Computing twisted conjugacy classes in free groups using nilpotent quotients

Computing t wisted conjugacy classes in free groups using nilp oten t quotien ts P . Christopher Staec k er ∗ Assistan t Professor, Dept. of Mathematical Sciences, Messiah College, Gran tham P A, 17027. No v em b er 21, 2018 Abstract There currently exists no alg ebraic algori th m fo r compu t ing twisted conjugacy classes in free gro up s. W e propose a new tec hnique for deciding tw isted conjugacy relations using nilp oten t qu otients. Our technique is a generalization of the common ab elianizati on method , but admits signifi- cantl y greater rates of success. W e present exp erimen tal results demon- strating the efficacy of the technique, and detail how it can b e ap p lied in the related settings of surface groups and doubly twisted conjugacy . 1 In tro duction Given tw o elemen ts x and y in a gro up G and an endomorphism ϕ : G → G , we say that x and y are t wiste d c onjugate if there is so me z such that x = ϕ ( z ) y z − 1 . Twisted conjugacy is a generaliza tion of the ordinary c onjugacy relation in groups, and the computation of twisted conjugacy classes is a pr oblem of con- siderable difficult y for many gro ups G . Computing twisted conjugac y classes (also called “Reidmeister classes”) is of in terest in v arious algebra ic contexts. Our own approa c h will b e motiv ated by Nielsen fixed point theory , though other motiv ations exist (see [Bo gopols ki et al. , 2006], in which the t wisted conjuga cy pr o blem in free groups arises naturally in the context of the ordinar y conjugacy problem in ce r tain other gr oups). Extant alg ebraic techniques for computing twisted conjugacy clas s es are ad ho c in na ture, and the goal o f this pap er is to pr esen t a new technique which is more genera lly applicable (thoug h still not in gener al algo rithmic), alo ng with exp erimen tal results demonstr ating its success. ∗ Email: cstaeck er@messiah.edu 1 Our technique is a n extension o f the co mmon ab elianization technique. View- ing the a belianizatio n a s the first nilp oten t quo tien t, we show tha t twisted con- jugacy classes can be distinguished with increa sing succ ess rates when pro jected int o nilpo ten t quo tien ts o f incr easing nilp otency cla ss. These pro jections, and the computations ne c essary to compute twisted conjuga cy in nilp oten t gr o ups, can be fairly intensiv e, and a s such we hav e implemented our technique in the Mag ma progr amming language. In Sectio n 2 we give the motiv ation for the twisted conjugacy pr o blem fr om Nielsen theory , Section 3 is an outline of our tec hnique, Section 4 gives some examples, Section 5 prese nts our experimental res ults, a nd Sections 6 and 7 show how the technique can b e adapted into tw o natural generaliz a tions of the main pr oblem. The bulk of this work was completed as part of the author’s doctor al disse r- tation, advised by Rob ert F. Brown. The co n tent in this pap er was a lso advised and suggested b y Peter W o ng. The author wishes to acknowledge their help and supp ort, as well as helpful input from Seungwon K im, Ar mando Martino, and Evelyn Hart. 2 Nielsen fi x ed p oin t theory Our principal motiv ation for studying the twisted conjugacy problem is Nielsen fixed p oint theory (standard re fer ences are [Jia ng, 1983] and [Kiang, 19 80 ]). Given a map f : X → X of a space with universal cov ering space e X and pro jection map p : e X → X , the fixed po in ts of f are partitioned into classe s o f the form p (Fix( e f )), where e f is a lift o f f . If a sing le lift e f is fixed once and for all, each fixed p oin t class can b e ex pressed as p (Fix α − 1 e f ) for some α ∈ π 1 ( X ). A central problem in Nielsen fixed p oint theor y is to determine the n umber of “essential” fixe d p oint classes o f a ma pping. This num b er gives a lower b ound for the minimal nu mber of fixed points for ma ps in the homotopy class o f f , and in the ca se when X is a manifold of dimension at least 3, this “ Niels en num b er” is in fact equal to the minimal num ber of fixed p oint s. A fundamental question whe n counting fixed point classes is the following: given tw o e lemen ts α, β ∈ π 1 ( X ), when is p (Fix ( α − 1 e f )) = p (Fix( β − 1 e f ))? The question can be answered with an elementary arg ument in cov er ing space theor y . Two such fixed po in t classes are equa l if and only if there is some γ ∈ π 1 ( X ) with α = ϕ ( γ ) β γ − 1 , where ϕ : π 1 ( X ) → π 1 ( X ) is the map induce d by f o n the fundamental g roup. If the above holds, we say that the elements α and β are twiste d c onjugate (if ϕ is the ident ity map, this is the ordinary conjuga cy relation). The twisted conjugacy classes of f are also called R eidemeister classes , a nd the set of s uc h classes is denoted R ( ϕ ). In the case where f is a selfmap of a compact surface with b oundary , the fundamen tal g roup π 1 ( X ) will b e a free gro up, say π 1 ( X ) = h g 1 , . . . , g k i . No 2 general algo rithm is known for co mputing the Nielsen num b er o f such a map (the case k = 1 is classically kno wn, and a n algorithm for the case k = 2 has recently bee n dev elo ped by Yi and Kim [Yi a nd Kim, 20 0 8 ], extending w or k b y J o yce W agne r [W agner, 1 999]). F adell and H uss eini, in [F adell and Husseini, 1983], prov ed: Theorem 2. 1 (F a dell, Huss eini, 1 983) . L et ϕ b e the m ap induc e d by f on π 1 ( X ) . Then the Nielsen numb er of f is the numb er of t erms with nonzer o c o efficie nt in R T ( ϕ ) = ρ 1 − n X i =1 ∂ ∂ g i ϕ ( g i ) ! , wher e ∂ is the F ox Calculus op er ator (se e [Cr owel l and F ox , 1963]), and ρ : Z π 1 ( X ) → Z R ( ϕ ) is the line ar ex t ension of the pr oje ction into twist e d c onjugacy classes. The theo rem a b ov e thus reduces the pro blem of computing the Niels en nu mber to the application of the pro jection ρ , which re q uires some algo rithm for distinguishing twisted conjuga cy classes. No such algo rithm exists in the literature, though Bog opolsk i, Martino, Masla k ov a, and V entura giv e an a l- gorithm in [Bo gopo lski et al. , 2006] for the case where ϕ is an isomorphism. Their a lgorithm inv olves the tra in tracks machinery of Bestvina and Handel [Bestvina and Handel, 19 92 ]. This pap er e xplores a purely algebr aic technique which ca n b e implemented by existing computer algebr a systems, and in so me cases can be done by hand. 3 Ab elian and nilp oten t quotien ts Throughout this and the next tw o sections, let G b e a finitely generated free group, and let ϕ : G → G b e an endomorphism. F or an element g ∈ G , let [ g ] denote the twisted conjugacy class o f g . Our go al is to co nfir m or deny the equality [ g ] = [ h ] fo r tw o gro up elements g and h . Existing alg ebraic tec hniques for deciding eq ua lit y of such classes are s ur- vey ed in [Hart, 200 5 ]. O ne such tec hnique is the algo rithm o f W agner [W agner, 1999], which is applicable if the mapping ϕ sa tis fies the combinatorial “remnant ” con- dition (this condition is sa tisfie d with probabilit y approaching 1 as the word lengths of the imag es o f generator s increas e). F or general mappings (without remnant), distinguishing cla sses is t ypica lly done by pro jecting in to the ab elian- ization ¯ G and solving the twisted conjug a cy rela tion there. Abelia nization is often sucessful at sho wing that t wo classes are dis tinct, but cannot b e used to show that tw o c la sses are equal. P ro jecting into nilp oten t quotients is a natur a l generaliz a tion o f the ab elianization technique, and w e show how it can als o furnish a technique for equating cla sses. W e will us e t he commutator notation [ x, y ] = xy x − 1 y − 1 . Let γ n ( G ) b e the terms of the low er central se r ies, a nd let b G n = G/γ n ( G ). Ea c h of the groups b G n are nilp otent of class n . The ab elianization is of c o urse ¯ G = b G 1 = 3 G/γ 1 ( G ). In general, we will use a bar to indicate pr o jection of elemen ts into the ab elianization, and a hat to indicate pro jection into b G n , with the v alue o f n to b e under stoo d by co n text. Computation in ¯ G is made easy by co mm utativity . F or n > 1, the groups b G n are not commutativ e, but p o werful commut atio n r ules mak e computation po ssible. The following e a sily verifiable commutator identities ho ld in any group: y x = [ y , x ] xy , [ y , x ] = [ x, y ] − 1 , [ xy , z ] = [ x, [ y , z ]][ y , z ][ x, z ] These rules ha ve a nicer form in a class 2 nilp oten t gro up, where all com- m utato r s will freely commute: Prop osition 3.1. If G is a class 2 nilp oten t gr oup, then, for any x, y , z ∈ G , we have y x = xy [ x, y ] − 1 , (1) [ y , x ] = [ x, y ] − 1 , (2) [ xy , z ] = [ x, z ][ y, z ] . (3) In a class 2 nilpo ten t gro up, we may use (1) to e x c hange the order of a n y non- commutator elements. Using (3) w e may write any co mm utator as a product of commutators of generators. Having reduced all commutators to commut a tors o f generator s, we may use (2) to ensure that the generators appear in a prescrib ed order. Viewed as a set of word rewriting rules, Pro p osition 3.1 sugg ests that there will be some sort of normal form in nilp oten t groups whic h can b e used to compare words. The desired normal form is pr ovided b y a theorem of P . Hall. Before stating the theorem, we give so me terminolog y and notation, following [Hall, 195 7 ]. F or a free group G , consider the elemen ts which can be formed by taking the closure of the generator set under the commutator op eration. Of these elements, the generato rs are referr ed to as weig ht 1 c ommutators , a nd the weight of any non-genera tor element is defined to be the sum of the w eig hts of the elements of which it is a co mm utator. Hall show ed that words in b G n can b e given in a nor mal form consisting of a pro duct of cer tain b asic c ommutators of weight n or less given in some pros cribed order. The construction of the ba sic comm utator s is so mewhat inv olved, and we refer to [Hall, 1957] a nd [Magnus et al. , 19 76 ] for the details. F or the purp ose of the examples in this paper, it is sufficien t to know that in a group of rank 2, say G = h a, b i : the basic weight 1 commutators are a and b , and the only basic w eight 2 co mm utator is [ a, b ] (the commutator [ b, a ] is not basic, since it is expressable as [ a, b ] − 1 ). W e also r e fer to Theo rem 5.1 1 of [Ma gn us et al. , 197 6 ] which gives a combinatorial form ula due to Witt for C n , the num b er o f ba s ic weigh t n co mm utators: C n = 1 n X d | n µ ( d ) k n/d , 4 where µ is the M¨ obius function, a nd k is the num b er of g enerators of G . Theorem 3.2 (P . Hall, 1957) . F or any x ∈ G , we c an write the pr oje ct io n b x ∈ b G n as b x = Y i b c k i i , wher e { c i } is the se quenc e of b asic c ommutators of weight less than or e qual t o n . This form for b x is unique up t o the or dering of the weight n b asic c ommu- tators, and we c al l this the Hall normal fo rm for b x . Since ϕ ( γ n ( G )) ⊂ γ n ( G ), there is a well defined quotient mapping b ϕ : b G n → b G n . Thus it makes sense to ask, for h, g ∈ G , whether or not [ b h ] = [ b g ] in b G n , that is, whether or not ther e is so me z ∈ b G n with b h = b ϕ ( z ) b g z − 1 . If [ b h ] 6 = [ b g ] in b G n , then we know that [ h ] 6 = [ g ] in G . 4 Some examples W e b egin with a sa mple co mputation b y hand, showing how the Hall normal form can b e used to solve twisted conjugacy r elations. Example 4.1 . W e will compute the Nielsen n umber of the map o n a surface with fundamental gr oup G = h a, b i w hich induces the homomo r phism: ϕ : a 7→ ab b 7→ b 2 a 4 Theorem 2.1 gives R T ( ϕ ) = ρ ( − 1 − b ) and th us we nee d only decide whe ther o r not [1] = [ b ]. Fir st we attempt to equate these clas ses in the ab elianization ¯ G . W riting elements additively , an element z ∈ ¯ G is of the form z = n ¯ a + m ¯ b , and we wish to so lv e 0 = ¯ ϕ ( z ) + ¯ b − z . W e compute ¯ ϕ ( z ) = n (¯ a + ¯ b ) + m (4 ¯ a + 2 ¯ b ) and − z = − n ¯ a − m ¯ b , and the a bov e equation b ecomes − ¯ b = 4 m ¯ a + ( n + m ) ¯ b, and we can solve for n a nd m to find that z = − ¯ a is a so lution. Remark 4.2. Given that 1 and b ar e twisted conjugate in the ab elianization b y the element − ¯ a , we migh t hop e that 1 and b are twisted conjugate in the gr oup G by the element a − 1 . This is not the case, how ever, as ϕ ( a − 1 ) ba = b − 1 a − 1 ba . The p ossibilit y rema ins, how ever, that these elements are twisted conjugate by some more complicated word which a belianizes to − ¯ a . 5 Having failed to decide the twisted conjuga cy a fter a chec k in the ab elian- ization, we pro ceed to the class 2 nilp otent quotient b G 2 . Any z ∈ G 2 is of the form z = b a n b b m [ b a, b b ] k . W e wish to so lv e 1 = b ϕ ( z ) b bz − 1 . W e alr eady know by the ab ov e calculation, how ever, tha t any s uc h elemen t z m ust ab elianize to b a ∈ b G in orde r to satisfy the t wisted conjuga cy equation. Thu s we may a ssume that z = b a − 1 [ b a, b b ] k . Now we compute z − 1 = [ b a, b b ] − k b a = b a [ b a, b b ] − k , and b ϕ ( z ) = b b − 1 b a − 1 [ b a b b, b b 2 b a 4 ] k = b a − 1 b b − 1 [ b a − 1 , b b − 1 ][ b a, b b ] 2 k [ b b, b a ] 4 k = b a − 1 b b − 1 [ b a, b b ] − 2 k +1 , where w e hav e used the rules of Prop osition 3.1, toge ther with the identit y [ x i , z ] = [ x, z ] i , which follows from setting x = y in identit y (3). W e a r e now r eady to tes t the twisted conjugac y equation ab ov e. The r ig h t hand s ide is: b ϕ ( z ) b bz − 1 = b a − 1 b b − 1 [ b a, b b ] − 2 k − 8 b b b a [ b a, b b ] − k = [ b a, b b ] − 3 k − 8 . Setting this equal to 1 requires that − 3 k − 8 = 0, which is impo ssible for k ∈ Z . Thu s there can b e no such z ∈ b G 2 , and so [1] 6 = [ b ] and the Nielsen num be r is 2. The example ab o ve inv olved a computation first in ¯ G and then in b G 2 . In e ac h step, ther e are ess en tially tw o types of co mputatio nal op erations inv olved. The first is the term rewriting to obtain the Hall normal form, whic h was fair ly ea s y in this ca s e but in general can b e quite tedious (though completely algor ithmic). The sec o nd is finding the so lution to a linear system, which was used to solve for n and m in the ¯ G step, and used to solve for k in the b G 2 step (in a free group with more gener ators, this would ha ve been a linea r system with more than o ne equa tion). As is to be ex p ected, co mputation of twisted co njuga cy clas ses in man y cas e s may requir e chec ks in b G n for n > 2. Such exa mples ca n easily be constructed by a computer search. Example 4.3. Let G = h a, b i , and let ϕ : a 7→ aba − 1 b 7→ a − 2 b 4 Theorem 2.1 gives R T ( ϕ ) = ρ ( aba − 1 − a − 2 − a − 2 b − a − 2 b 2 − a − 2 b 3 ) . It can be verified that all fiv e of the above terms are twisted conjugate in b G n for n ∈ { 1 , 2 , 3 } , but none are t wisted conjuga te in b G 4 . Thus the Nielsen nu mber is 5. 6 W e now tur n to the ques tion of verifying equality of twisted co njugacy classes. Applying the pro cess descr ibed ab o ve to tw o w ords which ar e in fact t wisted conjugate will result in a no n-terminating se quence o f computatio ns in the groups b G n , each time r esulting in solutions for the v ario us elements ab o ve lab eled z . T o av oid s uc h an infinite computation, we prop ose a technique in the spirit of Remark 4.2. In the a belianization ¯ G , if a solution element z is obtained, a sequence of “c andidates” for testing twisted conjugacy in G is constructed by pro ducing all p ossible reo rderings of the ge ne r ators app earing in z . Th us if we obtain an element z = ¯ a − 2 ¯ b , our list of candidates will be ab − 2 , b − 1 ab − 1 , b − 2 a. Each of these elements can b e tes ted by twisted c o njugation as a candidate for realizing the twisted co njuga cy in G . A similar pro cess can b e carr ie d out in b G n for n > 1: after obtaining z ∈ b G n , our list of candidates is o btained by inserting the weight n ba s ic commutators app earing in z in all p ossible or de r ings and in v ar ious forms int o each o f our candidates previous ly obtained in our chec k from b G n − 1 . Example 4.4. Let G = h a, b i , and ϕ : a 7→ a 2 ba b 7→ b 2 a W e will use the ab ov e candidates c hecking co nstruction to s ho w that [ a ] = [ a 2 b ] (the elements a 2 b and a app ear in R T ( ϕ )). Our c heck in ¯ G sho ws that the tw o elemen ts are t wisted conjugate b y − ¯ b . Thu s our only ca ndidate from ¯ G is the elemen t b − 1 , but a computation shows that ϕ ( b − 1 )( a 2 b ) b = a − 1 b − 2 a 2 b 2 6 = a. Now we do a check in b G 2 , a nd we find that the tw o elements are twisted conju- gate by b b − 1 [ b a, b b ] − 1 ∈ b G 2 . Now our lis t of c a ndidates is: g 1 = b − 1 [ a, b ] − 1 = ab − 1 a − 1 g 2 = b − 1 [ a − 1 , b − 1 ] − 1 = b − 2 a − 1 ba g 3 = b − 1 [ a − 1 , b ] = b − 1 a − 1 bab − 1 g 4 = b − 1 [ a, b − 1 ] = b − 1 ab − 1 a − 1 b g 5 = [ a, b ] − 1 b − 1 = b ab − 1 a − 1 b − 1 g 6 = [ a − 1 , b − 1 ] − 1 b − 1 = b − 1 a − 1 bab − 1 g 7 = [ a − 1 , b ] b − 1 = a − 1 bab − 2 g 8 = [ a, b − 1 ] b − 1 = ab − 1 a − 1 Computing ea c h of ϕ ( g i ) a 2 b ( g i ) − 1 reveals that ϕ ( g 1 ) a 2 bg − 1 1 = a , and s o [ a ] = [ a 2 b ]. 7 Note that in the ab ov e example we inserted the basic commutator [ a, b ] − 1 int o the word b − 1 in ea c h of tw o p ositions in one o f four forms, these b eing the four versions o f this commut a tor in G whic h b ecome [ b a, b b ] − 1 when pro jected into b G 2 . Constructing the v arious forms of a weight 3 commutator to use in a list of candidates would be cumbersome, and we do not attempt this co nstruction for nilpo tency cla ss higher than 2. As an alterna tiv e to the candidates co nstruction pro cedure, a more pedes - trian approach is a lw ays av ailable: a fter a c heck for twisted co njugacy in b G n , use a s list of ca ndidates all words in G of word length n . This pro duces a dif- ferent list of candidates from that describ ed above, but is guar a n teed to find an element realizing the twisted conjuga cy if n is s ufficie ntly high. 5 Success rates The technique exhibited in the examples ab o ve can be summar ized as follows: Starting with n = 1 , write an ex pr ession for a generic e le men t z ∈ b G n and the element b ϕ ( z ) in Hall nor mal form. Solve a linear system to decide if the elements are twisted conjugate in b G n . If the e le men ts are not t wisted conjugate in b G n , then the elements are not twisted co njugate in G . If the elemen ts are t wisted conjugate in b G n by a unique element z , then constr uct a finite list of candidates (either intelligen tly by using the structure of z or in a brute force manner by taking all words of le ng th n ) for testing t wisted conjugacy in G . If all of these candidates fail, then increment n and rep eat the ab ov e. This technique will decide an y given twisted conjuga cy proble m pro vided that the following statement is true: If ϕ : G → G is a map on a free g roup, and g and h are tw o elements of G which are not twisted conjuga te in G , then there is some n for whic h g and h are not t wisted conjugate in b G n . Thus any non-twisted-conjugate elements will be detected a s such in b G n for some n . Such a statement does no t hold in gener al, though, as the following argument s ho ws. Prop osition 5. 1. F or any ϕ : G → G , if g , h ∈ G ar e wor ds su ch that ϕ n ( g ) ∈ hγ n ( G ) for al l n , then [ b g ] = [ b h ] in b G n for al l n . Pr o of. The pro of is based on the fac t that [ ϕ ( x )] = [ x ] for a n y element x , since ϕ ( x ) = ϕ ( x ) x x − 1 . Itera tion of the map g iv es [ ϕ n ( x )] = [ x ] for any n . Now for o ur element g , we ha ve b ϕ n ( b g ) = b h in b G n , and so in particular [ b ϕ n ( b g )] = [ b h ]. But [ b ϕ n ( b g )] = [ b g ] by the ab o ve, and so we have [ b g ] = [ b h ]. W e ca n use the ab ov e to build homomor phisms ϕ : G → G with the prop erty that ther e are words g , h ∈ G with [ g ] 6 = [ h ] but [ b g ] = [ b h ] in every nilp oten t quotient b G n . Example 5.2. Let G = h a, b i , and let ϕ : G → G b e the map ϕ : a 7→ [ b, a ] b 7→ a − 1 b 8 Now w e have ϕ ( γ n ( G )) ⊂ γ n +1 ( G ) for n > 0, s ince ϕ r eplaces a with a w eight 2 co mm utator and all co mm utators in G inv olve the element a . Since ϕ ( a ) ∈ γ 1 ( G ), we hav e ϕ n ( a ) ∈ γ n ( G ) for all n , and thus that [ b a ] = [1] in b G n for all n . But ϕ is a mapping with remnant, and W agne r ’s algo rithm ca n b e used to show that in fact [ a ] 6 = [1] in G . Though the above exa mple shows that nilp otent quotients cannot b e used directly to solve t wisted conjugacy pro blems in all cases, the technique g iv es very go od success ra tes in exp erimental tes ting. W e hav e cr eated an implementation 1 of the pro cess in the computational algebr a system Magma [Bo sma et al. , 1994]. There are tw o wa ys that an implemen tation of this tec hinque can fail to decide any given twisted co njugacy pr oblem. One type of failur e is when the linear system which arises in the t wisted conjugacy computation in b G n has infinitely many solutions. In such a case the chec k in b G n +1 will require finding an integral solution of a polyno mial system, which will in general b e difficult. Such a failure can only occur when the co efficients in the linear system give a singular matrix, which we exp ect to o ccur relatively infrequently . A second type o f failur e is that the implemen tation ex hausts its resources in co mputing the require d Hall norma l for ms in b G n . Even fo r gro ups of 2 or 3 generator s, human computation of the Hall form in class 3 o r 4 is bar ely feasible. Since the num b er of basic commutators of weigh t n gr o ws expo nen tially in n we expect that any computer implementation could conceiv ably exhaust its resource s b efore detecting that t wo given twisted conjuga cy cla sses ar e indeed distinct. W e exp ect this type o f failure to occ ur increa singly in free groups with large num b ers of g e ne r ators. T ables 1 and 2 give ex p erimental results of application of the nilp oten t quo- tien ts tec hnique to 10 ,000 rando mly gener ated mapping s on the free g roup on k generators for k = 2 , 3 , 4 . In each ca se, a num b er l ∈ { 2 , 3 , 4 , 5 } is ch o s en and a mapping is generated by assigning the image o f e a c h generator to be a randomly chosen w ord of leng th at most l . Theorem 2.1 is applied to give a list of g r oup elements which must be divided in to their twisted conjuga cy cla sses. A sucessful computation is one which is able to decide the twisted co njugacy of these e lemen ts. T able 1 g iv es the r ates of each of the tw o t yp es of failur e ab ove. A “ma- trix failure” is dec lared when the linear system computation r esults in infinitely many solutions . A “complexity failure” is declar ed when the nilpotency class reaches 5 , as this is the level at which the computation of the Hall no r mal for m bec omes difficult for our implementation (run on a pe rsonal computer, current in 2005). Since a single mapping ca n trigger both t yp es o f err o r (computa- tion of R T ( ϕ ) in gener al req uir es several twisted conjuga cy decisions ), the row per cen tages may no t sum to 100%. The column lab eled “average depth” gives the average nilp otency class required to distinguish t wisted conjugacy classes in these random mappings. A depth of n indicates that a chec k in b G n was necessary . The column lab eled σ g iv es the standard deviatio ns of the depths. 1 Av ailable fr om the aut hor’s web site: http ://www.messiah .edu/ ~ cstaecke r 9 k l Success Matrix failure Complexity failur e Avg. depth σ 2 2 94.27% 4 .30% 2.97% 1.10 0.28 3 90.46% 6 .66% 4.09% 1.13 0.34 4 85.91% 7 .75% 9.06% 1.18 0.42 5 84.20% 9 .89% 8.66% 1.17 0.38 3 2 87.45% 11.26% 4.3 8% 1.18 0.45 3 85.60% 13.77% 4.7 4% 1.17 0.42 4 85.20% 12.38% 6.1 9% 1.21 0.47 5 84.46% 13.36% 6.2 9% 1.23 0.45 4 2 88.87% 10.95% 3.4 8% 1.13 0.41 3 85.60% 13.77% 4.7 4% 1.16 0.42 4 84.51% 14.56% 5.8 7% 1.21 0.45 5 84.13% 15.04% 5.6 4% 1.21 0.42 T able 1: Results o f testing for success ra tes on random mappings o f w or d le ng th l o n the free gro up on k g enerators. n l Nilp otent quotients Abelianiza tion W agner’s Alg. 2 2 94 . 27 % 82.75% 41.80% 3 90 . 46 % 70.45% 48.32% 4 85 . 91 % 59.07% 54.71% 5 84 . 20 % 51.14% 59.83% 3 2 90 . 98 % 77.96% 15.54% 3 87 . 45 % 65.99% 24.70% 4 85 . 20 % 55.17% 33.72% 5 84 . 46 % 47.43% 41.45% 4 2 88 . 87 % 76.50% 5.84% 3 85 . 60 % 64.76% 13.67% 4 84 . 51 % 54.63% 22.63% 5 84 . 13 % 47.81% 30.18% T able 2: Compariso n o f s uccess r ates of v ario us techniques on r a ndom mappings of word length l on the free gr oup on k gener ators. 10 T able 2 gives our sucess rates compared to the exis ting techniques of ab elian- ization and W agner’s algor ithm. The technique used for data in the “Ab elian- ization” column uses only the ab elianization for distinguishing classes, and uses only the identit y [ ϕ ( g )] = [ g ] for equating classes (this is the general stra teg y employ ed in [H a r t, 2005]). The column lab eled “W agner ’s a lg .” records the per cen tages of maps satisfying W ag ner’s remnant condition, for whic h her alg o- rithm will apply (that the p ercentages gr o w in l is expected in light of Theor em 3.7 o f [W agner, 19 9 9 ]). 6 Surfaces without b oundary The nilp oten t quotients pro cess can b e used with minimal modifications when G is the fundamental group of a compact hyperb olic surface without b oundary . W or k of F adell and Husseini in [F a dell and Husseini, 1983] together with a tech- nique by Da vey , Hart, and T rapp [Da vey et al. , 1 996] reduce the co mputation of the Nielsen n umber on compact hyperb olic surfaces without bo undary to the computation of twisted conjugac y classes in the fundamental group. W agne r ’s technique does not apply if G is not a free group, a nd neither will the techniques o f [Bog opolsk i et al. , 2 006]. Be c ause sur fa ce gro ups are eas ily expressed in ter ms of co mm utator rela tio ns, w e can use the nilp otent quotients techn ique with o nly triv ial mo difications in this setting. The only mo dification that must b e made is to the pr ecise structure of the Hall nor mal form. F o r instance, if G is the fundamental gr oup of the genus 2 compact surface, then G has group pres en tation G = h a, b, c, d | [ a, b ][ c, d ] = 1 i . The Hall nor mal form of an ele men t of (e.g.) b G 2 can b e o btained by applying commutation rules as if G w ere the free group on 4 gene r ators, along with an additional rule that [ c, d ] − 1 = [ a, b ]. It is conv enient for us that the group structure of G is so co mpatible with the Hall normal form. The ca se of surfaces without b oundary is somewhat mo r e difficult to imple- men t in Magma , as it inv olves computations in finitely-presented rather tha n free gro ups. The capa bilities of Ma gma ar e somewhat lacking in this regar d– in particular Magma (as of version 2.1 3-15) is unable to re liably so lv e the word problem in a surface group (altho ug h this word pro blem is solv able). This ca us es the candidates chec king pro cess to re turn false negatives, a s the implemen ta- tion ma y not recognize when t wo elements a re ac tually equal. Statistics such as those in T able 1 are also difficult to pro duce in this setting as it is difficult to generate random endomor phisms of surface gr oups. 7 Doubly t wisted conjugacy W e conclude with a brief discuss io n o f how our technique can be a pplied to the doubly twiste d c onjugacy relation: Given tw o maps ϕ, ψ : G → H and t wo elements h, k ∈ H , we say tha t h and k are (doubly) twisted conjugate (w e wr ite 11 [ h ] = [ k ]) if there is an e lemen t g ∈ G with h = ϕ ( g ) k ψ ( g ) − 1 . This re la tion is fundamen tal in Nielsen c oincidence theory (see [Gon¸ calves, 2005]), playing the same role as o rdinary twisted conjugac y in fixed po in t theory . F or any n , the maps ϕ and ψ will induce maps b ϕ, b ψ : b G n → b H n , a nd the doubly twisted co njugacy rela tion ca n in principle b e so lv ed by using Ha ll normal forms in b G n and b H n just as in the ordinary twisted conjugacy pr oblem. Example 7.1. Let G = H = h a, b i , a nd let our maps b e ϕ : a 7→ b 2 a b 7→ a − 2 ψ : a 7→ a 3 b 7→ a − 1 W e will decide the twisted conjuga cy of the elements b and b − 1 . W e b egin with check in the abelianiza tion, where any element z ∈ ¯ G has the form z = n ¯ a + m ¯ b . W e compute that ¯ ϕ ( z ) = ( n − 2 m )¯ a + 2 n ¯ b a nd − ¯ ψ ( z ) = ( − 3 n + m )¯ a , and thus we have ¯ ϕ ( z ) − ¯ b − ¯ ψ ( z ) = ( − 2 n − m ) ¯ a + (2 n − 1 ) ¯ b. Equating this with ¯ b and solv ing gives n = 1 and m = − 2. This solution in the ab elianization gives thr ee candidates for twisted conjugacy: ab − 2 , b − 1 ab − 1 , b − 2 a, but chec king each shows that none of these realize the twisted conjugacy in H . W e pr oceed to the class 2 nilpo ten t q uotien t, where a n y element z ∈ b G 2 has the form z = b a n b b m [ b a, b b ] k . Our co mputation in ¯ G s ho ws that n = 1 and m = − 2 , simplifying our element to z = b a b b − 2 [ b a, b b ] k . W e compute b ϕ ( z ) = b b 2 b a ( b a − 2 ) − 2 [ b b 2 b a, b a − 2 ] k = b a 5 b b 2 [ b a 5 , b b 2 ][ b b 2 , b a − 2 ] k = b a 5 b b 2 [ b a, b b ] 10+4 k , b ψ ( z ) = b a 3 ( b a − 1 ) − 2 [ b a 3 , b a − 1 ] k = b a 5 , and so b ϕ ( z ) b b − 1 b ψ ( z ) − 1 = b a 5 b b 2 [ b a, b b ] 10+4 k b b − 1 b a − 5 = b a 5 b b b a − 5 [ b a, b b ] 10+4 k = b b [ b a − 5 , b b ][ b a, b b ] 10+4 k = b b [ b a, b b ] 5+4 k . Equating this with b b g iv es 5 + 4 k = 0 which is imp ossible for integral k . Thus [ b ] 6 = [ b − 1 ]. There is in the liter a ture no ana lo gue of Theor em 2 .1 in co incidence theory , but presumably one may b e av ailable in the future, and our technique is cur- rently the only av a ilable technique for distinguishing doubly twisted conjugacy classes (no version of W agner ’s algor ithm is kno wn in co incidence theory , and 12 k 1 k 2 Success Matrix failure Co mplexit y failure Average depth σ 2 2 92.38% 4 .07% 3.55% 1.49 0.95 3 98.97% 1 .03% 0.% 1.09 0.28 4 99.78% 0 .22% 0.% 1.03 0.17 5 99.90% 0 .10% 0.% 1.01 0.12 3 2 30.53% 69.47% 0.% 1. 0. 3 92.32% 5 .83% 1.85% 1.41 0.80 4 98.80% 1 .20% 0.% 1.08 0.27 5 99.71% 0 .29% 0.% 1.03 0.18 4 2 14.56% 85.44% 0.% 1. 0. 3 32.88% 67.12% 0.% 1. 0. 4 91.98% 7 .13% 0.89% 1.33 0.66 5 98.50% 1 .15% 0.% 1.08 0.27 T able 3: Success rates for doubly twisted conjugacy relations. Random map- pings o f word length 3 from the free gro up on k 1 generator s to the free group on k 2 generator s were tested in deciding twisted conjugacy b et ween t wo random words of length at mos t 3. the metho ds of [Bogop olski et al. , 2 006] do not extend in an obvious wa y to doubly t wisted conjugacy). T able 3 gives succes s ra tes for the technique applied to 10,000 ra ndomly generated twisted conjugacy rela tio ns. In e a c h case, t wo random ma ppings of “word length” 3 (the quan tity lab eled l in T ables 1 and 2 ) ar e generated from the free group on k 1 generator s to the free group on k 2 generator s. Tw o r andom elements of the co domain group ar e generated with word length at mos t 3, and the implementation attempts to deter mine their twisted conjugacy . E n tries in the table with no digits to the right of the decimal p oin t a re exa c t figures, e.g. in the case where k 1 = 4 a nd k 2 = 3 the depth was exactly 1 in ea c h of the 10,000 test cases, and there were exactly 0 complexity failures . Note that the technique is muc h less succ essful if the rank of the do main is greater than the r ank o f the co domain. This is to b e exp ected, as the tec hnique will fail when our linear system computation (alwa ys having mo r e v ar iables than equations if k 1 > k 2 ) yields infinitely many solutions. Note that s uch cases ar e not handled by our Magma implementation, but could in principle b e do ne b y hand. These w ould require finding in teger solutions to p olynomial systems, and so we do not alwa ys exp ect the computation to b e s uccessful, but particular examples may b e computable. Esp ecially striking are the extremely hig h sucess ra tes when k 2 > k 1 . The v as t ma jorit y of these twisted conjugacy relations are decided in the ab elianiza- tion (and with negative result), as the ov erdetermined linear systems ar e unlikely to hav e an y so lutions. F or example in the case of k 1 = 3 and k 2 = 5 , an equiv a- lence in the abelia nization would require our random elemen ts to satisfy a linea r system of 5 equations a nd 3 unknowns. If the element s are indeed equiv a len t 13 in the abelia nization, a further equiv a lence in the class t wo nilp otent quo tien t would require our random da ta to satisfy a linea r system of 10 eq uations and 3 unknowns (the free group on 5 gener ators has 10 basic weigh t 2 commutators, and the free group on 3 genera tors has 3 basic weight 2 commutators). This is not, of course, to s a y that t wisted conjugate element s do not oc cur in suc h cases wher e k 2 > k 1 , but only that this occur s very infrequently in the case o f t wo words o f length 3. W e hav e omitted from o ur testing the cases where k 1 or k 2 is 1. The t wisted conjugacy relation is gener a lly solv able in these cas e s b y hand. Let ϕ, ψ : G → H ar e maps of free groups where G = h a i , and let h, k ∈ H be tw o words. Then the t wisted co njugacy problem is equiv a len t to finding some int eg e r n with 1 = h − 1 ϕ ( a ) n k ψ ( a ) − n , and this can typically b e confirmed or denied by ins p ection. If ϕ, ψ : G → H are maps of free groups a nd H has rank 1, let G = h a 1 , . . . , a n i , and note that ϕ ( γ 1 ( G )) ⊂ γ 1 ( H ) = 1 s ince H is ab elian. Thus we hav e ϕ ( uv ) = ϕ ( v u [ v , u ]) = ϕ ( vu ) for any words v , u ∈ G , a nd simila r ly ψ ( uv ) = ψ ( v u ). Thus, though ther e is no conv enient normal form for elements z ∈ G , we can say in genera l that ϕ ( z ) = ϕ ( a m 1 1 . . . a m n n ) , ψ ( z ) = ϕ ( m k 1 1 . . . m k n n ) by rearr anging the generato r s of z . 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