Hidden Symmetries and Commensurability of 2Bridge Link Complements
Abstract.
In this paper, we show that any nonarithmetic hyperbolic bridge link complement admits no hidden symmetries. As a corollary, we conclude that a hyperbolic bridge link complement cannot irregularly cover a hyperbolic manifold. By combining this corollary with the work of Boileau and Weidmann, we obtain a characterization of manifolds with nontrivial JSJdecomposition and rank two fundamental groups. We also show that the only commensurable hyperbolic bridge link complements are the figureeight knot complement and the link complement. Our work requires a careful analysis of the tilings of that come from lifting the canonical triangulations of the cusps of hyperbolic bridge link complements.
1. Introduction
Two manifolds are called commensurable if they share a common finite sheeted cover. Here, we focus on hyperbolic manifolds, that is, where is a discrete, torsionfree subgroup of . We are interested in analyzing the set of all manifolds commensurable with . Commensurability is a property of interest because it provides a method for organizing manifolds and many topological properties are preserved within a commensurability class. For instance, Schwartz [Sch] showed that two cusped hyperbolic manifolds are commensurable if and only if their fundamental groups are quasiisometric. In this paper, we restrict our attention to hyperbolic bridge link complements; see Section 2 for the definition of a bridge link. We use the word link to refer to a link in with at least one component. We use the word knot to only mean a single component link.
A significant challenge in understanding the commensurability class of a hyperbolic manifold is determining whether or not has any hidden symmetries. To understand hidden symmetries, we first need to introduce some terminology. The commensurator of is
.
It is a well known fact that two hyperbolic manifolds are commensurable if and only if their corresponding commensurators are conjugate in ; see [Wa, Lemma 2.3]. We denote by the restriction of to orientationpreserving isometries. We also denote by the normalizer of in and the restriction of to orientationpreserving isometries. Note that, . A symmetry of corresponds to an element of , and a hidden symmetry of corresponds to an element of that is not in . Geometrically, admits a hidden symmetry if there exists a symmetry of a finite cover of that is a not a lift of an isometry of . See Sections 2 and 3 of [Wa] for more details on commensurators and hidden symmetries.
In this paper, we give a classification of the hidden symmetries of hyperbolic bridge link complements. In [ReWa] Reid–Walsh used algebraic methods to determine that hyperbolic bridge knot complements (other than the figureeight knot complement) have no hidden symmetries. However, their techniques do not apply to hyperbolic bridge links with two components. Here, we use a geometric and combinatorial approach to prove the following theorem.
Theorem 1.1.
If is a nonarithmetic hyperbolic bridge link complement, then admits no hidden symmetries (both orientationpreserving and orientationreversing).
The only arithmetic hyperbolic bridge links are the figureeight knot, the Whitehead link, the link, and the link. Though it will not be needed in what follows, we refer the interested reader to [MaRe, Definition 8.2.1] for the definition of an arithmetic group .
We prove Theorem 1.1 by using the canonical triangulation of a hyperbolic bridge link complement, . This triangulation was first described by Sakuma–Weeks in [SaWe]. Guéritaud in his thesis [Gu] proved that this triangulation is geometrically canonical, i.e., topologically dual to the Ford–Voronoi domain for equal volume cusp neighborhoods. In addition, Akiyoshi–Sakuma–Wada–Yamashita in [AkSaWaYa] have announced a proof of this result where they analyze the triangulation via cone deformations of along the unknotting tunnel. Futer also showed that this triangulation is geometric by applying Rivin’s volume maximization principle; see the appendix of [GuFu]. By [GoHeHo, Theorem 2.6], if any such is nonarithmetic, then can be identified with the group of symmetries of the tiling of obtained by lifting , which we call . We prove that any nonarithmetic hyperbolic bridge link complement does not admit hidden symmetries by showing that any symmetry of actually corresponds to a composition of symmetries of and deck transformations of . In other words, .
Rather than analyze this tiling of , we drop down a dimension and instead analyze the (canonical) cusp triangulation of , induced by . By intersecting a cusp cross section of with its canonical triangulation , we obtain a canonical triangulation of the cusp(s). If has two components, we still end up with the same canonical triangulation on both components of since there is always a symmetry exchanging the two components, and we take equal volume cusp neighborhoods. We can lift to a triangulation of (or two copies of if has two components). We also place edge labels on which record edge valences of corresponding edges in the threedimensional triangulation. This labeling provides us with enough rigid structure in to rule out any hidden symmetries. Goodman–Heard–Hodgson use a similar approach to prove that nonarithmetic hyperbolic puncturedtorus bundles do not admit hidden symmetries [GoHeHo, Theorem 3.1].
If a hyperbolic manifold admits no hidden symmetries, then can not irregularly cover any hyperbolic orbifolds. A hyperbolic orbifold is any , where is a discrete subgroup of , possibly with torsion. All of the previous statements about commensurability of hyperbolic manifolds and the commensurator of also hold for hyperbolic orbifolds. Theorem 1.1 quickly gives us the following corollary about coverings of hyperbolic orbifolds by hyperbolic bridge link complements. For the arithmetic cases, volume bounds are taken into consideration to rule out irregular covers of manifolds.
Corollary 1.2.
Let be any hyperbolic bridge link complement. If is nonarithmetic, then does not irregularly cover any hyperbolic orbifolds (orientable or nonorientable). If is arithmetic, then does not irregulary cover any (orientable) hyperbolic manifolds.
By combining Corollary 1.2 with the work of Boileau–Weidmann in [BoWe], we get the following characterization of manifolds with nontrivial JSJdecomposition and rank two fundamental groups. For a more detailed description of this decomposition see Section 5.3.
Corollary 1.3.
Let be a compact, orientable, irreducible manifold with rank(. If has a nontrivial JSJdecomposition, then one of the following holds:
1. has Heegaard genus .
2. decomposes into a Seifert fibered manifold and hyperbolic manifold.
3. decomposes into two Seifert fibered manifolds.
The original characterization given by Bolieau–Weidmann included a fourth a possiblity: a hyperbolic piece of is irregularly covered by a bridge link complement. Corollary 1.2 eliminates this possibility.
Ruling out hidden symmetries also plays an important role in analyzing the commensurability class of a hyperbolic orbifold . By the commensurability class of a hyperbolic orbifold (or manifold) , we mean the set of all hyperbolic orbifolds commensurable with . A fundamental result of Margulis [Mar] implies that is discrete in (and is finite index in ) if and only if is nonarithmetic. Thus, in the arithmetic case, will have infinitely many hidden symmetries. In the nonarithmetic case, this result implies that the hyperbolic orbifold is the unique minimal (orientable) orbifold in the commensurability class of . So, in the nonarithmetic case, and are commensurable if and only if they cover a common minimal orbifold. Furthermore, when admits no hidden symmetries, , and so, is just the quotient of by its orientationpreserving symmetries.
By using Theorem 1.1 and thinking about commensurability in terms of covering a common minimal orbifold, we obtain the following result about commensurability classes of hyperbolic bridge link complements.
Theorem 1.4.
The only pair of commensurable hyperbolic bridge link complements are the figureeight knot complement and the link complement.
We prove Theorem 1.4 by analyzing the cusp of each minimal (orientable) orbifold, , in the commensurability class of a nonarithmetic hyperbolic bridge link complement. This orbifold always has one cusp since two component bridge links always have a symmetry exchanging the components. The cusp of this orbifold inherits a canonical cellulation from the canonical triangulation of the cusp(s) of . By comparing minimal orbifold cusp cellulations, we establish this result.
We now describe the organization of this paper. In Section 2, we provide some background on bridge links, including an algorithm for building any bridge link from a word in Ls and Rs. Section 3 describes how to build the canonical triangulation of a bridge link complement and the corresponding cusp triangulation based on this word . In this section we also prove some essential combinatorial properties of , the lift of to . Section 4 analyzes the possible symmetries of a bridge link complement in terms of the word , and describes the actions of these symmetries on . In Section 5, we prove Theorem 1.1, Corollary 1.2, and Corollary 1.3. In Section 6, we prove Theorem 1.4.
The authors would like to thank David Futer for helpful conversations and guidance on this work. We would also like to thank the referee for making a number of helpful suggestions.
2. Background on bridge links
In order to describe bridge links, we first need to define rational tangles. First, a tangle is a pair (, ), where is a pair of unoriented arcs embedded in the ball so that only intersects the boundary of in four specified marked points: SW, SE, NW, and NE (if we think of as the unit sphere centered at the origin in , then SW is the southwest corner , SE is the southeast corner , etc). Rational tangles are a special class of tangles. The simplest rational tangles are the tangle and the tangle. The tangle consists of two arcs that don’t twist about one another, with one arc connecting NW to NE, and the other arc connecting SW to SE. Similarly, the tangle consists of two unknotted arcs, with one arc connecting NE to SE and the other arc connecting NW to SW. Both of these tangles admit an obvious meridian curve contained on that bounds an embedded disk in the interior of . A rational tangle is constructed by taking one of these trivial tangles and alternating between twisting about the western endpoints (NW and SW) and twisting about the southern endpoints (SW and SE). This twisting process maps the meridian of the tangle (tangle) to a closed curve with rational slope , which determines this tangle, hence the name rational tangle. A bridge link is constructed by taking a rational tangle, connecting its western endpoints by an unknotted strand, and connecting its eastern endpoints by an unknotted strand.
Here, we describe a bridge link in terms of a word , which is a sequence of Ls and Rs: , (if is odd and the starting letter is ). The sequence gives the continued fraction expansion for the rational tangle used to construct a bridge link. Each corresponds to performing a lefthanded halftwist about the NW and SW endpoints of a tangle and each corresponds to performing a righthanded halftwist about the SW and SE endpoints of an tangle. Each syllable, i.e., each maximal subword or , corresponds to two strands wrapping around each other times. This word gives a procedure to construct an alternating string braid between two punctured spheres, and , where is exterior to the braid and is interior to the braid; see Figure 1. To construct a bridge link, we add a single crossing to the outside of , and we add a single crossing to the inside of . There is a unique way to add these crossings so that the resulting link diagram is alternating. Any bridge link can be constructed in this manner and we use the notation to designate the bridge link constructed by the word . The original source for this notation comes from the appendix of [GuFu], which contains more details of this construction.
The following are important facts about bridge links that we will use. From now on, we will state results in terms of and we assume that any bridge link has been constructed in the manner described above, unless otherwise noted.

Given a bridge link , we obtain a mirror image of the same link (with orientations changed on ) if we switch Ls and Rs in the word . Since we will only be considering unoriented link complements, we consider such links equivalent.

bridge links (and their complements) are determined by the sequence of integers up to inversion. Schubert gives this classification of bridge knots and links in [Schu], and Sakuma–Weeks [SaWe, Theorem II.3.1] give this classification of their complements by examining their (now known) canonical triangulations.

A bridge link is hyperbolic if and only if has at least two syllables. This follows from Menasco’s classification of alternating link complements [Men].

The only arithmetic hyperbolic bridge links are those listed below. This classification was given by Gehring–Maclachlan–Martin in [GeMaMa].

The figureeight knot given by or ,

The Whitehead link given by or ,

The link given by or , and

The link given by or .
We care about distinguishing between nonarithmetic and arithmetic hyperbolic link complements because different techniques have to be used for analyzing hidden symmetries and commensurability classes.

Throughout this paper, we will always assume that is hyperbolic, i.e., has at least two syllables. In Section 3, we will use the diagram of described above to build the canonical cusp triangulation of .
3. Cusp triangulations of bridge link complements
Let be a bridge link, defined as in Section 2, with a word in and , and its letter. We may assume that , as mentioned in Section 2. In this section we give a description of the construction of the triangulation of , and of the induced cusp triangulation , and its lift (if has two components, then the two cusp triangulations are identical). We then describe an algorithmic approach for constructing , and prove some facts about simplicial homeomorphisms . Our description of these triangulations follows that of [GuFu, Appendix A] and [SaWe, Chapter II], to which we refer the reader for further details.
To build the triangulation , we first place a punctured sphere at each crossing corresponding to a letter of , so that every crossing for is on one side of , and the remaining crossings are on the other side; see Figure 0(b). We will start by focusing on and . We triangulate both of them as shown in the first frame of Figure 1(a) (notice that the edge from the lowerleft to upperright puncture is in front for both). If we push along the link to the other side of the crossing , we see that some of its edges coincide with edges of (in particular, the horizontal edges coincide, and the diagonal edges of become vertical in , see Figure 1(a)). The vertical edges of , however, get pushed to diagonal edges that cannot be identified to the diagonal edges of . The top frame of Figure 1(b) shows and with appropriate edges identified, as seen lifted to (i.e., cut along top, bottom, and left edges then unfold). If we lift to in such a way that its triangulation has edge slopes , this choice forces to have edge slopes , as shown in the lower frame of Figure 1(b). This means that the triangulation of in is obtained by applying the matrix to the triangulation of . If the letter between and had been an , we would have found by the same analysis that the matrix taking us from the triangulation of to the triangulation of must be . This holds in general. If we know the edge slopes of the triangulation of , we can apply the appropriate matrix, depending on whether is an or an , to get the triangulation of (see Figure 3).
Remark 3.1.
Though we do not use this fact in what follows, the word can be viewed as a path in the Farey tesselation, with each letter corresponding to making a right (for ) or left (for ) turn from one Farey triangle to the next. In this case each four punctured sphere corresponds to a Farey triangle, and its slopes are given by the vertices of that triangle. For details of this approach, we again direct the interested reader to [GuFu] and [SaWe].
Coming back to and , we see in Figure 1(b) that between the (red) triangulation of and the (blue) triangulation of is a layer of two tetrahedra, which we denote . Similarly, between the punctured spheres and we get a layer of tetrahedra. This construction results in a “product region” , where and . We use quotation marks here because for , is not a true product since there will be an edge shared by all the .
To obtain from , we first “clasp” by folding along edges with slope and identifying pairs of triangles adjacent to those edges, as shown in Figure 4. We clasp in the same way, this time folding along either the edge with greatest slope or the edge with least slope, depending on whether the final letter of is or , respectively.
To understand the induced triangulation of a cusp cross section, we first consider a neighborhood of a single puncture in . For each layer of tetrahedra between and , we get a pair of triangles and going once around the puncture, as in Figure 4(a). In this figure vertices of are labelled according to the edges of that they are contained in, and edges of are labelled according to the edge of that they are across a face from. Notice in Figure 4(a) that has a vertex () meeting an edge of but not meeting , and has a vertex () meeting but not meeting . Thus is distinguished from .
To see how attaches to , we must consider how attaches to . Figure 5(a) shows and in (sandwiched between ) in the case where , and the corresponding triangles around the puncture. There is a unique edge of , corresponding to an edge of shared by both and , and with vertices and . This means that the edge moves us along the cusp cross section in the longitudinal direction, so it will be part of a longitude in . It makes sense then to adjust these edges to be horizontal, as we build the triangulation (see Figure 5(a)). Figure 5(b) shows the analogous adjustment when .
When we clasp , an edge of is identified to an edge of , and similarly for and when is clasped, as illustrated in Figure 7. We will call the triangles and clasping triangles. For , the triangulation around a puncture before clasping and after clasping is shown in Figures 6(b) and 6(c), respectively.
Before clasping, it is clear from the construction that the combinatorics around each of the four punctures is identical. Clasping identifies the punctures on in pairs, and identifies the punctures on in pairs, in an orientation preserving way. This means that for a component link, a cusp triangulation is obtained by gluing two puncture triangulations (as in Figure 6(c)) along their front edges, and along their back edges, in an orientation preserving way. For a knot, the situation is similar, except that we glue all four puncture triangulations, always identifying front edges to front edges, and back to back, with orientation preserved. In both cases the lifted triangulation of is the same, except that the fundamental region for a knot is twice as large as for a link. Note that when , the clasping triangle on the right is offset vertically from the clasping triangle on the left (as in Figure 6(c)), whereas if this will not be the case.
As a result of the above discussion, we can now give an algorithmic approach to constructing the lifted cusp triangulation for an arbitrary word (we will assume the last letter is for concreteness; the case where is similar). This follows the approach of SakumaWeeks in [SaWe, II.4], with some changes of notation. We start with a rectangle divided into triangles, each corresponding to a letter of , as in Figure 8. Vertices of are labelled as shown, with for , and . To fill out we first reflect in its top edge to get its mirror , so that is a triangulation of a puncture (with triangles in and triangles in ), as in 6(c). We then rotate by about (i.e., about the vertex labelled 1), and translate the resulting double of vertically and horizontally to fill . Finally, we remove all edges and , where if , and otherwise (i.e. all images of the red edges in Figure 8).
With this parametrization of the cusp triangulation in , deck transformations are generated by and , where if has two components, and if it has one component. We observe that the long edge of each clasping triangle goes all the way around the meridian of the cusp, and these edges are unique in this respect. For this reason we call these edges meridional edges (whether we are referring to them in or ), and we call each connected component of their union in a meridional line (i.e., any line , ). A strip of adjacent nonclasping triangles that all meet the lines and (in an edge or vertex), for some , is called a horizontal strip (see Figure 12).
We will now describe a correspondence between edges and vertices of . Given an edge in , meaning a truncated tip of an ideal triangle in , we have a corresponding edge in : this is just the edge of across from in the ideal triangle, as in Figure 4(a). Similarly, a vertex of corresponds to the edge in that it is contained in. We say that an edge and a vertex of correspond if their corresponding edges in have the same slope (when viewed in ). Edge and vertex correspondence in , for edges and vertices that do not come from or , can be read off of Figure 4(b), which shows the cusp cross section of a layer with vertices and edges of the same slope labelled the same. As for edges and vertices affected by clasping, we can easily read the correspondences off of the labellings in Figure 7 for the clasping of , and the clasping works similarly. This gives edge/vertex correspondences for , as shown in Figure 9 (as usual, we assume ). A fundamental region of is constructed by gluing together either two or four copies of by orientation reversing homeomorphisms and , as previously discussed. Hence, the algorithmic construction of by rotating by about then translating to tile the plane respects edge valence, and so edge/vertex correspondence for all of can be obtained in this way. From here forward we will consider the edges of to be labelled by the valence of a corresponding vertex, and we will refer to this number as the edge valence.
We summarize the preceding discussion in the following Lemma, part (d) of which corrects a minor error in the proof of Theorem II.3.1 in [SaWe] (this error does not, however, affect the validity of their proof). Note that the relevant notation in [SaWe] differs from ours in several ways: most importantly, what we call they denote , and we follow a different indexing convention for vertices of .
Lemma 3.2.
The lifted cusp triangulation for the link given by a word has the following description:
is obtained from the triangulated rectangle , described by Figure 8, as follows: reflect in to get , then rotate about , and translate the resulting two copies of by , , to tile .
The deck group of is generated by and , where is the number of components of the link .
Edge/vertex correspondence in is as follows (see Figure 9):
If is horizontal or is a meridional edge, then corresponds to the vertices across the two triangles adjacent to it.
If the lower endpoint of meets the line , and the upper endpoint meets , with k even (resp. odd), then corresponds to the vertex across the triangle to the left (resp. right) of .
If , then the vertices of , labelled as in Figure 8, have valence as follows (recall that if , otherwise):
In particular, note that for all , is odd if and only if . This fact is key to showing that nonarithmetic bridge links cannot have hidden symmetries. Since a hidden symmetry restricts to an isometry of , it is a simplicial automorphism of (i.e., a homeomorphism preserving the simplicial structure) and hence it is a simplicial automorphism of that preserves edge valence.
Definition 3.3.
We denote by the group of simplicial automorphisms of that preserve edge valence. Note that if we identify with the horoball centered at , then there is a natural injection .
By analyzing , which must preserve these odd valence vertices, we learn about the possible isometries of . The first step in this process is the following lemma:
Lemma 3.4.
If , then preserves clasping triangles and meridional edges.
Proof.
By the symmetry of the problem, we need only show that any triangle with vertex labels maps to a clasping triangle. Let , and let be the image of a triangle under , so that .
Case 1: . Since is odd if and only if , we must have . We will assume that ; the case is proved similarly. Then since and all other vertices that could share an edge with have valence .
If (i.e., ), then , since in this case no vertex or is connected to by an edge. If , then we must have , so that , which means that must have valence . But also implies that is the image of the valence vertex of the clasping triangle that shares a meridional edge with , giving a contradiction. Thus , and by the same argument we must also have .
If , then and , and we must have . Also, .
If , then , so . This implies that , so we must have . This determines the image of the two nonclasping triangles adjacent to , and we see that the vertex of one of these must be mapped to a vertex, which is impossible since .
If then and are connected by an edge, so , which forces the other labelled vertex of to map to , which is impossible by the above argument. Hence , and by the same argument we have .
That meridional edges map to meridional edges is immediate since implies that clasping triangles have a unique odd valence vertex, i.e., the vertex not meeting a meridional edge.
Case 2: and . If , then clasping triangles either have vertices with valences or , and they are the only triangles in with such a triple of valences. If then clasping triangles either have vertices with valences or , and they are the only triangles in with such a triple of valences. Furthermore, in every case two of these vertices have equal valence and the third has distinct valence, so meridional edges must be preserved.
Case 3: , . Then . If then , so and we must have vertices labelled mapping to vertices labelled or . But , so . If then clasping triangles all have vertices with valences , and they are the only triangles in with this triple of valences. Furthermore, meridional edges are preserved since even when (so that ), the vertices labelled are combinatorially distinct from the vertices labelled and : vertices labelled have four edges connecting them to valence vertices, while vertices labelled and have only two such edges. ∎
Corollary 3.5.
If , then preserves horizontal strips of .
Proof.
Let be the clasping triangle in the first quadrant of with a vertex at the origin. is adjacent to two horizontal strips; let be the one adjacent to the axis, and let be the other clasping triangle adjacent to . Let be the path directly across connecting the midpoints of the edges of adjacency with and . Consider the image of under a simplicial automorphism . Since crosses exactly triangles, so must . By Lemma 3.4, maps and to edges of clasping triangles, which are adjacent to distinct meridional lines since and are, and maps triangles crossed by to nonclasping triangles, so must be mapped into some number of vertically stacked horizontal strips. Since crosses all triangles transversely, if jumps from one horizontal strip to another the number of triangles it crosses must be one more that if it did not make the jump, as shown in Figure 10. Hence must be contained in one horizontal strip, the image of . ∎
Recall that in our algorithmic construction of , we chose coordinates so that the rectangle shown in Figure 8 is identified with . We have the following theorem:
Theorem 3.6.
If , then is generated by the deck transformations and a subset of the following:

OrientationPreserving: the rotations , , and about , , and , respectively, by an angle .

Orientation Reversing: the glide reflection given by the reflection across composed with .
Furthermore, we always have , and (resp. ) if and only if (resp. ) is a simplicial automorphism.
Proof.
Let , and let be the union of all edges of horizontal strips and clasping triangles, as shown in Figure 12. Since maps clasping triangles to clasping triangles, and horizontal strips to horizontal strips, it must map to itself. Since the simplicial structure of the triangulation within each horizontal strip must be preserved, and since we may assume all clasping triangles are congruent and triangles within each strip are uniformly sized, is forced to be a Euclidean isometry of . Let be the rotation by about the point , and let be the reflection about the line . We first consider the possible Euclidean isometries preserving :
translations: translations must preserve the integer lattice, so modulo deck transformations they have the form , , . Since , and do not preserve , and is trivial, we are left with
and their inverses.
rotations: since meridional lines and integer lattice points must be preserved, any rotation must be by an angle about a point , . The rotations about and do not preserve clasping triangles, so modulo deck transformations we are left with , and the rotations
about and , respectively.
reflections: reflections must preserve meridional lines and clasping triangles, so possible lines of reflection are or , . Modulo deck transformations, we get the reflection across , and the reflections across the lines , . We have
glide reflections: Since simplicial automorphisms preserve merdional lines and clasping triangles, the reflection component of the glide reflection must be across a line or , . If the reflection is across , then the translation must be , , so modulo deck transformations this is a pure reflection, and can be ruled out. Thus we are left with the glide reflection , given by the reflection across followed by the translation , and the compositions
all others being obtained by composing with deck transformations.
We show that by considering edge valences near a clasping triangle. Using the edge/vertex correspondences from Figure 9, we obtain the four pictures in Figure 11, which correspond to the cases , , , and , respectively (note that nonarithmetic implies ). For the first three pictures it is clear that does not preserve edge valence. For the last picture, if then , so that , which implies , a contradiction. Hence .
In order to rule out and the compositions above involving and , we will first need to establish the last assertion of the theorem, namely that we always have , and and are in if and only if they are simplicial automorphisms of . To see this first note that and are always simplicial automorphisms (by construction of ). Thus we need only show that if any of , , , or is a simplicial homeomorphism, then it is in . But this follows from the fact that each of , , , and preserve the edge/vertex correspondence given in Lemma 3.2(c) (shown graphically in Figure 9). In particular, each of these maps switches the parity of in part (c) of the Lemma, but also exchanges right and left. Thus, if is simplicial, it preserves vertex valence, and since it also preserves edge/vertex correspondence, it must preserve edge valence, i.e., . The same holds for , and , so the assertion is proved.
Now, suppose that . First, observe that . Since is always a simplicial automorphism (by construction of ), implies that is a simplicial automorphism. Thus by the above paragraph, . But implies that , a contradiction.
Thus we can rule out the compositions , and . For and , since is always a simplicial homeomorphism, the composition is simplicial if and only if is. But then by the above observation it follows that preserves edge valence, so the composition cannot preserve edge valence (becuase does not). Last, can now be ruled out since .
Since the only compositions we have not ruled out are generated by , and , and since compositions involving (resp. ) are in if and only if (resp. ) is, the result follows. ∎
4. Symmetries of bridge link complements
Let , and let denote the symmetries of , i.e., is the group of selfhomeomorphisms of up to isotopy. Here, we describe the action of on the triangulation . First, Theorem 4.1 gives a classification of the symmetries of in terms of the word . This theorem comes from combining Theorem II.3.2 and Lemma II.3.3 in [SaWe] and translating from to the word given by the following dictionary: , for , and . In [SaWe], these symmetries are called automorphisms of the triangulation of described in Section 3. Since by [Gu] this triangulation is now known to coincide with the canonical triangulation of , we know these automorphisms actually correspond to all of the symmetries of .
We let denote the subgroup of consisting of orientationpreserving symmetries. We say that is palindromic if for all .
Theorem 4.1 ([SaWe],[Gu]).
Let be any hyperbolic bridge link complement. Then if and only if is not palindromic. When is palindromic, then we have the following possibilities:

If is even, then and