The Word Prolem in Qundles Benjmin Fish Advisor: Ren Levitt April 5, 2013 1
1 Introdution A word over n lger A is finite sequene of elements of A, prentheses, nd opertions of A defined reursively: Given eh n-ry opertion of A, if 1, 2,, n re words, then ( 1, 2,, n ) is lso word. The word prolem over n lger A is the following deision prolem: Question. Given pir of words w 1,w 2 over A, is w 1 = w 2 in A? The relted isomorphism prolem is the deision prolem tht sks if two lgers re isomorphi. The word prolem ws first introdued y Dehn in 1911. He proved tht it ws solvle for some groups, ut it ws proved undeidle in the generl se y Novikov in 1955 [7]. Sine then, the word prolem hs een proved either deidle or undeidle for mny lgers, inluding mny types of groups [2] [4]. See [8] for more. However, fr littler is known out the word prolem in qundles. A qundle is n lgeri struture tht rises from the Reidemeister moves for knots. It is omplete invrint of knots, ut the isomorphism prolem is extremely diffiult in qundles. The word prolem over qundles is step in the diretion of the isomorphism prolem. While the word prolem in generl is undeidle in qundles, the word prolem for free qundles (nlogous to free groups) is deidle. In 1984, Winker introdued notion of Cyley grphs for qundles [10]. This ws motivted y the use of Cyley grphs for groups. A Cyley grph of group is representtion of tht group tht depits its struture: There is vertex for every element in the group nd n edge from elements x to y leled with genertor g of tht group if xg = y [5]. Cyley grphs for groups re importnt for the following reson: Theorem. The word prolem for group with generting set S is solvle if nd only if there exists n lgorithm to onstrut ny finite portion of the Cyley grph of the group with generting set S. Thus good trnsltion of Cyley grphs for groups into one for qundles my help solve the word prolem in qundles. In this pper, we improve the lgorithm introdued y Winker to rete Cyley grphs for qundles. We gurntee tht finite qundles Cyley grphs re onstruted in finite time, nd tht every element in the qundle is inluded in the grph, two gurntees tht the originl lgorithm does not mke. Using this improved lgorithm, we re le to 1) lssify ertin lsses of qundles nd 2) solve the word prolem for ertin lsses of qundles. In doing so, we introdue tehnique tht n e used to solve the word prolem for other qundles. Those lsses of qundles inlude qundles mde from Coxeter groups, nd qundles with single reltion. Coxeter groups nd singlereltor qundles were hosen euse of their nie presenttions nd the ft tht the word prolem in Coxeter groups nd single-reltor groups is solved [8] [9]. Our min results re the following: First, n lgorithm for onstruting Cyley grphs tht stisfies ll of our desired properties. It lwys onstruts the Cyley grph, nd will do so in finite time if the qundle is finite. Furthermore, if it onstruts ny finite portion of the grph in finite time for given qundle, it is gurnteed to solve the word prolem for tht qundle. Our seond min result is tht ll one-reltor involutory qundles on two genertors hve solvle word prolem. 2
Setion 1.1 introdues qundles nd qundle presenttions. Setion 1.2 introdues Cyley grphs for qundles. Setion 2 gives n improved lgorithm for finding Cyley grphs for qundles, nd then proves its orretness nd effetiveness. Setion 3 gives some exmples of qundle presenttions, nd how to use the lgorithm to lssify nd solve the word prolem. Speifilly, it introdues trnsltion of groups to qundles, nd solves the word prolem for the qundles ssoited with ertin Coxeter groups. In Setion 3.1, we lssify one-reltor involutory qundles on two genertors, nd give more generl results on how to solve the word prolem in other qundles using the lgorithm. 1.1 Qundles nd Presenttions A qundle is defined s the following: Definition 1. A qundle (Q,, /) is set Q together with two inry opertions, / suh tht I. Idempotene: x x = x IIA. Right Cnelltion A: (x y)/y = x IIB. Right Cnelltion B: (x/y) y = x III. Right Self-Distriutivity: (x y) z = (x z) (y z) for ll x, y, z Q. We ll the two opertions multiplition nd division respetively. Despite wht these nmes might imply, the two opertions re not t ll similr to the multiplition nd division found in the integers nd re simply lled so for the ske of onveniene. As slight use of nottion, we typilly ll Q the qundle. Note tht due to nelltion, the / opertion is uniquely defined y the opertion. In other words, we only need to define the opertion to well-define qundle. If n lgeri struture stisfies only xioms II nd III, it is lled rk. If the two opertions nd / define the sme opertion, then it is lled n involutory qundle. Tht is, Definition 2. An involutory qundle (Q, ) is set Q together with inry opertion suh tht I. Idempotene: x x = x II. Right Cnelltion: (x y) y = x III. Right Self-Distriutivity: (x y) z = (x z) (y z) for ll x, y, z Q. It is lwys possile to left-ssoite words in qundle. If word is left-ssoited, nd no idempotene or right nelltion rule n e pplied, it is sid to e in norml form. For exmple, (( ) ) = (((( ) ) ) ) = ((( ) ) ). When written in left-ssoited form, in involutory qundles nd only when it is ler, we typilly write it without expliitly showing the opertions. For exmple, ((( ) ) ) my e written s. There re mny wys to mke qundles, inluding from groups. The following re few exmples: Exmple 3. Let (G, ) e group. Then (G,, /) where is defined s x y y 1 xy nd / is defined s x/y yxy 1 is qundle lled the onjugtion qundle of G. 3
Proof. It is not diffiult to prove tht this is qundle y showing tht it stisfies the xioms of qundle: I. x x = x 1 xx = x. II. (x y)/y = (y 1 xy)/y = y(y 1 xy)y 1 = x. III. (x/y) y = (yxy 1 ) y = y 1 (yxy 1 )y = x. IV. nd Then (x y) z = (x z) (y z). (x y) z = (y 1 xy) z = z 1 (y 1 xy)z (x z) (y z) = (z 1 xz) (z 1 yz) = (z 1 yz) 1 (z 1 xz)(z 1 yz) = (z 1 y 1 z)(z 1 xz)(z 1 yz) = z 1 y 1 xyz. Another wy to rete qundles from groups is through the yli group: Exmple 4. Let Z p e the yli group of order p. Then the dihedrl qundle (Z p,, /) is the qundle defined y x y 2y x (mod p) nd x/y 2y x (mod p). Note this is n involutory qundle. For more exmples, we onsider qundle presenttions, nlogous to group presenttions: A qundle presenttion is set of genertors G nd set of reltions R, written G R. This is qundle whose underlying set re equivlene lsses of words. To define this, we first need to define some useful sets of words (lnguges), reltions, nd how to derive new reltions from them. First, n lphet is set of symols. If we let A e the lphet onsisting of the symols (, ),, / nd elements of set G, then word over A is sequene of those symols. Sine we don t wnt nonsense words, we define ouple of useful lnguges: Definition 5. L(G) is the lnguge over A defined reursively in the following mnner: 1. For every g in G, g is in L(G). 2. Given two words x nd y in L(G), (x) (y) nd (x)/(y) re lso in L(G). Typilly, prentheses re omitted when it is ler they re unneessry. For exmple, given, in G, () () is typilly written. As mentioned ove, words of qundle hve norml form: Definition 6. N(G) is defined s the suset of L(G) where x is in N(G) if x = ( ((g 1 g 2 ) g 3 ) g n 1 ) g n for g i in G nd in {, /, suh tht g 1 g 2 nd x ontins no suexpressions of the forms (y g i )/g i+1 or (y/g i ) g i+1 where y is suexpression nd g i = g i+1. 4
There is n onto normliztion funtion n : L(G) N(G). This funtion is defined y repeted uses of the xioms to first get n ritrry word into left-ssoited form, nd then use the first two xioms to nel ny repeted genertors. For this pper, it will lso e useful to onsider words tht hve een left-ssoited nd right-nelled, ut idempotene hs not een used. We will ll this form noniliztion: Definition 7. C(G), the set of nonilized words, is defined s the suset of L(G) where x is in C(G) if x = ( ((g 1 g 2 ) g 3 ) g n 1 ) g n for g i in G nd in {, /, suh tht x ontins no suexpressions of the forms (y g i )/g i+1 or (y/g i ) g i+1 where y is suexpression nd g i = g i+1. Given word x in L(G), we write x to e the nonilized form of x. Now we define how to derive reltions over these lnguge from other reltions: Definition 8. A derivtion (X) over lnguge L nd set of reltions R is finite sequene of reltions in L suh tht eh reltion p i = q i n e derived using the previous reltions p 1 = q 1, p 2 = q 2, p i 1 = q i 1 in the sequene nd R using set of rules X. In this pper, we will use few different rule sets. To desrie derivtion, it suffies to desrie the set of rules nd the lnguge L. For lnguge L, we ll X(R) the set of reltions for whih there is derivtion (X) tht inludes tht reltion. Exmple 9. We ll D the type of derivtion, over the lnguge L(G) for some set of genertors G, under the following rules: Reflexivity: For ny word w in L(G), derive w = w. Symmetry: Given reltion p = q derivle from R, derive q = p. Trnsitivity: Given reltions p = q nd q = r derivle from R, derive p = r. Right Multiplition: Given reltion p = q derivle from R, nd for ny word w in L(G), derive p w = q w. Left Multiplition: Given reltion p = q derivle from R, nd for ny word w in L(G), derive w p = w q. Using D nd given set of reltions, we n derive new reltions. For exmple, using the reltion = nd the reltions deriving from the qundle xioms, we n derive ( ( )) ( ( )) = : 1. = (primry reltion) 2. ( ) ( ) = ( ) (left multiplition of 1. y ) 3. ( ) = (involutory qundle xiom) 4. ( ) ( ) = (trnsitivity of 2. nd 3.) 5. ( ( )) ( ( )) = ( ) ( ) (involutory qundle xiom) 6. ( ( )) ( ( )) = (trnsitivity of 4. nd 5.) This is then derivtion (D) over the lnguge L(G) nd R = A G {(( ), ), where G = {,, nd A G is defined elow. 5
Definition 10. Given generting set G, A G is the set of reltions of the following forms: x x = x (x y) y = x (x y) z = (x z) (y z) for x, y, z in L(G). We now hve the tools to define presenttion of qundle. We strt with the presenttion of n involutory qundle: Definition 11. The involutory qundle given y the presenttion G R is defined to e L(G)/D(R A G ). Tht is, G R is the underlying set L(G) where equlity of words is given y the set of reltions D(R A G ). It is not diffiult to hek tht this is n involutory qundle. In the non-involutory se, we need to use the non-involutory qundle xioms insted: Definition 12. The qundle given y the presenttion G R is defined to e L(G)/D(R A G ) where L is L with the ddition of right division nd left division nd A G is the set of reltions of the following forms: x x = x (x y)/y = x (x/y) y = x (x y) z = (x z) (y z) for x, y, z in L(G). See [3] or [6] for other equivlent onstrutions. This will e our min soure of qundles. It is possile to rete qundle presenttions from knots. The underlying set will e the rs of given digrm for the knot, nd the opertions re defined y the rossings: Exmple 13. Let K e digrm of n oriented knot. Let A e the set of rs of K. Let R e the following set of reltions: Where is the overrossing r nd nd re the underrossing rs, let / = e in R if it is right-hnded rossing nd let = e in R if it is left-hnded rossing. Figure 1: Left-hnded rossing Figure 2: Right-hnded rossing Then the knot qundle is the qundle given y the presenttion A R. It turns out tht this qundle is omplete invrint of knots. This is primry motivtion for the use of qundles. In ft, the qundle xioms ome diretly from the Reidemeister moves: In 6
order to prove tht this qundle is n invrint, it must e invrint over the Reidemeister moves, whih in turn mens tht ny lgeri ojet reted in this mnner must stisfy numer of xioms, whih just so hppen to e the qundle xioms. The following is n exmple of n involutory knot qundle. Figure 3: The figure eight knot Exmple 14. First we lel the rs of figure eight knot, s shown in Figure 3. Then the involutory qundle presenttion is,,, d =, d =, = d, d =. Note sine this is for n involutory qundle, / is defined to e nd the hndedness of eh rossing does not mtter. Written s Cyley tle ( multiplition tle), the qundle looks like: * d d d d d d d d d d d d d d For eh row i nd olumn j the entry t (i, j) is i j. Rememer tht qundle is uniquely defined y its opertion, so there is no need to give tle for its / opertion. Or viewed s Cyley grph, whih will e disussed in more detil lter, it looks like Figure 4. d d d d d Figure 4: Involutory Cyley Grph for,,, d =, d =, = d, d = 7
But how do we know tht those re the only elements of the qundle? How do we know tht two of those elements ren t tully the sme? Formlly, we wnt to solve the word prolem: Definition 15. Let Q =< G R >. The word prolem over Q is the following deision prolem: Instne: A pir of words w 1, w 2 in L(G). Question: Is w 1 = w 2 in Q? This is equivlent to sking if the reltion w 1 = w 2 is in derivtion (D) over L(G) nd the reltions R. We will need dditionl tools to e le to nswer this question. We will ttempt to nswer this question for vriety of qundle presenttions. 1.2 Cyley Grphs In order to etter determine wht qundle looks like from its presenttion, it will e helpful to depit wht the genertors do to the words. Definition 16 (Winker). The Cyley grph of qundle Q = G R is direted, leled grph ssoited with Q suh tht there is extly one vertex leled x for every x in Q nd whenever x g = y for elements x, y Q nd genertor g G, there is direted edge from x (the vertex leled x) to y leled g. Note we sy tht word x lels vertex if x in nonil form n e found t tht vertex y following the pproprite edges. x g is found y following the direted edge from x leled g. x/g is found y following the direted edge to x leled g. In the se of n involutory qundle, we n use undireted edges insted, euse x g = y implies y g = x: Definition 17 (Winker). The Cyley grph of n involutory qundle Q = G R is n undireted, leled grph ssoited with Q suh tht there is extly one vertex leled x for every x in Q nd whenever x g = y for elements x, y Q nd genertor g G, there is n edge etween x (the vertex leled x) nd y leled g. The onstrution of these grphs n very diffiult nd my not lwys e possile. This diretly reltes to the diffiulty of the word prolem. When the word prolem isn t solvle, it is not possile to rete its Cyley grph, euse if it were then you ould use the grph to solve the word prolem. Nevertheless, the following onditions hold in Cyley grphs: 1. For every g G, there is vertex leled g (with n edge leled g tht direts to itself.) 2. For every vertex q nd g G, there is extly one edge leled g direted from q nd extly one edge leled g direted to q. (Note in the se of involutory qundles this just mens tht there is extly one edge leled g t q.) You n see tht these onditions hold in the exmple given in Figure 5. Note it is esy to determine wht element of the qundle ny word over the genertors is equl to y onverting the word to norml form nd then following the edges, strting t the first genertor in the word. It is not esy to tell, however, if this grph is in ft the Cyley grph for, =, =. Nmely, re there ny elements of the qundles tht pper t multiple verties? Winker desries n lgorithm tht ensures the grph reted is in ft the Cyley grph: no elements re duplited in the grph [10]. 8
Figure 5: Involutory Cyley Grph for IQ(I 3 ) =, =, = 2 Construting Cyley Grphs We need to ensure tht the reltions R given in presenttion G R re refleted in grph. Unfortuntely, there re other so-lled seondry reltions tht n e derived from R nd the qundle xioms tht my not e refleted in the grph tht the lgorithm lso must tke into ount. For exmple, in qundle with the reltion ( ) =, even if this reltion is refleted in grph y pth from to through, other reltions tht n e derived from it re not refleted y this pth, suh s (( ) ) = or (( ) ) ( ) = ( ). The question is then wht reltions you need to onsider to mke sure ll possile reltions derivle re refleted in the grph. It turns out tht two kinds of other reltions re suffiient: Definition 18. Let Q = G R. Let S(R), the seondry reltions of R, e the set of reltions of the form x s = x t for ll x in C(G) nd reltions s = t in R. Definition 19. Let Q = G R. Let I G, the idempoteny reltions, e the set of reltions of the form g g = g for ll g in G. Using these reltions, we n rete the Cyley grph of qundle G R y dding nd merging verties in prtilly finished grph until there re no more reltions to onsider. The proof of its orretness will e given for the Cyley grphs of involutory qundles, ut the generl proof for non-involutory qundles follows in the sme mnner. The lgorithm when the qundle is involutory is s follows: Algorithm 20. Let G R e qundle presenttion where G, R <. If R = 0, i.e. if G R is the free qundle, then to simplify the lgorithm nd its proof we set R to e G g = g for some g in G. (The proof tht ll elements of the qundle re represented in the qundle relies on the existene of t lest one reltion, nd this dded reltion does not hnge the qundle.) Strt with the prtil grph onsisting of vertex leled g for every g in G. Then tre the reltions R I G. To tre reltion is to do the following: For given reltion 1 2 i = 1 2 j, If 1 2 i or 1 2 j re not in the grph, rete the neessry edges nd verties. Tht is, rete the vertex 1 2 nd the edge leled 2 from 1 to 1 2, et. Then merge the two verties 1 2 i nd 1 2 j y removing one of them nd setting ll the edges tht used to go from tht vertex to the other. 9
Then repet the following until there re no more verties to merge: For ny newly merged verties p nd q nd for ll g in G, merge the verties p g nd q g if oth verties exist in the prtil grph. After R I G hs een tred, we will tre suset, whih we will ll T, of S(R) in the following mnner: For eh reltion p = q in R, tre x p = x q if x is lel of vertex in the grph or x = y g for some y whih is lel of vertex nd ll g in G. (Note tht it is not neessry to tre x p = x q if it hs lredy een tred.) After these reltions hve een tred, there my e new verties in the grph. Repet tring this set of seondry reltions until there re no new verties in the grph. Note tht this lgorithm will not terminte if the qundle is infinite. However, the lgorithm will nevertheless e helpful in showing wht the struture of these infinite qundles is. Now, in the generl se when the qundle isn t neessrily involutory, we modify this lgorithm in the following wys: The lgorithm insted retes direted edges. When two verties leled p nd q re merged, now not only p g nd q g need to e merged for eh g in G, ut p/g nd q/g lso need to e merged. Similrly, the seondry reltions x p = x q nd x/p = x/q need to e tred for eh primry reltion p = q where x is lel of vertex or x = y g for some y whih is lel of vertex nd ll g in G. Exmple 21. As n exmple, we show how the lgorithm runs on the involutory qundle presenttion, =. First, we rete vertex for every genertor, s shown in Figure 6. Figure 6: Prtil Cyley Grph for, = with the genertors Then the reltions I G re tred, s shown in Figure 7. Figure 7: Prtil Cyley Grph for, = with the genertors nd I G After, the primry reltion is tred, s shown in Figure 8. Figure 8: Prtil Cyley Grph for, = with the genertors, I G, nd the primry reltions Then, the seondry reltions re tred. There re two verties, nd, nd there is one word,, where = x g where x is either or nd g is genertor, ut x g is not lredy in the grph. Then the seondry reltions tht re tred re ( ) =, ( ) =, 10
Figure 9: Cyley Grph for, = nd ( ) ( ) = ( ). Converting these to nonil form, we get =, =, nd =. These re then tred, resulting in Figure 9. Sine the seondry reltions hve lredy een tred for every vertex in the grph nd for every x g, where x is lel of vertex nd g is genertor, x g is lredy leled y vertex in the grph, the lgorithm termintes. Then Figure 9 is the Cyley grph for, =. We prove tht the lgorithm works in the involutory se. In order to prove this lgorithm works, we will need to show tht ny reltion tht n e derived from the presenttion is refleted in the grph nd tht there is one nd only one vertex for every distint element in the qundle. We strt with the first of these. To do this, we show whih reltions re utomtilly implied y tring some given set of reltions: Proposition 22. Let Π e the prtil grph given y using the ove lgorithm to tre eh reltion in some set R over the genertors G. Two words v nd w lel the sme vertex if nd only if the reltion v = w is in D r (R) over the lnguge C(G), where D r is the following set of rules: Reflexivity: For ny word w in C(G), derive w = w. Symmetry: Given reltion p = q derivle from R, derive q = p. Trnsitivity: Given reltions p = q nd q = r derivle from R, derive p = r. Right Multiplition of genertors: Given reltion p = q derivle from R, nd for ny genertor g in G, derive (p g) = (q g). Proof. First ssume v = w is derivle over D r ut v nd w do not lel the sme vertex. If v = w is derivle, then there is series of steps resulting in v = w, eh of whih results in derived reltion. Let the reltion x = y e the first reltion in these steps where x nd y don t lel the sme vertex. Now there will e ontrdition no mtter wht kind of step derived x = y: If x = y ws reltion in R, then x nd y lel the sme vertex, s ditted y the onstrution of Π, ontrdition. Reflexivity: The reltion is in the form x = x. But x must lel the sme vertex s x, ontrdition. Symmetry: The reltion x = y is derived from y = x. But y nd x lel the sme vertex, s this reltion ws the first reltion whih doesn t lel the sme vertex, ontrdition. Trnsitivity: The reltion x = y is derived from x = z nd z = y. But then x lels the sme vertex s z nd z lels the sme vertex s y, so x lels the sme vertex s y, ontrdition. Right multiplition of genertors: The reltion is in the form p g = q g for some g in G, derived from some reltion p = q. But y ssumption p lels the sme vertex s q, nd y the onstrution of Π p g must lel the sme vertex s q g, ontrdition. 11
For the other diretion, ssume tht v nd w lel the sme vertex. Then t some point in the onstrution of Π, two verties leled v nd w were merged. In this se, either it ws reltion of R, in whih se it is ertinly derivle, or v = p g nd w = q g for some two verties leled p nd q tht were merged. Induting over the length of the word, p = q ws derivle, nd so p g = q g is lso derivle. Unfortuntely, this mens tht the Cyley grph doesn t reflet ll of the rules tht n e used to derive new reltions. To reflet left multiplition, we need to tre dditionl reltions. These re the seondry reltions S(R): Proposition 23. Let x = y e in D(R) for some set of reltions R over L(G) for some set of genertors G. Then x = y is in D r (R S(R)) over C(G). Proof. Sine x = y is in D(R), there is some derivtion (D) to x = y. For eh reltion in (D), reple it with its nonilized form. Let D e D with noniliztion. Tht is, D is the following set of rules: Reflexivity: For ny word w in L(G), derive w = w. Symmetry: Given reltion p = q derivle from R, derive q = p. Trnsitivity: Given reltions p = q nd q = r derivle from R, derive p = r. Right Multiplition: Given reltion p = q derivle from R, nd for ny word w in L(G), derive (p w) = (q w). Left Multiplition: Given reltion p = q derivle from R, nd for ny word w in L(G), derive (w p) = (w q). This new sequene will e vlid sequene in (D ), so x = y is in D (R). We now need to rest this derivtion to e derivtion (D r ). However, s it stnds, it my ontin right multiplitions y n ritrry word w in C(G): p = q implies (p w) = (q w). But (p w) = pg 1 g 2 g i nd similrly (q w) = qg 1 g 2 g i, so this derivtion my e expnded to series of right multiplitions y genertors, one of the rules of D r. It lso my ontin left multiplitions y n ritrry word w in C(G). We show tht when there re no left multiplitions efore given one tht derives the nth reltion of the derivtion, the first n reltions n e rest to sequene of reltions tht do not use left multiplitions ut insted my use reltions from S(R). This suffies, s for ny susequent left multiplitions this proess n then e repeted until there ren t ny left multiplitions left in the rest sequene. So ssume tht there re no left multiplitions efore the given one tht derives the nth reltion, x p = x q, of the derivtion, where p = q ws the reltion used to derive this one. This will e shown y indution on the numer of reltions in the derivtion. In other words, either the left multiplition will e moved erlier in the derivtion nd we repet, or it will e removed. In generl, the wy we do this is through the involutory qundle xiom III: We rest reltions derived from left multiplitions y first using seondry reltion nd then deriving wht we would hve derived from the left multiplitions y right multiplitions y genertors nd then tking dvntge of the involutory qundle xiom III. The se se is when n = 1. The lim is vuously true, s ny left multiplition requires using previous reltion in the derivtion, whih is ontrdition when there is only one reltion. Now suppose the lim is true for some n 1. It now suffies to show tht we n rest this sequene so the left multiplition ours no lter thn deriving the n 1th step. We do this with se-y-se nlysis on the n 1th step, whih is without loss of generlity one of the reltions used to derive the nth reltion: 12
If the n 1th reltion is p = q in R, then the nth reltion n remin x p = x q, s it is reltion in S(R). If the n 1th reltion is y p = y q in S(R), then the following sequene of reltions my e sustituted: (n 1) x p = x q (reltion in S(R)) (n) x p y = x q y (right multiplition) (n + 1) x p y p = x q y p (right multiplition) (n + 2) (x q y) p = (x q y) q (reltion in S(R)) (n + 3) x p y p = x q y q (trnsitivity) This lst reltion is the desired one, x (y p) = x (y q), in nonilized form. If the n 1th reltion is derived from the symmetry rule, i.e. this reltion is p = q, ws derived from some erlier reltion q = p, nd the nth reltion is x p = x q. Then we n sustitute the following reltions: (n 1) x q = x p (left multiplition) (n) p = q (symmetry) (n + 1) x p = x q (symmetry) Now pply the indution hypothesis to rest the first n 1 reltions. If the n 1th reltion is derived from trnsitivity, i.e. the n 1th reltion is p = q nd ws derived from erlier reltions p = r nd r = q. Now we need to rete two different derivtions whih hs een hnged only in the n 1th step. The new n 1th steps re x p = x r nd x r = x q. We n pply the indution hypothesis to oth of these new derivtions to rest them, nd then derive x p = x q from those derivtions, s desired. If the n 1th reltion is derived from right multiplition of genertor, i.e. the n 1th reltion is p g = q g nd ws derived from n erlier reltion p = q. Then we n sustitute the following reltions: (n 1) (x g) p = (x g) q (left multiplition) (n) p g = q g (right multiplition) (n + 1) x g p g = x g q g (right multiplition) This lst reltion is the desired one, x (p g) = x (q g), in nonilized form. Now pply the indution hypothesis to rest the first n 1 reltions. Now note tht we my ssume reflexivity is not used, s the only reltions tht n e derived from it is the reltion w = w, whih is derivtion lredy in the desired form. This is ll the possile ses, nd so we hve suessfully moved up left multiplitions until they ould e repled with reltions from S(R). Hene this new derivtion to x = y is derivtion (D r ) using the reltions R S(R). Thus x = y is in D r (R S(R)). Now we n prove tht two words re equl if nd only if they n e derived in the mnner pproprite for the Cyley grph using only the sets of reltions R, S(R), nd I G. 13
Corollry 24. p = q is in D(R A G ) if nd only if p = q is in D r (R S(R) I G ). Proof. Assume p = q is in D(R A G ). By Proposition 23, p = q is in D r (R A G S(R A G )). However in D r derivtions ll reltions re nonilized, in whih se reltions from A G orrelting to xioms II nd III re unneessry, s they re trivilized when nonilized to x = x, whih is lredy rule in D r. Then D r (R A G S(R A G )) = D r (R S(R) I G ). For the other diretion, sine there is some derivtion (D r ) to p = q, (D r ) n e rest y removing ll noniliztion nd y hnging ll instnes of xioms from S(R) to uses of the left multiplition rule. Also, I G A G, so this is now derivtion (D) to p = q using only the reltions R A G. Hene p = q is in D(R A G ), s desired. Sine R, S(R), nd I G re the reltions tred in the lgorithm nd D(R A G ) re the reltions tht n e derived from the presenttion, this shows tht ny reltions tht n e derived from the presenttion is refleted in the Cyley grph. Now we show tht there is one nd only one vertex for every distint element in the qundle. Proposition 25. Let Γ e the grph onstruted s y the ove lgorithm for given qundle presenttion Q = G R. For every x in C(G), x lels some vertex in Γ. Tht is, ll of the neessry verties re in the grph. Proof. Let x = g 1 g 2 g n in C(G). If x = z is reltion in R I G for some z in C(G), then y the definition of the lgorithm x would hve een tred, nd x indeed lels some vertex. Otherwise, we prove using indution over n tht x p = x q for p = q in R gets tred. g 1 is genertor nd therefore g 1 p = g 1 q is tred. Now ssume tht g 1 g 2 g i p = g 1 g 2 g i q is tred. Then g 1 g 2 g i is vertex in Γ. Thus either g 1 g 2 g i g i+1 is lredy vertex in Γ, or it isn t, ut then g i+1 is genertor nd g 1 g 2 g i is vertex, so g 1 g 2 g i g i+1 p = g 1 g 2 g i g i+1 q susequently gets tred. Sine y definition of the lgorithm, R > 0, g 1 g 2 g i g n lels some vertex. Finlly, we prove tht there is only one vertex for every unique element of the qundle. Corollry 26. Let Γ e the grph onstruted s y the ove lgorithm for given qundle presenttion Q = G R. Then p nd q re the sme element of the qundle if nd only if p nd q lel the sme vertex of Γ. Proof. By definition, p nd q re the sme element of the qundle if nd only if p = q is reltion in D(R A G ). By Corollry 24, p = q is reltion in D(R A G ) if nd only if p = q is in D r (R S(R) I G ). And y Proposition 22, p = q is in D r (R T I G ) if nd only if p nd q lel the sme vertex of Γ, where T is the suset of S(R) tred in the lgorithm. Rell T is omposed of one seondry reltion for eh vertex in Γ. It remins to show tht p = q is in D r (R S(R) I G ) if nd only if p = q is in D r (R T I G ). First of ll the kwrds diretion follows from T eing suset of S(R). Now let (D r ) e derivtion of the reltion p = q in D r (R S(R) I G ). For eh seondry reltion x = x in (D r ), let y = y e the reltion in T suh tht x nd y lel the sme vertex in Γ. From the definition of the lgorithm, there is t lest one reltion in T z = z suh tht z lels given vertex. Also, y Proposition 25, the vertex t z exists, so this reltion exists. x nd y lel the sme vertex in Γ, so x = y is in D r (R T I G ). Let δ(d r ) e the derivtion of x = y. Then we my reple the seondry reltion x = x with: 1. x = y (δ(d r )) 14
2. x = y (sequene of right multiplitions y genertors) 3. x = y (sequene of right multiplitions y genertors) 4. y = y (reltion in T ) 5. x = x (trnsitivity) This derivtion now uses only reltions in D r (R T I G ), so p = q is in D r (R T I G ), s desired. We now put together the ove proofs to show tht the onstrution for the Cyley grph is orret: Theorem 27. Let Γ e the grph onstruted s y the ove lgorithm for given qundle presenttion Q = G R. Then Γ is the Cyley grph for Q. We must show tht the ove lgorithm stisfies Definition 16. Proof. First we need to show tht if x g = y for x, y in Q nd g in G, there is direted edge from x to y leled g. But if x g = y in Q, then y Corollry 26, x g nd y lel the sme vertex, i.e. there is n edge from x to y leled g. Now we must show tht for every q in Q, there is vertex in Γ leled q. Write q in nonilized form. Then this follows diretly from Proposition 25. Finlly, we must show uniqueness of the verties, i.e. if p = q in Q then p nd q lel the sme vertex. This is implied y Corollry 26, so Γ is indeed the Cyley grph for Q. We now show tht not only is the lgorithm orret, ut tht it works quikly enough to e useful. Nmely, we wnt the lgorithm to tke finite time when the qundle is finite. Theorem 28. Let Q = G R where G nd R re finite. If Q is finite, then Algorithm 20 tkes finite time. Proof. It suffies to show tht R T I G is finite. Sine G nd R re finite, R I G is finite. Now in the tring of T, in eh itertion, R times the numer of new verties in the grph re tred. The numer of new verties is ounded y the totl numer of verties t the end of the lgorithm, whih must e finite if Q is finite. Then T is finite, s desired. 3 Using Cyley Grphs Using Cyley grphs llows us to simplify rgumenttion out qundles quite it. In order to show the desirility of this simplifition, we show how to ound the size of qundle with nd without qundles. The qundles we will do this for re qundles tht ome from Coxeter groups. See [1] for more on Coxeter groups. We introdue method to rete qundle presenttions from Coxeter groups: Definition 29. For Coxeter group H given y presenttion G R, we define Q(G, R) to e the qundle given y the presenttion G R, where R = {(r 1 r 2 ) r n 1 = r n r 1 r 2 r n R. 15
Sine for ll Coxeter groups, for every genertor r, r 2 = e, it is sfe to ssume tht there re no inverses in ny word in R. Furthermore, in this trnsltion, the ssoited reltion in the qundle to the reltion r 2 = e is r = r, trivil reltion. Then there is no need to inlude these trivil reltions in the set of reltions for the qundle presenttion. Similrly, Definition 30. For group H given y presenttion G R, we define IQ(G, R) to e the involutory qundle given y the presenttion G R, where R = {(r 1 r 2 ) r n 1 = r n r 1 r 2 r n R. These my not e well-defined opertions for given group. Tht is, two different presenttions for the sme group my yield different qundles. However, when it is ler whih presenttion we re using, we employ the nottion IQ(Q) for IQ(G, R). We will study the qundles IQ(I k ), where I k is the Coxeter group given y the Coxeter mtrix ( 1 k k 1 ). In other words, Ik =, 2, 2, () k, () k. Then IQ(I k ) =, =, =. First, we show the following ound on the size of IQ(I k ) without the use of Cyley grphs. Lemm 31. IQ(I k ) 2k 1. Here we use n indutive rgument insted. Proof. Let W e {,,,,.,,,,,.. k k (Note not ll of these words re neessrily unique.) We will show tht IQ(I k ) = W. Speifilly, we will show tht every word in IQ(I k ), in norml form, hs length less thn k. Rell tht the norml form of word is when the word hs een left-ssoited, nd instnes of xioms I nd II hve een used on it to shorten it s muh s possile (see Definition 6). Let w e word in IQ(I k ) e written in norml form. Then w = or w =. We will indut on w. When w < k, sine w is in norml form, w must e in W, s desired. Else w k. Without loss of generlity, w =. Sine w k, we n use the reltion =. Sustituting, we get w = () =. Repet until the length of w is no more thn k, in whih se it must e in W, proving the lim. W = 2k. Furthermore, = so =. Then k k. IQ(I k ) W 1 = 2k 1 However, Cyley grph for IQ(I k ) with verties will demonstrte the sme thing, without hving to go through n indutive rgument. 16
k 1.. k 1 k Figure 10: Involutory Cyley Grph for IQ(I k ) This grph, shown in Figure 10, will provide enough informtion to determine extly wht IQ(I k ) is. But first, we need to e le to trnslte the grph into qundle. The first step is to define homomorphism etween qundles tht needs only to e defined on the genertors tht n e then extended to the entire qundle: Proposition 32 (Winker). Let G R e presenttion of qundle Q. Let h : G Q e mp from the genertors to qundle Q. Let h : Q Q e n extension of h defined y h (t(g 1, g 2,, g n )) = t(h(g 1 ), h(g 2 ),, h(g n )) where t(g 1, g 2,, g n ) is term over the genertors in G (i.e. ny element of Q). h is well-defined homomorphism if h preserves every reltion of Q. Tht is, if w 1 = w 2 is reltion in R, then h (w 1 ) = h (w 2 ). With this homomorphism h, we n now give onditions for when it is n isomorphism: Corollry 33. If suh well-defined h : Q Q exists, h is onto, nd Q Q, then Q = Q. If suh n h : Q Q exists, then Q h(q). But if h is onto, then Q = h(q), in whih se Q Q Q, i.e. Q = Q. Thus h is lso injetive, nd hene is n isomorphism. Finlly, this llows us to trnslte the grph into qundle y relting the two through h : Corollry 34. Let Γ = (V, E) e the Cyley grph ssoited with qundle Q. Let Q e the qundle (V,, /) where is defined in the following mnner: x y is the element of V given y tring x y written in norml form. Tht is x y = ( ((g 1 g 2 ) g 3 ) g k 1 ) g k for some g 1, g 2,, g k in G. Strting with the vertex g 1, there exists pth given y tking edges g 2, g 3, et. in suession. x y is then defined s the vertex t the end of this pth. Then Q = Q. In other words, the Cyley grph of the qundle represents the qundle in n intuitive fshion. To prove it, we merely need to hek tht the onditions from Corollry 33 hold true. 17
Proof. First, it is strightforwrd to hek tht Q is well-defined. Let h : G Q e given y h(g) = g, where g is the vertex leled g. This is well-defined mp y Definition 16. Let h e given s in Proposition 32. h preserves reltions, s for ny reltion x = y in R, h (x) nd h (y) re verties in Γ nd must e the sme vertex y Definition 16. Then y Proposition 32, h is well-defined homomorphism. The ijetion etween elements of Q nd verties lso oth implies h is onto nd the numer of verties is equl to the size of the qundle. Then Q = Q. Thus we hve stisfied the onditions for Corollry 33, s desired. Corollry 35. IQ(I k ) = 2k 1. Insted of hving to use n indutive rgument, we n show tht the size of IQ(I k ) is 2k 1 muh more esily using Cyley grph. To show tht Figure 10 is the Cyley grph for IQ(I k ), it suffies to give the reltions needed to tre the grph for given k. = nd The primry reltions re = nd =. The seondry reltions re: =. The idempotene reltions re ( ) = () ( ) = ().( ) =.() l l.( ) =.() l l for 1 l < 2k 1. Tring these reltions result in Figure 10, whih hs 2k 1 verties. Thus IQ(I k ) = 2k 1. Furthermore, the Cyley grph gives wy to find out wht ny word in the qundle is: onvert the word to norml form nd then tre it through the grph. As n side, it turns out tht IQ(I k ) re relly nie lss of qundles, euse they re Ltin. Definition 36. A Ltin qundle is qundle Q where for ll x, y, z in Q, if x y = x z, then y = z. The nme is due to the ft tht the Cyley tle of finite Ltin qundle is lso Ltin squre, whih mens there re no repets in ny row or olumn. Proposition 37. IQ(I k ) =, =, = is Ltin. Proof. First, we show tht for ny finite involutory qundle Q, if for some x 0 in Q we hve x 0 y = x 0 z implies y = z for ll y, z in Q, then Q is Ltin. So suppose there exists n x 0 in Q suh tht we hve x 0 y = x 0 z implies y = z for ll y, z in Q. Now let x in Q nd suppose x y = x z. Sine Q is finite, there exists q in Q suh tht x 0 q = x. Then (x 0 q) y = (x 0 q) z. Multiplying on the right, ((x 0 q) y) q = ((x 0 q) z) q. 18
This is equivlent to x 0 (y q) = x 0 (z q). From our ssumption out x 0, y q = z q. Hene y = z, so Q is Ltin. Then it suffies to show tht for ll y, z in IQ(I k ), if y = z, then y = z. We do this y showing tht for some y in Q, y = q for ll q in IQ(I k ). Sine IQ(I k ) is finite, this implies tht y is unique for eh y, whih implies wht we need to show. Let y in Q. We know tht IQ(I k ) = {,,,,.,,,,,., so y =. for some l suh tht 1 l k or y = y =.. If l is odd, then l If l is even, then Now suppose tht y =.. If l is odd, then l If l is even, then l k for some l suh tht 1 l < k. First, suppose tht y = ( ) = =. l 2l 1 2l 1 y = ( ) = =. l 2l 1 2l y = ( ) = =. l 2l 1 2l y = ( ) = =. l 2l 1 2l 1 Thus there exists y in Q suh tht y is in {,,,,. But this set is IQ(I k ). So y = q for ll q in IQ(I k ), s desired. k l 3.1 One-Reltor Qundles Besides qundles reted from Coxeter groups, we re lso interested in one-reltor qundles, whih re qundles with single primry reltion. It turns out there is useful intersetion: Theorem 38. One-reltor involutory qundles on two genertors n e lssified into IQ(I k ) nd J k, where J k is defined s J k =, =. Furthermore, ll one-reltor involutory qundles on two genertors re finite. 2k Proof. Given reltion x = y over two genertors nd, we n ssume without loss of generlity tht x nd y re in norml form nd tht x strts with. x nd y must e of the form nd. Then we n move ll ut one genertor over from y to the left-hnd side, resulting in four possile reltions: =, =, =, or =. For those first two possile reltions, we then multiply oth sides on the right y or respetively, giving 19
two possiilities for this reltion: = for n even numer of genertors on the lefthnd side, or = for n odd numer of genertors on the left-hnd side. This yields IQ(I k ) =, =, nd J k =, =. 2k It remins to give the Cyley grph for J k. The Cyley grph is shown in Figure 11. k.. 2k Figure 11: Involutory Cyley Grph for J k The reltions neessry to tre this grph re the following: The primry reltion is, the idempotene reltions re = nd =. The seondry reltions re:.( ) =.() l 2k l ( ) = () 2l 2k 2l 2k = for 1 l k. Tring these reltions results in Figure 11, so J k = 3k. Hene ll one-reltor involutory qundles on two genertors re finite. Now tht we know how Cyley grphs n e used to explore qundles like the one-reltor involutory qundles, we return to the word prolem: Proposition 39. Let Q = G R. If ny finite portion of the Cyley grph for Q is onstrutile in finite time, then Q hs solvle word prolem. Proof. Let Q =< G R > e given. Let w 1, w 2 in L(G). Using Algorithm 20, onstrut finite portion of Q tht lels oth w 1 nd w 2. From Corollry 26, w 1 = w 2 if nd only if w 1 nd w 2 lel the sme vertex, solving the word prolem. 20
Sine our lgorithm onstruts the entire Cyley grph in finite time for finite qundles, it immeditely follows tht finite qundles hve solvle word prolem. This is very nturl result tht should e expeted to follow from this lgorithm. So fr, ll of the Cyley grphs we ve seen hve een finite. If the Cyley grph is infinite, however, the lgorithm ove to onstrut the grph will tke infinite time. However, the lgorithm niely llows n lterntive pproh to onstrut the Cyley grph: We n indut over the numer of itertions, merely showing tht eh itertion through the lgorithm doesn t ollpse verties introdued in previous itertions. For exmple, the Cyley grph for,, = is shown in Figure 12. In order to show tht this is the Cyley grph, we ssume s the indutive hypothesis tht the prtil Cyley grph fter itertion i hs the following verties: {,,, in the 2k 2k+1 omponent ontining nd {,, in the omponent ontining for eh 1 2k 2k+1 k i. And the only verties reted in step i were the lst four, i.e. nd. 2i+1 Then in itertion i + 1, the following reltions re tred: Normlizing, these reltions re: () = () 2i 2i () = () 2i+1 2i+1 () = () 2i 2i () = () 2i+1 2i+1 2i,,, 2i+1 2i 2i+3 2i 2 2i+3 2i 2 = 2i+1 = 2i+2 = 2i+1 = 2i+2 Tring these reltions, then, introdues four new verties:,,, nd 2i+2 2i+3 2i+2, s desired. It is esy to hek tht tring these reltions yields the struture of the grph 2i+3 seen in Figure 12. Thus the indution is omplete. 21
.. 4 Conlusion Figure 12: Involutory Cyley Grph for,, = In this pper, we hve given n improved lgorithm for solving the word prolem in qundles. This llowed us to solve the word prolem for ll finite qundles, inluding one-reltor involutory qundles on two genertors. It lso gives us method for solving the word prolem for non-finite qundles y tking dvntge of the struture of the lgorithm. This hs given us strong foothold into solving the word prolem for mny qundles. There re mny fronts for ontinuing work. First, there is work to e done on the lgorithm itself. We hve given nive onstrution of the lgorithm, ut lever one with fster symptoti runtime will estlish good ounds on how long it tkes to solve the word prolem when it is indeed solvle. Seond, it is quite possile to prove tht mny qundles hve solvle word prolem using this lgorithm. To find out whih qundles these re, we hve written n implementtion of the lgorithm to speed up this proess. One-reltor qundles ontinue to e good ndidtes, s they seem to ll hve solvle word prolem so fr. In ddition, the lgorithm tkes time proportionl to the numer of reltions, so these re fster to run. Other qundles tht my prove to e interesting re qundles whose presenttions ome from presenttions of groups. We give one wy of trnslting presenttions, ut there needs to e more work on finding well-defined trnsltion tht preserves the property tht the word prolem is solvle. Finlly, the existene nd usefulness of Cyley grphs for qundles suggests tht Cyley grphs my e possile for mny types of lgers. Extending notion of Cyley grphs to other lgers my then led to insight on the word prolem for other lgers. While there is muh more work to e done on the topi, this pper hs mde progress on solving 22
the word prolem in qundles. Referenes [1] M. Dvis. The Geometry And Topology of Coxeter Groups. London Mthemtil Soiety monogrphs: New Series. Prineton Univeristy Press, 2008. [2] Trevor Evns. The word prolem for strt lgers. Journl of the London Mthemtil Soiety, s1-26(1):64 71, 1951. [3] Roger Fenn nd Colin Rourke. Rks nd links in odimension two. Journl of Knot Theory nd Its Rmifitions, 01(04):343 406, 1992. [4] F. A. Grside. The rid group nd other groups. The Qurterly Journl of Mthemtis, 20(1):235 254, 1969. [5] W. Mgnus, A. Krrss, nd D. Solitr. Comintoril Group Theory: Presenttions Of Groups In Terms Of Genertors And Reltions. Dover Books on Mthemtis Series. Dover Pulitions, Inorported, 2004. [6] V.O. Mnturov. Knot Theory. Chpmn & Hll, 2004. [7] P.S. Novikov. On the lgorithmi unsolvility of the word prolem in group theory. Trudy Mt. Inst. Steklov., 44:3 143, 1955. [8] J. Stillwell. The word prolem nd the isomorphism prolem for groups. Bulletin (New Series) of the Amerin Mthemtil Soiety, 6(1):33 56, 1982. [9] J. Tits. Le proleme des mots dns les groupes de Coxeter. Sympos. Mth. Rome 1967/68. [10] Steven K. Winker. Qundles knot invrints nd the N-fold rnhed over. PhD thesis, University of Illinois t Chigo, 1984. 23