Various topics in rack and quandle homology
April 14, 2010
Master’s thesis in Mathematics Jorik Mandemaker
supervisor: Dr. F.J.-B.J. Clauwens second reader: Prof. Dr. F. Keune student number: 0314145 Faculty of Science (FNWI) Radboud University Nijmegen Contents
1 Introduction 5 1.1 Racks and quandles ...... 6 1.2 Quandles and knots ...... 7 1.3 Augmented quandles and the adjoint group ...... 8 1.3.1 The adjoint group ...... 9 1.4 Connected components ...... 9 1.5 Rack and quandle homology ...... 10 1.6 Quandle coverings ...... 10 1.6.1 Extensions ...... 11 1.6.2 The fundamental group of a quandle ...... 12 1.7 Historical notes ...... 12
2 The second cohomology group of alexander quandles 14 2.1 The adjoint group of Alexander quandles ...... 15 2.1.1 Connected Alexander quandles ...... 15 2.1.2 Alexander quandles with connected components . . . . . 18 2.2 The second cohomology group of connected Alexander quandles . 19
3 Polynomial cocycles 22 3.1 Introduction ...... 23 3.2 Polynomial cochains ...... 23 3.3 A decomposition and filtration of the complex ...... 24 3.4 The 3-cocycles ...... 25 3.5 Variable Reduction ...... 27 3.6 Proof of surjectivity ...... 28 3.6.1 t = s ...... 29 3.6.2 t < s ...... 30 3.7 Final steps ...... 33
4 Rack and quandle modules 35 4.1 Rack and Quandle Modules ...... 36 4.1.1 Definitions ...... 36 4.1.2 Some examples ...... 38 4.2 Reduced rack modules ...... 39
1 4.3 Right modules and tensor products ...... 40 4.4 The rack algebra ...... 42 4.5 Examples ...... 44 4.5.1 A free resolution for trivial quandles ...... 44 4.5.2 R3 ...... 44
5 Trunks 46 5.1 Introduction ...... 47 5.2 Trunks ...... 47 5.3 -sets and their realizations ...... 48 5.4 The nerve of a trunk ...... 49
6 Homology operators 51 6.1 The augmented rack space ...... 52 6.1.1 H1(BInnX) ...... 54 6.2 Homology operators of negative degree ...... 55 0 6.2.1 H (BGX X)...... 58 6.3 Variations ...... 59 6.3.1 A different G ...... 59 6.3.2 Quandle homology ...... 59 6.4 Relation to the literature ...... 60 6.4.1 Some computer calculations ...... 60 6.5 Some other operators ...... 61
7 The cup product of the rack space 63 7.1 The cup product of the realization of a -set ...... 64 7.2 A triangulation of the realization of a -set ...... 65 7.3 The cup product ...... 67
2 Summary
The goal of this thesis project was originally to survey the current literature on racks and quandles in order to comment on it or extend it. Eventually this resulted in this thesis being a mix of various loosely related topics. Since rack and quandle theory is a rather obscure part of mathematics this thesis contains a general introduction to racks and quandles covering all the ba- sics. The first chapter contains this introduction and can generally be regarded as ”common” knowledge about quandles. The second and third chapter are both concerned with computing the co- homology of certain Alexander quandles. The former being an application of the work by Eisermann [7]. The calculation in 2.1.1 I owe to my supervisor F. Clauwens. But the rest of the chapter is my own work. The third chapter is a correction of the work by Mochizuki [17, 18]. In these articles Mochizuki claims to have found a basis for the third cohomology group of some Alexander quandles. However this list contains a coboundary, so it is clearly incorrect. The next chapter focuses on the rack and quandle modules from the work by Jackson [12, 13]. Jackson studied these modules as an alternate way of defining rack and quandle homology. For connected quandles we’ll show that those modules coincide with modules over a subring of the group ring of the adjoint group of the quandle. Chapter 5 is another introductionary chapter which serves mostly to set the terminology used in the final two chapters. It covers the basics of trunks and their classifying spaces as found in [9]. Chapter 6 explores the homology of the augmented rack space which proves a useful tool to study the homology of racks. It also shines new light on some of the work by Niebrzydowski and Przytycki [19]. The orginal idea that the augmented rack spaces is a topological monoid that produces homology operators is once again due to F. Clauwens. Section 6.1.1 and further is my own work with the exception of formula 6.6. The final chapter studies the yet unexplored the cup product on the coho- mology of -sets which in particular covers that of racks and quandles.
3 Acknowledgements
I would like to thank my thesis supervisor Frans Clauwens who was very patient with me over the course of this thesis project. He worked tirelessly on racks and quandles himself. He often pointed me in the right direction and explained many interesting things to me. In particular I have to thank him for the computation of chapter 2.1.1 and the insight that the cycles and cocycles of BGX produce homology operators (chapter 6). I would also like to thank Frans Keune for being the second reader of my thesis.
4 Chapter 1
Introduction
5 1.1 Racks and quandles
Definition 1.1.1. A quandle (Q, B, C) is a set Q with two binary operators B and C such that the following axioms hold. (Q1) a B a = a for all a in Q (Q2) (a C b) B b = a = (a B b) C b for all a, b in Q
(Q3) (a B b) B c = (a B c) B (b B c) for all a, b, c in Q A set R with two binary operators satisfying axioms (Q2) and (Q3) but not necessarily (Q1) is called a rack.
If (Q, B, C) is a quandle or rack than the operation C is uniquely determined by B. Because of this we will usually omit it from the notation and simply write (Q, B) or even just Q. Definition 1.1.2. A quandle homomorphism between two quandles X and Y is a function φ : X → Y such that φ(aBb) = φ(a)Bφ(b) and φ(aCb) = φ(a)Cφ(b) for all a, b ∈ X. The composition of two quandle homomorphisms is again a quandle homomorphism so we can speak of the category Qnd with objects quandles and morphisms the quandle homomorphism. In the same way we obtain the category of racks denoted by Rack.
Convention 1.1.3. We let both B and C associate to the left. So when we write a B b C c B d we mean ((a B b) C c) B d
Theorem 1.1.1. If (Q, B, C) is a quandle then (Q, C, B) is also a quandle called the dual quandle. Definition 1.1.4. A quandle or rack Q is said to be involutive if it satisfies
(QInv) a B b B b = a for all a, b in Q
A quandle or rack Q is said to be abelian if it satisfies
(Qab) (a B x) B (y B b) = (a B y) B (x B b) for all a, b, x, y in Q −1 Example 1.1.1. Every group G becomes a quandle by setting a B b = b ab −1 and a C b = bab . This quandle is called the conjugation quandle of G denoted by Conj(G). Every group homomorphism between two groups G and H is also a quandle homomorphism between Conj(G) and Conj(H). Thus we obtain a functor Conj : Grp → Qnd.
Example 1.1.2. If X is any set then X becomes a quandle by aBb = aCb = a called the trivial quandle on X.
6 Example 1.1.3. If M is an abelian group and T : M → M is an automorphism of M. Then M becomes a quandle by setting
x B y = T (x) + (1 − T )(y) −1 x C y = T (x − (1 − T )(y)) This quandle is called the Alexander quandle Alex(M,T ).
Example 1.1.4. The dihedral quandle Rk is an Alexander quandle where the group is Zk and the automorphism is multiplication by −1. An alternate descrip- tion is that Rk is the subquandle of Conj(Dk) consisting of all the reflections. The automorphism group Aut(Q) of a quandle Q consists of all bijective quandle homomorphisms φ from Q to itself. We let this group act on Q from the right by evaluation notated by aφ = φ(a). This means that for φ, ψ ∈ Aut(Q) their composition φψ is given by a(φψ) = (aφ)ψ. For every element a ∈ Q we have an automorphism ρa defined by ρa(b) = b B a.
Definition 1.1.5. The subgroup of Aut(Q) generated by the ρa is called the inner automorphism group of Q. This group is notated by Inn(Q). We also have the map inn : Q → Inn(Q) defined by inn(a) = ρa. Convention 1.1.6. As a subgroup of Aut(Q) Inn(Q) acts on Q. We will often b inn(b) h −1 write a for a = a B b. We will also write g = h gh for conjugation in a group G in light of the quandle Conj(G).
1.2 Quandles and knots
Quandles are closely related to knots. Every oriented knot has a quandle asso- ciated with it, called the fundamental quandle of the knot. It’s easy to read of a knot diagram.
Take one generator for each arc and for each crossing like the one pictured above add a relation of the form a B b = c or equivalently c C b = a. Note that the orientation of the arc crossing under is irrelevant. It is easy to see that this construction is independent on the knot diagram. In fact the three quandle axioms correspond directly with the three Reidemeister moves. For more details on this construction and a topological interpretation see [14].
7 1.3 Augmented quandles and the adjoint group
In the last section we considered a quandle Q and we had a map inn : Q → Inn(Q). Also Inn(Q) acted on Q. This map and action have the property that inn(b) φ φ a B b = a and inn(a ) = inn(a) . This situation can be generalized to give rise to the notion of an augmented quandle. Definition 1.3.1. A representation of a quandle (or rack) Q in a group G is a −1 map φ : Q → G such that φ(aBb) = φ(b) φ(a)φ(b) for all a, b in Q. Essentially φ is a quandle homomorphism between Q and Conj(G). Definition 1.3.2. Let φ : Q → G be a representation and let α be a right action of G on Q denoted by α(a, g) = ag. We call the pair (φ, α) an augmentation if φ(b) g g a B b = a and φ(a ) = φ(a) for all a, b ∈ Q and g ∈ G. This means that the following diagram commutes.
id×φ φ×id Q × Q / Q × G / G × G
B α conj id φ Q / Q / G
g g gφ(bg ) φ(b)g g Since for g ∈ G and a, b ∈ X we have a B b = a = a = (a B b) , we can view the action α as a group homomorphism α : G → Aut(Q) When we have a representation φ : Q → G where φ(Q) generates G and there is an α : G → Aut(Q) such that α ◦ φ = inn we just say that φ : Q → G is an augmentation.
8 1.3.1 The adjoint group Definition 1.3.3. For a rack or quandle Q we define Adj(Q) the adjoint group of Q as follows. Adj(X) = he : x ∈ Q | e e = e e i x x y y xBy So we have one generator for every element of Q and relations encompassing the quandle operation. We also obtain a map adj : Q → Adj(Q) by adj(x) = ex Note that we can rewrite the defining relation of the adjoint group as
e = e−1e e (1.1) xBy y x y So we see that adj : Q → Adj(Q) is a representation. It is in fact the universal representation of Q. For every representation φ : Q → G there exists a unique h : Adj(Q) → G such that φ = h ◦ adj. In particular there is a map ρ : Adj(Q) → Inn(Q) that sends ex to inn(x). We let Adj(Q) act on Q through this map. Combining this action with (1.1) we find that for all g ∈ Adj(Q) we have
g −1 (ex) = g exg = exg (1.2) Definition 1.3.4. Let Q be a quandle (or rack). Define K = ker(ρ : Adj(Q) → Inn(Q)). In light of (1.2) we see that K is the center of Adj(Q). Definition 1.3.5. We can turn Adj into a functor from Qnd → Grp. For 0 0 f : Q → Q define Adj(f) : Adj(Q) → Adj(Q ) by Adj(f)(ex) = ef(x). The name adjoint group comes from the fact that: Theorem 1.3.1. The functor Adj is the left adjoint to the functor Conj. In some papers (e.g. [13]) the adjoint group is called the associated group As(Q).
1.4 Connected components
Definition 1.4.1. Let Q be a quandle (or rack) and x ∈ Q. Denote the orbit of x under the action of Inn(Q) by [x]. There orbits are called the connected components of Q. We call Q connected if Inn(Q) acts transitively on Q. Also define π0(Q) = {[x] | x ∈ Q}.
We can view π0(Q) as a trivial quandle. If we do so, the map x 7→ [x] becomes a quandle homomorphism. In fact every quandle homomorphism from Q to some trivial quandle X factors through this map. In particular if Q is connected and 0 0 f : Q → Q is a quandle homomorphism then the map Q → π0(Q ), x 7→ [f(x)] factors through π0(Q). So when Q is connected its image is also connected. Since Adj(Q) acts on Q through inner automorphisms the Adj(Q)-orbits are the same as the connected components of Q. Using (1.2) we see that if x and y are in the same connected component then ex and ey are conjugate.
9 Definition 1.4.2. Let Q be a quandle (or rack). Define : Adj(Q) → Z by ◦ (ex) = 1. Denote ker by Adj(Q) . When Q is a quandle (not a rack!) the orbits under the action of Adj(Q)◦ are still the connected components. Since
e−(g)g g a a = a
1.5 Rack and quandle homology
Most of the research into quandles and racks has at least some link to the homology of racks and quandles. Most of this thesis is also concerned with this homology
R Definition 1.5.1. For a rack X let Cn (X) be the free abelian group generated n R R by X . Define a map δn : Cn (X) → Cn−1(X) as follows 0 δi (x1, . . . , xn) = (x1, . . . , xi−1, xi+1, . . . , xn) 1 δi (x1, . . . , xn) = (x1 B xi, . . . , xi−1 B xi, xi+1, . . . , xn) n X i 0 1 δn = (−1) [δi − δi ] i=1
R A simple computation verifies that δn−1 ◦ δn = 0 so that {C∗ (X), δ∗} forms a chain complex. This chain complex is called the rack complex of X. We usually omit the subscript on δ.
D R Definition 1.5.2. Let Q be a quandle. Define Cn (Q) ⊂ Cn (Q) as the subgroup generated by the (x1, . . . , xn) with xi = xi+1 for some 1 ≤ i < n. D D A simple computation shows that δ(Cn (Q)) ⊂ Cn−1(Q). This is not true in general when Q is just a rack and not a quandle. So we get a subcomplex D C∗ (Q) called the degeneracy complex. Q R D Definition 1.5.3. Let Q be a quandle. Define Cn (Q) = Cn (Q)/Cn (Q). δ Q descends to this quotient so we get a chain complex C∗ called the quandle complex. Also define
W W Hn (X) = Hn(C∗ (X)) n n W HW (X) = H (C∗ (X)) For W = R,D,Q. Of course for W = R this is also defined for racks.
1.6 Quandle coverings
The notion of quandle coverings was studied by Eisermann in [6],[7]. All of the definitions and theorems we mention here can be found in [7]. For details and proofs look there.
10 Definition 1.6.1. Let p : Q˜ → Q be a surjective quandle homomorphism. If ˜ for all a, x, y ∈ Q p(x) = p(y) implies a B x = a B y then we call p a covering. Definition 1.6.2. When p : Q˜ → Q is a covering then Q acts on Q˜. If x ∈ Q and a ∈ Q˜ we can define ax = ax˜ wherex ˜ ∈ Q˜ is such that p(˜x) = x. This does not depend on the choice ofx ˜. This action extends in the obvious way to Adj(Q).
Definition 1.6.3. Let p1 : Q1 → Q and p2 : Q2 → Q be two coverings. A covering morphism from p1 to p2 is a quandle homomorphism φ : Q1 → Q2 such that p1 = p2 ◦ φ. That is: the following diagram is commutative.
φ Q1 / Q2 @ @@ ~~ @@ ~~ p1 @@ ~~ p2 @ ~ ~ Q
Definition 1.6.4. All coverings over a given quandle Q together with the cov- ering morphisms between them form a category Cov(Q). Definition 1.6.5. The group of all covering automorphisms of a covering p : Q˜ → Q is denoted by Aut(p) and called the deck transformation group of p. We let Aut(p) act on Q˜ from the left. This action commutes with the action of Adj(Q˜). Definition 1.6.6. We call a covering p : Q˜ → Q galois if Q˜ is connected and Aut(p) acts freely and transitively on each fibre.
1.6.1 Extensions The case of galois coverings can be generalized to the notion of quandle exten- sions.
Definition 1.6.7. An extension E :Λ y Q˜ → Q of a quandle Q by a group Λ consists of a group action Λ y Q˜ and a surjective quandle homomorphism p : Q˜ → Q such that the following two axioms hold. ˜ 1. (λx) B y = λ(x B y) and x B (λy) = x B y for all λ ∈ Λ and x, y ∈ Q. 2. Λ acts freely and transitively on each fibre p−1(x). The homomorphism p in every extension is a quandle covering and when Q˜ is connected it is galois. Also every galois covering becomes an extension by the action of Aut(p) on Q˜. Definition 1.6.8. Let Λ be a group and Q be a quandle. An equivalence be- p1 p2 tween two extensions E1 :Λ y Q1 /Q and E2 :Λ y Q2 /Q is a cover- ing isomorphism φ : p1 → p2 such that λφ = φλ for all λ ∈ Λ. The set of all equivalence classes of extensions is denoted by Ext(Q, Λ).
11 1.6.2 The fundamental group of a quandle Definition 1.6.9. A pointed quandle (Q, q) is a quandle Q along with an ele- ment q ∈ Q chosen as a base point. A pointed quandle homomorphism is just a quandle homomorphism that preserves the base point. We also have pointed quandle coverings defined in the same way as regular coverings. Definition 1.6.10. A pointed quandle covering p :(Q,˜ q˜) → (Q, q) is called universal if for every pointed quandle covering p0 :(Q0, q0) → (Q, q) there exists a unique homomorphism φ :(Q,˜ q˜) → (Q0, q0) with p = p0 ◦ φ. Of course every two universal coverings are isomorphic. Definition 1.6.11. The universal covering (Q,˜ q˜) of a connected quandle (Q, q) is given by Q˜ = {(a, g) ∈ Q × Adj(Q)◦ | qg = a} −1 Withq ˜ = (q, 1) and (a, g) B (b, h) = (a B b, g · adj(a) · adj(b)). The covering p :(Q,˜ q˜) → (Q, q) is given by p(a, g) = a. For more details on this definition as well as proof that it is indeed universal see [7].
Definition 1.6.12. The fundamental group π1(Q, q) of a connected quandle (Q, q) is defined to be
◦ g π1(Q, q) = {g ∈ Adj(Q) | q = q}
Theorem 1.6.1. For every connected quandle (Q, q) and every group Λ we have
2 ∼ ∼ HQ(Q, Λ) = Ext(Q, Λ) = Hom(π1(Q, q), Λ) and also Q ∼ H2 (Q) = π1(Q, q)ab
1.7 Historical notes
The first time racks were studied in general was in 1959 in unpublished corre- spondence between J.C. Conway and G.C. Wraith [5]. They used the term wrack for the concept. The name comes from the English phrase ”wrack and ruin” in relation to the ’wrecking’ of a group by throwing away the multiplication and only retaining conjugation. They studied racks and quandles as algebraic objects but also knew of the fundamental quandle of a knot. In a less general way racks and quandles have been studied earlier. In 1942 M. Takasaki studied involutive quandles which he called keis [20]. In particular he studied dihedral quandles arising from symmetries of regular polygons. Even earlier in 1929 Burnstin and Mayer studied ”distributive groups” which are semigroups with a self left and right distributive group operation [2]. Quandles of course generalize these objects.
12 The first published work on racks is from 1982 by Joyce [14]. In this paper the term quandle was coined. This paper introduced many of the basic notions such as the connected components of quandles as well as augmented quandles and the adjoint group. Joyce also studied racks in a topological way. His main result is that the fundamental quandle of a knot is a classifying invariant of knots. In 1995 Fenn, Rourke and Sanderson introduced trunks in [9] leading to the homology theory for racks. The refinement to quandle homology comes from Carter et al. [3, 4]. The majority of the current articles related to racks and quandles deal with this homology theory. Either computing parts of it [7, 8, 15, 17, 19] or using it to construct knot and link invariants [1, 4, 10].
13 Chapter 2
The second cohomology group of alexander quandles
14 2 ∼ Since HQ(Q, Λ) = Hom(π1(Q, q), Λ) computing the adjoint group of a quan- dle will help computing the second cohomology group. In this chapter we’ll give a description of Adj(Q) for alexander quandles Q and then apply the theory from [7] to compute the second cohomology group.
2.1 The adjoint group of Alexander quandles 2.1.1 Connected Alexander quandles Let M be an abelian group and T : M → M an automorphism. We give M the structure of a quandle by setting x B y = T x + (1 − T )y for x, y ∈ M. Define C0 as the connected component of M containing 0. Since 0 B x = (1 − T )x it follows that (1 − T )M ⊂ C0. Since T ((1 − T )x) + (1 − T )y = (1 − T )(T x + y) it follows that (1 − T )M is closed under Inn(M), so C0 = (1 − T )M. Thus M is connected iff (1 − T ) is surjective. We will assume (1 − T ) to be invertible, which is automatic when M is finite. Recall that Adj(M) is the group with generators ex for x ∈ M and relations e = e−1e e . Note that for all p ∈ M xBy y x y
e−1e e−1e e−1e p 0 x+y = p + x − T x + y − T y = p 0 x 0 y (2.1)
So −1 −1 −1 e0 ex+y = γ(x, y)e0 exe0 ey (2.2) for some γ(x, y) ∈ K = ker(ρ : Adj(M) → Inn(M)). Recall that K is the center of Adj(M). We find that γ(x, 0) = γ(0, x) = 1 directly from the definition. We −1 can expand e0 ex+y+z in two ways using formula (2.2) to find γ(x, y + z)γ(y, z) = γ(x + y, z)γ(x, y) (2.3)
−1 −1 We can rewrite (2.2) as ex+y = γ(x, y)exe0 ey = γ(x, y)e0 eT −1xey, which can be written as −1 eT −1xey = γ(x, y) e0ex+y (2.4) It follows that
−1 γ(x, y) e0ex+y = eT −1xey = eyex+(1−T )y −1 = γ(T y, x + (1 − T )y) e0eT y+x+(1−T )y −1 = γ(T y, x + (1 − T )y) e0ex+y (2.5)
Here the first and third equality follow from (2.4) and the second is the defining relation of Adj(M). So we get
γ(x, y) = γ(T y, x + (1 − T y)) (2.6)
Filling in x = 0 we find γ(T y, (1 − T )y) = 1 (2.7)
15 We can apply (2.6) to compute that
γ(x, y)γ((1 − T )y, x) = γ(T y, x + (1 − T )y)γ((1 − T )y, x) = γ(T y + (1 − T )y, x)γ(T y, (1 − T )y) = γ(y, x) (2.8)
The second equation follows from (2.3). One would hope that γ is symmetric. Unfortunately this isn’t the case. But by looking at how much γ falls short of being symmetric we can find out more. Therefore we define
λ(u, v) = γ(u, v)−1γ(v, u) = γ((1 − T )v, u) (2.9)
Using (2.3) twice we find
γ(u, v + w)γ(v, w) = γ(u + v, w)γ(u, v) γ(w + v, u)γ(w, v) = γ(w, v + u)γ(v, u)
Dividing the second relation by the first we get
λ(u, v + w)λ(v, w) = λ(u + v, w)λ(u, v) (2.10)
Using (2.9) we get λ((1 − T )−1x, y)−1 = γ(x, y). Combining this with (2.3) we find
λ((1 − T )−1(x + y), w)λ((1 − T )−1x, y) = λ((1 − T )−1x, y + w)λ((1 − T )−1y, w)
For x = (1 − T )u and y = (1 − T )v this reads
λ(u + v, w)λ(u, (1 − T )v) = λ(u, (1 − T )v + w)λ(v, w) (2.11)
Dividing this by (2.10) we get
λ(u, (1 − T )v)λ(u, v)−1 = λ(u, (1 − T )v + w)λ(u, v + w)−1 (2.12)
Note that the left hand side of this formula does not contain w. Specifying to w = −v we find λ(u, (1 − T )v)λ(u, v)−1 = λ(u, −T v) (2.13) Plugging this back into (2.12) we get
λ(u, −T v) = λ(u, (1 − T )v + w)λ(u, v + w)−1 (2.14)
For a = v + w and b = −T v it reads
λ(u, b)λ(u, a) = λ(u, a + b) (2.15)
So λ is linear in the second coordinate. Using that λ(x, y) = λ(y, x)−1 it follows that λ is in fact bilinear. Since λ(x, y) = γ((1 − T )x, y) and (1 − T ) is invertible it follows that γ is also bilinear.
16 Now we know that γ is bilinear we can restate (2.6) in the following way
γ(x, y) = γ(T y, x + (1 − T )y) = γ(T y, x)γ(T y, (1 − T )y) = γ(T y, x) (2.16)
Using this we can view γ as a group homomorphism from ST (M) to Adj(M) where ST (M) is defined as
ST (M) = M ⊗ M/ < x ⊗ y − T y ⊗ x > (2.17)
So in ST (M) we have the relation [x ⊗ y] = [T y ⊗ x]. Now we know enough to give a nice description of Adj(M). Define GT (M) by GT (M) = Z × M × ST (M) (2.18)
We make GT (M) into a group by defining
(k, x, µ) (l, y, ν) = (k + l, T l(x) + y, µ + ν + [T l(x) ⊗ y]) (2.19)
It’s easy to verify that this indeed defines a group structure on GT (M). In order to show associativity one has to use the defining relation of ST (M). Note that (k, x, µ)−1 = (−k, −T −kx, −µ + [x ⊗ x]). ∼ Theorem 2.1.1. Adj(M) = GT (M).
Proof. We start by defining a homomorphism φ : Adj(M) → GT (M) by setting −1 −1 φ(ex) = (1, x, 0) and φ(ex ) = (−1, −T x, [x ⊗ x]). In order to see that this is well defined we have to verify the following
−1 −1 φ(ex ) φ(ey) φ(ex) = (−1, −T x, [x ⊗ x]) (1, y, 0) (1, x, 0) = (−1, −T −1x, [x ⊗ x]) (2, T y + x, [T y ⊗ x]) = (1, −T x + T y + x, [x ⊗ x] + [T y ⊗ x] −[T x ⊗ T y + x]) = (1, y x, 0) = φ(e ) B yBx
We’ll now define another homomorphism ψ : GT (M) → Adj(M) by setting k−1 −1 ψ(k, x, µ) = e0 exγ(µ) . We first need to check that this indeed defines a group homomorphism.
k−1 −1 l−1 −1 ψ(k, x, µ)ψ(l, y, ν) = e0 exγ(µ) e0 eyγ(ν) k+l −1 −1 −1 = e0 e0 eT lxe0 eyγ(µ + ν) k+l −1 l −1 −1 = e0 e0 eT lx+yγ(T x, y) γ(µ + ν) = ψ(k + l, T lx + y, µ + ν + [T lx ⊗ y])
Now we’ll see that φ and ψ are each others inverses
ψ(φ(ex)) = ψ(1, x, 0) = ex (2.20)
17 The other way around takes more work. First we have
k−1 −1 −1 φ(ψ(k, x, µ)) = φ(e0 ex)φ(γ(µ) ) = (k, x, 0) φ(γ(µ) ) (2.21) It’s now enough to show that φ(γ([x ⊗ y])−1) = (0, 0, [x ⊗ y]) to complete the proof.
−1 −1 −1 φ(γ([x ⊗ y]) ) = φ(ex+yexe0 ey) = (−1, −T −1(x + y), [(x + y) ⊗ (x + y)]) (1, x, 0) (−1, 0, 0) (1, y, 0) = (−1, −T −1(x + y), [(x + y) ⊗ (x + y)]) (1, x + y, [x ⊗ y]) = (0, 0, [x ⊗ y])
2.1.2 Alexander quandles with connected components We’ll now tackle another class of Alexander quandles. This time assume that M 6= C0 but that the subquandle C0 = (1 − T )M is connected and therefore (1 − T ): C0 → C0 is invertible. An example of such a quandle would be R6, the Alexander quandle where M = Z6 and T is multiplication by −1. C0 is a connected component of M, but what do the other components of M look like? First note that the map f : x 7→ x + a for some fixed a ∈ M is a quandle homomorphism. We see that f maps C0 to the connected component of a. Since f is a bijection it follows that the connected component of a is equal to a + C0. Since (1 − T ): C0 → C0 is a bijection it also follows that every connected component has to contain exactly one element of the kernel of (1 − T ). So every connected component is of the form a + C0 with a ∈ ker(1 − T ). In the rest of this section let a, b, c etc. be variables over ker(1 − T ) and let x, y, z etc. be variables over C0. We hope that we can find Adj(C0) as a subgroup of Adj(M) since we already have described Adj(C0) in the previous section. In fact we hope to find one copy of Adj(C0) for each connected component of M since they are all isomorphic. −1 −1 Note that e0 ea ∈ K since (1 − T )a = 0. Therefore e0 ea is central in Adj(M). So we get −1 −1 −1 −1 e0 eaex = e0 eag e0g = g eag = ea+x (2.22) −1 where g = e0 eu with u such that (1 − T )u = x. The first and last equalities follow from 1.2. This is already enough to tie everything together. Let A = {a|a ∈ ker(1 − T ) \{0}} and set
GT (M) := Z × C0 × ST C0 × Z[A] = GT (C0) × Z[A] (2.23)
The group structure is simply the direct product of GT (C0) and Z[A]. We’ll write 0 = 0 in Z[A] to simplify notation.
18 ∼ Theorem 2.1.2. Adj(M) = GT (M) ∼ Proof. Since GT (M) = Adj(C0) × Z[A] we can define the following map φ : Adj(M) → GT (M) by setting φ(ea+x) = (ex, a). One easily checks that this is a well-defined. The inverse ψ : GT (M) → Adj(M) is given by
X Y −1 ka ψ(g, kaa) = g · (e0 ea) (2.24) a∈A a∈A
−1 Because e0 ea is central it’s easy to see that this is also a well defined homo- morphism. We have
−1 ψφ(ea+x) = ψ(ex, a) = exe0 ea = ea+x
X Y −1 ka Y X φψ(g, kaa) = φ(g) φ(e0 ea) = (g, 0) (0, kaa) = (g, kaa) a∈A a∈A a∈A a∈A
We now find a subgroup of Adj(M) isomorphic to Adj(C0) for every con- nected component of M. These subgroups are given by
Ga = {(k, x, µ, ka) ∈ GT (C0) × Z[A]}
2.2 The second cohomology group of connected Alexander quandles
In this section let M be a connected alexander quandle. Before we get to the explicit constructions of cocycles we’ll first alter the isomorphism between Adj(M) and GT (M) a bit. 0 Definition 2.2.1. Define φ : Adj(M) → GT (M) by φ(ex) = (1, (1 − T )x, 0). To see that φ0 is an isomorphism note that φ0 = φ ◦ Adj(1 − T ) where φ is the isomorphism from the previous section. Since (1 − T ) is an isomorphism so is Adj(1 − T ). Lemma 2.2.1. For g ∈ Adj(M) we have
xg = T (g)x + 0g (2.25)
Proof. We obviously have x1 = x = T 0(x) + 0. Now suppose for some g ∈ Adj(M) relation (2.25) holds. It suffices to prove that it also holds for gey and −1 gey . A simple computation shows:
ge g (g) g x y = x B y = (T (x) + 0 ) B y = T (g)+1(x) + T (0g) + (1 − T )(y) = T (gey )(x) + 0gey
−1 A similar computation shows the same for gey .
19 Thanks to this lemma we see that for g ∈ Adj(M) we have φ0(g) = ((g), 0g, ν) for some ν ∈ ST (M). In light of this we’ll take 0 as a base point for our quandle and find g ∼ π1(M, 0) = {g ∈ Adj(M)|(g) = 0, 0 = 0} = ST (M) The universal covering of M, M˜ , is given by
M˜ = {(a, g) ∈ M × Adj(M)◦|0g = a}
As noted in [7] the first coordinate is redundant since all information is contained in Adj(M)◦. The quandle operation from 1.6.11 is then transformed into
−1 −1 g B h = e0 gh e0h
◦ We know that Adj(M) = {(k, x, µ) ∈ GT (M)|k = 0}. So this formula becomes
(0, x, µ) B (0, y, ν) = (−1, 0, 0) (0, x, µ) (0, −y, −ν + [y ⊗ y]) (1, 0, 0) (0, y, ν) = (−1, 0, 0) (0, x − y, µ − ν + [y ⊗ y] − [x ⊗ y]) (1, y, ν) = (−1, 0, 0) (1, T x − T y + y, µ − ν + [y ⊗ y] − [x ⊗ y] +ν + [T x ⊗ y] − [T y ⊗ y]) = (0, x B y, µ + [y ⊗ x] − [x ⊗ y]) = (0, x B y, µ + [(1 − T )y ⊗ x]) (2.26) The covering p : M˜ = Adj(M)◦ → M is simply given by projection on the second coordinate. The action of Adj(M) on M˜ from 1.6.2 is given by
(0, x, µ)g = (−ε(g), 0, 0) (0, x, µ) (ε(g), 0g, ν) = (0,T ε(g)(x) + 0g, µ + ν + [T ε(g)(x) ⊗ 0g]) where g ∈ Adj(M) and φ0(g) = ((g), 0g, ν). We’ll drop the first coordinate of the elements of M˜ = Adj(M)◦ from now on since it’s 0 anyway. Note that when 0 g ∈ π1(M, 0) we can identify g with ν ∈ ST (M) where φ (g) = (0, 0, ν). The action for such g then becomes the simple formula.
(x, µ)g = (x, µ + ν)
Now we have the necessary preparations out of the way. Let Λ be a group ∼ and ξ : π1(M, 0) = ST (M) → Λ. This ξ defines an right action of ST (M) on Λ given by λg = λξ(g). Like in lemma 6.5 of [7] we now construct
g M˜ α := Λ × M/˜ h(λ , x, µ) ∼ (λ, x, µ + ν)i (2.27)
0 where again g ∈ π1(M, 0) and φ (g) = (0, 0, ν). The quandle operator on M˜ α is given by 0 (λ, x, µ) B (λ , y, ν) = (λ, x B y, µ + [(1 − T )y ⊗ x])
20 M˜ α covers M by projection on the second coordinate. Λ acts on M˜ α by 0 0 ˜ λ(λ , x, µ) = (λλ , x, µ). It’s easy to verify that this turns Λ y Mα → M into an extension of M by Λ. This completes the first half of the isomorphism from Hom(π1(M, 0), Λ) to 2 HQ(M). Now we create a 2-cocycle f like in lemma 9.6 of [7]. Define a section ˜ s : M → Mα by s(x) := (1Λ, x, 0). f is defined by s(a) B s(b) = f(a, b)s(a B b). We find
s(a) B s(b) = (1, a, 0) (1, b, 0) = (1, a B b, [(1 − T )b ⊗ a]) = (ξ([(1 − T )b ⊗ a], a B b, 0) = ξ([(1 − T )b ⊗ a])s(a B b)
The third equality follows from (2.27), the defining relation of M˜ α. So f(a, b) = ξ([(1 − T )b ⊗ a]). Since 1 − T is a bijection and ξ : M ⊗ M/ ∼→ Λ it follows that f is bilinear. Also note
f(a, b) = ξ([(1 − T )b ⊗ a]) = ξ([T a ⊗ (1 − T )b]) = ξ([T a ⊗ b − T a ⊗ T b] = ξ([(T − 1)a ⊗ b]) = −f(b, a)
Finally since [a ⊗ b] = [T a ⊗ T b] we also have f(a, b) = f(T a, T b). One can verify that any map M 2 → Λ with these properties is in fact a 2-cocycle.
21 Chapter 3
Polynomial cocycles
22 3.1 Introduction
In [17, 18] Mochizuki computes the third cohomology group of a collection of Alexander quandles. His proof contains some errors that result in his list of 3-cocycles being incorrect. However his basic methodology is useful. In this chapter we correct the proof given in [18].
3.2 Polynomial cochains
In this chapter we will consider Alexander quandles where M is a finite field and the automorphism is given by multiplication. Let q be a power of a prime p and let 0, 1 6= ω ∈ Fq, consider the Alexander quandle X = Fq where the quandle operation is given by a B b = ωa + (1 − ω)b. We are going to use the field structure of X to rewrite the cochain complex with polynomials.
Definition 3.2.1. Let k be any field extension of Fq. Define n C (k, q) = {f ∈ k[U1,...,Un]|degUi (f) < q} The obvious map from Cn(k, q) to the collection of all maps Xn → k is a bijection. Definition 3.2.2. We’ll turn C∗(k, q) into a chain complex by defining δ(f) ∈ Cn+1(q) for f ∈ Cn(q) by
δ(f)(U1,...,Un+1) = n X i−1 (−1) f(ωU1, . . . , ωUi−1, ωUi + Ui+1,Ui+2,...Un+1) i=1 n X i−1 − (−1) f(U1,...,Ui−1,Ui + Ui+1,Ui+2,...,Un+1) i=1 A simple computation shows that δ2 = 0 so this does in fact define a chain ∗ complex. This chain complex is equivalent to CR(X, k). Define the chain- ∗ ∗ homomorphism φ : C (k, q) → CR(X, k) by
φ(f)(x1, . . . , xn) = f(x1 − x2, . . . , xn−1 − xn, xn) It’s easy to verify that φδ = δφ, so φ is indeed a chain-homomorphism. It’s also simple to see that φ is a bijection.
Lemma 3.2.1. Let k1, k2 be two fields such that Fq ⊂ k1 ⊂ k2. We have n ∗ n ∗ dim(H (C (k1, q)) = dim(H (C (k2, q))).
Proof. Find a set I and ai ∈ k2 for i ∈ I such that the ai form a basis of k2 n n as a k1-vectorspace. Let f ∈ C (k2, q). Then we can find gi ∈ C (k1, q) such P P n that f = aigi. We find that δf = aiδgi. So if f is a cocycle in C (k2, q) n then it is a linear combination of cocycles in C (k1, q), and analogously for coboundaries.
23 Convention 3.2.3. Because of the last lemma we can restrict ourselves to n n k = Fq and we’ll simply write C for C (Fq, q).
3.3 A decomposition and filtration of the com- plex
If f is a polynomial of homogenous total degree d then so is δ(f). So we can decompose the complex by total degree.
Definition 3.3.1. For d ∈ N define
n n Cd = {f ∈ C |f is homogenously of total degree d}
n L n It’s clear that C = Cd . d
Definition 3.3.2. We’ll define an order on the monomials of k[U1,U2,...] by
X ei X ti Ui ≤ Ui ⇔ (e1, e2,...) ≤lex (t1, t2,...) (3.1) i i
Where ≤lex is the lexicographical order. Lemma 3.3.1. If m is a monomial then all the monomials occurring in δ(m) are smaller than or equal to m. The coefficient of m in δ(m) is equal to ωd − 1.
Proof. This follows from a trivial computation.
d ∗ Theorem 3.3.2. If ω 6= 1 then the complex Cd is acyclic. n Proof. Let f ∈ Cd with δf = 0. We’ll prove that f is a coboundary with e1 en induction to the largest monomial occurring in f. Let m = U1 ...Un be the largest monomial occurring in f and let A be its coefficient. Finally let k = max{i|ei 6= 0}. Suppose n − k is even. The coefficient of m in δf is equal to A(ωd − 1) 6= 0. n−1 d −1 So n − k is odd. Therefore en = 0 and m ∈ Cd . Now f − A · (ω − 1) δm only contains monomials strictly smaller than m. Proceeding with induction concludes the proof.
Convention 3.3.3. We’ll write Tn for Un when dealing with polynomials in n Cd to emphasize what the dimension of a polynomial is.
Definition 3.3.4. Let s ∈ N and define ( ) n(s) X aps n Cd = fa(U1,...,Un−1) · Tn ∈ Cd a n(∞) n \ n(s) Cd = {f0(U1,...,Un−1) ∈ Cd } = Cd s
24 n(s+1) n(s) s n(s) n(∞) It’s clear that Cd ⊂ Cd and when p > q then Cd = Cd because the degree of Tn has to be smaller than or equal to q. It’s also easy to check n(s) n+1(s) ∗ that δ(Cd ) ⊂ Cd . So we have a filtration of the complex Cd . Definition 3.3.5. Set
n n Zd = ker(δ) ∩ Cd n(s) n(s) Zd = ker(δ) ∩ Cd n n−1 Bd = δ(Cd ) n(s) n−1(s) Bd = δ(Cd )
(s) n(s) n(s) Definition 3.3.6. There is a chain homomorphism D∗ : Cd → Cd−ps given by ! (s) X aps X (a−1)ps Dn fa(U1,...,Un−1)Tn = a · fa(U1,...,Un−1)Tn a a
(s) (s) It can be easily checked that δ ◦ Dn = Dn+1 ◦ δ.
3.4 The 3-cocycles
In [18] Mochizuki gives a list of cocycles which he claims form a basis of H3(C∗). However it turns out that some items on this list are if fact not cocycles at all and others are coboundaries. This list is corrected below. First define the following polynomial in Fp[x, y]
p−1 X 1 χ(x, y) = (−1)i−1i−1xp−iyi ≡ ((x + y)p − xp − yp) (mod p) p i=1
Definition 3.4.1. Now for a, b ∈ N define
a b E0(a · p, b) = χ(ωU1,U2) − χ(U1,U2) T3 a −1 b E1(a, b · p) = U1 χ(U2,T3) − χ(U2, ω T3)
ps+ph s h Lemma 3.4.1. If ω = 1 and s > 0 then E0(p , p ) is a cocycle. pt+ps t s If ω = 1 and s > 0 then E1(p , p ) is a cocycle See lemmas 2.5 and 2.6 of [18] for the proofs.
a b c 3 Definition 3.4.2. For a, b, c < q define F (a, b, c) = U1 U2 T3 ∈ C .
q1+q2+q3 Lemma 3.4.2. Let q1, q2 and q3 be powers of p. If ω = 1 then F (q1, q2, q3) q1+q2 is a cocycle. If ω = 1 then F (q1, q2, 0) is also a cocycle. Proof. This follows from a simple direct computation.
25 Definition 3.4.3. Let Q be the set of all tuples (q1, q2, q3, q4) with each qi a q1+q3 q2+q4 power of p, such that qi < q, q1 < q3, q2 < q4, q2 ≤ q3 and ω = ω = 1 and one of the following conditions hold. If p = 2 we demand q2 < q3 on top of this.
1. ωq1+q2 = 1
q1+q2 2. ω 6= 1 and q3 > q4
q1+q2 3. ω 6= 1, q3 = q4 and p 6= 2
q1+q2 q1 q2 4. ω 6= 1, q2 ≤ q1 < q3 < q4 and p 6= 2 and ω = ω
q1+q2 q1 q2 5. ω 6= 1, q2 < q1 < q3 ≤ q4 and p = 2 and ω = ω
Note 3.4.4. Mochizuki has q3 < q4 in case 5. But there are in fact cocycles for q3 = q4 when p = 2 that are independent of the others. If p = 2 and q2 = q3 case 1 would apply but then Γ(q1, q2, q3, q4) = F (q1, 2 · q2, q4) which is redundant.
Definition 3.4.5. For each (q1, q2, q3, q4) ∈ Q we define a cocycle Γ by: Case 1 Γ(q1, q2, q3, q4) = F (q1, q2 + q3, q4) Case 2
Γ(q1, q2, q3, q4) = F (q1, q2 + q3, q4) − F (q2, q1 + q4, q3) q2 −1 q1+q2 −(ω − 1) (1 − ω ) · F (q1, q2, q3 + q4) − F (q1 + q2, q4, q3)
Case 3
−1 q1 Γ(q1, q2, q3, q4) = F (q1, q2 + q3, q4) − 2 (1 + ω )F (q1, q2, q3 + q4)
Case 4 and 5
Γ(q1, q2, q3, q4) = F (q1, q2 + q3, q4) + F (q2, q1 + q3, q4) q1 −1 q1+q2 −(ω − 1) (1 − ω )F (q1 + q2, q3, q4)
t s 3(s) Note 3.4.6. All the inequalities are chosen so that Γ(q1, p , q3, p ) ∈ Cd and ps 2(t) the coefficient of T3 is in Cd−ps .
Note 3.4.7. In [18] Mochizuki has incorrectly used (1−ω−q3 ) instead of (1+ωq1 ) in case 3. Note 3.4.8. Mochizuki’s list in [18] also contains a cocycle called Ψ. However this cocycle is in fact a coboundary and therefore clearly not an element of a basis for H3(C∗).
Lemma 3.4.3. For (q1, q2, q3, q4) ∈ Q the polynomial Γ(q1, q2, q3, q4) is a cocy- cle.
26 Proof. All the terms occurring in Γ(q1, q2, q3, q4) have very simple coboundaries. For example
q1+q2 q1 q2 q3 q4 δ((F (q1, q2 + q3, q4)) = −(ω − 1)U1 U2 U3 T4 q1+q3 q1 q3 q2 q4 −(ω − 1)U1 U2 U3 T4 q1+q2+q3+q4 q1 q2+q3 q4 +(ω − 1)U1 U2 U3 Using this relation and similar ones for the other terms it’s easy to verify that Γ(q1, q2, q3, q4) is a cocycle. Lemma 3.4.4. The constant function 1 is a rack cocycle.
Definition 3.4.9. In the following let qi denote a power of p. Define
q1+q2+q3 I = {F (q1, q2, q3)|ω = 1, q1 < q2 < q3 < q} q1+q2 ∪{F (q1, q2, 0)|ω = 1, q1 < q2 < q} p·q1+q2 ∪{E0(p · q1, q2)|ω = 1, q1 < q2 < q} q1+p·q2 ∪{E1(q1, p · q2)|ω = 1, q1 ≤ q2 < q}
∪{Γ(q1, q2, q3, q4)|(q1, q2, q3, q4) ∈ Q} ∪{1} and let H be the subgroup of C3 generated by I. Theorem 3.4.5. The natural map from H to H3(C∗) is a bijection. Proof. The proof will follow in the next sections.
3.5 Variable Reduction
In this section let d denote a non-negative integer with ωd = 1. Let s be an s s 3(s) P ap integer with p < q. Let f ∈ Zd be a 3 cocycle. Say f = a faps (U1,U2)T3 . ps By looking at the coefficient of T4 in δ(f) we find
−ps X (a−1)ps 0 = δ(fps )(U1,U2,U3) + (ω − 1) faps (U1,U2) · a · U3 a
−ps (s) Or shorter δ(fps ) = (ω −1)D3 (f). This implies that fps already determines 3(s+1) f modulo Cd . Definition 3.5.1. Define
(s) 2 (s) 3(s) Rd = {g ∈ Cd−ps |δ(g) ∈ D3 (Cd )}
3(s) (s) In light of the previous equation we get a function φ : Zd → Rd defined by φ(f) = fps
27 2(s) P aps If g ∈ Cd such that g = gaps (U1)T2 then
−ps (s) φ(δ(g)) = δ(gps ) + (1 − ω )D2 (g) This gives us a well defined homomorphism
Z3(s) R(s) φ : d → d 3(s) 3(s+1) (s) 2(s) 2 Bd + Zd D2 (Cd ) + Bd−ps
2(s) 2 Note 3.5.2. Mochizuki has Bd−ps instead of Bd−ps but these are too few coboundaries. It could explain why he included Ψ in his list of cocycles. Theorem 3.5.1. The homomorphism φ is an isomorphism. Proof. We’ll prove injectivity here, surjectivity will be tackled in the next sec- 3(s) (s) tion. Let f ∈ Zd and assume φ(f) = 0. So we have fps = δ(h) + D (g) for 1 2(s) 0 −ps −1 some h ∈ Cd−ps and g ∈ Cd . Consider f = f − (1 − ω ) δ(g). Obviously 0 0 (s) 0 0 3(s+1) φ(f) = φ(f ). But φ(f ) = δ(h) + δ(gps ) so D (f ) = 0, so f ∈ Zd .
3.6 Proof of surjectivity
We start with a useful lemma. d 2 2(s) t Lemma 3.6.1. If δ(T1 ) ∈ Cd is contained in Cd then d = p for some t < s s d (s) 2(s) s or d = b·p . If δ(T1 ) is even contained in Im(D ) ⊂ Cd then when d = b·p and b ≡ −1 (mod p) we have b = p − 1.
d 2(s) h Proof. Assume δ(T1 ) ∈ Cd . Let d = b · p with h maximal. Assume that h ph (b−1)p d ph h < s. The coefficient of Uq T2 in δ(T1 ) is (ω − 1)b 6= 0 so either b = 1 or b − 1 has at least s − h factors p. We’ll see that the second case is impossible. Suppose b = c · pu + 1 with u maximal. Now consider the following monomials d with their coefficients in δ(T1 ).
cpu+h ph cpu+h U1 T2 has coef. (ω − 1)b u u+h u+h h u+h cp + 1 U (c−1)p T p +p has coef. (ω(c−1)p − 1) 1 2 pu + 1
u+h Because ωp 6= 1 it’s not possible that both these coefficients are zero. This d 2(s) h contradicts the fact that δ(T1 ) ∈ Cd . So b = 1 and d = p . d (s) Now assume that δ(T1 ) ∈ Im(D ), h = s and b = ap − 1. We need to show that a = 1. Assume a > 1 and let u be the largest integer such that pu < a. ps pu+1 ap−1−pu+1 The coefficient of U1 T2 is equal to a · p − 1 u+1+s (ωp − 1) pu+1 which is nonzero. This gives another contradiction so a = 1 and b = p − 1.
28 Definition 3.6.1. For t ≤ s define
(t) (s) 2(t) K = Rd ∩ Cd−ps We’ll prove surjectivity by downwards induction on t. We’ll eliminate the terms of a g ∈ K(t) by repeatedly applying the same technique; we compute a coefficient in δg that has to be zero, which allows us to apply lemma 3.6.1. But when we know the degree of g it’s easy to eliminate it’s terms.
3.6.1 t = s (s) (s) Let g ∈ K since we are working modulo Im(D2 ) we can assume that
X (ap−1)ps g(U1,T2) = g(ap−1)ps (U1)T2 a
s (s) (p−1)p Since δ(g) ∈ Im(D ) the coefficient of T3 in δ(g) has to be zero. Direct computation of this coefficient gives us that −ps+1 X ap − 1 (a−1)ps+1 0 = δ(g s )(U ,U ) + (ω − 1) g s (U )U (p−1)p 1 2 p − 1 (ap−1)p 1 2 a (3.2) ap−1 Since p−1 ≡ 1 (mod p) we can rephrase this as
−ps+1 −1 (p−1)ps g(U1,T2) = −(ω − 1) δg(p−1)ps (U1,T2)T2 (3.3)
So we see that when g 6= 0 then also g(p−1)ps 6= 0. Now from (3.2) we 2(s+1) see δ(g(p−1)ps ) ∈ C so applying lemma 3.6.1 we get that deg(g(p−1)ps ) = d − ps+1 is either ph for some h < s + 1 or b · ps+1.
s+1 s+1 2 • Assume that d − p = b · p . We are working modulo Bd−ps and this bps+1 (p−1)ps (b+1)ps+1−ps kills g since the coefficient of U1 T2 in δ(T1 ) is equal to s+1 s s+1 bp + (p − 1)p (ωd−p − 1) (p − 1)ps which is nonzero. Combining this with (3.3) we see that we can kill g by (b+1)ps+1−ps A · δ(T1 ) for some A ∈ k. • Now suppose d − ps+1 = ph for some h ≤ s. We have
h s −ps ph (p−1)ps φ(E1(p , p · p )) = (1 − ω )U1 T2
So we can get rid of g(p−1)ps using φ(E1) and by relation (3.3) then g = 0.
29 3.6.2 t < s (t+1) 3(s) Assume that for g ∈ K we can find a cocycle f ∈ Zd such that g +φ(f) ∈ 2 s (t) Bd−ps + Im(D ). We’ll now prove the same for g ∈ K . So let g ∈ K(t). Say X apt g = gapt (U1)T2 a
t (s) p Since δ(g) ∈ Im(D ) the coefficient of T3 in δ(g) has to be zero. Direct computation of this coefficient gives
−pt−ps X (a−1)pt δ(gpt )(U1,U2) + (ω − 1) gapt (U1) · a · U2 = 0 (3.4) a
t s This prompts us the make a distinction between whether ωp +p = 1 or not.
pt+ps • First assume that ω 6= 1. Now relation (3.4) implies that gpt defines (t+1) g modulo K . We can also apply lemma 3.6.1 to find that deg(gpt ) = d − pt − ps is either ph for h < t or b · pt with b such that b = p − 1 or b 6≡ −1 (mod p). If d = ph + pt + ps then we can use φ(F (ph, pt, ps)) to t s t t s kill g. And if d − p − p = (p − 1)p we can kill g using φ(E0(p · p , p )) since t s (p−1)pt (p−1)pt (φ(E0(p · p , p )))pt = (ω − 1)U1 t s t Finally when d − p − p = b · p with b 6≡ −1 (mod p) we can kill gpt (and t (t+1) (b+1)p thereby the rest of g modulo K ) using δ(T1 ).
pt+ps • Next assume that ω = 1. This time relation (3.4) shows that δ(gpt ) = pt t t 0. So deg(gpt ) = 0 or gp = 0. If deg(gp ) = 0 we see that g = A · δ(T1 )
for some A ∈ k. The case where gpt = 0 takes some more work.
Continuing from the second case, first assume p 6= 2. This time considering 2pt the coefficient of T3 in δ(g) we find. −pt X a (a−2)pt 0 = δ(g t )(U ,U ) + (ω − 1) g t (U )U (3.5) 2p 1 2 2 ap 1 2 a>1
(t) When g2pt 6= 0 we get δ(g2pt ) ∈ ImD and can apply lemma 3.6.1. This gives s t h t us that deg(g2pt ) = d − p − 2p is equal to either p with h < t or (p − 1)p or b · pt with b 6≡ −1 (mod p).
h t h p 2p • If deg(g2pt ) is equal to p then g = AU1 T2 for some A 6= 0 ∈ k. But this is equal to Aφ(Γ(ph, pt, pt, ps)), since this is case 1 for Γ.
t t+1 • The case where deg(g2pt ) = (p − 1)p is impossible since then d = p + t s t+1 pt + ps. But combining this with ωd = 1 = ωp +p would give ωp = 1 which is a contradiction.
30 t (b+1)pt • Finally when deg(g2pt ) = b·p with b 6≡ −1 (mod p), we have ω = 1. t bp bpt The coordinate of U1 in δ(g2pt )(U1,U2) is equal to A(ω − 1) for some A 6= 0 ∈ k. This is nonzero. Using (3.5) we find that b 6≡ −2 (mod p). t t t b·p 2p (b+2)p bpt−1 b+2 But the coefficient of U1 T2 in δ(T1 ) is equal to (ω −1) 2 2pt which is nonzero. So we can get rid of the the g2pt (U1)T monomial in t t b+1p p (b+1)pt g. And since the coefficient of U1 T2 is (ω − 1)(b + 2) = 0 we won’t introduce unwanted monomials.
After this we can assume that g2pt = 0 and so relation (3.5) implies that P (ap+1)pt 2(t+1) g = h + g(ap+1)pt (U1)T2 for some h ∈ Cd−ps . When p = 2 this is a>0 also obviously the case. Repeatedly applying the following lemma concludes the proof of theorem 3.5.1.
i t P (ap +1)p 0 (t) Lemma 3.6.2. For i > 0 and g = g(api+1)pt (U1)T2 + g ∈ K with a>0 0 2(t+1) g ∈ Cd−ps . We have for some q1
t t+i s X (api+1+1)pt 00 g = Aφ(Γ(q1, p , p , p )) + g(api+1+1)pt (U1)T2 + g + j a>0
00 2(t+1) 2 where g ∈ Cd−ps , j ∈ Bd−ps and A ∈ k.
(pi+1)pt Proof. Considering the coefficient of T3 in δ(g) we find i −pt+i X (a−1)pt+i ap + 1 0 = δ(g i t )(U ,U ) − (ω − 1) g i t U (3.6) (p +1)p 1 2 (ap +1)p 2 pi + 1 a
If g(pi+1)pt 6= 0 then we can use lemma 3.6.1 to find that the degree of g(pi+1)pt is either ph for h < t + i or b · pt+i.
h t+i t h p p +p • Suppose deg(g(pi+1)pt ) = p with h < t + i. So g(pi+1)pt = AU1 T2 for some A 6= 0 ∈ k. We’ll prove that (ph, pt, pt+i, ps) ∈ Q. We already t s h t+i have the required inequalities and we also have ωp +p = 1 = ωp +p . h t If ωp +p = 1 we are in case 1. Otherwise we’ll have to use one of the other cases. Only when cases 2 and 3 don’t apply can there be some trouble. Since for case 4 and 5 we have to check some extra relations. So assume pt+1 < ps when p 6= 2 or assume pt+1 ≤ ps when p = 2. We need to show t h pt ≤ ph and when p = 2 the stronger pt < ph as well as ωp = ωp . h t t+i 0 p +p p 0 Let A be the coefficient of U1 T2 in g . We want to show that A0 6= 0.
h t t+i h t p p p – When p = p then the coefficient of U1 U2 T3 in δg is equal to h t h (ωp +p − 1)A − 2(ωp − 1)A0. But it also has to be 0 since δg ∈ Im(D(s)). So when p = 2 this is a contradiction and when p 6= 2 it shows A0 is nonzero.
31 h t t+i h t p p p – When p 6= p then the coefficient ofU1 U2 T3 in δg is equal to h t h (ωp +p − 1)A − (ωp − 1)A0. So this time it directly shows A0 is nonzero regardless of p.
pt ph pt+i But then the coefficient of U1 U2 T3 is equal to
h h t (ωp − 1)A0 + (ωp +p − 1)A00
t h t+i 00 p p +p 00 where A is the coefficient of U1 T2 . This shows that A is nonzero. If ph < pt then g can’t contain this monomial since g ∈ K(t), so pt ≤ ph. h t If ph = pt then obviously ωp = ωp and if ph > pt then the coefficient of pt pt+i ph U1 U2 T3 is equal to
t t+i t (ωp +p − 1)A00 + (ωp − 1)A000 (3.7)
t t+1 h 000 p +p p 000 Where A is the coefficient of U1 T2 . However A has to be zero t+1 t h p p p pt+i 000 because the coefficient of U1 U2 T3 in δg is equal to (ω − 1)A . t t+i But when A000 = 0 formula 3.7 shows that ωp +p = 1 and therefore h t ωp = ωp . So we finally have (ph, pt, pt+i, ps) ∈ Q. We have that φ(Γ(ph, pt, pt+i, ps)) ph pt+pt+i is equal to U1 T2 in cases 1-3 and equal to
ph pt+pt+i pt ph+pt+i ph −1 pt+ph ph+pt pt+i U1 T2 + U1 T2 − (ω − 1) (ω − 1)U1 T2 in cases 4 and 5. But the extra terms in cases 4 and 5 are all contained 2(t+1) in Cd−ps so they don’t create any trouble. So now we indeed have g = Aφ(Γ(ph, pt, pt+i, ps)) + g00 where g00 ∈ K(t+1) which is what we needed to prove.
t+i • Next assume deg(g(pi+1)pt ) = b·p . First take b = p−1. Now g(pi+1)pt = (p−1)pt+i pt+pt+i AU1 T2 for some A 6= 0 ∈ k. But then the coefficient of t+i t+i t (p−1)p p p pt+i+1 (s) U1 U2 T3 in δg is A(ω − 1). But since δg ∈ Im(D ) it should be zero which is a contradiction. • Finally suppose b 6≡ −1 (mod p). Since we have d = pt +ps +(b+1)pt+i we t+i t t+i t (b+1)pt+i (b+1)p p (b+1)p +p have ω = 1. So the coefficient of U1 T2 in δ(T1 ) 2(t+1) is zero. All the other monomials in this coboundary are either in Cd−ps (b+1−a)pt+i (api+1)pt or are of the form U1 T2 . In particular the coefficient of bpt+i pt+pt+i U1 T2 is nonzero. So we can use this coboundary to get rid of g(pi+1)pt so we can assume it is zero.
But when g(pi+1)pt = 0 relation (3.6) implies that g(api+1)pt is zero unless a ≡ 0 (mod p) which means that g is in the form we were required to prove.
32 3.7 Final steps
(s) (s) 3(s) Definition 3.7.1. Define the subset Id ⊂ I by Id = Cd ∩ I.
(s) Note that the set Id contains exactly the cocycles we used in the proof of theorem 3.5.1.
d 3(s) 3 3(s) 3(∞) 3 Lemma 3.7.1. When ω = 1 we have Bd = Bd ∩ Cd and Bd = Bd ∩ 3(∞) Cd . Proof. The proof is a very simple application of the same technique we used repeatedly during the last section. For details see [18] lemma 3.11.
(s) Lemma 3.7.2. All cocycles in the set Id are independent. Proof. Combining the previous lemma with theorem 3.5.1 we see that it’s enough to show that the φ images of the cocycles are independent. Recall the order on (s) the monomials defined in 3.3.2. For f ∈ Id we’ll look that the largest mono- d−ps mial of φ(f). We’ll also look at that the largest monomial of δ(T1 ) that is not in Im(D(s)). These monomials will be different. Since this holds for all f it must also hold for every linear combination of them which gives the desired result. a b In the following table < a, b > stands for U1 T2 . The first column contains f, the second contains the largest monomial of φ(f) and the third contains a s d−p (s) monomial occurring in δ(T1 ) that is not in Im(D ) and that is larger than the monomial in the second column. s F (q1, q2, p ) < q1, q2 > < q2, q1 > s E0(p · q1, p ) * * s s s s E1(q1, p · p ) < q1, (p − 1)p > < p , (p − 2)p + q1 > s Γ(q1, q2, q3, p ) cases 1,2 and 3 < q1, q2 + q3 > < q3, q1 + q2 > s Γ(q1, q2, q3, p ) cases 4 and 5 < q1 + q2, q3 > < q3, q1 + q2 >
(p−1)q1 *: The largest monomial occuring in φ(E0) depends on wether ω is s d−p p·q1 equal to 1 or not. However only monomial ocurring in δ(T1 ) is U1 , so there are no problems. (s) The proof of theorem 3.5.1 also shows that all the cocycles in Id are inde- (s+1) pendent of all the cocycles in Id . All that remains to be done is tackle the 3(∞) case of Zd .
(∞) 3(∞) (∞) Definition 3.7.2. Define Id = I ∩ Cd and let Hd be the subgroup (∞) generated by Id .
3(∞) 3(∞) 3(∞) Lemma 3.7.3. Zd = Hd ⊕ Bd
33 Proof. Once again the proof is a simple application of the technique we have been using in the previous section. 3(∞) 2 Let f ∈ Zd . We have f = f0(U1,U2) where f0 ∈ Zd . So we are essentially 2(s) dealing with 2 cocycles. Suppose f0 ∈ Zd with s minimal. Let g = f0 and P ps ps say g = a gaps (U1)T2 . As usual consider the coordinate of T3 in δg. This gives us
−ps X (a−1)ps 0 = δ(gps )(U1,U2) + (ω − 1) a · gaps (U1)U2 a
pt ps So again we apply lemma 3.6.1 and find that g = AU1 U2 for some A ∈ k t s s and t < s, which means that f = A · F (p , p , 0), or deg(gps ) = b · p . b = p − 1 is impossible since then d = ps+1 which contradicts ωd = 1. And when b 6≡ −1 (b+1)ps (mod p) we can use δ(T1 ) to get rid of gps . Doing this as usual pushes g ∈ Z2(s+1) and we repeat the process. The only case this doesn’t cover is the case with d = 0 but then f is just constant.
We can also quickly see that the F (q1, q2.0) are independent of the rest by the same argument as before. So summing up
Theorem 3.7.4. The set I forms a basis of H3(C). Proof. This follows directly from theorem 3.5.1 and the lemmas in this section.
34 Chapter 4
Rack and quandle modules
35 4.1 Rack and Quandle Modules
In the articles [12, 13], Jackson introduced the notion of rack and quandle modules. These modules form a rich enough category to use them to define a homology theory. This theory is however different from the usual homology of racks and quandles. This chapter we will present a different approach to these modules. We use some different notation that makes the definition of free rack modules easier. The main result is that for a connected quandle X the X-modules coincide with the modules over a subring of Z[Adj(X)].
4.1.1 Definitions Let X be a rack. Define a directed graph G(X) as follows. Take the underlying set of the rack as the vertices of G(X) and add the following labeled edges for all x, y ∈ X
−1 θx,y θx,y x / xy xy / x
ηy,x y / xy
Next form the free category generated by this graph, call it C0(X). In this category now replace Hom(x, y) by Z[Hom(x, y)] and extend the composition linearly. Call this category C1(X). Note that the hom-sets of C1(X) are abelian groups, in these groups divide out the following relations:
−1 θx,yθx,y = idxy (4.1) −1 θx,yθx,y = idx (4.2)
θxy ,zθx,y = θxz ,yz θx,z (4.3)
θxy ,zηy,x = ηyz ,xz θy,z (4.4)
ηz,xy = θxz ,yz ηz,x + ηyz ,xz ηz,y (4.5)
We can define the composition on these quotient groups in the obvious way. The result is another category which we’ll call CR(X). The relations (4.3) and (4.4) are equivalent to saying that the following squares are commutative:
θx,y ηy,x x / xy y / xy
θx,z θxy ,z θy,z θxy ,z z yz yz yz x / x η z z / x θxz ,yz y ,x
Relation (4.5) says that in the following diagram the diagonal arrow is the sum of the two outer compositions.
36 η z z,y yz NN / NN y NN ηz,x ηz,x NNN ηyz ,xz NNN θxz ,yz N' xz / xyz Note 4.1.1. We’ll mostly focus on racks in the next few sections. When X is a quandle there is a notion of a quandle module. The only difference is that instead of working with CR(X) one uses CQ(X) where CQ(X) is obtained by dividing out ηx,x + θx,x = idx in HomCR(X)(x, x).
Convention 4.1.2. We’ll simply write Hom(x, y) for HomCR(X)(x, y) in this chapter. Note that the θ functions capture the action of the quandle on itself. This extends to the adjoint group.
g Definition 4.1.3. Define θx,g ∈ Hom(x, x ) for g ∈ Adj(X) recursively. First define
θx,ey = θx,y −1 θ −1 = (θx y,y) x,ey C Then proceed with recursion using the following rule
θx,gh = θxg ,h ◦ θx,g
This is well defined thanks to relations (4.1)-(4.3). What exactly does Hom(x, y) look like? It’s still an abelian group and the composition respects the addition. If f ∈ Hom(x, y) then f is a sum of compositions of θ’s, θ−1’s and η’s. If such a composition contains one or more η’s, we can use relation (4.4) to move all those to the front of the composition. If there are multiple η’s we can then use relation (4.5) to replace it by a sum of compositions with less η’s. In the end we can make sure that every element of Hom(x, y) is a sum of functions in either of the following two forms.
g θx,g g ∈ Adj(X), x = y x g θzx,gηx,z z ∈ X, g ∈ Adj(X), (z ) = y
Definition 4.1.4. A (left) rack module over X or X-module is a functor A : CR(X) → Ab that respects the addition of morphisms. Ab denotes the category of abelian groups with their homomorphisms.
An X-module A = (A, φ, ψ) consists of abelian groups Ax for each x ∈ X along with isomorphisms φx,y : Ax → Axy and homomorphisms ψy,x : Ay → Axy such that the φ and ψ satisfy the above relations.
37 Definition 4.1.5. An X-module homomorphism k : A = (A, φ, ψ) → B = (B, χ, ω) is just a natural transformation between the functors A and B, i.e. it consists of a series of homomorphisms kx : Ax → Bx such that
χx,ykx = kxy φx,y (4.6)
ωy,xky = kxy ψy,x (4.7) Convention 4.1.6. If A is an X-module, then for f ∈ Hom(x, y) we’ll simply write f instead of A(f) for the corresponding map in Hom(Ax,Ay). With this notation we can rephrase (4.6) and (4.7) as
kyf = fkx f ∈ Hom(x, y) Since X-modules are just functors and since X-module maps are just natural transformations we immediately see that all X-modules form a category which we’ll denote by RModX .
4.1.2 Some examples Example 4.1.1. Take an abelian group A. We can form an X-module A by setting Ax = A, φx,y = idA and ψy,x = 0. These modules are called trivial. Example 4.1.2. Now let A be an Adj(X)-module. We can then turn A into an X-module by setting Ax = A, φx,y(a) = ex · a and ψy,x = 0. Example 4.1.3. A more interesting example where ψ 6= 0. Let A be an abelian group and let T : A → A be an automorphism. If X is the Alexander-quandle associated with this group and morphism then we can make an X module by setting Ax = A, φx,y = T and ψy,x = 1 − T . We would also like to define what a free X-module is. Before we do this we first need some notion of a forgetful functor on X-modules to which the construction of free modules should be adjointed. Let P (X) be the category with one object for each element of X and no morphisms besides the identities. Now the category SetP (X) - i.e. the category of collections of sets indexed by X - is P (X) a good target category for the forgetful functor. So let U : RModX → Set be the functor that sends each Ax to it’s underlying set and forgets about the φ and ψ maps. P (X) Definition 4.1.7. Let S = (Sx) ∈ Set . Define F (S) ∈ RModX the free X-module on S as follows. Say F (S) = (A, φ, ψ), then [ Ax := Z Sy × Hom(y, x) ∼ y∈X Where ∼ is given by n · (s, f) ∼ (s, n · f) (s, f) + (s, g) ∼ (s, f + g)
38 for n ∈ Z, s ∈ Sy and f, g ∈ Hom(y, x). Define φ and ψ by
φx,y(s, f) = (s, θx,yf) (4.8)
ψy,x(s, f) = (s, ηy,xf) (4.9)
Clearly F (S) is an X − module. To turn F into a proper functor let 0 0 k = (kx : Sx → Sx). Define F (k): F (S) → F (S ) by F (k)x(s, f) = (kx(s), f). Theorem 4.1.1. F is a left adjoint to the forgetful functor U. Proof. Let S ∈ SetP (X) and let A be an X −module. The required isomorphism α : Hom(S, U(A)) → Hom(F (S), A) is given by
α(k)x(s, f) = f(ky(s)) (4.10)
4.2 Reduced rack modules
If A = (A, φ, ψ) is an rack module and x, y ∈ X are in the same connected component then Ax and Ay are isomorphic. We’ll now introduce the notion of reduced rack modules to strip away this redundancy. For connected quandles the notion of a reduced rack module will just coincide with the normal notion of a module over a ring.
Definition 4.2.1. Let X be a rack. For every connected component c ∈ π0(X) R R choose a pc ∈ c. Let D (X) be the full subcategory of C (X) with as objects only the pc. Definition 4.2.2. Let X be a rack. A reduced (left) rack module over X is a functor A : DR(X) → Ab that respects addition of morphisms. A homo- morphism between two reduced rack modules is again a natural transformation. The reduced rack modules over X form a category RRMODX .
A reduced X-module B exists of groups Bpc for every pc and homomorphisms
Bp1 → Bp2 for every element of Hom(p1, p2). Like before we have a map
Hom(p1, p2) → Hom(Bp1 ,Bp2 ): f 7→ f
Theorem 4.2.1. The categories RMODX and RRMODX are equivalent. Proof. DR(X) is the skeleton of CR(X) so the inclusion I : DR(X) → CR(X) is an equivalence of categories. Let H : CR(X) → DR(X) be the functor and : id → I ◦ H and δ : X ◦ I → id be the two natural isomorphism that show that I is an equivalence. We can choose H in such a way that it respects the addition of morphisms.
39 Now define F : RMODX → RRMODX by setting F (A) = A ◦ I and also define G : RRMODX → RMODX by G(B) = B ◦ H. We now also have natural isomorphisms 0 : id → G ◦ F and δ0 : F ◦ G → id given by
0 B : B → B ◦ I ◦ H = idB ◦ 0 δA : A ◦ H ◦ I → A = idB ◦ δ
We’ll make the functors F and G explicit. If A = (A, φ, ψ) is an X-module then if F (A) = B = (Bpc ) we have Bpc = Apc and the maps between the Bpc 0 are just the compositions of φ’s and ψ’s. When f : A → A the F (f)pc = fpc .
G takes a bit more work. Let B = (Bpc ) be a reduced X-module. Say G(B) = A = (A, φ, ψ) then we have [ Ax := Z (Bpc × Hom(pc, x)) / ∼
c∈π0(X) Where ∼ is given by the following relations (n · b, f) ∼ n · (b, f) ∼ (b, n · f) (b, f) + (b0, f) ∼ (b + b0, f) (b, f) + (b, g) ∼ (b, f + g) (b, fh) ∼ (h(b), f)
0 0 y For n ∈ Z, b, b ∈ Bpc , f, g ∈ Hom(pc, x) and h ∈ Hom(pc, pc ). φx,y : Ax → Ax and ψy,x : Ay → Axy are given by
φx,y(b, f) := (b, θx,yf)
ψy,x(b, f) := (b, ηy,xf) The relations that should hold for the φ and ψ hold because they hold for θ and η. If k : B → B0 is an homomorphism between two reduced X-modules then
G(k)x(b, f) := (kpc (b), f) We of course also have free reduced modules defined in a similar way to free modules. The only difference is that instead of P (X) one should use DP (X) where DP (X) is the category with one point for each connected component and no morphisms besides the identities.
4.3 Right modules and tensor products
Definition 4.3.1. Let X be a rack. A right rack module over X is a contravari- ant functor CR(X) → Ab that respects addition of morphisms. Of course we do the same for reduced right rack modules. The categories of (reduced) right rack modules over X are denoted by RMODX resp. RRMODX .
40 Once again these categories are equivalent. We’ll again write F for the func- tor from RMODX to RRMODX that is induced by the inclusion of DR(X) in CR(X). Convention 4.3.2. When dealing with (reduced) right X-modules we’ll write op the indices of θ, η, φ, ψ etc. with superscript instead of subscript. So θx,y = θx,y : xy → x etc. Definition 4.3.3. Let A = (A, φ, ψ) be a right X-module and let B = (B, χ, ω) be a left X-module. Define the tensor product A ⊗ B as follows. " # [ A ⊗ B = Z Ax × Bx ∼ (4.11) x∈X Where ∼ is given by
(a1 + a2, b) ∼ (a1, b) + (a2, b) (4.12)
(a, b1, b2) ∼ (a, b1) + (a, b2) (4.13) (na, b) ∼ n(a, b) ∼ (a, nb) (4.14) x,y (φ c, b) ∼ (c, χx,yb) (4.15) y,x (ψ c, d) ∼ (c, ωy,xd) (4.16)
For all x, y ∈ X a, a1, a2 ∈ Ax c ∈ Axy b, b1, b2 ∈ Bx and d ∈ By. Relations (4.15)-(4.16) can be summarized as op (f (a), b) ∼ (a, f(b)) (4.17) We’ll denote the equivalence class of (a, b) by a ⊗ b We define the tensor product for reduced X-modules in the same way except that in (4.11) we take the union over only the pc. The tensor product on rack modules and reduced rack modules is compatible in the following sense. Theorem 4.3.1. Let X be a rack, let A = (A, φ, ψ) be a right X-module and let B = (B, χ, ω) be a left X-module. We have A ⊗ B =∼ F (A) ⊗ F (B) Proof. Let P = A ⊗ B and Q = F (A) ⊗ F (B). Define f : Q → P by f(a ⊗ b) =
(a ⊗ b) for a ∈ Apc and b ∈ Bpc . It’s trivial that this is well defined. wx To define g : P → Q first choose wx ∈ Adj(X) for x ∈ X such that p[x] = x. Now for a ∈ Ax and b ∈ Bx note that a ⊗ b = φp[x],wx a ⊗ χ−1 b p[x],wx thanks to relation (4.17). So we use this to define
g(a ⊗ b) = φp[x],wx a ⊗ χ−1 b p[x],wx Once again (4.17) makes this well defined and also shows that f ◦ g = id and g ◦ f = id.
41 4.4 The rack algebra
The main advantage of using reduced modules is that when X is a connected rack or quandle, DR(X) only has one point p. As a result R(X) := Hom(p, p) is a ring and reduced X-modules are just modules over the R(X). This is compatible with the tensor product. We call R(X) the rack algebra of X. Definition 4.4.1. This differs slightly from what Jackson calls the rack algebra in [12] but is essentially the same thing. The rack algebra from [12] is denoted by ZX and is the free X-module on S where every set Sx has only one point. R(X) is essentially equal to the free reduced X-module on a single point. So what does R(X) look like? It consists of sums of products of θ’s and η’s. As noted after definition 4.1.3 we can move all η’s in a product to the left and keep only one of them. So
R(X) = θ , η θ |x ∈ X g, h ∈ Adj(X): pg = p and ph = x Z p,g x,pCx p,h Convention 4.4.2. We’ll simply write θ for θ and η for η . g p,g x x,pCx Multiplication is given by
θg · θh = θhg
ηxθg · θh = ηxθhg
θg · ηxθh = ηxg θhg
g ge ηxθg · ηyθh = ηy θhg − ηy x θhgex for g, h ∈ Adj(X) and x, y ∈ X. Note 4.4.3. For quandles X we have a quandle algebra RQ(X) The only dif- ference is that besides the above relations we also have ηxθ1 + θep = 1. The map i defined below also works for RQ(X).
There is a map from R(X) to Z[Adj(X)]. This however is not a ring homo- morphism but a ring anti-homomorphism. This is somewhat unfortunate but not a real issue. The cause is that R(X) as a function ring naturally acts on the left on X while Adj(X) acts from the right.
Definition 4.4.4. Define a ring anti-homomorphism i : R(X) → Z[Adj(X)] by
i(θg) = g
i(ηxθh) = h(1 − ex)
42 We’ll verify this in fact is a ring anti-homomorphism.
i(θh) · i(θg) = hg = i(θhg) = i(θg · θh)
i(θh) · i(ηxθg) = hg(1 − ex) = i(ηxθhg) = i(ηxθg · θh) −1 i(ηxθh) · i(θg) = h(1 − ex)g = h(1 − gexg g )g = hg(1 − exg ) g = i(ηxθhg) = i(θg · ηxθh)
i(ηyθh) · i(ηxθg) = h(1 − ey)g(1 − ex) = hg − heyg − hgex + heygex −1 −1 −1 = hg(1 − g eyg) − hgex(1 − ex g eygex)
= hg(1 − eyg ) − hgex(1 − eygex )
g ge = i(ηy θhg − ηy x θhgex ) = i(ηxθg · ηyθh)
Theorem 4.4.1. Let X be a quandle. i : RQ(X) → Z[Adj(X)] is injective. Proof. Look at the subrings g B := Z[θg|g ∈ Adj(X): p = p] ⊂ R(X) g C := Z[g ∈ Adj(x)|p = p] ⊂ Z[Adj(X)] It’s clear that i maps B bijectively onto C. We’ll regard R(X) as a right B- module and Z[Adj(X)] as a left C-module. Again choose wx ∈ Adj(X) such that pwx = x. We have M R(X) = B ⊕ ηxθwx B x∈X\{p} M Z[Adj(X)] = C ⊕ Cwx x∈X\{p} P If v = fp + ηxθwx · fx ∈ R(X) then x∈X\{p} X i(v) = i(fp) + i(fx) · wx(1 − ex) x∈X\{p} X −1 = i(fp) + i(fx) · wx(1 − wx epwx) x∈X\{p} X = i(fp) + i(fx)(1 − ep) · wx x∈X\{p}
So as long as (1 − ep) is not a zero divisor in C this shows that i is injective. Define M : C → Z as follows
M(a0g0 + ... + angn) = max((g0), . . . , (gn))
gi for 0 6= ai ∈ Z and gi ∈ Adj(X) with p = p. Now we find M(f ·ep) = M(f)+1 for 0 6= f ∈ C. So 1 − ep is not a zero divisor. So how close is i to being surjective? If X is a finite quandle then Inn(X) is a finite group. Since Inn(X) =∼ Adj(X)/K and Z[K] ⊂ Im(i) we see that the cokernel of i is finitely generated.
43 4.5 Examples 4.5.1 A free resolution for trivial quandles
In [11] Jackson gives a free resolution of Z viewed as a trivial X-module for a trivial rack X. However it is incorrect. This fault is based on the fact that he uses what is here called θx,g for g ∈ Inn(X) instead of Adj(X). In his later article [13] he does use the correct definition but doesn’t correct the example from [11]. So let X = {1, . . . , m} be a trivial quandle with m points. Turn Z into a trivial X-module like in example 4.1.1. Let ZX be the rack algebra of X i.e. the free module on S where Sx = ∗ for x ∈ X. Recall that ZXx is generated by elements of the form
(∗, θxg−1 ,g) for g ∈ Adj(X)
(∗, θ g−1 η g−1 ) for y ∈ X, g ∈ Adj(X) x ,g y,x Cy
Now define : X → by sending the (∗, θ g−1 η g−1 ) to zero and the Z Z x ,g y,x Cy other generators to 1. Now in [11] Jackson claimed that the kernel of was the same as ZX. This indeed true if you work with g ∈ Inn(X) since Inn(X) is trivial but with g ∈ Adj(X) it’s clearly not the case. Now if X = {x} is trivial quandle with only one point then Adj(X) = n {ex|n ∈ Z} and so RQ(X) = [θn ] ∼ [t, t−1] Z ex = Z So RQ(X) is just a Laurent polynomial ring in one variable. Notice that RQ(X) =∼ Z[Adj(X)] in this case. : RQ(X) → ZX is given by (t) = 1 and (1) = 1. It’s kernel is generated by t − 1. So we get a free resolution of Z
(1−t) 0 /Z[t, t−1] /Z[t, t−1] /Z /0 This still shows that the homology theory we can derive from these modules R is different form the usual theory since Hn (X) = Z for all n.
4.5.2 R3 Q ∼ Now we’ll study R (X) for the dihedral quandle X = R3. Recall that Adj(X) = Z × Z3 × ST (Z3). ST (Z3) is the trivial group so an element of Adj(X) is pair (n, x) that acts on X by y(n,x) = T n(y) + x = x + (−1)ny. It’s obvious that t = (1, 0) and r = (0, 1) generate Adj(X) and that the following relations hold
r3 = 1, trt−1 = r−1
So Z[Adj(X)] = Z[t, t−1, r]/I where I is generated by the above relations. We again choose 0 as the base point of X. Now 0(n,x) = x so all the Q θg in R (X) are powers of θ(1,0). Also note that (n, 0) · (0, x) = (n, x) so
44 Q ηxθ(n,x) = ηxθ(0,x) · θ(n,0). So only the ηxθ(0,x) are needed to generated R (X). Furthermore
η1θ(0,1) · η1θ(0,1) = η2θ(0,2) − η0θ(1,0) = η2θ0,2 − (1 − θ(1,0)) · θ(1,0)
So we only need η1θ(0,1). We also find the following relationship
θ(1,0) · η1θ(0,1) = η2θ(1,2) = η2θ(0,2) · θ(1,0)
Combining the last two relations gives us
η1θ(0,1) · η1θ(0,1) = θ(1,0) · η1θ(0,1)θ(−1,0) − θ(1,0) + θ(2,0)
Now consider that
(η1θ(0,1) · θ(1,0)) · η1θ(0,1) = η1θ(1,1) · η1θ(0,1)
= η0θ(1,0) − η2θ(2,2)
= θ(1,0) − θ(2,0) − θ(1,0) · η1θ(0,1) · θ(1,0)
Finally note that (2, 0) acts trivially on X so θ(2,0) commutes with everything. Now consider the ring A = Z[T,T −1,R]/J where J is generated by the relations
R2 = TRT −1 − T + T 2 RTR = T − T 2 − TRT T 2R = RT 2
It can easily be checked that we have an isomorphism RQ(X) =∼ A that sends −1 θ(1,0) to T, η1θ(0,1) to R and η2θ(0,2) to TRT . The map i : A → Z[Adj(X)] is now given by i(T ) = t and i(R) = r(1 − tr2).
45 Chapter 5
Trunks
46 5.1 Introduction
In this section we’ll give a short overview of trunks and there cubical nerve. These concepts were introduced in [9] and give rise to a homology theory of racks and quandles. This section just serves as an introduction to the ideas and most details are left out. For more details and more explicit constructions see [9].
5.2 Trunks
Trunks are objects loosely related to categories. A trunk also consists of points and arrows between those points. But instead of composition there is a collection of preferred squares of arrows. Composition could be described as a set of preferred triangles to further to analogy. Definition 5.2.1. A trunk consists of a set of points, a set of arrows between these points and a collection of preferred squares, where preferred squares are oriented squares of arrows of the following shape.
c C / D O O
b x d
a A / B
The curved arrow in the center indicates the (counter-clockwise) orientation. We’ll now give some examples of trunks.
Example 5.2.1. For every category C we have a trunk T (C) with the same points and arrows. The preferred squares are given by the commutative squares in C with both possible orientations. Example 5.2.2. Let X be a rack. We define the rack trunk T (X) as follows. T (X) has a single point • and one arrow for every element of x. The preferred squares are given by a b • B / • O O
b x b
a • / •
47 Example 5.2.3. Let X be a rack and let Y be a set on which X acts from the right such that yab = yb(aBb) for y ∈ Y and a, b ∈ X. We define the action rack trunk TY (X) as follows. Take one point for every element of y and all arrows a of the form y /ya . The preferred squares are given by
a b yb B / yab = yb(aBb) O O
b x b
a y / ya
Example 5.2.4. If φ : X → G is an augmentation of a rack X then X acts on x G by g = gφ(x). So we can make a trunk TG(X) like in the previous example. This trunk is called the augmented rack trunk. It will be studied extensively in the next chapter. Both the rack trunk and the action rack trunks are examples of so called corner trunks (c.f. [9]). Basically this means that if you have a diagram like
B O
a x
b A / C it can be uniquely filled in to a preferred square. And if you start with 3 arrows with the same origin you could fill it in to a cube etc.
Definition 5.2.2. A Trunk map between two trunks T1 and T2 is the analog of a functor for trunks. It consists of map from the points of T1 to the points of T2 and a mapping between the arrows. Of course in such a way that the base- and endpoints of the arrows are mapped to the base- resp. endpoint of the image of the arrow. Finally these mappings have to map preferred squares to preferred squares. The set of all trunk maps between two trunks is denoted by Hom(T1, T2).
5.3 -sets and their realizations Like a category has a nerve which is a simplicial set, a trunk has also a nerve but it’s not a simplicial set but a -set (pronounced square-set). We’ll glance over many details here so please refer to [9].
48 Definition 5.3.1. A -set X = {Xn|n ∈ N} is a collection of sets Xn and a ε set of face maps δi : Xn → Xn−1. Such that the maps respect the following η ε ε η relation δj−1δi = δi δj for 1 ≤ i < j ≤ n and η, ε ∈ {0, 1}.
The sets Xn represent the n-cubes of the -set and the face maps tell you what n − 1-cubes are what faces of a given n-cube. Definition 5.3.2. Let In = [0, 1]n ⊂ Rn the unit n-cube. For i = 1, . . . , n and ε n−1 n ε ∈ {0, 1}. Define the following inclusions δi : I → I by ε δi (t1, . . . , tn−1) = (t1 . . . , ti−1, ε, ti, . . . , tn−1) (5.1) Naturally these maps satisfy the same relations as we demanded of the face maps of a -set. We make the realization of a -set by taking an n-cube for every element of Xn and glueing them together along their faces as described by the face maps.
Definition 5.3.3. Let X = {Xn} be a -set. Form a topological space ||X|| -the realization of X- as follows. Start by taking the disjoint union ` Xn × In. n Where I has its usual topological meaning and Xn is given the discrete topol- ogy. On this space make the identifications given by ε ε (δi x, t) ∼ (x, δi t) (5.2) n−1 Where x ∈ Xn, t ∈ I , i ∈ 1 . . . , n and ε ∈ {0, 1}. We can view ||X|| as a CW-complex with one n-cell for each element of n ∼ X . The homology of ||X|| has an easy description. Cn(||X||) = Xn and the boundary operator is given by n X i 0 1 δn = (−1) [δi − δi ] (5.3) i=1 5.4 The nerve of a trunk
With every trunk we’ll associate a -set called the nerve of the trunk, similar to the nerve of a category. n Definition 5.4.1. For n ∈ N define a trunk IT runk as follows. Let the points be all subsets of {1, . . . , n} and let there be an arrow from A to B if and only if B = A ∪ {i} for some i 6∈ A. The preferred squares are given by
A ∪ {j} / A ∪ {i, j} O O
x
A / A ∪ {i} for i < j.
49 k n We can consider all trunk maps from IT runk to IT runk. It’s easy to see that these are all compositions of the following ones. Definition 5.4.2. Define for i ∈ {1, . . . , n} and ε ∈ {0, 1} the trunk map ε n−1 n δi : IT runk → IT runk by mapping the points as follows:
ε A 7→ {j ∈ A|j < i} ∪ {j + 1|j ∈ A|j ≥ i} ∪ Vi (5.4)
ε where Vi is empty if ε = 0 and {i} otherwise. These maps are of course the analogues of the maps from In−1 to In.
Definition 5.4.3. For a trunk T define a -set NT the nerve of T by setting n NTn = Hom(IT runk, T ). The face maps are given by
ε ε δi (F ) = F ◦ δi (5.5)
n where F : IT runk → T is a trunk map.
Example 5.4.1. Let X be a rack and let TX be the associated rack trunk. We n have (NTX )n = X with boundary maps given by
0 δi (x1, . . . , xn) = (x1, . . . , xi−1, xi+1, . . . , xn) 1 xi xi δi (x1, . . . , xn) = (x1 , . . . , xi−1, xi+1, . . . , xn)
The realization of this nerve ||NTX || is called BX, the classifying space of X. A typical point of BX looks like (t1, x1, . . . , tn, xn) where the ti ∈ [0, 1]. These points are subject to the relations
(t1, x1, . . . , xi−1, 0, xi, ti+1, . . . , tn, xn) ∼ (t1, x1, . . . , xi−1, ti+1, . . . , tn, xn) xi xi (t1, x1, . . . , xi−1, 1, xi, ti+1, . . . , tn, xn) ∼ (t1, x1 , . . . , xi−1, ti+1, . . . , tn, xn)
50 Chapter 6
Homology operators
51 6.1 The augmented rack space
Let X be a rack and φ : X → G an augmentation. We can make a trunk called the augmented rack trunk TGX that is essentially a generalization of the action rack trunk described early. This trunk already appears in [9]. It has one x point for each element of G and arrows of the form g /gφ(x) and preferred squares like in the case of the action rack trunk. In [9] it’s noted that G acts in two different ways on TGX but we are not interested in these actions here. Instead we’ll see that for certain G the homology of BG(X) = ||NTGX|| defines homology operators on the homology of the rack. n We have Cn(BG(X)) = G × X with boundary operators given by
0 δi (g, x1, . . . , xn) = (g, x1, . . . , xi−1, xi+1, . . . , xn) 1 δi (g, x1, . . . , xn) = (gφ(xi), x1 B xi, . . . , xi−1 B xi, xi+1, . . . , xn)
For the rest of this chapter we’ll focus on G = Inn(X) with the natural augmentation.
Convention 6.1.1. For x ∈ X we’ll write ρ(x) = inn(x) ∈ Inn(X). This won’t cause confusion with the map ρ : Adj(X) → Inn(X) since ρ(x) = ρ(ex).
G n th Convention 6.1.2. We’ll write Cn = G × X for the n chain group of BGX. Definition 6.1.3. Let G = Inn(X). We will define a product on the homology G G of BG(X) as follows. For (g, x1, . . . , xn) ∈ Cn and (h, y1, . . . , yk) ∈ Ck define the following n + k chain
h h (g, x1, . . . , xn) ∗ (h, y1, . . . , yk) = (gh, x1 , . . . , xn, y1, . . . , yk) (6.1)
G G Theorem 6.1.1. Let G = Inn(X). For c ∈ Cn and d ∈ Ck we have δ(c ∗ d) = (δc) ∗ d + (−1)n(c ∗ δd) (6.2)
52 Proof. Let c = (g, x1, . . . , xn) and d = (h, y1, . . . , yk). We have h h δ(c ∗ d) = δ(gh, x1 , . . . , xn, y1, . . . , yk) n X ih h h h h = (−1) (gh, x1 , . . . , xi−1, xi+1, . . . , xn, y1, . . . , yk) i=1 h h h h h h h i −(ghρ(xi ), x1 B xi , . . . , xi−1 B xi , xi+1, . . . , xn, y1, . . . , yk) k n X ih h h +(−1) (−1) (gh, x1 , . . . , xn, y1, . . . , yi−1, yi+1, . . . , yk) i=1 i h yi h yi yi yi −(ghρ(yi), (x1 ) ,..., (xn) , (y1) ,..., (yi−1) , yi+1 . . . yk) n X ih 0 = (−1) δi ((g, x1 . . . , xn)) ∗ (h, y1, . . . , yk) i=1 h h h h i −(gρ(xi)h, (x1 B xi) ,..., (xi−1 B xi) , xi+1, . . . , xn, y1, . . . , yk) +(−1)nc ∗ (δd) = (δc) ∗ d + (−1)n(c ∗ δd)
Because of this the product descends to homology, since if d is a cycle then δ(c ∗ d) = (δc) ∗ d. So products with boundaries produce boundaries again. The product of two cycles is again a cycle. So we can view it as a product on homology ∗ : Hn(BGX) × Hk(BGX) → Hn+k(BGX) Since the product works on homology one would suspect that the space itself is equipped with a product. This is in fact the case. Recall that a point of BGX typically looks like (g, t1, x1, . . . , tn, xn) with ti ∈ [0, 1]. The obvious formula defines the product
(g, t1, x1, . . . , tn, xn) ∗ (h, s1, y1, . . . , skyk) h h = (gh, t1, x1 , . . . , tn, xn, s1, y1, . . . , sk, yk)
It’s easy to see that this respects the relations on the points in BGX and that it does define a continuous function BGX × BGX → BGX. This product is strictly associative and has the 0-cell (1) as a unit. Thus BGX is a topological monoid. As already noted in [9] the augmented rack space BGX covers the rack space BX. The covering map just forgets the G coordinate. This covering is nicely compatible with this product as a result we’ll get that every homology class of BGX defines a homology operator on the rack homology. Definition 6.1.4. Once again let G = Inn(X). Extend the product ∗ to BX × BGX → BX by
(t1, x1, . . . , tn, xn) ∗ (h, s1, y1, . . . , skyk) h h = (t1, x1 , . . . , tn, xn, s1, y1, . . . , sk, yk)
53 G On chain level we get a map ∗ : Cn(X) × Ck → Cn+k(X) given by
g g (x1, . . . , xn) ∗ (g, y1, . . . , yk) = (x1, . . . , xn, y1, . . . , yk) (6.3)
R Inn Theorem 6.1.2. For c ∈ Cn (X) and d ∈ Ck we have δ(c ∗ d) = (δc) ∗ d + (−1)n(c ∗ δd) (6.4)
Proof. The proof is obtained from that of theorem 6.1.1 by ignoring the first coordinate of most terms.
This shows that a k-cycle of BGX produces a homology operator of degree k on the rack homology and this operator only depends on the homology class of this cycle.
6.1.1 H1(BInnX)
In this section let G = Inn(X) The first homology group of BGX is easily computed using the Hurewicz theorem. We’ll choose the 0-cell (1) as a basepoint of BGX. Every path from this basepoint to some other 0-cell is homotopic to a path in the 1-skeleton of BGX . But since all edges are labeled by X and can be directed we see that such a path corresponds to an element of the free group on X. The 2-cells of BGX are sufficient to determine what paths are homotopic.
x y gρ(y) B / gρ(x)ρ(y) O O y y
x g / gρ(x)
Above is a picture of typical 2-cell of BGX. We see that this produces exactly the relations of the adjoint group. So every path from (1) to (g) is up to homotopy determined by some h ∈ Adj(X) with ρ(h) = g. In particular we find that π1(BGX, (1)) = K, where K = ker(ρ) : Adj(X) → Inn(X) is the center of Adj(X). So we find: ∼ Theorem 6.1.3. H1(BGX) = K ∼ Proof. Since π1(BGX, (1)) is abelian the Hurewicz theorem states that H1(BGX) = π1(BGX, (1)). We’ll make the isomorphism explicit.
Definition 6.1.5. As we saw above all the elements of πq(BGX, (1)) are cycles along the 1-cubes of BGX and so we only need to sum over those on a chain level. Let g ∈ K. Say g = et1 . . . etn where t ∈ {−1, 1}. Now the corresponding x1 xn i cycle n X c(g) = ti · ci i=1
54 t Where c is equal to (ρ(et1 , . . . e i−1 ), x ) if t = 1 and equal to (ρ(et1 , . . . eti ), x ) i x1 xi−1 i i x1 xi i if ti = −1. To see that this is indeed a cycle note that
δ(t · c ) = (ρ(et1 , . . . eti−1 )) − (ρ(et1 , . . . eti )) (6.5) i i x1 xi−1 x1 xi And since ρ(g) = 1, c(g) is indeed a cycle.
6.2 Homology operators of negative degree
In the last section we found that the homology of BInnX produced a lot of homology operators of positive degree. It’s also possible to use the cochain complex of BInnX to define operators going down in dimension.
R Definition 6.2.1. Let G = Inn(X). For x = (x1, . . . , xn, y1, . . . yk) ∈ Cn+k(X) G and f : Ck → Z define a rack n-chain by
X g−1 x ∗ f = f(g, y1, . . . yk) · (x1, . . . , xn) (6.6) g∈G
One would hope that this would define a homology operator preferably one that would only depend on the cohomology class of f. While this does define a homology operator, it sadly does not just depend on the cohomology class. However by switching to a slightly different chain complex it can be salvaged. The space BGX comes from the trunk TGX, a special case of the action rack trunk. We can also let X act on G × X and get another trunk with its own nerve and homology. We’ll denote this trunk by TGX X and its associated space with BGX X.
G Definition 6.2.2. Let G = Inn(X). Denote the chain groups of BGX X by Dn . That is
G n Dn = Z[(G × X) × X ] 0 δi (g, a, x1, . . . , xn) = (g, a, x1, . . . , xi−1, xi+1, . . . , xn) 1 xi xi xi δi (g, a, x1, . . . , xn) = (gρ(xi), a , (x1) ,..., (xi−1) , xi+1, . . . , xn)
k−1 R G ∼ G Definition 6.2.3. Let f ∈ DG and x ∈ Cn+k(X). Since Dk = Ck+1 we can still use formula (6.6) to define x ∗ f.
R Theorem 6.2.1. Let G = Inn(X). For x = (x1, . . . , xn, y1, . . . , yk) ∈ Cn+k(X) k−1 and f ∈ DG we have δ(x ∗ f) = (δx) ∗ f + (−i)n−1(x ∗ δf)
55 Proof. First observe that:
n X X i g−1 δ(x ∗ f) = f(g, y1, . . . , yk) · (−1) (x1, . . . , xi−1, xi+1, . . . , xn) g∈G i=1 g−1 −(x1 B xi, . . . xi−1 B xi, xi+1, . . . , xn) X g−1 = f(g, y1, . . . , yk) · (δ(x1, . . . , xn)) g∈G
We also have
56 n X ih (δx) ∗ f = (−1) (x1, . . . , xi−1, xi+1, . . . , xn, y1, . . . , yk) ∗ f i=1 i −(x1 B xi, . . . , xi−1 B xi, xi+1, . . . , xn, y1, . . . , yk) ∗ f k n X h +(−1) (x1, . . . , xn, y1, . . . , yi−1, yi+1, . . . , yk) ∗ f i=1 i −(x1 B yi, . . . , xn B yi, y1 B yi, . . . , yi−1 B yi, yi+1, . . . , yk) ∗ f X g−1 = f(g, y1, . . . , yk) · (δ(x1, . . . , xn)) g∈G k n X X i h g−1 +(−1) (−1) · f(g, xn, y1, . . . , yi−1, yi+1, . . . , yk) · (x1, . . . , xn−1) i=1 g∈G −1 i yi yi yi yi yi g −f(g, (xn) , (y1) ,..., (yi−1) , yi+1, . . . , yk) · ((x1) ,..., (xn−1) )
X g−1 = f(g, y1, . . . , yk) · (δ(x1, . . . , xn)) g∈G k n X X i g−1 +(−1) (−1) · f(g, xn, y1, . . . , yi−1, yi+1, . . . , yk) · (x1, . . . , xn−1) i=1 g∈G k n X X i yi yi yi −(−1) (−1) f(gρ(yi), (xn) , (y1) ,..., (yi−1) , yi+1, . . . , yk) i=1 g∈G −1 yi yi (gρ(yi)) ·((x1) ,..., (xn−1) ) X g−1 = f(g, y1, . . . , yk) · (δ(x1, . . . , xn)) g∈G k n X X i g−1 +(−1) (−1) · f(g, xn, y1, . . . , yi−1, yi+1, . . . , yk) · (x1, . . . , xn−1) i=1 g∈G k n X X i yi yi yi −(−1) (−1) f(gρ(yi), (xn) , (y1) ,..., (yi−1) , yi+1, . . . , yk) i=1 g∈G g−1 ·(x1, . . . , xn−1) n X g−1 = δ(x ∗ f) + (−1) δ(f)(g, xn, y1, . . . , yk) · (x1, . . . , xn−1) g∈G = δ(x ∗ f) + (−1)n(x ∗ δf)
Once again this shows that for every k-cocycle f we have an homology operator on the rack homology of degree −k − 1 that only depends on the
57 cohomology class of f.
0 6.2.1 H (BGX X) G Homology operators of degree -1 are given by 0-cocycles. So let f : D0 → Z be such a cocycle. Since δ(f) = 0 it follows that 0 = δ(f)(g, x, y) = f(g, x) − f(gρ(y), x B y) and therefore f(g, x) = f(gρ(y), x B y). Since the ρ(x) generate Inn(X) we see that for all h ∈ Inn(X) we have f(g, x) = f(gh, xh). So it follows that f is completely determined by its values on f(1, x) for all x ∈ X and that 0 G these values can be freely chosen. So dim H (D∗ ) = #X. G g Definition 6.2.4. Define fx : D0 → Z by setting fx(g, y) = 1 whenever x = y and fx(g, y) = 0 otherwise.
0 G As mentioned above the fx form a basis of H (D∗ ). Theorem 6.2.2. Let x, y ∈ X be two elements of the same connected compo- nent. Now fx and fy induce the same homology operator.
h Proof. First find a h ∈ Inn(X) with x = y. Now notice that fy(g, a) = G fx(hg, a) for all g ∈ G and a ∈ X. Now for any (x1, . . . , xn) ∈ Cn this gives us
X g−1 (x1, . . . , xn) ∗ fy = fy(g, xn) · (x1, . . . xn−1) g∈G X g−1 = fx(hg, xn) · (x1, . . . , xn−1) g∈G X (hg)−1h = fx(hg, xn) · (x1, . . . , xn−1) g∈G h = ((x1, . . . , xn) ∗ fx)
Going back to the homology of BGX for a while. We have simple boundaries of the form δ(g, x) = (g) − (gρ(x)). This shows that (1) − (g) is a boundary in G G g C0 for all g ∈ G. Now for a n-cycle c ∈ Cn we have that c ∗ ((1) − (g)) = c − c . g h So c and c are homologous for all g. Since c ∗ fy = (c ∗ fx) we conclude that fy and fx do in fact induce the same homology operator. Definition 6.2.5. For a connected component c of a rack X define a function c : Adj(X) → Z by setting c(ex) = 1 when x ∈ c and c(ex) = 0 otherwise. Theorem 6.2.3. Let x ∈ X and let [x] be the connected component containing R x. For g ∈ K and d ∈ Cn (X) an n-cycle we have that (d∗c(g))∗fx is homologous to [x](g) · c.
Proof. Like in definition 6.1.5 let g = et1 . . . etn and x1 xn
n X c(g) = ti · ci i=1
58 Where ci is as in definition 6.1.5. We’ll prove that (d ∗ ti · ci) ∗ fx is homologous to ti · d when xi ∈ [x] and zero h0 otherwise. Recall that ci = (h, xi) for some h ∈ Inn(X). So (d∗ci)∗fx = d for 0 some h ∈ Inn(X) if and only if xi ∈ [x] and (d ∗ ci) ∗ fx = 0 otherwise. Like in 0 the last proof dh is homologous to d. Summing over i concludes the proof.
6.3 Variations 6.3.1 A different G
In the last few sections we focused on G = Inn(X). But the spaces BGX and BGX X exist for all augmentations. In order for the proofs to go smoothly, all φy that is required is that x = x B y which holds in every augmentation. The augmentation adj : X → Adj(X) is the universal augmentation of X (c.f. [7]). The map ρ : Adj(X) → Inn(X) induces a continuous function BAdjX → BInnX. Since ρ is surjective it in facts induces a covering. For more ∼ details about this see [9]. In the same way we argued that π1(BInnX, (1)) = K we can see that the fundamental group of BAdjX vanishes and that BAdjX is the universal covering of BInnX. Since the action of Adj(X) on X factors through Inn(X) the homology of BAdjX won’t provide any more homology operators than BInnX already did. Using Aut(X) offers a few more operators than Inn(X) but essentially only in dimension 0. BAutX has one connected component for every element of Aut(X)/Inn(X) and all of these components are homeomorphic to BInnX. So the only extra in dimension 0 is an operator for every element of Aut(X)/Inn(X). But in higher dimensions the only extra we get are cycles of the form (g) ∗ c where g ∈ Aut(X) and c is a cycle of BInnX viewed as a subspace of BAutX. Every other augmentation with an arbitrary group G has the action of G factor through Aut(X). And therefore it doesn’t offer more than Aut(X) already does.
6.3.2 Quandle homology We’ve only been concerned with rack homology so far. Like the rack complex R D G G C∗ (X) has a degeneracy subcomplex C∗ (X) so do the C∗ and D∗ . Definition 6.3.1. G(D) G Cn = Z[(g, x1, . . . , xn) ∈ Cn |xi = xi+1 for some i] G(D) G Dn = Z[(g, x1, . . . , xn+1) ∈ Dn |xi = xi+1 for some i] G The product on C∗ is compatible with this. If either c or d is degenerate then so is c∗d. So we could study the quandle homology operators coming from G G(D) the homology of C∗ /C∗ . G G G(D) For D∗ this doesn’t work. Even if f : Dk → Z is zero on Dk it does not D D mean that if c ∈ Cn+k+1 then c ∗ f ∈ Cn . This goes wrong when xn = xn+1 is the only place where (x1, . . . , xn+k+1) has two equal terms in a row.
59 6.4 Relation to the literature
In [19] M. Niebrzydowski and J.H. Przytycki have studied the homology of dihedral quandles. In particular they gave some homology operators on the rack and quandle homology of these quandles. 0 The first operator they’ve described is ha for a ∈ Rk. It is defined as follows Definition 6.4.1.
0 ha(x1, . . . , xn) = (x1, . . . , xn, a) + (x1 B a, . . . , xn B a, a) (6.7) 2 In our terms this is the operator coming from c(ea) ∈ H1(BInnRk). Niebrzy- 0 dowski and Przytycki also provide a homology operator ha going down one 0 0 dimension. They find that ha ◦ ha is chain homotopic to 4 · id. They note that 0 ha only works on rack homology. This is to be be expected since some study 0 0 Inn shows that ha is equal to twice the map we get from fa ∈ H (D∗ ). Combining this with theorem 6.2.3 gives the same result. Since we know that Rp is a connected alexander quandle for odd primes p, ∼ we know that Adj(Rp) = Z × Rp × GT (Zp). Since ∼ GT (Zp) = Zp ⊗ Zp/hx ⊗ y ∼ −y ⊗ xi = 1 (6.8) 2 we see that ea indeed generates K so that the homology BInnRp does not provide any other operators of degree 1. The second operator is one of degree 2.
Definition 6.4.2. Let x0, . . . , xk−1 ∈ X such that xi+2 = xi B xi+1 for all i where the indices are taken modulo k. Define hs : Cn(X) → Cn+2(X) by
k+1 X hs(y1, . . . , yn) = (y1, . . . , yn, xi, xi+1) (6.9) i=0 where again xk stands for x0. Again we can use our techniques to describe this. It’s not hard to check that
k−1 X Inn (1, xi, xi+1) ∈ C2 (6.10) i=0 is a cycle. Although Niebrzydowski and Przytycki claim that hs is injective for some examples they don’t provide an operator of degree −2 to prove so in general. A 1 Inn little further on we’ll see that H (D∗ ) doesn’t provide this either for R3.
6.4.1 Some computer calculations
I’ve done some computer calculations on the homology of BInnR3 and the co- homology of BInnR3 R3 in the hope of finding more homology operators. In [19] the authors expected to find an operator of degree 4 in particular.
60 Q ∼ Conjecture 6.4.1. (Niebrzydowski and Przytycki) For any odd prime p, tor(Hn (Rp)) = fn Zp . Where fn is given by the recursive relation fn+3 = fn+2 + fn with f1 = f2 = 0 and f3 = 1. Q Q If an operator h4 : Hn (Rp) → Hn+4(Rp) exists that is sufficiently nice this could prove the conjecture. I’ve computed Hn(BInnR3; A) for A = Z and n ≤ 5 and for A = F3 and 0 0 n ≤ 6. For A = Z it turns out that ha and hs and their products generate every homology group up to n = 5. When A = F3 there is an extra cycle in dimension 3. Everything up to dimension 6 is again generated by the products 0 0 of this cycle, ha and hs. Either way there is nothing to be found in dimension 4. This doesn’t mean that there can be no such operator but if it exists it will not come from the homology of BGX. 0 In the last section we already saw that ha is injective for odd p. Since 1 Inn H (D∗ ) provides us with more operators going down in dimension we could 0 1 Inn hope to be able to prove hs injective. The computer tells us H (D∗ ) is three dimensional. One of the generators is given by f : Z[Inn × R3 × R3] → Z. f is 1 on the following list and zero on everything else. (id, 0, 0) (id, 0, 1) (id, 0, 2)
(g1, 1, 0) (g1, 1, 1) (g1, 1, 2)
(g2, 2, 0) (g2, 2, 1) (g2, 2, 2)
Where g1 = inn(0)inn(2) and g2 = inn(0)inn(1). The other two generators are given by cyclicly permuting id, g1 and g2. We see that −1 −1 0 g1 g2 (c ∗ hs) ∗ f = c + c + c (6.11) But this is homologous to three times c when c is a cycle. But since 3 kills all the torsion of the homology of R3 (c.f. [19]) this isn’t very useful.
6.5 Some other operators
Another special case of the action rack space comes from the action of X on itself. The trunk TX X we obtain from this is called the extended rack trunk in [9]. n+1 Definition 6.5.1. The chain group Cn(BX X) is generated by X with boundary operators given by
0 δi (y, x1, . . . , xn) = (y, x1, . . . , xi−1, xi+1, . . . , xn) 1 δi (y, x1, . . . , xn) = (y B xi, x1 B xi, . . . , xi−1 B xi, xi+1, . . . xn) This is very similar to the chain complex of the rack itself. It’s immediately R obvious that Cn+1(X) = Cn(BX X). The only differences between the boundary R R operators is that δ : Cn+1(X) → Cn (X) contains 2 more terms than the other. 0 1 However these terms are δ1 and −δ1 and they cancel.
61 R Definition 6.5.2. Define a chainmap h : Cn+1(X) → Cn(BX X) by hn = (−1)nid. It’s trivial to very that this is in fact a chain map.
R ∼ So we find that Hn+1(X) = Hn(BX X). On the other the hand BX X covers BX . This covering is given by the map that forgets the first coordinate. This gives us the following chainmap
R Definition 6.5.3. Define p : Cn(BX X) → Cn (X) by
p(y, x1, . . . , xn) = (x1, . . . , xn)
Combining these two maps gives us an operator p0 of degree −1 given by p0 = p ◦ h
R Definition 6.5.4. Let C be a connected component of X. Define sC : Cn (X) → R Cn+1(X) by X sC (x1, . . . , xn) = (c, x1, . . . , xn) c∈C
Theorem 6.5.1. sC is a chain map. Proof.
n X i−1 X h δsC (x1, . . . , xn) = (−1) (c, x1, . . . , xi−1, xi+1, . . . , xn) i=1 c∈C i −(c B xi, x1 B xi, . . . , xi−1 B xi, xi+1, . . . , xn) n X X i−1 = (−1) (c, x1, . . . , xi−1, xi+1 . . . , xn) c∈C i=1 n X X i−1 (−1) (c, x1 B xi, . . . , xi−1 B xi, xi+1 . . . , xn) c∈C i=1
= sC δ(x1, . . . , xn)
The second equality follows from the fact that inn(xi)(C) = C since C is a connected component of X.
D D Like before sC sends Cn (X) into Cn+1(X) and therefore also works on quandle homology, whereas p fails to do so. It’s also clear that
0 n p sC (x) = (−1) #C · x
62 Chapter 7
The cup product of the rack space
63 7.1 The cup product of the realization of a -set In this section we’ll study the cup product on realizations of -sets such as the rack space. We’ll give a formula for the cup product and will prove this correct.
ε Definition 7.1.1. For any -set {Xn, δi } and A ⊂ {1, . . . , n} define δ0 = δ0 ◦ ... ◦ δ0 A a1 ak δ1 = δ1 ◦ ... ◦ δ1 A a1 ak
Where A = {a1, . . . , ak} and a1 < . . . < ak.
0 Example 7.1.1. Let X be a rack. If #A = k then δA(x1, . . . , xn) is an n − k 1 front face of (x1, . . . , xn) and using δA we’ll get a back face. If B = {1, . . . , n}\ A = {b1, . . . , bn−k} with b1 < . . . < bn−k then we get
0 δA(x1, . . . , xn) = (xb1 , . . . , xbn−k )
1 g1 gn−k δA(x1, . . . , xn) = ((xb1 ) ,..., (xbn−k ) ) where g = (e )ti,1 · ... · (e )ti,k and t is zero when a < b and one i xa1 xak i,j j i otherwise. 0 1 In words: δA removes the x’s with an index in A and δA also removes these x’s but first lets them operate on the ones with a smaller index. Example 7.1.2. Let A = {2, 4}. We have
0 δA(a, b, c, d, e) = (a, c, e) 1 δA(a, b, c, d, e) = (a B b B d, c B d, e) Definition 7.1.2. For A ⊂ {1, . . . , n} we’ll define (A), the sign of A as follows. Let A = a1, . . . , ak with a1 < . . . < ak and let B = {1, . . . , n}\ A. Again set B = {b1, . . . , bn−k} with b1 < . . . < bn−k. Now define (A) to be the sign of the permutation that maps (1, . . . , n) to (b1, . . . , bn−k, a1, . . . , ak). We’ll now give the formula for the cup product.
ε k Definition 7.1.3. Let X = {Xn, δi } be a -set and let f ∈ C (X) and g ∈ Cm(X) be cocycles. Set n = k + m. Define f ∪ g by
X 0 1 f ∪ g = (A)(f ◦ δA) · (g ◦ δB) A
Where B = {1, . . . , n}\ A and the sum goes over all subsets of {1, . . . , n} of cardinality m.
64 7.2 A triangulation of the realization of a -set ε In this section let X = {Xn, δi } be a -set and let B = ||X||. In order to prove that the formula for the cup product is correct we’ll triangulate B and use the well known formula for the cup product on simplicial sets and show it coincides with definition 7.1.3. We’ll triangulate every n-cube of the B in the same way. We’re not worried about doing this in an efficient manner and will use n! n-simplices per cube. So how do we triangulate the n-cube? We can walk along the edges of the cube from (0,..., 0) to (1,..., 1) in n steps in n! different ways. All of these paths have a convex hull that is an n-simplex. These are the simplices that we need to fill up the n-cube. To see this take one of these paths. We start at (0,..., 0) then go to (0,..., 1,..., 0) where the 1 is at position i1 and next we add a second one in position i2 etc. until we finally add a 1 at position in. Now the convex hull of this contains all points (x1, . . . , xn) with xi1 ≥ xi2 ≥ ... ≥ xin . So the simplices fill the entire cube. The intersection of one or more of them them are only the points with 2 or more equal coordinates. These intersections make up the lower dimensional simplices used in this triangulation. To make these simplices all we have to do is fill in the diagonals of the cube. We apply this triangulation to every n-cube of B. Definition 7.2.1. Define n = {1, . . . , n}
Definition 7.2.2. An ordered k-partition of n is a sequence of sets (S1| ... |Sk) S with Si ∩Sj = ∅ if i 6= j and Si = n. We’ll omit the curly brackets when writ- ing down explicit ordered partitions. So instead of ({1}|{3}|{2, 4}) we simply write (1|3|2, 4).
` n Definition 7.2.3. Recall that B = Xn×I / ∼. For every ordered k-partition of n and x ∈ Xn the triangulation of B will contain a k-simplex (x; S1| ... |Sk) that is given by the set of points (x, t1, . . . , tn) ∈ B with ti ≥ tj iff ti ∈ Si0 and 0 0 tk ∈ Sj0 with i ≤ j . The boundary operator is given as follows
δ (x; S | ... |S ) = (δ1 (x); f(S )| ... |f(S )) 0 1 k S1 2 k δi(x; S1| ... |Sk) = (x; S1| ... |Si−1|Si ∪ Si+1|Si+1| ... |Sk) for 0 < i < k δ (x; S | ... |S ) = (δ0 (x); g(S )| ... |g(S )) k 1 k Sk 1 k−1 k X i δ = (−1) δi i=0
Where f is the unique order preserving function from S2 ∪ ... ∪ Sk to n − #S1. And g is the unique order preserving function from S1 ∪ ... ∪ Sk−1 to n − #Sk.
Definition 7.2.4. For x ∈ Xn and σ ∈ Sn define
σ(x) = (x; σ(1)| ... |σ(n))
65 Note that for all σ ∈ Sn the simplex σ(x) is exactly one of the simplices of the triangulation of the cube. So the cube consists of all these simplices. On the chain level we need to keep track of the orientations of the simplices.
Definition 7.2.5. For x ∈ Xn define X S(x) = (σ)σ(x)
σ∈Sn We’ll verify that this is ok with respect to the boundaries. Lemma 7.2.1. δ(S(x)) = S(δ(x))
Proof. For all σ ∈ Sn and 0 < i < n one of σ(x) and (i i + 1)σ(x) looks like (... |a|b| ...)x while the other looks like (... |b|a| ...)x. Therefore the δi of these two terms cancels in δ(S(x)). So we get that
n δ(S(x)) = δ0(S(x)) + (−1) δn(S(x))
Now consider the following disjoint partition of Sn. [ Sn = {σ ∈ Sn|σ(1) = i} i
We’ll call these sets Ti. Now for x ∈ Xn and σ ∈ Ti we have
1 1 δ0(σ(x)) = (δi (x); fiσ(2)| ... |fiσ(n)) = τ(δi (x)) where ( j j < i fi(j) = j − 1 j > i and τ ∈ Sn−1 is the permuation that maps j to fiσ(j + 1). In order to find out the sign of τ we can consider τ to be in Sn and then observe
τ = (i i + 1 . . . n)−1σ(1 2 . . . n)
So we find that (τ) = (−1)n−i(σ)(−1)n−1 = (−1)i−1(σ). Therefore X δ0 (σ)σ(x)
σ∈Ti X = δ0 (x; σ)(i|σ(2)| ... |σ(n))
σ∈Ti i−1 X 1 = (−1) (τ)(δi (x); τ(1)| ... |τ(n − 1))
τ∈Sn−1 i 1 = −(−1) S(δi (x))
66 Now we split up Sn up by looking what σ(n) is. So set
Ri = {σ ∈ Sm|σ(n) = i}
The union of these sets is again Sn. This time notice that for x ∈ Xn and σ ∈ Ri we have
0 0 0 δn(σ(x)) = (δi (x); fiσ(1)| ... |fiσ(n − 1)) = τ (δi (x))
0 where fi is as above and τ ∈ Sn−1 is the permuation that maps j to fiσ(j). This time we find (τ 0) = (−1)n−i(σ) and so X δn (σ)σ(x)
σ∈Ri X = δn (σ)(x; σ(1)| ... |σ(n − 1)|i)
σ∈Ri n−i X 0 0 0 = (−1) (τ )τ (δi (x)) 0 τ ∈Sn−1 n i 0 = (−1) (−1) S(δi (x))
Summing over i concludes the proof.
7.3 The cup product
ε Again let X = {Xm, δi } and let B = ||X||. Also let T denote the simplical set of the triangluation of B we created in the last section and let B0 = ||T || denote it’s realization. For simplicial sets the formula for the cup product is well known (e.g. [16]). In our case f is a k-cochain and g an m-cochain and n = k + m. Then
0 1 (f ∪g)((x; S1| ... |Sn)) = f((δAx; p(S1)| ... |p(Sk)))·g((δBx; q(Sk+1)| ... |q(Sn))) where A = Sk+1 ∪ ... ∪ Sn and B = S1 ∪ ... ∪ Sk = n \ A and p and q are the required order preserving functions. Since the spaces B and B0 are homeomorphic and the formula above comes from a continuous function homotopic to the diagonal, we need to fill in S(x) into the formula above to verfiy that definition 7.1.3 is correct. Theorem 7.3.1. The formula given in 7.1.3 is the cup product of the rack space. Proof. Let f be a k-cochain and g an m-cochain and set n = k + m. The proof again relies on partition of Sn. For an m-subset A ⊂ n define n o TA = σ ∈ Sn|{σ(k + 1), . . . , σ(n)} = A
67 S Again we have that Sn = A TA where the union goes over all m-subsets of {1, . . . , n}. Let A = {ai} with a1 < . . . < am and let n\A = B = {bj} with b1 < . . . < bk. Let τ be the permutation that sends (1, . . . , n) to (b1, . . . , bk, a1, . . . , am). Now note TA = {σ1σ2τ|σ1 ∈ S(B), σ2 ∈ S(A)} Now for any fixed A we have ! X (f ∪ g) (σ)σ(x)
σ∈TA X X = (f ∪ g) (A)(σ1)(σ2)σ1σ2τ(x)
σ1∈S(B) σ2∈S(A) X X 0 1 = (A) (σ1)f(σ1(δA(x)))(σ2)g(σ2(δB(x)))
σ1∈Sk σ2∈Sm 0 1 = (A)f(S(δA(x1, . . . , xn)))g(S(δB(x1, . . . , xn)))
Summing over all A completes the proof.
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