 1

For n > 0,

∂n+1Bn + Bn−1∂n = ∂n+1(1 − λn+1)snNn + (1 − λn)sn−1Nn−1∂n

0 = (1 − λn)∂n+1snNn + (1 − λn)sn−1Nn−1∂n by Lemma 5.1.8

0 0 = (1 − λn)∂n+1snNn + (1 − λn)sn−1∂nNn by Lemma 5.1.13

0 0 = (1 − λn) ∂n+1sn(Nn) + sn−1∂n(Nn) 137

= (1 − λn)Nn as consequence of Proposition 5.1.5

= 0 by Lemma 5.1.11,

which was what we wanted.

We conclude that the above, denoted DC••(A, B, ε), is in fact a bicomplex by Deﬁnition 2.4.1.

We can then consider the homology of the total complex associated to DC••(A, B, ε), denoted

Tot DC••(A, B, ε). That is,

Hn(Tot DC••(A, B, ε)).

5.2 Proofs of the Equivalences

The goal of this section is to unite all three constructions of secondary cyclic homology. That is, we will show that

λ ∼ ∼ Hn(C• (A, B, ε)) = Hn(Tot CC••(A, B, ε)) = Hn(Tot DC••(A, B, ε)).

The proofs are very similar with those seen in [28]. We start by proving the ﬁrst isomorphism, where we suppose k is of characteristic zero and contains Q.

Lemma 5.2.1. Every row of the secondary cyclic bicomplex CC••(A, B, ε) is acyclic.

⊗n+1 ⊗ n(n+1) th Proof. Recalling that CCp,n(A, B, ε) = Cn(A, B, ε) = A ⊗ B 2 , the n row has the form

1−λn Nn 1−λn Nn Cn(A, B, ε) ←−−−− Cn(A, B, ε) ←−− Cn(A, B, ε) ←−−−− Cn(A, B, ε) ←−− ...

To see this is exact, we will show that the identity map is null-homotopic. Deﬁne

fn : Cn(A, B, ε) −→ Cn(A, B, ε) 138 1 by f (x) = · x, where x ∈ C (A, B, ε). Likewise, deﬁne n n + 1 n

gn : Cn(A, B, ε) −→ Cn(A, B, ε)

λ + 2λ2 + 3λ3 + ··· + nλn by g (x) = − n n n n (x), where again x ∈ C (A, B, ε). n n + 1 n

Every time a map precedes or succeeds (1 − λn), we apply gn, and likewise every time a map precedes or succeeds Nn, we apply fn. See the diagram below.

1 − λn Nn 1 − λn Nn Cn(A, B, ε) Cn(A, B, ε) Cn(A, B, ε) Cn(A, B, ε) ...

gn fn gn

1 − λn Nn 1 − λn Nn Cn(A, B, ε) Cn(A, B, ε) Cn(A, B, ε) Cn(A, B, ε) ...

Now we compute:

1 λ + 2λ2 + ··· + nλn N f + g (1 − λ ) = N − (1 − λ ) n n n n n n n n + 1 n n n + 1 1 λ + 2λ2 + ··· + nλn − λ2 − 2λ3 − · · · − nλn+1 = N − n n n n n n n + 1 n n + 1 N λ + λ2 + ··· + λn − n = n − n n n n + 1 n + 1 N 1 + λ + λ2 + ··· + λn − (n + 1) = n − n n n n + 1 n + 1 N N − (n + 1) = n − n n + 1 n + 1 n + 1 = = 1. n + 1

Notice that by the exact same argument, fnNn + (1 − λn)gn = 1. Therefore, the identity map is null-homotopic, and by Lemma 2.2.6 our proof is established.

λ There is a natural projection map π : Tot CC••(A, B, ε) −→ C• (A, B, ε). It is determined in 139 the following way:

We recall that Tot CC••(A, B, ε) consists of

C0(A, B, ε) ⊕ C1(A, B, ε) ⊕ · · · ⊕ Cn(A, B, ε)

in each dimension for n ≥ 0. Then πn will project the last component onto Cn(A, B, ε)/ Im(1−λ). This is exactly n ! M Cn(A, B, ε) π C (A, B, ε) = = Cλ(A, B, ε). n i Im(1 − λ ) n i=0 n Proposition 5.2.2. The natural projection π is a quasi-isomorphism. In particular,

∼ λ Hn(Tot CC••(A, B, ε)) = Hn(C• (A, B, ε))

for all n ≥ 0.

Proof. Each row is exact by Lemma 5.2.1. Therefore, because

C•(A, B, ε) λ Coker(1 − λ) = = C• (A, B, ε) Im•(1 − λ)

and by Theorem 2.4.3, we get what we want.

Lemma 5.2.3 (Killing Contractible Complexes). ([28]) Let

  α β   dn =   γ δ dn+1 dn−1 ... An ⊕ Bn An−1 ⊕ Bn−1 ...

be a complex of k-modules such that (B•, δ) is split exact, and hence has the contracting homotopy

hn : Bn −→ Bn+1. 140 Then the following inclusion of complexes is a quasi-isomorphism:

(id, −hγ):(A•, α − βhγ) −→ (A• ⊕ B•, d).

There is an injective map ι : Tot DC••(A, B, ε) −→ Tot CC••(A, B, ε) determined by

ι(x) = (x, snNn(x)),

where x ∈ Tot DC••(A, B, ε)p,q = Cq−p(A, B, ε), and therefore (x, snNn(x)) ∈ Cq−p(A, B, ε) ⊕

Cq−p+1(A, B, ε) = CC2p,q−p(A, B, ε) ⊕ CC2p−1,q−p+1 ⊆ Tot CC••(A, B, ε)p+q.

Proposition 5.2.4. The natural inclusion map ι is a quasi-isomorphism. In particular

∼ Hn(Tot DC••(A, B, ε)) = Hn(Tot CC••(A, B, ε)) for all n ≥ 0.

Proof. We want to apply the Killing Contractible Complexes Lemma 5.2.3. Notice that keeping in line with the notation, we take

(A• ⊕ B•, d) = (Tot CC••(A, B, ε), d), where     α β ∂ 1 − λ     d =   =   . γ δ N −∂0

We take B• to be what is leftover after taking out A• = Tot DC••(A, B, ε) from CC••(A, B, ε) in every dimension. For example, what is in parentheses in the below diagram makes up the complex B• for each respective dimension, while what is outside the parentheses are exactly the 141

modules from Tot DC••(A, B, ε):

... −→ A⊗5 ⊗ B⊗10 ⊕ A⊗3 ⊗ B⊗3 ⊕ A ⊕ (A⊗4 ⊗ B⊗6 ⊕ A⊗2 ⊗ B)

−→ A⊗4 ⊗ B⊗6 ⊕ A⊗2 ⊗ B ⊕ (A⊗3 ⊗ B⊗3 ⊕ A)

−→ A⊗3 ⊗ B⊗3 ⊕ A ⊕ (A⊗2 ⊗ B) −→ A⊗2 ⊗ B ⊕ (A) −→ A ⊕ (0) −→ 0.

Notice then that this B• is split exact by Proposition 5.1.5 when endowed with the morphisms δ = −∂0 in each component. Thus, again using the notation from the Killing Contractible Complexes Lemma 5.2.3, we get that h = −s. Therefore,

(id, −hγ):(A•, α − βhγ) −→ (A• ⊕ B•, d) becomes

(id, sN) : (Tot DC••(A, B, ε), ∂ + (1 − λ)sN) −→ (CC••(A, B, ε), d) where of course B = (1 − λ)sN. Thus the inclusion of complexes

ι = (id, sN) : (Tot DC••(A, B, ε), ∂ + B) −→ (Tot CC••(A, B, ε), d)

is a quasi-isomorphism.

Remark 5.2.5. By Proposition 5.2.2 and Proposition 5.2.4, we get that

λ ∼ ∼ Hn(C• (A, B, ε)) = Hn(Tot CC••(A, B, ε)) = Hn(Tot DC••(A, B, ε)).

Hence, HCn(A, B, ε) is the homology of any of the three complexes. 142 5.3 The Connes’ Long Exact Sequence for the Secondary Case

Connes ﬁrst discovered a long exact sequence relating Hochschild and cyclic (co)homology. One can see [28] or [42] for details. It was shown in [27] that there is a secondary analogue of Connes’ long exact sequence uniting the secondary Hochschild and cyclic (co)homologies associated to the triple (A, B, ε).

Theorem 5.3.1. For a triple (A, B, ε) we have a long exact sequence

B∗ I∗ S∗ B∗ I∗ ... −−→ HHn(A, B, ε) −→ HCn(A, B, ε) −−→ HCn−2(A, B, ε) −−→ HHn−1(A, B, ε) −→ ...

In this section we will prove the secondary analogue of Connes’ long exact sequence for homology (Proposition 5.3.1) in three different ways: each corresponding to one of the deﬁnitions.

λ One can be done with the secondary Connes’ complex C• (A, B, ε), one with the secondary cyclic bicomplex CC••(A, B, ε), and one with the diagonal cyclic bicomplex DC••(A, B, ε). We will approach these in order so as to stay consistent.

5.3.1 Proof 1

First we assume that k contains Q and k is of characteristic zero. Next we consider the chain complex

∂n+2 ∂n+1 ∂n ∂n−1 ... −−−→ Im(1 − λn+1) −−−→ Im(1 − λn) −−→ Im(1 − λn−1) −−−→ ...

2 0 Notice that ∂ = 0 as before, and this is indeed well-deﬁned because ∂n(1 − λn) = (1 − λn−1)∂n by Lemma 5.1.8. (That is, ∂n sends images of 1 − λn to images of 1 − λn−1.) Denote the above complex as Im•(1 − λ) with Imn(1 − λ) := Im(1 − λn) and note that it is a subcomplex of

C•(A, B, ε) λ C•(A, B, ε). Moreover, observe that = C• (A, B, ε). Im•(1 − λ) Proof 1. Observe that we have the following short exact sequences of complexes:

C•(A, B, ε) 0 −→ Im•(1 − λ) ,→ C•(A, B, ε) −→ 0. Im•(1 − λ) 143 By Theorem 2.2.11 this induces a long exact sequence

I∗ ... −→ Hn(Im•(1 − λ)) −→ HHn(A, B, ε) −−→ HCn(A, B, ε) −→ Hn−1(Im•(1 − λ)) −→ ...

If we can show that ∼ Hn−1(Im•(1 − λ)) = HCn−2(A, B, ε)

for all n, then we will be done. To do this, we consider the following sequence:

C•(A, B, ε) N 0 1−λ 0 −→ −−→ (C•(A, B, ε), ∂ ) −−−→ Im•(1 − λ) −→ 0. Im•(1 − λ)

0 We ﬁrst note that N and 1 − λ are chain maps. This is so because ∂nNn = Nn−1∂n by Lemma

0 5.1.13 and ∂n(1 − λn) = (1 − λn−1)∂n by Lemma 5.1.8. We further claim that this is short exact. It is enough to show that

Cn(A, B, ε) Nn 0 1−λn 0 −→ −−→ (Cn(A, B, ε), ∂ ) −−−−→ Im(1 − λn) −→ 0 Im(1 − λn)

is short exact. Notice that (1 − λn)Nn = 0 by Lemma 5.1.11 and so Im(Nn) ⊆ Ker(1 − λn).

Furthermore, 1 − λn is clearly onto because if x ∈ Im(1 − λn), then (1 − λn)(y) = x for some

y ∈ Cn(A, B, ε).

In order to see that Nn is one-to-one, we need to show that Ker(Nn) consists of all elements in

Im(1 − λn). So notice again by Lemma 5.1.11 we have that Ker(Nn) ⊇ Im(1 − λn). Therefore, to

see Ker(Nn) ⊆ Im(1 − λn), we take x ∈ Cn(A, B, ε) such that Nn(x) = 0, and we want to show

that (1 − λn)(y) = x for some y ∈ Cn(A, B, ε). Deﬁne y as follows:

1 y = − (λ + 2λ2 + ··· + nλn)(x). n + 1 n n n 144 Thus,

λ + 2λ2 + ··· + nλn (1 − λ )(y) = (1 − λ ) − n n n (x) n n n + 1 λ + 2λ2 + ··· + nλn − λ2 − 2λ3 − · · · − (n − 1)λn − nλn+1 = − n n n n n n n (x) n + 1 λ + λ2 ··· + λn − n = − n n n (x) n + 1 1 + λ + λ2 ··· + λn − (n + 1) = − n n n (x) n + 1 N (x) − (n + 1)(x) = − n n + 1 = x.

Finally, to see that Ker(1−λn) ⊆ Im(Nn), we take x ∈ Cn(A, B, ε) such that (1−λn)(x) = 0.

2 n This implies that x = λn(x) = λn(x) = ··· = λn(x). Then notice that

x x N = (1 + λ + λ2 + ··· + λn) n n + 1 n n n n + 1 x + λ (x) + λ2 (x) + ··· + λn(x) = n n n n + 1 x + x + x + ··· + x = n + 1 (n + 1)x = n + 1 = x.

This is what we wanted, and hence our sequence is short exact. That short exact sequence of complexes induces a long exact sequence (again by Theorem 2.2.11) as follows:

0 ... −→ HCn(A, B, ε) −→ Hn(C•(A, B, ε), ∂ ) −→ Hn(Im•(1 − λ)) −→ HCn−1(A, B, ε) −→ ... 145 0 Then by Proposition 5.1.5 we note that Hn(C•(A, B, ε), ∂ ) = 0 for all n, and therefore

∼ Hn(Im•(1 − λ)) = HCn−1(A, B, ε) for all n.

5.3.2 Proof 2

The process of this proof is similar to that seen done in [28]. We start by setting notation.

{i,j,...,l} Let CC••(A, B, ε) denote only the columns i through l in the original secondary cyclic

{i,j,...,l} bicomplex CC••(A, B, ε). Put zeros everywhere else. In particular, CC••(A, B, ε) is still a bicomplex.

{0,1} Remark 5.3.2. By above, note that CC••(A, B, ε) denotes the zeroth and ﬁrst columns of

{0} CC••(A, B, ε) with zeros as every remaining column. Note that we can view CC••(A, B, ε)

{1} and CC••(A, B, ε) in a similar way.

{0} ∼ Lemma 5.3.3. Hn(Tot CC••(A, B, ε) ) = HHn(A, B, ε) for all n ≥ 0.

{0} Proof. Observe that CC••(A, B, ε) has the following form:

. . . .

∂3 A⊗3 ⊗ B⊗3 0 ···

∂2 A⊗2 ⊗ B 0 ···

∂1 A 0 ···

{0} Therefore the total complex Tot CC••(A, B, ε) is

... −−→∂4 A⊗4 ⊗ B⊗6 −−→∂3 A⊗3 ⊗ B⊗3 −−→∂2 A⊗2 ⊗ B −−→∂1 A −→ 0. 146

This is exactly the complex C•(A, B, ε), and hence

{0} ∼ ∼ Hn(Tot CC••(A, B, ε) ) = Hn(C•(A, B, ε)) = HHn(A, B, ε).

This establishes the lemma.

{1} Lemma 5.3.4. Hn(Tot CC••(A, B, ε) ) = 0 for all n ≥ 0.

{1} Proof. Observe that CC••(A, B, ε) has the following form:

...... 0 −∂3 0 A⊗3 ⊗ B⊗3 0 ··· 0 −∂2 0 A⊗2 ⊗ B 0 ··· 0 −∂1 0 A 0 ···

{1} Therefore the total complex Tot CC••(A, B, ε) is

... −−−→−∂4 A⊗4 ⊗ B⊗6 −−−→−∂3 A⊗3 ⊗ B⊗3 −−−→−∂2 A⊗2 ⊗ B −−−→−∂1 A −→ 0 −→ 0.

0 Shifted by one, this is exactly the complex (C•(A, B, ε), ∂ ) which is exact by Proposition 5.1.5. Therefore, {1} ∼ 0 Hn(Tot CC••(A, B, ε) ) = Hn((C•(A, B, ε), ∂ )) = 0.

This is what we wanted.

{0,1} ∼ Corollary 5.3.5. Hn(Tot CC••(A, B, ε) ) = HHn(A, B, ε) for all n ≥ 0.

Proof. Consider the short exact sequence of bicomplexes:

{0} {0,1} {1} 0 −→ CC••(A, B, ε) ,→ CC••(A, B, ε) CC••(A, B, ε) −→ 0. 147 By Theorem 2.2.11 this induces a long exact sequence

{0} {0,1} ... −→ Hn(Tot CC••(A, B, ε) ) −→ Hn(Tot CC••(A, B, ε) ) −→

{1} {0} Hn(Tot CC••(A, B, ε) ) −→ Hn−1(Tot CC••(A, B, ε) ) −→ ...

{1} {0} ∼ By Lemma 5.3.4, Hn(Tot CC••(A, B, ε) ) = 0, and this implies Hn(Tot CC••(A, B, ε) ) =

{0,1} Hn(Tot CC••(A, B, ε) ). By applying Lemma 5.3.3 we get what we want.

Next let CC••(A, B, ε)[m,n] denote a shift of m dimensions horizontally and n dimensions

vertically. That is, CCp,q(A, B, ε)[m,n] = CCp−m,q−n(A, B, ε).

Remark 5.3.6. By above, CC••(A, B, ε)[2,0] will have all zeros in the zeroth and ﬁrst columns.

Observe that since CC••(A, B, ε)[2,0] is just shifted by two, we get that

∼ ∼ Hn(Tot CC••(A, B, ε)[2,0]) = Hn−2(Tot CC••(A, B, ε)) = HCn−2(A, B, ε).

Proof 2. Observe that we have the following short exact sequences of bicomplexes:

{0,1} 0 −→ CC••(A, B, ε) ,→ CC••(A, B, ε) CC••(A, B, ε)[2,0] −→ 0.

By Theorem 2.2.11 this induces a long exact sequence

{0,1} I∗ ... −→ Hn(Tot CC••(A, B, ε) ) −−→ Hn(Tot CC••(A, B, ε)) −→

{0,1} I∗ Hn(Tot CC••(A, B, ε)[2,0]) −→ Hn−1(Tot CC••(A, B, ε) ) −−→ ...

We then apply Corollary 5.3.5 and Remark 5.3.6 to get what we want.

5.3.3 Proof 3

This proof follows [28].

Let Tot DC••(A, B, ε)[m] denote a shift by m dimensions. 148 Proof 3. Observe that we have the following short exact sequence of complexes:

ι S 0 −→ C•(A, B, ε) −→ Tot DC••(A, B, ε) −→ Tot DC••(A, B, ε)[2] −→ 0.

The first map ι is the identification of C•(A, B, ε) with the first column of DC••(A, B, ε). The second map S is obtained by factoring out this first column. ∼ Note Hn(Tot DC••(A, B, ε)[2]) = Hn−2(Tot DC••(A, B, ε)). Thus, by Theorem 2.2.11, this induces the long exact sequence that we want. 149

CHAPTER 6 PROBLEMS FOR FUTURE RESEARCH

Here we list some problems that we encountered along the way. We look forward to investi- gating these matters, and others, in the future.

Problem 6.1. Can we view the simplicial module B(A, B, ε) as a projective object in some category? This would permit us to see the secondary Hochschild (co)homology as a Tor (respectively Ext) functor.

Problem 6.2. Is Hn((A, A, id); M) = 0 for all n > 0? We know this is valid for n = 1 and n = 2.

Problem 6.3. Can we compute HHn(A, A, id) for all n > 0? This seems unlikely, but would have far reaching consequences. It is hopeful this would behave like HHn(k). Similarly, can we compute HCn(A, A, id) for all n > 0? This depends on HHn(A, A, id), but one hopes this behaves as HCn(k). Finding one can lead to the other by use of the secondary long exact sequence.

Problem 6.4. Can we deﬁne a secondary de Rham cohomology? This could lead to interesting geometrical results. Assuming that (A, B, ε) satisﬁes some “regular” conditions, can we get an analogous result similar to the Hochschild-Kostant-Rosenberg Theorem?

Problem 6.5. Does HHn(A, B, ε) have a shufﬂe-product as well? This seems likely since a shufﬂe- product property exists for the usual Hochschild homology.

Problem 6.6. Can we deﬁne a secondary Hochschild homology for when we take B to be noncommutative. This could lead to a secondary K-theory and secondary Chern character. With this concept, would the secondary Hochschild and cyclic (co)homologies preserve Morita equivalence? In particular, are they Morita invariant under matrices? 150

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