# Geometric triangulations of a family of hyperbolic 3-braids

Barbara Nimershiem  Department of Mathematics, Franklin & Marshall College,
Lancaster, PA 17604<EMAIL_ADDRESS>
###### Abstract.

We construct topological triangulations for complements of $(-2,3,n)$-pretzel
knots and links with $n\geq 7$. Following a procedure outlined by Futer and
Guéritaud, 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.

###### 2020 Mathematics Subject Classification:

57K32

##  1\. 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, $w\in B_{n}$, 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, $\sigma_{1}$
and $\sigma_{2}$ shown in Figure 1, and one relation
$\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$. 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
$C^{k}\sigma_{1}^{p_{1}}\sigma_{2}^{-q_{1}}\cdots\sigma_{1}^{p_{s}}\sigma_{2}^{-q_{s}}$
with $k\in\mathbb{Z}$ and $p_{i}$, $q_{i}$, and $s$ positive integers, and $w$
is not conjugate to $\sigma_{1}^{p_{0}}\sigma_{2}^{q_{0}}$ 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 $k=2$, $s=1$, $p_{1}\geq 1$, and $q_{1}=1$ thus offering a direct
proof of the “if” direction of their theorem in these cases.

\begin{overpic}[unit=.25in]{B3gen.pdf} \put(1.8,0.3){\small{$\sigma_{1}$}}
\put(8.35,0.3){\small{$\sigma_{2}$}} \end{overpic} Figure 1. The generators of
$B_{3}$, the braid group on three strands.

Let $L_{p}$ be the braid closure of $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ with
$p\geq 1$, which has one component if $p$ is odd and two if $p$ is even. The
complement of $L_{p}$ is homeomorphic to the complement of a
$(-2,3,p+6)$-pretzel (Corollary 5.6), so proving
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ is hyperbolic will cover those pretzel
knots and links as well. To show $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ 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
[18]. It turns out that, as in the 2-bridge case detailed by Futer in [9,
Appendix], 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], 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
finite-volume 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], 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 $n\geq 7$ 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 structure — is presented in
Section 5 with some comments about extending the construction in Section 6.

### Acknowledgments

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 at
https://tinyurl.com/ReginaFileC2sigma1psigma2inv 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 [20].

David Futer kindly gave feedback on an early draft of Appendix A 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 picky-ness. I am
also grateful for several useful suggestions offered by the referee. That
said, any remaining errors are mine.

##  2\. Ideal triangulations

Let $X_{p}=S^{3}-L_{p}$, the complement of the closure of the braid
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$. 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 Appendix A, which contains a slight re-visualization of the geometric
triangulations of 2-bridge link complements presented by Futer in [9,
Appendix]. 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, $S^{2}\times I$, and its complement in this region is also a product,
$S\times I$, where $S$ is a 4-punctured sphere. In [9, Appendix], the product
region, $S\times I$, appears between two nested pillowcases and Futer showed
that the region can be triangulated using a sequence of layers of ideal
tetrahedra. In Appendix A, 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 $S\times I$ with a 3-ball from which the clasp has
been removed.

Analogously, we observe that the closure of
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ 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 $\sigma_{1}^{p}\sigma_{2}^{-1}$ can be replaced by
$\sigma_{1}^{p}\sigma_{4}^{-1}$ because $\sigma_{2}^{-1}$ can slide past the
central full twist $C$ on the bottom to become $\sigma_{4}^{-1}$ 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
$\sigma_{1}^{p}\sigma_{4}^{-1}$ 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.

\begin{overpic}[unit=.25in]{6braids.pdf}
\put(-0.4,2.9){\small{$\sigma_{1}^{p}\sigma_{2}^{-1}$}}
\put(7.2,2.8){\small{$\approx$}}
\put(8.3,2.9){\small{$\sigma_{1}^{p}\sigma_{4}^{-1}$}} \end{overpic} Figure 2.
The closure of $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ is the 6-braid
$\sigma_{1}^{p}\sigma_{2}^{-1}$ — or equivalently
$\sigma_{1}^{p}\sigma_{4}^{-1}$ — “closed up” with a three-stranded full twist
on both the top and the bottom.

###  2.1. 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 $\triangle 025$ is in front of the page and $\triangle 035$ 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:

$\triangle 025$ is identified to $\triangle 035$

white:

$\triangle 023$ is identified to $\triangle 523$

yellow:

$\triangle 012$ in the front is identified to $\triangle 543$ in the front

pink:

$\triangle 012$ in the back is identified to $\triangle 543$ in the back

\begin{overpic}[unit=.25in]{fulltwistbright.pdf} \put(0.57,14.5){\tiny
0}\put(1.9,14.5){\tiny 1}\put(3.23,14.5){\tiny 2}\put(4.56,14.5){\tiny
3}\put(5.89,14.5){\tiny 4}\put(7.22,14.5){\tiny 5} \par\put(10.57,12.0){\tiny
0}\put(11.24,12.0){\tiny 1}\put(11.91,12.0){\tiny 2}\put(15.88,12.0){\tiny
3}\put(16.55,12.0){\tiny 4}\put(17.22,12.0){\tiny 5} \par\put(10.57,6.0){\tiny
0}\put(11.24,6.0){\tiny 1}\put(11.91,6.0){\tiny 2}\put(15.88,6.0){\tiny
3}\put(16.55,6.0){\tiny 4}\put(17.22,6.0){\tiny 5} \par\put(10.57,0.0){\tiny
0}\put(11.24,0.0){\tiny 1}\put(11.91,0.0){\tiny 2}\put(15.88,0.0){\tiny
3}\put(16.55,0.0){\tiny 4}\put(17.22,0.0){\tiny 5} \par\put(0.57,0.0){\tiny
0}\put(1.24,0.0){\tiny 1}\put(1.91,0.0){\tiny 2}\put(5.88,0.0){\tiny
3}\put(6.55,0.0){\tiny 4}\put(7.22,0.0){\tiny 5} \par\put(0.0,15.0){\small(a)}
\put(9.9,16.5){\small(b)} \put(9.9,10.5){\small(c)} \put(9.9,4.5){\small(d)}
\put(0.0,4.5){\small(e)} \par\end{overpic} Figure 3. Forming a full twist $C$.

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 ones — will help us follow the
triangles through each step.

  * •

$\triangle 025$ is identified to $\triangle 035$: 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 $\triangle 023$ and both $\triangle 012$’s (appearing in the front). The
right side is formed by $\triangle 523$ and both $\triangle 543$’s (in the
back). This diagram is analogous to the third step in Figure 14 in the
appendix.

  * •

$\triangle 023$ is identified to $\triangle 523$: By pushing the 0 and 5
strands into the pillow until they meet, $\triangle 023$ can be identified to
$\triangle 523$ 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 $\triangle 012$’s
and the two $\triangle 543$’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 $\triangle 012$’s are identified to the $\triangle 543$’s: Bringing
strands 1 and 4 together and identifying each $\triangle 012$ to its
corresponding $\triangle 543$ 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 $\triangle 023$ is in the front
and is identified to $\triangle 523$ 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 $\tau_{p}$ 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.

###  2.2. A foundational ideal triangulation

To specify the triangulation of $X_{1}$, we will place tetrahedra in a product
region from which the 6-braid, $\sigma_{1}\sigma_{4}^{-1}$, 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 Appendix A suggests using three layers of tetrahedra (one on
either side of the two crossings). Label the top, middle, and bottom layers
$\Delta_{t}$, $\Delta_{m}$, and $\Delta_{b}$ (with $\Delta_{t}$ split into
$\Delta_{tt}$ and $\Delta_{tb}$). The six ideal tetrahedra that will
triangulate the complement are shown in Figure 4: $t_{1}^{\prime}$ and
$t_{2}^{\prime}$ in $\Delta_{t}$, $m_{1}$ and $m_{2}$ in $\Delta_{m}$, and
$b_{1}$ and $b_{2}^{\prime}$ in $\Delta_{b}$, 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 $t_{1}^{\prime}$ and $t_{2}^{\prime}$ 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
$b_{2}^{\prime}$.

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 $\hat{\tau}_{1}$.

\begin{overpic}[unit=.25in]{p=1light.pdf}\put(0.7,1.1){\tiny
0}\put(2.03,1.1){\tiny 1}\put(3.36,1.1){\tiny 2}\put(4.69,1.1){\tiny
3}\put(6.02,1.1){\tiny 4}\put(7.35,1.1){\tiny 5} \put(0.7,3.1){\tiny
0}\put(2.03,3.1){\tiny 1}\put(3.36,3.1){\tiny 2}\put(4.69,3.1){\tiny
3}\put(6.02,3.1){\tiny 4}\put(7.35,3.1){\tiny 5} \put(0.7,5.1){\tiny
0}\put(2.03,5.1){\tiny 1}\put(3.36,5.1){\tiny 2}\put(4.69,5.1){\tiny
3}\put(6.02,5.1){\tiny 4}\put(7.35,5.1){\tiny 5} \put(0.7,7.1){\tiny
0}\put(2.03,7.1){\tiny 1}\put(3.36,7.1){\tiny 2}\put(4.69,7.1){\tiny
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0}\put(2.03,9.1){\tiny 1}\put(3.36,9.1){\tiny 2}\put(4.69,9.1){\tiny
3}\put(6.02,9.1){\tiny 4}\put(7.35,9.1){\tiny 5} \put(0.7,11.1){\tiny
0}\put(2.03,11.1){\tiny 1}\put(3.36,11.1){\tiny 2}\put(4.69,11.1){\tiny
3}\put(6.02,11.1){\tiny 4}\put(7.35,11.1){\tiny 5} \put(0.7,13.1){\tiny
0}\put(2.03,13.1){\tiny 1}\put(3.36,13.1){\tiny 2}\put(4.69,13.1){\tiny
3}\put(6.02,13.1){\tiny 4}\put(7.35,13.1){\tiny 5}
\par\put(7.8,1.9){\small$\Delta_{b}=$}\put(7.8,5.9){\small$\Delta_{m}=$}\put(7.8,9.9){\small$\Delta_{tb}=$}\put(7.8,11.9){\small$\Delta_{tt}=$}
\par\put(13.1,1.9){\small$=$}\put(13.1,5.9){\small$=$}\put(13.1,9.9){\small$=$}\put(13.1,11.9){\small$=$}
\par\put(14.4,11.1){\small$t_{1}^{\prime}$}
\put(16.4,9.1){\small$t_{2}^{\prime}$}
\put(14.4,5.1){\small$m_{1}$}\put(16.4,5.1){\small$m_{2}$}
\put(14.4,1.1){\small$b_{1}$}\put(16.4,1.1){\small$b_{2}^{\prime}$}
\par\put(9.1,11.1){\tiny 0}\put(9.766,11.1){\tiny 1}\put(10.432,11.1){\tiny
2}\put(11.098,11.1){\tiny 3}\put(11.764,11.1){\tiny 4}\put(12.43,11.1){\tiny
5} \put(9.1,9.1){\tiny 0}\put(9.766,9.1){\tiny 1}\put(10.432,9.1){\tiny
2}\put(11.098,9.1){\tiny 3}\put(11.764,9.1){\tiny 4}\put(12.43,9.1){\tiny 5}
\put(9.1,7.1){\tiny 0}\put(9.766,7.1){\tiny 1}\put(10.432,7.1){\tiny
2}\put(11.098,7.1){\tiny 3}\put(11.764,7.1){\tiny 4}\put(12.43,7.1){\tiny 5}
\put(9.1,2.43){\tiny 0}\put(9.766,2.43){\tiny 1}\put(10.432,2.43){\tiny
2}\put(11.098,2.43){\tiny 3}\put(11.764,2.43){\tiny 4}\put(12.43,2.43){\tiny
5} \par\put(13.75,13.1){\tiny 0}\put(15.35,13.1){\tiny 5}
\put(13.75,11.4){\tiny 2}\put(15.35,11.4){\tiny 3} \par\put(15.75,11.1){\tiny
0}\put(17.35,11.1){\tiny 5} \put(15.75,9.4){\tiny 1}\put(17.35,9.4){\tiny 2}
\par\put(13.75,7.1){\tiny 1}\put(15.35,7.1){\tiny 2}\put(15.75,7.1){\tiny
3}\put(17.35,7.1){\tiny 4} \put(13.75,5.4){\tiny 0}\put(15.35,5.4){\tiny
5}\put(15.75,5.4){\tiny 2}\put(17.35,5.4){\tiny 5} \par\put(13.75,3.1){\tiny
2}\put(15.35,3.1){\tiny 3}\put(15.75,3.1){\tiny 2}\put(17.35,3.1){\tiny 5}
\put(13.75,1.4){\tiny 0}\put(15.35,1.4){\tiny 5}\put(15.75,1.4){\tiny
3}\put(17.35,1.4){\tiny 4} \par\par\end{overpic} Figure 4. A triangulation of
the complement of the closure of $C^{2}\sigma_{1}\sigma_{2}^{-1}$.

When comparing to Appendix A, the reader should be concerned that the pairs of
tetrahedra do not fill the entire layer. In $\Delta_{m}$, 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
$t_{1}^{\prime}$ can slide down the braid through $\Delta_{tb}$ and
$\Delta_{m}$ and past the $\sigma_{4}^{-1}$ half-twist to be identified with
the 245 face of $b_{2}^{\prime}$. We denote such faces and face pairings by:

$t_{1}^{\prime}(235)\sim b_{2}^{\prime}(245),$

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 afore-mentioned additional
gray segments and show that the back 235 region of $\Delta_{m}$ is indeed
covered.

As indicated in Figure 4, $t_{1}^{\prime}$ sits on top of $t_{2}^{\prime}$
with their 025 faces identified. The remaining faces on the top of the
tetrahedra in $\Delta_{t}$ and those on the bottom of the tetrahedra in
$\Delta_{b}$ are triangles appearing in the previous section (2.1) and are
identified by the full-twist identifications described there. For example, the
face $b_{1}(023)$ on the bottom of $\Delta_{b}$ is the triangle $\triangle
023$, which is identified to $\triangle 523$, which is the 523 face of
$b_{2}^{\prime}$, another face on the bottom of $\Delta_{b}$, so
$b_{1}(023)\sim b_{2}^{\prime}(523)$. 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 $t_{2}^{\prime}$. After traveling through the half-twist that
interchanges 0 and 1, $t_{2}^{\prime}(125)$ becomes $\triangle 025$ on the top
of $\Delta_{m}$ still in the back. Then it can travel through $\Delta_{m}$ and
$\Delta_{b}$ all the way to $\triangle 025$ on the bottom of $\Delta_{b}$ in
the back. This triangle is identified to $\triangle 035$ in the front, i.e.,
the 035 face of $b_{1}$, so $t_{2}^{\prime}(125)\sim b_{1}(035)$.

The remaining face pairings of $\hat{\tau}_{1}$ 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. The reader may enjoy
tracing the long and twisted journey of $t_{2}^{\prime}(012)$ to its mate,
$m_{1}(021)$. All twelve face pairings of $\hat{\tau}_{1}$ are listed in Table
1.

Table 1. Face pairings for $\hat{\tau}_{1}$, an initial triangulation of
$X_{1}$.
$t_{1}^{\prime}(023)$ | $\sim$ | $m_{2}(523)$
---|---|---
$t_{1}^{\prime}(025)$ | $\sim$ | $t_{2}^{\prime}(025)$
$t_{1}^{\prime}(035)$ | $\sim$ | $m_{1}(125)$
$t_{1}^{\prime}(235)$ | $\sim$ | $b_{2}^{\prime}(245)$
$t_{2}^{\prime}(015)$ | $\sim$ | $m_{1}(105)$
$t_{2}^{\prime}(012)$ | $\sim$ | $m_{1}(021)$
$t_{2}^{\prime}(125)$ | $\sim$ | $b_{1}(035)$
$m_{1}(025)$ | $\sim$ | $b_{1}(025)$
$m_{2}(345)$ | $\sim$ | $b_{2}^{\prime}(354)$
$m_{2}(234)$ | $\sim$ | $b_{2}^{\prime}(243)$
$m_{2}(245)$ | $\sim$ | $b_{1}(235)$
$b_{1}(023)$ | $\sim$ | $b_{2}^{\prime}(523)$

For reasons explained in 4.1, the triangulation $\hat{\tau}_{1}$ is not quite
what we want for our triangulation of $X_{1}$. Instead, we will simplify
$\hat{\tau}_{1}$ by applying Pachner moves (also known as bistellar flips,
introduced in [15]). Table 1 shows the following identifications of the 24
edge of $m_{2}$:

$24\text{ in }m_{2}\sim 23\text{ in }b_{1}\sim 23\text{ in
}b_{2}^{\prime}\sim_{\text{back to}}24\text{ in }m_{2}.$

Whenever three tetrahedra surround an edge, we can perform a 3-2 Pachner move
that replaces them with two tetrahedra sharing a face. Figure 5 shows how to
replace $m_{2}$, $b_{1}$, and $b_{2}^{\prime}$ with $\bar{s}$ 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 $b_{2}^{\prime}$; $b$
represents 3 in $b_{1}$ and $b_{2}^{\prime}$ and 4 in $m_{2}$; $c$ represents
5 in $m_{2}$ and $b_{1}$; and $d$ represents 0 in $b_{1}$ and 5 in
$b_{2}^{\prime}$. The surviving faces of $m_{2}$, $b_{1}$, and
$b_{2}^{\prime}$ are renamed as follows: $m_{2}(235)\mapsto\bar{s}(2ac)$,
$m_{2}(345)\mapsto s(abc)$, $b_{1}(025)\mapsto\bar{s}(d2c)$,
$b_{1}(035)\mapsto s(dbc)$, $b_{2}^{\prime}(245)\mapsto\bar{s}(2ad)$, and
$b_{2}^{\prime}(345)\mapsto s(bad)$. Making these substitutions and adding
$\bar{s}(acd)\sim s(acd)$ yields the face pairings listed in Figure 5.

\begin{overpic}[unit=.25in]{3-2move.pdf}
\put(1.0,1.7){\small$s$}\put(0.9,5.0){\small$\bar{s}$}
\put(1.8,3.2){\small$m_{2}$}
\put(3.2,4.2){\begin{turn}{60.0}\small$b_{2}^{\prime}$\end{turn}}
\put(1.6,4.0){\begin{turn}{40.0}\reflectbox{\small$b_{1}$}\end{turn}}
\par\put(4.2,3.0){\tiny$a$}\put(2.2,0.15){\tiny$b$}\put(0.15,3.0){\tiny$c$}\put(3.6,5.73){\tiny$d$}
\par\put(2.2,6.45){\tiny 2} \par\put(2.0,0.6){\begin{turn}{-55.0}\tiny
4\end{turn}} \put(2.3,0.4){\begin{turn}{60.0}\tiny 4\end{turn}}
\put(1.89,0.85){\begin{turn}{-55.0}\tiny 3\end{turn}}
\put(2.43,0.65){\begin{turn}{60.0}\tiny 3\end{turn}}
\par\put(3.9,3.25){\begin{turn}{60.0}\tiny 4\end{turn}}
\put(3.8,2.95){\begin{turn}{60.0}\tiny 3\end{turn}}
\par\put(0.4,3.4){\begin{turn}{-55.0}\tiny 5\end{turn}}
\put(0.53,3.1){\begin{turn}{-55.0}\tiny 5\end{turn}}
\par\put(3.2,5.6){\begin{turn}{-215.0}\tiny 0\end{turn}}
\put(3.4,5.45){\begin{turn}{-215.0}\tiny 5\end{turn}} \par\par\end{overpic}

$t_{1}^{\prime}(023)$ | $\sim$ | $\bar{s}(c2a)$
---|---|---
$t_{1}^{\prime}(025)$ | $\sim$ | $t_{2}^{\prime}(025)$
$t_{1}^{\prime}(035)$ | $\sim$ | $m_{1}(125)$
$t_{1}^{\prime}(235)$ | $\sim$ | $\bar{s}(2ad)$
$t_{2}^{\prime}(015)$ | $\sim$ | $m_{1}(105)$
$t_{2}^{\prime}(012)$ | $\sim$ | $m_{1}(021)$
$t_{2}^{\prime}(125)$ | $\sim$ | $s(dbc)$
$m_{1}(025)$ | $\sim$ | $\bar{s}(d2c)$
$s(abc)$ | $\sim$ | $s(bda)$
$\bar{s}(acd)$ | $\sim$ | $s(acd)$

Figure 5. A 3-2 move eliminates the 24 edge of $m_{2}$, which is also the 23
edge in $b_{1}$ and $b_{2}^{\prime}$, and yields the indicated face pairings.

Next, observe that the 23 edge of $t_{1}^{\prime}$ is identified to the $2a$
edge of $\bar{s}$, which is taken back to the 23 edge of $t_{1}^{\prime}$.
This very short cycle of edges means $t_{1}^{\prime}$ and $\bar{s}$ are
identified across an edge, so we can perform a 2-0 Pachner move to collapse
$t_{1}^{\prime}$ and $\bar{s}$. This move identifies $t_{1}^{\prime}(025)$
with $\bar{s}(c2d)$ and $t_{1}^{\prime}(035)$ with $\bar{s}(cad)$. See Figure
6, which shows how $t_{1}^{\prime}$ and $\bar{s}$ are stacked before they are
collapsed and eliminated. The 2-0 move results in a triangulation of $X_{1}$
comprising three tetrahedra — $t_{2}^{\prime},$ $m_{1},$ and $s$ — whose face
pairings are in Figure 6. Define $\tau_{1}$ to be this triangulation.

\begin{overpic}[unit=.25in]{2-0move.pdf}
\put(0.6,1.2){\small$t_{1}^{\prime}$}\put(5.0,1.2){\small$\bar{s}$}
\par\put(1.75,5.7){\tiny 5}\put(6.15,5.7){\tiny$d$} \put(0.0,2.9){\tiny
2}\put(3.4,2.9){\tiny 3}\put(4.4,2.9){\tiny 2}\put(7.8,2.9){\tiny$a$}
\put(1.75,0.1){\tiny 0}\put(6.15,0.1){\tiny$c$} \par\end{overpic}

$t_{2}^{\prime}(015)$ | $\sim$ | $m_{1}(105)$
---|---|---
$t_{2}^{\prime}(012)$ | $\sim$ | $m_{1}(021)$
$t_{2}^{\prime}(025)$ | $\sim$ | $m_{1}(520)$
$t_{2}^{\prime}(125)$ | $\sim$ | $s(dbc)$
$m_{1}(125)$ | $\sim$ | $s(cad)$
$s(abc)$ | $\sim$ | $s(bda)$

Figure 6. A 2-0 move eliminates $t_{1}^{\prime}$ and $\bar{s}$, stacked as
indicated, and yields face pairings for $\tau_{1}$, a triangulation of
$X_{1}$.

###  2.3. Adding layers to extend to $p=2$ and $3$

It will be more instructive to first extend the construction of the previous
section to $p=3$. To triangulate the complement of the closure of
$C^{2}\sigma_{1}^{3}\sigma_{2}^{-1}$, where there are four half-twists in the
braid, begin with five layers: $\Delta_{t}$, $\Delta_{m}$, and $\Delta_{b}$
and two layers inserted between $\Delta_{t}$ and $\Delta_{m}$, labelled
$\Delta_{w1}$ and $\Delta_{w2}$. Place two tetrahedra in each layer as
indicated in Figure 7. Though $\Delta_{t}$ is depicted slightly differently
(as one layer with the shared face of $t_{1}^{\prime}$ and $t_{2}^{\prime}$,
025, positioned in the middle), all tetrahedra in $\Delta_{t}$, $\Delta_{m}$,
and $\Delta_{b}$ are named and situated exactly as in Figure 4. The new
tetrahedra, those in the $\Delta_{wi}$ layers, will be called $w_{i}$ and
$w_{i}^{\prime}$, where a prime still indicates the tetrahedron is behind the
plane of the page.

\begin{overpic}[unit=.25in]{p=3.pdf} \put(0.7,1.1){\tiny
0}\put(2.03,1.1){\tiny 1}\put(3.36,1.1){\tiny 2}\put(4.69,1.1){\tiny
3}\put(6.02,1.1){\tiny 4}\put(7.35,1.1){\tiny 5} \put(0.7,3.1){\tiny
0}\put(2.03,3.1){\tiny 1}\put(3.36,3.1){\tiny 2}\put(4.69,3.1){\tiny
3}\put(6.02,3.1){\tiny 4}\put(7.35,3.1){\tiny 5} \put(0.7,5.1){\tiny
0}\put(2.03,5.1){\tiny 1}\put(3.36,5.1){\tiny 2}\put(4.69,5.1){\tiny
3}\put(6.02,5.1){\tiny 4}\put(7.35,5.1){\tiny 5} \put(0.7,7.1){\tiny
0}\put(2.03,7.1){\tiny 1}\put(3.36,7.1){\tiny 2}\put(4.69,7.1){\tiny
3}\put(6.02,7.1){\tiny 4}\put(7.35,7.1){\tiny 5} \put(0.7,9.1){\tiny
0}\put(2.03,9.1){\tiny 1}\put(3.36,9.1){\tiny 2}\put(4.69,9.1){\tiny
3}\put(6.02,9.1){\tiny 4}\put(7.35,9.1){\tiny 5} \put(0.7,11.1){\tiny
0}\put(2.03,11.1){\tiny 1}\put(3.36,11.1){\tiny 2}\put(4.69,11.1){\tiny
3}\put(6.02,11.1){\tiny 4}\put(7.35,11.1){\tiny 5} \put(0.7,13.1){\tiny
0}\put(2.03,13.1){\tiny 1}\put(3.36,13.1){\tiny 2}\put(4.69,13.1){\tiny
3}\put(6.02,13.1){\tiny 4}\put(7.35,13.1){\tiny 5} \put(0.7,15.1){\tiny
0}\put(2.03,15.1){\tiny 1}\put(3.36,15.1){\tiny 2}\put(4.69,15.1){\tiny
3}\put(6.02,15.1){\tiny 4}\put(7.35,15.1){\tiny 5} \put(0.7,17.1){\tiny
0}\put(2.03,17.1){\tiny 1}\put(3.36,17.1){\tiny 2}\put(4.69,17.1){\tiny
3}\put(6.02,17.1){\tiny 4}\put(7.35,17.1){\tiny 5} \put(0.7,19.1){\tiny
0}\put(2.03,19.1){\tiny 1}\put(3.36,19.1){\tiny 2}\put(4.69,19.1){\tiny
3}\put(6.02,19.1){\tiny 4}\put(7.35,19.1){\tiny 5}
\par\put(8.0,2.0){\small$\Delta_{b}=$}\put(8.0,6.0){\small$\Delta_{m}=$}\put(8.0,10.0){\small$\Delta_{w2}=$}\put(8.0,14.0){\small$\Delta_{w1}=$}\put(8.0,18.0){\small$\Delta_{t}=$}
\par\put(10.4,17.1){\small$t_{1}^{\prime}$}\put(12.4,17.1){\small$t_{2}^{\prime}$}
\put(10.4,13.1){\small$w_{1}$}\put(12.4,13.1){\small$w_{1}^{\prime}$}
\put(10.4,9.1){\small$w_{2}$}\put(12.4,9.1){\small$w_{2}^{\prime}$}
\put(10.4,5.1){\small$m_{1}$}\put(12.4,5.1){\small$m_{2}$}
\put(10.4,1.1){\small$b_{1}$}\put(12.4,1.1){\small$b_{2}^{\prime}$}
\par\put(9.75,19.1){\tiny 0}\put(11.35,19.1){\tiny 5}\put(11.75,19.1){\tiny
0}\put(13.35,19.1){\tiny 5} \put(9.75,17.4){\tiny 2}\put(11.35,17.4){\tiny
3}\put(11.75,17.4){\tiny 1}\put(13.35,17.4){\tiny 2} \par\put(9.75,15.1){\tiny
1}\put(11.35,15.1){\tiny 2}\put(11.75,15.1){\tiny 0}\put(13.35,15.1){\tiny 5}
\put(9.75,13.4){\tiny 0}\put(11.35,13.4){\tiny 5}\put(11.75,13.4){\tiny
1}\put(13.35,13.4){\tiny 2} \par\put(9.75,11.1){\tiny 1}\put(11.35,11.1){\tiny
2}\put(11.75,11.1){\tiny 0}\put(13.35,11.1){\tiny 5} \put(9.75,9.4){\tiny
0}\put(11.35,9.4){\tiny 5}\put(11.75,9.4){\tiny 1}\put(13.35,9.4){\tiny 2}
\par\put(9.75,7.1){\tiny 1}\put(11.35,7.1){\tiny 2}\put(11.75,7.1){\tiny
3}\put(13.35,7.1){\tiny 4} \put(9.75,5.4){\tiny 0}\put(11.35,5.4){\tiny
5}\put(11.75,5.4){\tiny 2}\put(13.35,5.4){\tiny 5} \par\put(9.75,3.1){\tiny
2}\put(11.35,3.1){\tiny 3}\put(11.75,3.1){\tiny 2}\put(13.35,3.1){\tiny 5}
\put(9.75,1.4){\tiny 0}\put(11.35,1.4){\tiny 5}\put(11.75,1.4){\tiny
3}\put(13.35,1.4){\tiny 4} \par\par\end{overpic} Figure 7. A triangulation of
the complement of the closure of $C^{2}\sigma_{1}^{3}\sigma_{2}^{-1}$, with
$\Delta_{t}$ depicted as one layer with the shared face of $t_{1}^{\prime}$
and $t_{2}^{\prime}$, 025, positioned in the middle.

The face pairings are determined in the same manner as they were for $p=1$,
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, $t_{1}^{\prime}(035)$ is
identified to $\triangle 025$ in the front, which, after passing through the
half-twist, becomes $\triangle 125$, a top face of $w_{1}$, so
$t_{1}^{\prime}(035)$ hits $w_{1}$ before it gets to $m_{1}$. We will call
this initial triangulation $\hat{\tau}_{3}$, and, as in the $p=1$ case, will
perform simplifying Pachner moves on $\hat{\tau}_{3}$ to create $\tau_{3}$.
The shadings in Table 2 will help us determine the resulting identifications.

Table 2. Face pairings for $\hat{\tau}_{3}$, an initial triangulation of
$X_{3}$.

$t_{2}^{\prime}(015)$ | $\sim$ | $w_{1}(105)$
---|---|---
$t_{2}^{\prime}(125)$ | $\sim$ | $w_{1}^{\prime}(025)$
$w_{1}(025)$ | $\sim$ | $w_{2}(125)$
$w_{1}(012)$ | $\sim$ | $w_{2}^{\prime}(102)$
$w_{1}^{\prime}(015)$ | $\sim$ | $w_{2}(105)$
$w_{1}^{\prime}(125)$ | $\sim$ | $w_{2}^{\prime}(025)$
$w_{2}(025)$ | $\sim$ | $m_{1}(125)$
$w_{2}^{\prime}(015)$ | $\sim$ | $m_{1}(105)$
$m_{1}(025)$ | $\sim$ | $b_{1}(025)$
$m_{2}(234)$ | $\sim$ | $b_{2}^{\prime}(243)$
$m_{2}(245)$ | $\sim$ | $b_{1}(235)$

$t_{1}^{\prime}(023)$ | $\sim$ | $m_{2}(523)$
---|---|---
$t_{1}^{\prime}(025)$ | $\sim$ | $t_{2}^{\prime}(025)$
$t_{1}^{\prime}(235)$ | $\sim$ | $b_{2}^{\prime}(245)$
$t_{2}^{\prime}(012)$ | $\sim$ | $m_{1}(021)$
$b_{1}(023)$ | $\sim$ | $b_{2}^{\prime}(523)$

$t_{1}^{\prime}(035)$ | $\sim$ | $w_{1}(125)$
---|---|---
$m_{2}(345)$ | $\sim$ | $w_{1}^{\prime}(201)$
$b_{1}(035)$ | $\sim$ | $w_{2}^{\prime}(125)$
$b_{2}^{\prime}(345)$ | $\sim$ | $w_{2}(201)$

Again, three tetrahedra in $\hat{\tau}_{3}$ surround the 24 edge of $m_{2}$
(which is identified to the 23 edges of $b_{1}$ and $b_{2}^{\prime}$),
allowing for the same 3-2 move shown in Figure 5. Thus $m_{2}$, $b_{1}$, and
$b_{2}^{\prime}$ are replaced by $\bar{s}$ and $s$. As before, we can then
perform a 2-0 move collapsing $t_{1}^{\prime}$ and $\bar{s}$ because they are
identified across the edge $23=2a$. Again, $t_{1}^{\prime}(025)$ and
$t_{1}^{\prime}(035)$ are identified with $\bar{s}(c2d)$ and $\bar{s}(cad)$ as
in Figure 6. Let $\tau_{3}$ be the resulting triangulation of $X_{3}$,
consisting of the seven remaining tetrahedra: $t_{2}^{\prime}$, $w_{1}$,
$w_{1}^{\prime}$, $w_{2}$, $w_{2}^{\prime}$, $m_{1}$, and $s$.

The face pairings in the darkest cells in Table 2 involve only the four
replaced or collapsed tetrahedra ($m_{2}$, $b_{1}$, $b_{2}^{\prime}$,
$t_{1}^{\prime}$), so none of them will survive the Pachner moves defining
$\tau_{3}$. 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 $\tau_{3}$ come from those in the light gray
cells in Table 2. They involve the faces of $m_{2}$, $b_{1}$,
$b_{2}^{\prime}$, and $t_{1}^{\prime}$ that survive the Pachner moves. In
other words, these faces do appear in $\tau_{3}$. They are just relabelled by
the Pachner moves. Consider first $m_{1}(025)\sim b_{1}(025)$. Under the 3-2
move shown in Figure 5, $b_{1}(025)$ becomes $\bar{s}(d2c)$, which is
identified by the 2-0 move in Figure 6 to $t_{1}^{\prime}(520)$. This face is
paired with $t_{2}^{\prime}(520)$, so, after the two Pachner moves,
$m_{1}(025)$ is identified with $t_{2}^{\prime}(520)$, which we choose to
write as $t_{2}^{\prime}(025)\sim m_{1}(520)$.

The final four faces to be relabelled — $t_{1}^{\prime}(035)$, $m_{2}(345)$,
$b_{1}(035)$, and $b_{2}^{\prime}(345)$ — are all faces that belong to the new
tetrahedra $s$. More specifically, the 2-0 move identifies
$t_{1}^{\prime}(035)$ with $\bar{s}(cad)$, which is the new face introduced by
the 3-2 move, the face shared by $\bar{s}$ and $s$. Thus, after the two
Pachner moves, $t_{1}^{\prime}(035)$ 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
$\tau_{3}$, 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.

Table 3. Face pairings for $\tau_{3}$, a triangulation of $X_{3}$.

$t_{2}^{\prime}(015)$ | $\sim$ | $w_{1}(105)$
---|---|---
$t_{2}^{\prime}(125)$ | $\sim$ | $w_{1}^{\prime}(025)$
$w_{1}(025)$ | $\sim$ | $w_{2}(125)$
$w_{1}(012)$ | $\sim$ | $w_{2}^{\prime}(102)$
$w_{1}^{\prime}(015)$ | $\sim$ | $w_{2}(105)$
$w_{1}^{\prime}(125)$ | $\sim$ | $w_{2}^{\prime}(025)$
$w_{2}(025)$ | $\sim$ | $m_{1}(125)$
$w_{2}^{\prime}(015)$ | $\sim$ | $m_{1}(105)$

$t_{2}^{\prime}(012)$ | $\sim$ | $m_{1}(021)$
---|---|---
$t_{2}^{\prime}(025)$ | $\sim$ | $m_{1}(520)$

$s(cad)$ | $\sim$ | $w_{1}(125)$
---|---|---
$s(abc)$ | $\sim$ | $w_{1}^{\prime}(201)$
$s(dbc)$ | $\sim$ | $w_{2}^{\prime}(125)$
$s(bad)$ | $\sim$ | $w_{2}(201)$

The reader likely anticipates that we will create the desired triangulation of
$X_{p}$ in two steps: First form an initial triangulation, $\hat{\tau}_{p}$,
consisting of the tetrahedra in Figure 4 together with some $w$-layers between
$\Delta_{t}$ and $\Delta_{m}$ 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 $p=3$ construction and address the $p=2$
case.

Figure 7 shows that $t_{2}^{\prime}$ is situated and labelled in exactly the
same way as $w_{1}^{\prime}$ and $w_{2}^{\prime}$, so the labels on the bottom
faces of $t_{2}^{\prime}$, 015 and 125, are the same as those on
$w_{1}^{\prime}$ and $w_{2}^{\prime}$. Furthermore, as the face
$t_{2}^{\prime}(015)$ slides through the half-twist, it rotates to the front,
matching with $w_{1}(105)$, just as the 015 face of $w_{1}^{\prime}$ matches
with $w_{2}(105)$. Similarly, as the face $t_{2}^{\prime}(125)$ slides through
the half-twist, it stays in the back, matching with $w_{1}^{\prime}(025)$,
just as the the 125 face of $w_{1}^{\prime}$ matches with
$w_{2}^{\prime}(025)$. In these identifications, $t_{2}^{\prime}$ is acting as
a tetrahedron labelled $w_{0}^{\prime}$ 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 labelled
$w_{3}$ would.

Using this relabelling, we make several observations about the face pairings
in $\tau_{3}$.

  * •

By relabelling $t_{2}^{\prime}$ as $w_{0}^{\prime}$ and $m_{1}$ as $w_{3}$,
the unshaded face pairings in the first column of Table 2 can be summarized as
follows:

$w_{i}^{\prime}(015)$ | $\sim$ | $w_{i+1}(105)$ | for $i=0,1,2$;
---|---|---|---
$w_{i}^{\prime}(125)$ | $\sim$ | $w_{i+1}^{\prime}(025)$ | for $i=0,1$;
$w_{i}(025)$ | $\sim$ | $w_{i+1}(125)$ | for $i=1,2$; and
$w_{i}(012)$ | $\sim$ | $w_{i+1}^{\prime}(102)$ | for $i=1$.

These inter-$w$ identifications of $\hat{\tau}_{3}$ are not affected by
Pachner moves involving $m_{2}$, $b_{1}$, $b_{2}^{\prime}$, and
$t_{1}^{\prime}$, so they are also face pairings of $\tau_{3}$, as shown in
the first column of Table 3.

  * •

The top faces of $w_{0}^{\prime}$ 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,
$w_{0}^{\prime}(012)\sim w_{3}(021)$ and $w_{0}^{\prime}(025)\sim w_{3}(520)$.
Both of these pairings also occur in $\tau_{1}$ (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 $w_{1}^{\prime}(201)$, and the other
two are identified to bottom faces in the last $w$-layer, $w_{2}(201)$ and
$w_{2}^{\prime}(125)$.

Properly interpreted, these observations also apply when $p=2$. Let the
triangulation $\tau_{2}$ of $X_{2}$, the complement of the closure of
$C^{2}\sigma_{1}^{2}\sigma_{2}^{-1}$, consist of $w_{0}^{\prime}$, $w_{1}$,
$w_{1}^{\prime}$, $w_{2}$, 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 $\Delta_{w1}$.
Harvesting the remaining face pairings from the first two observations, we
obtain the face pairings for $\tau_{2}$ given in Table 4.

Table 4. Face pairings for $\tau_{2}$, a triangulation of $X_{2}$.

$w_{0}^{\prime}(015)$ | $\sim$ | $w_{1}(105)$
---|---|---
$w_{0}^{\prime}(125)$ | $\sim$ | $w_{1}^{\prime}(025)$
$w_{1}(025)$ | $\sim$ | $w_{2}(125)$
$w_{1}^{\prime}(015)$ | $\sim$ | $w_{2}(105)$

$w_{0}^{\prime}(012)$ | $\sim$ | $w_{2}(021)$
---|---|---
$w_{0}^{\prime}(025)$ | $\sim$ | $w_{2}(520)$

$s(cad)$ | $\sim$ | $w_{1}(125)$
---|---|---
$s(abc)$ | $\sim$ | $w_{1}^{\prime}(201)$
$s(dbc)$ | $\sim$ | $w_{1}^{\prime}(125)$
$s(bad)$ | $\sim$ | $w_{1}(201)$

###  2.4. An ideal triangulation for any $p$

Guided by the previous examples along with Figure 7, form an initial
triangulation, $\hat{\tau}_{p}$, of $X_{p}$, the complement of the closure of
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$, that consists of $2p+4$ tetrahedra:
$t_{1}^{\prime}$ and $w_{0}^{\prime}$ (née $t_{2}^{\prime}$) in $\Delta_{t}$;
$w_{p}$ (née $m_{1}$) and $m_{2}$ in $\Delta_{m}$; $b_{1}$ and
$b_{2}^{\prime}$ in $\Delta_{b}$; and $w_{i}$ and $w_{i}^{\prime}$ in
$\Delta_{wi}$ where $i=1,\dots,p-1$. The face pairings of $\hat{\tau}_{p}$ 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
2.2 and 2.3. As before, the addition of the $w$-layers does not affect
identifications amongst $m_{2}$, $b_{1}$, $b_{2}^{\prime}$, and
$t_{1}^{\prime}$, so the Pachner moves depicted in Figures 5 and 6 can be
applied. Define $\tau_{p}$ to be the resulting triangulation. As with
$\tau_{2}$, the observations about $\tau_{3}$ from the previous section apply
to $\tau_{p}$, yielding the face pairings listed in Table 5 for $p>1$. (For
$p=1$, see Figure 6.)

Table 5. Face pairings for $\tau_{p}$, a triangulation of $X_{p}$ for $p>1$. $w_{i}^{\prime}(015)$ | $\sim$ | $w_{i+1}(105)$ | $i=0,\dots,p-1$
---|---|---|---
$w_{i}^{\prime}(125)$ | $\sim$ | $w_{i+1}^{\prime}(025)$ | $i=0,\dots,p-2$
$w_{i}(025)$ | $\sim$ | $w_{i+1}(125)$ | $i=1,\dots,p-1$
$w_{i}(012)$ | $\sim$ | $w_{i+1}^{\prime}(102)$ | $i=1,\dots,p-2$
$w_{0}^{\prime}(012)$ | $\sim$ | $w_{p}(021)$ |
$w_{0}^{\prime}(025)$ | $\sim$ | $w_{p}(520)$ |
$s(cad)$ | $\sim$ | $w_{1}(125)$ |
$s(abc)$ | $\sim$ | $w_{1}^{\prime}(201)$ |
$s(dbc)$ | $\sim$ | $w_{p-1}^{\prime}(125)$ |
$s(bad)$ | $\sim$ | $w_{p-1}(201)$ |

The triangulation $\hat{\tau}_{p}$ had $2p+4$ 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 $\tau_{p}$ has $2p+1$ tetrahedra and thus $4p+2$ face
pairings. There are exactly $4p+2$ identifications in Table 5 (and six in
Figure 6). Thus, the definition of the triangulation $\tau_{p}$ is complete.

Note that most of the tetrahedra in $\tau_{p}$ (all but $s$) are visible in
the complement of the 6-braid $\sigma_{1}^{p}\sigma_{4}^{-1}$. Using this
visualization can help us quickly deduce relationships between the tetrahedra.
For example, Figure 8 shows how $w_{0}^{\prime}$, $w_{1}$, $w_{1}^{\prime}$,
$\dots$, $w_{p-1}$, $w_{p-1}^{\prime}$, and $w_{p}$ are situated in the braid
complement, and all but six face pairings of $\tau_{p}$ can be derived
directly from following their faces through the half-twists in the braid.

\begin{overpic}[unit=.25in]{anypRev.pdf} \put(0.7,1.1){\tiny
0}\put(2.03,1.1){\tiny 1}\put(3.36,1.1){\tiny 2}\put(4.69,1.1){\tiny
3}\put(6.02,1.1){\tiny 4}\put(7.35,1.1){\tiny 5} \put(0.7,3.1){\tiny
0}\put(2.03,3.1){\tiny 1}\put(3.36,3.1){\tiny 2}\put(4.69,3.1){\tiny
3}\put(6.02,3.1){\tiny 4}\put(7.35,3.1){\tiny 5} \put(0.7,5.1){\tiny
0}\put(2.03,5.1){\tiny 1}\put(3.36,5.1){\tiny 2}\put(4.69,5.1){\tiny
3}\put(6.02,5.1){\tiny 4}\put(7.35,5.1){\tiny 5} \put(0.7,7.1){\tiny
0}\put(2.03,7.1){\tiny 1}\put(3.36,7.1){\tiny 2}\put(4.69,7.1){\tiny
3}\put(6.02,7.1){\tiny 4}\put(7.35,7.1){\tiny 5} \put(0.7,9.1){\tiny
0}\put(2.03,9.1){\tiny 1}\put(3.36,9.1){\tiny 2}\put(4.69,9.1){\tiny
3}\put(6.02,9.1){\tiny 4}\put(7.35,9.1){\tiny 5} \put(0.7,11.1){\tiny
0}\put(2.03,11.1){\tiny 1}\put(3.36,11.1){\tiny 2}\put(4.69,11.1){\tiny
3}\put(6.02,11.1){\tiny 4}\put(7.35,11.1){\tiny 5} \put(0.7,13.1){\tiny
0}\put(2.03,13.1){\tiny 1}\put(3.36,13.1){\tiny 2}\put(4.69,13.1){\tiny
3}\put(6.02,13.1){\tiny 4}\put(7.35,13.1){\tiny 5} \put(0.7,15.1){\tiny
0}\put(2.03,15.1){\tiny 1}\put(3.36,15.1){\tiny 2}\put(4.69,15.1){\tiny
3}\put(6.02,15.1){\tiny 4}\put(7.35,15.1){\tiny 5} \put(0.7,17.1){\tiny
0}\put(2.03,17.1){\tiny 1}\put(3.36,17.1){\tiny 2}\put(4.69,17.1){\tiny
3}\put(6.02,17.1){\tiny 4}\put(7.35,17.1){\tiny 5} \put(0.7,19.1){\tiny
0}\put(2.03,19.1){\tiny 1}\put(3.36,19.1){\tiny 2}\put(4.69,19.1){\tiny
3}\put(6.02,19.1){\tiny 4}\put(7.35,19.1){\tiny 5}
\par\put(12.4,17.1){\small$w_{0}^{\prime}$}
\put(10.4,13.1){\small$w_{1}$}\put(12.4,13.1){\small$w_{1}^{\prime}$}
\put(10.4,9.1){\small$w_{2}$}\put(12.4,9.1){\small$w_{2}^{\prime}$}\put(14.5,9.1){\small$s$}
\put(10.4,5.1){\small$w_{p-1}$}\put(12.4,5.1){\small$w_{p-1}^{\prime}$}
\put(10.4,1.1){\small$w_{p}$} \par\put(11.75,19.1){\tiny
0}\put(13.35,19.1){\tiny 5} \put(11.75,17.4){\tiny 1}\put(13.35,17.4){\tiny 2}
\par\put(9.75,15.1){\tiny 1}\put(11.35,15.1){\tiny 2}\put(11.75,15.1){\tiny
0}\put(13.35,15.1){\tiny 5} \put(9.75,13.4){\tiny 0}\put(11.35,13.4){\tiny
5}\put(11.75,13.4){\tiny 1}\put(13.35,13.4){\tiny 2} \par\put(9.75,11.1){\tiny
1}\put(11.35,11.1){\tiny 2}\put(11.75,11.1){\tiny 0}\put(13.35,11.1){\tiny
5}\put(13.75,11.1){\tiny d}\put(15.35,11.1){\tiny c } \put(9.75,9.4){\tiny
0}\put(11.35,9.4){\tiny 5}\put(11.75,9.4){\tiny 1}\put(13.35,9.4){\tiny
2}\put(13.75,9.4){\tiny a}\put(15.35,9.4){\tiny b} \par\put(9.75,7.1){\tiny
1}\put(11.35,7.1){\tiny 2}\put(11.75,7.1){\tiny 0}\put(13.35,7.1){\tiny 5}
\put(9.75,5.4){\tiny 0}\put(11.35,5.4){\tiny 5}\put(11.75,5.4){\tiny
1}\put(13.35,5.4){\tiny 2} \par\put(9.75,3.1){\tiny 1}\put(11.35,3.1){\tiny 2}
\put(9.75,1.4){\tiny 0}\put(11.35,1.4){\tiny 5} \par\par\end{overpic} Figure
8. The tetradedra of $\tau_{p}$ with $[w_{0}^{\prime}(01)]$ 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 $w_{0}^{\prime}$ are
identified to the bottom faces of $w_{p}$ and $s$ is glued to the top faces of
$w_{1}$ and $w_{1}^{\prime}$ and the bottom faces of $w_{p-1}$ and
$w_{p-1}^{\prime}$ as in Table 5.

###  2.5. Equivalence classes of edges in the triangulation

Knowing the relationships between the edges of $\tau_{p}$ 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
$w_{0}^{\prime}$. In the braid picture, $w_{0}^{\prime}(01)$ slides down
through the first $\sigma_{1}$ half-twist to be identified with $w_{1}(10)$
and $w_{1}^{\prime}(10)$, so the equivalence class of $w_{0}^{\prime}(01)$,
$[w_{0}^{\prime}(01)]$, contains the 01 edge of $w_{1}$ and of
$w_{1}^{\prime}$. Continuing to move the 01 edge through more half-twists
shows the 01 edges of $w_{i}$ and $w_{i}^{\prime}$ belong to
$[w_{0}^{\prime}(01)]$ for all $i$. Two edges of $s$ also appear in this
equivalence class, which can be seen most easily in Table 5:
$w_{1}^{\prime}(01)\sim s(bc)$ and $w_{p-1}(01)\sim s(ad)$. These edges of
$s$, $s(bc)$ and $s(ad)$, are also identified to $w_{p-1}^{\prime}(25)$ and
$w_{1}(25)$, respectively, which can themselves be slid along along the braid
to show that the 25 edges of each $w_{i}$ and $w_{i}^{\prime}$ are in
$[w_{0}^{\prime}(01)]$. We have not yet considered face pairings between
$w_{0}^{\prime}$ and $w_{p}$. The first in Table 5, $w_{0}^{\prime}(012)\sim
w_{p}(021)$ shows $w_{p}(02)$ and $w_{0}^{\prime}(02)$ are also in
$[w_{0}^{\prime}(01)]$. Based on how the tetrahedra are positioned in the
braid complement, 02 is the top edge in $w_{0}^{\prime}$ and the bottom edge
in $w_{p}$, which allows us to describe $[w_{0}^{\prime}(01)]$ as follows.

######  Observation 1.

The equivalence class $[w_{0}^{\prime}(01)]$ contains a total of $4p+4$ edges:
the 01 edges and 25 edges of each $w_{i}$ and $w_{i}^{\prime}$; $s(bc)$ and
$s(ad)$; the top edge of $w_{0}^{\prime}$ and the bottom edge of $w_{p}$.
These edges are colored red in Figure 8.

######  Observation 2.

A similar analysis shows that $X_{p}$ has two edges with degree 5. The edge
$[w_{0}^{\prime}(05)]$ arises from

$w_{0}^{\prime}(05)\sim w_{1}(15)\sim s(cd)\sim w_{p-1}^{\prime}(51)\sim
w_{p}(05)\sim_{\text{back to}}w_{0}^{\prime}(05)$

and, when $p>1$, the edge $[w_{0}^{\prime}(12)]$ from

$w_{0}^{\prime}(12)\sim w_{1}^{\prime}(02)\sim s(ba)\sim w_{p-1}(20)\sim
w_{p}(21)\sim_{\text{back to}}w_{0}^{\prime}(12).$

When $p=1$, $w_{1}^{\prime}(02)$ and $w_{p-1}(20)$ are undefined and are
replaced by $s(db)$ and $s(ac)$.

It turns out that the remaining $2p-2$ edges have degree 4.

##  3\. Angle structures and the Casson-Rivin program

We will prove that the triangulation $\tau_{p}$ of $X_{p}$ defined in 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].

###  3.1. 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 $\tau$ of a 3-manifold is an
assignment of angles to the edges of the tetrahedra in $\tau$ satisfying the
following conditions:

  1. (1)

Angles assigned to opposite edges of a tetrahedron are equal, meaning it is
enough to specify three angles $\theta_{3i-2}$, $\theta_{3i-1}$, and
$\theta_{3i}$, for each tetrahedron $T_{i}\in\tau$;

  2. (2)

$\theta_{3i-2}+\theta_{3i-1}+\theta_{3i}=\pi$ for all $i$; and

  3. (3)

The sum of the angles surrounding an edge of the 3-manifold equals $2\pi$.

This paper utilizes both _taut angle structures_ , where
$\theta_{k}\in\\{0,\pi\\}$ for all $k$, and _positive angle structures_ ,
where $\theta_{k}>0$ for all $k$. Note that Condition (2) of the definition
implies that a positive angle structure must have all angles strictly between
0 and $\pi$. In this case, the three angles that sum to $\pi$ specify a
_positively oriented ideal hyperbolic tetrahedron_. The tetrahedron is
isometric to one whose vertices are at $0$, $1$, $\infty$, and a complex
number $z$ with $\text{Im}(z)>0$, where $0$, $1$, and $z$ form a Euclidean
triangle having the assigned angles. If $\text{Im}(z)=0$, the tetrahedron is
said to be _degenerate_ , and, if $\text{Im}(z)<0$, _negatively oriented_. If
an angle structure $\theta$ assigns the same angles to two tetrahedra, $T_{1}$
and $T_{2}$, then we will say they are _isometric with respect to $\theta$_
and write $T_{1}\cong_{\theta}T_{2}$.

We use $\mathcal{A}(\tau)$ to denote the space of all positive angle
structures on a triangulation $\tau$ and note that, if $\tau$ has $n$
tetrahedra, $\mathcal{A}(\tau)\subset(0,\pi)^{3n}$ is a convex polytope with
compact closure. As such, the points obtained through coordinate-by-coordinate
averaging of elements of $\mathcal{A}(\tau)$ are again in $\mathcal{A}(\tau)$.
Also, real-valued functions defined on $\overline{\mathcal{A}(\tau)}$ attain
their extreme values. See [5, Proposition 3.2] for more details about the
space of angle structures.

As will be described in 3.3, the initial step in the Casson-Rivin program is
to argue that $\mathcal{A}(\tau)$ is nonempty, i.e., that the triangulation
admits a positive angle structure. We accomplish this in our case by first
showing (in 4.1) that $\tau_{p}$ 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 $\tau$ 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
$\pi$ edge can be either red or blue)

  * •

when viewed from any of the four ideal vertices, the $\pi$-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 $\pi$-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_.

Figure 9. A tetrahedron in a veering triangulation.

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 $2\pi$.
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] [16], so one should not expect completeness for an arbitrary point in
$\mathcal{A}(\tau)$. It turns out that the independent work of Casson and
Rivin [17], which forms the basis of the Casson-Rivin program in [5] and
outlined in 3.3, connects the question of completeness to the volume
associated to an angle structure in $\mathcal{A}(\tau)$, i.e., the sum of the
volumes of the corresponding hyperbolic tetrahedra.

###  3.2. Volumes of angle structures

Recall ([13], for example) that the volume of an ideal hyperbolic tetrahedron,
$T$, with dihedral angles $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$ is
given by

$V(T)={\textcyr{L}}(\theta_{1})+{\textcyr{L}}(\theta_{2})+{\textcyr{L}}(\theta_{3}),$

where the Lobachevsky function, L, is defined as

${\textcyr{L}}(x)=-\int_{0}^{x}\log|2\sin t|\,dt.$

######  Definition 3.3.

For a point $\theta\in\mathcal{A}(\tau)$, which specifies angles for the $n$
tetrahedra $T_{i}\in\tau$ and has all $\theta_{i}\in(0,\pi)$, define the
_volume associated to $\theta$_, $\mathcal{V}(\theta)$, by

$\mathcal{V}(\theta)=\displaystyle{\sum_{i=1}^{n}V(T_{i})}=\displaystyle{\sum_{i=1}^{3n}{\textcyr{L}}(\theta_{i})}$

and note that $\mathcal{V}(\theta)>0$ for $\theta\in\mathcal{A}(\tau)$.

As summarized in [5], L is well-defined and continuous on $\mathbb{R}$. Thus,
the definition of $\mathcal{V}$ can be extended to the closure of
$\mathcal{A}(\tau)$, which may include points with coordinates equal to 0 or
$\pi$ where the integral itself is improper. If
$\theta\in\overline{\mathcal{A}(\tau)}$, then $\mathcal{V}(\theta)\geq 0$.

Straightforward tetrahedra-by-tetrahedra calculations show that for a positive
angle structure $\theta$ and a tangent vector $\vec{v}\in
T_{\theta}\mathcal{A}(\tau)$,

$\frac{\partial\mathcal{V}}{\partial\vec{v}}=\displaystyle{\sum_{i=1}^{3n}-v_{i}\log\sin\theta_{i}}\quad\text{
and }\quad\frac{\partial^{2}\mathcal{V}}{\partial\vec{v}\,^{2}}<0.$ (1)

See, for example, [5, Lemma 5.3].

Because $\mathcal{V}$ is strictly concave down on $\mathcal{A}(\tau)$, any
critical point in $\mathcal{A}(\tau)$ 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 $\mathcal{A}(\tau)\neq\emptyset$, the point at which the volume
functional $\mathcal{V}:\overline{\mathcal{A}(\tau)}\rightarrow[0,\infty)$
attains its maximum is unique.

###### Proof.

If there is a critical point $\alpha$ in $\mathcal{A}(\tau)$, the strict
concavity of $\mathcal{V}$ guarantees that $\alpha$ is unique and that
$\mathcal{V}(\alpha)$ is the maximum volume. If, on the other hand, there is
no critical point on the interior, the compactness of
$\overline{\mathcal{A}(\tau)}$ ensures that there is a point
$\alpha\in\overline{\mathcal{A}(\tau)}-\mathcal{A}(\tau)$ with
$\mathcal{V}(\alpha)$ maximal. To show such an angle structure is unique, we
first examine its coordinates. Because $\alpha\notin\mathcal{A}(\tau)$, there
is at least one tetrahedron in $\tau$ with an assigned angle of $0$ or $\pi$.
The maximality of $\alpha$ further constrains the angles of such a tetrahedron
as noted in the following lemma.

######  Lemma 3.2.

When $\mathcal{V}(\alpha)$ is a maximum, any tetrahedron assigned an angle of
$0$ or $\pi$ must have angles of $0$, $0$, and $\pi$ in some order.

###### Proof of Lemma 3.2.

By Condition (2) of Definition 3.1, the angles in a tetrahedron must sum to
$\pi$, so this statement is obviously true if one angle in a tetrahedron is
assigned $\pi$ or if two 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 $\alpha$.

Consider, for example, a tetrahedron, $T_{0}\in\tau$, with nonnegative angles
$0$, $\theta_{0}$, and $\pi-\theta_{0}$. For any $0<\lambda<1$, define a
family of tetrahedra $T(t)|_{t\geq 0}$ with angles $\theta_{1}=t$,
$\theta_{2}=\theta_{0}-\lambda t$, and $\theta_{3}=\pi-(t+\theta_{0}-\lambda
t)$, so $T(0)=T_{0}$ and, as $t$ increases, the angles of the tetrahedron
change — $\theta_{1}$ is increased, while $\theta_{2}$ and $\theta_{3}$ are
decreased. The first partial in (1) restricted to this family of tetrahedra
yields

$\frac{dV(T(t))}{dt}\Big{|}_{t=0^{+}}=\displaystyle{\sum_{i=1}^{3}-\frac{d\theta_{i}}{dt}\log\sin\theta_{i}}\Big{|}_{t=0^{+}}=+\infty.$

In other words, increasing the solitary $0$ angle by a small amount would
increase $T_{0}$’s volume (and thus the total volume $\mathcal{V}$) by a much
larger amount, violating the maximality of $\alpha$. Thus, if a maximal angle
structure assigns any angles in $\\{0,\pi\\}$ to a tetrahedron, two of its
angles are assigned $0$ and one $\pi$, thus proving the lemma. We call a
tetrahedron with angles in $\\{0,\pi\\}$ _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
$\alpha\in\overline{\mathcal{A}(\tau)}-\mathcal{A}(\tau)$ to prove that
$\alpha$ is unique. Let $\beta$ be in
$\overline{\mathcal{A}(\tau)}-\mathcal{A}(\tau)$ with
$\mathcal{V}(\beta)=\mathcal{V}(\alpha)$ and let $\mu\in[0,\pi]^{3n}$ be the
point whose coordinates are determined by averaging those of $\alpha$ and
$\beta$. Then $\mu\in\overline{\mathcal{A}(\tau)}$, because the set
$\overline{\mathcal{A}(\tau)}$ is convex. We will examine the volume of each
tetrahedron $T$ under the angle structure assigned by $\mu$ and will denote
this volume by $\mathcal{V}(\mu|_{T})$.

Case 1:

If $\alpha$ and $\beta$ assign the same angles to a tetrahedron $T$, then
their average $\mu$ does as well, so $\mathcal{V}(\mu|_{T})$ is equal to
$\mathcal{V}(\alpha|_{T})$ (and also $\mathcal{V}(\beta|_{T})$).

Case 2:

If $\alpha$ and $\beta$ assign different angles to a tetrahedron $T$ and both
$\alpha|_{T}$ and $\beta|_{T}$ are flat, then, because $\alpha$ and $\beta$
are maximum points, the angles must be $0$, $0$, and $\pi$ (in different
orders), and the coordinates of $\mu|_{T}$ are $0$, $\frac{\pi}{2}$, and
$\frac{\pi}{2}$ in some order. In this case, $\mathcal{V}(\mu|_{T})$ equals
$0$ just as $\mathcal{V}(\alpha|_{T})$ and $\mathcal{V}(\beta|_{T})$ do.

Case 3:

If $\alpha$ and $\beta$ assign different angles to a tetrahedron $T$ and at
least one of $\alpha|_{T}$ or $\beta|_{T}$ is not flat, we will argue that
$\mathcal{V}(\mu|_{T})$ is strictly greater than the average of
$\mathcal{V}(\alpha|_{T})$ and $\mathcal{V}(\beta|_{T})$. Without loss of
generality suppose $\alpha|_{T}$ is not flat. Define $\gamma(t)$ to be the
line segment joining $\alpha|_{T}$ and $\beta|_{T}$:

$\gamma(t)=t\alpha|_{T}+(1-t)\beta|_{T}\text{ for }t\in[0,1].$

Let $V(t)$ be the volume of the tetrahedron whose angles are given by
$\gamma(t)$. As a restriction of $\mathcal{V}$, the function $V$ is continuous
on $[0,1]$. Also, because $\alpha|_{T}$ is not flat, $\gamma$ is in
$(0,\pi)^{3}$ for $0<t\leq 1$, so $V$ is differentiable on $(0,1]$ and by the
second partial in (1), $V^{\prime\prime}<0$. Applying the Mean Value Theorem
to $V$ yields points $c_{1}\in\left(0,\frac{1}{2}\right)$ and
$c_{2}\in\left(\frac{1}{2},1\right)$ with

$\displaystyle V^{\prime}(c_{1})$
$\displaystyle=2\left(V\left(\textstyle{\frac{1}{2}}\right)-V\left(0\right)\right)$
$\displaystyle V^{\prime}(c_{2})$
$\displaystyle=2\left(V\left(1\right)-V\left(\textstyle{\frac{1}{2}}\right)\right)$

and negative second derivatives, so $V^{\prime}(c_{1})>V^{\prime}(c_{2})$,
which means

$2V\left(\textstyle{{\frac{1}{2}}}\right)>V(1)+V(0).$

Because $V\left(\textstyle{\frac{1}{2}}\right)=\mathcal{V}(\mu|_{T})$,
$V(1)=\mathcal{V}(\alpha|_{T})$, and $V(0)=\mathcal{V}(\beta|_{T})$, this
inequality implies
$\mathcal{V}(\mu|_{T})>\frac{1}{2}(\mathcal{V}(\alpha|_{T})+\mathcal{V}(\beta|_{T}))$.

In all cases, the volume of a tetrahedron $T$ computed using $\mu$, the
average of the angle structures, is greater than or equal to the average of
the volumes of $T$ computed using the angle structures $\alpha$ and $\beta$,
so summing the individual volumes over all tetrahedra in $\tau$ yields

$\mathcal{V}(\mu)\geq\textstyle{\frac{1}{2}}(\mathcal{V}(\alpha)+\mathcal{V}(\beta))=\mathcal{V}(\alpha)$

with equality only when there are no tetrahedra in Case 3. But
$\mathcal{V}(\alpha)$ is the maximum volume, so
$\mathcal{V}(\mu)\leq\mathcal{V}(\alpha)$, meaning
$\mathcal{V}(\mu)=\mathcal{V}(\alpha)$ and there are no Case 3 tetrahedra. The
only way $\alpha$ and $\beta$ can assign different angles to a tetrahedron $T$
is if $\alpha|_{T}$ and $\beta|_{T}$ are both flat (Case 2). In this case,
$\mathcal{V}(\mu)=\mathcal{V}(\alpha)$, so $\mu$ is also maximal, but, as we
have seen, the coordinates of $\mu|_{T}$ are $0$, $\frac{\pi}{2}$, and
$\frac{\pi}{2}$, which is not allowed for maximal angle structures. Therefore,
$\alpha$ and $\beta$ cannot assign different angles to any of the tetrahedra
and $\beta=\alpha$, which is the unique maximum point. ∎

Positive angle structures that maximize the total volume play an important
role in the Casson-Rivin program.

###  3.3. 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 [19]) is satisfied. In
doing so, they prove their main result, a theorem based on independent work of
Rivin and Casson usually cited as [17].

######  Casson-Rivin Theorem ([5, Theorem 1.2]).

Let $M$ be an orientable 3-manifold with boundary consisting of tori, and let
$\tau$ be an ideal triangulation of $M$. Then a point
$\theta\in\mathcal{A}(\tau)$ corresponds to a complete hyperbolic metric on
the interior of $M$ if and only if $\theta$ is a critical point of the
functional $\mathcal{V}:\mathcal{A}(\tau)\rightarrow[0,\infty)$.

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
$\tau_{p}$.

The Casson-Rivin Theorem will enable us to conclude $\tau_{p}$ is geometric as
follows. In 4.1, we establish that the space of positive angle structures on
our triangulations, $\mathcal{A}(\tau_{p})$, is nonempty. Consequently, the
volume functional
$\mathcal{V}:\overline{\mathcal{A}(\tau_{p})}\rightarrow[0,\infty)$ will
attain its maximum. Then, in Section 5, we show that the maximum point of
$\mathcal{V}$ must belong to $\mathcal{A}(\tau_{p})$. Then, by the Casson-
Rivin Theorem, there is a positive angle structure on our triangulation
$\tau_{p}$ that yields the complete hyperbolic structure on the link
complement, $X_{p}$. Thus, the constructed triangulations are geometric and
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ is hyperbolic for $p\geq 1$.

###  3.4. Symmetries of angle structures

When arguing that the maximum of the volume cannot appear on the boundary of
$\mathcal{A}(\tau_{p})$, a certain symmetry of the space of positive angle
structures will be helpful.

######  Definition 3.4.

Any symmetry, $\rho$, of a triangulation, $\tau$, induces a map on the space
of positive angle structures by assigning the angles of $T$ to $\rho(T)$ for
all $T\in\tau$. When such an assignment results in a positive angle structure
for all points in $\mathcal{A}(\tau)$, we say _$\rho$ induces a symmetry of
the space of positive angle structures_, which we will also denote by $\rho$.
More specifically, given a symmetry $\rho:\tau\rightarrow\tau$ with
$\rho(T_{i})=T_{j}$, let $\theta\in\mathcal{A}(\tau)$ and define the
coordinates of $\rho\theta$ by $\rho\theta_{3j-2}=\theta_{3i-2}$,
$\rho\theta_{3j-1}=\theta_{3i-1}$, and $\rho\theta_{3j}=\theta_{3i}$. The
angles of $\rho\theta$ are positive and meet Conditions (1) and (2) of
Definition 3.1, so $\rho$ induces a symmetry of $\mathcal{A}(\tau)$ whenever
Condition (3) is also met.

######  Observation 3.

A symmetry $\rho:\mathcal{A}(\tau)\rightarrow\mathcal{A}(\tau)$ rearranges the
angles of $\theta$ triple by triple, so
$\mathcal{V}(\theta)=\mathcal{V}(\rho\theta)$.

##  4\. The space of positive angle structures is nonempty

We now return to the triangulations, $\tau_{p}$, defined in 2.4. Our immediate
goal is to complete the first step of the Casson-Rivin program by showing that
$\mathcal{A}(\tau_{p})$ is nonempty, i.e., that $\tau_{p}$ admits a positive
angle structure. We will do so by showing that $\tau_{p}$ has a veering angle
structure, which can be deformed to a positive angle structure [12] [6]. We
then use the veering structure to describe an explicit coordinate system for
$\mathcal{A}(\tau_{p})$ and conclude the section by writing the edge equations
in these coordinates and using them to describe a useful symmetry.

###  4.1. The triangulations are veering

While the initial triangulations $\hat{\tau}_{p}$ were not veering, the
Pachner moves eliminated the obstructions and

######  Proposition 4.1.

Each triangulation $\tau_{p}$ is veering.

###### Proof.

With the exception of $s$, the tetrahedra in $\tau_{p}$ have an upward
orientation induced by the product region containing the 6-braid, which is how
the non-$s$ tetrahedra of $\tau_{p}$ are flattened in Figure 8. Flatten $s$ as
indicated in the same figure. In each tetrahedron, assign an angle of $\pi$ 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 $w_{i}^{\prime}$; the edges $s(bc)$ and
$s(ad)$; and the top diagonal of $w_{0}^{\prime}$ 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 $\tau_{p}$, which is also transverse taut.

By Observation 1, the edges in the equivalence class $[w_{0}^{\prime}(01)]$
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
$\\{0,\pi\\}$. Only diagonals are assigned an angle of $\pi$, so, if there are
exactly two diagonals in each equivalence class, Condition (3) will also be
met (the angle sum around each edge is $2\pi$).

The class $[w_{0}^{\prime}(01)]$, whose members are listed in Observation 1,
contains exactly two diagonals, $w_{0}^{\prime}(02)$ and $w_{p}(02)$. The
degree 5 edges of $X_{p}$ are completely described in Observation 2. The edges
$[w_{0}^{\prime}(05)]$ and $[w_{0}^{\prime}(12)]$ each contain exactly two
diagonals: The top diagonal $w_{1}(15)$ is matched with the bottom diagonal
$w_{p-1}^{\prime}(51)$ and similarly for $w_{1}^{\prime}(02)$ and
$w_{p-1}(20)$ (or $s(db)$ and $s(ac)$ when $p=1$, which completes this case).

To check the remaining $2p-2$ edges of $X_{p}$ for $p>1$, i.e., those of
degree 4, we examine $w_{i-1}(02)$ and $w_{i-1}^{\prime}(15)$, the bottom
diagonals of $w_{i-1}$ and $w_{i-1}^{\prime}$ 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 $\sigma_{1}$ half-twist, the first
of these faces is rotated to the back and identified with
$w_{i}^{\prime}(102)$, whereas the second stays in front and is identified
with $w_{i}(125)$. These faces share the edge 12, which, after another
$\sigma_{1}$ is twisted to the back and identified to $w_{i+1}^{\prime}(02)$.
A similar argument applies to $w_{i-1}^{\prime}(15)$. Symbolically,

$\displaystyle\begin{split}w_{i-1}(02)\sim w_{i}(12)=w_{i}^{\prime}(12)\sim
w_{i+1}^{\prime}(02)\quad&\text{ for }i=2,\dots p-2.\\\
w_{i-1}^{\prime}(15)\sim w_{i}(05)=w_{i}^{\prime}(05)\sim
w_{i+1}(15)\quad&\text{ for }i=1,\dots p-1.\end{split}$ (2)

The only diagonals are the 02 and 15 edges, so the edge classes, $[w_{i}(12)]$
and $[w_{i}(05)]$, contain exactly two $\pi$ 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:

$\displaystyle\begin{split}s(ac)\sim w_{1}^{\prime}(21)&=w_{1}(21)\sim
w_{2}^{\prime}(20)\\\ s(bd)\sim w_{p-1}^{\prime}(21)&=w_{p-1}(21)\sim
w_{p-2}(20).\end{split}$ (3)

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.
∎

######  Corollary 4.2.

The triangulation $\tau_{p}$ admits a positive angle structure.

###### Proof.

In 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]. ∎

###  4.2. Coordinates for the space of positive angle structures

The veering structure on $\tau_{p}$ allows us to introduce convenient
coordinates for a point $\theta$ in $\mathcal{A}(\tau_{p})$. For $T$ in
$\tau_{p}$, let $\theta_{D}T$ denote the angle assigned to the diagonals of
$T$, $\theta_{R}T$ the angle assigned to the red edges, and $\theta_{B}T$ to
the blue edges. Thus, we can write
$\theta\in\mathcal{A}(\tau_{p})\subset(0,\pi)^{3(2p+1)}$ as

$(\theta_{R}w_{0}^{\prime},\theta_{B}w_{0}^{\prime},\theta_{D}w_{0}^{\prime},\dots,\theta_{R}w_{p-1}^{\prime},\theta_{B}w_{p-1}^{\prime},\theta_{D}w_{p-i}^{\prime},\\\
\theta_{R}w_{1},\theta_{B}w_{1},\theta_{D}w_{1},\dots,\theta_{R}w_{p},\theta_{B}w_{p},\theta_{D}w_{p},\theta_{R}s,\theta_{B}s,\theta_{D}s).$

######  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_ :

$\displaystyle\left(\sum_{i=0}^{p-1}2\theta_{R}w_{i}^{\prime}\right)+\left(\sum_{i=1}^{p}2\theta_{R}w_{i}\right)+2\theta_{R}s+\theta_{D}w_{0}^{\prime}+\theta_{D}w_{p}$
$\displaystyle=2\pi$
$\displaystyle\theta_{B}w_{0}^{\prime}+\theta_{D}w_{1}+\theta_{B}s+\theta_{D}w_{p-1}^{\prime}+\theta_{B}w_{p}$
$\displaystyle=2\pi$
$\displaystyle\theta_{B}w_{0}^{\prime}+\theta_{D}w_{1}^{\prime}+\theta_{B}s+\theta_{D}w_{p-1}+\theta_{B}w_{p}$
$\displaystyle=2\pi$
$\displaystyle\theta_{D}w_{i-1}+\theta_{B}w_{i}+\theta_{B}w_{i}^{\prime}+\theta_{D}w_{i+1}^{\prime}$
$\displaystyle=2\pi$ $\displaystyle\text{ for }i=2,\dots p-2$
$\displaystyle\theta_{D}w_{i-1}^{\prime}+\theta_{B}w_{i}+\theta_{B}w_{i}^{\prime}+\theta_{D}w_{i+1}$
$\displaystyle=2\pi$ $\displaystyle\text{ for }i=1,\dots p-1$
$\displaystyle\theta_{D}s+\theta_{B}w_{1}^{\prime}+\theta_{B}w_{1}+\theta_{D}w_{2}^{\prime}$
$\displaystyle=2\pi$
$\displaystyle\theta_{D}s+\theta_{B}w_{p-1}^{\prime}+\theta_{B}w_{p-1}+\theta_{D}w_{p-2}$
$\displaystyle=2\pi.$

These equations hold for $p>1$ as long as each term is defined. (Recall that
there are no tetrahedra labelled $w_{0}$ or $w_{p}^{\prime}$.)

The first two equations also hold when $p=1$, and, after substituting $s(db)$
and $s(ac)$ for the undefined terms $w_{1}^{\prime}(02)$ and $w_{p-1}(20)$ as
in Observation 2, so does the third:

$\theta_{B}w_{0}^{\prime}+\theta_{D}s+\theta_{B}s+\theta_{D}s+\theta_{B}w_{1}=2\pi.$

######  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:

$\displaystyle\theta_{D}w_{1}-\theta_{D}w_{1}^{\prime}=\theta_{D}w_{3}-\theta_{D}w_{3}^{\prime}=\theta_{D}w_{5}-\theta_{D}w_{5}^{\prime}=$
$\displaystyle\cdots$
$\displaystyle\theta_{D}w_{2}-\theta_{D}w_{2}^{\prime}=\theta_{D}w_{4}-\theta_{D}w_{4}^{\prime}=\theta_{D}w_{6}-\theta_{D}w_{6}^{\prime}=$
$\displaystyle\cdots.$

In addition,

$\displaystyle\theta_{D}w_{p-2}-\theta_{D}w_{p-2}^{\prime}$
$\displaystyle=\theta_{D}w_{p}-\theta_{D}s$
$\displaystyle\theta_{D}s-\theta_{D}w_{0}^{\prime}$
$\displaystyle=\theta_{D}w_{2}-\theta_{D}w_{2}^{\prime}$
$\displaystyle\theta_{D}w_{p-1}-\theta_{D}w_{p-1}^{\prime}$
$\displaystyle=\theta_{D}w_{1}-\theta_{D}w_{1}^{\prime}.$

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,
$\theta_{D}w_{j}-\theta_{D}w_{j}^{\prime}$, with $j$ even are also equal to
$\theta_{D}s-\theta_{D}w_{0}^{\prime}$ and $\theta_{D}w_{p}-\theta_{D}s$, but
not necessarily to $\theta_{D}w_{1}-\theta_{D}w_{1}^{\prime}$, etc.

The corresponding equation for $p=1$ is:

$\theta_{D}s-\theta_{D}w_{0}^{\prime}=\theta_{D}w_{1}-\theta_{D}s.$

###  4.3. A symmetry of the space of angle structures

Consider the closure of $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ as shown on the
right of Figure 2. Moving the lone half-twist up so that it occurs halfway
along the $\sigma_{1}$ 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 $\tau_{p}$ — 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 $\iota:\tau_{p}\rightarrow\tau_{p}$ that fixes $s$ and takes
$w_{i}^{\prime}$ to $w_{p-i}$ by matching up their $R$, $B$, and $D$ edges
induces a symmetry of $\mathcal{A}(\tau_{p})$.

###### Proof.

Let $\theta\in\mathcal{A}(\tau_{p})$. According to Definition 3.4, to verify
that $\iota\theta$ is also in $\mathcal{A}(\tau_{p})$, we need to check that
the sum of the dihedral angles at each edge of of $X_{p}$ is $2\pi$, when the
angle measure is given by $\iota\theta$. Thus, it is enough to verify that
$\iota\theta$ satisfies the edge equations listed in Observation 4 where

$\iota\theta_{\star}s=\theta_{\star}s,\quad\iota\theta_{\star}w_{i}^{\prime}=\theta_{\star}w_{p-i},\quad\text{and}\quad\iota\theta_{\star}w_{i}=\theta_{\star}w_{p-i}^{\prime}$

for $\star=R,B$, 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 $w_{0}^{\prime}$ and
$w_{p}$. The involution $\iota$ permutes these edges, thereby permuting the
terms of the sum:

$\displaystyle\left(\sum_{i=0}^{p-1}2\iota\theta_{R}w_{i}^{\prime}\right)+\left(\sum_{i=1}^{p}2\iota\theta_{R}w_{i}\right)+2\iota\theta_{R}s+\iota\theta_{D}w_{0}^{\prime}+\iota\theta_{D}w_{p}$
$\displaystyle=$
$\displaystyle\left(\sum_{i=1}^{p}2\theta_{R}w_{i}\right)+\left(\sum_{i=0}^{p-1}2\theta_{R}w_{i}^{\prime}\right)+2\theta_{R}s+\theta_{D}w_{p}+\theta_{D}w_{0}^{\prime}$
$\displaystyle=2\pi,$

so the first edge equation is satisfied by $\iota\theta$. The terms in the
next two sums, those for the degree-5 edges, are also permuted by $\iota$, so
those equations are satisfied by $\iota\theta$ 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

$\iota\theta_{D}s+\iota\theta_{B}w_{p-1}^{\prime}+\iota\theta_{B}w_{p-1}+\iota\theta_{D}w_{p-2},$

but this sum is just

$\theta_{D}s+\theta_{B}w_{1}+\theta_{B}w_{1}^{\prime}+\theta_{D}w_{2}^{\prime},$

which equals $2\pi$ 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. ∎

######  Corollary 4.4.

Let $\iota:\tau_{p}\rightarrow\tau_{p}$ be as in Proposition 4.3 and let $k$
be such that $p=2k$ if $p$ is even and $p=2k+1$ if $p$ is odd. Positive angle
structures $\theta\in\mathcal{A}(\tau_{p})$ with the property that
$\iota\theta=\theta$ have $w_{i}^{\prime}\cong_{\theta}w_{p-i}$ and thus
satisfy a simpler list of edge equations:

$\displaystyle\left(\sum_{i=1}^{p}4\theta_{R}w_{i}\right)+2\theta_{R}s+2\theta_{D}w_{p}$
$\displaystyle=2\pi$ (4) $\displaystyle
2\theta_{B}w_{p}+2\theta_{D}w_{1}+\theta_{B}s$ $\displaystyle=2\pi$ (5)
$\displaystyle 2\theta_{B}w_{p}+2\theta_{D}w_{p-1}+\theta_{B}s$
$\displaystyle=2\pi$ (6)
$\displaystyle\theta_{D}w_{i-1}+\theta_{B}w_{i}+\theta_{B}w_{p-i}+\theta_{D}w_{p-(i+1)}$
$\displaystyle=2\pi$ $\displaystyle\text{ for }i=2,\dots k$ (7)
$\displaystyle\theta_{D}w_{p-(i-1)}+\theta_{B}w_{i}+\theta_{B}w_{p-i}+\theta_{D}w_{i+1}$
$\displaystyle=2\pi$ $\displaystyle\text{ for }i=1,\dots k$ (8)
$\displaystyle\theta_{D}s+\theta_{B}w_{p-1}+\theta_{B}w_{1}+\theta_{D}w_{p-2}$
$\displaystyle=2\pi.$ (9)

In addition,

$\displaystyle\theta_{D}w_{p}=\theta_{D}s\quad\text{ and
}\quad\theta_{D}w_{i}=\theta_{D}w_{p-i}\quad\text{ for }i=1,\dots k.$ (10)

These equations hold for all $p\geq 1$, with one exception. When $p=1$,
Equation (6) reads: $2\theta_{B}w_{1}+2\theta_{D}s+\theta_{B}s=2\pi.$

###### Proof.

By assumption,
$\iota\theta_{\star}w_{i}^{\prime}=\theta_{\star}w_{i}^{\prime}$ for
$\star=R,B$, and $D$, and, by definition, $\iota\theta_{\star}w_{i}^{\prime}$
also equals $\theta_{\star}w_{p-i}$, so
$\theta_{\star}w_{i}^{\prime}=\theta_{\star}w_{p-i}$, and the tetrahedra
$w_{i}^{\prime}$ and $w_{p-i}$ have the same angle assignments under $\theta$,
so $w_{i}^{\prime}\cong_{\theta}w_{p-i}$. This isometry allows us to obtain
Equations (4)–(9), by replacing $w_{i}^{\prime}$ with $w_{p-i}$ in the edge
equations of Observation 4 and eliminating redundancies.

To obtain Equation (10) for $p=1$, replace $w_{0}^{\prime}$ with $w_{1}$ in
the $p=1$ equation of Observation 5, which shows

$\theta_{D}s-\theta_{D}w_{1}=\theta_{D}w_{1}-\theta_{D}s,$

so $\theta_{D}w_{1}=\theta_{D}s$. If $p>1$, Equations (5) and (6) imply that
$\theta_{D}w_{1}$ equals $\theta_{D}w_{p-1}$, which, because of the isometry,
equals $\theta_{D}w_{1}^{\prime}$, so the difference
$\theta_{D}w_{1}-\theta_{D}w_{1}^{\prime}$ equals 0 as do all odd-index
differences listed in Observation 5. Replacing $w_{i}^{\prime}$ with
$w_{p-i}$, yields Equation (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 $w_{i}^{\prime}$ with $w_{p-i}$, Equation (10) holds for all
$i$ when $p$ is odd. To see that Equation (10) also holds when $p$ and $i$ are
even, note that

$\theta_{D}w_{p}-\theta_{D}s=\theta_{D}w_{2}-\theta_{D}w_{2}^{\prime}=\theta_{D}w_{p-2}-\theta_{D}w_{p-2}^{\prime}=\theta_{D}s-\theta_{D}w_{0}^{\prime}=\theta_{D}s-\theta_{D}w_{p}$

so $\theta_{D}s=\theta_{D}w_{p}$. Thus, all differences in Observation 5 with
even index are also 0 and replacing $w_{i}^{\prime}$ with $w_{p-i}$ results in
Equation (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 $\tau_{p}$ is nonempty, so we have accomplished the first step
of the Casson-Rivin program described by Futer and Guéritaud in [5] and
summarized in 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 $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$
and $X_{p}$ be its complement in the 3-sphere. Then there is an ideal
triangulation of $X_{p}$ that is geometric.

###### Proof.

Let $\tau_{p}$ be the triangulation defined in 2.4. By Corollary 4.2,
$\mathcal{A}(\tau_{p})\neq\emptyset$, which means the volume functional
$\mathcal{V}$ attains its maximum on the compact set
$\overline{\mathcal{A}(\tau_{p})}$, say at the point $\alpha$. A maximal point
has two important properties. The first was proved in Lemma 3.2.

######  Property 1.

If $\alpha$ is maximal and assigns an angle in $\\{0,\pi\\}$ to a tetrahedron,
then the tetrahedron must be flat, i.e., its angles are $0$, $0$, $\pi$ in
some order.

The second important property derives from a symmetry of the coordinates of a
maximal point. Let the involution $\iota:\tau_{p}\rightarrow\tau_{p}$ be as in
Proposition 4.3, so $\iota$ fixes $s$ and takes $w_{i}^{\prime}$ to $w_{p-i}$.
Then $\iota$ induces a symmetry on $\overline{\mathcal{A}(\tau_{p})}$, which
rearranges the coordinates of an angle structure tetrahedron by tetrahedron,
so, just as in Observation 3, $\mathcal{V}(\iota\alpha)=\mathcal{V}(\alpha)$.
Thus, if $\alpha$ is maximal, so is $\iota\alpha$, but maximal points are
unique by Proposition 3.1, so $\iota\alpha=\alpha$, and Corollary 4.4 together
with Definition 3.1 guarantee the following.

######  Property 2.

If $\alpha$ is maximal, then the angles assigned by $\alpha$ belong to
$[0,\pi]$; sum to $\pi$ within a tetrahedron; and are determined by the angles
assigned to $w_{1},w_{2},\dots,w_{p}$ and $s$, which must satisfy Equations
(4)–(10).

Our goal is to use these properties to show that the maximal point, $\alpha$,
is in $\mathcal{A}(\tau_{p})$, not its boundary. Then we can apply the Casson-
Rivin Theorem from 3.3 [5, Theorem 1.2] and conclude that $\alpha$ corresponds
to the complete hyperbolic structure on $X_{p}$, thus proving Theorem 5.1. We
will show that $\alpha$ is in $\mathcal{A}(\tau_{p})$ by showing that points
in $\mathcal{B}=\overline{\mathcal{A}(\tau_{p})}-\mathcal{A}(\tau_{p})$ will
never maximize the volume. We do so by proving the following.

######  Proposition 5.2.

Any $\beta\in\mathcal{B}$ satisfying Properties 1 and 2 must have
$\mathcal{V}(\beta)=0$.

###### Proof of Proposition 5.2.

Let $\beta\in\mathcal{B}$ satisfy both Property 1 and Property 2. Because
$\beta$ is in the boundary of the space of positive angle structures, there is
a tetrahedron in $\tau_{p}$ to which $\beta$ assigns angles from the set
$\\{0,\pi\\}$. By Property 1, this tetrahedron is flat (has angles $0$, $0$,
and $\pi$). Using Property 2, we can conclude that $\beta$ must assign $\pi$
to at least one of the edges of $w_{1},w_{2},\dots w_{p}$ 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 $\pi$ to any $R$ edge of a $w$ tetrahedron violates Equation
(4) of Corollary 4.4 and thus Property 2. Therefore, no $\beta_{R}w_{i}$ can
equal $\pi$. We will use this result often, so we record it as a lemma and
follow with another useful lemma.

######  Lemma 5.3.

If $\beta\in\mathcal{B}$ satisfies both Property 1 and 2, $\beta_{R}w_{i}$
cannot equal $\pi$ for $i=1,\dots,p$.∎

######  Lemma 5.4.

If $\beta\in\mathcal{B}$ satisfies both Property 1 and 2, and either $w_{p}$
or $s$ is flat, then all tetrahedra are flat, so $\mathcal{V}(\beta)=0$.

###### Proof of Lemma 5.4.

If a tetrahedron is flat, one of its edges is assigned an angle of $\pi$. By
Lemma 5.3, $\beta_{R}w_{p}$ cannot equal $\pi$. However, $\beta_{R}s$ could
equal $\pi$, and, if this is the case, Equation (4) forces $\beta_{R}w_{i}$ to
equal $0$ for all $i=1,\dots,p$, so, by Property 1, the non-$s$ tetrahedra are
also flat. Because, $\beta_{D}w_{p}$ also appears in Equation (4), a similar
argument shows that if $\beta_{D}w_{p}=\pi$, all tetrahedra are flat. By
Equation (10), $\beta_{D}w_{p}=\beta_{D}s$, so the same holds if
$\beta_{D}s=\pi$. Thus, it only remains to check what happens if the angle at
a $B$ edge of $w_{p}$ or $s$ equals $\pi$.

  * •

If $\beta_{B}w_{p}=\pi$, Equation (5) implies that $\beta_{B}s=0$, so, by
Property 1, there is a $\pi$ angle at either the $R$ or $D$ edge of $s$, and,
in either case, as observed above, all tetrahedra are flat.

  * •

If $\beta_{B}s=\pi$, then $\beta_{D}s=0$, so $\beta_{D}w_{p}$ also equals $0$
(Equation (10)), and, by Property 1, either $\beta_{R}w_{p}$ or
$\beta_{B}w_{p}$ equals $\pi$. Because of Lemma 5.3, $\beta_{B}w_{p}$ must be
$\pi$, so, by the previous case, all tetrahedra are flat.

Therefore, if either $w_{p}$ or $s$ has an assigned angle of $\pi$,
$\mathcal{V}(\beta)=0$. ∎

Returning to the proof of Proposition 5.2, recall that $\beta\in\mathcal{B}$
must assign $\pi$ to at least one of the edges of $w_{1},w_{2},\dots w_{p}$ or
$s$. Lemmas 5.3 and 5.4 cover the $R$ edges and the tetrahedra $w_{p}$ and
$s$, so the only edges left to consider are the $B$ and $D$ edges of
$w_{1},w_{2},\dots w_{p-1}$. We start with the $B$ edges and first examine the
case when $\beta_{B}w_{1}=\pi$.

If $\beta_{B}w_{1}=\pi$, then $\beta_{D}w_{1}$ equals $0$ and, by Equation
(10)i=1, so does $\beta_{D}w_{p-1}$. Property 1 implies that in $w_{p-1}$ one
of the other angles must be $\pi$, but, because of Lemma 5.3, it cannot be the
angle at the $R$ edge. Consequently, $\beta_{B}w_{p-1}$ equals $\pi$ and
Equation (9) implies $\beta_{D}s$ must equal $0$, so $s$ is flat and, by Lemma
5.4, $\mathcal{V}(\beta)=0$.

Having shown that if $\beta$ assigns the angle $\pi$ to a $B$ edge of $w_{1}$,
then $\mathcal{V}(\beta)=0$, we now consider the $B$ edges of $w_{j}$ with
$1<j\leq k$. By repeatedly applying Equations (10) and (7) together with
Property 1 and Lemma 5.3, we will push the flatness of $w_{j}$ all the way
down to $w_{1}$, allowing us to once again conclude $\mathcal{V}(\beta)=0$. In
particular, if $\beta_{B}w_{j}=\pi$, then $\beta_{D}w_{j}$ equals $0$ and, by
Equation (10)i=j, so does $\beta_{D}w_{p-j}$. There is a $\pi$ angle in
$w_{p-j}$ (Property 1) and it cannot occur at the $R$ edge (Lemma 5.3), so
$\beta_{B}w_{p-j}$, will equal $\pi$. With both $\beta_{B}w_{j}$ and
$\beta_{B}w_{p-j}$ equal to $\pi$, Equation (7)i=j implies that
$\beta_{D}w_{j-1}=0$, so, by Equation (10)i=j-1, $\beta_{D}w_{p-(j-1)}=0$
also. Another application of Property 1 and Lemma 5.3 shows both
$\beta_{B}w_{j-1}$ and $\beta_{B}w_{p-(j-1)}$ are equal to $\pi$ and we can
apply Equation (7)i=j-1 to conclude $\beta_{D}w_{j-2}=0$. Continuing in this
manner and eventually applying Equation (7)i=2 when both $\beta_{B}w_{2}$ and
$\beta_{B}w_{p-2}$ are equal to $\pi$, allows us to conclude that
$\beta_{D}w_{1}=0$. Thus (by Property 1 and Lemma 5.3), $\beta_{B}w_{1}=\pi$,
so $\mathcal{V}(\beta)=0$.

It remains to examine the $B$ edges of $w_{j}$ where $j>k$. If
$\beta_{B}w_{j}$ equals $\pi$, then $\beta_{D}w_{j}=0$, so applying Equation
(10)i=p-j, Property 1, and Lemma 5.3 shows that $\beta_{B}w_{p-j}=\pi$.
Because $p-j\leq k$, this case has already been covered. Therefore, if any of
the $B$ edges of $w_{1},w_{2},\dots w_{p-1}$ have angle equal to $\pi$, then
$\mathcal{V}(\beta)=0$.

The proof for the $D$ edges is a little simpler, because there is no need to
appeal to Lemma 5.3. If $\beta_{D}w_{1}=\pi$, then, by Equation (10)i=1,
$\beta_{D}w_{p-1}=\pi$, so both $\beta_{B}w_{1}$ and $\beta_{B}w_{p-1}$ are
equal to $0$. In this situation, applying Equation (9) yields
$\beta_{D}s=\pi$, so $s$ is flat and, by Lemma 5.4, $\mathcal{V}(\beta)=0$. If
$j\leq k$ and $\beta_{D}w_{j}$ equals $\pi$, then, by Equation (10)i=j, so
does $\beta_{D}w_{p-j}$, and both $\beta_{B}(w_{j})$ and $\beta_{B}(w_{p-j})$
equal $0$, so Equation (7)i implies $\beta_{D}w_{j-1}=\pi$. Continuing in this
manner shows that if $\beta_{D}w_{j}$ equals $\pi$, so does $\beta_{D}w_{1}$,
and thus $\mathcal{V}(\beta)=0$. If $j>k$ and $\beta_{D}w_{j}=\pi$, then, by
Equation (10)i=p-j, $\beta_{D}w_{p-j}=\pi$ with $p-j\leq k$, 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, $\beta$, on the
boundary of $\mathcal{A}(\tau_{p})$ satisfying these properties has
$\mathcal{V}(\beta)=0$. Therefore, the maximal point $\alpha$ must be on the
interior of $\mathcal{A}(\tau_{p})$, and, by the Casson-Rivin Theorem (3.3 and
[5, Theorem 1.2]), $\alpha$ corresponds to the complete hyperbolic structure
on $X_{p}$. Thus, the triangulation $\tau_{p}$ is geometric, which concludes
the proof of Theorem 5.1. ∎

######  Corollary 5.5.

In the braid group, $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ is not conjugate to
$\sigma_{1}^{p_{0}}\sigma_{2}^{q_{0}}$ for integers $p_{0}$ and $q_{0}$.

###### Proof.

Closures of braids of the form $\sigma_{1}^{p_{0}}\sigma_{2}^{q_{0}}$ are not
hyperbolic, but Theorem 5.1 shows that the closure of
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ is. ∎

######  Corollary 5.6.

The complements of the $(-2,3,n)$-pretzel knots and links admit geometric
triangulations for $n\geq 7$.

###### Proof.

Recall that the braid group has the single relation
$\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$ 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 $C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$ with $p\geq
1$, is also the closure of

$\displaystyle(\sigma_{2}\sigma_{1}\sigma_{2})(\sigma_{1}\sigma_{2}\sigma_{1})C\sigma_{1}^{p}\sigma_{2}^{-1}$
$\displaystyle\sim\sigma_{2}\sigma_{1}\sigma_{2}C\sigma_{1}^{p+2}$
$\displaystyle=\sigma_{1}\sigma_{2}C\sigma_{1}^{p+3}$
$\displaystyle\sim\sigma_{2}(\sigma_{1}\sigma_{2}\sigma_{1})(\sigma_{1}\sigma_{2}\sigma_{1})\sigma_{1}^{p+4}$
$\displaystyle=\sigma_{1}\sigma_{2}\sigma_{1}\sigma_{1}\sigma_{1}\sigma_{2}\sigma_{1}^{p+5}$
$\displaystyle\sim\sigma_{1}^{3}\sigma_{2}\sigma_{1}^{p+6}\sigma_{2};$

here $\sim$ denotes the equivalence relation of conjugacy.

As indicated in Figure 10, the closure of
$\sigma_{1}^{3}\sigma_{2}\sigma_{1}^{p+6}\sigma_{2}$ is a pretzel. Using our
definitions, $\sigma_{1}$ and $\sigma_{2}$ generate left-handed (negative)
twists, so $L_{p}$ is a pretzel with $-(p+6)$, $2$, and $-3$ half-twists, or
equivalently — after rolling the $p+6$ half-twists to the right — $L_{p}$ is
the $(2,-3,-(p+6))$-pretzel, and $X_{p}$ is its complement. By Theorem 5.1,
$X_{p}$ admits a geometric triangulation. But the complement of the
$(-2,3,p+6)$-pretzel is homeomorphic (via a reflection) to $X_{p}$, so it also
has a geometric triangulation. ∎

\begin{overpic}[unit=.25in]{arepretzels.pdf} \put(-0.5,3.2){\small{$p+6$}}
\put(0.7,3.2){$\Bigg{\\{}$} \end{overpic} Figure 10. The
$(-(p+6),2,-3)$-pretzel is also the closure of the braid
$\sigma_{1}^{3}\sigma_{2}\sigma_{1}^{p+6}\sigma_{2}$. Start at the dashed line
and read the braid counterclockwise (strands numbered from the outside in).

The braid relation
$\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$ and the
centrality of $C=(\sigma_{1}\sigma_{2})^{3}$ can also be used to show that
$L_{p}$ is equivalent to the braid closure of
$\sigma_{1}^{p+1}(\sigma_{1}\sigma_{2})^{5}$:

$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}=\sigma_{1}^{p}(\sigma_{1}\sigma_{2})^{6}\sigma_{2}^{-1}=\sigma_{1}^{p}(\sigma_{1}\sigma_{2})^{5}\sigma_{1}\sim\sigma_{1}^{p+1}(\sigma_{1}\sigma_{2})^{5}.$

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.

######  Corollary 5.7.

The complements of the T-links/Lorenz links formed as closures of the braids
$\sigma_{1}^{p+1}(\sigma_{1}\sigma_{2})^{5}$ admit geometric triangulations
for $p>1$. ∎

##  6\. Extending the construction

Having constructed geometric triangulations for the braid closure of
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-1}$, 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
$C^{k}\sigma_{1}^{p}\sigma_{2}^{-1}$ with $p>0$ and $k\geq 2$, 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.

Another possible extension would be to add more $\sigma_{2}^{-1}$ half-twists.
The construction of $\hat{\tau}_{p}$ in Section 2.4 can be extended, without
difficulty, to $C^{2}\sigma_{1}^{p}\sigma_{2}^{-q}$ for all $q>0$, and, in
each case, there is also a sequence of Pachner moves that results in a veering
triangulation. Thus, the first step of the Casson-Rivin program can be carried
out. However, the final push for geometricity (the analog of Proposition 5.2)
would require additional arguments. After completing
$C^{2}\sigma_{1}^{p}\sigma_{2}^{-q}$, it may be possible to further extend the
construction to
$C^{2}\sigma_{1}^{p_{1}}\sigma_{2}^{-q_{1}}\cdots\sigma_{1}^{p_{s}}\sigma_{2}^{-q_{s}}$,
moving closer to the general form in the Futer-Kalfagianni-Purcell
characterization of hyperbolic 3-braids [7] stated in Section 1.

##  Appendix A Visualizing triangulations of 2-bridge link complements

Sakuma and Weeks constructed topological triangulations of 2-bridge link
complements [18], which Futer showed were geometric [9, Appendix]. Futer’s
approach follows that of Guéritaud, whose development of geometric
triangulations of once-punctured torus bundles and 4-punctured sphere bundles
forms the bulk of [9]. This appendix presents another way to visualize the
triangulations of 2-bridge link complements.

Futer’s description of these triangulations is based on the fact that, with
the exception of the unknot and the trivial link with two components, 2-bridge
links can be constructed from certain 4-braids whose ends are connected. More
specifically, these 4-braids run between nested pillowcases; are formed by a
sequence of $R$ and $L$ moves on the strands; and are closed off by adding
crossing strands inside the inner pillowcase and outside the outer one. Please
see [9, Appendix] for the details and a very clear exposition.

As Futer mentions, his 2-bridge link diagrams can be isotoped so the
pillowcases are horizontal, that is, they are perpendicular to the plane of
the page and contain the point at infinity. The result is a diagram in which
the braid strands run vertically between the bounding pillowcases, constrained
to the plane of the page, except near the crossings. See Figure 11 for a
comparison of the defining $R$ and $L$ moves that occur between two
pillowcases.

\begin{overpic}[unit=.25in]{RandL.pdf} \put(3.0,1.5){\small$L$}
\put(11.5,1.5){\small$L$} \put(3.0,5.5){\small$R$} \put(11.5,5.5){\small$R$}
\end{overpic} Figure 11. The actions of $R$ and of $L$ on strands between
pillowcases. Left side is Figure 14 in [9, Appendix]. Right side is version
with vertical strands.

Just as in Futer’s Appendix, the 4-braid lives in a product region,
$S^{2}\times I$, and its complement in the region is also a product, $S\times
I$, where $S$ is a 4-punctured sphere. As an example (left side of Figure 18
in [9, Appendix]), Futer uses the 2-bridge link $K(\Omega)$ where
$\Omega=R^{3}L^{2}R$, so we will too. The left side of Figure 12 shows how to
visualize $K(\Omega)$ — 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 labelled $\Delta_{i}$ and they each contain ideal tetrahedra
$T_{i}$ and $T_{i}^{\prime}$. The layer $\Delta_{i}$ is bounded by 4-punctured
spheres, below by $S_{i}$ and above by $S_{i+1}$, with $S_{i+1}$ in
$\Delta_{i}$ identified to $S_{i+1}$ in $\Delta_{i+1}$ by passing through a
half-twist. Numbering the punctures 0–3 will help in describing the resulting
face pairings.

\begin{overpic}[unit=.25in]{2-bridgelink.pdf}
\put(0.25,19.8){\small$S_{6}$}\put(0.25,17.8){\small$S_{5}$}\put(0.25,15.8){\small$S_{5}$}\put(0.25,13.8){\small$S_{4}$}\put(0.25,11.8){\small$S_{4}$}\put(0.25,9.8){\small$S_{3}$}\put(0.25,7.8){\small$S_{3}$}\put(0.25,5.8){\small$S_{2}$}\put(0.25,3.8){\small$S_{2}$}\put(0.25,1.8){\small$S_{1}$}
\par\put(8.25,20.2){\small$R$}\put(8.25,16.8){\small$L$}\put(8.25,12.8){\small$L$}\put(8.25,8.8){\small$R$}\put(8.25,4.8){\small$R$}\put(8.25,1.2){\small$R$}
\par\put(1.7,20.1){\tiny 0}\put(3.7,20.1){\tiny 1}\put(5.7,20.1){\tiny
2}\put(7.7,20.1){\tiny 3} \put(1.7,18.1){\tiny 0}\put(3.7,18.1){\tiny
1}\put(5.7,18.1){\tiny 2}\put(7.7,18.1){\tiny 3} \put(1.7,16.1){\tiny
0}\put(3.7,16.1){\tiny 1}\put(5.7,16.1){\tiny 2}\put(7.7,16.1){\tiny 3}
\put(1.7,14.1){\tiny 0}\put(3.7,14.1){\tiny 1}\put(5.7,14.1){\tiny
2}\put(7.7,14.1){\tiny 3} \put(1.7,12.1){\tiny 0}\put(3.7,12.1){\tiny
1}\put(5.7,12.1){\tiny 2}\put(7.7,12.1){\tiny 3} \put(1.7,10.1){\tiny
0}\put(3.7,10.1){\tiny 1}\put(5.7,10.1){\tiny 2}\put(7.7,10.1){\tiny 3}
\put(1.7,8.1){\tiny 0}\put(3.7,8.1){\tiny 1}\put(5.7,8.1){\tiny
2}\put(7.7,8.1){\tiny 3} \put(1.7,6.1){\tiny 0}\put(3.7,6.1){\tiny
1}\put(5.7,6.1){\tiny 2}\put(7.7,6.1){\tiny 3} \put(1.7,4.1){\tiny
0}\put(3.7,4.1){\tiny 1}\put(5.7,4.1){\tiny 2}\put(7.7,4.1){\tiny 3}
\put(1.7,2.1){\tiny 0}\put(3.7,2.1){\tiny 1}\put(5.7,2.1){\tiny
2}\put(7.7,2.1){\tiny 3} \par\put(11.7,20.1){\tiny 1}\put(14.1,20.1){\tiny
2}\put(14.7,20.1){\tiny 0}\put(17.1,20.1){\tiny 3} \put(11.7,17.7){\tiny
0}\put(14.1,17.7){\tiny 3}\put(14.7,17.7){\tiny 1}\put(17.1,17.7){\tiny 2}
\par\put(11.7,16.1){\tiny 1}\put(14.1,16.1){\tiny 2}\put(14.7,16.1){\tiny
0}\put(17.1,16.1){\tiny 3} \put(11.7,13.7){\tiny 0}\put(14.1,13.7){\tiny
3}\put(14.7,13.7){\tiny 1}\put(17.1,13.7){\tiny 2} \par\put(11.7,12.1){\tiny
1}\put(14.1,12.1){\tiny 2}\put(14.7,12.1){\tiny 0}\put(17.1,12.1){\tiny 3}
\put(11.7,9.7){\tiny 0}\put(14.1,9.7){\tiny 3}\put(14.7,9.7){\tiny
1}\put(17.1,9.7){\tiny 2} \par\put(11.7,8.1){\tiny 1}\put(14.1,8.1){\tiny
2}\put(14.7,8.1){\tiny 0}\put(17.1,8.1){\tiny 3} \put(11.7,5.7){\tiny
0}\put(14.1,5.7){\tiny 3}\put(14.7,5.7){\tiny 1}\put(17.1,5.7){\tiny 2}
\par\put(11.7,4.1){\tiny 1}\put(14.1,4.1){\tiny 2}\put(14.7,4.1){\tiny
0}\put(17.1,4.1){\tiny 3} \put(11.7,1.7){\tiny 0}\put(14.1,1.7){\tiny
3}\put(14.7,1.7){\tiny 1}\put(17.1,1.7){\tiny 2}
\par\put(9.2,18.9){\small$\Delta_{5}\quad=\quad$}\put(9.2,14.9){\small$\Delta_{4}\quad=\quad$}\put(9.2,10.9){\small$\Delta_{3}\quad=\quad$}\put(9.2,6.9){\small$\Delta_{2}\quad=\quad$}
\put(9.2,2.9){\small$\Delta_{1}\quad=\quad$}
\par\put(12.8,17.4){\small$T_{5}$}\put(15.8,17.4){\small$T_{5}^{\prime}$}
\put(12.8,13.4){\small$T_{4}$}\put(15.8,13.4){\small$T_{4}^{\prime}$}
\put(12.8,9.4){\small$T_{3}$}\put(15.8,9.4){\small$T_{3}^{\prime}$}
\put(12.8,5.4){\small$T_{2}$}\put(15.8,5.4){\small$T_{2}^{\prime}$}
\put(12.8,1.4){\small$T_{1}$}\put(15.8,1.4){\small$T_{1}^{\prime}$}
\par\end{overpic} Figure 12. The link $K(\Omega)$ where $\Omega=R^{3}L^{2}R$.
Compare left side to left side of Figure 18 in [9, Appendix]. Also similar to
the top of Figure II.3.3 in [18].

The right side of Figure 12 shows the tetrahedra forming the complement of
$K(\Omega)$. 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 $\Delta_{1}$, 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 $T_{1}^{\prime}$ — live 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 Figure 15
[9, Appendix]):

  * •

$\triangle 013$ in the front (blue) is half twisted to $\triangle 023$ in the
front;

  * •

$\triangle 123$ in the front (green) is half twisted to $\triangle 213$ in the
back;

  * •

$\triangle 023$ in the back (orange) is half twisted to $\triangle 013$ in the
back; and

  * •

$\triangle 012$ in the back (pink) is half twisted to $\triangle 021$ in the
front.

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 $\Delta_{1}$, 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. $T_{1}^{\prime}$ is behind the page with faces 012
and 023 on top. The next two diagrams also show $T_{1}$ and $T_{1}^{\prime}$,
using arrows to indicate edge identifications. The last one is our version of
the two tetrahedra in Figure 16 [9, Appendix], 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 associated with the flattened 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.

\begin{overpic}[unit=.25in]{firstlayerLight.pdf}
\put(0.25,11.2){\small$S_{2}$}\put(0.25,9.2){\small$S_{2}$}\put(0.25,7.2){\small$S_{1}$}
\par\put(8.25,8.2){\small$\Delta_{1}$} \par\put(8.25,10.2){\small$R$}
\par\put(1.7,11.4){\tiny 0}\put(3.7,11.4){\tiny 1}\put(5.7,11.4){\tiny
2}\put(7.7,11.4){\tiny 3} \put(1.7,9.4){\tiny 0}\put(3.7,9.4){\tiny
1}\put(5.7,9.4){\tiny 2}\put(7.7,9.4){\tiny 3} \put(1.7,7.4){\tiny
0}\put(3.7,7.4){\tiny 1}\put(5.7,7.4){\tiny 2}\put(7.7,7.4){\tiny 3}
\par\put(11.4,9.4){\tiny 0}\put(13.4,9.4){\tiny 1}\put(15.4,9.4){\tiny
2}\put(17.4,9.4){\tiny 3} \put(11.4,3.4){\tiny 0}\put(13.4,3.4){\tiny
1}\put(15.4,3.4){\tiny 2}\put(17.4,3.4){\tiny 3} \par\put(3.7,5.4){\tiny
0}\put(6.1,5.4){\tiny 3} \put(3.7,3.25){\tiny 1}\put(6.1,3.25){\tiny 2}
\put(3.7,1.1){\tiny 0}\put(6.1,1.1){\tiny 3}
\par\par\put(11.2,10.1){\small$T^{\prime}_{1}$}\put(11.2,8.3){\small$T_{1}$}
\put(11.2,4.8){\small$T^{\prime}_{1}$}\put(11.2,1.8){\small$T_{1}$}
\put(3.1,4.0){\small$T^{\prime}_{1}$}\put(3.1,2.0){\small$T_{1}$}
\par\end{overpic} Figure 13. The layer $\Delta_{1}$. Compare first figure to
Figure 15 in [9, Appendix] and last figure to Figure 16 in [9, Appendix].

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
$\Omega$ starts with $R$, identify bottom layer triangles $\triangle
012\sim\triangle 013$ and $\triangle 023\sim\triangle 123$; If $\Omega$ starts
with $L$, identify bottom layer triangles $\triangle 012\sim\triangle 312$ and
$\triangle 023\sim\triangle 013$. The top layer triangles are identified in
the same way. For example, if $\Omega$ ends with $R$, $\triangle
012\sim\triangle 013$ and $\triangle 023\sim\triangle 123$.

It should not be surprising that these identifications yield diagrams looking
very much like Figure 17 in [9, Appendix] and Figure II.2.7 in [18]. For
example, the steps forming the bottom clasp of $K(\Omega)$ where
$\Omega=R^{3}L^{2}R$ are shown in Figure 14. While making the specified
identifications, we can follow the punctures (their paths are drawn as
strands) and see that they sweep out a clasp. Folding $S_{1}$ down along the
12 edge forms a pillow. Bringing the 2 and 3 strands together identifies the
gray triangles, $\triangle 012$ and $\triangle 013$. Placing this triangle in
the plane of the page means the blue 23 edge forms a belt around a new pillow
with half in the front, like the 2 strand, and half behind as the 3 strand is.
The top of the pillow is a cone formed by $\triangle 123$ and the bottom by
$\triangle 023$, the white triangles with the cyan markings. Pushing the 1 and
0 strands into the pillow flattens the cones and identifies $\triangle 123$ to
$\triangle 023$, thus completing the identifications.

\begin{overpic}[unit=.25in]{claspBright.pdf} \put(0.25,7.2){\small$S_{1}$}
\par\put(1.7,7.4){\tiny 0}\put(3.7,7.4){\tiny 1}\put(5.7,7.4){\tiny
2}\put(7.7,7.4){\tiny 3} \par\put(10.9,8.4){\tiny 0}\put(12.9,8.4){\tiny
1}\put(14.9,8.4){\tiny 2}\put(16.9,8.4){\tiny 3} \put(12.7,7.4){\tiny
1}\put(15.1,7.4){\tiny 2} \put(12.7,5.0){\tiny 0}\put(15.1,5.0){\tiny 3}
\par\put(10.9,3.4){\tiny 0}\put(12.9,3.4){\tiny 1}\put(14.9,3.4){\tiny
2}\put(16.9,3.4){\tiny 3} \put(12.7,2.4){\tiny 1} \put(12.7,0.0){\tiny 0}
\par\put(1.9,3.4){\tiny 0}\put(3.9,3.4){\tiny 1}\put(5.9,3.4){\tiny
2}\put(7.9,3.4){\tiny 3} \par\end{overpic} Figure 14. Forming a clasp. Compare
to Figure 17 [9, Appendix] and Figure II.2.7 in [18].

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.

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