Geometric triangulations of a family of hyperbolic 3-braids

We construct topological triangulations for complements of $(-2,3,n)$-pretzel knots and links with $n\ge7$. Following a procedure outlined by Futer and Gu\'eritaud, we use a theorem of Casson and Rivin to prove the constructed triangulations are geometric. Futer, Kalfagianni, and Purcell have shown (indirectly) that such braids are hyperbolic. The new result here is a direct proof.


Introduction
A knot or link in S 3 is hyperbolic if its complement admits a complete hyperbolic structure.A braid on n strands, which is represented by a word in the braid group, w 2 B n , is hyperbolic if its braid closure, L w , is.Hyperbolic braids formed from three strands have been characterized by Futer, Kalfagianni and Purcell [7], a characterization that depends on their representation in the braid group.B 3 has two generators, 1 and 2 shown in Figure 1, and one relation, 1 2 1 D 2 1 2 .The square of this element is commonly denoted by C , which is central.In [7,Theorem 5.5], Futer, Kalfagianni and Purcell have shown that S 3 L w is hyperbolic if and only if w is conjugate to C k p 1 1 q 1 2 p s 1 q s 2 with k 2 Z and p i , q i , and s positive integers, and w is not conjugate to p 0 1 q 0 2 for some integers p 0 and q 0 (closures of such braids are either unknots, torus knots, or connected sums of torus knots).The proof is by contradiction; they show the required hyperbolic structures exist without constructing associated triangulations.In this paper, we construct geometric triangulations for braids with k D 2, s D 1, p 1 1, and q 1 D 1, thus offering a direct proof of the "if" direction of their theorem in these cases. 2 is hyperbolic will cover those pretzel knots and links as well.To show C 2 p 1 1 2 is hyperbolic, we will first decompose its complement into ideal tetrahedra whose faces are identified in pairs (Section 2).These topological triangulations (one for each p) are very much in the spirit of the canonical decompositions first described for 2-bridge knot and link complements by Sakuma and Weeks in [19].It turns out that, as in the 2-bridge case detailed by Futer in [9, Appendix A], our triangulations are not just topological; they are also geometric (though not canonical).In other words, each ideal tetrahedron can be given a hyperbolic shape of positive volume and, when the faces are paired by hyperbolic isometries, the resulting hyperbolic structure on S 3 L p is metrically complete.
To prove that our topological triangulations are geometric, we will employ the "Casson-Rivin program" laid out by Futer and Guéritaud in [5] -their survey of Casson and Rivin's technique for finding the hyperbolic structure on a 3-manifold whose boundary consists of tori.Guéritaud has implemented this program for punctured torus bundles and 4-punctured sphere bundles [9] and Futer for 2-bridge knots [9, Appendix A], whereas Guéritaud and Schleimer have applied it to layered solid tori and Dehn fillings [10] and Ham and Purcell used the same procedure to determine geometric triangulations on highly twisted links [11].
Geometric triangulations can lead to better understandings of the geometry of finitevolume hyperbolic 3-manifolds, including simpler proofs.However, it is not known whether every such 3-manifold admits a geometric triangulation.Currently the list of infinite families of finite-volume hyperbolic 3-manifolds that have been shown to admit geometric triangulations is short: punctured torus bundles and 4-punctured sphere bundles [9], 2-bridge knots and links [9, Appendix A], and certain Dehn fillings of fully augmented 2-bridge knots and links [11].The main result of this paper is that the .2; 3; n/-pretzel knots and links with n 7 can be added to the list.

Organization
After constructing the topological triangulations in Section 2, we offer definitions and a basic outline of the Casson-Rivin program in Section 3. The first step of the program is carried out in Section 4, where we show that the space of positive angle structures for our triangulations (Definition 3.1) is nonempty.The final step -arguing that the critical point of the volume functional (Definition 3.3) is a positive angle structureis presented in Section 5 with some comments about extending the construction in Section 6.
The explicit triangulations appearing here, as well as several more in the infinite family, were confirmed using the software packages Regina [3] and SnapPy [4] and checked against the census of veering triangulations created by Giannopolous, Schleimer and Segerman [8].See the Regina data file [15] for the triangulations constructed in Section 2.
To make the graphics more accessible to those with color vision deficiencies, I have used two of Paul Tol's qualitative color schemes -bright (Figures 3 and 14) and light (Figures 4 and 13).His palettes are mathematically designed to appear distinct to all [21].
David Futer kindly gave feedback on an early draft of the appendix for which I am grateful.I especially want to thank Bill Dunbar for many helpful conversations.There is no aspect of this paper -the structure, the notation, the figures, the tables, the exposition, the results -that has not been improved by his insights, careful reading, and self-described pickiness.I am also grateful for several useful suggestions offered by the referee.That said, any remaining errors are mine.

Ideal triangulations
Let X p D S 3 L p , the complement of the closure of the braid C 2 p 1 1 2 .In this section, we construct a triangulation of X p .In other words, we decompose the space into tetrahedra with face pairings.The decomposition will be developed using ideas from the appendix, which contains a slight revisualization of the geometric triangulations of 2-bridge link complements presented by Futer in [9, Appendix A].There he uses the fact that a 2-bridge link can be described as a 4-braid "closed up" with a clasp at each end.The 4-braid lives in a product region, S 2 I , and its complement in this region is also a product, S I , where S is a 4-punctured sphere.In [9, Appendix A], the product region, S I , appears between two nested pillowcases and Futer showed that the region can be triangulated using a sequence of layers of ideal tetrahedra.In the appendix, we position the product region vertically.As a result, the faces of Futer's layered tetrahedra can be easily seen in the braid complement, and the bottom of one layer is identified to the top of the next according to the half-twist between them.(See, for example, Figures 12 and 13.)The full link complement can be obtained by identifying the top of the product region to itself in a way that forms the clasp needed at the top and similarly for the bottom (Figure 14).These identifications amount to capping off each end of the S I with a 3-ball from which the clasp has been removed.
Analogously, we observe that the closure of C 2 p 1 1 2 can be formed from a 6-braid "closed up" with a three-stranded full twist on both the top and the bottom.In this context, the 6-braid can slide past the central full twist C on the bottom to become 1  4 as indicated in Figure 2. In this section, we will triangulate the complement, X p , by placing ideal tetrahedra in the product region containing the 6-braid p 1 1 4 and identifying the topmost layer to itself in a way that forms a full twist on three strands and similarly at the bottom.We begin with the challenge of how to identify a 6-punctured sphere to itself to form a full twist.

Forming a three-stranded full twist
Let S 0 be the top 6-punctured sphere.Label the punctures 0 through 5 and divide S 0 into eight triangles as shown in Figure 3(a).Think of the punctures -and the strands descending from them -as living in the plane of the page, so 4025 is in front of the To see that identifying S 0 to itself in this way is the same as attaching a 3-ball with a full twist removed, first fold S 0 along the 23 edge into a cylindrical pillow shown in Figure 3(b).As we go step-by-step through the identifications, we will follow the punctures (their paths are drawn as strands) and see they sweep out a full twist.The extra markings -cyan on the white triangles and gray on the yellow and pink oneswill help us follow the triangles through each step.
4025 is identified to 4035: Figure 3(c) shows the result of bringing the 2 and 3 strands together to identify the gray triangles.Placing this triangle in the plane of the page means the 23 (blue) edge forms a belt around a new pillow with half in front, as the 1 and 2 strands are, and half behind as the 3 and 4 strands are.The left side of the pillow consists of 4023 and both 4012's (appearing in the front).The right side is formed by 4523 and both 4543's (in the back).This diagram is analogous to the third step in Figure 14 in the appendix.
4023 is identified to 4523: By pushing the 0 and 5 strands into the pillow until they meet, 4023 can be identified to 4523 as shown in Figure 3(d).The 0 strand goes behind 1 and 2 and the 5 strand passes in front of 3 and 4. The result is not the same as the final step in Figure 14, because of the presence of the two 4012's and the two 4543's.While these triangles themselves are not identified yet, their 02 edges and 53 edges (dark red and green) have come together, and they form a sphere with the two pink triangles in front and the two yellow in back.The interlocked gray and white faces meet along the dark red and green longitudes.These faces appear behind the sphere on the left and in front on the right.The 4012's are identified to the 4543's: Bringing strands 1 and 4 together and identifying each 4012 to its corresponding 4543 collapses the sphere.The result, shown in Figure 3(e), is the desired positive full twist on three strands (read right to left, the positive direction for the top of the 3-braid with a counterclockwise orientation).
To obtain the triangles for the 6-punctured sphere at the bottom of the product region, rotate S 0 about the horizontal line through the punctures, so now the strands of the braid go up, and use the same identifications.For example, at the bottom 4023 is in the front and is identified to 4523 in the back.
Having capped off the ends, we turn our attention to placing ideal tetrahedra in the product region, determining their face pairings, and thus defining triangulations p on the manifolds X p for all p.We start by constructing a triangulation of X 1 upon which the other triangulations will be built.

A foundational ideal triangulation
To specify the triangulation of X 1 , we will place tetrahedra in a product region from which the 6-braid, 1 1 4 , has been removed.The left diagram in Figure 4 shows this region with the braid in the plane of the page, except near the crossings.
The method of the appendix suggests using three layers of tetrahedra (one on either side of the two crossings).Label the top, middle, and bottom layers t , m , and b (with t split into t t and t b ).The six ideal tetrahedra that will triangulate the complement are shown in Figure 4: t 0 1 and t 0 2 in t , m 1 and m 2 in m , and b 1 and b 0 2 in b , where a prime indicates that the tetrahedron is behind the plane of the page; the labels on the ideal vertices come from the labels on their punctures (0 through 5 when read left to right); and the "top" faces of a tetrahedron are above the "bottom" faces relative to the product region.
The edges of the tetrahedra are drawn both in the braid complement on the left and in flattened versions in the center and on the right.On the left, layers are expanded by making two copies of the edges where the top of a tetrahedron meets the bottom, allowing the faces of the tetrahedra and their remaining edges to be seen more easily.
The flattened versions show just one copy of each edge.
The tetrahedra t 0 1 and t 0 2 share the face 025, shaded orange on the left, and their flattened versions are drawn separately.The pair m 1 and m 2 share only an edge (thickened in the center diagram), so their interiors do not overlap when flattened, and they can both be seen when drawn side-by-side.The same is true for b 1 and b 0 2 .The left figure contains additional segments that divide the boundaries of each layer into triangles.These curves appear gray in the diagram and will be useful in determining the face pairings for this initial triangulation, which we shall refer to as O 1 .
When comparing to the appendix, the reader should be concerned that the pairs of tetrahedra do not fill the entire layer.In m , for example, the tetrahedra are only in front of the plane of the page.Nonetheless, if we slide the faces of these six tetrahedra around the link -up and down the braid and through the full twists at the top and bottom -they will fill out the entire product region.For example, face 235 on the bottom of t 0 1 can slide down the braid through t b and m and past the 1 4 half-twist to be identified with the 245 face of b 0 2 .We denote such faces and face pairings by t 0 1 .235/b 0 2 .245/;taking care that the order of the vertices indicates the correct match.Images of the intersection of this triangle's path with each layer's boundary are shaded pink in the diagram.They make use of the aforementioned additional gray segments and show that the back 235 region of m is indeed covered.
As indicated in Figure 4, t 0 1 sits on top of t 0 2 with their 025 faces identified.The remaining faces on the top of the tetrahedra in t and those on the bottom of the tetrahedra in b are triangles appearing in Section 2.1 and are identified by the full-twist identifications described there.For example, the face b 1 .023/on the bottom of b is the triangle 4023, which is identified to 4523, which is the 523 face of b 0 2 , another face on the bottom of b , so b 1 .023/b 0 2 .523/.These faces are shaded green.
Sometimes the full-twist identifications will combine with sliding along the braid.To see such a combination, look at triangles shaded blue, starting with the 125 face of t 0 2 .After traveling through the half-twist that interchanges 0 and 1, t 0 2 .125/becomes 4025 on the top of m still in the back.Then it can travel through m and b all the way to 4025 on the bottom of b in the back.This triangle is identified to 4035 in the front, ie the 035 face of b 1 , so t 0 2 .125/b 1 .035/.
The remaining face pairings of O 1 are obtained by sliding the remaining faces along the braid, twisting through half-twists and using full-twist identifications at the top and bottom as needed.known as bistellar flips, introduced in [16].Table 1 shows the following identifications of the 24 edge of m 2 : Whenever three tetrahedra surround an edge, we can perform a 3-2 Pachner move that replaces them with two tetrahedra sharing a face.Next, observe that the 23 edge of t 0 1 is identified to the 2a edge of N s, which is taken back to the 23 edge of t 0 1 .This very short cycle of edges means t 0 1 and N s are identified across an edge, so we can perform a 2-0 Pachner move to collapse t 0 1 and N s.This move identifies t 0 1 .025/with N s.c2d / and t 0 1 .035/with N s.cad /.See Figure 6, which shows how t 0 1 and N s are stacked before they are collapsed and eliminated.The 2-0 move results in a triangulation of X 1 comprising three tetrahedra -t 0 2 , m 1 , and s -whose face pairings are in Figure 6.Define 1 to be this triangulation.

Adding layers to extend to p D 2 and 3
It will be more instructive to first extend the construction of the previous section to p D 3. To triangulate the complement of the closure of C 2 3 1 1 2 , where there are four half-twists in the braid, begin with five layers: t , m , and b and two layers inserted between t and m , labeled w1 and w2 .Place two tetrahedra in each layer as indicated in Figure 7. Though t is depicted slightly differently (as one layer with the shared face of t 0 1 and t 0 2 , 025, positioned in the middle), all tetrahedra in t , m , and b are named and situated exactly as in Figure 4.The new tetrahedra, those in the wi layers, will be called w i and w 0 i , where a prime still indicates the tetrahedron is behind the plane of the page.
The face pairings are determined in the same manner as they were for p D 1, but now over half of the twenty face pairings result from moving through a half-twist from the bottom of one layer to the top of the next.These appear in the left column of Table 2.The middle column contains the face pairings that come directly from Table 1 and are unaffected by adding the w-layers.The final column contains more complicated face pairings that do not occur in Table 1 because these triangles hit tetrahedra in the w-layers before encountering their mates from Table 1.For example, 2 , with t depicted as one layer with the shared face of t 0 1 and t 0 2 , 025, positioned in the middle.identified to 4025 in the front, which, after passing through the half-twist, becomes 4125, a top face of w 1 , so t 0 1 .035/hits w 1 before it gets to m 1 .We will call this initial triangulation O 3 , and, as in the p D 1 case, will perform simplifying Pachner moves on O 3 to create 3 .The shadings in Table 2 will help us determine the resulting identifications.
Again, three tetrahedra in O 3 surround the 24 edge of m 2 (which is identified to the 23 edges of b 1 and b 0 2 ), allowing for the same 3-2 move shown in Figure 5. Thus m 2 , b 1 , and b 0 2 are replaced by N s and s.As before, we can then perform a 2-0 move collapsing t 0 1 and N s because they are identified across the edge 23 D 2a.Again, t 0 1 .025/and t 0 1 .035/are identified with N s.c2d / and N s.cad / as in Figure 6.Let 3 be the resulting triangulation of X 3 , consisting of the seven remaining tetrahedra: t 0 2 , w 1 , w 0 1 , w 2 , w 0 2 , m 1 , and s.
The face pairings in the darkest cells in Table 2 involve only the four replaced or collapsed tetrahedra (m 2 , b 1 , b 0 2 , t 0 1 ), so none of them will survive the Pachner moves defining 3 .The face pairings in the unshaded cells, on the other hand, do not involve any of these tetrahedra, and are, thus, unaffected by the two Pachner moves.
The remaining face pairings of 3 come from those in the light gray cells in Table 2 s.d 2c/, which is identified by the 2-0 move in Figure 6 to t 0 1 .520/.This face is paired with t 0 2 .520/,so, after the two Pachner moves, m 1 .025/ is identified with t 0 2 .520/,which we choose to write as t 0 2 .025/m 1 .520/.
The final four faces to be relabeled -t 0 1 .035/,m 2 .345/,b 1 .035/,and b 0 2 .345/-are all faces that belong to the new tetrahedra s.More specifically, the 2-0 move identifies t 0 1 .035/with N s.cad /, which is the new face introduced by the 3-2 move, the face shared by N s and s.Thus, after the two Pachner moves, t 0 1 .035/becomes s.cad /.The other three faces form the rest of s as indicated in Figure 5 and are thus the faces s.abc/, s.dbc/, and s.bad /, respectively.Table 3 contains all identifications of 3 , where the first column still lists face pairings that result from moving from the bottom of one layer to the top of the next and the final column now shows how the new tetrahedron, s, glues in.
The reader likely anticipates that we will create the desired triangulation of X p in two steps: first form an initial triangulation, O p , consisting of the tetrahedra in Figure 4 together with some w-layers between t and m , and then perform two Pachner moves -a correct guess that will be made explicit in the next section.Before moving on, we make some observations about the p D 3 construction and address the p D 2 case.
Figure 7 shows that t 0 2 is situated and labeled in exactly the same way as w 0 1 and w 0 2 , so the labels on the bottom faces of t 0 2 , 015 and 125, are the same as those on w 0 1 and w 0 2 .Furthermore, as the face t 0 2 .015/slides through the half-twist, it rotates to the front, matching with w 1 .105/,just as the 015 face of w 0 1 matches with w 2 .105/.Similarly, as the face t 0 2 .125/slides through the half-twist, it stays in the back, matching with w 0 1 .025/,just as the 125 face of w 0 1 matches with w 0 2 .025/.In these identifications, t 0 2 is acting as a tetrahedron labeled w 0 0 would.A similar analysis at the bottom of the w-layers shows that the top faces of m 1 glue to the bottom faces of w 2 in the same way those in a tetrahedron labeled w 3 would.
Using this relabeling, we make several observations about the face pairings in 3 .
By relabeling t 0 2 as w 0 0 and m 1 as w 3 , the unshaded face pairings in the first column of Table 2 can be summarized as follows:  2 , and t 0 1 , so they are also face pairings of 3 , as shown in the first column of Table 3.
The top faces of w 0 0 are not included in the identifications listed above, and neither are the bottom faces of w 3 .These faces are identified to each other.More specifically, after the Pachner moves, w 0 0 .012/w 3 .021/and w 0 0 .025/w 3 .520/.Both of these pairings also occur in 1 (see the table in Figure 6) and were, thus, unaffected by the addition of the w-layers.These face pairings appear in the second column of Table 3.
The face pairings in the third column of Table 3, those involving s, can be characterized as follows: two of the faces of s are identified to top faces in the first w-layer, w 1 .125/and w 0 1 .201/,and the other two are identified to bottom faces in the last w-layer, w 2 .201/and w 0 2 .125/.
Properly interpreted, these observations also apply when p D 2. Let the triangulation 2 of X 2 , the complement of the closure of C 2 2 1 1 2 , consist of w 0 0 , w 1 , w 0 1 , w 2 and s.Because there is only one w-layer, it is both the first w-layer and the last w-layer, and the final observation tells us that the faces of s are identified to four faces in w1 .Harvesting the remaining face pairings from the first two observations, we obtain the face pairings for 2 given in Table 4.

An ideal triangulation for any p
Guided by the previous examples along with Figure 7, form an initial triangulation, O p , of X p , the complement of the closure of C 2 p 1 1 2 , that consists of 2p C 4 tetrahedra: t 0 1 and w 0 0 (née t 0 2 ) in t ; w p (née m 1 ) and m 2 in m ; b 1 and b 0 2 in b ; and w i and w 0 i in wi where i D 1; : : : ; p 1. The face pairings of O p are determined by sliding faces along the braid and using the full-twist identifications at the top and bottom of the product region as described in Sections 2.2 and 2.3.As before, the addition of the w-layers does not affect identifications amongst m 2 , b 1 , b 0 2 , and t 0 1 , so the Pachner moves depicted in Figures 5 and 6  triangulation.As with 2 , the observations about 3 from the previous section apply to p , yielding the face pairings listed in Table 5 for p > 1. (For p D 1, see Figure 6.) The triangulation O p had 2p C 4 tetrahedra.The 3-2 Pachner move reduced this number by one and the 2-0 Pachner move further reduced the number of tetrahedra by two, so p has 2p C 1 tetrahedra and thus 4p C 2 face pairings.There are exactly 4p C 2 identifications in Table 5 (and six in Figure 6).Thus, the definition of the triangulation p is complete.
Note that most of the tetrahedra in p (all but s) are visible in the complement of the 6-braid p 1 1 4 .Using this visualization can help us quickly deduce relationships between the tetrahedra.For example, Figure 8 shows how w 0 0 , w 1 , w 0 1 , : : : , w p 1 , w 0 p 1 and w p are situated in the braid complement, and all but six face pairings of p can be derived directly from following their faces through the half-twists in the braid.

Equivalence classes of edges in the triangulation
Knowing the relationships between the edges of p will prove helpful because their equivalence classes form the edges of X p and the (internal) dihedral angles along these edges will dictate its geometry.In determining equivalence classes, we can use Figure 8 or Table 5, whichever is easier.
As an example, we will determine all edges identified to the 01 edge of w 0 0 .In the braid picture, w 0 0 .01/slides down through the first 1 half-twist to be identified with w 1 .10/and w 0 1 .10/,so the equivalence class of w 0 0 .01/,OEw 0 0 .01/,contains the 01 edge 1 and the bottom faces of w p 1 and w 0 p 1 as in Table 5.
of w 1 and of w 0 1 .Continuing to move the 01 edge through more half-twists shows the 01 edges of w i and w 0 i belong to OEw 0 0 .01/for all i .Two edges of s also appear in this equivalence class, which can be seen most easily in Table 5: w 0 1 .01/s.bc/ and w p 1 .01/s.ad /.These edges of s, s.bc/ and s.ad /, are also identified to w 0 p 1 .25/and w 1 .25/,respectively, which can themselves be slid along the braid to show that the 25 edges of each w i and w 0 i are in OEw 0 0 .01/.We have not yet considered face pairings between w 0 0 and w p .The first in Table 5, w 0 0 .012/w p .021/ shows w p .02/ and w 0 0 .02/are also in OEw 0 0 .01/.Based on how the tetrahedra are positioned in the braid complement, 02 is the top edge in w 0 0 and the bottom edge in w p , which allows us to describe OEw 0 0 .01/as follows.
Observation 1 The equivalence class OEw 0 0 .01/contains a total of 4p C 4 edges: the 01 edges and 25 edges of each w i and w 0 i ; s.bc/ and s.ad /; the top edge of w 0 0 and the bottom edge of w p .These edges are colored red in Figure 8.
It turns out that the remaining 2p 2 edges have degree 4.

Angle structures and the Casson-Rivin program
We will prove that the triangulation p of X p defined in Section 2.4 is geometric by applying the Casson-Rivin program.This section provides some definitions, useful results, and a basic outline of the program.For more details and a beautiful exposition, including an elementary proof of the Casson-Rivin theorem, the reader is encouraged to read Futer and Guéritaud's paper [5].

Angle structures
To move from a topological triangulation to a geometric one, we need a way to impose a hyperbolic structure on each tetrahedron where the face pairings are hyperbolic isometries.A start is to assign angles to the edges of the triangulation.
Definition 3.1 An angle structure on an ideal triangulation of a 3-manifold is an assignment of angles to the edges of the tetrahedra in satisfying the following conditions: (1) Angles assigned to opposite edges of a tetrahedron are equal, meaning it is enough to specify three angles Â 3i 2 , Â 3i 1 , and Â 3i , for each tetrahedron T i 2 . ( (3) The sum of the angles surrounding an edge of the 3-manifold equals 2 .
This paper utilizes both taut angle structures, where Â k 2 f0; g for all k, and positive angle structures, where Â k > 0 for all k.Note that condition (2) of the definition implies that a positive angle structure must have all angles strictly between 0 and .In this case, the three angles that sum to specify a positively oriented ideal hyperbolic tetrahedron.The tetrahedron is isometric to one whose vertices are at 0, 1, 1, and a complex number z with Im.z/ > 0, where 0, 1, and z form a Euclidean triangle having the assigned angles.If Im.z/ D 0, the tetrahedron is said to be degenerate, and, if Im.z/ < 0, negatively oriented.If an angle structure Â assigns the same angles to two tetrahedra, T 1 and T 2 , then we will say they are isometric with respect to Â and write We use A. / to denote the space of all positive angle structures on a triangulation and note that, if has n tetrahedra, A. / .0;/ 3n is a convex polytope with compact closure.As such, the points obtained through coordinate-by-coordinate averaging of elements of A. / are again in A. /.Also, real-valued functions defined on A. / attain their extreme values.See [5, Proposition 3.2] for more details about the space of angle structures.
As will be described in Section 3.3, the initial step in the Casson-Rivin program is to argue that A. / is nonempty, ie that the triangulation admits a positive angle structure.We accomplish this in our case by first showing (in Section 4.1) that p admits a special type of taut angle structure -a veering angle structure.The notion of "veering", first introduced by Agol [1], has several equivalent formulations.The simplest to apply to our triangulations is from [12].Definition 3.2 A veering angle structure on an ideal triangulation of an oriented 3-manifold M is a taut angle structure meeting an additional condition: The edges of M can be colored red and blue in such a way that, within each tetrahedron, two of the edges with angle 0 are red and two are blue (the color on an angle edge can be either red or blue); when viewed from any of the four ideal vertices, the -angled edge is followed in the counterclockwise direction by a blue edge and then a red edge.Each tetrahedron in a veering triangulation appears as the one in Figure 9, where the -angled edges on the diagonal can be red or blue.If there is also a consistent "upward" orientation on the four faces (for example out of the page in Figure 9), the angle structure is transverse taut.A triangulation that admits a veering angle structure is said to be veering.
Positive angle structures give geometric structures on the tetrahedra in a triangulation.In turn, these structures determine the face-pairing isometries.Condition (3) of Definition 3.1 guarantees that, under these isometries, the tetrahedra fit together to fill the space around the edges of the manifold, because the sum of dihedral angles around any edge is 2 .However, there is no guarantee that the result is metrically complete.In fact, if a 3-manifold has a complete hyperbolic structure of finite volume, Mostow-Prasad rigidity ensures that the structure is unique up to isometry [14; 17], so one should not expect completeness for an arbitrary point in A. /.It turns out that the independent work of Casson and Rivin [18], which forms the basis of the Casson-Rivin program in [5] and outlined in Section 3.3, connects the question of completeness to the volume associated to an angle structure in A. /, ie the sum of the volumes of the corresponding hyperbolic tetrahedra.

Volumes of angle structures
Recall -see [13], for example -that the volume of an ideal hyperbolic tetrahedron T , with dihedral angles Â 1 , Â 2 and Â 3 , is given by where the Lobachevsky function, L, is defined as log j2 sin tj dt: Definition 3.3 For a point Â 2 A. /, which specifies angles for the n tetrahedra T i 2 and has all Â i 2 .0;/, define the volume associated to Â, V.Â /, by and note that V.Â / > 0 for Â 2 A. /.
As summarized in [5], L is well-defined and continuous on R. Thus, the definition of V can be extended to the closure of A. /, which may include points with coordinates equal to 0 or where the integral itself is improper.If Â 2 A. /, then V.Â / 0.
Straightforward tetrahedra-by-tetrahedra calculations show that for a positive angle structure Â and a tangent vector Because V is strictly concave down on A. /, any critical point in A. / is unique and an absolute maximum.In fact, the first and second partials (1) are enough to prove the same for a maximal point on the boundary.
Proposition 3.1 Whenever A. / ¤ ∅, the point at which the volume functional V W A. / !OE0; 1/ attains its maximum is unique.
Proof If there is a critical point ˛in A. /, the strict concavity of V guarantees that ˛is unique and that V.˛/ is the maximum volume.If, on the other hand, there is no critical point on the interior, the compactness of A. / ensures that there is a point ˛2 A. / A. / with V.˛/ maximal.To show such an angle structure is unique, we first examine its coordinates.Because ˛… A. /, there is at least one tetrahedron in with an assigned angle of 0 or .The maximality of ˛further constrains the angles of such a tetrahedron as noted in the following lemma.
Lemma 3.2 When V.˛/ is a maximum, any tetrahedron assigned an angle of 0 or must have angles of 0, 0, and in some order.
Proof By condition (2) of Definition 3.1, the angles in a tetrahedron must sum to , so this statement is obviously true if one angle in a tetrahedron is assigned or if two Algebraic & Geometric Topology, Volume 23 (2023) angles are assigned 0. We now argue, as also noted in [9, Proposition 7.1] and [10, Proposition 7], that assigning exactly one 0 angle contradicts the maximality of ˛.
Consider, for example, a tetrahedron, T 0 2 , with nonnegative angles 0, Â 0 , and Â 0 .For any 0 < < 1, define a family of tetrahedra T .t/j t 0 with angles Â 1 D t, Â 2 D Â 0 t, and Â 3 D .tC Â 0 t /, so T .0/D T 0 and, as t increases, the angles of the tetrahedron change -Â 1 is increased, while Â 2 and Â 3 are decreased.The first partial in (1) restricted to this family of tetrahedra yields In other words, increasing the solitary 0 angle by a small amount would increase T 0 's volume (and thus the total volume V) by a much larger amount, violating the maximality of ˛.Thus, if a maximal angle structure assigns any angles in f0; g to a tetrahedron, two of its angles are assigned 0 and one , thus proving the lemma.We call a tetrahedron with angles in f0; g flat and observe that its volume is 0.
Returning to the proof of Proposition 3.1, we will use the constraints Lemma 3.2 imposes on the coordinates of a maximal point ˛2 A. / A. / to prove that ˛is unique.Let ˇbe in A. / A. / with V.ˇ/ D V.˛/ and let 2 OE0; 3n be the point whose coordinates are determined by averaging those of ˛and ˇ.Then 2 A. /, because the set A. / is convex.We will examine the volume of each tetrahedron T under the angle structure assigned by and will denote this volume by V. j T /.
Case 1 If ˛and ˇassign the same angles to a tetrahedron T , then their average does as well, so V. j T / is equal to V.˛j T / (and also V.ˇj T /).
Case 2 If ˛and ˇassign different angles to a tetrahedron T and both ˛jT and ˇjT are flat, then, because ˛and ˇare maximum points, the angles must be 0, 0, and (in different orders), and the coordinates of j T are 0, 2 , and 2 in some order.In this case, V. j T / equals 0 just as V.˛j T / and V.ˇj T / do.
Case 3 If ˛and ˇassign different angles to a tetrahedron T and at least one of ˛jT or ˇjT is not flat, we will argue that V. j T / is strictly greater than the average of V.˛j T / and V.ˇj T /.Without loss of generality suppose ˛jT is not flat.Define .t/ to be the line segment joining ˛jT and ˇjT , .t/ D t ˛jT C .1 t /ˇj T for t 2 OE0; 1: Let V .t/be the volume of the tetrahedron whose angles are given by .t/.As a restriction of V, the function V is continuous on OE0; 1. Also, because ˛jT is not flat, is in .0;/ 3 for 0 < t Ä 1, so V is differentiable on .0; 1 and by the second partial in (1), V 00 < 0. Applying the mean value theorem to V yields points c 1 2 0;1 2 and c 2 2 1 2 ; 1 with and negative second derivatives, so In all cases, the volume of a tetrahedron T computed using , the average of the angle structures, is greater than or equal to the average of the volumes of T computed using the angle structures ˛and ˇ, so summing the individual volumes over all tetrahedra in yields V.
with equality only when there are no tetrahedra in case 3.But V.˛/ is the maximum volume, so V. / Ä V.˛/, meaning V. / D V.˛/ and there are no case 3 tetrahedra.The only way ˛and ˇcan assign different angles to a tetrahedron T is if ˛jT and ˇjT are both flat (case 2).In this case, V. / D V.˛/, so is also maximal, but, as we have seen, the coordinates of j T are 0, 2 , and 2 , which is not allowed for maximal angle structures.Therefore, ˛and ˇcannot assign different angles to any of the tetrahedra and ˇD ˛, which is the unique maximum point.
Positive angle structures that maximize the total volume play an important role in the Casson-Rivin program.

The Casson-Rivin program
In their proof of Casson and Rivin's theorem, Futer and Guéritaud relate the critical point of the volume (if it exists) and the complete hyperbolic structure (if it exists).More specifically, they show that the derivative of the volume vanishes in every direction exactly when Thurston's system of gluing equations guaranteeing completeness (given in [20]) is satisfied.In doing so, they prove their main result, a theorem based on independent work of Rivin and Casson usually cited as [18].
Casson-Rivin theorem [5, Theorem 1.2] Let M be an orientable 3-manifold with boundary consisting of tori , and let be an ideal triangulation of M .Then a point Â 2 A. / corresponds to a complete hyperbolic metric on the interior of M if and only if Â is a critical point of the functional V W A. / !OE0; 1/.
In our situation, the manifold M is the complement of an open neighborhood of the braid closure, L p .We seek a complete hyperbolic metric on X p , which is homeomorphic to the interior of M and is triangulated by p .
The Casson-Rivin theorem will enable us to conclude p is geometric as follows.In Section 4.1, we establish that the space of positive angle structures on our triangulations, A. p /, is nonempty.Consequently, the volume functional V W A. p / !OE0; 1/ will attain its maximum.Then, in Section 5, we show that the maximum point of V must belong to A. p /.Then, by the Casson-Rivin theorem, there is a positive angle structure on our triangulation p that yields the complete hyperbolic structure on the link complement, X p .Thus, the constructed triangulations are geometric and C 2 p 1 1 2 is hyperbolic for p 1.

Symmetries of angle structures
When arguing that the maximum of the volume cannot appear on the boundary of A. p /, a certain symmetry of the space of positive angle structures will be helpful.Definition 3.4 Any symmetry, , of a triangulation, , induces a map on the space of positive angle structures by assigning the angles of T to .T / for all T 2 .When such an assignment results in a positive angle structure for all points in A. /, we say induces a symmetry of the space of positive angle structures, which we will also denote by .More specifically, given a symmetry W ! with .T i / D T j , let Â 2 A. / and define the coordinates of Â by Â 3j 2 D Â 3i 2 , Â 3j 1 D Â 3i 1 , and Â 3j D Â 3i .The angles of Â are positive and meet conditions (1) and (2) of Definition 3.1, so induces a symmetry of A. / whenever condition (3) is also met.

The space of positive angle structures is nonempty
We now return to the triangulations, p , defined in Section 2.4.Our immediate goal is to complete the first step of the Casson-Rivin program by showing that A. p / is nonempty, ie that p admits a positive angle structure.We will do so by showing that p has a veering angle structure, which can be deformed to a positive angle structure [6; 12].We then use the veering structure to describe an explicit coordinate system for A. p / and conclude the section by writing the edge equations in these coordinates and using them to describe a useful symmetry.

The triangulations are veering
While the initial triangulations O p were not veering, the Pachner moves eliminated the obstructions and: Proof With the exception of s, the tetrahedra in p have an upward orientation induced by the product region containing the 6-braid, which is how the non-s tetrahedra of p are flattened in Figure 8. Flatten s as indicated in the same figure.In each tetrahedron, assign an angle of to the diagonal edges and an angle of 0 to the others.Use red to color the 01 and 25 edges of each w i and w 0 i ; the edges s.bc/ and s.ad /; and the top diagonal of w 0 0 along with the bottom diagonal of w p .(These are exactly the edges colored red in Figure 8.) Color the remaining edges blue.We claim this coloring forms a veering angle structure on p , which is also transverse taut.
By Observation 1, the edges in the equivalence class OEw 0 0 .01/are exactly the edges colored red.Thus, the blue edges compose the remaining equivalence classes, and the colorings are consistent with the face pairings that form M .
Next we show that the assignment of angles forms a taut angle structure.The angle assignments themselves guarantee conditions (1) and (2) of Definition 3.1 as well as the requirement that taut angle structures have angles in f0; g.Only diagonals are assigned an angle of , so, if there are exactly two diagonals in each equivalence class, condition (3) will also be met (the angle sum around each edge is 2 ).
The class OEw 0 0 .01/,whose members are listed in Observation 1, contains exactly two diagonals, w 0 0 .02/and w p .02/.The degree 5 edges of X p are completely described in Observation 2. The edges OEw 0 0 .05/and OEw 0 0 .12/each contain exactly two diagonals: the top diagonal w 1 .15/ is matched with the bottom diagonal w 0 p 1 .51/and similarly for w 0 1 .02/and w p 1 .20/(or s.db/ and s.ac/ when p D 1, which completes this case).To check the remaining 2p 2 edges of X p for p > 1, ie those of degree 4, we examine w i 1 .02/and w 0 i 1 .15/,the bottom diagonals of w i 1 and w 0 i 1 in the braid complement in Figure 8.The edge w i 1 .02/bounds both w i 1 .012/and w i 1 .025/.After passing through the 1 half-twist, the first of these faces is rotated to the back and identified with w 0 i .102/,whereas the second stays in front and is identified with w i .125/.These faces share the edge 12, which, after another 1 is twisted to the back and identified to w 0 i C1 .02/.A similar argument applies to w 0 i 1 .15/.Symbolically, (2) w i 1 .02/w i .12/D w 0 i .12/w 0 i C1 .02/for i D 2; : : : ; p 2; w 0 i 1 .15/w i .05/D w 0 i .05/w i C1 .15/for i D 1; : : : ; p 1: The only diagonals are the 02 and 15 edges, so the edge classes, OEw i .12/and OEw i .05/,contain exactly two angles as required.The remaining degree-4 edges contain the diagonals of s, s.ac/ and s.bd /.The identifications in Table 5 show that their equivalence classes contain exactly one other diagonal: (3) s.ac/ w 0 1 .21/D w 1 .21/w 0 2 .20/;s.bd / w 0 p 1 .21/D w p 1 .21/w p 2 .20/:With exactly two diagonals in each equivalence class, the angle structure is taut.Each tetrahedron appears as in Figure 9, and the gluings of Table 5 identify bottom faces to top faces, so the "upward" orientations are consistent, forming a transverse taut angle structure, which is also layered.

Coordinates for the space of positive angle structures
The veering structure on p allows us to introduce convenient coordinates for a point Â in A. p /.For T in p , let Â D T denote the angle assigned to the diagonals of T , Â R T the angle assigned to the red edges, and Â B T to the blue edges.Thus, we can write Â 2 A. p / .0;/ 3.2pC1/ as .Â R w 0 0 ; Â B w 0 0 ; Â D w 0 0 ; : : : ; Â R w 0 p 1 ; Â B w 0 p 1 ; Â D w 0 p 1 ; Â R w 1 ; Â B w 1 ; Â D w 1 ; : : : ; Â R w p ; Â B w p ; Â D w p ; Â R s; Â B s; Â D s/: Observation 4 Using this notation, the edge identifications in Observations 1 and 2 and equations (2) and (3) together with condition (3) of Definition 3.1 yield the following edge equations: for i D 2; : : : ; p 2; These equations hold for p > 1 as long as each term is defined.(Recall that there are no tetrahedra labeled w 0 or w 0 p .) The first two equations also hold when p D 1, and , after substituting s.db/ and s.ac/ for the undefined terms w 0 1 .02/and w p 1 .20/as in Observation 2, so does the third : Observation 5 Several pairs of edge equations above share angles coming from the blue edges.These angles will cancel when one equation is subtracted from the other, yielding the following equalities for the diagonals: These equations hold when p > 1 and each term is defined.The final equation, which is derived by taking the difference between the equations for the degree-5 edges implies that, if p is odd (so p 1 is even), then all of the differences listed above are equal.If p is even, all differences, Â D w j Â D w 0 j , with j even are also equal to The corresponding equation for p D 1 is

A symmetry of the space of angle structures
Consider the closure of C 2 p 1 1 2 as shown on the right of Figure 2. Moving the lone half-twist up so that it occurs halfway along the 1 half-twists reveals an order 2 symmetry of X p -rotate about a horizontal line through the lone half-twist.This involution induces a symmetry on the triangulation p -rotate the tetrahedra in Figure 8 about a horizontal line and take s to itself.This symmetry will respect angle structures: Proposition 4.3 An involution ÃW p ! p that fixes s and takes w 0 i to w p i by matching up their R, B, and D edges induces a symmetry of A. p /.
Proof Let Â 2 A. p /.According to Definition 3.4, to verify that ÃÂ is also in A. p /, we need to check that the sum of the dihedral angles at each edge of of X p is 2 , when the angle measure is given by ÃÂ.Thus, it is enough to verify that ÃÂ satisfies the edge equations listed in Observation 4 where ÃÂ ?s D Â ?s; ÃÂ ?w 0 i D Â ?w p i ; ÃÂ ?w i D Â ?w 0 p i for ?D R; B, and D.
The first equation describes the angle sum around the red edge of X p , which consists of all R edges and the red diagonals of w 0 0 and w p .The involution Ã permutes these edges, thereby permuting the terms of the sum: Algebraic & Geometric Topology, Volume 23 (2023) so the first edge equation is satisfied by ÃÂ.The terms in the next two sums, those for the degree-5 edges, are also permuted by Ã, so those equations are satisfied by ÃÂ as well.
Something a little different happens with the degree-4 edges (only present when p > 1).
To confirm the final equation of Observation 4, for example, we need to examine but this sum is just which equals 2 by the penultimate equation in Observation 4. In this instance, the involution takes the terms in one sum to the terms in another, effectively permuting the edge equations themselves.The same happens for the remaining degree-4 equations.
for i D 2; : : : ; k; In addition, (10) Â D w p D Â D s and Â D w i D Â D w p i for i D 1; : : : ; k: These equations hold for all p 1, with one exception.When p D 1, (6) reads: Proof By assumption, ÃÂ ?w 0 i D Â ?w 0 i for ?D R; B, and D, and, by definition, ÃÂ ?w 0 i also equals Â ?w p i , so Â ?w 0 i D Â ?w p i , and the tetrahedra w 0 i and w p i have the same angle assignments under Â, so w 0 i Š Â w p i .This isometry allows us to obtain (4)-( 9), by replacing w 0 i with w p i in the edge equations of Observation 4 and eliminating redundancies.
To obtain (10) for p D 1, replace w 0 0 with w 1 in the p D 1 equation of Observation 5, which shows and ( 6) imply that Â D w 1 equals Â D w p 1 , which, because of the isometry, equals Â D w 0 1 , so the difference Â D w 1 Â D w 0 1 equals 0 as do all odd-index differences listed in Observation 5. Replacing w 0 i with w p i yields (10) for odd i .The remaining equations are derived differently depending on the parity of p.As noted in Observation 5, when p is odd, all differences are equal (in this case, equal to 0), so, after replacing w 0 i with w p i , (10) holds for all i when p is odd.To see that (10) also holds when p and i are even, note that Thus, all differences in Observation 5 with even index are also 0 and replacing w 0 i with w p i results in (10).
5 The maximal volume occurs at a positive angle structure Corollary 4.2 shows that the space of positive angle structures on each triangulation p is nonempty, so we have accomplished the first step of the Casson-Rivin program described by Futer and Guéritaud in [5] and summarized in Section 3.3.In this section, we complete the program, proving the main result.
Theorem 5.1 Let L p be the closure of the 3-braid C 2 p 1 1 2 and X p be its complement in the 3-sphere.Then there is an ideal triangulation of X p that is geometric.
Proof Let p be the triangulation defined in Section 2.4.By Corollary 4.2, A. p / ¤ ∅, which means the volume functional V attains its maximum on the compact set A. p /, say at the point ˛.A maximal point has two important properties.The first was proved in Lemma 3.2.
Property 1 If ˛is maximal and assigns an angle in f0; g to a tetrahedron, then the tetrahedron must be flat, ie its angles are 0, 0, and in some order.
The second important property derives from a symmetry of the coordinates of a maximal point.Let the involution ÃW p ! p be as in Proposition 4.3, so Ã fixes s and takes w 0 i to w p i .Then Ã induces a symmetry on A. p /, which rearranges the coordinates of an angle structure tetrahedron by tetrahedron, so, just as in Observation 3, V.Ã˛/ D V.˛/.
Thus, if ˛is maximal, so is Ã˛, but maximal points are unique by Proposition 3.1, so Ã˛D ˛, and Corollary 4.4 together with Definition 3.1 guarantee the following: Property 2 If ˛is maximal , then the angles assigned by ˛belong to OE0; , sum to within a tetrahedron, and are determined by the angles assigned to w 1 ; w 2 ; : : : ; w p and s, which must satisfy (4)- (10).
Our goal is to use these properties to show that the maximal point, ˛, is in A. p /, not its boundary.Then we can apply the Casson-Rivin theorem from Section 3.3 [5, Theorem 1.2] and conclude that ˛corresponds to the complete hyperbolic structure on X p , thus proving Theorem 5.1.We will show that ˛is in A. p / by showing that points in B D A. p / A. p / will never maximize the volume.We do so by proving the following: Proposition 5.2 Any ˇ2 B satisfying Properties 1 and 2 must have V.ˇ/ D 0.
Proof Let ˇ2 B satisfy both Properties 1 and 2. Because ˇis in the boundary of the space of positive angle structures, there is a tetrahedron in p to which ˇassigns angles from the set f0; g.By Property 1, this tetrahedron is flat (has angles 0, 0, and ).Using Property 2, we can conclude that ˇmust assign to at least one of the edges of w 1 ; w 2 ; : : : ; w p or s.We now explore which edges can have such assignments and determine the resulting volume.
Observe that, because all angle assignments are nonnegative, assigning an angle measure of to any R edge of a w tetrahedron violates (4) of Corollary 4.4 and thus Property 2. Therefore, no ˇRw i can equal .We will use this result often, so we record it as a lemma and follow with another useful lemma.Proof If a tetrahedron is flat, one of its edges is assigned an angle of .By Lemma 5.3, ˇRw p cannot equal .However, ˇRs could equal , and, if this is the case, (4) forces ˇRw i to equal 0 for all i D 1; : : : ; p, so, by Property 1, the non-s tetrahedra are also flat.Because, ˇDw p also appears in (4), a similar argument shows that if ˇDw p D , all tetrahedra are flat.By (10), ˇDw p D ˇDs, so the same holds if ˇDs D .Thus, it only remains to check what happens if the angle at a B edge of w p or s equals .
The proof for the D edges is a little simpler, because there is no need to appeal to Lemma 5.3.If ˇDw 1 D , then, by (10) with i D 1, ˇDw p 1 D , so both ˇB w 1 and ˇB w p 1 are equal to 0. In this situation, applying (9) yields ˇDs D , so s is flat and, by Lemma 5.4, V.ˇ/ D 0. If j Ä k and ˇDw j equals , then, by (10) with i D j , so does ˇDw p j , and both ˇB .wj / and ˇB .wp j / equal 0, so (7) with i D j implies ˇDw j 1 D .Continuing in this manner shows that if ˇDw j equals , so does ˇDw 1 , and thus V.ˇ/ D 0. If j > k and ˇDw j D , then, by (10) with i D p j , ˇDw p j D with p j Ä k, a previous case.
Any point maximizing the volume functional must satisfy Properties 1 and 2. By proving Proposition 5.2, we have shown that any point, ˇ, on the boundary of A. p / satisfying these properties has V.ˇ/ D 0. Therefore, the maximal point ˛must be on the interior of A. p /, and, by the Casson-Rivin theorem (Section 3.3 and [5, Theorem 1.2]), ˛corresponds to the complete hyperbolic structure on X p .Thus, the triangulation p is geometric, which concludes the proof of Theorem 5.1.Corollary 5.6 The complements of the .2; 3; n/-pretzel knots and links admit geometric triangulations for n 7.
Proof Recall that the braid group has the single relation 1 2 1 D 2 1 2 and that C , which is this element's square, is central.Conjugate words in the braid group yield equivalent braid closures, so, using the relation and the centrality of C , L p , the closure of C 2 p 1 1 2 with p 1, is also the closure of As indicated in Figure 10, the closure of 3 1 2 pC6 1 2 is a pretzel.Using our definitions, 1 and 2 generate left-handed (negative) twists, so L p is a pretzel with .pC 6/, 2, and 3 half-twists, or equivalently -after rolling the p C 6 half-twists to the right -L p is the .2;3; .pC6//-pretzel,and X p is its complement.By Theorem 5.1, X p admits a geometric triangulation.But the complement of the .2; 3; pC6/-pretzel is homeomorphic (via a reflection) to X p , so it also has a geometric triangulation.
The braid relation 1 2 1 D 2 1 2 and the centrality of C D . 1 2 / 3 can also be used to show that L p is equivalent to the braid closure of In this form, L p is a T-link as defined by Birman and Kofman in [2], where they show that T-links are in one-to-one correspondence with Lorenz links. . 1 2 / 5 admit geometric triangulations for p > 1.

Extending the construction
Having constructed geometric triangulations for the braid closure of C 2 p 1 1 2 , it is natural to ask whether this construction can be extended to cover more cases.For example, hyperbolic T-links on three strands must be equivalent to closures of braids of the form C k p 1 1 2 with p > 0 and k 2, so adding more full twists to the construction would provide triangulations of all possible hyperbolic T-links on three strands.However, a useful combinatorial triangulation is not obvious.D R 3 L 2 R, so we will too.The left side of Figure 12 shows how to visualize K. / -the thickened curves -in a vertical product region.(The thinner curves will be explained later as will the right side.)Following Futer's techniques, we triangulate the product region using a sequence of layers of ideal tetrahedra where the top of one layer is identified to the bottom of the next.Adopting his notation, the layers in Figure 12 are labeled i and they each contain ideal tetrahedra T i and T 0 i .The layer i is bounded by 4-punctured spheres, below by S i and above by S i C1 , with S i C1 in i identified to S iC1 in i C1 by passing through a half-twist.Numbering the punctures 0-3 will help in describing the resulting face pairings.
The right side of Figure 12 shows the tetrahedra forming the complement of K. /.These tetrahedra also appear in the left side of the figure.The thin curves are their edges.To see how this works, examine the bottom layer, redrawn in Figure 13.The first diagram of Figure 13 (in the top left) shows an expanded version of the tetrahedra that compose 1 , one in which there are two copies of each edge in the plane of the page.This expansion makes it easier to see the triangles in the bounding punctured spheres, S 1 and S 2 , and how these triangles -faces of the tetrahedra T 1 and T 0 1live in the link complement.The coloring indicates how the triangles in the lower S 2 pillowcase are identified to those in the upper one after passing through the R half-twist (our version of [9,  The face pairings defined by passing through the L half-twist can be determined in the same manner, using, for example, the pillowcase S 4 in Figure 12.
The second diagram in Figure 13 shows a collapsed version of 1 , one consisting of two tetrahedra identified along the edges 01, 12, 23, and 30, which contains the point at infinity.T 1 is in front of the page with faces 013 and 123 on top.T 0 1 is behind the page with faces 012 and 023 on top.The next two diagrams also show T 1 and T 0 1 , using arrows to indicate edge identifications.The last one is our version of the two tetrahedra in [9, Figure 16], which can be separated to form the pair of tetrahedra on the bottom right in Figure 12.
In each layer, the expanded version of the pair of tetrahedra on the left of Figure 12 can be similarly with the versions on the right.Thus, the tetrahedra that triangulate the product region appear in our visualization and the half-twists show how to obtain the face pairings between them.
All that remains to specify a triangulation of the link complement is to show how to close off the top and bottom.This amounts to saying how the triangles in the bounding pillowcases are identified to themselves to form clasps: If starts with R, identify bottom layer triangles 4012 4013 and 4023 4123.If starts with L, identify bottom layer triangles 4012 4312 and 4023 4013.The top layer triangles are identified in the same way.For example, if ends with R, 4012 4013 and Because the strands form a clasp, identifying S 1 to itself in such a way is equivalent to attaching a 3-ball with the needed clasp removed.Doing the same at the top results in a triangulation of the 2-bridge link complement.This triangulation is the same as Sakuma and Weeks' as described by Futer and is one in which the faces of the tetrahedra are easy to see in the braid diagram, as are the face pairings induced by the half-twists.

Figure 2 :
Figure 2: The closure of C 2 p 1

Figure 3 :
Figure 3: Forming a full twist C .page and 4035 is behind the page.Their shared edge, 05, passes through the point at infinity.The colors indicate the identifications on S 0 that will yield a full twist: gray 4025 is identified to 4035.white 4023 is identified to 4523.yellow 4012 in the front is identified to 4543 in the front.pink 4012 in the back is identified to 4543 in the back.

Figure 4 :
Figure 4: A triangulation of the complement of the closure of C 2 1 1 2 .

Figure 5 shows how to replace m 2 , b 1 , 2 ; b represents 3 in b 1 and b 0 2 and 4 in m 2 ; c represents 5 in m 2 and b 1 ; and d represents 0 in b 1 and 5 in b 0 2 .
and b 0 2 with N s on the top and s on the bottom.Label the vertices of s with abcd (circled in the diagram), where a represents 3 in m 2 and 4 in b 0 The surviving faces of m 2 , b 1 , and b 0 2 are renamed as follows: m 2 .235/7 !N s.2ac/, m 2 .345/7 !s.abc/, b 1 .025/7 !N s.d 2c/, b 1 .035/7 !s.dbc/, b 0 2 .245/7 !N s.2ad /, and b 0 2 .345/7 !s.bad /.Making these substitutions and adding N s.acd / s.acd / yields the face pairings listed in Figure 5.

Figure 7 :
Figure 7: A triangulation of the complement of the closure of C 2 3 1 . They involve the faces of m 2 , b 1 , b 0 2 , and t 0 1 that survive the Pachner moves.In other words, these faces do appear in 3 .They are just relabeled by the Pachner moves.Consider first m 1 .025/b 1 .025/.Under the 3-2 move shown in Figure 5, b 1 .025/becomes N These inter-w identifications of O 3 are not affected by Pachner moves involving m 2 , b 1 , b 0

Figure 8 :
Figure8: The tetrahedra of p with OEw 0 0 .01/colored red.All but s are shown in the braid complement on the left.The dashed arrows on the right indicate that a bottom face of the originating tetrahedron is identified through a half-twist to a top face of the target tetrahedron.Among faces unpaired by half-twists, the top faces of w 0 0 are identified to the bottom faces of w p and s is glued to the top faces of w 1 and w 0 1 and the bottom faces of w p 1 and w 0 p 1 as in Table5.

Figure 9 :
Figure 9: A tetrahedron in a veering triangulation.

Corollary 4 . 2
The triangulation p admits a positive angle structure.ProofIn their main result, [12, Theorem 1.5], Hodgson, Rubinstein, Segerman and Tillmann prove that veering triangulations admit positive angle structures (which they call strict angle structures).A constructive proof showing how to deform a veering angle structure to a positive angle structure has also been given by Futer and Guéritaud[6, Theorem 1.3].

Lemma 5 . 3
If ˇ2 B satisfies both Properties 1 and 2, ˇRw i cannot equal for i D 1; : : : ; p. Lemma 5.4 If ˇ2 B satisfies both Properties 1 and 2, and either w p or s is flat, then all tetrahedra are flat, so V.ˇ/ D 0.

Corollary 5 . 5
In the braid group,

Corollary 5 . 7
The complements of the T-links/Lorenz links formed as closures of the braids pC1 1

Figure 11 :
Figure 11: The actions of R and of L on strands between pillowcases.The left side is [9, Figure 14].The right side is a version with vertical strands.

1 Figure 12 :
Figure 12: The link K. / where D R 3 L 2 R. Compare the left side to the left side of [9, Figure 18].This is also similar to the top of [19, Figure II.3.3].

Table 1 :
The reader may enjoy tracing the long and twisted journey of t 0 2 .012/ to its mate, m 1 .021/.All twelve face pairings of O 1 are listed in Table1.For reasons explained in Section 4.1, the triangulation O 1 is not quite what we want for our triangulation of X 1 .Instead, we will simplify O 1 by applying Pachner moves -also Face pairings for O 1 , an initial triangulation of X 1 .
Figure 6: A 2-0 move eliminates t 0 1 and N s, stacked as indicated, and yields face pairings for 1 , a triangulation of X 1 .

Table 2 :
Face pairings for O 3 , an initial triangulation of X 3 .

Table 3 :
Face pairings for 3 , a triangulation of X 3 .

Table 4 :
Face pairings for 2 , a triangulation of X 2 .

Table 5 :
can be applied.Define p to be the resulting Face pairings for p , a triangulation of X p for p > 1.
Algebraic & Geometric Topology, Volume 23 (2023) Corollary 4.4 Let ÃW p ! p be as in Proposition 4.3 and let k be such that p D 2k if p is even and p D 2k C 1 if p is odd.Positive angle structures Â 2 A. p / with the property that ÃÂ D Â have w 0 i Š Â w p i and thus satisfy a simpler list of edge equations: