# Essential loops in taut ideal triangulations

Saul Schleimer  Mathematics Institute
University of Warwick
Coventry CV4 7AL, United Kingdom<EMAIL_ADDRESS>and  Henry
Segerman  Department of Mathematics
Oklahoma State University
Stillwater, OK, 74078, USA<EMAIL_ADDRESS>
###### Abstract.

In this note we combinatorialise a technique of Novikov. We use this to prove
that, in a three-manifold equipped with a taut ideal triangulation, any
vertical or normal loop is essential in the fundamental group.

This work is in the public domain.

##  1\. Introduction

The notion of a taut ideal triangulation of a three-manifold is due to
Lackenby [7]. He combinatorialised the angle structures introduced
independently by Casson and by Rivin [12]. They in turn linearised the
geometric triangulations of Thurston [13]. Each of these structures plays an
important role in modern low-dimensional topology. In particular, taut ideal
triangulations have a strong connection to the subject of taut foliations,
introduced by Gabai [5], and to that of taut branched surfaces, due to Oertel
[10]. In addition to the results of Lackenby, taut ideal triangulations play a
central role in the theory of layered triangulations. One spectacular
contribution been as a prerequisite for Agol’s theory of veering
triangulations [1].

Novikov [9, Theorem 6.1] gives one of the early applications of foliations to
the study of the fundamental group of a manifold. He starts with a loop
$\delta$ in good position with respect to a foliation $\mathcal{F}$. He
further supposes that $H\colon D\to M$ is a null-homotopy of $\delta$, also in
good position. Pulling back, he obtains a singular foliation
$H^{-1}(\mathcal{F})$ on the disk $D$. The Poincaré–Hopf theorem gives
combinatorial control of the singularities, which translates to topological
control over the homotopy. Morally, the positivity of the Euler characteristic
of the disk constrains the position of $\delta$. We refer to [4, Chapter 9]
for a history of the subject and for detailed proofs.

We introduce a combinatorial version of the Novikov technique; instead of
pulling back a foliation we pull back a taut ideal triangulation. This gives a
train track with stops in the disk $D$. We obtain a new, and very simple,
proof of one of Novikov’s results. That is, suppose that $M$ is a three-
manifold, equipped with a taut ideal triangulation $\mathcal{T}$. Let
$\mathcal{B}=\mathcal{T}^{(2)}$ be the resulting branched surface in $M$.

Any loop $\delta$ in $M$ which is vertical with respect to $\mathcal{B}$ is
essential in $\pi_{1}(M)$.

There is an indirect proof of this using Novikov’s original technique [3,
Theorem 4.35(3)], once we observe that $\mathcal{B}$ carries an essential
lamination ([6, Example 5.1], [7, page 373]) which extends to a taut foliation
of $M$. We also obtain a new result, as follows.

Any loop $\gamma$ in $M$ which is normal with respect to $\mathcal{B}$ is
essential in $\pi_{1}(M)$.

The proof of Theorem 5.1 is more delicate than that of Theorem 3.2; new
behaviour near the boundary of $D$ must be dealt with.

From Theorems 3.2 and 5.1 we deduce that vertical and normal loops are
infinite order in the fundamental group. This is, however, still a bit weaker
than the situation of a train track $\sigma$ in a surface – there loops dual
to or carried by $\sigma$ are again essential but also non-peripheral.

We have a simple corollary of Theorem 5.1. Let ${\widetilde{M}}$ be the
universal cover of $M$ and let ${\widetilde{\mathcal{B}}}$ be the resulting
branched surface.

######  Corollary 1.1.

Suppose that $F$ is a connected surface (perhaps with boundary) carried by
${\widetilde{\mathcal{B}}}$ and realised as a (perhaps finite) union of faces
of ${\widetilde{\mathcal{B}}}$. Then $F$ is a disk. ∎

### Previous work

The idea of pulling back a branched surface by a null-homotopy has previously
appeared. In Gabai–Oertel’s proof of their Lemma 2.7 in [6] they use this to
show that laminations carried by essential branched surfaces are
$\pi_{1}$–injective. Our Theorem 5.1 is both more general and less – our loop
$\gamma$ need not lie in a leaf however we do not deal with the case where $M$
is closed. Calegari [2] gives another combinatorial version of results of
Novikov, in the closed case. He introduces the notion of a _local orientation_
on a closed triangulated three-manifold $M$. This is in a sense dual to having
a transverse taut branched surface $\mathcal{B}\subset M$ where all components
of $M-\mathcal{B}$ are taut balls.

### Acknowledgements

We thank Marc Lackenby for helpful conversations. The second author was
supported in part by National Science Foundation grant DMS-1708239.

##  2\. Background

Throughout the paper we will use $M$ to denote a compact connected manifold
with non-empty boundary. All boundary components will be tori or Klein
bottles. Suppose that $\mathcal{T}$ is an abstract three-dimensional
triangulation. Let $|\mathcal{T}|$ be the realisation space of $\mathcal{T}$
and let $\mathcal{T}^{(k)}$ be its $k$–skeleton. Let $n(\mathcal{T}^{(0)})$ be
an open regular neighbourhood of the vertices of $\mathcal{T}$. We call
$\mathcal{T}$ a _ideal triangulation_ of $M$ if
$|\mathcal{T}|-n(\mathcal{T}^{(0)})$ is homeomorphic to $M$.

A _taut angle structure_ on $\mathcal{T}$ is an assignment of dihedral angles,
zero or $\pi$, to each edge of each model tetrahedron in $\mathcal{T}$. The
assignment is required to obey two conditions. The _edge equalities_ state
that, for an edge $e\in\mathcal{T}^{(1)}$, the sum of the dihedral angles of
its models is $2\pi$. The _triangle equalities_ state that, for a vertex of a
model tetrahedron, the sum of the dihedral angles of the three adjacent edges
is $\pi$. We say that the tetrahedra of $\mathcal{T}^{(3)}$ are _taut_. See
Figure 2.1a.

2pt 0 at 20 130 0 at 240 120 0 at 135 27 0 at 135 217 $\pi$ at 125 140 $\pi$
at 125 87 [width=0.35]Figures/TautTet

(a) A taut tetrahedron.

[width=0.55]Figures/branched_surface

(b) All faces meeting a single edge in $\mathcal{B}$.

Figure 2.1.

We deduce that every taut tetrahedron has four edges with dihedral angle zero.
We call the union of the four edges the _equator_ of the taut tetrahedron.

Suppose now that $e$ is an edge of $\mathcal{T}^{(1)}$. There are exactly two
model edges for $e$ with angle $\pi$; all others are zero. Obeying these
dihedral angles, we isotope the two-skeleton $\mathcal{T}^{(2)}$ to obtain a
smooth _branched surface_ $\mathcal{B}$. See Figure 2.1b. Some references
would call $\mathcal{B}$ a non-generic branched surface without vertices. See
for example [3, Section 6.3].

######  Definition 2.1.

Suppose that $\delta$ is a smooth embedded loop in $M$, transverse to
$\mathcal{B}$. Suppose that for every tetrahedron $t$ we have that every arc
$d$ of $\delta\cap t$ _links_ the equator of $t$. (That is, the endpoints of
$d$ are separated in $\partial t$ by the equator of $t$.) Then we say that the
loop $\delta$ is _vertical_ with respect to $\mathcal{B}$.

######  Definition 2.2.

Suppose that $\gamma$ is a smooth loop immersed in $\mathcal{B}$ and
transverse to the edges of $\mathcal{B}$. Suppose that, for every face $f$ of
$\mathcal{B}$ and for every component $J$ of $\gamma^{-1}(f)$, the arc
$\gamma(J)$ is _normal_ in $f$. (That is, the endpoints of $\gamma(J)$ lie in
distinct edges of $f$.) Then we say that the loop $\gamma$ is _normal_ with
respect to $\mathcal{B}$.

##  3\. Combinatorics of null-homotopies

Suppose that $\delta$ is a loop in $M$ which is transverse to the branched
surface $\mathcal{B}$. Let $D=D^{2}$ be the unit disk with the usual
orientation. Suppose that $H\colon D\to M$ is a null-homotopy of $\delta$. We
homotope $H$ relative to $\partial D$ to make $H$ transverse to $\mathcal{B}$.

We define $\tau=H^{-1}(\mathcal{B})$. Thus $\tau$ is a _train track_ in
$\partial D$. The _switches_ of $\tau$ are exactly the points of
$H^{-1}(\mathcal{B}^{(1)})$. The _stops_ of $\tau$ are exactly the points of
$(H|\partial D)^{-1}(\mathcal{B})$. The standard reference for train tracks is
[11]; we also rely on [8]. We note that our track $\tau$ does not satisfy the
so-called “geometry-condition” [11, page 5], [page 52][8].

We call a connected component $R$ of $D-\tau$ a _region_. Let
$\operatorname{cusps}(R)$ and $\operatorname{corners}(R)$ count the number of
(necessarily outwards) cusps and corners on the boundary of $R$. As a bit of
terminology, we divide $\partial R$ into _sides_ : these are the components of
$\partial R$ minus all outward cusps and corners. Note that a side $s$ of $R$
may be a union of several branches of $\tau$.

We define the _index_ of $R$ to be

$\operatorname{ind}(R)=\chi(R)-\frac{\operatorname{cusps}(R)}{2}-\frac{\operatorname{corners}(R)}{4}$

In Table 1 we give pictures of, and names to, all possible disk regions with
non-negative index. Note that index is additive under taking the union of
regions [8, page 57]. Thus the sum of the indices of the regions of $D-\tau$
is exactly $\chi(D)$; that is, one. We deduce from this that there is at least
one region $R$ with positive index.

$\operatorname{corners}\backslash\textrm{index}$ |  $1$ |  $1/2$ |  $0$
---|---|---|---
|  |  |
0 |  [width=0.16]Figures/D_00 |  [width=0.16]Figures/D_10 |  [width=0.16]Figures/D_20
2 |  |  [width=0.16]Figures/D_02 |  [width=0.16]Figures/D_12
4 |  |  |  [width=0.16]Figures/D_04
Table 1. Disk regions with non-negative index, organised by the number of
corners. These are named as follows: _nullgon_ , _cusped monogon_ , _cusped
bigon_ , _boundary bigon_ , _boundary trigon_ , and _rectangle_.

Let $r(H)$ be the number of regions of $D-\tau$. Over all null-homotopies of
$\delta$, transverse to $\mathcal{B}$, we choose $H$ to minimise $r(H)$. We
call such an $H$ _minimal_.

######  Lemma 3.1.

Suppose that $\delta$ is a loop in $M$ transverse to $\mathcal{B}$. Suppose
that $H\colon D\to M$ is a minimal null-homotopy of $\delta$. Then we have the
following.

  1. (1)

All regions of $D-\tau$ are disks.

  2. (2)

If $s$ is a side of a region $R$, then the interior of $s$ meets at most one
switch.

  3. (3)

No region $R$ of $D-\tau$ is a nullgon.

  4. (4)

No region $R$ of $D-\tau$ is a monogon.

Thus, all positive index regions of $D$ are boundary bigons.

###### Proof.

  1. (1)

If there were a region with topology then we could compress it into the
containing tetrahedron and reduce $r(H)$.

  2. (2)

Suppose that the interior of $s$ meets at least two switches. Then there is a
branch $b\subset\tau$ so that $H(b)$ is a non-normal arc. We homotope $b$ to
be simple and then across its bigon. This does not increase $r(H)$. If $r(H)$
does not decrease, then this move disconnects $\tau$, and creates a region
with topology, contradicting (1).

  3. (3)

Suppose that $R$ is a nullgon. If $H(\partial R)$ is disjoint from
$\mathcal{B}^{(1)}$ then the region adjacent to $R$ is not a disk,
contradicting (1). It follows that $\partial R$ consists of an even number of
branches of $\tau$ (alternating between the two faces of a tetrahedron $t$ on
either side of a $\pi$-edge of $t$). But this contradicts (2).

  4. (4)

Suppose that $t$ is the taut tetrahedron containing $H(R)$. Let $s$ be the
boundary of $R$. We deduce that the loop $s$ crosses the equator of $t$
exactly once, a contradiction.

Since there are no nullgons or monogons, the only possible positive index
regions are boundary bigons. ∎

Equipped with this we can now prove the following.

######  Theorem 3.2.

Any loop $\delta$ in $M$ which is vertical with respect to $\mathcal{B}$ is
essential in $\pi_{1}(M)$.

###### Proof.

Suppose that $H\colon D\to M$ is a minimal null-homotopy of the vertical loop
$\delta$. Applying Lemma 3.1, there must be a region $R$ of $D-\tau$ which is
a boundary bigon. Let $t$ be the tetrahedron containing $H(R)$. Let
$d=\partial R\cap\partial D$ and let $s=\partial R-d^{\circ}$. From the
definition of vertical, we have that $H(d)$ links the equator of $t$.
Therefore $H(s)$ crosses the equator of $t$ an odd number of times, and so at
least once, a contradiction. ∎

##  4\. Transverse taut

In order to prove Theorem 5.1, we will use the following strengthening of the
notion of a taut structure. A _transverse taut structure_ on $\mathcal{T}$ is
a taut structure together with a co-orientation on $\mathcal{B}$ with the
following property. If faces $f$ and $f^{\prime}$ of any model tetrahedron
share a common model edge $e$, then

  * •

the edge $e$ is part of the equator of $t$ if and only if exactly one of the
co-orientations on $f$ and $f^{\prime}$ points into $t$.

See Figure 4.1a. It follows that the co-orientations on faces incident to an
edge change direction precisely twice as we go around an edge. See Figure
4.1b.

2pt 0 at 20 130 0 at 240 120 0 at 135 27 0 at 135 217 $\pi$ at 125 140 $\pi$
at 125 87 [width=0.35]Figures/TransverseTet

(a) Co-orientations and angles in a transverse taut tetrahedron.

[width=0.55]Figures/transverse_branched_surface

(b) Co-orientations around an edge.

Figure 4.1.

Suppose that $\mathcal{T}$ is an ideal triangulation of a manifold $M$
equipped with a taut structure. We now construct a triangulation
${\widetilde{\mathcal{T}}}$ of a double cover ${\widetilde{M}}$ of $M$. By
construction, the lift of the taut structure on $\mathcal{T}$ to
${\widetilde{\mathcal{T}}}$ will support a transverse taut structure.

For each taut tetrahedron $t$ of $\mathcal{T}$, we arbitrarily label the two
model edges with dihedral angle $\pi$ as $e^{\prime}$ and $e^{\prime\prime}$.
In ${\widetilde{\mathcal{T}}}$, we have two taut tetrahedra $t^{\prime}$ and
$t^{\prime\prime}$ corresponding to $t$. We assign a co-orientation to the
model faces of $t^{\prime}$ and $t^{\prime\prime}$ in such a way that the co-
orientation points into the tetrahedron on the two model faces of $t^{*}$
incident to $e^{*}$. Now suppose that $t_{i}$ and $t_{j}$ are tetrahedra of
$\mathcal{T}$, glued to each other along model faces $f_{i}$ and $f_{j}$. In
${\widetilde{\mathcal{T}}}$ we have tetrahedra $t^{\prime}_{i}$,
$t^{\prime\prime}_{i}$, $t^{\prime}_{j}$, and $t^{\prime\prime}_{j}$, with
model faces $f^{\prime}_{i}$, $f^{\prime\prime}_{i}$, $f^{\prime}_{j}$, and
$f^{\prime\prime}_{j}$, respectively.

We glue $t^{\prime}_{i}$ to either $t^{\prime}_{j}$ or $t^{\prime\prime}_{j}$
as the co-orientation on $f^{\prime}_{i}$ agrees with $f^{\prime}_{j}$ or
$f^{\prime\prime}_{j}$. We similarly glue $t^{\prime\prime}_{i}$ to the
remaining copy of $t_{j}$. Having made all such gluings, the resulting
triangulation ${\widetilde{\mathcal{T}}}$ has a transverse taut structure by
construction. It has one component if and only if the taut structure on
$\mathcal{T}$ does not support a transverse taut structure.

##  5\. Proof of the main result

######  Theorem 5.1.

Any loop $\gamma$ in $M$ which is normal with respect to $\mathcal{B}$ is
essential in $\pi_{1}(M)$.

###### Proof.

Suppose for a contradiction, that the normal loop $\gamma$ is null-homotopic.
Thus $\gamma$ lifts to a normal loop in any cover. Thus, without loss of
generality, we may assume that the taut structure on $\mathcal{T}$ supports a
transverse taut structure. This gives us a local notion of _upwards_. In
particular, every model tetrahedron has two lower faces and two upper faces,
separated by its equator.

Lemma 3.1 does not apply directly to a normal loop $\gamma$. Instead, we
choose a small smooth homotopy $G\colon A\to M$ which moves $\gamma$ slightly
upwards. Thus $A$ is an annulus with $G(\partial_{0}A)=\gamma$. Furthermore
$G$ is transverse to $\mathcal{B}$ away from $\partial_{0}A$ and for each
vertical interval $J$ in $A$, the tangents to $G(J)$ point upwards. Let
$\delta=G(\partial_{1}A)$. We will apply Lemma 3.1 to $\delta$.

We call $\delta$ a _raised curve_. We call the components of
$\delta-\mathcal{B}$ _raised arcs_. There are six _types_ of raised arc. These
are shown in Figure 5.1. There is a cellulation of $A$ with one-skeleton
$\partial A\cup G^{-1}(\mathcal{B})$. Suppose that $C$ is a two-cell. Let
$c=C\cap\partial_{0}A$ and $d=C\cap\partial_{1}A$. Thus $G(c)\subset\gamma$
and $G(d)\subset\delta$. We say that $G(c)$ is the _lowering_ of the raised
arc $G(d)$. We record this by the _lowering map_ , $L$, where $L(G(d))=G(c)$.
Note that $G(c)$ may be either a single vertex, a single normal arc, or two
normal arcs. Again, see Figure 5.1.

2pt $\textsc{a}_{3}$ at 405 152 $\textsc{a}_{2}$ at 363 224 $\textsc{a}_{1}$
at 342 280 $\textsc{b}_{1}$ at 63 172 $\textsc{b}_{2}$ at 70 98 c at 215 98
[height = 5.5 cm]Figures/raised_arcs

Figure 5.1. A taut tetrahedron containing the six possible types of raised
arcs of $\delta$. These are drawn in (solid) green. The normal arcs (or
points) of $\gamma$, namely the lowerings of the raised arcs, are drawn in
(dashed) blue. Images of the two-cells of the annulus $A$ are shaded in light
blue. Filled green dots indicate endpoints of raised arcs on the top two faces
of the tetrahedron; open green dots indicate endpoints on the bottom two
faces.

Suppose that $H\colon D\to M$ is a minimal null-homotopy of $\delta$. Recall
that $\tau=H^{-1}(\mathcal{B})$. Applying Lemma 3.1 implies that $D-\tau$ has
at least two boundary bigons. Applying another small homotopy, we can retain
minimality and also make $H$ transverse to $\gamma$.

Pulling back the transverse taut structure on $\mathcal{B}$ by $H$ gives a
transverse orientation on the branches of $\tau$ which is consistent across
switches. Thus, for any region $R$ of $D-\tau$ and for any side $s$ of $R$,
the transverse orientation on $s$ points either into, or out of, $R$. This
gives us a classification of boundary bigons. Suppose that $R$ is a boundary
bigon and $s=\partial R-\partial D$ is its side in $\tau$. If the transverse
orientation on $s$ points out of $R$ then we call $R$ a _min-bigon_. If it
points into $R$ we call $R$ a _max-bigon_.

###  5.1. Min-bigons

Suppose that $R$ is a min-bigon. We use $R$ to tell us how to move $\gamma$ up
across $H(R)$, producing a new normal loop $\gamma^{\prime}$, a new raised
loop $\delta^{\prime}$, and a new null homotopy $H^{\prime}$. See Figure 5.2.
The loop $\gamma^{\prime}$ may be shorter than, the same length as, or longer
than $\gamma$ (see types $\textsc{a}_{1},\textsc{a}_{2}$ and $\textsc{a}_{3}$
in Figure 5.1). However, $H^{\prime}$ has exactly one fewer region. That is,
$r(H^{\prime})=r(H)-1$. We repeat this process until there are no more min-
bigons.

[width=]Figures/push_over_min_bigon

Figure 5.2. Pushing over a min-bigon of type $A_{3}$.

###  5.2. Max-bigons

Suppose that $R_{0}$ is a max-bigon. Our plan is to uniquely associate to
$R_{0}$ two small subregions of $D-\tau$, each with index $-1/4$. This will
imply that the index of $D$ is at most zero. This contradiction finally proves
Theorem 5.1.

We begin as follows. Let $s$ be the side of $R_{0}$ in $\tau$. Let
$d_{0}=\partial R_{0}-s\subset\partial D$. We give $d_{0}$ the (tangential)
orientation it receives from $D$. In Figure 5.3, this orientation will point
left. Note that $H(d_{0})\subset\delta$ is a raised arc. Let
$c_{0}=L(H(d_{0}))$ be its lowering.

######  Claim 5.2.

  * •

The raised arc $H(d_{0})$ has type c.

  * •

The side $s$ meets exactly one switch $c^{\prime}_{0}$ of $\tau$.

  * •

The vertices $c_{0}$ and $H(c^{\prime}_{0})$ cobound a sub-edge
$e_{0}\subset\mathcal{B}^{(1)}$.

###### Proof.

Let $t_{0}$ be the tetrahedron containing $H(R_{0})$. By the definition of a
max-bigon the transverse orientation on $s$ points into $R_{0}$. Thus each
corner of $H(R_{0})$ is contained in a lower face of $t_{0}$. Consulting
Figure 5.1 we deduce that $H(d_{0})$ is of type c. Thus each corner of
$H(R_{0})$ is contained in its own lower face of $t_{0}$. We deduce that $s$
meets at least one switch of $\tau$. By Lemma 3.1(2) the side $s$ meets
exactly one switch, which we call $c^{\prime}_{0}$.

Since $H$ is transverse to $\gamma$, the vertices $c_{0}$ and
$H(c^{\prime}_{0})$ are distinct. They are contained in the same edge of
$\mathcal{B}^{(1)}$: namely the bottom edge $\epsilon_{0}$ of $t_{0}$. In
$\epsilon_{0}$ they cobound a sub-edge, which we call $e_{0}$. ∎

2pt $\delta$ at 0 85 $\gamma$ at 9 34 $a$ at 9 68 $\gamma^{\prime}$ at -2 9
$d_{0}$ at 50 92 $d_{1}$ at 95 92 $d_{2}$ at 112 92 $d_{3}$ at 130 92 $d_{4}$
at 148 92 $d_{5}$ at 166 92 $\cdots$ at 194 92 $d_{K}$ at 222 92 $b_{0}$ at 77
68 $b_{1}$ at 94 68 $b_{2}$ at 114 68 $b_{3}$ at 131 68 $b_{4}$ at 148 68
$b_{5}$ at 166 68 $b_{K-1}$ at 214 45 $c_{0}$ at 57 41 $s_{\triangleright}$ at
238 45 $e_{0}$ at 46 22 $c^{\prime}_{0}$ at 46 0 $c^{\prime}_{2}$ at 65 0
$c^{\prime}_{3}$ at 83 0 $c^{\prime}_{5}$ at 101 0 $\cdots$ at 136 0
$c^{\prime}_{K}$ at 176 0 [width = ]Figures/downhill_from_max_bigon2

Figure 5.3. A possible picture of part of the annulus $A$ (in back), the
bigons $B(k)$ (in front), and the homotopies $E(k)$ (bottom). To lighten the
notation in this figure, we have omitted applying $H$ to labels of subsets of
$D$. Transverse orientations on the branches $b_{i}$ are shown with arrows.
Note that $e_{0}=e_{1}$, $c^{\prime}_{0}=c^{\prime}_{1}$,
$c^{\prime}_{3}=c^{\prime}_{4}$, and so on.

Let $a$ and $b_{0}$ be the components of $s-c^{\prime}_{0}$, where $b_{0}$
meets the right endpoint of $d_{0}$. See the leftmost region on Figure 5.3.

We now induct. Consider a sequence of regions $R_{0},R_{1},\ldots,R_{N}$ that
meet $\partial D$ in the sides $d_{0},d_{1},\ldots d_{N}$ as we move along
$\partial D$ to the right. Define $c_{k}=L(H(d_{k}))\subset\gamma$: the
lowering of the raised arc $H(d_{k})$. Define
$\gamma(k)=\cup_{i=0}^{k}c_{i}\subset\gamma$.

Let $b_{i}$ be the branch of $\tau$ that meets $\partial D$ at the right
corner of $R_{i}$. We arrange matters so that $b_{i-1}$ and $b_{i}$ have the
same transverse orientation for $1\leq i<N$, while they have opposite
transverse orientations for $i=N$. Thus, $R_{N}$ is at a _local minimum_ of
$\gamma$, and we are going downhill to it from the _local maximum_ at $R_{0}$.
This downhill condition implies that for $i\in[1,N)$, the raised arc
$H(d_{i})$ is either of type $\textsc{b}_{1}$ or $\textsc{b}_{2}$. Again, see
Figure 5.1.

Recall that all positive index regions are now max-bigons. Thus none of the
$R_{i}$ can have positive index for $i>0$. Let $K$ be the smallest number for
which $R_{K}$ has negative index, or if there is none, then set $K=N$.

######  Claim 5.3.

The region $R_{K}$ is not a boundary trigon.

###### Proof.

If $K<N$ then by definition, $R_{K}$ has negative index, and so is not a
boundary trigon. If $K=N$ then $R_{N}$ cannot be a boundary trigon since the
transverse orientations on the two sides of a boundary trigon must agree, yet
$R_{N}$ is at a local minimum of $\gamma$. ∎

For all $k\in[0,K)$ we define the union $B(k)=\cup_{i=0}^{k}R_{i}$. Define
$\gamma^{\prime}(k)=\partial B(k)-(a^{\circ}\cup b^{\circ}_{k}\cup\partial
D)$.

######  Definition 5.4.

Suppose that $g,h\colon[0,1]\to\mathcal{B}$ are paths. Suppose that
$F\colon[0,1]\times[0,1]\to\mathcal{B}$ is a homotopy from $g$ to $h$. Thus
$g(x)=F(x,0)$ and $h(x)=F(x,1)$. We say that $F$ is _transverse_ if whenever
$F(x_{0},t_{0})$ is contained in a (1- or 2-) cell $C$ of $\mathcal{B}$, we
have that the _trace_ $F(x_{0},[0,1])$ lies in $C$.

######  Claim 5.5.

For all $k\in[1,K)$:

  1. (1)

The region $R_{k}$ is a boundary trigon.

  2. (2)

The union $B(k)$ has exactly two corners and no cusps.

  3. (3)

There is a transverse homotopy $F_{k}$ taking $\gamma(k)$ to
$H(\gamma^{\prime}(k))$.

###### Proof.

We will prove this by induction. Claim 5.2 implies the base case (for $k=1$)
in a manner essentially identical to the general inductive step, so we omit
its proof.

Suppose that the hypotheses hold at step $k$. Recall that $H(d_{k})$ has type
$\textsc{b}_{1}$ or $\textsc{b}_{2}$, so it has precisely one lower endpoint.
Let $f_{k}$ be the face that contains the lower endpoint. Let $p$ be the
endpoint of $\gamma(k)$, and let $\epsilon_{k}$ be the edge of $f_{k}$
containing $p$. Let $\beta$ be the normal arc of $\gamma$ immediately after
$p$. Let $f_{\beta}$ be the face containing $\beta$. Viewed in a small
neighbourhood of $\epsilon_{k}$, the faces $f_{\beta}$ and $f_{k}$ are on the
same side (say the right side) of $\epsilon_{k}$, and $f_{\beta}$ is below
$f_{k}$.

Let $p^{\prime}$ be the endpoint of $\gamma^{\prime}(k)$ meeting $b_{k}$. By
hypothesis (3), the transverse homotopy $F_{k}$ takes $p$ to $H(p^{\prime})$,
with trace lying in $\epsilon_{k}$. Since $H$ is transverse to $\epsilon_{k}$
at $H(p^{\prime})$, we deduce that $H(D)$ meets both $f_{k}$ and $f_{\beta}$
at $H(p^{\prime})$. Thus $p^{\prime}$ is a switch of $\tau$ with a cusp
immediately below $b_{k}$, to the right of $p^{\prime}$, pointing at
$\gamma^{\prime}(k)$ (which extends to the left of $p^{\prime}$). This cusp
lies in $R_{k+1}$, since $b_{k}$ is part of the boundary of $R_{k+1}$. See
Figure 5.3.

If $R_{k+1}$ has negative index then $k+1=K$ and we have nothing to prove. So
suppose that $R_{k+1}$ has index zero. Consulting Table 1 we deduce that
$R_{k+1}$ is a boundary trigon. This proves hypothesis (1). Note that
hypothesis (2) follows because $B(k)$ meets $R_{k+1}$ along $b_{k}$.

Let $s_{k+1}=\partial R_{k+1}-(d_{k+1}\cup b^{\circ}_{k})$ be the remaining
side of the boundary trigon $R_{k+1}$. By Lemma 3.1(2) there is at most one
switch in the interior of $s_{k+1}$. Let
$c^{\prime}_{k+1}=s_{k+1}-b^{\circ}_{k+1}-\partial D$. Note that
$\gamma^{\prime}(k+1)=\gamma^{\prime}(k)\cup c^{\prime}_{k+1}$.

The path $H(s_{k+1})$ has endpoints $H(p^{\prime})$ and the lower endpoint of
$H(d_{k+1})$. The point $H(p^{\prime})$ lies on the edge $\epsilon_{k}$.
Recall that $f_{k+1}$ is the face containing the lower endpoint of
$H(d_{k+1})$. There are two cases, depending on the type of $H(d_{k+1})$.

  * •

Suppose that $H(d_{k+1})$ has type $\textsc{b}_{1}$. Then
$\gamma(k+1)=\gamma(k)$. In this case, $\epsilon_{k}$ is a boundary edge of
$f_{k+1}$. Since there is at most one switch in the interior of $s_{k+1}$,
there are in fact no such switches. So $s_{k+1}=b_{k+1}$ and
$c^{\prime}_{k+1}$ is a single switch, equal to $p^{\prime}$. We deduce that
$\gamma^{\prime}(k+1)=\gamma^{\prime}(k)$. Since $\gamma(k+1)=\gamma(k)$ and
$\gamma^{\prime}(k+1)=\gamma^{\prime}(k)$, we set $F_{k+1}=F_{k}$.

  * •

Suppose that $H(d_{k+1})$ has type $\textsc{b}_{2}$. Then
$\gamma(k+1)=\gamma(k)\cup c_{k+1}$. Let $t_{k+1}$ be the tetrahedron
containing $H(R_{k+1})$. In this case, the path $H(s_{k+1})$ must cross the
bottom edge of $t_{k+1}$ in order to get into $f_{k+1}$. Since there is at
most one switch in the interior of $s_{k+1}$, there is exactly one. Let $f$ be
other lower face of $t_{k+1}$. Thus $c_{k+1}$ is a normal arc in $f$. Note
that $H(c^{\prime}_{k+1})$ is a properly immersed arc in $f$, with endpoints
on the same edges as those of $c_{k+1}$. Thus there is a transverse homotopy
$E$ taking $c_{k+1}$ to $H(c^{\prime}_{k+1})$. Reparametrising $E$, we set
$F_{k+1}=F_{k}\cup E$.

This proves hypothesis (3). ∎

Let $B_{\triangleright}=B(K-1)$. This is the _right-bigon_ for $R_{0}$. We
rerun the argument of Claim 5.5 to the left to obtain the _left-bigon_ for
$R_{0}$, denoted $B_{\triangleleft}$.

The induction in the proof of Claim 5.5 extends to show that $R_{K}$ contains
a cusp pointing at $\gamma^{\prime}(K-1)$. The cusp lies between $b_{K-1}$ and
another branch on the boundary of $R_{K}$, which we call $c^{\prime}_{K}$. See
the far right of Figure 5.3. Let $Q_{\triangleright}$ be a small closed
regular neighbourhood of $b_{K-1}$ in $R_{K}$. The boundary of the subregion
$Q_{\triangleright}$ has four sides; we call it a _right-quadrilateral_. The
four sides are $d_{K}\cap N,b_{K-1},c_{K}\cap N,$ and a fourth side,
$s_{\triangleright}$ say. Note that $s_{\triangleright}$ is properly embedded
in $R_{K}$. The quadrilateral $Q_{\triangleright}$ therefore has one cusp and
three corners, and so it has index $-1/4$. An identical argument builds the
_left-quadrilateral_ $Q_{\triangleleft}$.

Let $S(R_{0})=Q_{\triangleleft}\cup B_{\triangleleft}\cup
B_{\triangleright}\cup Q_{\triangleright}$.

######  Claim 5.6.

For any max-bigons $R$ and $R^{\prime}$

  1. (1)

$S(R)$ is embedded in $D$,

  2. (2)

if $R\neq R^{\prime}$ then $S(R)$ and $S(R^{\prime})$ are disjoint, and

  3. (3)

$S(R)$ is a rectangle.

###### Proof.

Let $B_{\triangleleft}$ and $B_{\triangleright}$ be the right- and left-bigons
for $R$; define $B^{\prime}_{\triangleleft}$ and $B^{\prime}_{\triangleright}$
similarly for $R^{\prime}$. Note that boundary trigons in $B_{\triangleleft}$
have transverse orientations on their branches that disagree with the
tangential orientation on $\partial D$. On the other hand, boundary trigons in
$B^{\prime}_{\triangleright}$ have transverse orientations that agree with the
tangential orientation.

This proves that $B_{\triangleleft}$ and $B_{\triangleright}$ share only one
region: the max-bigon itself, and so $B=B_{\triangleleft}\cup
B_{\triangleright}$ is again a boundary bigon. The same argument shows that
$B$ and $B^{\prime}=B^{\prime}_{\triangleleft}\cup
B^{\prime}_{\triangleright}$ have no regions in common if $R\neq R^{\prime}$.

We claim that $\partial B$ and $\partial B^{\prime}$ are disjoint. To see
this, note that $\partial B$ consists of an arc in $\partial D$, and an arc in
$\tau$. The transverse orientation on the arc in $\tau$ points into $B$, and
similarly for $B^{\prime}$.

Let $Q_{\triangleleft}$, $Q_{\triangleright}$, $Q^{\prime}_{\triangleleft}$
and $Q^{\prime}_{\triangleright}$ be the quadrilaterals for $R$ and
$R^{\prime}$. Since these are obtained by taking subsets of small regular
neighbourhoods of branches in $\partial B$ and $\partial B^{\prime}$, these
are all pairwise disjoint (if $R\neq R^{\prime}$). This proves parts (1) and
(2).

Adding the subregions $Q_{\triangleleft}$ and $Q_{\triangleright}$ replaces
the two corners of $B$ with four corners, and thus $S(R)$ is a rectangle, and
we obtain (3). ∎

Let $D^{\prime}=D-\cup S(R)$, where the union ranges over all max-bigons $R$.

######  Claim 5.7.

The induced cellulation of $D^{\prime}$ has no regions of positive index.

###### Proof.

Suppose that $R^{\prime}$ is a region of $D^{\prime}$. If $R^{\prime}$ were a
nullgon or monogon, then it would be a nullgon or monogon in $D$,
contradicting Lemma 3.1. The region $R^{\prime}$ is not a boundary bigon since
we removed all of them. Thus $R^{\prime}$ was created by cutting
quadrilaterals out of some region $R$ of $D-\tau$. Note that $R^{\prime}$
meets $\tau$, meets $\partial D$ and meets $\partial Q_{\triangleright}$ (say)
along some side $s_{\triangleright}$. So $R^{\prime}$ has at least three
corners. If $R^{\prime}$ has positive index, it must have exactly three
corners. Thus $R=R^{\prime}\cup Q_{\triangleright}$ is a boundary trigon,
contradicting Claim 5.3. ∎

Note that $D^{\prime}$ has both outward and _inward_ corners (a combinatorial
version of the exterior angle being $3\pi/2$). Generalizing our definition of
index to also deal with inward corners [8, page 57] implies that $D^{\prime}$
has non-positive index. Since rectangles have index zero, from the additivity
of index we deduce that $D$ has non-positive index, a contradiction. This
concludes the proof of Theorem 5.1. ∎

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