Andruskiewitch, Grana - From Racks to Pointed Hopf Algebras.pdf

http://www.elsevier.com/locate/aimAdvances in Mathematics 178 (2003) 177243

From racks to pointed Hopf algebras$

Nicolas Andruskiewitscha,* and Matas Granab,c

aFacultad de Matematica, Astronoma y Fsica, Universidad Nacional de Cordoba,

CIEM-CONICET, Ciudad Universitaria, 5000 Cordoba, ArgentinabMIT, Mathematics Department, 77 Mass. Ave., 02139 Cambridge, MA, USA

cFCEyN-UBA, Pab. I-Ciudad Universitaria, 1428 Buenos Aires, Argentina

Received 9 February 2002; accepted 22 May 2002

Communicated by Pavel Etingof

Abstract

A fundamental step in the classication of nite-dimensional complex pointed Hopf

algebras is the determination of all nite-dimensional Nichols algebras of braided vector

spaces arising from groups. The most important class of braided vector spaces arising from

groups is the class of braided vector spaces CX ; cq; where X is a rack and q is a 2-cocycle onX with values in C: Racks and cohomology of racks appeared also in the work oftopologists. This leads us to the study of the structure of racks, their cohomology

groups and the corresponding Nichols algebras. We will show advances in these three

directions. We classify simple racks in group-theoretical terms; we describe projections of

racks in terms of general cocycles; we introduce a general cohomology theory of racks

containing properly the existing ones. We introduce a Fourier transform on racks

of certain type; nally, we compute some new examples of nite-dimensional Nichols

algebras.

r 2003 Elsevier Inc. All rights reserved.

MSC: Primary 16W30; Secondary 17B37; 52M27

Keywords: Pointed Hopf algebras; Racks; Quandles

ARTICLE IN PRESS

$This work was partially supported by ANPCyT, CONICET, Agencia Cordoba Ciencia, Fundacion

Antorchas, Secyt (UBA) and Secyt (UNC).

*Corresponding author.

E-mail address: [email protected] (N. Andruskiewitsch).

0001-8708/03/$ - see front matter r 2003 Elsevier Inc. All rights reserved.

doi:10.1016/S0001-8708(02)00071-3

0. Introduction

1. This paper is about braided vector spaces arising from pointed Hopf algebras,and their Nichols algebras. Our general reference for pointed Hopf algebras is[AS2]. wE SHALL work over the eld C of complex numbers; many resultsbelow are valid over more general elds. We denote by GN the group of roots ofunity of C:

2. The determination of all complex nite-dimensional pointed Hopf algebras Hwith group of group-likes GH isomorphic to a xed nite group G is stillwidely open. Even the existence of such Hopf algebras H (apart from thegroup algebra CG) is unknown for many nite groups G: If G is abelian,substantial advances were done via the theory of quantum groups at roots ofunit [AS1]. The results can be adapted to a non-necessarily abelian group G: ifaij1pi;jpy is a nite Cartan matrix, g1;y; gy are central elements in G; andw1;y; wy are multiplicative characters of G; such that wigjwjgi wigiaij (plussome technical hypotheses on the orders of wigj), then a nite-dimensionalpointed Hopf algebra H with group GHCG can be constructed from thisdatum. Besides these, only a small number of examples have appeared in print;in these examples, G is S3; S4; D4 (see [MS]), S5 (using [FK]), A4 Z=2(see [G1]).

3. An important invariant of a pointed Hopf algebra H is its infinitesimal braiding;this is a braided vector space, that is, a pair V ; c; where V is a vector space andcAAutV#V is a solution of the braid equation: c#idid#cc#id id#cc#idid#c: Let BV be the Nichols algebra of V ; c; see [AS2]. Ifdim H is nite, then dimBV is nite and divides dim H: Conversely, given abraided vector space V ; c; where V is a YetterDrinfeld module over the groupalgebra CG; then the Radfords biproduct, or bosonization, H BV#CG is apointed Hopf algebra with GHCG: Thus, a fundamental problem is todetermine the dimension of the Nichols algebra of a nite-dimensional braidedvector space. We remark that the same braided vector space can arise as theinnitesimal braiding of pointed Hopf algebras H with very different GH; as inthe examples with Cartan matrices above. Analogously, there are innitely manynite-dimensional pointed Hopf algebras H; with non-isomorphic groups GHand the same innitesimal braiding as, respectively, the examples above relatedto the groups S3; S4; D4; S5; A4 Z=2; and such that they are link-indecomposable in the sense of [MS,M].

4. It is then more convenient to study in a rst stage Nichols algebras ofbraided vector spaces of group-type [G1, Denition 1.4.10]. However, thereare strong constraints on a braided vector space of group-type to have anite-dimensional Nichols algebra [G1, Lemma 3.1]. Briey, we shall considerin this paper the class of braided vector spaces of the form CX ; cq;where X ; x is a nite rack, q is a 2-cocycle with values on the multiplicativegroup of invertible elements of C; and cq is given by cqi#j qij ix j#i (seethe precise denitions in the main part of the text). See also the discussion in[A, Chapter 5].

ARTICLE IN PRESSN. Andruskiewitsch, M. Gra *na / Advances in Mathematics 178 (2003) 177243178

We are faced with the following questions: to determine the general structureof nite racks; to compute their second cohomology groups; and to decidewhether the dimensions of the corresponding Nichols algebras are nite.

5. We now describe the contents of this paper. The notions of rack andquandle, sets provided with a binary operation like the conjugation of agroup, have been considered in the literature, mainly as a way to produce knotinvariants (cf. [B,CJKLS,De,FR,J1,K,Ma]). Section 1 is a short survey of thetheory of racks and quandles, addressed to non-specialists on these structures,including a variety of examples relevant for this paper.The determination of all nite racks is a very hard task. There are two

successive approximations to this problem. First, any nite rack is a union ofindecomposable components. However, indecomposable racks can be puttogether in many different ways, and the description of all possible ways isagain very difcult. In other words, even the determination of all indecompo-sable nite racks would not solve the general question.In Section 2, we describe epimorphisms of racks and quandles by general

cocycles. We then introduce modules over a quandle, resp. a rack, X : We showthat modules over X are in one-to-one correspondence with the abelian groupobjects in the category of arrows over X ; if X is indecomposable. Our denitionof modules over X generalizes those in [CES,CENS].We say that a non-trivial rack is simple if it has no proper quotients. Then any

indecomposable nite rack with cardinality41 is an extension of a simple rack.We study simple racks in Section 3. One of our main results is the explicitclassication of all nite simple racks, see Theorems 3.9 and 3.12. The proof isbased on a group-theoretical result kindly communicated to us by Guralnick.After acceptance of this paper, it became to our attention Joyces article [J2],

where results similar to those in Section 3.2 are obtained. However, notice thatour classication in Theorem 3.9 includes more quandles than that of [J2,Theorem 7(2)]. This is one of the reasons why we decided to leave our results.The other reason is that, by using a result in [EGS] (which depends upon theclassication of simple groups), we can split the simple quandles into two classesregarding their cardinality. These classes appear naturally in [J2] also, but thefact that they are split by cardinality was impossible to prove in 1982.It is natural to dene homology and cohomology theories of racks and

quandles as standard homology and cohomology theories for abelian groupobjects in the category of arrows over X [Q]. We propose in Section 4 a complexthat, conjecturally, would be suitable to compute these homology andcohomology theories. We show that the homology and cohomology theoriesknown so far (see [CENS,CJKLS,FR,G1]) are special cases of ours. We discussas well non-abelian cohomology theories.A braided vector space of the form CY ; cq does not determine the rack Y and

the 2-cocycle q: We provide a general way of constructing two braided vectorspaces CY ; cq and CY; cq where the racks Y and Y are not isomorphic ingeneral, but such that the corresponding Nichols algebras have the samedimension. In our construction, Y is an extension X a A; where A is an

ARTICLE IN PRESSN. Andruskiewitsch, M. Gra *na / Advances in Mathematics 178 (2003) 177243 179

X -module; Y is an extension X b A; where A; the group of characters of A; is alsoan X -module. The construction can be thought as a Fourier transform. We showhow to use this construction to obtain new examples of pointed Hopf algebras withnon-abelian group of group-like elements. This is the content of Section 5.In Section 6, we present several new examples of nite-dimensional Nichols

algebras BV over nite groups. First, we show that some Nichols algebras canbe computed by reduction to Nichols algebras of diagonal type, via Fouriertransform. Next we use Fourier transform again to compute a Nichols algebrarelated to the faces of the cube, starting from a Nichols algebra related to thetranspositions of S4 computed in [MS]. Finally, we establish some relations thathold in Nichols algebras related to afne racks, and use them to compute Nicholsalgebras related to the vertices of the tetrahedron (a result announced in [G1]) andthe afne rack Z=5; x 2: Support to our proofs is given by Theorem 6.4 whichgives criteria to insure that a nite dimensional braided Hopf algebra is a Nicholsalgebra.In most of the paper, we shall only consider nite racks, or quandles, or crossed

sets, and omit the word nite when designing them, unless explicitly stated.6. In conclusion, we remark that the next natural step in the classication of nite-

dimensional pointed Hopf algebras is to deal with Nichols algebras of braidedvector spaces arising from simple racks.

1. Preliminaries

1.1. Racks, quandles and crossed sets

Denition 1.1. A rack is a pair X ; x where X is a non-empty set and x :X X-X is a function, such that

fi : X-X ; fi j ix j; is a bijection for all iAX ; 1:1

ix jx k ix jx ix k 8i; j; kAX : 1:2

A quandle is a rack X ; x which further satises

ix i i; for all iAX : 1:3

A crossed set is a quandle X ; x which further satises

jx i i whenever ix j j: 1:4

A morphism of racks is dened in the obvious way: c : X ; x -Y ; x is amorphism of racks if cix j cixc j; for all i; jAX : Morphisms of quandles(resp. crossed sets) are morphisms of racks between quandles (resp. crossed sets).


In particular, a subrack of a rack X ; x is a non-empty subset Y such thatY xY Y : If X is a crossed set and Y is a subrack, then, clearly, it is a crossedsubset; same for quandles.

Denition 1.2. If G is a group, any non-empty subset XDG stable under conjugationby G (i.e, a union of conjugacy classes) is a crossed set with the structure ix j iji1:A crossed set isomorphic to one of these shall be called standard.

A primary goal is to compare arbitrary crossed sets with standard ones.

Denition 1.3. Let X ; x be a rack and let SX denote the group of symmetries ofX : By (1.1), we have a map f : X-SX : Let X ; x be a rack. We set

Autx X : fgASX : gix j gix g jg;

Innx X : the subgroup of SX generated by fX:

By (1.2), Innx X is a subgroup of Autx X : On the other hand, it is easy to seethat

gfxg1 fgx; 8gAAutx X; xAX : 1:5

Therefore, fX CAutx X is a standard crossed subset, f : X-Autx X is amorphism of racks, and Innx X is a normal subgroup of Autx X: It is not true ingeneral that Innx X Autx X:

Example 1.4. Let X f7i;7jg; a standard subset of the group of units of thequaternions. Then Innx X aAutx X:

Proof. It is easy to compute Autx X and Innx X: One sees that Innx X /fi;fjSCC2 C2 and Autx X /fi;fj; s0S has order 8, where s07i 7jand s07j 7i: &

Another basic group attached to X is the following one:

Denition 1.5 (Brieskorn [B], Fenn and Rourke [FR], Joyce [J1]). Let X ; x be arack. We dene the enveloping group of X as

GX FX=/xyx1 xx y; x; yAXS;

where FX denotes the free group generated by X : The assignment x/fx extendsto a group homomorphism pX : GX-Innx X; the kernel of pX is called the defectgroup of X in the literature and coincides with the subgroup G considered in [So,Theorem 2.6].


The name enveloping group is justied by the following fact, containedessentially in [J1]:

Lemma 1.6. The functor X/GX is left adjoint to the forgetful functor H/FH fromthe category of groups to that of racks. That is,

HomgroupsGX ; HCHomracksX ;FH

by natural isomorphisms.

Proof. Easy. &

The denition of rack was proposed a long time ago by Conway and Wraith, seethe historical account in [FR]. Quandles were introduced independently in [J1,Ma]and studied later in [B] and other articles. They are being extensively studiednowadays in relation with knot invariants, see [CS] and references therein. In [G1], itwas proposed to consider crossed sets with the normalizing conditions (1.3) and(1.4); the conditions also appear in [So]. It is worth noting that in most of thesearticles the quandle structure is the opposite to the one here (i.e., x*y for our

f1y x). Racks, quandles and crossed sets are related as follows.

1.1.1. From racks to quandles

We follow [B]. Let X be a rack and let Cx X be the centralizer in Autx X ofInnx X: For cACx X; dene x c by

a x cb axcb cax b caxcb: 1:6

Then X ; x c is again a rack; we say that it is conjugated to X ; x via c:Let now i : X-X be given by ax ia a; which is well dened by (1.1). Then, by

(1.2),

ax b ax bx ib ax bx ax ib;

hence ax ib iax b: In particular, a iax a; and i is surjective. Also,ax iax ib ax iax ax ib ax ax ib;

so that, by (1.1), iax ib ax ib iax b: Suppose now that ia ib:Then

a ax ia iax ia ibx ib b:

That is, i is injective, and belongs to Cx X : We can then consider X ; x i; whichis a quandle.In conclusion, any rack X ; x is conjugated to a unique quandle, called the

quandle associated to X ; x . If F : X-Y is a morphism of racks, then F i iF ;


hence F is a morphism of the associated quandles. It follows thatAutx XCAut x iX : (But Innx X and Inn x iX may be very different).

1.1.2. From quandles to crossed sets

We exhibit a functor Q from the category of nite quandles to that of crossed sets,which assigns to a quandle X a quotient crossed set QX with the expected universalproperty: any morphism of quandles X-Y is uniquely factorized through QXwhenever Y is a crossed set. To see this, take on X B as the equivalence relationgenerated by

xBx0 if (y s:t: xx y y and yx x x0: 1:7Then B coincides with the identity relation if and only if X is a crossed set. TakeX1 : X=B: We must see that X1 inherits the structure of a quandle. First, supposex; x0; y are as in (1.7). Then x0x y yx xx yx y yx xx y y: Next, forzAX ; we have

yx x0x z x0x yx x0x z x0x yx z

yx xx yx z yx xx z;whence we see that fx fx0 ; and then fx fx00 for any x00Bx: Thus, it makes senseto consider x : X=B X-X : Finally, we have for zAX ;

zx xx zx y zx xx y zx yand

zx yx zx x zx yx x zx x0:Then zx xBzx x0; and it makes sense to consider x : X=B X=B-X=B: If X1 is not a crossed set then take X2 X1=B; and so on. SinceX is nite, we must eventually arrive to a crossed set. The functoriality and universalproperty are clear.

1.2. Basic definitions

We collect now a number of denitions and results; many of them appear alreadyin previous papers on racks or quandles, see [B,CJKLS,FR,J1].We shall say that X is trivial if ix j j for all i; jAX :

Lemma 1.7. Let X ; x be a rack, H a group and j : X-H an injective morphism ofracks such that the image is invariant under conjugation in H (thus X ; x is actually acrossed set). Then the map *j : H-SX ; given by *jhx j1hjxh1; is a grouphomomorphism and its image is contained in Autx X :

Proof. Left to the reader. &


The assignment X/Innx X is not functorial in general; just take iAX withfiaid; the inclusion figCX does not extend to a morphism Innx fig-Innx Xcommuting with f: But we have:

Lemma 1.8. If p : X-Y is a surjective morphism of racks, then it extends to a grouphomomorphism Innx p : Innx X-Innx Y :

Proof. Let Innx p be dened on fX by Innx pfx fpx: It is welldened and it extends to a morphism of groups since p is surjective. &

We determine now the structure of Innx X when X is standard.

Lemma 1.9. Let H be a group and let XCH be a standard subset.

(1) Innx X CC=ZC; where C /XS is the subgroup of H generated by X ; whichis clearly normal.

(2) If X generates H then Innx XCH=ZH:(3) If H is simple non-abelian and Xaf1g; then H Innx X:

Proof. We prove (2); (1) and (3) will follow. By Lemma 1.7, we have a morphism

c : H-Autx X; whose image is Innx X : Now, hAkerc3hxh1 x 8xAX3hAZH: &

Corollary 1.10. Let X be a rack, let H be a group and let c : X-H be a morphism. IfZ/cXS is trivial, then c extends to a morphism C : Innx X -H:

Proof. If Y cX; then C : Innx X -Innx c

Innx Y C/YS+H: &

The map f : X-Innx X is not injective, in general.

Denition 1.11. We shall say that the rack X ; x is faithful when the correspondingf is injective. Observe that in this case X is a crossed set, since it is standard.

Remark 1.12. If X ; x is faithful then the center of Innx X is trivial. Moregenerally, if zAZInnx X ; then fzi fi; for all iAX :

Denition 1.13. A decomposition of a rack X ; x is a disjoint union X Y,Zsuch that Y and Z are both subracks of X : (In particular, both Y and Z are non-empty). X is decomposable if it admits a decomposition, and indecomposableotherwise.

The image of an indecomposable rack under a morphism is again in-decomposable.


We shall occasionally denote inx j : fni j; nAZ: The orbit of an element xAX isthe subset

Ox : fi71s x i71s1x yi711 x xy j i1;y; isAXg:That is, Ox is the orbit of x under the natural action of the group Innx X : (If X isnite, then Ox fisx is1x yi1x xy j i1;y; isAXg).

Lemma 1.14. Let X ; x be a rack, YaX a non-empty subset and Z X Y : Thenthe following are equivalent:

(1) X Y,Z is a decomposition of X :(2) Y xZDZ and ZxYDY :(3) X xYDY :

Proof. Easy. &

Lemma 1.15. Let X ; x be a rack. Then the following are equivalent:(1) X is indecomposable.(2) X Ox for all (for some) xAX :

Proof. Easy. &

Note that a standard crossed set XCH need not be indecomposable, even if itconsists of only one H-orbit. However, it is so when H is simple by Lemma 1.9.

Example 1.16. If XCH is a conjugacy class with two elements, then it is trivial ascrossed set. As another example, take A an abelian group and G the group ofautomorphisms of A; let H AsG; and let XCA be any orbit for the action of G:Let 1GAG be the unit and consider XCH as Xs1G: Then X is trivial.

Proposition 1.17. Any rack X is the disjoint union of maximal indecomposablesubracks.

Proof. Given YCX a subset, consider

Y 0 Y,Y xY ,Y1xY Y,fyx z j y; zAYg,fy1x z j y; zAYg:

Then Y 0*Y and any subrack of X containing Y contains Y 0: The subrack generatedby Y is thus

SnAN Y

n; where Y n1 Y n0 and Y 1 Y : This is the smallest subrackof X containing Y :For YCX ; we say that it is connectable if for any two elements y1; y2AY

there exist ue11 ;y; uenn ; where uiAY and eiAf71g 8i; such that y2

ue11 x ue22 x?uenn x y1 (here the intermediate elements ueii x uei1i1x?uenn x y1may not belong to Y ). Then it is easy to see that if Y is connectable then so is Y 0:


Hence, for Y connectable, the subrack generated by Y is connectable and, being asubrack, it is indecomposable. Also, since the union of intersecting indecomposablesubracks is connectable, we see that they generate an indecomposable subrack.Hence the indecomposable component of xAX ; the union of all indecomposablesubracks containing x; is an indecomposable subrack. Now, X is the disjoint unionof such components. &

Unlike the situation of Lemma 1.14, the indecomposable components may not bestable under the action of X : The case of two components is more satisfactorybecause we can describe how to glue two racks.

Lemma 1.18.

(1) Let Y ; Z be two racks and X Y0Z be their disjoint union. The following areequivalent:(a) Structures of rack on X such that X Y,Z is a decomposition.(b) Pairs s; t of morphisms of racks s : Y-Autx Z; t : Z-Autx Y such

that

yx tzu tsyzyx u; 8y; uAY ; zAZ; i:e:; fytz tsyzfy; 1:8

zx syw stzyzxw; 8yAY ; z; wAZ; i:e:; fzsy stzyfz: 1:9

(2) Assume that Y and Z are crossed sets and (1.8), (1.9) hold. Then X is a crossed setexactly when

syz z if and only if tzy y; 8yAY ; zAZ: 1:10

Proof. Left to the reader. &

If the conditions of the lemma are satised, we shall say that X is the amalgamatedsum of Y and Z: If s and t are trivial, we say that X is the disjoint sum of Y and Z:Clearly, one can dene the disjoint sum of any family of racks (resp. quandles,crossed sets).For example, let X be a rack (resp. quandle, crossed set) and set X 2 2X

X f1; 2g; this is a rack (resp. quandle, crossed set) with x; ix y; j xx y; j;and 2X X1,X2 is a decomposition, where Xi X fig: Note that fx;i fx;j;f is not injective. In an analogous way, we dene the crossed set nX ; for any positiveinteger n: More generally, we have

Example 1.19. Let X ; Y be two racks (resp. quandles, crossed sets). Then X Y is arack (resp. quandle, crossed set), with x; yx u; v xx u; yx v; this is the directproduct of X and Y in the category of racks (resp. quandles, crossed sets).


Lemma 1.20. Let X ; Y be two racks (resp. quandles, crossed sets). If X Y isindecomposable then X and Y are indecomposable. The converse is true if X or Y is aquandle.

Proof. Since the canonical projections X Y-X and X Y-Y are rackhomomorphisms, the rst statement is immediate. For the second one, let

x1; y1; x2; y2AX Y : As X ; Y are indecomposable, there exist ue11 ;y; uenn ;ve01

1 ;y; ve0mm where uiAX ; vjAY and ei; e0jAf71g 8i; j; such that x2

ue11 x ue22 x?uenn x x1 and y2 ve01

1 x ve02

2 x?ve0m

m x y1: Suppose X is a quandle.Adding if necessary at the end of the sequence in Y pairs y1; y

11 ; we may suppose

that mXn: Adding if necessary at the end of the sequence in X elements x1; we maysuppose that m n: Then,

x2; y2 ue11 ; ve01

1 x ue22 ; ve02

2 x?uenn ; ve0n

n x x1; y1: &

Let X be a rack, take f : X-Innx X as usual. Let us abbreviate Fy : f1y; aber of f: If X is a quandle, any ber Fy is a trivial subquandle of X :

Lemma 1.21. (1) For x; yAfX the fibers Fy and Fxx y have the same cardinality.(2) If X is an indecomposable crossed set, then the fibers of f all have the same

cardinality.

Proof. We claim that ixFyDFfi x y: Indeed, if jAFy; then fix j fifjf1i fiyf

1i fix y; the claim follows. Similarly, i1xFyDFf1i x y; hence (1). Now (2)

follows from (1). &

We give nally some denitions of special classes of crossed sets, following[J1].

Denition 1.22. Let X ; x be a quandle. We shall say that X is involutory if f2x idfor all xAX : That is, if xx xx y y for all x; yAX :We shall say that X is abelian if xxwx yx z xx yx wx z for all

x; y; w; zAX :

1.3. Examples

1.3.1. A rack which is not a quandle

Take X any set and fASX any function. Let xx y f y: This is a rack, and it isnot a quandle if faidX : This rack is called permutation rack.

1.3.2. A quandle which is not a crossed set

Take X fx;;g; xx7 8; f7 idX :


1.3.3. Amalgamated sums

Let Z be a rack; we describe all the amalgamated sums X Y,Z for Y f0; 1gthe trivial rack. Denote AutY f id;g: Let s; t be as in Lemma 1.18. First,Z should decompose as a disjoint union of subracks Z Z,Z; where tZ7 7: Second, (1.8) is equivalent to Z7 being stable by s0 and s1; and condition (1.9)reads

fzs0 s0fz; fzs1 s1fz; 8zAZ;

fzs0 s1fz; fzs1 s0fz; 8zAZ:

Another way to describe the situation is: Z Z,Z a disjoint union; let C /fx; xAZS be the group generated by fZ ; C /fxfy; x; yAZS; and letC /C; CS: Then s0; C 1ASZ; s1 fxs0f1x for any xAZ: If Z is aquandle then X is a quandle. If Z is a crossed set then X is a crossed set iffZ Zs0 ffixed points of s0g Zs1 :

1.3.4. Polyhedral crossed sets

Let PCR3 be a regular polyhedron with vertices X fx1;y; xng and center in 0.For 1pipn; let Ti be the orthogonal linear map which xes xi and rotates theorthogonal plane by an angle of 2p=r with the right-hand rule (pointing the thumb toxi), where r is the number of edges ending in each vertex. Then X ; x dened byxix xj Tixj is a crossed set. To see this, simply take G as the group of orthogonaltransformations of P with determinant 1 and notice that fT1;y; Tng is a conjugacyclass of G; whose underlying (standard) crossed set is isomorphic to X : It is evidentthat Innx X is isomorphic to the group generated by fT1;y; Tng: It is clear that toeach polyhedron also corresponds an analogous crossed set given by the faces; it isisomorphic to the crossed set given by the vertices of the dual polyhedron. It followsby inspection that the crossed sets of the vertices of the tetrahedron, the octahedron,the dodecahedron and the icosahedron are indecomposable, while that of the cubehas two components, each of which is isomorphic to the crossed set of the vertices ofthe tetrahedron. It is easy to see that in the indecomposable cases the group Innx Xcoincides with G: See Fig. 1.

1.3.5. Coxeter racks

Let V ;/;S be a vector space over a eld k; provided with an anisotropicsymmetric bilinear form /;S: Let vx u u 2/u;vS/v;vS v: Then V f0g; x is arack, as well as any subset closed for this operation. Particular cases of this arethe root systems of semisimple Lie algebras; the action fa coincides with theaction of waAW ; the Weyl group. To turn this into a quandle one can either quotientout by the relation vB v; or take the conjugated quandle V f0g; x i as in 1.1.1(see (1.6)). It is easy to see that v x i u 2/u;vS/v;vS v u; and then this also is acrossed set.


1.3.6. Homogeneous crossed sets

Let G be a nite group and let f :SG-FunG;SG be the function given by

fsxy syx1x; sASG; x; yAG:

Then fsxftx fstx ; i.e., f is a morphism of groups, with the pointwise multiplication

in FunG;SG:If, in addition, s : G-G is a group automorphism, then dene xx y fsxy

syx1x: It is easy to see that this makes G; x into a crossed set. For instance, letus check (1.4):

xx y y 3 y syx1x 3 yx1 syx1 3 xy1 sxy1 3 yx x x:

We shall say that G; x is a principal homogeneous crossed set, and we will denote itby G; s:Let t :G-G; tx sx1x; so that xx y sy tx: It is clear that

fx fz 3 tx tz; 1:11

whence the bers as a crossed set are the same as the bers of t: Note that t is a grouphomomorphism if and only if ImtDZG; the center of G:More generally, let HCGs be a subgroup, where Gs is the subgroup of elements of

G xed by s: Then H\G is a crossed set, with HxxHy Hsyx1x; it is called ahomogeneous crossed set.It can be shown that a crossed set X is homogeneous if and only if it is a single

orbit under the action of Autx X [J1].

1.3.7. Twisted homogeneous crossed sets

In the same vein, let G be a group and sAAutG: Take xx y xsyx1: This is acrossed set which is different, in general, to the previous one. Any orbit of this iscalled a twisted homogeneous crossed set.

ARTICLE IN PRESS

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Fig. 1. Polyhedral crossed set of the tetrahedron.

N. Andruskiewitsch, M. Gra *na / Advances in Mathematics 178 (2003) 177243 189

1.3.8. Affine crossed sets

Let A be a nite abelian group and g : A-A an isomorphism of groups. Thecorresponding principal homogeneous crossed set is called an affine crossed set, anddenoted by A; g: (They are also called Alexander quandles, see e.g. [CJKLS]).Let f id g; then xx y x gy x f x gy:Let us compute the orbits of A; g: Since g id f ; we have xx y f x y

y: It is clear then by induction that x1x x2x?xnx yAy f A; whence Oy y Im f : In this case, by (1.11), indecomposable is equivalent to being faithful(since A is nite).We compute now when two indecomposable afne crossed sets are isomorphic.

Lemma 1.23. Two indecomposable affine crossed sets A; g and B; h are isomorphicif and only if there exists a linear isomorphism T : A-B such that Tg hT :

If this happens, any isomorphism of crossed sets U : A; g-B; h can be uniquelywritten as U tbT ; where tb : B-B is the translation tbx x b and T : A-B isa linear isomorphism such that Tg hT :

Proof. Let U : A; g-B; h be an isomorphism of crossed sets. Since thetranslations are isomorphisms of crossed sets, we decompose U tbT ; where T isan isomorphism of crossed sets with T0 0: Let f id g; k id h: We have

T f x gy kTx hTy:

Letting x 0; we see that Tg hT ; letting y 0; we see that Tf kT : HenceT f x gy T f x Tgy: But A; g is indecomposable, thus f isbijective; we conclude that T is linear. &

Remark 1.24. After we nished (the rst version of) the paper, Nelson gave in [Ne] aclassication of non-indecomposable afne crossed sets.

Corollary 1.25. If A; g is an indecomposable affine crossed set, then Autx A; g isthe semidirect product AsAutAg; where AutAg is the subgroup of all linearautomorphisms of A such that Tg gT :

Proof. Easy. &

Notice that when A; g is not indecomposable, the corollary does not hold. Forinstance, if g id; then f 0 and A; id is trivial, whence Autx A; g SA:The group Innx A; g is usually smaller: it can be easily shown that Innx A; g

Im fs/gS: In fact, if aAA then fa f a; gAAsAutAg: In particular,Innx A; g is solvable.

Remark 1.26. We conclude that a standard crossed set O; where O is a single non-trivial orbit of a simple non-abelian group, cannot be afne; cf. Lemma 1.9.


Similarly, let us discuss when a polyhedral crossed set X is afne. The crossedset of vertices of the tetrahedron is afne, actually isomorphic to A Z2 Z2;g 0

111

: It can be seen by hand that the crossed set of the vertices of thecube is not afne. Indeed, it is easy to see that the underlying group A should

be either Z=2 Z=4 or Z=23; but since AutZ=2 Z=4 has order 8 andInnx X has 3-torsion, the case A Z=2 Z=4 is impossible. Furthermore, onecan exclude the case A Z=23 by looking at automorphisms gAAutA s.t.g3 id: For the other polyhedral crossed sets, we have that Innx X is S4for the octahedron, and A5 for the icosahedron and the dodecahedron, so thatX is not afne.As a particular case, let A Zn Z=nZ; and let qAZn such that 1 qAZn :

Then we have a structure on Zn given by

Zn; x q; x x q y 1 qx qy:

This is an indecomposable crossed set, and it can be seen that

Zn; x qCZn; x q0 3q q0 (by Lemma 1.23, since AutZn is abelian).It is immediate to see that Z3; x 2 is the only indecomposable crossed set with

three elements; and that any crossed set with three elements is either trivial or

isomorphic to Z3; x 2:It is proved in [EGS] that any indecomposable rack of prime order p is either

isomorphic to Zp; x q for qAZp or it is isomorphic to Zp; x with xx y y 1:Indecomposable racks of order p2 are classied in [G3], in particular it is proved

there that any indecomposable quandle of order p2 is afne.

1.3.9. Affine racks

These are a generalization of afne quandles. Let A be an abelian group,gAAutA; fAEndA be such that fg gf and f id g f 0: We dene then onA the structure of rack given by xx y f x gy: It is clear that A is a quandle ifff id g: Notice that id f id g1f id; thus id f is an automorphism ofA: We can consider then the afne quandle A; id f : It is easy to see that thisquandle is the associated quandle for the rack just dened (see (1.6)). As an example,one can take A Zp2 ; g id; f x px:

1.3.10. Amalgamated unions of affine crossed sets

Let A; g; A; h be two indecomposable afne crossed sets; let f ; k be given byf id g; k id h; and let s : A; g-Autx A; h; t : A; h-Autx A; g havethe form sx ax b (i.e., sxy ax by) and ty g dy (i.e.,tyx gx dy) for certain b; gAAutA; a; dAEndA: Then by Lemma 1.23we must have bAAutAh; gAAutAg and one can verify that

1:8 is equivalent to f f g da 0; db gd 0;


1:9 is equivalent to k kb ad 0; ag ha 0;

1:10 is equivalent to ax by y3gx dy x:

Thus, the disjoint union X A; g0A; h is a decomposable crossed set. Forinstance, let h gf 1 id f 1: We dene s and t by taking a id; b g; g h;d id:If A; g; B; h are non-isomorphic indecomposable afne crossed sets, then any

amalgamated sum A,B is not afne. This is a consequence of the following easylemma:

Lemma 1.27. If A; g is an affine crossed set, then its orbits are isomorphic as crossedsets.

Proof. Let fx0 0;y; xng be a full set of representatives of coclasses in A=Im f :The orbits are Im f ; Im f x1;y; Im f xn: Thus, ti : Im f xi-Im f ;tiy y xi are isomorphisms. &1.3.11. Involutory crossed sets

This is a discretization of symmetric spaces, which we shall roughly present (see[J1] for the full explanation). Let S be a set provided with a collection of functionsg : Z-S; called geodesics, such that any two points of S belong to the image of someof them. Consider the afne crossed set Z;1: Assume that the following conditionholds: if x; yAS belong to two geodesics g and g0; say gn g0n0 x; gm g0m0 y; then gnxm g0n0xm0: Then we can dene on S a unique binaryoperation x in such a way that the geodesics respect x ; namely, xx y gnxm g2n m for any geodesic g such that gn x and gm y: Thisoperation furnishes S with the structure of an involutory crossed set if it mapsgeodesics to geodesics; that is, if xx g is a geodesic for any x and any g: It can beshown that any involutory crossed set arises in this way [J1].

1.3.12. Core crossed sets

Let G be a group. The core of G is the crossed set G; x ; where xx y xy1x:The core is an involutory crossed set. If G is abelian, its core is the afne crossed setwith g id: More generally, one can dene the core of a Moufang loop.

1.3.13. The free quandle of a set

Let C be a set and let FC be the free group generated by C; let Oc denote theorbit of cAC in FC: We claim that the standard crossed set XC :

ScAC Oc is the

free quandle on the set C:For, let c : C-X ; x be any function and let C : FC-Innx X be the unique

group homomorphism extending c/fcc: We dene then #c : XC-X by

#cy Cuc; if y ucu1; cAC:


The well-deniteness of #c is a consequence of the following facts about orbits in freegroups:

(1) If c; dAC and Oc Od ; then c d:(2) The centralizer of cAC is /cS:

It is not difcult to see that #c is indeed a morphism of quandles extending c; and theunique one. It is easy to see that XC is also the free crossed set generated by C: It isnot, however, a free rack.

2. Extensions

2.1. Extensions with dynamical cocycle

We now discuss another way of constructing racks (resp. quandles, crossed sets),generalizing Example 1.19. The proof of the following result is essentiallystraightforward.

Lemma 2.1. Let X be a rack and let S be a non-empty set. Let a : X X-FunS S; S be a function, so that for each i; jAX and s; tAS we have an element aijs; tAS:We will write aijs : S-S the function aijst aijs; t: Then X S is a rack withrespect to

i; sx j; t ix j; aijs; tif and only if the following conditions hold:

aijs is a bijection; 2:1

ai;jx ks; ajkt; u aix j;ix kaijs; t; aiks; u 8i; j; kAX ; s; t; uAS: 2:2in other words, ai;jx ksaj;kt aix j;ix kai;js; tai;ks:

If X is a quandle, then X S is a quandle iff furtheraiis; s s for all iAX and sAS: 2:3

If X is a crossed set, then X S is a crossed set iff furtherajit; s s whenever ix j j and aijs; t t 8i; jAX ; s; tAS: 2:4

Proof. Easy. &

Denition 2.2. If these conditions hold we say that a is a dynamical cocycle and thatX S is an extension of X by S; we shall denote it by X a S: When necessary, weshall say that a is a rack (resp. quandle, crossed set) dynamical cocycle to specify that


we require it to satisfy (2.1)+(2.2) (resp. (2.1)+(2.2)+(2.3), (2.1)+(2.2)+(2.3)+(2.4)). The presence of the parameter s justies the name of dynamical.

Assume that X is a quandle and let a be a quandle dynamical cocycle. For iAXconsider sx it : aiis; t: It is immediate to see that S; x i becomes a quandle8iAX : Then (2.2), when j k; says:

aijs; tx ju aijs; tx ix jaijs; u; 8s; t; uAS: 2:5

In other words, the map aijs is an isomorphism of quandles aijs : S; x j-(S,x ix j).The projection X a S-X is clearly a morphism. Conversely, it turns out that

projections of indecomposable racks (resp. quandles, crossed sets) are alwaysextensions. Before going over this, we state a technical lemma for further use.

Lemma 2.3. Let X ; x be a quandle which is a disjoint union X iAY Xi such thatthere exists %x :Y Y-Y with XixXj Xi %x j: Suppose that cardXi cardXj 8i; j (this holds for instance if X is indecomposable). Then Y ; %x is aquandle.

Furthermore, take S a set such that cardS cardXi and for each iAY setgi : Xi-S a bijection. Let a : Y Y-FunS S; S be given by aijs; t :gi %x jg1i sx g1j t: Then a is a dynamical cocycle and XCY a S:

Proof. This follows without troubles from Lemma 2.1. &

Remark 2.4. (1) Within the hypotheses of the lemma, if X is indecomposable then sois Y :(2) The whole lemma can be stated in terms of racks.(3) In order to state the lemma in terms of crossed sets, it is necessary to further

assume that Y ; %x is a crossed set, i.e., that it satises (1.4).

Corollary 2.5. Let X ; x ; Y ; %x be quandles (resp. racks, crossed sets). Letf : X-Y be a surjective morphism such that the fibers f 1y all have the samecardinality (this happens for instance if X is indecomposable). Then X is an extensionX Y a S:

Proof. Easy. &

Let X be a rack. Let a : X X-FunS S; S be a dynamical cocycle and letg : X-SS be a function. Dene a0 : X X-FunS S; S by

aijs; t gix jaijg1i s; g1j t; i:e:; a0ijs gix jaijg1i sg1j : 2:6


Then a0 is a dynamical cocycle and we have an isomorphism of racksT : X a S-X a0 S given by Ti; s i; gis: Conversely, if there is anisomorphism of racks T : X a S-X a0 S which commutes with the canonicalprojection X S-X then there exists g : X-SS such that a and a0 are relatedas in (2.6).

Denition 2.6. We say that a and a0 are cohomologous if and only if there existsg : X-SS such that a and a0 are related as in (2.6).

Example 2.7. Let Y be the crossed set given by the faces of the cube. Then Y is thedisjoint union of the subsets made out of the pairs of opposite faces. This unionsatises the hypotheses of Lemma 2.3 and, being indecomposable, the quotient

X ; %x is isomorphic to the crossed set Z3; x 2:

Example 2.8. Let A; g be an afne crossed set. Suppose that there exists a subgroupBDA invariant by g; let %g be the induced automorphism of A=B: Consider the afne

crossed set A=B; %g; the projection A; g-p A=B; %g is a morphism of crossed sets.Corollary 2.5 applies and we see that A is an extension of A=B:

More examples appear in [CHNS] by means of group extensions. They are used tocolor twist-spun knots.

2.2. Extensions with constant cocycle

Let X be a rack. Let b : X X-SS: We say that b is a constant rack cocycle ifbi;jx kbj;k bix j;ix kbi;k: 2:7

If X is a quandle, we say that b is a constant quandle cocycle if it further satises

bii id; 8iAX : 2:8

If X is a crossed set, we say that b is a constant crossed set cocycle if it further satises

bji id whenever ix j j and bijt t for some tAS: 2:9

We have then an extension X b S : X a S; taking aijs; t bijt: Note that x iis trivial for all i; and the ber Ffi;s Ffi S:We shall say in this case that the extension is non-abelian. It is clear that an

extension X a S is non-abelian if and only if aijs aijt 8s; tAS; 8i; jAX :

Denition 2.9. Let g : X-SS be a function and let b be a constant cocycle. Deneb0 : X X-SS by

bij gix j bij g1j :


Then we have an isomorphism T : X b S-X b0 S given by Ti; s i; gis:In this case, we shall say that b and b0 are cohomologous.

The use of the word cocycle is not only suggested by its analogy with group2-cocycles, which describe extensions: there is a general denition of abeliancohomology (see Section 4) for which this is its natural non-abelian counterpart. Theuse of the word cocycle in the phrase dynamical cocycle stands on the samebasis.For X b S a non-abelian extension, let c : X-Innx X b S be given by ci

fi;t for an arbitrary tAS; that is, ci j; s ix j; bijs: Then cX generatesInnx X b S: Let Hi fhAInnx X b S j hi; sAi S 8sASg:

Denition 2.10. Assume that X is indecomposable. A constant cocycle b : X X-SS is transitive if for some iAX ; the group Hi acts transitively on i S: Notethat this denition does not depend on i:

We have seen that all the bers of an indecomposable rack (resp. quandle, crossedset) have the same cardinality; we provide now a precise description of anindecomposable rack (resp. quandle, crossed set). Recall the map f : Y-Innx Yfrom Denition 1.3.

Proposition 2.11. Let Y be an indecomposable rack (resp. quandle, crossed set), letX fY and let S be a set with the cardinality of the fibers of f: Then we have anisomorphism T : Y-X b S for some transitive constant cocycle b:

Conversely, a non-abelian extension X b S is indecomposable if and only if X isindecomposable and b is transitive.

Proof. Choose, for each xAX ; a bijection gx : Fx f1x-S: We have then abijection T : Y-X S; Ti fi; gfii: We dene, for x; yAX andsAS; bxys gxx yT1x; sxT1y; s: It is straightforward to see that b is atransitive constant cocycle and that T is an isomorphism.The second part is clear. &

Example 2.12. Let X f1; 2; 3; 4g be the tetrahedral crossed set dened in Section1.3 and let S fa; bg: Then SS fid; sgCC2 f71g: There is a non-trivial 2-cocycle b : X X-S given by

bxy 1 if x 1 or y 1 or x y;bxy 1 otherwise:

(

Let c :X-Innx X b S be as in the paragraph preceding Denition 2.10. It isclear that c1c2AH4; and c1c24 a c13 b 4 b; whence b istransitive and X b S is indecomposable.


A way to construct cocycles, which resembles the classication of YetterDrinfeldmodules over group algebras, is the following one:

Example 2.13. Let X be an indecomposable ( nite) rack, x0AX a xed element,G Innx X; and let H Gx0 be the subgroup of the inner isomorphisms which xx0: Let Z be a ( nite) set and r : H-SZ a group homomorphism. There is then abijection G=H-X given by g/gx0: Fix a (set theoretical) section s : X-G; i.e.,sx x0 x 8xAX : This determines, for each x; yAX ; an element tx;yAH such thatfxsy sxx ytx;y: To see this, we compute

sxx y1fxsy x0 sxx y1fx y sxx y1 xx y x0:

Then it is straightforward to see that b : X X-SZ; bx;y rtx;y is a constantcocycle. Explicitly, this denes an extension X b Z as

x; zx x0; z0 xx x0; rsxx x01fxsx0z0:

Even if X is a quandle, this is not a quandle in general, since it does not necessarilysatisfy (2.8). Let us compute when (2.8) is satised (suppose that X is a quandle)

bx;x rsxx x1fxsx rsx1fxsx rfsx1x rfx0:

Then X b Z is a quandle iff X is a quandle and rfx0 1ASZ:

Remark 2.14. It is easy to see that for polyhedral quandles and afne quandles thegroup H is generated by fx0 ; and then we cannot construct non-trivial quandleextensions in this way.

2.3. Modules over a rack

Throughout this subsection, R will denote the category of racks. All theconstructions below can be performed in the category Q of quandles, or in thecategory CrS of crossed sets.It is clear that nite direct products exist in the category R; cf. Example 1.19. Then

we can consider group objects in R; they are determined by the followingProposition.

Proposition 2.15. A group object in R is given by a triple G; s; t; where G is a group,sAAutG; t : G-ZG is a group homomorphism, and* st ts;* sxx1txAkert 8xAG:The rack structure is given by xx y txsy:


A group object G; s; t in R is a quandle iff tx sx1x 8xAG; iff it is anhomogeneous quandle (hence crossed set).

A group object G; s; t in R is abelian iff it is an affine rack.

Proof. The second and third statements follow from the rst without difculties(notice that the image of t lies in the center).A group object in R is a triple G; ; x with G; a group and G; x a rack,

such that the multiplication is a morphism of racks. Let s; t :G-G; sx 1x x; tx xx 1: Then both s and t are group homomorphisms since sxy 1x xy 1 1x x y 1x x 1x y sxsy; and analogously for t:Furthermore,

xx y 1 xx y 1 1x y xx 1 sytx

xx 1 1x y txsy:Then, s must be an isomorphism, and then tx is central 8x: Last,

xx yx z tx sty s2z;

xx yx xx z t2x tsy stx s2z;

whence tx sty t2x tsy stx: Taking x 1 we see that st ts: Takingy 1 we see that ttxsxx1 1: The converse is not difcult. &

Let us now consider the comma category RjX over a xed rack X ; recall that theobjects of RjX are the maps f : Y-X and the arrows between f : Y-X andg : Z-X are the commutative triangles, i.e., the maps h : Y-Z such that gh f :Since equalizers exist in R (they are just the set-theoretical equalizers with the

induced x ), R has nite limits. It follows that RjX also has nite limits. We arewilling to determine all abelian group objects in RjX :

Denition 2.16. Let X be a rack and let A be an abelian group. A structure of X-module on A consists of the following data:

* a family Ziji;jAX of automorphisms of A; and* a family tiji;jAX of endomorphisms of A;such that the following axioms hold:

Zi;jx k Zj;k Zix j;ix k Zi;k; 2:10

ti;jx k Zix j;ix k ti;k tix j;ix k ti;j; 2:11

Zi;jx k tj;k tix j;ix k Zi;j: 2:12


If X is a quandle, a quandle structure of X-module on A is a structure of X -modulewhich further satises

Zii tii id: 2:13If X is a crossed set, a crossed set structure of X-module on A is a quandle structure ofX -module which further satises

if ix j j and id Zijt tijs for some s; t then id Zjis tjit: 2:14

An X -module is an abelian group A provided with a structure of X -module.

Remark 2.17. Taking j k in (2.10), one gets in presence of (1.3) the suggestiveequality:

Zi;jZj;j Zix j;ix jZi;j: 2:15

Let A be an X -module. We dene aij : A A-A byaijs; t : Zijt tijs:

Theorem 2.18. (1) aij is a dynamical cocycle, hence we can form the rack Y X a A:(2) The canonical projection p : Y-X is an abelian group in RjX :(3) If p : Y-X is an abelian group in RjX and X is indecomposable, then

YCX a A for some X-module A and p is the canonical projection.(4) If X is a quandle and A is a quandle X-module, then the preceding statements are

true in the category of quandles. Same for crossed sets.

Proof.

(1) Condition (2.1) follows since Zij is an automorphism. The left-hand side of (2.2)is

Zi;jx kZj;ku Zi;jx ktj;kt ti;jx ks

and the right-hand side of (2.2) is

Zix j;ix kZi;ku tix j;ix kZi;jt Zix j;ix kti;ks tix j;ix kti;js:

Thus, (2.2) follows from (2.10)(2.12).(2) Let s : X-X a A; si i; 0; iAX ; it is clearly a morphism of racks. Let

:Y X Y-Y ; i; a i; b i; a b; it is clearly a morphism in RjX : It isnot difcult to verify that Y ; is an abelian group in RjX with identityelement s:

(3) Let p : Y-X be an abelian group in RjX ; X an arbitrary rack. We havemorphisms in RjX : Y X Y-Y and s : X-Y satisfying the axioms of


abelian group. In particular, we have:(a) The existence of s implies that p is surjective; if iAX ; Ai : p1i is an

abelian group with identity si:(b) a bx c d ax c bx dAAix j; if a; bAAi; c; dAAj:(c) The map hij : Aj-Aix j; hij the restriction of fsi; is an isomorphism of

abelian groups, for all i; jAX :

Assume now that X is indecomposable. Then all the abelian groups Ai areisomorphic, by (3c). Fix an abelian group A provided with group isomorphismsgi : Ai-A: Dene aij : A A-A; i; jAX ; by

aijs; t : gix jg1i sx g1j t:

We claim that aijs; t Zijt tijs; where Zijt aij0; t and tijs aijs; 0; thisfollows without difculty from (3b), since the gs are linear. Now, Zijt gix jhijg1jis a linear automorphism, whereas tij is linear by (3b). We need nally to verifyconditions (2.10)(2.12); this is done reversing the arguments in part (1).

(4) Condition (2.3) amounts in the present case to (2.13), whereas condition (2.4)amounts to (2.14). &

Remark 2.19. Assume now that X is a non-indecomposable quandle and keep thenotation of the proof. Then Ai is a subquandle of Y ; indeed an abelian group in Q:By Proposition 2.15, Ai is afne, with respect to some giAAutAi:

Example 2.20. If A; g is an afne crossed set, then it is an X -module over any rackX with Zij g; tij f id g: We shall say that A is a trivial X -module if g id;that is when it is trivial as crossed set.

Example 2.21. Let X be a trivial quandle, let AiiAX be a family of afne quandlesand let Y be the disjoint sum of the Ais, with the evident projection Y-X : ThenY-X is an abelian group in QjX which is not an extension of X by any X -module, ifthe Ais are not isomorphic.

We now show that the category of X -modules, X an indecomposable quandle orrack, is abelian with enough projectives. Actually, it is equivalent to the category ofmodules over a suitable algebra.

Denition 2.22. The rack algebra of a rack X ; x is the Z-algebra ZfXg presentedby generators Z71ij i;jAX and tiji;jAX ; with relations ZijZ1ij Z1ij Zij 1; (2.10),(2.11) and (2.12).The quandle algebra of a quandle X ; x is the Z-algebra ZX presented by

generators Z71ij i;jAX and tiji;jAX ; with relations ZijZ1ij Z1ij Zij 1; (2.10), (2.11),(2.12) and (2.13).


It is evident that the category of X -modules, X a rack, is equivalent to the categoryof modules over ZfXg; therefore, it is abelian with enough injective and projectiveobjects. Same for quandle X -modules and ZX:The algebra ZX is augmented with augmentation e : ZX-Z; eZij 1;

etij 0: The algebra ZfXg is also augmented, composing e with the projectionZfXg-ZX:There are various interesting quotients of the algebra ZX:First, the quotient of ZX by the ideal generated by the tij s is isomorphic to the

group algebra of the group LX presented by generators Ziji;jAX with relations(2.10).Next, consider the following elements of the group algebra ZGX (see Denition

1.5):

Zij : i; tij : 1 ix j: 2:16

It is not difcult to see that this denes a surjective algebra homomorphismZX -ZGX ; in particular any GX -module has a natural structure of X -module.

Denition 2.23. If X is a crossed set and M is any GX -module then M also satisesthe extra condition (2.14). We shall say that M is a restricted X-module.

Denition 2.24. Let X be a rack and let A be an X -module. A 2-cocycle on X withvalues in A is a collection kiji;jAX of elements in A such that

Zi;jx kkjk ki;jx k Zix j;ix kkik tix j;ix kkij kix j;ix k 8i; j; kAX : 2:17

Two 2-cocycles k and k0 are cohomologous iff (f : X-A such that

k0ij kij Zij f j f ix j tij f i: 2:18

As remarked earlier, the reader can nd in Section 4 a complex which justiesthese names.

Proposition 2.25.

(1) Let X be a rack and consider functions Z : X X-AutA; t : X X-EndA; k : X X-A: Let us define a map a by

aija; b Zijb tija kij ; a; bAA: 2:19

Then the following conditions are equivalent:(a) a is a dynamical cocycle.(b) Z; t define a structure of an X-module on A and kij is a 2-cocycle, i.e., it

satisfies (2.17).(2) Let k and k0 be 2-cocycles and let a; a0 be their respective dynamical cocycles. If k

and k0 are cohomologous then a and a0 are cohomologous.


Proof. Straightforward. For the second part, if f is as in (2.18), then take g : X-SAby gis f i s: It is easy to verify (2.6). &

Denition 2.26. If X is a rack, A is an X -module and kiji;jAX is a 2-cocycle on Xwith values in A; then the extension X a A; where a is given by (2.19), is called anaffine module over X. By abuse of notation, this extension will be denoted X k A:

All the constructions and results in this section are valid more generally over axed commutative ring R:

3. Simple racks

3.1. Faithful indecomposable crossed sets

To classify indecomposable racks, we may rst consider faithful indecomposablecrossed sets (recall that a faithful rack is necessarily a crossed set), and next computeall possible extensions.

Proposition 3.1. Let X be an indecomposable finite rack. Then X is isomorphic to anextension Y a S; where Y is a faithful indecomposable crossed set. Furthermore, Ycan be chosen uniquely with the property that any surjection X-Z of racks, with Zfaithful, factorizes through Y.

Proof. Consider the sequence X-X1 : fX-X2 : fX1y : Since X is nitethe sequence stabilizes, say at Y Xn; which is clearly faithful and indecomposable.By Corollary 2.5, XCY a S: Now let c : X-Z be a surjection, with Z faithful. ByLemma 1.8, it gives a surjection c1 : X1-Z; and so on. &

Faithful indecomposable crossed sets can be characterized as follows.

Proposition 3.2.

(1) If X is a faithful indecomposable crossed set, then there exists a group G and aninjective morphism of crossed sets j : X-G; such that ZG is trivial and jX isa single orbit generating G as a group. Furthermore, G is unique up toisomorphisms with these conditions.

(2) If X is a single orbit in a group G with ZG trivial and X generates G, then X is afaithful indecomposable crossed set.

(3) There is an equivalence of categories between(a) The category of faithful indecomposable crossed sets, with surjective

morphisms.(b) The category of pairs G;O; where G is a group with trivial center and O is an

orbit generating G; a morphism f : G;O-K ;U is a group homomorphismf : G-K such that f O U:


Proof.

(1) Existence: take G Innx X ; uniqueness: by Lemma 1.9.(2) Immediate.(3) Follows from (1), (2) and Lemma 1.8. &

We shall say that a projection (surjective homomorphism) p : X-Y of racks istrivial if card Y is either 1 or card X :

Denition 3.3. A rack X is simple if it is not trivial and any projection of racksp : X-Y is trivial.

A decomposable rack has a projection onto the trivial rack with two elements; itfollows that a simple rack is indecomposable.Let X be a simple rack with n elements; then fX has only one point, or f is a

bijection. In the rst case, X is a permutation rack dened by a cycle of length n; inthe second case X is a crossed set (and necessarily card X42).It is not difcult to see that a permutation rack with n elements dened by a cycle

of length n is simple if and only if n is prime.

Proposition 3.4. Let X be an indecomposable faithful crossed set, which corresponds to apair G;O: Then X is simple if and only if any quotient of G different from G is cyclic.

Proof. Assume X is simple. If p : G-K is an epimorphism of groups, then eitherpX is a single point or pjX is bijective. If pX is a point, this point generates K;and then K must be cyclic. If pjX is bijective, take hAkerp; then hxh1 x 8xAX ;whence hAZG: This means that p is bijective.Assume now that any non-trivial quotient of G is cyclic. Let p : X-Y be a

surjective morphism of crossed sets, then GCInnx X-Innx Y is an epimorph-ism, so that either it is a bijection (then XCY ) or Innx Y is cyclic. In the last case,fY ; being indecomposable, is a point; hence Y ; being also indecomposable, is apoint. &

Example 3.5. Let X be the crossed set of the faces of the cube. We can realize X asthe orbit given by 4-cycles inside S4; since X is faithful and Innx X CS4:Considering the quotient S4-S4=KCS3; where K is the Klein subgroup, we seethat X is not simple. Indeed, this gives X as the same extension Z3 a Z2 as inExample 2.7. Also, if X 0 is the orbit given by the six transpositions in S4; we seetaking the same quotient S4=K that X

0 is an extension Z3 a0 Z2:

Example 3.6. Since the only proper quotient of Sn nX5) is cyclic, we see that if X isa non-trivial orbit of Sn then either

* X generates Sn and then it is simple, or* XCAn:


If XCAn; then it might fail to be an orbit in An: This would happen if and onlyif the centralizers of the elements of X lie inside An (because the order of theorbits is the ratio between the order of the group and the order of the centralizers).In this case, thus, X decomposes as a union of two orbits, which are isomorphicvia conjugation by any element in Sn\An (an example of this case arises forn 5 and X the 5-cycles). On the other hand, if the centralizers of the elements of Xare not included in An then X is indecomposable, and hence simple by theproposition.

3.2. Classification of simple racks

We characterize now simple racks in group-theoretical terms. We rst classifynite groups G such that ZG is trivial and G=N is cyclic for any normal non-trivialsubgroup of G: We are grateful to R. Guralnick for help in this question.To x notation, if G acts on H; we put HsG the semidirect product with

structure h; gh0; g0 hg h0; gg0:

Theorem 3.7 (Guralnick). Let G be a non-trivial finite group such that ZG is trivialand G=H is cyclic for any normal non-trivial subgroup H. Then there are a simplegroup L, a positive integer t and a finite cyclic group C /xS in AutN; whereN : Lt L ? L (t times), such that one of the following possibilities hold.(1) L is abelian, so that N is elementary abelian of order pt; x is not trivial and it acts

irreducibly on N. Furthermore, GCNsC:(2) L is simple non-abelian, G NCCNsC=ZNsC and x acts by

x l1;y; lt ylt; l1;y; lt1 3:1

for some yAAutL:Conversely, all the groups in (1) or (2) have the desired properties. Furthermore, two

groups in either of the lists are isomorphic if and only if the corresponding groups L are

isomorphic, the corresponding integers t are equal and the corresponding automorph-isms x define, up to conjugation, the same element OutN AutN=IntN:

Before proving the Theorem, we observe that:

* If G is a group which is not abelian and such that G=H is abelian for any normalnon-trivial subgroup H; then G has a unique minimal non-trivial normalsubgroup, namely G; G:

* If G is a nite group such that G=N is cyclic for any normal non-trivial subgroupN; then ZG is trivial is equivalent to G is non-abelian.

* Case (2) covers the case where G is non-abelian simple (t 1; C is trivial).* In case (1), identify L with Fp and the automorphism x with TAGLt; Fp: Then x

acts irreducibly if and only if the characteristic polynomial of T is irreducible,hence equals the minimal polynomial of T : In this case, if n ord x and if ddivides n; 1adan; then kerTd id 0; this implies that N 0 is a union of


copies of C: Hence jCj divides pt 1: Clearly, we may assume that x acts by thecompanion matrix of an irreducible polynomial in FpX of degree t:

* Let N; C be two groups, C acting on N by group automorphisms. The

center ZNsC is given by ZNsC fn; c j cAZC; nANC ; c m n1mn 8mANg: In particular, if p : NsC-NsC=ZNsC is the projec-tion then pNCN=ZNC :

Proof. Step I: We rst show that the groups described in the Theorem have thedesired properties.Let L; N; C; G be as in case (1). By the irreducibility of the action of C; being x

non-trivial, we see that ZG is trivial. We claim that any non-trivial normalsubgroup M of G contains N: For, M-N is either trivial or M-N N: If aAG andmAM; mae; then a; mAM-G; GDM-N: Thus M-N is non-trivial, sinceotherwise mAZG; proving the claim. Hence any non-trivial quotient of G; being aquotient of C; is cyclic, and G satises the requirements of the theorem.Let now L; N; C; G be as in case (2). We identify N with its image in G: We

claim next that ZG is trivial. Let n; cANsC be such that pn; c ncAZG: Itis easy to see that nANx and c acts on N by conjugation by n1: Thusn; cAZNsC and nc 1 in G:Any normal subgroup P of N is of the form

QjAJ Lj; for some subset J of

f1;y; tg; if P is also x-stable then either P is trivial or equals N; because x permutesthe copies of L: As in case (1), we conclude that any non-trivial normal subgroup Mof G contains N; hence any non-trivial quotient of G is cyclic.

Step II: Let G be a nite group with a minimal normal non-trivial subgroup N;and assume that G=N is cyclic. Then there exists a simple subgroup L of N; and asubgroup C /xS of G such that N L ? L t copies) and G NC:Indeed, let xAG be such that its class generates G=N and let C be the subgroup

generated by x; then G NC: Let L be a minimal normal non-trivial subgroup of N:Then Li : xi1Lxi1 is also a minimal normal subgroup of N and L1?Lj-Lj1is either trivial or Lj1: Let t be the smallest positive integer such thatL1?Lt-Lt1 Lt1: Then L1?LtCL1 ? Lt is a normal subgroup of Nand is stable by conjugation by x; so it is normal in G and therefore equal to N: If Sis a normal subgroup of L; then SCS 1? 1 is normal in N; and byminimality of L; L is simple.

Step III: We now show that any group G satisfying the requirements of thetheorem is either as in case (1) or (2). We may assume that G is not simple. We keepthe notation of Step II and assume then that ZG is trivial and that any properquotient of G is cyclic.If N is abelian then N-CDZG is trivial, whence G NsC: Furthermore, x

should act irreducibly since any subgroup PCN which is x-stable is normal. Hence,G is as in case (1).

If L is not abelian and t41 we have, for i41; xLtx1; Li xLt; Li1x1 1;from where x sends Lt isomorphically to L1; and x acts as in case (2) of thestatement. Consider the projection p : NsC-G: Since ZG is trivial,


ZNsCCkerp: Let now n; cAkerp; then c n1 in G; whence c acts byconjugation on N by n1: Thus n; cAZNsC; and we are done.

Step IV: We prove the uniqueness statement. N is unique since, as remarked,N G; G: Now, L is unique by JordanHolder, then t is unique. Since x waschosen modulo N; x and x0 give rise to the same group if they coincide in OutN:Furthermore, since any automorphism G-G must leave N invariant, x is unique upto conjugation in OutN: &

Remark 3.8. Let N; C be nite groups with C acting on N by groupautomorphisms, and let G NsC: If m; z; n; yAG; then

m; zn; ym; z1 mz nzyz1 m1; zyz1:

When C is abelian, it follows that

Cn; y [zAC

Oyz n fyg; 3:2

where C stands for conjugacy class, and Oy for the orbit under the action of N on

itself given by

m,yn : mny m1: 3:3

Note thatS

zA/yS Oyz n Oyn: For, n1,yn y n; and the claim follows.Note also that m,yn is not the same as mx n:

We can now state the classication of simple racks. We begin by the followingimportant theorem. The proof uses [EGS, Lemma 8], which is in turn based on theclassication of simple nite groups.

Theorem 3.9. Let X ; x be a simple crossed set and let p be a prime number. Thenthe following are equivalent.

(1) X has pt elements, for some tAN:(2) Innx X is solvable.(3) Innx X is as in case (1) of Theorem 3.7.(4) X is an affine crossed set Ftp; T where TAGLt; Fp acts irreducibly.

Proof.

1 ) 2 This is [EGS, Lemma 8].2 ) 3 This is Proposition 3.4 plus Theorem 3.7.3 ) 4 This follows from the preceding discussion. Since fX CInnx X is a

conjugacy class, by (3.2) we have fX N fxrg for some r: SincefX generates Innx X ; r must be coprime to the order of x: Takey xr and call TAGLt; Fp the action of y (which is also irreducible).


We have

m; yx n; y m; yn; ym; y1 Tn id Tm; y:

4 ) 1 Clear. &

Corollary 3.10. The classification of simple racks with pt elements, for some primenumber p and tAN; is the following:

(1) Affine crossed sets Ftp; T; where T is the companion matrix of an irreduciblemonic polynomial in FpX different from X 1 and X.

(2) The permutation rack corresponding to the cycle 1; 2;y; p if t 1:

Proof. Easy. &

Remark 3.11. Keep the notation of Step III in Theorem 3.7. If L is abelian, then xdoes not necessarily send Lt to L1; as wrongly stated in [J2, Lemma 4(ii)]. Thisexplains why in [J2, Lemma 6], only irreducible polynomials of the form X t aappear; while, as we have seen, this restriction is not necessary.

Theorem 3.12. Let X ; x be a crossed set whose cardinality is divisible by at leasttwo different primes. Then the following are equivalent.

(1) X is simple.(2) There exist a non-abelian simple group L, a positive integer t and xAAutLt;

where x acts by (3.1) for some yAAutL; such that X Oxn is an orbit of theaction ,x of N Lt on itself as in (3.3) (nam1 if t 1 and x is inner,xp mpm1). Furthermore, L and t are unique, and x only depends on itsconjugacy class in OutLt: If m; pAX then

mx p mxpm1: 3:4

Proof.

1 ) 2 By Theorem 3.9, Innx X is as in case (2) of Theorem 3.7. Therefore, wehave L; t and xAAutLt: Let G NsC=ZNsCC Innx X ;and G NsC and let p : G-G be the projection. If Cn; y is aconjugacy class in G then pCn; y is a conjugacy class in G; and it isnot difcult to see that p :Cn; y-pCn; y is an isomorphism ofcrossed sets. Then X has the structure of Cn; y given by (3.3). Now,nyAG must be such that pCn; y generates G: Since the subgroupgenerated by pCn; y is invariant, we know by the proof of Theorem3.7 that it is either trivial or it contains N: It is trivial if

n; yAZNsC; i.e., if t 1 and y acts on N by conjugation by n1;thus we must exclude this case. This case excluded, y must generate C;


and then we can take x y in the proof of Theorem 3.7, whence X is asin (3.4).

2 ) 1 Similar. &

We restate the previous results as: if X is a simple rack then either

(1) jX j p a prime, XCFp a permutation rack, xx y y 1:(2) jX j pt; XCFtp; T is afne, as in Corollary 3.10(1).(3) jX j is divisible by at least two different primes, and X is twisted homogeneous, as

in Theorem 3.12.

Compare this with [J2, Theorem 7].

The simple crossed sets in (2) can be alternatively described as X ; x a whereXCFq; q pt; aAFq generates Fq over Fp and x x ay 1 ax ay: It followseasily that AutX ; x a is the semidirect product FqsFq :It is natural to ask how many different simple crossed sets with q pt

elements there are. This is a well-known elementary result. For, if Indenotes the number of monic irreducible polynomials in FpX with degree n; thenP

djn dId pn: Thus

In 1n

Xdjn

mn

d

pd ;

where m is the Mobius function.

4. Cohomology

4.1. Abelian cohomology

Let X ; x be a rack. We dene now a cohomology theory which contains allcohomology theories of racks known so far. We think that this cohomology can becomputed by some cohomology theory in the category of modules over X :For a sequence of elements x1; x2;yxnAX n we will denote

x1?xn x1x x2x ?xn1x xn?:

Notice that if ion then

x1?xix x1? #xi?xn x1?xn:

Denition 4.1. Let *AX be a xed element (which is important only in degrees 0 and1). Let ZfXg be the rack algebra of X (see Denition 2.22) and let, fornX0; CnX ZfXgX n; i.e., the free left ZfXg-module with basis X n X 0 f*g


is a singleton). Let @ @n : Cn1X-CnX be the ZfXg-linear map dened on thebasis byFor nX1:

@x1;y; xn1

Xni1

1iZx1?xi ;x1? #xi?xn1x1;y; #xi;y; xn1

Xni1

1ix1;y; xi1; xix xi1;y; xix xn1

1n1tx1?xn;x1?xn1xn1x1;y; xn:

For n 0:@x t

* ;*1x x* : 4:1

Lemma 4.2. CX; @ is a complex.

Proof. We decompose @n Pn1

i1 1i@in; where

@inx1;y; xn1 Zx1?xi ;x1?xi?xn1x1;y; xi;y; xn1

x1;y; xi1; xix xi1;y; xix xn1 for ipn

@n1n x1;y; xn1 tx1?xn;x1?xn1xn1x1;y; xn for i n 141;

@10x t* ;*1x x * for n 0:

Then, it is straightforward to verify that

@in1@jn @j1n1@in for 1piojpn 1;

and thus

@n1@n X

1piojpn11ij@i@j

X1pjpipn

1ij@i@j

X

1piojpn11ij@j1@i

X1pjpipn

1ij@i@j 0: &

We are now in position to dene rack (co)homology.

Denition 4.3. Let X be a rack. Let A A; Z; t be a left X -module and takeCnX ; A HomZfXgCnX; A; and the differential d @n: By the lemma, this is a


cochain complex. We dene then

HnX ; A HnCX ; A:

If A is a right X -module (i.e., a right ZfXg-module), we deneCnX ; A A#ZfXgCnX and HnX ; A HnCX ; A:

Remark 4.4. Low degree cohomology can be interpreted in terms of extensions: a2-cocycle is the same as a function k : X X-A which satises (2.17); and two2-cocycles are cohomologous if and only if they satisfy (2.18).

Remark 4.5. Let X be a quandle; replacing the rack algebra ZfXg by the quandlealgebra ZX in Denition 4.1 one can dene quandle cohomology theory that hasas a particular case the quandle cohomology HQX ; A in [CJKLS].

We consider now particular cases of this denition.

4.2. Cohomology with coefficients in an abelian group

Recall that ZfXg has an augmentation ZfXg-Z given by Zi;j/1; ti;j/0: Then,any abelian group A becomes an X -module. The complexes CX ; A; CX ; Acoincide thus with previous complexes found in the literature (see for instance[CJKLS,FR,G1]). We recall them for later use:

CnX ; A A#ZZX n; CnX ; A HomZZX n; ACFunX n; A

@a#x1;y; xn1 Xni1

1ia#x1;y; xi;y; xn1

a#x1;y; xi1; xix xi1;y; xix xn1

df x1;y; xn1 Xni1

1i f x1;y; xi;y; xn1

f x1;y; xi1; xix xi1;y; xix xn1: 4:2

Here, @0 : C1X ; A-C0X ; A vanishes. Notice that H1X ; A Ap0X; wherep0X is the set of Innx X-orbits in X :

Example 4.6. Let X ; x be a crossed set, let A be an abelian group (denotedadditively), and let f be a 2-cocycle with values in A: Let B : X X-SA be given by

Bija a fij ; i; jAX ; aAA:


Then B is a constant rack cocycle (i.e., it satises (2.7)). It is a constant quandlecocycle (i.e., it satises (2.8)) iff fii 0 8iAX : The denition of quandle cocycle isthus seen to be the same for these kind of extensions as that in [CJKLS]. The cocycleBij satises (2.9) iff fji 0 whenever ix j j and fij 0:

Lemma 4.7. H2X ; GCHomH2X ;Z; G for any abelian group G:

Proof. This follows from the Universal Coefcient Theorem since H1X ;Z is free(cf. [CJKS, Proposition 3.4]). &

Remark 4.8. Some second and third cohomology groups are computed in [LN];some others in [Mo]. See [O] for tables of the computations done so far. In

particular, we excerpt from [Mo] that H2X ;Z Z if X Z=p; x q where p is aprime and 1aqAZ=p:

Lemma 4.9. Let X be the disjoint sum of the indecomposable crossed sets Y and Z:

Then H2X ;ZCH2Y ;Z"H2Z;Z"Z2:

Proof. Let fAZ2X ;Z; i.e., fj;k fi;jx k fix j;ix k fi;k 8i; j; kAX : Then one hasfi;jx k fi;k ftx i;k for all i; tAY ; j; kAZ; since the actions of Z on Y and viceversaare trivial. Therefore, fi;k ft;j for all i; tAY ; j; kAZ; being both Y and Zindecomposable. We conclude that Z2X ;ZCZ2Y ;Z"Z2Z;Z"Z2: Now, ifg : X-Z then dg is really only a function on Y Y,Z Z; and the claimfollows. &

4.3. Restricted modules

Consider a restricted X -module A; see Denition 2.23. This is the same as aGX -module, GX the enveloping group of X : We get then the complexCX ; A; dCFunX ; A; d; with

df x1;y; xn1 Xni1

1ix1?xif x1;y; xi;y; xn1

Xni1

1if x1;y; xi1; xix xi1;y; xix xn1

1n11 x1?xn1f x1;y; xn:

As a particular case, suppose that X is any quandle and A; g is an afne crossed set.Take L ZT ; T1 the ring of Laurent polynomials. Then A becomes a L-module,and a fortiori a GX module by x a Ta ga 8xAX ; aAA (since for any quandlethere is a unique algebra map ZGX-L; x/T). Then Zi;j acts by g and ti;j acts byf 1 g on A: We get in this way the complex considered in [CES].


Lemma 4.10. If X is a rack and A is an abelian group with trivial action then

H2X ; A H1GX ;FunX ; A:

Here the space FunX ; A of all functions from X to A is a trivial left GX -module; theright action is given by f xz f xx z:

Proof (Sketch, see [EG]). Consider the map F : X-GX ; x/x: Letf : GX-FunX ; A and take r f : X X-A; where r f x; y f Fxy: It iseasy to see that this map gives a morphism H1GX ;FunX ; A-H2X ; A: On theother hand, let gAZ2X ; A: This gives a map g0 : X-FunX ; A; g0xy gx; y:Recall that for a right GX -module M; a map p : GX-M is a 1-cocycle iff the map#p : GX-GXrM given by z/z; pz is a homomorphism of groups. Denote M FunX ; A: We have a map xg : X-GXrM given by xgx x; g0x: So we needto show that xg extends to a homomorphism GX-GXrM: But the group GX isgenerated by X with relations xy xx yx: Thus, we only need to check that thexgxs satisfy the same relations, and it is straightforward to see that this isequivalent to dg 0: Now, it is easy to see that this map is the inverse of r: &

4.4. Non-principal cohomology

Let X TiAI Xi be a decomposition of the rack X : It is possible then todecompose the complex CX ;Z of (4.2) into a direct sum

CnX "iAI

CinX; CinX ZX n1 XiCZX n1#ZXi:

For each iAI ; let Ai be an abelian group. We denote by AI this collection. Then wetake CX ; AI the complex

CX ; AI "iAI

HomZCinX; Ai

Notice that if A Ai 8iAI ; then this complex is the same as CX ; A in (4.2).

4.5. Non-abelian cohomology

Let X ; x be a rack and let G be a group. We dene:H1X ;G Z1X ;G fg :X-G j gix j gj; 8i; jAXg; 4:3

Z2X ;G fb :X X-G j bi;jx kbj;k bix j;ix kbi;k 8i; j; kAXg: 4:4

The elements of Z2X ;G shall be called non-abelian 2-cocycles with co-efcients in G: If b; *b : X X-G we set bB *b if and only if there exists g : X-G


such that

*bij gix j1bijgj: 4:5

It is easy to see that B is an equivalence relation, and that Z2X ;G is stable underB: We dene

H2X ;G Z2X ;G=B: 4:6

Example 4.11. Let S be a non-empty set; then H2X ;SS parameterizes isomorph-ism classes of constant extensions of X by S; as in Section 2.2.

Remark 4.12. Though obvious, we point out that H2X ;SS H2X ;Z2 when Shas only two elements.

4.6. Non-abelian non-principal cohomology

We combine the theory of non-principal cohomology in Section 4.4 with that of

non-abelian cohomology: let X TiAI Xi be a decomposition of the rack X and foreach iAI let Gi be a group. Let us denote GI this collection. We consider

Z2X ;GI f Tifi : X Xi-Gijnfix; yx zfiy; z fixx y; xx zfix; z 8x; yAX ; zAXi

o:

As usual, if f ; fAZ2X ;GI; we say that fBf iff (g Tigi : Xi-Gi such that

fix; y gixx y1fix; ygiy 8xAX ; yAXi:

The importance of such a theory becomes apparent in Theorem 4.14.

Denition 4.13. For iAI ; let be given a positive integer ni and a subgroup

GiCGLC; ni: Let f Tifi : X Xi-Gi: Take V "iAI

Xi Cni a vector space andconsider the linear isomorphism

cf : V#V-V#V ; cf x; a#y; b xx y; fix; yb#x; a;

xAXj; yAXi; aACnj ; bACni :

Theorem 4.14. (1) Let X ;GI ; V ; cf be as in the definition. Then cf satisfies the BraidEquation if and only if fAZ2X ;GI :


(2) Furthermore, if fAZ2X ;GI ; then there exists a group G such that V is aYetterDrinfeld module over G: In particular, the braiding of V as an object in GGYD

coincides with cf : If Gi is finite 8iAI and X is finite, then G can be chosen to be finite.(3) Conversely, if G is a finite group and VAGGYD; then there exist X TXi; finite

groups GI ; fAZ2X ;GI such that V is given as in the definition and the braidingcAAutV#V in the category GGYD coincides with cf : Here, X can be chosen to be acrossed set.

(4) If fBf and V ; cf ; V; cf are the spaces associated to f ; f; then they areisomorphic as braided vector spaces, i.e., there exists a linear isomorphism g : V-V

such that g#gcf cf g#g : V#V-V#V :

Proof.

(1) Straightforward.(2) It follows from [G2, 2.14]. The niteness of G follows from the fact that the

group GCGLV can be chosen to be the group generated by the mapsy; b/xx y; f x; yb; which is contained in the product QiAI Gi; where Gi SXi Gi:

(3) It follows from the structure of the modules in GGYD: Indeed, if V "iMgi; ri(see [G1] for the notation), giAG; Gi fxgi gixg the centralizer of gi;fhi1;y; hisig a set of representatives of left coclasses G=Gi; and tiujvAGu dened byhijgihij1huv huv0tiujv ; then take X TiXi; Xi fhi1;y; hisig; and f hij; huv rutiujv:

(4) It is straightforward to verify that if fix; y gixx y1fix; ygiy8xAX ; yAXi; then the map x; a/x; gixa xAXi is an isomorphism ofbraided vector spaces. &

Notice that any YetterDrinfeld module over a group algebra can be constructedby means of a crossed set, and one does not need the more general setting ofquandles, nor racks for it. However, racks may give easier presentations than crossedsets for some braided vector spaces.

5. Braided vector spaces

We have seen that it is possible to build a braided vector space CX ; cq from arack X ; x and a 2-cocycle q; cf. Theorem 4.14 and [G1]. It turns out that thebraided vector space does not determine the rack. We now present a systematic wayof constructing examples of different racks, with suitable cocycles, giving rise toequivalent braided vector spaces. We consider afne modules over a rack X ; that isextensions of the form X k A; where A is an abelian X -module; see Denition 2.26.If the cocycle q is chosen in a convenient way, we can change the basis a` la Fourier


and obtain a braided vector space arising from a set-theoretical solution of theQYBE. This solution is in turn related to the braided vector space arising from thederived rack of the set-theoretical solution of the QYBE.Section 5.1 is an exposition of the relevant facts about set-theoretical solutions

needed in this paper. In Section 5.2 we discuss braided vector spaces arising from set-theoretical solutions. In Section 5.3 we present the general method and discussseveral examples.

5.1. Set-theoretical solutions of the QYBE

There is a close relation between racks and set-theoretical solutions of the YangBaxter equation, or, equivalently, of the braid equation. It was already observed byBrieskorn [B] that racks provide solutions of the braid equation. On the other hand,certain set-theoretical solutions of the braid equation produce racks. This is provedin [LYZ1,So], which belong to a series of papers (see [EGS,ESS,LYZ2,LYZ3])devoted to set-theoretical solutions of the braid equation and originated in aquestion by Drinfeld [Dr]. We give here the denitions necessary to us.Let X be a non-empty set and let S : X X-X X be a bijection. We say that S

is a set-theoretical solution of the braid equation if S idid SS id id SS idid S: We shall briey say that S is a solution or that X ; S is abraided set. A trivial example of a solution is the transposition t : X X-X X ;x; y/y; x: It is well-known that S is a solution if and only if R tS : X X-X X is a solution of the set-theoretical quantum YangBaxter equation. IfX ; S is a braided set, there is an action of the braid group Bn on X n; the standardgenerators si acting by Si;i1; which means, as usual, that S acts on the i; i 1entries.In particular, a nite braided set gives rise to a nite quotient of Bn for any n;

namely the image of the group homomorphism rn :Bn-SX n induced by the action.

Lemma 5.1. [B]. Let X be a set and let x : X X-X be a function. Let

c : X X-X X ; ci; j ix j; i: 5:1

Then c is a solution if and only if X ; x is a rack.

Proof. It is easy to check that c is a bijection if and only if (1.1) holds, and that itsatises the braid equation if and only if (1.2) holds. &

Denition 5.2. Let X ; X be two non-empty sets and let S : X X-X X ; S : X X-X X be two bijections. We say that X ; S and X; S are equivalent if thereexists a family of bijections Tn : X n-Xn such that TnSi;i1 Si;i1Tn; for all nX2;1pipn 1:


If X ; S and X; S are equivalent and X ; S is a solution, then X; S isalso a solution and the Tns intertwine the corresponding actions of the braidgroup Bn:

Denition 5.3 ([ESS,LYZ1]). Let X ; S be a solution and let f ; g : X-FunX ; Xbe given by

Si; j gi j; fji: 5:2

The solution (or the braided set) is non-degenerate if the images of f and g lieinside SX :

Proposition 5.4 ([So,LYZ1]). Let S be a non-degenerate solution with the notation in(5.2) and define x by

ix j figf 1j

i j: 5:3

(1) One has

f preserves x ; i:e: fi jx k fi jx fik; 5:4

fifj ffi jfgji; 8i; jAX : 5:5

(2) If c is given by (5.1), then c is a solution; we call it the derived solution of S:The solutions S and c are equivalent, and X ; x is a rack.(3) Let X ; x be a rack and let f : X-SX : We define g : X-SX by

gi j f 1fji fjix j: 5:6

Let S : X X-X X be given by (5.2). Then S is a solution if and only if (5.4), (5.5)hold. If this happens, the solutions S and c are equivalent, and S is non-degenerate.

Proof.

(1) It is not difcult.(2) It is enough to show that S and c are equivalent; automatically, c is a

solution and a fortiori X ; x is a rack. Let Tn : X n-X n be denedinductively by

T2i; j fji; j; Tn1 QnTn id;where Qni1;y; in1 fin1i1;y; fin1in; in1: One veries using (5.4) and(5.5) that TnSi;i1 ci;i1Tn; as needed.

(3) Straightforward. &

Note that (5.6) is equivalent to

gf 1j

h j f 1h hx j: 5:7


Remark 5.5. Let X ; S be a non-degenerate solution and let x be dened by (5.3).Then X ; x is a quandle if and only if

f 1i i gf 1i ii; 8iAX ; 5:8

if this holds, it is a crossed set if and only if

f 1i j gf 1j i j ) f1

j i gf 1i ji 8i; jAX : 5:9

Let X ; S; X; S be two non-degenerate braided sets, with correspondingmaps f ; g; resp. f; *g: A function j : X-X is a morphism of braided sets if andonly if

*gjij j jgi j; 5:10

fjij j jfi j; 8i; jAX : 5:11

It can be shown that j is a morphism of braided sets if and only j is a morphism ofthe associated racks and (5.11) holds. One may say that a non-degenerate braided setis simple if it admits no non-trivial projections. It follows that any solution associatedto a simple crossed set is simple, but the converse is not true as the following exampleshows: take a set X with p elements, p a prime, and a cycle m of length p: ThenSi; j m j; m1i is simple but the associated rack is trivial.

Denition 5.6 ([ESS,SO]). Let X ; S be a solution and let S2i; j Gi j; Fji:The group GX ; resp. AX ; is the quotient of the free group generated by X by therelations ij gi jfji; resp. fjij Fjifji; for all i; jAX :If X ; x is a rack and c is the corresponding solution, then GX is the group

already dened in Denition 1.5, and coincides with AX :

5.2. Braided vector spaces of set-theoretical type

We now describe how set-theoretical solutions of the QYBE plus a 2-cocycle giverise to braided vector spaces. We begin by the case of solutions arising from a rack.

Let X ; x be a rack and let qAZ2X ;C; so thatqi;jx kqj;k qix j;ix kqi;k 8i; j; kAX : 5:12

Then, by Theorem 4.14, the space V CX has a structure of a YetterDrinfeldmodule over a group whose braiding is given by cq :CX#CX-CX#CX ;

cqi#j qi;j ix j#i; i; jAX :

Let X fx1;y; xng be a set, let S : X X-X X be a bijection and let F : X X-C be a function. Let CX denote the vector space with basis X and dene


SF :CX#CX-CX#CX by

SF i#j Fi;jSi; j Fi;jgij#fji; i; jAX ; 5:13

where we use the notation in (5.2).

Lemma 5.7. (1) SF is a solution of the braid equation if and only if X ; S a solutionand

Fi;jFfj i;kFgij;gfj ik Fj;kFi;gjkFfgj ki;fkj; i; j; kAX : 5:14

(2) Assume that (5.14) holds. Then SF is rigid if and only if S is non-degenerate.

Proof.

(1) Straightforward.(2) Rigidity is equivalent to cw : Vn#V-V#Vn being an isomorphism, where

cw evV#idV#VnidVn#c#idVnidVn#V#evnV :

Assume for simplicity that F 1: Let diiAX be the basis of Vn dual to X : Thencwdi#j

Ph:gjhi fh j#h: Hence, if cw is an isomorphism then gj is bijective

for all j; for cwdi#j 0 if i is not in the image of gj: Now, if fh j fhk thencwdgjh#j cwdgjk#j; which implies j k:

Conversely, if S is non-degenerate then cw is an isomorphism with inverse

cw

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Andruskiewitch, Grana - From Racks to Pointed Hopf Algebras.pdf