Shapes of hyperbolic triangles and once-punctured torus groups
Sang-hyun Kim, Thomas Koberda, Jaejeong Lee, Ken'ichi Ohshika, Ser Peow Tan, with an appendix by Xinghua Gao
SShapes of hyperbolic triangles and once-punctured torus groups
SANG-HYUN KIM, THOMAS KOBERDA, JAEJEONG LEE, KEN’ICHI OHSHIKA,AND SER PEOW TAN
WITH AN APPENDIX BY
XINGHUA GAOA bstract . Let ∆ be a hyperbolic triangle with a fixed area ϕ . We prove that forall but countably many ϕ , generic choices of ∆ have the property that the groupgenerated by the π –rotations about the midpoints of the sides of the triangle ad-mits no nontrivial relations. By contrast, we show for all ϕ P p , π qz Q π , a denseset of triangles does a ff ord nontrivial relations, which in the generic case map tohyperbolic translations. To establish this fact, we study the deformation space C θ of singular hyperbolic metrics on a torus with a single cone point of angle θ “ p π ´ ϕ q , and answer an analogous question for the holonomy map ρ ξ of sucha hyperbolic structure ξ . In an appendix by X. Gao, concrete examples of θ and ξ P C θ are given where the image of each ρ ξ is finitely presented, non-free andtorsion-free; in fact, those images will be isomorphic to the fundamental groupsof closed hyperbolic 3–manifolds.
1. I ntroduction
Take a geodesic triangle in the hyperbolic plane, and consider the rotations ofangle π about the midpoints of the three sides, which we call the side involutions .It is natural to wonder whether or not some nontrivial compositions of the sideinvolutions will move the triangle exactly back to itself.For a regular pentagon in the Euclidean plane, there are many “unexpected” co-incidences of this type [19]. On the other hand, a regular tetrahedron in R cannever be moved back to its original position [1].For a hyperbolic triangle the answer depends on, among other things, its area.For instance, if the area is a rational multiple of π , then so is the interior angle sum.In this case, a suitable power of a composition of the three side involutions will betrivial. Even if the area is not a rational multiple of π , torsion relations still appearfor a dense choice of hyperbolic triangles (cf. Theorem 5.2 below). So, we areled to address the question whether or not relations that are not “consequences” oftorsion relations can still be found for a dense set of triangles; we refer the readerto Definition 1.4 for a precise formulation. Mathematics Subject Classification.
Primary: 30F35, 30F40; Secondary: 20E05, 11J70.
Key words and phrases.
Fuchsian groups, Kleinian groups. a r X i v : . [ m a t h . G T ] J un S. KIM, T. KOBERDA, J. LEE, K. OHSHIKA, AND S. P. TAN
Question A.
In the space of hyperbolic triangles whose area is fixed as ϕ P p , π q ,what are the necessary and su ffi cient conditions for the side involutions to generatethe Coxeter group Z ˚ Z ˚ Z ? Do the involutions for a dense set of hyperbolictriangles admit relations that are not consequences of torsion relations? Here, the space of hyperbolic triangles
ABC with fixed area ϕ is identified withits parameter space of the interior angles: T ϕ : “ tp θ A , θ B , θ C q P p , π q | θ A ` θ B ` θ C “ π ´ ϕ u . Our take on Question A will be via the space of marked incomplete hyperbolicstructures on a punctured torus. Throughout this paper, we let M be a torus minus apuncture p . Let us fix an angle θ P p , π q . We consider the deformation space C θ ,called a Fricke–Klein space [12], of marked incomplete hyperbolic structures on M having exactly one conical singularity at the puncture p with a cone angle θ . Wewill also consider the Fricke–Klein space C : “ ď θ Pp , π q C θ of all marked hyperbolic structures with cone angles in p , π q on M .We fix an oriented meridian X and an oriented longitude Y of M , and identify π p M q with a rank-two free group F : “ x X , Y , Z | XYZ “ y . It is convenient to regard F as an index-two subgroup of a free Coxeter group W : “ Z ˚ Z ˚ Z “ x P , Q , R | P “ Q “ R “ y using the embedding X ÞÑ QR , Y ÞÑ RP , Z ÞÑ PQ . Each hyperbolic structure ξ P C θ defines a conjugacy class of holonomy maps , arepresentative of which we denote by ρ ξ : π p M q Ñ PSL p , R q – Isom ` p H q . These holonomy maps are defined via the developing map ˜ M Ñ H . The isome-tries t ρ ξ p X q , ρ ξ p Y q , ρ ξ p Z qu are hyperbolic elements with pairwise intersecting axes.Because the group Im ρ ξ contains two hyperbolic elements with intersecting axes,it is always non-elementary. Since F is Hopfian, we have that ρ ξ is faithful if andonly if Im ρ ξ nonabelian and free.As is well-known, the space C θ can be identified with its parameter space tp x , y , z q P p , | x ` y ` z ´ xyz ´ “ ´ p θ { qu by the character map ξ ÞÑ ` tr ` ρ ξ p X q , tr ` ρ ξ p Y q , tr ` ρ ξ p Z q ˘ . hapes of hyperbolic triangles and once-punctured torus groups 3 Here, tr ` p¨q is understood as the absolute value of the trace of an element in PSL p , R q .To see the connection between C θ and Question A, pick a hyperbolic structure ξ P C θ on M for some fixed θ P p , π q . Let γ X be the simple closed geodesicrealizing X P π p M q . There exists a unique incomplete geodesic δ X on M disjointfrom γ X , and that starts and ends at p . Construct the geodesics t γ Y , γ Z , δ Y , δ Z u analogously. By cutting M along δ X Y δ Y Y δ Z , we obtain two isometric hyperbolictriangles whose interior angle sums are θ {
2. The shape of this triangle determinesa point in T π ´ θ { . A lift of this picture to H is illustrated in Figure 1. γ X γ Y γ Z h X O P O Q O R δ X O A O B O C R p O C q θ A θ B θ C F igure
1. Developing M to H .As in the figure, we let t O P , O Q , O R u denote the midpoints of the sides fromone of the triangles. We write t ˆ ρ ξ p P q , ˆ ρ ξ p Q q , ˆ ρ ξ p R qu for the corresponding sideinvolutions. Then we have that ρ ξ p X q “ ˆ ρ ξ p Q q ¨ ˆ ρ ξ p R q , ρ ξ p Y q “ ˆ ρ ξ p R q ¨ ˆ ρ ξ p P q , ρ ξ p Z q “ ˆ ρ ξ p P q ¨ ˆ ρ ξ p Q q . Thus ρ ξ extends to a surjectionˆ ρ ξ : W Ñ x ˆ ρ ξ p P q , ˆ ρ ξ p Q q , ˆ ρ ξ p R qy , called a Coxeter extention of ρ ξ , which is determined uniquely from the holonomyrepresentation ρ ξ [9]. Moreover, the spaces T π ´ θ { and C θ are homeomorphic via ahomeomorphism which conjugates this extension (Lemma 4.2).Through Coxeter extensions, we will approach Question A by studying the fol-lowing closely related question. Question B.
For a fixed angle θ P p , π q , what are the necessary and su ffi cientconditions for ξ P C θ to yield a faithful representation ρ ξ : π p M q Ñ PSL p , R q ? Isthere a dense choice of ξ P C θ for which the representation ρ ξ admits relations thatare not consequences of torsion relations? S. KIM, T. KOBERDA, J. LEE, K. OHSHIKA, AND S. P. TAN
Let A be a topological space. We say that a certain statement holds for a verygeneral (or, generic ) ξ P A if the statement holds for all ξ in some dense G δ subsetof A . If A is an open subset of a Euclidean space, then it follows that the statementholds on a full-measure subset of A .In this paper, we are primarily interested in establishing the faithfulness part ofQuestion B for a very general point ξ P C θ , and in answering the density questionfor a suitable interpretation of the phrase “consequences of torsion relations”. Remark . One may also consider side reflections instead of side involutions ofa hyperbolic triangle ∆ with angles p θ A , θ B , θ C q , although we will not pursue thisdirection. We briefly note in this case that two representations ˆ ρ ξ and ρ ξ are stillwell-defined on W and on F respectively. The representation ρ ξ is a holonomymap of an incomplete hyperbolic three-punctured sphere S , with three conicalsingularities of angles p θ A , θ B , θ C q . This approach could be useful when one isinterested in periodic billiard orbits on ∆ .The faithfulness question is easier when the cone angle θ is not fixed. Namely,we have that ρ ξ is faithful for a very general point ξ P C (cf. Proposition 3.1). So,we will be actually interested in the case when θ is fixed.Our first main result answers the first half of Question B with probability one: Theorem 1.2 (cf. Theorem 3.3) . If θ P p , π q has the property that cos θ is tran-scendental, then a very general point ξ in C θ corresponds to a faithful holonomymap. Note that the hypothesis holds for all but countably many values of θ . The readermay compare Theorem 1.2 to other generic phenomena in PSL p , R q ; see [5, 7,15], the first two of which deal with the faithfulness question in the case of closedsurface groups. One can trace at least back to Hausdor ff [11] the idea of using thetranscendency of cos θ in order to produce faithful group actions.It is a well-known consequence of the Gelfond–Schneider theorem that cos θ istranscendental if θ { π is irrational and algebraic. Thus, we deduce more concretecases when ρ ξ is faithful as follows. Corollary 1.3. If θ P p , π q is an irrational algebraic multiple of π , then ρ ξ isfaithful for a very general point ξ P C θ . We have briefly mentioned the abundance of torsion relations at the beginningof this introduction. Indeed, the image of ρ ξ is dense in PSL p , R q unless ρ ξ isdiscrete (and hence Fuchsian), as is true of all Zariski dense subgroups of simpleLie groups [14]. Since elliptic isometries form an open subset in PSL p , R q , it isreasonable to expect that an elliptic element in the image of ρ ξ would not haveconstant rotation angle under perturbations in C θ near ξ . Thus, a torsion relationwould appear under small deformations of ξ having a fixed cone angle θ ; we direct hapes of hyperbolic triangles and once-punctured torus groups 5 the reader to Theorem 5.2 for a concrete formulation of this intuition. Moreover,in the presence of torsion there are generally many more elements of ker ρ ξ that areconsequences of such torsion relations.We aim to find relations of ρ ξ which are not “consequences” of torsion relations.To state the result rigorously, we make the following definition. Definition 1.4.
A word u P F is of torsion-type if u belongs to the set ď m ě ď r P F zt u xx r m yy . A word is of non-torsion-type if it is not of torsion-type.For a word u P F , being a non-torsion-type word is a stronger condition thannot equal to a proper power. In the special case when θ “ π { n , the group ρ ξ p F q isisomorphic to a Fuchsian orbifold group x X , Y | r X , Y s n “ y for all choices of ξ P C θ ; in this case, every kernel element must be of torsion-type.When θ R Q π , we can produce non-torsion-type kernel elements as addressed in thesecond half of Question B. Theorem 1.5 (cf. Theorem 5.5) . If θ P p , π q is an irrational multiple of π , then ker ρ ξ ă F contains a palindromic non-torsion-type word for a dense choice of ξ P C θ . In particular, when ker ρ ξ contains a non-torsion-type word, the image ρ ξ p F q cannot be isomorphic to a one-relation group with torsion with respect to the stan-dard generating set t ρ ξ p X q , ρ ξ p Y qu ; see Definition 5.3 and Proposition 5.4.We will actually produce a dense subset S θ Ď p , such that each point on thecoordinate curves t x “ s u , t y “ s u , t z “ s u in C θ for s P S θ corresponds to holonomy maps admitting non-torsion-type kernelelements. It follows that the same property holds on (densely many) doubles pointsof such coordinate curves. Remark . A consequence of Theorem 1.5 is that a free indiscrete representa-tion is algebraically unstable , even when restricted to C θ ; roughly speaking thismeans that such a representation is a limit of non-free representations. In the ab-sence of such a “relativizing” restriction, this type of instability is well-known fornon-abelian free groups, even in a much more general setting of connected Liegroups [8]. On the other hand, Sullivan [20] established algebraic stability of allconvex–cocompact subgroups in PSL p , C q . S. KIM, T. KOBERDA, J. LEE, K. OHSHIKA, AND S. P. TAN
Let us direct our attention back to Question A. The faithfulness part has the sameanswer as Question B; see Proposition 4.3(1). Regarding the density part, it isunclear to the authors whether the non-torsion-type kernel elements in F found inTheorem 1.5 can still be non-torsion-type in W . However, their additional propertyof being palindromic has a simple interpretation in W : they are products of twoinvolutions in W (Lemma 4.1). Thus we establish the following partial answer toQuestion A. Corollary 1.7.
The following conclusions hold:(1) For all but countably many ϕ P p , π q , the side involutions of a very generaltriangle with area ϕ generates the Coxeter group W.(2) If ϕ is an irrational multiple of π , then for a dense choice of hyperbolic tri-angles with area ϕ the side involutions admit a relation which is the productof two involutions in W. We remark that if u P W is a product of two involutions then, for all ξ P C ,the image ˆ ρ ξ p u q is either trivial or hyperbolic (Proposition 4.3(2)). In fact, it ishyperbolic for a very general point in C θ (Remark 5.8).As far as the authors are aware, no examples of ξ P C were previously docu-mented such that ρ ξ p F q is non-free and torsion-free. Question 1.8. If θ P p , π q is an irrational multiple of π , under what conditions on ξ is ρ ξ non-faithful with torsion-free image? In the appendix written by X. Gao, computational heuristics for approachingQuestion 1.8, together with their implementation, are given. The examples of pairs p θ, ξ q exhibited therein have the property that the images of ρ ξ are actually isomor-phic to fundamental groups of closed hyperbolic 3–manifolds.2. F ricke –K lein space Most of the material in the section is well-known; we direct the reader to [9] for astandard reference. We adopt the standing convention that group and matrix actionsare on the left, unless stated otherwise.The group SL p , R q acts on H by M¨obius transforms, with kernel given by thecenter, and with image PSL p , R q – Isom ` p H q . Let tr p A q denote the trace of a matrix A . For each word w “ w p X , Y q P F “ x X , Y y there exists a trace polynomial g w P Z r x , y , z s such that whenever U , V P SL p , R q we havetr w p U , V q “ g w p tr U , tr V , tr UV q . hapes of hyperbolic triangles and once-punctured torus groups 7 The existence of this polynomial is one of the simplest instances of invariant theoryon character varieties [18]; see also [13] for a concrete formula which computes g w .For example, the trace polynomial of r X , Y s is easily seen as g r X , Y s p x , y , z q “ κ p x , y , z q : “ x ` y ` z ´ xyz ´ P Z r x , y , z s . This means that for each pair of matrices U , V P SL p , R q we have thattr r U , V s “ κ p tr U , tr V , tr UV q . Lemma 2.1.
For each w P F zt u , the polynomial g w p x , y , z q is not constant.Proof. Let w “ w p X , Y q P F “ x X , Y y . If g w were constant, we would have g w p x , y , z q “ g w p , , q “ tr w p Id , Id q “ tr Id “ . However, there exist two–generated nonelementary purely loxodromic Fuchsiangroups which are nonabelian and free. This implies g w cannot be identically equalto 2. (cid:3) In this paper, we are mostly concerned with matrices in PSL p , R q . The trace ofsuch a matrix A P PSL p , R q is only determined up to sign, so as mentioned in theintroduction, we often use the quantitytr ` A : “ | tr A | . Recall we let C θ denote the deformation space ( Fricke–Klein space ) of markedincomplete hyperbolic metrics on M with a fixed cone angle θ P p , π q at p . Eachpoint ξ P C θ corresponds to a conjugacy class of a holonomy map ρ : π p M q “ x X , Y , Z | XYZ “ y Ñ PSL p , R q so that ρ r X , Y s is a rotation of angle θ . The chosen representation ρ has a unique liftto ˜ ρ : π p M q Ñ SL p , R q (1)which satisfies tr ˜ ρ p X q , tr ˜ ρ p Y q , tr ˜ ρ p Z q ą . It turns out [9] that the lift ˜ ρ satisfiestr ˜ ρ r X , Y s “ ´ p θ { q . Using the parametrization ξ ÞÑ p tr ˜ ρ p X q , tr ˜ ρ p Y q , tr ˜ ρ p Z qq “ p tr ` ρ p X q , tr ` ρ p Y q , tr ` ρ p Z qq , we can therefore identify C θ “ tp x , y , z q P p , | κ p x , y , z q “ ´ p θ { qu . S. KIM, T. KOBERDA, J. LEE, K. OHSHIKA, AND S. P. TAN
Remark . The surjectivity of the above parametrization can be seen by defininga normal form ˜ ρ ξ of ξ “ p x , y , z q P p , . We let ζ ě ζ ` ζ ´ “ z and define ˜ ρ ξ via˜ ρ ξ p X q : “ ˆ x ´
11 0 ˙ , ˜ ρ ξ p Y q : “ ˆ ζ ´ ´ ζ y ˙ . From an easy computation [10, Section 2.2.3] one readily sees thattr ˜ ρ ξ p Z q “ tr ρ ξ pp XY q ´ q “ z . We let ρ ξ : π p M q Ñ PSL p , R q be the projection of ˜ ρ ξ . These concrete formula for˜ ρ ξ and ρ ξ are not actually needed in this paper.3. V ery general representations are free Recall that C : “ ď θ Pp , π q C θ “ tp x , y , z q P p , | κ p x , y , z q P p´ , qu . The following is well-known to experts; we include a proof as motivation for thenext theorem.
Proposition 3.1.
For a very general point ξ P C , the representation ρ ξ is faithful.Proof. For each w P F , let us set Y w : “ t ξ P C | ρ ξ p w q “ Id u Ď t ξ P C | g w p ξ q “ ˘ u . By Lemma 2.1, we see that Y w is contained in a proper algebraic subset of R whenever w ‰
1. In particular, Y w has no interior inside the open subset C of R . Itfollows that C z ¨˝ ď w P F zt u Y w ˛‚ is a G δ –set. (cid:3) Remark . The above proof actually implies that a very general point ξ P C corresponds to representations ρ ξ without any nontrivial parabolic elements in theimage. Theorem 3.3. If θ P p , π q has the property that cos θ is transcendental, then fora very general point ξ in C θ the representation ρ ξ is faithful.Remark . Note that the conclusion of Theorem 3.3 holds for all but countablymany θ . Observe furthermore that the theorem does not extend to all θ P p , π q .For instance, if θ P p Q z Z q π then ρ ξ p F q has nontrivial torsion. hapes of hyperbolic triangles and once-punctured torus groups 9 We let ¯ Q denote the algebraic closure of Q in C . Proof of Theorem 3.3.
Let θ be given as in the hypothesis. Recall for w P F wedefined the trace polynomial g w p x , y , z q . The following claim is the key observationin the proof. Claim.
For each w P F zt u and for each algebraic number c, the setY w p c q : “ t ξ P C θ | g w p ξ q “ c u has empty interior in C θ . Let us assume the claim for now. Then each ξ in the G δ -set C θ z ¨˝ ď w P F zt u Y w p˘ q ˛‚ has the property that ρ ξ is injective, which completes the proof.We now establish the claim. Write g w p x , y , z q “ n ÿ i “ f i p x , y q z i for suitable f i P Z r x , y s . We first consider the case n ě f n ‰
0. We define S : “ tp x , y , z q P C θ X p Q ˆ Q ˆ R q | f n p x , y q ‰ u . Since Q is dense in R , the set S is dense in C θ . For all ξ “ p x , y , z q P S , we seethat z R ¯ Q from the transcendency hypothesis on cos θ . So, we have g w p ξ q R ¯ Q . Inparticular, we obtain S X Y w p c q “ ∅ . Since S is dense, the claim is proved in thiscase.We now consider the case that n “
0. We have g w “ f p x , y q . By Lemma 2.1,we know that f is not constant. Mimicking the previous argument, we consider thefollowing two dense subsets of C θ : S : “ tp x , y , z q P C θ X p Q ˆ R ˆ Q q | f p x , y q ‰ u , S : “ tp x , y , z q P C θ X p R ˆ Q ˆ Q q | f p x , y q ‰ u . Again by considering transcendency of coordinates, we obtain that g w p S i q X ¯ Q “ ∅ for i “ i “
2; that is, we have S i X Y “ ∅ . The conclusion follows from thedensity of S i . (cid:3) If θ is a rational multiple of π , then the image of ρ ξ contains nontrivial torsionelements, so that Im ρ ξ is non-free regardless of the choice of ξ P C θ . On theother hand, if θ is an algebraic irrational multiple of π then the Gelfond–Schneidertheorem implies that cos θ is transcendental, so that ρ ξ is faithful for a very general point ξ P C θ (Theorem 3.3). Hence, the only remaining case in the faithfulness partof Question B is the following. Question 3.5. If θ P p , π q is a transcendental multiple of π , and if cos θ is alge-braic, then is ρ ξ faithful for a very general point ξ P C θ ? The general case of this question seems mysterious. For instance, one may askwhether or not a very general point ξ in C arccos p { q correspond to a faithful holonomymap. 4. C oxeter extensions It will be computationally convenient for us to consider an embedding from ρ ξ p F q to a bigger group generated by involutions (as in Question B). Such an em-bedding will be also used here to see the connection between Questions A and B.Let us begin with an algebraic discussion. We say a word w p X , Y q is palin-dromic in t X , Y u if it reads the same forward and backward, that is, w p X , Y q ´ “ w p X ´ , Y ´ q . We say w P F “ x X , Y , Z y is palindromic if it can be expressed asbeing palindromic in either t X , Y u , t Y , Z u or t Z , X u .Recall we are regarding F “ x X , Y , Z y as an index-two subgroup of W “x P , Q , R y using the embedding X ÞÑ QR , Y ÞÑ RP , Z ÞÑ PQ . Note, in particular, that
RXR “ X ´ and RYR “ Y ´ . Lemma 4.1.
A word u P F is palindromic if and only if it is a product of twoinvolutions in W, with one of the two being P, Q, or R.Proof. The word u is palindromic in t X , Y u if and only if u p X , Y q ´ “ u p X ´ , Y ´ q “ u p RXR , RYR q “ Ru p X , Y q R , which is true if and only if p Ru q “
1. This last expression holds if and only if Ru “ gIg ´ for some I P t P , Q , R u and g P W , as follows from the characterizationof torsion in right-angled Coxeter groups. It follows that u “ R ¨ gIg ´ .Similarly, one shows that u is palindromic in t Y , Z u (resp. t Z , X u ) if and only if u “ P ¨ gIg ´ (resp. u “ Q ¨ gIg ´ ) for some I P t P , Q , R u and g P W . (cid:3) Turning to the geometric side of the embedding F ă W , let us fix θ P p , π q and ξ “ p x , y , z q P C θ . Since κ p x , y , z q ‰
2, the holonomy map ρ ξ is irreducible [10,Proposition 2.3.1]. It follows from [10, Theorem B and Theorem 3.2.2] that ρ ξ uniquely extends to the Coxeter extension ˆ ρ ξ of ρ ξ as shown in the commutative hapes of hyperbolic triangles and once-punctured torus groups 11 diagram below: W “ x P , Q , R y ˆ ρ ξ (cid:39) (cid:39) F “ x X , Y y ρ ξ (cid:47) (cid:47) PSL p , R q Explicitly, one can define the desired extension by projecting the following for-mula [10] in SL p , R q to PSL p , R q :ˆ ρ ξ p R q “ ˜ ρ ξ p X q ˜ ρ ξ p Y q ´ ˜ ρ ξ p Y q ˜ ρ ξ p X q a ´ κ p x , y , z q , where ˜ ρ ξ is the lift in (1). The formulae for P and Q are analogous.From now on, we will often suppress the symbols ρ ξ and ˆ ρ ξ , and simply write w P F or w P W for ˆ ρ ξ p w q when the meaning is clear.We let 2 h X , h Y , h Z be the translation lengths of the hyperbolic isometries X , Y , Z P PSL p , R q , so that x “ tr X “ h X , (2)and similarly for y and z . In the introduction we noted that the punctured torus M can be cut along geodesics and developed to H as illustrated in Figure 1.Let us continue to use the notation from the same figure. In particular, we let t θ A , θ B , θ C u be the interior angles of the triangle O A O B O C . We let d A be the lengthof the geodesic segment O B O C . The isometry X is a hyperbolic translation alongthe oriented geodesic O R O Q , and similar statements hold for the pairs p Y , O P O R q and p Z , O Q O P q . Moreover, the three isometries A : “ QPR , B : “ RQP , C : “ PRQ are (counterclockwise) rotations of angle θ { O A , O B and O C , respec-tively.The midpoints of the sides of this triangle are centers of the π -rotations P , Q , R .The quadrilateral R p O C q O B O C O A can be regarded as a “pseudo-fundamental do-main” for the incomplete hyperbolic structure ξ P C θ on M . We refer the readerto [9, 10] for more details. We have that θ { “ θ A ` θ B ` θ C , and K “ r X , Y s “ p QPR q “ A is a rotation by an angle θ . We also record the identities B “ r Y , Z s , C “ r Z , X s . If we denote by d p¨ , ¨q the distances in H , then the three lengths t d p O A , O R O Q q , d p O B , O R O Q q , d p O C , O R O Q qu are all equal. Lemma 4.2.
Let θ P p , π q . Then there exists a homeomorphism Φ from T π ´ θ { to C θ such that for each η P T π ´ θ { , the Coxeter extension of Φ p η q is precisely therepresentation W Ñ PSL p , R q coming from the side involutions. More precisely, we have cos θ A “ yz p x ` q ´ p y ` z q x ´ x a p xyz ´ x ´ y qp xyz ´ x ´ z q . Proof.
Given p θ A , θ B , θ C q P T π ´ θ { , it is straightforward to define Φ using the hyper-bolic law of cosines: first find t d A , d B , d C u , then t h A , h Y , h C u and then t x , y , z u using(2).Now, given p x , y , z q P C θ , the inverse Φ ´ can be computed as follows. If wecut M along the images of the incomplete geodesic δ X “ O B O C and the completegeodesic γ X extending O Q O R , then we obtain a cylinder, which consists of twocopies of a tri-rectangle with a non-right angle θ { h X and d A {
2, so that the edge of length d A { x { “ cosh h X “ cosh p d A { q sin p θ { q . See [2], for example. Once t d A , d B , d C u is found, using the hyperbolic law of cosineswe can compute cos θ A “ cosh d B cosh d C ´ cosh d A sinh d B sinh d C . The remainder of the proof is straightforward. (cid:3)
We now exhibit a connection between the main questions in the introduction.
Proposition 4.3.
For θ P p , π q and ξ P C θ , we have the following.(1) The representation ρ ξ is faithful if and only if ˆ ρ ξ is faithful.(2) If u P F is palindromic, then the image ˆ ρ ξ p u q P PSL p , R q is either trivialor hyperbolic.Proof. (1) Suppose ρ ξ is faithful. If 1 ‰ g P ker ˆ ρ ξ , then we have that g P F X ker ˆ ρ ξ “ ker ρ ξ “ t u . Hence g P W has order two; this implies that g is conjugate to P , Q or R . Withoutloss of generality, we may assume g “ P . Then we have a contradiction sinceˆ ρ ξ p P q “ ρ ξ p Y q “ ρ ξ p RPRP q “ ˆ ρ ξ p R q “ . The converse is obvious. hapes of hyperbolic triangles and once-punctured torus groups 13 (2) By Lemma 4.1 if u P F is palindromic, then u “ I I for some involutions I and I in W . Since the only nontrivial involutions in PSL p , R q are π -rotations,so are the images ˆ ρ ξ p I q and ˆ ρ ξ p I q . Thus ˆ ρ ξ p u q “ ˆ ρ ξ p I q ˆ ρ ξ p I q is trivial (if ˆ ρ ξ p I q “ ˆ ρ ξ p I q ) or hyperbolic (if ˆ ρ ξ p I q ‰ ˆ ρ ξ p I q ). (cid:3)
5. D ensity of non - faithful representations In this section we give geometric constructions of non-faithful holonomy maps.The relations will be torsion elements as well as non-torsion-type. The idea ofconstruction is to consider the Coxeter extension of F and deform a given repre-sentation ρ slightly by twist .5.1. Twist deformations.
Let us describe a deformation of a given structure ξ P C θ along one of the three coordinate curves t x “ const. u , t y “ const. u , t z “ const. u passing through ξ in C θ .For this, let us fix θ P p , π q and ξ P C θ . We will use the notation from Figure 1for the pseudo-fundamental domain O A O C O B R p O C q of ρ ξ . Consider the equidis-tance locus L to the geodesic O R O Q containing the point O A . Take an arbitrarypoint O A on L and let t O R u “ O A O B X O R O Q , t O Q u “ O A O C X O R O Q as in Figure 2. Then we have the following. Lemma 5.1.
Regardless of the position of O A on L, we have that d p O R , O Q q “ d p O R , O Q q , and the areas of O A O B O C and O A O B O C are the same. We will denote the rotation of angle α at the point V P H as Rot α p V q . Proof of Lemma 5.1.
Let Q and R denote the π –rotations at O Q and O R respec-tively. The geodesic O R O Q is the common invariant geodesic for the hyperbolictranslations Q ¨ R and Q ¨ R , both of which map O B to O C . Thus we have Q ¨ R “ Q ¨ R and hence d p O R , O Q q “ d p O R , O Q q . Moreover, we have thatRot θ { p O C q “ C “ P ¨ R ¨ Q “ P ¨ R ¨ Q “ Rot θ { p O C q . This implies that θ “ θ , which is the twice of the interior angle sum of eachtriangle. (cid:3) Recall our convention that a word w P W also denotes its image ˆ ρ ξ p w q P PSL p , R q when ξ is fixed. The deformation from x P , Q , R y to x P , Q , R y is called an X-twist .Since the distance h X “ d p R , Q q remains constant by the above lemma, we see from(2) that the X -twist occurs along the coordinate curve x “ const. “ tr X . Moreover,such a twist occurs inside C θ since the interior angle sum is preserved. O P O Q O R O A O B O C O Q O R O A L F igure
2. Twist deformation: X “ Q ¨ R “ Q ¨ R .Note that the distance d A “ d p O B , O C q is invariant as well under X -twists. In fact,the whole bi-infinite sequence of π -rotations t X i BX ´ i u i P Z (as well as their variousconjugates by B , C and P ) remain unchanged under X -twists. Lastly, observe that,by a suitable X -twist, we can always increase either d p O A , O B q or d p O A , O C q by anarbitrary amount.Analogously, we define Y-twists and
Z-twists .5.2.
Torsion relations.
In this subsection we will show that holonomy maps con-taining torsion in their image are dense in C θ . For this, let us first recall the followingdefinition due to Lobachevskii.Let h P R ą be a positive real number, and let s Ď H be a geodesic segmentof length h . Choose a bi-infinite geodesic γ which meets an endpoint p of s at aright angle. We then connect the other endpoint q of s with an endpoint at infinityof γ via a geodesic ray δ . The parallel angle α p h q is defined as the acute angle at q between s and δ . It is straightforward to check that α p h q is independent of all thechoices that are made. Theorem 5.2.
Let θ P p , π qz Q π . Then there exists a dense subset S θ Ď p , such that every point ξ P C θ X p S θ ˆ R ˆ R q corresponds to the holonomy map ρ ξ satisfying ρ ξ p r m q “ for some r P π p M qzt u and m ě . Analogous results also hold for R ˆ S θ ˆ R and R ˆ R ˆ S θ .Proof of Theorem 5.2. We use the notation in Figure 1. Fix an arbitrary point ξ P C θ . Since θ R Q π , we can find an integer N ą N θ { π of B N and C N is between 0 and α p d A { q . Consider the isosceles triangle OO B O C as in Figure 3 for which the common angle at O B and O C is N θ {
2. Let J , J B , and hapes of hyperbolic triangles and once-punctured torus groups 15 O P O B O C O F igure
3. Proof of Theorem 5.2. J C denote the reflections along the geodesics O B O C , OO B , and OO C , respectively.Then we have B N “ J J B , C N “ J C J , C N B N “ J C J B “ Rot β p O q , where ´ π ă β “ ´ > p O B , O , O C q ă
0. Since the common angle N θ { O B and O C is constant in C θ , the angle β depends only on the distance d A .Now we perturb the representation (by a Y -twist or Z -twist, for example) in orderto increase the distance d A slightly, so that the deformed angle β satisfies β P Q π and thus C N B N “ r Z , X s N r Y , Z s N becomes a torsion element. Since d A is constantunder X -twists, we have just found an element s of S θ such that every point in the X -coordinate curve C θ X pt s u ˆ R ˆ R q yields a holonomy map with a torsion relation of the form pr Z , X s N r Y , Z s N q m forsome m ě
2. Since we found such a curve which is arbitrarily close to any givenpoint ξ P C θ , the set S θ must be dense in p , . (cid:3) Non-torsion-type kernel elements.
For an element g in a group G , we let xx g yy G denote the normal closure of g in G . When the meaning is clear from thecontext we will simply write xx g yy suppressing a reference to the ambient group G .Let ρ : F Ñ PSL p , R q be a representation. Recall from Definition 1.4 that aword in the set F z ´ď txx r m yy | r P F , m ě ρ p r m q “ u ¯ is called a non-torsion-type word.Being of non-torsion-type is an intrinsic concept for a free group, independentlyof any representation. If a relation (that is, a kernel element of some representation ρ ξ ) is of non-torsion-type, then we may roughly say it is not a consequence of asingle relation that is a proper power. To make this statement more precise, let usintroduce the following notion and note the proposition below. Definition 5.3.
Let G and G be groups with fixed generating sets S and S ,respectively. We say G and G are isomorphic with respect to the generating setsS and S , if there exists a commutative diagram of group homomorphisms F p S q (cid:15) (cid:15) (cid:47) (cid:47) F p S q (cid:15) (cid:15) G (cid:47) (cid:47) G such that the vertical maps are natural quotients extending S i ã Ñ G i and the hori-zontal maps are isomorphisms. Proposition 5.4.
Let ξ P C . If u P ker ρ ξ is a non-torsion-type word, then for anarbitrary one-relator group with torsion G “ x a , b | r m y , the image ρ ξ p F q is notisomorphic to G with respect to the given generating sets t ρ ξ p X q , ρ ξ p Y qu and t a , b u . The following theorem now obviously implies Theorem 1.5.
Theorem 5.5. If θ P p , π qz Q π , then there exists a dense subset S θ Ď p , suchthat for each s P S θ every point ξ on the coordinate curve C θ X t z “ s u corresponds to a holonomy map ρ ξ that has a palindromic non-torsion-type wordin the kernel. By symmetry, the conclusion z “ s can be replaced by x “ s or y “ s . Remark . The theorem implies that there exists a dense subset S θ “ C θ X pp R ˆ S θ ˆ S θ q Y p S θ ˆ R ˆ S θ q Y p S θ ˆ S θ ˆ R qq of C θ consisting of doubles points of coordinate curves such that each point ξ P S θ corresponds to a holonomy map with non-torsion-type kernel elements. O C O A O B O R Z p O R q “ B ´ p O R q B p O R q F igure
4. Construction of the proof of Theorem 5.5. hapes of hyperbolic triangles and once-punctured torus groups 17
Proof of Theorem 5.5.
We use the notation from Figure 1 and from Section 4. Since Z p O R q “ B ´ p O R q , we have that d p O A , Z p O R qq “ d p O A , B ´ p O R qq “ d p O A , B p O R qq . In Figure 4, two circles passing through Z p O R q and B p O R q are drawn (using theKlein projective model of H ); they are centered at O A and O B , respectively.We claim that this configuration can be deformed by a suitable arbitrarily small X -twist so that two conjugates of the π -rotation R coincide, that is, A N p BRB ´ q A ´ N “ ZRZ ´ . To see this claim, let us first recall that the angle > p O A , O B , Z p O R qq “ > p O A , O B , B p O R qq “ θ { C θ . Let β “ > p Z p O R q , O A , B p O R qq , and we can choose N ą N θ “ p N qp θ { q mod 2 π of A N is arbitrarily close to but less than β . Take a suitable small X -twist thatincreases the length d p O A , O B q ; it will also increase the length d p O B , O R q “ d p O B , Z p O R qq “ d p O B , B p O R qq . Then the angle β decreases slightly and we twist until N θ “ β mod 2 π. (We note that this argument works for any θ P p , π qz Q π .) In the end, we obtainthe relation A N p BRB ´ q A ´ N “ ZRZ ´ , which is invariant under Z -twists. The proof of the claim is complete.If we define u N : “ r Z ´ A N B , R s P F , then u N is a product of two involutions and the above relation is equivalent to ρ ξ p u N q “
1. Thus for a dense choice of s in p , there exists an integer N “ N p s q ą ξ on the coordinate curve t z “ s u satisfies u N P ker ρ ξ . We can write u N as a word in F from the computation below. u N “p Z ´ A N B q R p B ´ A ´ N Z q R “ Z ´ r X , Y s N p RQP q R p PQR qp RPQRPQ q N p PQ q R “ Z ´ r X , Y s N p RQ qp PR qp PQ qr R p RPQRPQ q N P s QR “ Z ´ r X , Y s N X ´ Y ´ Z r R p R qp PQRPQR q N ´ p PQRPQ q P s X “ Z ´ r X , Y s N X ´ Y ´ Z p ZY X q N ´ ZYZ ´ X “ XY r X , Y s N X ´ Y ´ Y ´ X ´ r Y ´ , X ´ s N ´ Y ´ X ´ Y XY X “ XY r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s N Y X . As can be expected from Lemma 4.1, u N is indeed palindromic (but not a properpower). To finish the proof it su ffi ces to have the following proposition, which isproved in the next section. (cid:3) Proposition 5.7.
For each N ě , we have that u N is of non-torsion-type.Remark . In the above proof, by Proposition 4.3, the image ρ ξ p u N q is trivial for ξ P C θ X t z “ s u , while it is a nontrivial hyperbolic translation for ξ P C θ zt z “ s u .Recall we denote by g w p x , y , z q P Z r x , y , z s the trace polynomial of a word w P F .From the above proof and the remark following it, we see for each s P S θ thereexists a positive integer N “ N p θ, s q such that the coordinate curve C θ X t z “ s u can be expressed as the intersection of two surfaces C θ X t z “ s u “ C θ X t g u N “ ˘ u . In general, two words U and V may admit the same trace polynomial. Forinstance, we have g U “ g V if U and V are conjugate, or if U “ s . . . s (cid:96) and V “ s (cid:96) s (cid:96) ´ ¨ ¨ ¨ s for s i P t X ˘ , Y ˘ u . In our case, the word u N is palindromicand so, it seems rare that another surface t g w “ ˘ u contains the above coordinatecurve for a di ff erent word w P F not conjugate to u N . This raises the followingquestion: Question 5.9.
Let θ P p , π qz Q π . Does there exist a dense subset R Ă S θ suchthat for each r P R and for N “ N p θ, r q as above, a very general point ξ in thecoordinate curve C θ X t z “ r u corresponds to the monodromy ρ ξ whose image isisomorphic to the one-relator group x X , Y | u N “ y ?6. P roof of P roposition ff er is somewhat tech-nical by nature, but based on a beautiful solution by Newman of the word problemin one-relator groups with torsion (Corollary 6.2) as we describe now. hapes of hyperbolic triangles and once-punctured torus groups 19 Magnus (1932) discovered that the word problem for a (general) one-relatorgroup is solvable. In the case of a one-relator with torsion, the following theoremdue to Newman (1968) gives a particularly simple solution. For a reduced word w in a free group, we let | w | denote its word length and we let r w s denote its cyclicconjugacy class. Theorem 6.1 (Newman Spelling Theorem) . Let G “ x x , . . . , x n | r k y be a one-relator group for some cyclically reduced word r in F n : “ F p x , . . . , x n q , wherek ě . Suppose that a nontrivial reduced word w P F n belongs to the kernel ofthe natural quotient map F n Ñ G. Then w contains a subword u, of length strictlylarger than p k ´ q| r | , such that u is a subword of r ˘ k . If u is a subword of v , we write u ď v . By the Newman Spelling Theorem,we have the following solution to the word problem in two-generator one-relatorgroup with torsion, which we call the Dehn–Newman algorithm , as it is based onthe Dehn’s solution to the word problem for surface groups. Recall our conventionthat F “ x X , Y y . Corollary 6.2 (Dehn–Newmann Algorithm) . Let u and r be cyclically reducedwords in F , and let m ě . Then the truth value of the statement u P xx r m yy can be decided in the following steps, within a finite time. Step 1
If there exists a cyclic conjugation R “ R R of r ˘ such that | R | “ and such that R m ´ R is a subword of u, then proceed to Step 2 ; otherwise,conclude that u
R xx r m yy . Step 2
Let U be the word obtained from u by substituting R ´ for R m ´ R . If U ‰ ,then repeat Step 1 ; otherwise, conclude that u
P xx r m yy . Let us first see a simple application of the algorithm.
Lemma 6.3.
Let N ě .(1) If r P t Y , XY , r X , Y su and m ě , then u N R xx r m yy .(2) If r P F and m ě , then u N R xx r m yy .Proof. (1) We let G : “ F {xx r m yy . Assume for contradiction that u N “ G .Suppose first that r “ Y . By the Dehn–Newman Algorithm (Corollary 6.2), weshould have Y ˘ m ď u N . This implies m “
2. In G “ F {xx Y yy , the element u N coincides with the word u : “ XY r X , Y s N X ´ r Y ´ , X ´ s N Y X . Since Y ˘ ę u , we see that u N “ u ‰ G .Now, consider the case r “ XY . Since a cyclic conjugation of a length fivesubword of p XY q does not appear in u N as a subword, we see that m “
2. Wecancel out all the occurrences of p XY q ˘ , p Y X q ˘ in u N to obtain u : “ X ´ Y ´ r X , Y s N ´ XYY X r Y ´ , X ´ s N ´ Y ´ X ´ . Since u does not contain a length three subword of p XY q ˘ , p Y X q ˘ we conclude from the Dehn–Newman algorithm that u ‰ G .Finally, assume r “ r X , Y s . Writing N “ mq ` t for t P r , m ´ s , the element u N can be rewritten in G as u : “ XY r X , Y s t X ´ Y ´ X ´ r Y ´ , X ´ s t Y X . It is obvious that u does not contain a subword of r m with length 4 p m ´ q ` u ‰ G .(2) Suppose not. The Dehn–Newman Algorithm implies that for some cyclicallyreduced word r that is not a proper power we have r ď u . This considerationrestricts the choices of r to be a cyclic conjugation of a word in t Y , XY , r X , Y su ˘ . By part (1), this is impossible. (cid:3)
We are now ready to give a proof of the proposition.
Proof of Proposition 5.7.
Suppose u N P xx r m yy for some r P F and m ě
2. ByLemma 6.3, we need only consider the case m “
2. Consider the following proce-dure, which takes u “ u N as an input.(i) Pick a cyclically reduced word r “ r r in F such that | r | “ r r r ď u ; here, we require that r r is not cyclically conjugate to a word in t Y , XY , r X , Y su ˘ . Proceed to Step (ii).(ii) Replace an occurrence of r r r in u by r ´ to obtain a reduced word u . Pro-ceed to Step (iii).(iii) Pick a cyclic conjugation r r of r ˘ such that | r | “ r r r ď u . Plug u Ð u and r i Ð r i . Repeat Step (ii).This process is a simplified version of Dehn–Newman Algorithm. It is obviousthat if the procedure terminates at step (i) or (iii), we can conclude that u N is ofnon-torsion-type.We can enumerate all the possible choices of r r r in step (i) as follows, ex-cluding the obvious case r r r “ u N . Verifying all of the following cases, we canconclude that u N R xx r m yy for all r P F and m ě XY r X , Y s t X for 1 ď t ď N ´ XY r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s t for 1 ď t ď N ; hapes of hyperbolic triangles and once-punctured torus groups 21 (3) r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t for 1 ď s , t ď N ;(4) r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s N Y X for 1 ď s ď N ;(5) X r Y ´ , X ´ s t Y X for 1 ď t ď N ´ X ´ Y ´ r X , Y s t X ´ for 1 ď t ď N ´ X ´ Y ´ r X , Y s t X ´ Y ´ X ´ for 0 ď t ď N ´ X ´ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ for 0 ď s , t ď N ´ X ´ Y ´ X ´ ;(10) X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ for 0 ď t ď N ´ X ´ r Y ´ , X ´ s t Y ´ X ´ for 1 ď t ď N ´ Y r X , Y s t XY for 1 ď t ď N ´ Y r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ Y for 0 ď t ď N ´ Y r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s N Y ;(15) Y X ´ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ Y for 0 ď s , t ď N ´ Y X ´ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s N Y for 0 ď s , t ď N ´ Y X r Y ´ , X ´ s t Y for 1 ď t ď N ´ Y ´ r X , Y s t X ´ Y ´ for 1 ď t ď N ´ Y ´ r X , Y s t X ´ Y ´ for 0 ď t ď N ´ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ for 0 ď s , t ď N ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ for 0 ď t ď N ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ for 1 ď t ď N ´ r : ‚ (1) and (12); ‚ (6) and (18); ‚ (11) and (22).So, for each of the 19 cases remaining, we verify that the above algorithm termi-nates. Case (1).
This case corresponds to r “ X , r “ Y r X , Y s t for 1 ď t ď N . Replacing r r r “ XY r X , Y s t X in u N by r ´ “ r Y , X s t Y ´ , we obtain u “ r Y , X s t X ´ Y ´ r X , Y s N ´ t ´ X ´ Y ´ X ´ r Y ´ , X ´ s N Y X “ r Y , X s t X ´ Y ´ r X , Y s N ´ t ´ X ´ Y ´ X ´ p Y ´ X ´ qr Y , X s t r Y , X s N ´ t ´ p Y X q We now continue with r “ X ´ , r “ Y ´ X ´ r Y , X s t ´ Y XY ´ . One sees that r r r “ X ´ Y ´ X ´ r Y , X s t ď u . Plug u Ð u and r i Ð r i . We replace r r r in u by r ´ to obtain u : “ r Y , X s t X ´ Y ´ r X , Y s N ´ t ´ X ´ Y ´ X ´ Y ´ r X , Y s t ´ XY r Y , X s N ´ t ´ p Y X q It is then clear that Step (iii) cannot be executed for u and r . We conclude that u “ u ‰ F {xx r yy . Case (2).
In this case we have r “ X , r “ Y r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s t ´ Y ´ X ´ Y for 1 ď t ď N . Then r ´ “ Y ´ XY r X ´ , Y ´ s t ´ XY X r Y , X s N Y ´ . We replace r r r in u N by r ´ to obtain u “ Y ´ XY r X ´ , Y ´ s t ´ XY X r Y , X s N Y ´ r Y ´ , X ´ s N ´ t Y X , which is reduced. It is clear that no cyclic conjugate of p r r q ˘ occurs as a subwordof u , as is seen from aligning r Y , X s N in u with the corresponding copy in r ´ . Case (3).
In this case we have r “ X , r “ Y X ´ Y ´ r X , Y s s ´ X ´ Y ´ X ´ r Y ´ , X ´ s t ´ Y ´ X ´ Y for 1 ď s , t ď N . Then r ´ “ Y ´ XY r X ´ , Y ´ s t ´ XY X r Y , X s s ´ Y XY ´ . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ s Y ´ XY r X ´ , Y ´ s t ´ XY X r Y , X s s ´ Y XY ´ r Y ´ , X ´ s N ´ t Y X . This word is reduced unless N “ s or N “ t , in which case a subword X appears.It is clear that no cyclic conjugate of p r r q ˘ occurs as a subword of u . Case (4).
In this case we have r “ X , r “ Y X ´ Y ´ r X , Y s s ´ X ´ Y ´ X ´ r Y ´ , X ´ s N Y for 1 ď s ď N . Then r ´ “ Y ´ r X ´ , Y ´ s N XY X r Y , X s s ´ Y XY ´ . Replacing r r r by r ´ givesus u “ XY r X , Y s N ´ s Y ´ r X ´ , Y ´ s N XY X r Y , X s s ´ Y XY ´ , which is reduced. There is exactly one occurrence of Y in u and in p r r q ´ , and ifthese are aligned then a further reduction can be made. Clearly such an alignmentcannot be made, as is checked by reading to the left or right from Y in u . Thus,we must be able to find a subword of u which is a cyclic conjugate of p r r q ˘ andwhich begins and ends with the same power of Y , that is Y ´ X ´ r Y ´ , X ´ s N Y XY X ´ Y ´ r X , Y s s ´ X ´ Y ´ or its inverse. It is clear that this does not happen. Case (5).
In this case we have r “ X , r “ r Y ´ , X ´ s t Y for 1 ď t ď N ´ . Then r ´ “ Y ´ r X ´ , Y ´ s t . Replacing r r r by r ´ gives us u “ XY r X , Y s t X ´ Y ´ X ´ r Y ´ , X ´ s Y ´ X ´ YY ´ r X ´ , Y ´ s t , hapes of hyperbolic triangles and once-punctured torus groups 23 which reduces to XY r X , Y s t X ´ Y ´ X ´ r Y ´ , X ´ s Y ´ X ´ r X ´ , Y ´ s t . This latter word is both reduced and cyclically reduced. In order to performfurther reductions, u must contain subwords of the form Y X r Y ´ , X ´ s t or X ´ Y ´ r X ´ , Y ´ s t . For the first of these cases to occur we would need t “
1, and then it is clear that nocyclic permutation of u contains a cyclic permutation of this word. In the secondcase, we similarly see that no cyclic conjugate of this word occurs as a subword ofa cyclic permutation of u . Case (6).
In this case we have r “ X ´ , r “ Y ´ r X , Y s t for 1 ď t ď N ´ . Then r ´ “ r Y , X s t Y . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ t ´ XY r Y , X s t Y X ´ r Y ´ , X ´ s N Y X , which is reduced. In order to perform further reductions, we must align r X , Y s t X ´ Y ´ or Y X r Y , X s t in u and p r r q ˘ . By inspection, this is not possible. Case (7).
In this case we have r “ X ´ , r “ Y ´ r X , Y s t X ´ Y ´ for 0 ď t ď N ´ . Then r ´ “ Y X r Y , X s t Y . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ t ´ XY X r Y , X s t Y r Y ´ , X ´ s N Y X , which reduces to XY r X , Y s N ´ t ´ XY X r Y , X s t X ´ Y X r Y ´ , X ´ s N ´ Y X . There is only one power of Y of absolute value more than one in u , and it is clearthat Y X r Y , X s t Y X “ p r r q ´ does not occur as a subword of u or of its cyclic conjugates, nor do any of the cyclicconjugates of p r r q ´ which keep the Y intact. We therefore consider subwords ofthe form Y X r Y , X s t Y XY and its inverse. It is immediate to verify that there are nosuch subwords, even in the degenerate case where t “ Case (8).
In this case we have r “ X ´ , r “ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ for 0 ď s , t ď N ´ . Then r ´ “ Y r X ´ , Y ´ s t XY X r Y , X s s Y . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ t ´ XY r X ´ , Y ´ s t XY X r Y , X s s Y X r Y ´ , X ´ s N ´ t ´ Y X , which is reduced and cyclically reduced. Up to cyclic permutation, we must alignsubwords of p r r q ˘ and u of the form X ´ Y ´ X ´ or XY X , or we must set r “ Y and r r r “ Y X r Y , X s s Y XY r X ´ , Y ´ s t XY . Reading to the left or to the right in u from this copy of Y quickly yields a contra-diction. Case (9).
In this case we have r “ X ´ , r “ Y ´ . Then r ´ “ Y . Replacing r r r by r ´ gives us u “ XY r X , Y s N Y r Y ´ , X ´ s N Y X “ XY r X , Y s N ´ XY X ´ Y X r Y ´ , X ´ s N ´ Y X , where the latter expression is both reduced and cyclically reduced. In order toreduce further, the word u must have a subword of the form Y or Y ´ . Indeed,such a subword would arise from either p r r q ˘ or p r r q ˘ as a subword of u , orby setting r “ Y ˘ and r r r P t
Y XY , Y ´ X ´ Y ´ u . Clearly none of these cases occur.
Case (10).
In this case we have r “ X ´ , r “ Y ´ X ´ r Y ´ , X ´ s t Y ´ for 0 ď t ď N ´ . Then r ´ “ Y r X ´ , Y ´ s t XY . Replacing r r r by r ´ gives us u “ XY r X , Y s N Y r X ´ , Y ´ s t XY Y X r Y ´ , X ´ s N ´ t ´ Y X , which reduces to XY r X , Y s N ´ XY X ´ r X ´ , Y ´ s t XY X r Y ´ , X ´ s N ´ t ´ Y X , which is both reduced and cyclically reduced. Up to cyclic permutation, we musthave that u has a subword of the form X ´ Y ´ X ´ or XY X , or a cyclic permutationof a word of the form r r r “ Y XY r X ´ , Y ´ s t XY or r r r “ Y ´ X ´ r Y ´ , X ´ s Y ´ X ´ Y ´ in order to perform further reductions. Clearly this is not the case. Case (11).
In this case we have r “ X ´ , r “ r Y ´ , X ´ s t Y ´ for 1 ď t ď N ´ . Then r ´ “ Y r X ´ , Y ´ s t . Replacing r r r by r ´ gives us u “ XY r X , Y s N X ´ Y ´ Y r X ´ , Y ´ s t Y X r Y ´ , X ´ s N ´ t ´ Y X , which reduces to XY r X , Y s N X ´ Y ´ r X ´ , Y ´ s t Y X r Y ´ , X ´ s N ´ t ´ Y X . hapes of hyperbolic triangles and once-punctured torus groups 25 This last expression is both reduced and cyclically reduced. In order to performfurther reductions, a cyclic permutation of u must contain a subword which is cyclicpermutation of a word of the form XY r X ´ , Y ´ s t or r Y ´ , X ´ s t Y ´ X ´ . The firstof these does not, since t ą r “ Y ´ , r “ X ´ Y X r Y ´ , X ´ s t ´ Y ´ X ´ . Then, r r r “ r Y ´ , X ´ s t Y ´ X ´ Y ´ , which does not occur as a subword of u up to cyclic permutation. Case (13).
In this case we have r “ Y , r “ r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ for 0 ď t ď N ´ . Then r ´ “ XY r X ´ , Y ´ s t XY X r Y , X s N . Replacing r r r by r ´ gives us u “ X Y r X ´ , Y ´ s t XY X r Y , X s N X r Y ´ , X ´ s N ´ t ´ Y X , which reduces to X Y r X ´ , Y ´ s t XY X r Y , X s N ´ Y XY ´ r Y ´ , X ´ s N ´ t ´ Y X . This last expression is reduced and cyclically reduced. In order to perform fur-ther reductions, we must have that after performing a cyclic permutation, there is asubword of u of the form Y r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ “ r r or its inverse. By considering the necessary alignment of product of commutators r Y , X s N , we quickly obtain a contradiction. Case (14).
In this case we have r “ Y , r “ r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s N . Then r ´ “ r X ´ , Y ´ s N XY X r Y , X s N . Replacing r r r by r ´ gives us u “ X r X ´ , Y ´ s N XY X r Y , X s N X , which reduces to Y ´ XY r X ´ , Y ´ s N ´ XY X r Y , X s N X , which is both reduced and cyclically reduced. Note that the words p r r q ˘ havelength 4 N ` u has length 4 N `
4, whence no cyclic permutation of p r r q ˘ occurs as a subword of u . Case (15).
In this case we have r “ Y , r “ X ´ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t Y ´ X ´ for 0 ď s , t ď N ´ . Then r ´ “ XY r X ´ , Y ´ s t XY X r Y , X s s Y X . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ s ´ X Y r X ´ , Y ´ s t XY X r Y , X s s Y X r Y ´ , X ´ s N ´ t ´ Y X , which is both reduced and cyclically reduced. In order to perform further reduc-tions, we would require a cyclic permutation of Y ´ XY r X ´ , Y ´ s t XY X r Y , X s s Y X or its inverse to occur as a subword of u , which it does not. Case (16).
In this case we have r “ Y , r “ X ´ Y ´ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s N for 0 ď s ď N ´ . Then r ´ “ r X ´ , Y ´ s N XY X r Y , X s s Y X . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ s ´ X r X ´ , Y ´ s N XY X r Y , X s s Y X , which reduces to XY r X , Y s N ´ s ´ Y ´ XY r X ´ , Y ´ s N ´ XY X r Y , X s s Y X . This last expression is reduced and cyclically reduced. In order to perform furtherreductions, we must have a subword of u which is a cyclic permutation of Y ´ r X ´ , Y ´ s N XY X r Y , X s s Y X or its inverse. This is not the case.
Case (17).
In this case we have r “ Y , r “ X r Y ´ , X ´ s t for 0 ď t ď N ´ . Then r ´ “ r X ´ , Y ´ s t X ´ . Replacing r r r by r ´ gives us u “ XY r X , Y s N X ´ Y ´ X ´ r Y ´ , X ´ s N ´ t ´ Y ´ X ´ r X ´ , Y ´ s t after one free reduction, and this word is both reduced and cyclically reduced. Toperform further reductions, we would need u to have a subword which is a cyclicconjugate of r X ´ , Y ´ s t X ´ Y ´ or its inverse Y X r Y ´ , X ´ s t , the first of which is not the case.The second of these does occur as a subword of u if N ´ t ´ ą t . In this casewe set r “ r and r “ r and observe that Y X r Y ´ , X ´ s t Y “ r r r is not a subword of u . Case (19).
In this case we have r “ Y ´ , r “ r X , Y s t X ´ Y ´ for 0 ď t ď N ´ . hapes of hyperbolic triangles and once-punctured torus groups 27 Then r ´ “ Y X r Y , X s t . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ t ´ XY X ´ Y X r Y , X s t X ´ r Y ´ , X ´ s N Y X , which is both reduced and cyclically reduced. For further reductions to be possible,we would require a subword of u which is a cyclic conjugate of Y X r Y , X s t or its inverse, which up to cyclic permutation is Y ´ r X , Y s t X ´ Y ´ . Clearly this is not the case if t ě t “ u “ XY r X , Y s N ´ XY X ´ Y XX ´ r Y ´ , X ´ s N Y X , which reduces to XY r X , Y s N ´ XY X ´ Y X r Y ´ , X ´ s N ´ Y X , which is both reduced and cyclically reduced. Now, r r “ Y ´ X ´ Y ´ , whichdoes not appear as a subword of u up to cyclic permutation, but its inverse Y XY does. In order to perform further reductions, we would have to either set r “ Y and r r r “ Y XY or r “ X and r r r “ XY X . Neither of these words appear,even after cyclic permutation. Case (20).
In this case we have r “ Y ´ , r “ r X , Y s s X ´ Y ´ X ´ r Y ´ , X ´ s t for 0 ď s , t ď N ´ . Then r ´ “ r X ´ , Y ´ s t XY X r Y , X s s . Replacing r r r by r ´ gives us u “ XY r X , Y s N ´ s ´ XY X ´ r X ´ , Y ´ s t XY X r Y , X s s X ´ Y X r Y ´ , X ´ s N ´ t ´ Y X , which is both reduced and cyclically reduced. To perform further reductions, wemust have a subword of u which is a cyclic permutation of Y r X ´ , Y ´ s t XY X r Y , X s s or its inverse, which is not the case. Case (21).
In this case we have r “ Y ´ , r “ Y ´ X ´ r Y ´ , X ´ s t for 0 ď t ď N ´ . Then r ´ “ r X ´ , Y ´ s t XY . Replacing r r r by r ´ gives us u “ XY r X , Y s N X ´ r X ´ , Y ´ s t XY X ´ Y X r Y ´ , X ´ s N ´ t ´ Y X , which is both reduced and cyclically reduced. Supposing the possibility of furtherreductions, we would need a subword which is a cyclic permutation of r X ´ , Y ´ s t XY or its inverse, which up to cyclic permutation is given by Y ´ X ´ r Y ´ , X ´ s t Y ´ . We easily see that this is not the case. (cid:3) A ppendix A. A computational approach , by X inghua G ao As mentioned in the introduction, it was not known whether there exist points ξ P C such that ρ ξ has a non-free torsion-free image. In this section, we provide andimplement computational heuristics that can find such examples of ρ ξ , which turnout to have closed hyperbolic 3-manifold groups as images.A.1. An arithmetic formulation of the problem.
The starting point is a hyper-bolic 3-manifold T with a single cusp, whose fundamental group G is generated bytwo elements a and b . As an explicit example, one can take a hyperbolic 2-bridgeknot complement. Then, consider a hyperbolic Dehn filling T q of T for some q P Q ,and a number field Q p α q such that G “ x a , b y “ π p T q q ď PSL p , Q p α qq . We abuse language slightly and also write G “ x a , b y for a lift of G to SL p , Q p α qq (cf. [16, 3]). Question A.1.
Under what conditions on T and q do the following conclusionshold?(i) There is a Galois automorphism σ : α ÞÑ β with β P R , .(ii) t tr a σ , tr b σ , tr p ab q σ u Ă R zr´ , s ;(iii) tr r a , b s σ P p´ , q . Computationally, Question A.1 suggests that we enumerate such possible T and q , and verify all of the above three conditions. As we have the hyperbolic structureof T q and the algebraic number α P C , the verification step should be computa-tionally straightforward up to the computation of Galois conjugates. Once we havesuch an example of T and q , then the resulting point ξ : “ p tr a σ , tr b σ , tr p ab q σ q is a point in the character variety of C θ for θ satisfying ´ p θ { q “ tr r a , b s σ . Moreover, the image of the monodromy ρ ξ is isomorphic to π p T q q as we desire. Remark
A.2 . Note that a parabolic generator of π p T q will remain parabolic afterGalois conjugation. hapes of hyperbolic triangles and once-punctured torus groups 29 A.2.
An algorithm for producing explicit examples.
Examples of p T , q q withthe desired properties can be produced using SnapPy [4, 6]. In order to construct arepresentation with the image in PSL p , R q , we will need the following fact. Lemma A.3. [17, Corollary 3.2.5] If Γ is a nonelementary subgroup of SL p , C q such that Q p tr Γ q is a subset of R , then Γ is conjugate to a subgroup of SL p , R q . Let T r be a closed hyperbolic 3-manifold obtained by applying Dehn filling toone-cusped hyperbolic 3-manifold T along a curve of slope r . We fix a triangula-tion of T by ideal tetrahedra, which also gives an ideal triangulation of T r . Then theedge gluing equations of T together with the Dehn filling equation determine thehyperbolic structure of T r . The solution to the this system of equations is a set ofcomplex numbers, which parameterize the shape of tetrahedra in T r . These param-eters generate the tetrahedral field K of T r . We want to find a real embedding σ ofthe number field K so that all the tetrahedra of T have real shapes after applying σ .The associated holonomy representation then gives a faithful representation ρ R : π p T r q Ñ PSL p , C q with all matrices in the image having real trace. Therefore by Lemma A.3, thegroup ρ R p π p T r qq is conjugate into PSL p , R q . Since a conjugacy leaves the traceunchanged, we obtain a desired hyperbolic 3-manifold T r giving an a ffi rmative an-swer to Question A.1.The process of finding a suitable T and corresponding tetrahedral field can beformulated with the following steps:Step 1) Let T be a hyperbolic knot complement with the default triangulation inSnapPy and apply Dehn filling of slope r . Compute the tetrahedral field K of T r using the SnapPy manifold class tetrahedra field gens() . Wecan then use the SageMath number field class find field() to find thedefining polynomial of K .Step 2) Find a real embedding σ of K , if there exits one, using the SageMath num-ber field class real embeddings() .Step 3) Apply the real embedding σ and set up the new triangulation with real shapeparameters, using the SnapPy manifold class set tetrahedra shapes() .Step 4) Computes the associated holonomy representation ρ R and the image of a , b , ab and r a , b s under ρ R , using the SnapPy fundamental group classe SL2C() .Finally, compute the resulting traces.
A.3.
Explicit examples.
Here we produce several examples giving a ffi rmative an-swer to Question A.1 . The traces are truncated to four places after the decimalpoint. tr ρ R p a q tr ρ R p b q tr ρ R p ab q tr ρ R pr a , b sq p q p q p q p q -2.2535 2.1399 -2.2535 1.868610 p q -3.7588 -3.0575 9.0343 -0.7349In this table, a and b are the generators of the corresponding fundamental group,with the default triangulation in SnapPy. Presentations of the fundamental groupof 7 p q and 8 p q are included below, as well as the matrix representatives of a and b . Presentations of the fundamental groups of the other three manifolds areunwieldy due to their size, so we have omitted them. The interested readers mayuse the manifold class fundamental group() to verify our claims.For the first manifold, π p p qq “ (cid:104) a , b | abABBAbABabbaBabbaBAbABBAbaBabbaB , aaaaabABBAbABabbaBabAbaBabbaBAbABBAb (cid:105) ,ρ R p a q “ ˆ . ´ . . ˙ , ρ R p b q “ ˆ . . . ˙ . Here, upper case and lower case versions of a letter are inverses of each other.Again, we truncate matrix entries to four places. We note however that they are allalgebraic numbers in a real embedding of the tetrahedra field K of 7 p q , with thedefining polynomial p p x q “ x ´ x ´ x ` x ´ x ` x ´ x ` x ´ x ` x ´ x ` x ´ . For the second manifold, π p p qq “ (cid:104) a , b | aaabABBAAABabbaaabABBAbaaabbaBAAABBAbaaabbaBAAABabb , aaabABBAAABabbaaabABBAbABBAbaaabbaBAAABBAbaaabABAb (cid:105) ,ρ R p a q “ ˆ . . i ´ . i ˙ , ρ R p b q “ ˆ . ´ . i ´ . i ´ . ˙ . hapes of hyperbolic triangles and once-punctured torus groups 31 Note that these two matrices are not in PSL p , R q , but they are simultaneously con-jugate into PSL p , R q , according to Lemma A.3. The defining polynomial of thetetrahedra field K of 8 p q is p p x q “ x ´ x ´ x ´ x ` x ` x ´ x ´ x ´ x ` x ` x ´ x ´ x ´ x ´ . For now, we are only able to produce closed hyperbolic 3-manifolds via 0-Dehnfilling to two-bridge knot complements as examples.A cknowledgements
The authors thank A. Reid and R. Schwartz for helpful discussions. X. Gaowould like to thank K. Ohshika for suggesting to her the question in the appendixand thank N. Dunfield for helping her with the code. S. Kim and X. Gao are sup-ported by Mid-Career Researcher Program (2018R1A2B6004003) through the Na-tional Research Foundation funded by the government of Korea. T. Koberda is par-tially supported by an Alfred P. Sloan Foundation Research Fellowship and by NSFGrant DMS-1711488. J. Lee is supported by the grant NRF-2019R1F1A1047703.K. Ohshika is partially supported by the JSPS Grant-in-aid for Scientific ResearchNo 17H02843. Tan is partially supported by the National University of Singaporeacademic research grant R-146-000-289-114.R eferences
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