Shapes of hyperbolic triangles and once-punctured torus groups

Let $\Delta$ be a hyperbolic triangle with a fixed area $\varphi$. We prove that for all but countably many $\varphi$, generic choices of $\Delta$ have the property that the group generated by the $\pi$--rotations about the midpoints of the sides of the triangle admits no nontrivial relations. By contrast, we show for all $\varphi\in(0,\pi)\setminus\mathbb{Q}\pi$, a dense set of triangles does afford nontrivial relations, which in the generic case map to hyperbolic translations. To establish this fact, we study the deformation space $\mathfrak{C}_\theta$ of singular hyperbolic metrics on a torus with a single cone point of angle $\theta=2(\pi-\varphi)$, and answer an analogous question for the holonomy map $\rho_\xi$ of such a hyperbolic structure $\xi$. In an appendix by X.~Gao, concrete examples of $\theta$ and $\xi\in\mathfrak{C}_\theta$ are given where the image of each $\rho_\xi$ is finitely presented, non-free and torsion-free; in fact, those images will be isomorphic to the fundamental groups of closed hyperbolic 3--manifolds.


Introduction
Take a geodesic triangle in the hyperbolic plane, and consider the rotations of angle 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 side involutions will move the triangle exactly back to itself.
For a regular pentagon in the Euclidean plane, there are many "unexpected" coincidences of this type [19]. On the other hand, a regular tetrahedron in R 3 can never 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 be trivial. Even if the area is not a rational multiple of π, torsion relations still appear for a dense choice of hyperbolic triangles (cf. Theorem 5.2 below). So, we are led to address the question whether or not relations that are not "consequences" of torsion relations can still be found for a dense set of triangles; we refer the reader to Definition 1.4 for a precise formulation.
Question A. In the space of hyperbolic triangles whose area is fixed as ϕ P p0, πq, what are the necessary and sufficient conditions for the side involutions to generate the Coxeter group Z 2˚Z2˚Z2 ? Do the involutions for a dense set of hyperbolic triangles admit relations that are not consequences of torsion relations?
Here, the space of hyperbolic triangles ABC with fixed area ϕ is identified with its parameter space of the interior angles: T ϕ :" tpθ A , θ B , θ C q P p0, πq 3 | θ A`θB`θC " π´ϕu.
Our take on Question A will be via the space of marked incomplete hyperbolic structures on a punctured torus. Throughout this paper, we let M be a torus minus a puncture p. Let us fix an angle θ P p0, 2πq. 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. We will also consider the Fricke-Klein space C θ of all marked hyperbolic structures with cone angles in p0, 2πq on M.
We fix an oriented meridian X and an oriented longitude Y of M, and identify π 1 pMq with a rank-two free group F 2 :" xX, Y, Z | XYZ " 1y.
It is convenient to regard F 2 as an index-two subgroup of a free Coxeter group W :" Z 2˚Z2˚Z2 " xP, Q, R | P 2 " Q 2 " R 2 " 1y using the embedding X Þ Ñ QR, Y Þ Ñ RP, Z Þ Ñ PQ. Each hyperbolic structure ξ P C θ defines a conjugacy class of holonomy maps, a representative of which we denote by ρ ξ : π 1 pMq Ñ PSLp2, Rq -Isom`pH 2 q.
These holonomy maps are defined via the developing mapM Ñ H 2 . The isometries tρ ξ pXq, ρ ξ pYq, ρ ξ pZqu 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 2 is Hopfian, we have that ρ ξ is faithful if and only if Im ρ ξ nonabelian and free.
Here, tr`p¨q is understood as the absolute value of the trace of an element in PSLp2, Rq.
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 p0, 2πq. Let γ X be the simple closed geodesic realizing X P π 1 pMq. There exists a unique incomplete geodesic δ X on M disjoint from γ X , and that starts and ends at p. Construct the geodesics tγ 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 hyperbolic triangles whose interior angle sums are θ{2. The shape of this triangle determines a point in T π´θ{2 . A lift of this picture to H 2 is illustrated in Figure 1. As in the figure, we let tO P , O Q , O R u denote the midpoints of the sides from one of the triangles. We write tρ ξ pPq,ρ ξ pQq,ρ ξ pRqu for the corresponding side involutions. Then we have that ρ ξ pXq "ρ ξ pQq¨ρ ξ pRq, ρ ξ pYq "ρ ξ pRq¨ρ ξ pPq, ρ ξ pZq "ρ ξ pPq¨ρ ξ pQq.
Through Coxeter extensions, we will approach Question A by studying the following closely related question.
Question B. For a fixed angle θ P p0, 2πq, what are the necessary and sufficient conditions for ξ P C θ to yield a faithful representation ρ ξ : π 1 pMq Ñ PSLp2, Rq? Is there a dense choice of ξ P C θ for which the representation ρ ξ admits relations that are not consequences of torsion relations?
Let A be a topological space. We say that a certain statement holds for a very general (or, generic) ξ P A if the statement holds for all ξ in some dense G δ subset of A. If A is an open subset of a Euclidean space, then it follows that the statement holds on a full-measure subset of A.
In this paper, we are primarily interested in establishing the faithfulness part of Question B for a very general point ξ P C θ , and in answering the density question for a suitable interpretation of the phrase "consequences of torsion relations". Remark 1.1. One may also consider side reflections instead of side involutions of a hyperbolic triangle ∆ with angles pθ A , θ B , θ C q, although we will not pursue this direction. We briefly note in this case that two representationsρ ξ and ρ ξ are still well-defined on W and on F 2 respectively. The representation ρ ξ is a holonomy map of an incomplete hyperbolic three-punctured sphere S 0,3 with three conical singularities of angles p2θ A , 2θ B , 2θ C q. This approach could be useful when one is interested 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: . If θ P p0, 2πq has the property that cos θ is transcendental, then a very general point ξ in C θ corresponds to a faithful holonomy map.
Note that the hypothesis holds for all but countably many values of θ. The reader may compare Theorem 1.2 to other generic phenomena in PSLp2, Rq; see [5,7,15], the first two of which deal with the faithfulness question in the case of closed surface groups. One can trace at least back to Hausdorff [11] the idea of using the transcendency of cos θ in order to produce faithful group actions.
It is a well-known consequence of the Gelfond-Schneider theorem that cos θ is transcendental if θ{π is irrational and algebraic. Thus, we deduce more concrete cases when ρ ξ is faithful as follows. Corollary 1.3. If θ P p0, 2πq is an irrational algebraic multiple of π, then ρ ξ is faithful for a very general point ξ P C θ .
We have briefly mentioned the abundance of torsion relations at the beginning of this introduction. Indeed, the image of ρ ξ is dense in PSLp2, Rq unless ρ ξ is discrete (and hence Fuchsian), as is true of all Zariski dense subgroups of simple Lie groups [14]. Since elliptic isometries form an open subset in PSLp2, Rq, it is reasonable to expect that an elliptic element in the image of ρ ξ would not have constant rotation angle under perturbations in C θ near ξ. Thus, a torsion relation would appear under small deformations of ξ having a fixed cone angle θ; we direct 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 are consequences 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.
A word is of non-torsion-type if it is not of torsion-type.
For a word u P F 2 , being a non-torsion-type word is a stronger condition than not equal to a proper power. In the special case when θ " 2π{n, the group ρ ξ pF 2 q is isomorphic to a Fuchsian orbifold group xX, Y | rX, Ys n " 1y for all choices of ξ P C θ ; in this case, every kernel element must be of torsion-type. When θ R Qπ, we can produce non-torsion-type kernel elements as addressed in the second half of Question B. Theorem 1.5 (cf. Theorem 5.5). If θ P p0, 2πq is an irrational multiple of π, then ker ρ ξ ă F 2 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 pF 2 q cannot be isomorphic to a one-relation group with torsion with respect to the standard generating set tρ ξ pXq, ρ ξ pYqu; see Definition 5.3 and Proposition 5.4.
We will actually produce a dense subset S θ Ď p2, 8q such that each point on the coordinate curves tx " su, ty " su, tz " su in C θ for s P S θ corresponds to holonomy maps admitting non-torsion-type kernel elements. It follows that the same property holds on (densely many) doubles points of such coordinate curves. Remark 1.6. A consequence of Theorem 1.5 is that a free indiscrete representation is algebraically unstable, even when restricted to C θ ; roughly speaking this means that such a representation is a limit of non-free representations. In the absence of such a "relativizing" restriction, this type of instability is well-known for non-abelian free groups, even in a much more general setting of connected Lie groups [8]. On the other hand, Sullivan [20] established algebraic stability of all convex-cocompact subgroups in PSLp2, Cq.
Let us direct our attention back to Question A. The faithfulness part has the same answer as Question B; see Proposition 4.3 (1). Regarding the density part, it is unclear to the authors whether the non-torsion-type kernel elements in F 2 found in Theorem 1.5 can still be non-torsion-type in W. However, their additional property of being palindromic has a simple interpretation in W: they are products of two involutions in W (Lemma 4.1). Thus we establish the following partial answer to Question A.
Corollary 1.7. The following conclusions hold: (1) For all but countably many ϕ P p0, πq, the side involutions of a very general triangle with area ϕ generates the Coxeter group W. (2) If ϕ is an irrational multiple of π, then for a dense choice of hyperbolic triangles with area ϕ the side involutions admit a relation which is the product of two involutions in W.
We remark that if u P W is a product of two involutions then, for all ξ P C, the imageρ ξ puq is either trivial or hyperbolic (Proposition 4.3(2)). In fact, it is hyperbolic 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 documented such that ρ ξ pF 2 q is non-free and torsion-free. Question 1.8. If θ P p0, 2πq 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 approaching Question 1.8, together with their implementation, are given. The examples of pairs pθ, ξq exhibited therein have the property that the images of ρ ξ are actually isomorphic to fundamental groups of closed hyperbolic 3-manifolds.

Fricke-Klein space
Most of the material in the section is well-known; we direct the reader to [9] for a standard reference. We adopt the standing convention that group and matrix actions are on the left, unless stated otherwise.
The group SLp2, Rq acts on H 2 by Möbius transforms, with kernel given by the center, and with image PSLp2, Rq -Isom`pH 2 q. Let trpAq denote the trace of a matrix A. For each word w " wpX, Yq P F 2 " xX, Yy there exists a trace polynomial g w P Zrx, y, zs such that whenever U, V P SLp2, Rq we have tr wpU, Vq " g w ptr U, tr V, tr UVq.
The existence of this polynomial is one of the simplest instances of invariant theory on character varieties [18]; see also [13] for a concrete formula which computes g w .
This means that for each pair of matrices U, V P SLp2, Rq we have that trrU, Vs " κptr U, tr V, tr UVq.
Lemma 2.1. For each w P F 2 zt1u, the polynomial g w px, y, zq is not constant.
However, there exist two-generated nonelementary purely loxodromic Fuchsian groups which are nonabelian and free. This implies g w cannot be identically equal to 2.
In this paper, we are mostly concerned with matrices in PSLp2, Rq. The trace of such a matrix A P PSLp2, Rq is only determined up to sign, so as mentioned in the introduction, we often use the quantity tr`A :" | tr A|.
Recall we let C θ denote the deformation space (Fricke-Klein space) of marked incomplete hyperbolic metrics on M with a fixed cone angle θ P p0, 2πq at p. Each point ξ P C θ corresponds to a conjugacy class of a holonomy map ρ : π 1 pMq " xX, Y, Z | XYZ " 1y Ñ PSLp2, Rq so that ρrX, Ys is a rotation of angle θ. The chosen representation ρ has a unique lift toρ which satisfies trρpXq, trρpYq, trρpZq ą 2.
We let ρ ξ : π 1 pMq Ñ PSLp2, Rq be the projection ofρ ξ . These concrete formula for ρ ξ and ρ ξ are not actually needed in this paper.

Very general representations are free
Recall that The following is well-known to experts; we include a proof as motivation for the next theorem. Proof. For each w P F 2 , let us set Y w :" tξ P C | ρ ξ pwq " Idu Ď tξ P C | g w pξq "˘2u.
By Lemma 2.1, we see that Y w is contained in a proper algebraic subset of R 3 whenever w ‰ 1. In particular, Y w has no interior inside the open subset C of R 3 . It follows that Remark 3.2. The above proof actually implies that a very general point ξ P C corresponds to representations ρ ξ without any nontrivial parabolic elements in the image. Theorem 3.3. If θ P p0, 2πq has the property that cos θ is transcendental, then for a very general point ξ in C θ the representation ρ ξ is faithful.
Remark 3.4. Note that the conclusion of Theorem 3.3 holds for all but countably many θ. Observe furthermore that the theorem does not extend to all θ P p0, 2πq. For instance, if θ P pQzZqπ then ρ ξ pF 2 q has nontrivial torsion.
We letQ 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 2 we defined the trace polynomial g w px, y, zq. The following claim is the key observation in the proof.
Claim. For each w P F 2 zt1u and for each algebraic number c, the set Y w pcq :" tξ P C θ | g w pξq " cu has empty interior in C θ .
Let us assume the claim for now. Then each ξ in the G δ -set has the property that ρ ξ is injective, which completes the proof.
We now establish the claim. Write We first consider the case n ě 1 and f n ‰ 0. We define S :" tpx, y, zq P C θ X pQˆQˆRq | f n px, yq ‰ 0u.
Since Q 2 is dense in R 2 , the set S is dense in C θ . For all ξ " px, y, zq P S , we see that z RQ from the transcendency hypothesis on cos θ. So, we have g w pξq RQ. In particular, we obtain S X Y w pcq " ∅. Since S is dense, the claim is proved in this case.
We now consider the case that n " 0. We have g w " f 0 px, yq. By Lemma 2.1, we know that f 0 is not constant. Mimicking the previous argument, we consider the following two dense subsets of C θ : Again by considering transcendency of coordinates, we obtain that g w pS i q XQ " ∅ for i " 1 or i " 2; that is, we have S i X Y " ∅. The conclusion follows from the density of S i .
If θ is a rational multiple of π, then the image of ρ ξ contains nontrivial torsion elements, so that Im ρ ξ is non-free regardless of the choice of ξ P C θ . On the other hand, if θ is an algebraic irrational multiple of π then the Gelfond-Schneider theorem 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 part of Question B is the following.
Question 3.5. If θ P p0, 2πq is a transcendental multiple of π, and if cos θ is algebraic, 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 ask whether or not a very general point ξ in C arccosp1{3q correspond to a faithful holonomy map.

Coxeter extensions
It will be computationally convenient for us to consider an embedding from ρ ξ pF 2 q to a bigger group generated by involutions (as in Question B). Such an embedding 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 wpX, Yq is palindromic in tX, Yu if it reads the same forward and backward, that is, wpX, Yq´1 " wpX´1, Y´1q. We say w P F 2 " xX, Y, Zy is palindromic if it can be expressed as being palindromic in either tX, Yu, tY, Zu or tZ, Xu.
Recall we are regarding F 2 " xX, Y, Zy as an index-two subgroup of W " xP, Q, Ry using the embedding Note, in particular, that RXR " X´1 and RYR " Y´1. Proof. The word u is palindromic in tX, Yu if and only if upX, Yq´1 " upX´1, Y´1q " upRXR, RYRq " RupX, YqR, which is true if and only if pRuq 2 " 1. This last expression holds if and only if Ru " gIg´1 for some I P tP, Q, Ru and g P W, as follows from the characterization of torsion in right-angled Coxeter groups. It follows that u " R¨gIg´1.
Similarly, one shows that u is palindromic in tY, Zu (resp. tZ, Xu) if and only if u " P¨gIg´1 (resp. u " Q¨gIg´1) for some I P tP, Q, Ru and g P W.
Turning to the geometric side of the embedding F 2 ă W, let us fix θ P p0, 2πq and ξ " px, y, zq P C θ . Since κpx, y, zq ‰ 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 extensionρ ξ of ρ ξ as shown in the commutative diagram below: Explicitly, one can define the desired extension by projecting the following formula [10] in SLp2, Rq to PSLp2, Rq: whereρ 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 andρ ξ , and simply write w P F 2 or w P W forρ ξ pwq when the meaning is clear.
We let 2h X , 2h Y , 2h Z be the translation lengths of the hyperbolic isometries X, Y, Z P PSLp2, Rq, so that 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 2 as illustrated in Figure 1.
Let us continue to use the notation from the same figure. In particular, we let The midpoints of the sides of this triangle are centers of the π-rotations P, Q, R. The quadrilateral RpO C qO B O C O A can be regarded as a "pseudo-fundamental domain" for the incomplete hyperbolic structure ξ P C θ on M. We refer the reader to [9, 10] for more details. We have that θ{2 " θ A`θB`θC , and K " rX, Ys " pQPRq 2 " A 2 is a rotation by an angle θ. We also record the identities B 2 " rY, Zs, C 2 " rZ, Xs.
If we denote by dp¨,¨q the distances in H 2 , then the three lengths Lemma 4.2. Let θ P p0, 2πq. Then there exists a homeomorphism Φ from T π´θ{2 to C θ such that for each η P T π´θ{2 , the Coxeter extension of Φpηq is precisely the representation W Ñ PSLp2, Rq coming from the side involutions. More precisely, we have Proof. Given pθ A , θ B , θ C q P T π´θ{2 , it is straightforward to define Φ using the hyperbolic law of cosines: first find td A , d B , d C u, then th A , h Y , h C u and then tx, y, zu using (2). Now, given px, y, zq P C θ , the inverse Φ´1 can be computed as follows. If we cut M along the images of the incomplete geodesic δ X " O B O C and the complete geodesic γ X extending O Q O R , then we obtain a cylinder, which consists of two copies of a tri-rectangle with a non-right angle θ{4 and opposite edge lengths h X and d A {2, so that the edge of length d A {2 is adjacent to the non-right angle. In this case we have a formula See [2], for example. Once td A , d B , d C u is found, using the hyperbolic law of cosines we can compute The remainder of the proof is straightforward.
We now exhibit a connection between the main questions in the introduction.
Proposition 4.3. For θ P p0, 2πq and ξ P C θ , we have the following.
(1) The representation ρ ξ is faithful if and only ifρ ξ is faithful.
(2) If u P F 2 is palindromic, then the imageρ ξ puq P PSLp2, Rq is either trivial or hyperbolic. Proof.
(1) Suppose ρ ξ is faithful. If 1 ‰ g P kerρ ξ , then we have that Hence g P W has order two; this implies that g is conjugate to P, Q or R. Without loss of generality, we may assume g " P. Then we have a contradiction sincê ρ ξ pPq " 1 and thus ρ ξ pY 2 q " ρ ξ pRPRPq "ρ ξ pR 2 q " 1.
The converse is obvious.

Density 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 of construction is to consider the Coxeter extension of F 2 and deform a given representation ρ slightly by twist.

Twist deformations.
Let us describe a deformation of a given structure ξ P C θ along one of the three coordinate curves tx " const.u, ty " const.u, tz " const.u passing through ξ in C θ .
For this, let us fix θ P p0, 2πq and ξ P C θ . We will use the notation from Figure Figure 2. Then we have the following.
This implies that θ " θ 1 , which is the twice of the interior angle sum of each triangle.
Recall our convention that a word w P W also denotes its imageρ ξ pwq P PSLp2, Rq when ξ is fixed. The deformation from xP, Q, Ry to xP, Q 1 , R 1 y is called an X-twist. Since the distance h X " dpR, Qq 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.
Note that the distance d A " dpO B , O C q is invariant as well under X-twists. In fact, the whole bi-infinite sequence of π-rotations tX i BX´iu iPZ (as well as their various conjugates by B, C and P) remain unchanged under X-twists. Lastly, observe that, by a suitable X-twist, we can always increase either dpO A , O B q or dpO A , O C q by an arbitrary amount.
Analogously, we define Y-twists and Z-twists.

Torsion relations.
In this subsection we will show that holonomy maps containing torsion in their image are dense in C θ . For this, let us first recall the following definition due to Lobachevskii. Let h P R ą0 be a positive real number, and let s Ď H 2 be a geodesic segment of length h. Choose a bi-infinite geodesic γ which meets an endpoint p of s at a right angle. We then connect the other endpoint q of s with an endpoint at infinity of γ via a geodesic ray δ. The parallel angle αphq is defined as the acute angle at q between s and δ. It is straightforward to check that αphq is independent of all the choices that are made.
Theorem 5.2. Let θ P p0, 2πqzQπ. Then there exists a dense subset S θ Ď p2, 8q such that every point ξ P C θ X pS θˆRˆR q corresponds to the holonomy map ρ ξ satisfying ρ ξ pr m q " 1 for some r P π 1 pMqzt1u and m ě 2. Analogous results also hold for RˆS θˆR and RˆRˆS θ .
Proof of Theorem 5.2. We use the notation in Figure 1. Fix an arbitrary point ξ P C θ . Since θ R Qπ, we can find an integer N ą 0 so that the common rotation angle Since the common angle Nθ{2 at 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 order to increase the distance d A slightly, so that the deformed angle β 1 satisfies β 1 P Qπ and thus C 2N B 2N " rZ, Xs N rY, Zs N becomes a torsion element. Since d A is constant under X-twists, we have just found an element s of S θ such that every point in the X-coordinate curve C θ X ptsuˆRˆRq yields a holonomy map with a torsion relation of the form prZ, Xs N rY, Zs N q m for some m ě 2. Since we found such a curve which is arbitrarily close to any given point ξ P C θ , the set S θ must be dense in p2, 8q.

5.3.
Non-torsion-type kernel elements. For an element g in a group G, we let xxgyy G denote the normal closure of g in G. When the meaning is clear from the context we will simply write xxgyy suppressing a reference to the ambient group G.
Let ρ : F 2 Ñ PSLp2, Rq be a representation. Recall from Definition 1.4 that a word in the set F 2 z´ďtxxr m yy | r P F 2 , m ě 2 and ρpr m q " 1uī s called a non-torsion-type word.
Being of non-torsion-type is an intrinsic concept for a free group, independently of 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 a single relation that is a proper power. To make this statement more precise, let us introduce the following notion and note the proposition below. Definition 5.3. Let G 1 and G 2 be groups with fixed generating sets S 1 and S 2 , respectively. We say G 1 and G 2 are isomorphic with respect to the generating sets S 1 and S 2 , if there exists a commutative diagram of group homomorphisms such that the vertical maps are natural quotients extending S i ãÑ G i and the horizontal maps are isomorphisms.
Proposition 5.4. Let ξ P C. If u P ker ρ ξ is a non-torsion-type word, then for an arbitrary one-relator group with torsion G " xa, b | r m y, the image ρ ξ pF 2 q is not isomorphic to G with respect to the given generating sets tρ ξ pXq, ρ ξ pYqu and ta, bu.
The following theorem now obviously implies Theorem 1.5.
Theorem 5.5. If θ P p0, 2πqzQπ, then there exists a dense subset S θ Ď p2, 8q such that for each s P S θ every point ξ on the coordinate curve C θ X tz " su corresponds to a holonomy map ρ ξ that has a palindromic non-torsion-type word in the kernel.
By symmetry, the conclusion z " s can be replaced by x " s or y " s.
Remark 5.6. The theorem implies that there exists a dense subset S 1 θ " C θ X ppRˆS θˆS θ q Y pS θˆRˆS θ q Y pS θˆS θˆR qq of C θ consisting of doubles points of coordinate curves such that each point ξ P S 1 θ corresponds to a holonomy map with non-torsion-type kernel elements. Proof of Theorem 5.5. We use the notation from Figure 1 and from Section 4. Since ZpO R q " B´1pO R q, we have that dpO A , ZpO R qq " dpO A , B´1pO R qq " dpO A , BpO R qq.
In Figure 4, two circles passing through ZpO R q and BpO R q are drawn (using the Klein projective model of H 2 ); 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, To see this claim, let us first recall that the angle is constant under any deformation in C θ . Let β " >pZpO R q, O A , BpO R qq, and we can choose N ą 0 so that the rotation angle Nθ " p2Nqpθ{2q mod 2π of A 2N is arbitrarily close to but less than β. Take a suitable small X-twist that increases the length dpO A , O B q; it will also increase the length dpO B , O R q " dpO B , ZpO R qq " dpO B , BpO R qq.
Then the angle β decreases slightly and we twist until Nθ " β mod 2π.
(We note that this argument works for any θ P p0, 2πqzQπ.) In the end, we obtain the relation which is invariant under Z-twists. The proof of the claim is complete.
If we define then u N is a product of two involutions and the above relation is equivalent to ρ ξ pu N q " 1. Thus for a dense choice of s in p2, 8q there exists an integer N " Npsq ą 0 such that the holonomy map of every point ξ on the coordinate curve tz " su satisfies u N P ker ρ ξ .
We can write u N as a word in F 2 from the computation below.
"XYrX, Ys N X´1Y´1Y´1X´1rY´1, X´1s N´1 Y´1X´1Y XY X "XYrX, Ys N X´1Y´2X´1rY´1, X´1s N Y X. As can be expected from Lemma 4.1, u N is indeed palindromic (but not a proper power). To finish the proof it suffices to have the following proposition, which is proved in the next section.
Proposition 5.7. For each N ě 1, we have that u N is of non-torsion-type.
Remark 5.8. In the above proof, by Proposition 4.3, the image ρ ξ pu N q is trivial for ξ P C θ X tz " su, while it is a nontrivial hyperbolic translation for ξ P C θ ztz " su.
Recall we denote by g w px, y, zq P Zrx, y, zs the trace polynomial of a word w P F 2 . From the above proof and the remark following it, we see for each s P S θ there exists a positive integer N " Npθ, sq such that the coordinate curve C θ X tz " su can be expressed as the intersection of two surfaces C θ X tz " su " C θ X tg u N "˘2u.
In general, two words U and V may admit the same trace polynomial. For instance, we have g U " g V if U and V are conjugate, or if U " s 1 . . . s and V " s s ´1¨¨¨s1 for s i P tX˘1, Y˘1u. In our case, the word u N is palindromic and so, it seems rare that another surface tg w "˘2u contains the above coordinate curve for a different word w P F 2 not conjugate to u N . This raises the following question: Question 5.9. Let θ P p0, 2πqzQπ. Does there exist a dense subset R Ă S θ such that for each r P R and for N " Npθ, rq as above, a very general point ξ in the coordinate curve C θ X tz " ru corresponds to the monodromy ρ ξ whose image is isomorphic to the one-relator group xX, Y | u N " 1y? 6. Proof of Proposition 5.7 In this section, we prove Proposition 5.7. The proof we offer is somewhat technical by nature, but based on a beautiful solution by Newman of the word problem in one-relator groups with torsion (Corollary 6.2) as we describe now.
Magnus (1932) discovered that the word problem for a (general) one-relator group is solvable. In the case of a one-relator with torsion, the following theorem due 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 rws denote its cyclic conjugacy class. Theorem 6.1 (Newman Spelling Theorem). Let G " xx 1 , . . . , x n | r k y be a onerelator group for some cyclically reduced word r in F n :" Fpx 1 , . . . , x n q, where k ě 2. Suppose that a nontrivial reduced word w P F n belongs to the kernel of the natural quotient map F n Ñ G. Then w contains a subword u, of length strictly larger than pk´1q|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-relator group with torsion, which we call the Dehn-Newman algorithm, as it is based on the Dehn's solution to the word problem for surface groups. Recall our convention that F 2 " xX, Yy. Corollary 6.2 (Dehn-Newmann Algorithm). Let u and r be cyclically reduced words in F 2 , and let m ě 2. Then the truth value of the statement u P xxr m yy can be decided in the following steps, within a finite time.
Step 1 If there exists a cyclic conjugation R " R 1 R 2 of r˘1 such that |R 1 | " 1 and such that R m´1 R 1 is a subword of u, then proceed to Step 2; otherwise, conclude that u R xxr m yy.
Step 2 Let U be the word obtained from u by substituting R´1 2 for R m´1 R 1 . If U ‰ 1, then repeat Step 1; otherwise, conclude that u P xxr m yy.
Let us first see a simple application of the algorithm. Lemma 6.3. Let N ě 1.
(2) If r P F 2 and m ě 3, then u N R xxr m yy.
Proof. (1) We let G :" F 2 {xxr m yy. Assume for contradiction that u N " 1 in G. Suppose first that r " Y. By the Dehn-Newman Algorithm (Corollary 6.2), we should have Y˘m ď u N . This implies m " 2. In G " F 2 {xxY 2 yy, the element u N coincides with the word u 1 :" XYrX, Ys N X´2rY´1, X´1s N Y X.
Since Y˘2 ę u 1 , we see that u N " u 1 ‰ 1 in G.
Now, consider the case r " XY. Since a cyclic conjugation of a length five subword of pXYq 3 does not appear in u N as a subword, we see that m " 2. We cancel out all the occurrences of pXYq˘2, pY Xq˘2 in u N to obtain u 1 :" X´1Y´1rX, Ys N´2 XYY XrY´1, X´1s N´2 Y´1X´1.
Since u 1 does not contain a length three subword of pXYq˘2, pY Xq˘2 we conclude from the Dehn-Newman algorithm that u 1 ‰ 1 in G.
Finally, assume r " rX, Ys. Writing N " mq`t for t P r0, m´1s, the element u N can be rewritten in G as It is obvious that u 1 does not contain a subword of r m with length 4pm´1q`1. Hence, we see u 1 ‰ 1 in G.
(2) Suppose not. The Dehn-Newman Algorithm implies that for some cyclically reduced word r that is not a proper power we have r 2 ď u. This consideration restricts the choices of r to be a cyclic conjugation of a word in tY, XY, rX, Ysu˘1.
By part (1), this is impossible.
We are now ready to give a proof of the proposition.
Proof of Proposition 5.7. Suppose u N P xxr m yy for some r P F 2 and m ě 2. By Lemma 6.3, we need only consider the case m " 2. Consider the following procedure, which takes u " u N as an input.
(i) Pick a cyclically reduced word r " r 1 r 2 in F 2 such that |r 1 | " 1 and such that r 1 r 2 r 1 ď u; here, we require that r 1 r 2 is not cyclically conjugate to a word in tY, XY, rX, Ysu˘1.

Proceed to
Step (ii). (ii) Replace an occurrence of r 1 r 2 r 1 in u by r´1 2 to obtain a reduced word u 1 . Proceed to Step (iii). (iii) Pick a cyclic conjugation r 1 1 r 1 2 of r˘1 such that |r 1 1 | " 1 and such that r 1 1 r 1 2 r 1 1 ď u 1 . Plug u Ð u 1 and r i Ð r 1 i . Repeat Step (ii). This process is a simplified version of Dehn-Newman Algorithm. It is obvious that if the procedure terminates at step (i) or (iii), we can conclude that u N is of non-torsion-type.
We can enumerate all the possible choices of r 1 r 2 r 1 in step (i) as follows, excluding the obvious case r 1 r 2 r 1 " u N . Verifying all of the following cases, we can conclude that u N R xxr m yy for all r P F 2 and m ě 2.
Plug u Ð u 1 and r i Ð r 1 i . We replace r 1 r 2 r 1 in u by r´1 2 to obtain u 1 :" rY, Xs t X´1Y´1rX, Ys N´t´1 X´1Y´1X´1Y´1rX, Ys t´1 XYrY, Xs N´t´1 pY Xq 2 It is then clear that Step (iii) cannot be executed for u 1 and r. We conclude that u 1 " u ‰ 1 in F 2 {xxr 2 yy.
Then r´1 2 " Y´1XYrX´1, Y´1s t´1 XY 2 XrY, Xs N Y´1. We replace r 1 r 2 r 1 in u N by r´1 2 to obtain which is reduced. It is clear that no cyclic conjugate of pr 1 r 2 q˘1 occurs as a subword of u 1 , as is seen from aligning rY, Xs N in u 1 with the corresponding copy in r´1 2 .
This word is reduced unless N " s or N " t, in which case a subword X 2 appears. It is clear that no cyclic conjugate of pr 1 r 2 q˘1 occurs as a subword of u 1 . Case (4). In this case we have r 1 " X, r 2 " Y X´1Y´1rX, Ys s´1 X´1Y´2X´1rY´1, X´1s N Y for 1 ď s ď N.
Then r´1 2 " Y´1rX´1, Y´1s N XY 2 XrY, Xs s´1 Y XY´1. Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N´s Y´1rX´1, Y´1s N XY 2 XrY, Xs s´1 Y XY´1, which is reduced. There is exactly one occurrence of Y 2 in u 1 and in pr 1 r 2 q´1, and if these are aligned then a further reduction can be made. Clearly such an alignment cannot be made, as is checked by reading to the left or right from Y 2 in u 1 . Thus, we must be able to find a subword of u 1 which is a cyclic conjugate of pr 1 r 2 q˘1 and which begins and ends with the same power of Y, that is Y´1X´1rY´1, X´1s N Y XY X´1Y´1rX, Ys s´1 X´1Y´1 or its inverse. It is clear that this does not happen.
This latter word is both reduced and cyclically reduced. In order to perform further reductions, u 1 must contain subwords of the form Y XrY´1, X´1s t or X´1Y´1rX´1, Y´1s t .
For the first of these cases to occur we would need t " 1, and then it is clear that no cyclic permutation of u 1 contains a cyclic permutation of this word. In the second case, we similarly see that no cyclic conjugate of this word occurs as a subword of a cyclic permutation of u 1 .
Then r´1 2 " rY, Xs t Y. Replacing r 1 r 2 r 1 by r´1 2 gives us which is reduced. In order to perform further reductions, we must align rX, Ys t X´1Y´1 or Y XrY, Xs t in u 1 and pr 1 r 2 q˘. By inspection, this is not possible. Case (7). In this case we have Then r´1 2 " Y 2 XrY, Xs t Y. Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N´t´1 XY 3 XrY, Xs t YrY´1, X´1s N Y X, which reduces to XYrX, Ys N´t´1 XY 3 XrY, Xs t X´1Y XrY´1, X´1s N´1 Y X.
There is only one power of Y of absolute value more than one in u 1 , and it is clear that Y 2 XrY, Xs t Y X " pr 1 r 2 q´1 does not occur as a subword of u 1 or of its cyclic conjugates, nor do any of the cyclic conjugates of pr 1 r 2 q´1 which keep the Y 2 intact. We therefore consider subwords of the form Y XrY, Xs t Y XY and its inverse. It is immediate to verify that there are no such subwords, even in the degenerate case where t " 0.
Then r´1 2 " YrX´1, Y´1s t XY 2 XrY, Xs s Y. Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N´t´1 XY 2 rX´1, Y´1s t XY 2 XrY, Xs s Y 2 XrY´1, X´1s N´t´1 Y X, which is reduced and cyclically reduced. Up to cyclic permutation, we must align subwords of pr 1 r 2 q˘1 and u 1 of the form X´1Y´2X´1 or XY 2 X, or we must set r 1 1 " Y and r 1 1 r 1 2 r 1 1 " Y XrY, Xs s Y XYrX´1, Y´1s t XY 2 . Reading to the left or to the right in u 1 from this copy of Y 2 quickly yields a contradiction.
Case (9). In this case we have Then r´1 2 " Y 2 . Replacing r 1 r 2 r 1 by r´1 2 gives us where the latter expression is both reduced and cyclically reduced. In order to reduce further, the word u 1 must have a subword of the form Y 2 or Y´2. Indeed, such a subword would arise from either pr 1 r 2 q˘1 or pr 2 r 1 q˘1 as a subword of u 1 , or by setting r 1 1 " Y˘1 and r 1 1 r 1 2 r 1 1 P tY XY 2 , Y´1X´1Y´2u. Clearly none of these cases occur.
Case (10). In this case we have Then r´1 2 " YrX´1, Y´1s t XY 2 . Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N YrX´1, Y´1s t XY 2 Y XrY´1, X´1s N´t´1 Y X, which reduces to XYrX, Ys N´1 XY X´1rX´1, Y´1s t XY 3 XrY´1, X´1s N´t´1 Y X, which is both reduced and cyclically reduced. Up to cyclic permutation, we must have that u 1 has a subword of the form X´1Y´2X´1 or XY 2 X, or a cyclic permutation of a word of the form r 1 1 r 1 2 r 1 1 " Y XYrX´1, Y´1s t XY 2 or r 1 1 r 1 2 r 1 1 " Y´1X´1rY´1, X´1sY´1X´1Y´2 in order to perform further reductions. Clearly this is not the case.
This last expression is both reduced and cyclically reduced. In order to perform further reductions, a cyclic permutation of u 1 must contain a subword which is cyclic permutation of a word of the form XYrX´1, Y´1s t or rY´1, X´1s t Y´1X´1. The first of these does not, since t ą 0. Thus we are reduced to the case where r 1 1 " Y´1, r 1 2 " X´1Y XrY´1, X´1s t´1 Y´1X´1. Then, r 1 1 r 1 2 r 1 1 " rY´1, X´1s t Y´1X´1Y´1, which does not occur as a subword of u 1 up to cyclic permutation.
Then r´1 2 " XYrX´1, Y´1s t XY 2 XrY, Xs N . Replacing r 1 r 2 r 1 by r´1 2 gives us which reduces to This last expression is reduced and cyclically reduced. In order to perform further reductions, we must have that after performing a cyclic permutation, there is a subword of u 1 of the form YrX, Ys N X´1Y´2X´1rY´1, X´1s t Y´1X´1 " r 1 r 2 or its inverse. By considering the necessary alignment of product of commutators rY, Xs N , we quickly obtain a contradiction. Case (14). In this case we have r 1 " Y, r 2 " rX, Ys N X´1Y´2X´1rY´1, X´1s N .
Then r´1 2 " rX´1, Y´1s N XY 2 XrY, Xs N . Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XrX´1, Y´1s N XY 2 XrY, Xs N X, which reduces to Y´1XYrX´1, Y´1s N´1 XY 2 XrY, Xs N X, which is both reduced and cyclically reduced. Note that the words pr 1 r 2 q˘have length 4N`5 and that u 1 has length 4N`4, whence no cyclic permutation of pr 1 r 2 q˘occurs as a subword of u 1 . Case (15). In this case we have r 1 " Y, r 2 " X´1Y´1rX, Ys s X´1Y´2X´1rY´1, X´1s t Y´1X´1 for 0 ď s, t ď N´1.
Then r´1 2 " XYrX´1, Y´1s t XY 2 XrY, Xs s Y X. Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N´s´1 X 2 YrX´1, Y´1s t XY 2 XrY, Xs s Y X 2 rY´1, X´1s N´t´1 Y X, which is both reduced and cyclically reduced. In order to perform further reductions, we would require a cyclic permutation of Y´1XYrX´1, Y´1s t XY 2 XrY, Xs s Y X or its inverse to occur as a subword of u 1 , which it does not. Case (16). In this case we have r 1 " Y, r 2 " X´1Y´1rX, Ys s X´1Y´2X´1rY´1, X´1s N for 0 ď s ď N´1.
Then r´1 2 " rX´1, Y´1s N XY 2 XrY, Xs s Y X. Replacing r 1 r 2 r 1 by r´1 2 gives us This last expression is reduced and cyclically reduced. In order to perform further reductions, we must have a subword of u 1 which is a cyclic permutation of Y´1rX´1, Y´1s N XY 2 XrY, Xs s Y X or its inverse. This is not the case.
Then r´1 2 " rX´1, Y´1s t X´1. Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N X´1Y´2X´1rY´1, X´1s N´t´1 Y´1X´1rX´1, Y´1s t after one free reduction, and this word is both reduced and cyclically reduced. To perform further reductions, we would need u 1 to have a subword which is a cyclic conjugate of rX´1, Y´1s t X´1Y´1 or its inverse Y XrY´1, X´1s t , the first of which is not the case.
The second of these does occur as a subword of u 1 if N´t´1 ą t. In this case we set r 1 1 " r 1 and r 1 2 " r 2 and observe that Y XrY´1, X´1s t Y " r 1 1 r 1 2 r 1 1 is not a subword of u 1 . Case (19). In this case we have r 1 " Y´1, r 2 " rX, Ys t X´1Y´1 for 0 ď t ď N´1.
Then r´1 2 " Y XrY, Xs t . Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N´t´1 XY X´1Y XrY, Xs t X´1rY´1, X´1s N Y X, which is both reduced and cyclically reduced. For further reductions to be possible, we would require a subword of u 1 which is a cyclic conjugate of Y 2 XrY, Xs t or its inverse, which up to cyclic permutation is Y´1rX, Ys t X´1Y´1.
Clearly this is not the case if t ě 1.
If t " 0 then which reduces to XYrX, Ys N´1 XY X´2Y XrY´1, X´1s N´1 Y X, which is both reduced and cyclically reduced. Now, r 1 r 2 " Y´1X´1Y´1, which does not appear as a subword of u 1 up to cyclic permutation, but its inverse Y XY does. In order to perform further reductions, we would have to either set r 1 1 " Y and r 1 1 r 1 2 r 1 1 " Y XY 2 or r 1 1 " X and r 1 1 r 1 2 r 1 1 " XY 2 X. Neither of these words appear, even after cyclic permutation.
Then r´1 2 " rX´1, Y´1s t XY 2 XrY, Xs s . Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N´s´1 XY X´1rX´1, Y´1s t XY 2 XrY, Xs s X´1Y XrY´1, X´1s N´t´1 Y X, which is both reduced and cyclically reduced. To perform further reductions, we must have a subword of u 1 which is a cyclic permutation of YrX´1, Y´1s t XY 2 XrY, Xs s or its inverse, which is not the case. Case (21). In this case we have r 1 " Y´1, r 2 " Y´1X´1rY´1, X´1s t for 0 ď t ď N´1.
Then r´1 2 " rX´1, Y´1s t XY. Replacing r 1 r 2 r 1 by r´1 2 gives us u 1 " XYrX, Ys N X´1rX´1, Y´1s t XY X´1Y XrY´1, X´1s N´t´1 Y X, which is both reduced and cyclically reduced. Supposing the possibility of further reductions, we would need a subword which is a cyclic permutation of rX´1, Y´1s t XY 2 or its inverse, which up to cyclic permutation is given by Y´1X´1rY´1, X´1s t Y´1.
We easily see that this is not the case.
Appendix A. A computational approach, by Xinghua Gao 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 and implement computational heuristics that can find such examples of ρ ξ , which turn out to have closed hyperbolic 3-manifold groups as images.
A.1. An arithmetic formulation of the problem. The starting point is a hyperbolic 3-manifold T with a single cusp, whose fundamental group G is generated by two elements a and b. As an explicit example, one can take a hyperbolic 2-bridge knot complement. Then, consider a hyperbolic Dehn filling T q of T for some q P Q, and a number field Qpαq such that G " xa, by " π 1 pT q q ď PSLp2, Qpαqq.
We abuse language slightly and also write G " xa, by for a lift of G to SLp2, Qpαqq (cf. [16,3]).
Question A. 1. Under what conditions on T and q do the following conclusions hold?
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 structure of T q and the algebraic number α P C, the verification step should be computationally straightforward up to the computation of Galois conjugates. Once we have such an example of T and q, then the resulting point ξ :" ptr a σ , tr b σ , trpabq σ q is a point in the character variety of C θ for θ satisfyinǵ 2 cospθ{2q " trra, bs σ .
Moreover, the image of the monodromy ρ ξ is isomorphic to π 1 pT q q as we desire.
Remark A.2. Note that a parabolic generator of π 1 pT q will remain parabolic after Galois conjugation.
A.2. An algorithm for producing explicit examples. Examples of pT, qq with the desired properties can be produced using SnapPy [4,6]. In order to construct a representation with the image in PSLp2, Rq, we will need the following fact.
Lemma A.3. [17, Corollary 3.2.5] If Γ is a nonelementary subgroup of SLp2, Cq such that Qptr Γq is a subset of R, then Γ is conjugate to a subgroup of SLp2, Rq.
Let T r be a closed hyperbolic 3-manifold obtained by applying Dehn filling to one-cusped hyperbolic 3-manifold T along a curve of slope r. We fix a triangulation of T by ideal tetrahedra, which also gives an ideal triangulation of T r . Then the edge gluing equations of T together with the Dehn filling equation determine the hyperbolic structure of T r . The solution to the this system of equations is a set of complex numbers, which parameterize the shape of tetrahedra in T r . These parameters generate the tetrahedral field K of T r . We want to find a real embedding σ of the 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 1 pT r q Ñ PSLp2, Cq with all matrices in the image having real trace. Therefore by Lemma A.3, the group ρ R pπ 1 pT r qq is conjugate into PSLp2, Rq. Since a conjugacy leaves the trace unchanged, we obtain a desired hyperbolic 3-manifold T r giving an affirmative answer to Question A.1.
The process of finding a suitable T and corresponding tetrahedral field can be formulated with the following steps: Step 1) Let T be a hyperbolic knot complement with the default triangulation in SnapPy and apply Dehn filling of slope r. Compute the tetrahedral field K of T r using the SnapPy manifold class tetrahedra field gens(). We can then use the SageMath number field class find field() to find the defining polynomial of K.
Step 2) Find a real embedding σ of K, if there exits one, using the SageMath number field class real embeddings().
Step 3) Apply the real embedding σ and set up the new triangulation with real shape parameters, 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 ra, bs under ρ R , using the SnapPy fundamental group classe SL2C( In this table, a and b are the generators of the corresponding fundamental group, with the default triangulation in SnapPy. Presentations of the fundamental group of 7 6 p0q and 8 13 p0q are included below, as well as the matrix representatives of a and b. Presentations of the fundamental groups of the other three manifolds are unwieldy due to their size, so we have omitted them. The interested readers may use the manifold class fundamental group() to verify our claims.
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 all algebraic numbers in a real embedding of the tetrahedra field K of 7 6 p0q, with the defining polynomial ppxq " x 12´2 x 11´2 x 10`1 4x 9´2 5x 8`3 2x 7 35x 6`3 8x 5´3 8x 4`3 0x 3´1 7x 2`6 x´1.
For now, we are only able to produce closed hyperbolic 3-manifolds via 0-Dehn filling to two-bridge knot complements as examples.