# Reshaping Convex Polyhedra

Joseph O’Rourke and Costin Vîlcu

###### Contents

  1. Abstract
  2. Preface
  3. I Tailoring for Every Body
    1. 1 Introduction to Part I
      1. 1.1 Alexandrov’s Gluing Theorem
      2. 1.2 Tailoring Examples
      3. 1.3 Summary of Part-I Results
    2. 2 Preliminaries
      1. 2.1 Cut locus properties
        1. 2.1.1 Star-Unfolding and Cut Locus
        2. 2.1.2 Fundamental Triangles
        3. 2.1.3 Cut Locus Partition
      2. 2.2 Cauchy’s Arm Lemma
      3. 2.3 A Rigidity Result
      4. 2.4 Vertex-Merging
    3. 3 Domes and Pyramids
      1. 3.1 Domes
      2. 3.2 Cube/Tetrahedron Example
        1. 3.2.1 Slice $\to$ G-Domes for Cube/Tetrahedron
      3. 3.3 Proof: G-dome $\to$ Pyramids
        1. 3.3.1 G-domes $\to$ Pyramids for Cube/Tetrahedron
    4. 4 Tailoring via Sculpting
      1. 4.1 Slice $\to$ G-domes
      2. 4.2 Pyramid $\to$ Tailoring
        1. 4.2.1 Notation
      3. 4.3 Small volume slices
        1. 4.3.1 Pyramid case
        2. 4.3.2 General case
      4. 4.4 Cube/Tetrahedron: Completion
        1. 4.4.1 Pyramid Removals
        2. 4.4.2 Pyramid Reductions by Tailoring
        3. 4.4.3 Seals
      5. 4.5 Hexagonal Pyramid Example
      6. 4.6 Tailoring is finer than sculpting
    5. 5 Pyramid Seal Graph
      1. 5.1 Pyramid Digon Removal
        1. 5.1.1 Notation I
      2. 5.2 Cone Viewpoint
        1. 5.2.1 Notation II
      3. 5.3 Examples
      4. 5.4 Preliminary Lemmas
      5. 5.5 Pyramid Seal Graph is a Tree
        1. 5.5.1 Other Digon Orderings
    6. 6 Algorithm for Tailoring via Sculpting
      1. 6.1 Algorithm 1: slice$~{}\to~{}$g-domes
        1. 6.1.1 Complexity of sculpting
      2. 6.2 Algorithm 2: g-dome$~{}\to~{}$pyramids
      3. 6.3 Algorithm 3: pyramid$~{}\to~{}$digons
      4. 6.4 Overall Tailoring Algorithm
    7. 7 Crests
      1. 7.1 Examples
      2. 7.2 Proofs
      3. 7.3 Algorithm 4: pyramid $\to$ crest
    8. 8 Tailoring via Flattening
      1. 8.1 Proofs
        1. 8.1.1 Digon-tailor $P\to P_{\textrm{flat}}$
        2. 8.1.2 Vertex-merge $Q\to Q_{\textrm{flat}}$
        3. 8.1.3 Scale $Q_{\textrm{flat}}\to Q^{s}_{\textrm{flat}}$
        4. 8.1.4 Trim $P_{\textrm{flat}}\to P^{s}_{\textrm{flat}}$
        5. 8.1.5 Reverse $P^{s}_{\textrm{flat}}\to Q^{t}$
        6. 8.1.6 Theorem: Tailoring via Flattening
      2. 8.2 Algorithm for Tailoring via Flattening
    9. 9 Enlarging and $P$-Unfoldings
      1. 9.1 Enlarging and Reshaping
      2. 9.2 $P$-unfoldings
        1. 9.2.1 $P$-unfoldings and Reshaping
        2. 9.2.2 $P$-unfoldings and the WBG Theorem
  4. II Vertex Merging and Convexity
    1. 10 Introduction to Part II
    2. 11 Vertex-Merging Reductions and Slit Graphs
      1. 11.1 Slit Graphs for Vertex Mergings
      2. 11.2 Example: Reductions of Flat Hexagon
      3. 11.3 Example: Reductions of Cube
      4. 11.4 Example: Icosahedron
      5. 11.5 Example: Hexagonal Shape with Cycle
      6. 11.6 Vertex Merging and Unfoldings
      7. 11.7 Unfolding Irreducible Surfaces
        1. 11.7.1 $S$: Doubly-covered Triangle
        2. 11.7.2 Net and Overlap
        3. 11.7.3 $S$: Isosceles Tetrahedron
    3. 12 Planar Spiral Slit Tree
      1. 12.1 Sequential Spiral Merge
      2. 12.2 Notation
      3. 12.3 Algorithm Description
      4. 12.4 Planar Proof
    4. 13 Convexity on Convex Polyhedra
      1. 13.1 Convex Curves
      2. 13.2 Notions of Convexity
      3. 13.3 Ag-convexity
      4. 13.4 Geodesic Segments and Convex Sets
      5. 13.5 Relative Convexity
      6. 13.6 Convex Hull
      7. 13.7 Relative Convex Hull
      8. 13.8 Extreme Points
      9. 13.9 Relative Convex Hull of Vertices
      10. 13.10 Summary of Properties
    5. 14 Minimal-length Enclosing Polygon
      1. 14.1 Properties of the Minimal Enclosing Polygon
      2. 14.2 Shortening Algorithm
        1. 14.2.1 Curve-Shortening Flow
        2. 14.2.2 Algorithm Overview
        3. 14.2.3 Finding an Enclosing Geodesic Polygon
        4. 14.2.4 Algorithm for Curve Shortening
      3. 14.3 AG-convexity and $Z$
      4. 14.4 Algorithm for ${\rm{rconv}}(V)=R(W)$
    6. 15 Spiral Tree on Polyhedron
      1. 15.1 Notation
      2. 15.2 Icosahedron Example
      3. 15.3 Spiraling Algorithm for ${\rm{rconv}}$
      4. 15.4 Proof: Slit Graph is a Tree
      5. 15.5 Spiraling Algorithm for $Z(V)=\min\ell[V]$
    7. 16 Unfoldings via Slit Trees
      1. 16.1 Notation
      2. 16.2 Unfoldings via spiraling algorithms
        1. 16.2.1 Two Cones
        2. 16.2.2 Reduction to Cylinder
        3. 16.2.3 Cube Example
        4. 16.2.4 Icosahedron Example
    8. 17 Vertices on Quasigeodesics
      1. 17.1 Notation
      2. 17.2 ${\cal Q}_{k}$ Theorem
        1. 17.2.1 $|V(Q)|=1$
        2. 17.2.2 $|V(Q)|=2$
        3. 17.2.3 $|V(Q)|=k$, with $3\leq k\leq n$
    9. 18 Conclusions and Open Problems
      1. 18.1 Part I
      2. 18.2 Part II

### Abstract

Given a convex polyhedral surface $P$, we define a _tailoring_ as excising
from $P$ a simple polygonal domain that contains one vertex $v$, and whose
boundary can be sutured closed to a new convex polyhedron via Alexandrov’s
Gluing Theorem. In particular, a _digon-tailoring_ cuts off from $P$ a _digon_
containing $v$, a subset of $P$ bounded by two equal-length geodesic segments
that share endpoints, and can then zip closed.

In the first part of this monograph, we primarily study properties of the
tailoring operation on convex polyhedra. We show that $P$ can be reshaped to
any polyhedral convex surface $Q\subset{\rm conv}(P)$ by a sequence of
tailorings. This investigation uncovered previously unexplored topics,
including a notion of _unfolding of_ $Q$ _onto_ $P$—cutting up $Q$ into pieces
pasted non-overlapping onto $P$.

In the second part of this monograph, we study _vertex-merging_ processes on
convex polyhedra (each vertex-merge being in a sense the reverse of a digon-
tailoring), creating embeddings of $P$ into enlarged surfaces. We aim to
produce non-overlapping polyhedral and planar unfoldings, which led us to
develop an apparently new theory of convex sets, and of minimal length
enclosing polygons, on convex polyhedra.

All our theorem proofs are constructive, implying polynomial-time algorithms.

##### MSC Classifications

###### Primary:

52A15, 52B10, 52C45, 53C45, 68U05.

###### Secondary:

52-02, 52-08, 52A37, 52C99.

##### Keywords and phrases

convex polyhedron
Alexandrov’s Gluing Theorem
polyhedron truncation
cut locus
unfolding polyhedra
net for polyhedron; star-unfolding
vertex-merging
convex sets and convex hulls on convex polyhedra
minimal length enclosing polygon on convex polyhedra

### Preface

The research reported in this monograph emerged from exploring a simple
question: Given two convex polyhedra $P$ and $Q$, with $Q$ inside $P$, can one
reshape $P$ to $Q$ by repeatedly “snipping” off vertices? We call this
snipping-off operation _tailoring_. A precise definition is deferred to the
introductory Chapter 1, but here we contrast it with _vertex truncation_ ,
which slices off a vertex with a plane and replaces it with a new facet lying
in that plane. This is, for example, one way to construct the truncated cube:
see Fig. 1.

Figure 1: Truncated cube. [Image by Tilman Piesk, Wikipedia].

Tailoring differs from vertex truncation in two ways: first, it does not slice
by a plane but instead uses a digon bound by a pair of equal-length geodesics
(again, definition deferred), and second, rather than filling the hole with a
new facet, the boundary of the hole is “sutured” closed without the addition
of new surface.

Our first experiment started with a paper cube and tailored its $8$ vertices,
producing the shape shown in Fig. 2.

Figure 2: Cube after tailoring $8$ vertices.

Although not evident from this crude model, this shape is actually a convex
polyhedron of $16$ vertices, an implication of Alexandrov’s Gluing Theorem.
This led us imagine that continuing the process on the $16$ vertices might
allow reshaping the cube into a roughly spherical polyhedron—“whittling” a
cube to a sphere. And indeed, one of our main theorems is that any $P$ can be
reshaped to any $Q\subset P$ by a finite sequence of tailorings (Theorem 4.6).
This holds whether $Q$ has fewer vertices than $P$ or more vertices than $P$:
see Fig. 3.

Figure 3: Cube inside dodecahedron. Icosahedron inside cube.

Investigating tailoring in Part I uncovered previously unexplored topics,
including a notion of “unfolding” $Q$ onto $P$—cutting up $Q$ into pieces
pasted non-overlapping onto $P$. This led us to a systematic study of _vertex-
merging_ , a technique introduced by Alexandrov that can be viewed as a type
of inverse operation to tailoring, a study we carry out in Part II. Here we
start with $P$ and gradually enlarge it via vertex-merges until $P$ is
embedded onto a nearly flat shape $S$: a doubly-covered triangle, an isosceles
tetrahedron, a pair of cones, or a cylinder. The first two “vertex-merge
irreducible shapes” (doubly-covered triangle or isosceles tetrahedron) are the
result of vertex-merging in arbitrary order, and can result in disconnecting
an $n$-vertex $P$ into $O(n^{2})$ pieces (Corollary 11.2). In contrast we
pursue the goal of minimizing the disconnection of $P$, with the ultimate goal
of achieving a planar _net_ for $P$—a nonoverlapping simple polygon.

Toward this end, we partition $P$ into two half-surfaces by a quasigeodesic
$Q$, a simple closed curve on $P$, convex to both sides.111 In Part I, $Q$ is
the target polyhedron resulting from reshaping $P$. In Part II, there is no
equivalent fixed “target” polyhedron, and we use $Q$ to denote a
quasigeodesic. Then we can prove that a particular spiral vertex-merge
sequence avoids disconnecting $P$ in each half. (Theorems 15.7 and 15.14.)
This approach requires a notion of convexity and convex hull on the surface of
a convex polyhedron, apparently novel topics, explored in our longest chapter,
Chapter 13.

We can achieve a net for $P$ (Theorem 16.5) assuming the truth of a conjecture
concerning quasigeodesics (Open Problem 18.13). This is among the many avenues
for future work uncovered by our investigations.

##### Acknowledgements

We benefitted from discussions with Jin-ichi Itoh and with Anna Lubiw, on some
topics of this work.

Costin Vîlcu’s research was partially supported by UEFISCDI, project no. PN-
III-P4-ID-PCE-2020-0794.

##  Part I Tailoring for Every Body

###  Chapter 1 Introduction to Part I

Let $P$ and $Q$ be convex polyhedra, each the convex hull of finitely many
points in ${\mathbb{R}^{3}}$. If $Q\subset P$, it is easy to see that $Q$ can
be _sculpted_ from $P$ by “slicing $P$ with planes.” By this we mean
intersecting $P$ with half-spaces each of whose plane boundary contains a face
of $Q$. If $Q\not\subset P$, we can shrink $Q$ until it fits inside. So a
_homothet_ of any given $Q$ can be sculpted from any given $P$, where a
homothet is a copy possibly scaled, rotated, and translated. Main results of
Part I are similar claims but via “tailorings”: a homothet of any given $Q$
can be tailored from any given $P$ (Theorems 4.6, 7.5, 8.1).111A preliminary
version of Part I is [OV20].

With some abuse of notation,222 Justified by Alexandrov’s Gluing Theorem, see
ahead. we will use the same symbol $P$ for a polyhedral hull and its boundary.
We define two types of tailoring. A _digon-tailoring_ cuts off a single vertex
of $P$ along a digon, and then sutures the digon closed. A _digon_ is a subset
$D(x,y)$ of $P$ bounded by two equal-length geodesic segments that share
endpoints $x$ and $y$. A _geodesic segment_ is a shortest geodesic between its
endpoints. A _crest-tailoring_ cuts off a single vertex of $P$ but via a more
complicated shape we call a “crest.” Again the hole is sutured closed. We
defer discussion of crests to Chapter 7. Meanwhile, we often shorten “digon-
tailoring” to simply _tailoring_.

Figure 1.1:  (a) A digon $D(x,y)$ on the regular tetrahedron $P=abcd$,
surrounding vertex $d$. (b) Digon excised. (c) A digon $D(o,c)$ with one
endpoint at vertex $c$, the other the centroid $o$.

Cutting out a digon means excising the portion of the surface between the
geodesics, including the vertex they surround.333 An informal view (due to
Anna Lubiw) is that one could pinch the surface flat in a neighborhood of the
vertex, and then snip-off the flattened vertex with scissors. Once removed,
the digon hole is closed by naturally identifying the two geodesics along
their lengths. This identification is often called “gluing” in the literature,
although we also call it “zipping” or “suturing” or “sealing.” Fig. 1.1
illustrates digons on a regular tetrahedron $P$. In (a) of the figure, neither
digon endpoint $x$ nor $y$ is a vertex of $P$; (b) shows $P$ after excision
but before sealing. After sealing, both $x$ and $y$ become vertices. (See
ahead to Fig. 2.6 for a depiction of the polyhedron that results after
sealing.) In (c), one endpoint of the digon coincides with a vertex. Here,
after excision and suturing, $o$ becomes a vertex and $c$ remains a vertex (of
sharper curvature).

####  1.1 Alexandrov’s Gluing Theorem

Throughout, we make extensive use of Alexandrov’s Gluing Theorem [Ale05,
p.100], which guarantees that the surface obtained after a tailoring of $P$
corresponds uniquely to a convex polyhedron $P^{\prime}$. A precise statement
of this theorem, which we will abbreviate to AGT, is as follows.

######  Theorem AGT.

Let $S$ be a topological sphere obtained by gluing planar polygons (i.e.,
naturally identifying pairs of sides of the same length) such that at most
$2\pi$ surface angle is glued at any point. Then $S$, endowed with the
intrinsic metric induced by the distance in ${\mathbb{R}^{2}}$, is isometric
to a convex polyhedron $P\subset{\mathbb{R}^{3}}$, possibly degenerated to a
doubly-covered convex polygon. Moreover, $P$ is unique up to rigid motion and
reflection in ${\mathbb{R}^{3}}$.

The case of doubly-covered convex polygon is important and we will encountered
it often.

Because the sides of the digon are geodesics, gluing them together to seal the
hole leaves $2\pi$ angle at all but the digon endpoints. The endpoints lose
surface angle with the excision, and so have strictly less than $2\pi$ angle
surrounding them. So AGT applies and yields a new convex polyhedron.

This shows that tailoring is possible and alters the given $P$ to another
convex polyhedron. How to “aim” the tailoring toward a given target $Q$ is a
long story, told in subsequent sections.

Alexandrov’s proof of his celebrated theorem is a difficult existence proof
and gives little hint of the structure of the polyhedron guaranteed by the
theorem. And as-yet there is no effective procedure to construct the three-
dimensional shape of the polyhedron guaranteed by his theorem. It has only
been established that there is a theoretical pseudopolynomial-time algorithm
[KPD09], achieved via an approximate numerical solution to the differential
equation established by Bobenko and Izmestiev [BI08] [O’R07]. But this remains
impractical in practice. Only small or highly symmetric examples can be
reconstructed, for example [ADO03] for the foldings of a square, and more
recently [ALZ20] for polyhedra built from regular pentagons. Figs. 1.3 and 2.6
(ahead) were reconstructed through ad hoc methods.

AGT is a fundamental tool in the geometry of convex surfaces and, at a
theoretical level, our work helps to elucidate its implications. While AGT has
proved useful in several investigations, our application here to reshaping
has, to our knowledge, not been considered before as the central object of
study.

####  1.2 Tailoring Examples

Before discussing background context, we present several examples. Throughout
we let $xy$ denote the line segment between points $x$ and $y$,
$x,y\in{\mathbb{R}^{3}}$. Also we make extensive use of vertex curvature. The
_discrete_ (or _singular_) _curvature_ ${\omega}(v)$ at a vertex $v\in P$ is
the angle deficit: $2\pi$ minus the sum of the face angles incident to $v$.

######  Example 1.1.

Let $T=abcd$ be a regular tetrahedron, and let $o$ be the center of the face
$abd$; see Fig. 1.1(c). Cut out the digon on $T$ between $c$ and $o$
“encircling” $d$, and zip it closed.

The unfolding $U(T)$ of $T$ with respect to $c$ is a planar regular triangle
$c_{a}c_{b}c_{d}$ with center $o$. Cutting out that digon from $T$ is
equivalent to removing from $T$ the isosceles triangle $oc_{a}c_{b}$. See Fig.
1.2(a). We zip it closed by identifying the digon-segments $c_{a}o$ and
$c_{b}o$, and refolding the remainder of $T$ by re-identifying $c_{a}b$ and
$bc_{d}$, and $c_{b}a$ and $ac_{d}$. The result is the doubly-covered kite
$K=aobc_{d}$, shown in Fig. 1.2(b).

Figure 1.2: Illustrations for Examples 1.1–1.2. In (c), $e$ and $e^{\prime}$
are joined when the $D(x,y)$ digon is excised.

######  Example 1.2.

We further tailor the doubly-covered kite $K$ obtained in Example 1.1.
Excising a digon encircling $o$, between points $x\in oa$ and $y\in ob$, and
zipping closed, yields a doubly-covered pentagon, as in Fig. 1.2 (b).

If instead we excise a digon encircling $o$, between corresponding points
$x,y\in oc$ on different sides of $K$ (see Fig. 1.2 (c)), and seal closed, the
result is a non-degenerate pentahedron, illustrated in Fig. 1.3. This is among
the rare cases where the polyhedron guaranteed by AGT can be explicitly
reconstructed.

Figure 1.3: The non-degenerate pentahedron obtained in Example 1.2. The point
marked $e=e^{\prime}$ is the join of the points $e$ and $e^{\prime}$ in Fig.
1.2(c), the midpoint of the edge $ab$.

######  Example 1.3.

A digon may well contain several vertices, but for our digon-tailoring we only
consider those containing at most one vertex. The limit case of a digon
containing no vertex is an edge between two vertices. (This will play a role
in Chapter 8.) In this case, gluing back along the cut would produce the
original polyhedron, but we can as well seal the cut closed from another
starting point. For example, cutting along an edge of an isosceles tetrahedron
(a polyhedron composed of four congruent faces)444 Also called a
tetramonohedron, or an isotetrahedron. and carefully choosing the gluing leads
to a doubly-covered rectangle. See Fig. 1.4.

By AGT, this limit-case tailoring of an empty-interior digon only applies
between vertices $v_{1},v_{2}$ of curvatures
${\omega}_{1},{\omega}_{2}\geq\pi$, because both $v_{1}$ and $v_{2}$ will be
identified with interior points of the edge.555 In [DO07, Sec. 25.3.1] this
structure is called a “rolling belt.”

Figure 1.4: (a) An isosceles tetrahedron: All four faces are congruent.
Cutting along $v_{3}v_{4}$ and regluing the two halves of that slit
differently, creasing at the blue segments, yields a doubly-covered rectangle,
as shown in (b), folding as shown.

In view of Fig. 1.2(c) and Fig. 1.3, it is clear that, even though tailoring
is area decreasing, it is not necessarily volume decreasing.

Let a tailoring step remove vertex $v$ inside digon $D=(x,y)$. In general,
neither $x$ nor $y$ is a vertex before tailoring, but they become vertices
after removing $D$, thereby increasing the number of vertices of $P$ by $1$.
This is illustrated in Fig. 1.1(ab). If one of $x$ or $y$ is a vertex and the
other not, then the total number of vertices remains fixed, as in Fig. 1.1(c).
And if both $x$ and $y$ are vertices, then the number of vertices of $P$ is
decreased by $1$. The challenge answered in our work is to direct tailoring to
“aim” from one polyhedon $P$ to the target $Q$, which may have a quite
different number of vertices.

####  1.3 Summary of Part-I Results

Here we list our main theorems in Part I, each with a succinct (and at this
stage, quite approximate) summary of their claims.

  * •

Theorem 4.6: $Q$ may be digon-tailored from $P$, tracking a sculpting of $P$
to $Q$.

  * •

Theorem 7.5: $P$ may be crest-tailored to $Q$, again tracking a sculpting.

  * •

Theorem 8.1: A different proof of a similar result, that $P$ may be digon-
tailored to a homothet of $Q$, but this time without relying on sculpting.

  * •

Theorems 6.2 and 7.5 and 8.3: Tailoring algorithms have time-complexity
$O(n^{4})$.

  * •

Theorem 9.1: Reversing tailoring yields procedures for enlarging $Q$ inside
$P$ to match $P$. As a consequence, $Q$ may be cut up and “unfolded”
isometrically onto $P$.

Along the way to our central theorems, we obtain results not directly related
to AGT:

  * •

Theorem 2.9: If two convex polyhedra with the same number of vertices match on
all but the neighborhoods of one vertex, then they are congruent.

  * •

Theorem 3.2: Every “g-dome” can be partitioned into a finite sequence of
pyramids by planes through its base edges.

The above results raise several open problems of various natures, either
scattered along the text or presented in the last section of Part I. [Revisit
later.]

Finally, we sketch the logic behind the first result listed above, whose
statement in Chapter 4 we repeat here:

######  Theorem 4.6.

Let $P$ be a convex polyhedron, and $Q\subset P$ a convex polyhedron resulting
from repeated slicing of $P$ with planes. Then $Q$ can also be obtained from
$P$ by tailoring. Consequently, for any given convex polyhedra $P$ and $Q$,
one can tailor $P$ “via sculpting” to obtain any homothetic copy of $Q$ inside
$P$.

Start with $Q$ inside $P$, and imagine a sequence of slices by planes that
sculpt $P$ to $Q$. Lemma 4.4 shows how to digon-tailor one such slice, which
then establishes the claim that we can tailor $P$ to $Q$. Theorem 4.6 is
achieved by first slicing off shapes we call “g-domes,” and then showing in
Theorem 3.2 that every g-dome can be reduced to its base by slicing off
pyramids, i.e., by vertex truncations. Lemma 4.4 shows that such vertex
truncations can be achieved by tailoring. And the proof of Lemma 4.4 relies on
the rigidity established by Theorem 2.9. So the path of logic is:

$\displaystyle\textrm{plane
slice}\;\to\;\textrm{g-domes}\;\to\;\textrm{pyramids}\;\to\;$
$\displaystyle\textrm{digon removals}\;.$ $\displaystyle\uparrow$ rigidity

###  Chapter 2 Preliminaries

In this chapter we present basic properties of cut loci on convex polyhedra,
the star-unfolding, prove a rigidity result, and describe the technique of
vertex-merging. All of these geometric tools will be needed subsequently. The
reader might skip this section and return to it as the tools are deployed.

####  2.1 Cut locus properties

The _cut locus_ $C(x)$ _of the point $x$_ on a convex polyhedron $P$ is the
closure of the set of points to which there is more than one shortest path
from $x$. This concept goes back to Poincaré [Poi05], and has been studied
algorithmically since [SS86] (there, the cut locus is called the “ridge
tree”). The cut locus is one of our main tools throughout this work. The next
lemma establishes notation and lists several known properties.

######  Lemma 2.1 (Cut Locus Basics).

The following hold for the cut locus $C(x)$:

  1. (i)

$C(x)$ has the structure of a finite $1$-dimensional simplicial complex which
is a tree. Its leaves (endpoints) are vertices of $P$, and all vertices of
$P$, excepting $x$ (if it is a vertex) are included in $C(x)$. All points
interior to $C(x)$ of tree-degree $3$ or more are known as _ramification
points_ of $C(x)$.111In some literature, these points are called “branch
points” or “junctions” of $C(x)$. All vertices of $P$ interior to $C(x)$ are
also considered as ramification points.

  2. (ii)

Each point $y$ in $C(x)$ is joined to $x$ by as many geodesic segments as the
number of connected components of $C(x)\setminus{y}$. For ramification points
in $C(x)$, this is precisely their degree in the tree.

  3. (iii)

The edges of $C(x)$ are geodesic segments on $P$.

  4. (iv)

Assume the geodesic segments ${\gamma}$ and ${\gamma}^{\prime}$ (possibly
${\gamma}={\gamma}^{\prime}$) from $x$ to $y\in C(x)$ are bounding a domain
$D$ of $P$, which intersects no other geodesic segment from $x$ to $y$. Then
there is an arc of $C(x)$ at $y$ which intersects $D$ and it bisects the angle
of $D$ at $y$.

###### Proof.

The statements (i)-(ii) and (iv) are well known. The statement (iii) is Lemma
2.4 in [AAOS97]. ∎

The following is Lemma 4 in [INV12].

######  Lemma 2.2 (Path Cut Locus).

If $C(x)$ is a path, the polyhedron is a doubly-covered (flat) convex polygon,
with $x$ on the rim.

The following lemma will be invoked in Chapter 5.

######  Lemma 2.3 (Angle $<\pi$).

Let $x$ be a point on a convex polyhedron $P$ and let $y$ be a ramification
point of $C(x)$ of degree $k\geq 2$. Let $e_{1},...,e_{k}$ be the edges of
$C(x)$ incident to $y$, ordered counterclockwise, and let ${\alpha}_{j}$ be
the angle of the sector between $e_{j}$ and $e_{j+1}$ at $y$ (with $k+1\equiv
1$). Then ${\alpha}_{j}<\pi$, for all $j=1,...,k$.

###### Proof.

Let ${\gamma}_{1},...{\gamma}_{k}$ be the geodesic segments from $x$ to $y$,
say with ${\gamma}_{j}$ between $e_{j}$ and $e_{j+1}$ (again with $k+1\equiv
1$). See Fig. 2.1. By Lemma 2.1 (iv), $e_{j}$ is the bisector of
$\angle({\gamma}_{j},{\gamma}_{j+1})$. Put
$2{\beta}_{j+1}=\angle({\gamma}_{j},{\gamma}_{j+1})$. Then the total surface
angle ${\theta}_{y}$ incident to $y$ satisfies
${\theta}_{y}=\sum_{j=1}^{k}2{\beta}_{j}\leq 2\pi$ if $k\geq 3$, and
${\theta}_{y}<2\pi$ if $k=2$. Therefore, when $k\geq 3$,
$\sum_{j=1}^{k}{\beta}_{j}=\pi$ and so
${\alpha}_{j}={\beta}_{j}+{\beta}_{j+1}<\pi$. And when $k=2$,
${\alpha}_{1}={\alpha}_{2}={\beta}_{1}+{\beta}_{2}<\pi$. ∎

Figure 2.1: Edge $e_{2}$ of $C(x)$ bisects
$\angle({\gamma}_{1},{\gamma}_{2})=2{\beta}_{2}$.

#####  2.1.1 Star-Unfolding and Cut Locus

Next we introduce a general method for unfolding any convex polyhedron $P$ to
a simple (non-overlapping) polygon in the plane. We use this subsequently
largely because of its connection to the cut locus.

To form the _star-unfolding_ of a $P$ with respect to $x$, one cuts $P$ along
the geodesic segments (unique if $x$ is “generic”) from $x$ to every vertex of
$P$. The idea goes back to Alexandrov [Ale05]; the non-overlapping of the
unfolding was established in [AO92], where the next result was also proved.
See Fig. 2.2.

######  Lemma 2.4 ($S_{P}$ Voronoi Diagram).

Let $S_{P}=S_{P}(x)$ denote the star-unfolding of $P$ with respect to $x\in
P$. Then the image of $C(x)$ in $S_{P}$ is the restriction to $S_{P}$ of the
Voronoi diagram of the images of $x$.

Figure 2.2: (a) Cut segments to the $8$ vertices of a cube from a point $x$ on
the top face. (No cut is interior to the bottom face.) T, F, R, K, L, B $=$
Top, Front, Right, Back, Left, Bottom. (b) The star-unfolding from $x$. The
cut locus $C(x)$ (red) is the Voronoi diagram of the $8$ images of $x$
(green). Two pairs of fundamental triangles are shaded.

Notice that several properties of cut loci, presented in the previous section,
could easily be derived from Lemma 2.4.

#####  2.1.2 Fundamental Triangles

A geodesic triangle on $P$ (i.e., with geodesic segments as sides) is called
_flat_ if its curvature vanishes.

######  Lemma 2.5 (Fundamental Triangles [INV12]).

For any point $x\in P$, $P$ can be partitioned into flat triangles whose bases
are edges of $C(x)$, and whose lateral edges are geodesic segments from $x$ to
the ramification points or leaves of $C(x)$. Moreover, those triangles are
isometric to plane triangles, congruent by pairs.

See Fig. 2.2(b) (and ahead to Fig. 8.1).

#####  2.1.3 Cut Locus Partition

Another tool we need (in Chapter 4) is a cut locus partition lemma, a
generalization of lemmas in [INV12]. On a polyhedron $P$, connect a point $x$
to a point $y\in C(x)$ by two geodesic segments ${\gamma},{\gamma}^{\prime}$.
This partitions $P$ into two “half-surface” digons $H_{1}$ and $H_{2}$. If we
now zip each digon separately closed by joining ${\gamma}$ and
${\gamma}^{\prime}$, AGT leads to two convex polyhedra $P_{1}$ and $P_{2}$.
The lemma says that the cut locus on $P$ is the “join” of the cut loci on
$P_{i}$. See Fig. 2.3.

######  Lemma 2.6 (Cut Locus Partition).

Under the above circumstances, the cut locus $C(x,P)$ of $x$ on $P$ is the
_join_ of the cut loci on $P_{i}$: $C(x,P)=C(x,P_{1})\sqcup_{y}C(x,P_{2})$,
where $\sqcup_{y}$ joins the two cut loci at point $y$. And starting instead
from $P_{1}$ and $P_{2}$, the natural converse holds as well.

###### Proof.

Notice first that a straightforward induction and Lemma 2.5 on fundamental
triangles shows that the cut locus of $x$ on $P_{i}$ is indeed the truncation
of $C(x,P)$. Therefore, $C(x,P)=C(x,P_{1})\sqcup_{y}C(x,P_{2})$.

Assume we start now from $P_{1}$ and $P_{2}$, having vertices $x_{1},y_{1}\in
P_{1}$ and $x_{2},y_{2}\in P_{2}$ such that

  * •

${\rho}_{P_{1}}(x_{1},y_{1})={\rho}_{P_{2}}(x_{2},y_{2})$, where
${\rho}_{P_{i}}(\;)$ is the geodesic distance between the indicated points on
$P_{i}$.

  * •

${\theta}_{x_{1}}+{\theta}_{x_{2}}\leq 2\pi$, where ${\theta}_{x}$ is the
total surface angle incident to $x$, and

  * •

${\theta}_{y_{1}}+{\theta}_{y_{2}}\leq 2\pi$.

Then we can cut open $P_{i}$ along a geodesic segment ${\gamma}_{i}$ from
$x_{i}$ to $y_{i}$, $i=1,2$, and join the the two halves by AGT, such that
$x_{1},x_{2}$ have a common image $x$, and $y_{1},y_{2}$ have a common image
$y$.

Now, all geodesic segments starting at $x$ into $H_{i}$ remain in $H_{i}$,
because geodesic segments do not branch. Therefore, $H_{1}$ has no influence
on $C(x,P_{2})$ and $H_{2}$ has no influence on $C(x,P_{1})$. ∎

Figure 2.3: Geodesic segments ${\gamma}$ and ${\gamma}^{\prime}$ (purple)
connect $x{=}x_{1}{=}x_{2}$ to $y{=}y_{1}{=}y_{2}$. $P_{1}$ folds to a
tetrahedron, and $P_{2}$ to an $8$-vertex polyhedron, with $x$ and $y$
vertices in each. $P_{1}$ and $P_{2}$ are cut open along geodesic segments
from $x_{i}$ to $y_{i}$ and glued together to form $P$. Based on the cube
unfolding in Fig. 2.2(b).

####  2.2 Cauchy’s Arm Lemma

In several proofs, we will need an extension of Cauchy’s Arm Lemma, which we
now describe.

Let $C=x_{1},\ldots,x_{n}$ be a directed polygonal chain in the plane, with
left angles ${\theta}_{i}=\angle x_{i-1}x_{i}x_{i+1}$, where $x_{n}=x_{1}$ if
$C$ is closed, and ${\theta}_{1}$ and ${\theta}_{n}$ undefined if $C$ is open.
Let $d=|x_{1}x_{n}|$ be the distance between the endpoints, with $d=0$ if $C$
is closed. If ${\theta}_{i}\leq\pi$ for all $i$, then $C$ is a _convex chain_.
_Cauchy’s Arm Lemma_ applies to reconfiguration of $C$ while all the edge
lengths remain fixed (the edges are _bars_). We use primes to indicate the
reconfigured chain. The lemma says that if the ${\theta}_{i}$ angles are
“straightened” while remaining convex, in the sense that
${\theta}_{i}\leq{\theta}^{\prime}_{i}\leq\pi$, then the distance
$d^{\prime}=|x^{\prime}_{1}x^{\prime}_{n}|$ only increases: $d^{\prime}\geq
d$.

For the extension needed later, we reformulate in terms of turn angles
$\tau_{i}=\pi-{\theta}_{i}\geq 0$, as described in [O’R01]. Now the
straightening condition says that $0\leq\tau^{\prime}_{i}\leq\tau_{i}$. The
extended version of Cauchy’s arm lemma says that, as long as, in the
reconfiguration, $-\tau_{i}\leq\tau^{\prime}_{i}\leq\tau_{i}$, then the same
conclusion holds: $d^{\prime}\geq d$. Thus $C^{\prime}$ might no longer be a
convex chain, but its reflexivities are bounded by the original turn angles.
One way to interpret $d^{\prime}\geq d$ is that there is a “forbidden” disk of
radius $d=|x_{1}x_{n}|$ centered on $x_{1}$ that $x^{\prime}_{n}$ cannot
penetrate. See Fig. 2.4.

Figure 2.4: Any turns within the indicated $\pm\tau_{i}$ angle ranges
constitute straightening. Then $x_{n}$ will not penetrate the $|x_{1}x_{n}|$
disk.

We summarize in a theorem:

######  Theorem 2.7.

(Cauchy’s Arm Lemma.) A straightening reconfiguration of a planar convex chain
$C$—retaining edge-lengths fixed and confining new turn angles
$\tau^{\prime}_{i}$ to lie in $[-\tau_{i},\tau_{i}]$—only increases the
distance between the endpoints:
$|x^{\prime}_{1}x^{\prime}_{n}|\geq|x_{1}x_{n}|$.

The next elementary result assures the angle increase, in the frameworks in
which we will apply Cauchy’s Arm Lemma.

######  Lemma 2.8.

Consider three rays $r_{1},r_{2},r_{3}$ in ${\mathbb{R}}^{3}$, emanating from
the point $w$, and put ${\tau}_{i}=\angle(r_{i},r_{i+1})$, with $3+1\equiv 1$
mod $3$. Then $\theta_{1}\leq\theta_{2}+\theta_{3}$.

###### Proof.

Imagine a unit sphere $S$ centered on $w$ and let $\\{s_{i}\\}=r_{i}\cap S$,
and use ${\rho}$ to indicate spherical distance. Then the claim of the lemma
is the triangle inequality for spherical distances:
${\rho}(s_{1},s_{2})\leq{\rho}(s_{1},s_{3})+{\rho}(s_{2},s_{3})$. ∎

####  2.3 A Rigidity Result

In this section we present a technical result for later use, which may be of
independent interest. The theorem says that two convex polyhedra that are
isometric on all but the neighborhoods of one vertex, are in fact congruent.
We also show this result cannot be strengthened: two convex polyhedra can
differ in the neighborhoods of just two vertices.

######  Theorem 2.9.

Assume $P,Q$ are convex polyhedra with the same number of vertices, such that
there are vertices $p\in P$ and $q\in Q$, and respective small neighborhoods
$N_{p}\subset P$, $N_{q}\subset Q$ not containing other vertices, and an
isometry $\iota:P\setminus N_{p}\to Q\setminus N_{q}$. Then $P$ is congruent
to $Q$.

###### Proof.

The existence of $\iota$ on all but neighborhoods of $p$ and $q$ yields, in
particular, that the curvatures ${\omega}_{P}(p)$ of $P$ at $p$ and
${\omega}_{Q}(q)$ of $Q$ at $q$ are equal, to satisfy the curvature sum of
$4\pi$ (by Gauss-Bonnet).

Take a point $x\in P$ joined to each vertex of $P$ by precisely one geodesic
segment, a _generic point_ $x$. Such an $x$ maybe be found in a “ridge-free”
region of $P$ [AAOS97]; it is equivalent to the fact that no vertex of $P$ is
interior to $C(x)$. Moreover, we may choose $x$ such that $\iota(x)$ has the
same property on $Q$.

Denote by $u$ the ramification point of $C(x)$ neighboring $p$ in $C(x)$,
i.e., the ramification point of degree $\geq 3$ closest to $p$. Let $v$ be the
similar ramification point of $C(\iota(x))$ neighboring $q$ in $C(\iota(x))$.
Since $N_{p}$ and $N_{q}$ are small, we may assume they are disjoint from $u$
and $v$ and all the segments described above.

Star unfold $P$ with respect to $x$, and $Q$ with respect to $\iota(x)$, and
denote by $S_{P}$ and $S_{Q}$ the resulting planar polygons. We’ll continue to
use the symbols $p$ and $q$, $u$ and $v$ to refer to the corresponding points
in $S_{P}$ and $S_{Q}$ respectively. Let $x_{i}$, $i=1,2$ be the images of $x$
surrounding $p$ in $S_{P}$, and $\iota(x_{i})$ the similar images in $S_{Q}$.
See Fig. 2.5(a,b).

By hypothesis, we have respective neighborhoods $\bar{N}_{p}\subset S_{P}$ and
$\bar{N}_{q}\subset S_{Q}$ and an isometry $\bar{\iota}$ induced by $\iota$,
with $\bar{\iota}:S_{P}\setminus\bar{N_{p}}\to S_{Q}\setminus\bar{N_{q}}$.
Thus in Fig. 2.5(b), all of $S_{P}$ outside of the wedge $(x_{1},u,x_{2})$ is
identical in $S_{Q}$. Therefore the triangles ${x}_{1}{u}{x}_{2}$ and
${x}^{\prime}_{1}{v}{x}^{\prime}_{2}$ are congruent. Moreover, ${p}$ lies on
the bisector of the angle $\angle{x}_{1}{u}{x}_{2}$, and ${q}$ lies on the
bisector of the angle $\angle{x}^{\prime}_{1}{v}{x}^{\prime}_{2}$. Since
$\angle{x}_{1}{p}{u}=\angle{x}^{\prime}_{1}{q}{v}=\pi-\frac{1}{2}{\omega}(p)=\pi-\frac{1}{2}{\omega}(q)$,
$p$ and $q$ are uniquely determined. Consequently, $S_{P}$ and $S_{Q}$
coincide, and refolding according to the same gluing identifications leads to
congruent $P$ and $Q$. ∎

Figure 2.5:  (a) A $6$-vertex polyhedron $P$. The $F$ and $K$ faces are unit
squares; $B$ is a $1\times\frac{1}{2}$ rectangle, with $x\in B$. (b) Star-
unfolding $S_{P}$ of $P$. $\angle{x}_{1}{p}{u}$ is marked. (c) Moving
$x_{\alpha}$ on the circle arc moves the bisectors incident to $u$, and so
moves $p_{1}$ and $p_{2}$. Refolding results in a polyhedron incongruent to
(a).

######  Remark 2.10.

It is perhaps surprising that the above result cannot be extended to claim
that isometries excluding neighborhoods of two vertices always imply
congruence.

###### Proof.

If the points $p_{1},p_{2}\in P$ and $q_{1},q_{2}\in Q$ do not have a common
neighbor in $C(x)$ and $C(\iota(x))$ respectively, the above proof establishes
rigidity.

Next we focus on $P$, and try to find positions for $p_{1},p_{2}\in P$
determined by the hypotheses. Assume, in the following, that $p_{1},p_{2}\in
P$ have a common degree-$3$ ramification neighbor $u$ in $C(x)$.

Star unfold $P$ with respect to some $x\in P$, to $S_{P}$. The region of
$S_{P}$ exterior to the wedge $(x_{1},u,x_{2})$ is uniquely determined and
identical in $S_{Q}$. See Fig. 2.5(c).

Take a point $x_{\alpha}$ on the circle of center ${u}$ and radius
$|x_{1}u|=|x_{2}u|$. We now argue that positions of $x_{\alpha}$ on this
circle allow $p_{1}$ and $p_{2}$ to vary while maintaining all outside of the
$(x_{1},u,x_{2})$ wedge fixed.

Let $\angle{x}_{\alpha}{u}{x}_{1}=2{\alpha}$. On the bisector of that angle
incident to $u$, one can uniquely determine a point ${p}_{1}$ such that
$\angle{x}_{1}{{p_{1}}}{u}=\pi-\frac{1}{2}{\omega}(p_{1})$. Similarly, one can
uniquely determine a point ${p}_{2}$ on the bisector of that angle
$\angle{x}_{\alpha}{u}{x}_{2}$, such that
$\angle{x}_{2}{{p_{2}}}{u}=\pi-\frac{1}{2}{\omega}(p_{2})$.

Thus we have identified a continuous $1$-parameter family of star-unfoldings,
and consequently of convex polyhedra, verifying the hypotheses. ∎

####  2.4 Vertex-Merging

Digon-tailoring is, in some sense, the opposite of _vertex-merging_ , a
technique introduced by A. D. Alexandrov [Ale05, p. 240], and subsequently
used later by others, see e.g. [Zal07], [OV14], [O’R20]. We will employ
vertex-merging in Chapters 8 and 9, and focus on it in Part II.

Consider two vertices $v_{1},v_{2}$ of $P$ of curvatures
${\omega}_{1},{\omega}_{2}$, with ${\omega}_{1}+{\omega}_{2}<2\pi$, and cut
$P$ along a geodesic segment ${\gamma}$ joining $v_{1}$ to $v_{2}$. Construct
a planar triangle $T=\bar{v}^{\prime}\bar{v}_{1}\bar{v}_{2}$ of base length
$|\bar{v}_{1}-\bar{v}_{2}|=|{\gamma}|$ and the base angles equal to
${\omega}_{1}/2$ and ${\omega}_{2}/2$ respectively. Glue two copies of $T$
along the corresponding lateral sides, and further glue the two bases of the
copies to the two “banks” of the cut of $P$ along ${\gamma}$. By Alexandrov’s
Gluing Theorem (AGT), the result is a convex polyhedral surface $P^{\prime}$.
On $P^{\prime}$, the points (corresponding to) $v_{1}$ and $v_{2}$ are no
longer vertices because exactly the angle deficit at each has been sutured-in;
they have been replaced by a new vertex $v^{\prime}$ of curvature
${\omega}^{\prime}={\omega}_{1}+{\omega}_{2}$. So vertex-merging always
reduces the number of vertices of $P$ by one. See Fig. 2.6.

Figure 2.6: (a) $Q$ is a $5$-vertex, $6$-face polyhedron, symmetric about a
vertical plane through $xy$. Its base $abc$ is an equilateral triangle. (b)
Vertex merging $x$ and $y$ by gluing a pair of $xyz$ triangles. (c) The
merging reduces $Q$ to a regular tetrahedron (not to same scale). Cf. Fig.
1.1(a,b).

In order to repeat vertex-merging, we need to know when there is a pair of
vertices that can be merged. This is answered in the following lemma.

######  Lemma 2.11.

Every convex polyhedron $Q$ has at least one pair of vertices admitting
merging, unless it is an isosceles tetrahedron or a doubly-covered triangle.

Recall from Chapter 1 that an _isosceles tetrahedron_ $T$ is a tetrahedron
with four congruent faces, and each vertex of curvature $\pi$. See Fig. 1.4.

###### Proof.

If there is a pair of vertices whose sum of curvatures is strictly less than
$2\pi$, then vertex-merging is possible, as just described. So assume that,
for any two vertices of $Q$, their sum of curvatures is at least $2\pi$. In
this case, it must be that $n\leq 4$. Indeed, since $\sum_{v\in
Q}{\omega}(v)=4\pi$ (by the Gauss-Bonnet theorem), if the sum of at least $5$
positive numbers is $4\pi$ then the smallest two have sum $<2\pi$.

If $n=3$, the cut locus of any vertex is a line-segment, by Lemma 2.1(i), so
$Q$ is a doubly-covered triangle (see Lemma 2.2).

If $n=4$ then necessarily all vertex curvatures of $Q$ are $\pi$. Indeed, if
the sum of $4$ positive numbers is $4\pi$ then the smallest two have the sum
$\leq 2\pi$, with equality if and only if all are $\pi$. So $Q$ is an
isosceles tetrahedron. ∎

###  Chapter 3 Domes and Pyramids

One of our goals in this work, achieved in Theorem 4.6, is to show that if $Q$
can be obtained from $P$ by sculpting, then it can also be obtained from $P$
by tailoring. A key step (Lemma 4.1) repeatedly slices off shapes we call
g-domes. Each g-dome slice can itself be achieved by slicing off pyramids,
i.e., by suitable vertex truncations. Lemma 4.4 will show that slicing off a
pyramid can be achieved by tailoring, and thus leading to Theorem 4.6.

Our main goal in this chapter is to prove that g-domes can be partitioned into
“stacked” pyramids, a crucial part of the above process. We illustrate the
slice $\to$ g-domes and g-dome $\to$ pyramids of the process on a simple
example, a tetrahedron $Q$ inside a cube $P$. Later (in Chapter 4) this
example will be completed by tailoring the pyramids.

####  3.1 Domes

As usual, a _pyramid_ $P$ is the convex hull of a convex polygon _base_ $X$,
and one vertex $v$, the _apex_ of $P$, that does not lie in the plane of $X$.
The degree of $v$ is the number of vertices of $X$.

A _dome_ is a convex polyhedron $G$ with a distinguished face $X$, the _base_
, and such that every other face of $G$ shares a (positive-length) edge with
$X$. Domes have been studied primarily for their combinatorial [Epp20], [EL13]
or unfolding [DO07] properties. In [Epp20] they are called “treetopes” because
removing the base edges from the $1$-skeleton leaves a tree, which the author
calls the _canopy_.111 These polyhedra are not named in [EL13]. Here we need a
slight generalization.

A _generalized-dome_ , or _g-dome_ $G$, has a base $X$, with every other face
of $G$ sharing either an edge or a vertex with $X$. Every dome is a g-dome,
and it is easy to obtain every g-dome as the limit of domes. An example is
shown in Fig. 3.1, which also shows that removing base edges from the
$1$-skeleton does not necessarily leave a tree: $(v_{1},x_{2},v_{2})$ forms a
cycle. Let us define the _top-canopy_ $T$ of a g-dome $G$ as the graph that
results by deleting from the $1$-skeleton of $G$ all base vertices and their
incident edges. In Fig. 3.1 the top-canopy is $v_{1}v_{2}$.

Figure 3.1: A g-dome with base $\triangle x_{1}x_{2}x_{3}$ and top-canopy
$v_{1}v_{2}$.

######  Lemma 3.1.

The top-canopy $T$ of a g-dome $G$ is a tree.

###### Proof.

If $G$ is a dome, the claim follows, because even including the edges incident
to the base $X$ results in a tree, and removing those edges leaves a smaller
tree.

If $G$ is not a dome, then slice it with a plane parallel to, and at small
distance above, the base. The result is a dome, and we can apply the previous
reasoning. ∎

####  3.2 Cube/Tetrahedron Example

In this section we start to track an example, tailoring a cube to a
tetrahedron. $P$ is a triangulated unit cube. $Q$ is a right tetrahedron in
the corner of $P$; see Fig. 3.2(a). $Q$ can be obtained from $P$ by a single
slice by plane $\Pi=p_{1}p_{3}p_{8}$. The algorithm to be described in Chapter
6 will first produce two g-domes $G_{1}$ and $G_{2}$, and reducing those
g-domes will lead to four pyramids $P_{1},P_{2},P_{3},P_{4}$. Later (in
Chapter 4) we will show how each of those pyramids is reduced by digon-
tailoring, finally yielding $Q$.

#####  3.2.1 Slice $\to$ G-Domes for Cube/Tetrahedron

Now we describe at an intuitive level what will be proved in Lemma 4.1: that
the slice by $\Pi$ can be effected by partitioning the sliced portion of $P$
into g-domes.

We start by rotating the plane $\Pi$ about the edge $p_{1}p_{3}$ until it
encounters a face of $P$; call that plane $\Pi_{0}=p_{1}p_{2}p_{3}$. Now
continue rotating until we reach $\Pi_{1}$ when (a) The portion of $P$ between
$\Pi_{0}$ and $\Pi_{1}$ is a g-dome with base on $\Pi_{1}$, and (b) Any
further rotation would render the portion between the planes to a polyhedron
no longer a g-dome. $\Pi_{1}=p_{1}p_{3}p_{7}p_{5}$, because any further
rotation would isolate the face $p_{5}p_{6}p_{7}$ of $P$, no longer touching
$\Pi_{1}$. Note here the top face of the cube is triangulated with the
diagonal $p_{5}p_{7}$. See Fig. 3.2(bc). Continuing to rotate, we reach the
original $\Pi$, partitioning the g-dome illustrated in Fig. 3.2(d). So $G_{1}$
has base $X_{1}=p_{1}p_{3}p_{7}p_{5}$ and top-canopy $p_{2}p_{6}$, and $G_{2}$
has base $X_{2}=p_{1}p_{2}p_{3}$ and top-canopy $p_{5}p_{7}$. It is clear that
removing $G_{1}\cup G_{2}$ would reduce $P$ to $Q$.

Figure 3.2: (a) $Q$ in corner of cube $P$. (b) Partitioning sliced portion of
$P$ into two g-domes. (c,d) G-domes $G_{1}$ and $G_{2}$. Figure 3.3:  (a) A
dome $G$. Vertex $v_{1}$ of degree $k_{1}=6$ is adjacent to $X$, i.e., there
are edges $v_{1}x_{i}$. (b) $G^{\prime}$ after removal of $v_{1}$.

####  3.3 Proof: G-dome $\to$ Pyramids

We now prove the reduction of a g-dome to pyramids, using Fig. 3.3 to
illustrate the proof details. Afterward we will return to the cube/tetrahedron
example to apply the constructive proof to $G_{1}$ and $G_{2}$.

######  Theorem 3.2.

Every g-dome $G$ of base $X$ can be partitioned into a finite sequence of $n$
pyramids $P_{i}$ with the following properties:

  * •

Each $P_{i}$ has a common edge with $X$.

  * •

Each $G_{j}=G\setminus\bigcup_{i=1}^{j}P_{i}$ is a g-dome, for all
$j=1,...,n$.

  * •

The last pyramid $P_{n}$ in the sequence has the same base $X$ as $G$.

The partition into pyramids specified in this theorem has the special
properties listed. Without these properties, it would be easier to prove that
every g-dome may be partitioned into pyramids. But the properties are needed
for the subsequent digon-tailoring described in Chapter 4. In particular, the
pyramids are “stacked” in the sense that each can be sliced-off in sequence,
with each slice retaining $X$ unaltered.

The proof is a double induction, and a bit intricate. One induction simply
removes one vertex $v_{1}$ from the top-canopy. The second induction, inside
the first one, reduces the degree of $v_{1}$ to achieve removal of $v_{1}$, at
the cost of increasing the degree of $v_{2}$.

Figure 3.4:  (a) After slicing by the plane $\Pi_{1}=x_{1}x_{2}v_{2}$. (b)
Removing $a_{3}$. (c) Removing $a_{4}$ creates $a^{\prime}_{4}$. (d) Removing
$a^{\prime}_{4}$. (e) $a_{6}$ has replaced $v_{1}$ with one lower degree. (f)
Intersection points with $\Pi_{2}=x_{2}x_{3}v_{2}$.

###### Proof.

Let $m$ be the number of vertices in the top-canopy $T$ of $G$. If $m=1$, $G$
is already a pyramid, and we are finished. So assume $G$’s top-canopy $T$ has
at least two vertices. Choose $v_{1}$ to be a leaf of $T$ given by Lemma 3.1,
and $v_{2}$ its unique parent. Let $v_{1}$ be adjacent to $k$ vertices of $X$.
If $G$ is a dome, $v_{1}$ has degree $k+1$; if $G$ is a g-dome, then possibly
$v_{1}$ has degree $k+2$. Since the later case changes nothing in the proof,
we assume for the simplicity of exposition that $G$ is a dome. Our goal is to
remove $v_{1}$ through a series of pyramid subtractions.

Let the vertices of $X$ adjacent to $v_{1}$ be $x_{1},x_{2},\ldots,x_{k}$. Let
$\Pi_{1}$ be the plane $x_{1}x_{2}v_{2}$. This plane cuts into $G$ under
$v_{1}$, and intersects the edges $v_{1}x_{i}$, $i\geq 3$, in points $a_{i}$.
In Fig. 3.4(a), those points are $a_{3},a_{4},a_{5},a_{6}$. Remove the pyramid
whose apex is $v_{1}$ and whose base (in our example) is
$x_{1},x_{2},a_{3},a_{4},a_{5},a_{6},v_{2}$.

We now proceed to reduce the chain of new vertices $a_{3},\ldots,a_{k}$ one-
by-one until only $a_{k}$ remains.

First, with the plane $x_{2}x_{3}a_{4}$, we slice off the tetrahedron whose
degree-$3$ apex is $a_{3}$; Fig. 3.4(b). Next, with the plane
$x_{2}x_{3}a_{5}$, we slice off the pyramid with apex $a_{4}$. Unfortunately,
because $a_{4}$ has degree-$4$, this introduces a new vertex $a^{\prime}_{4}$;
Fig. 3.4(c). So next we slice with $x_{3}x_{4}a_{5}$ to remove the tetrahedron
whose degree-$3$ apex is $a^{\prime}_{4}$; Fig. 3.4(d). Continuing in this
manner, alternately removing a tetrahedron followed by a pyramid with a
degree-$4$ apex, we reach Fig. 3.4(e).

Note that $v_{1}$ was connected to $k=6$ vertices of $X$, but $a_{6}$ is only
connected to $5$: the connection of $v_{1}$ to $x_{1}$ has in a sense been
transferred to $v_{2}$. In general, $a_{k}$ has degree one less than $v_{1}$’s
degree, and the degree of $v_{2}$ has increased by $1$.

Now we repeat the process, starting by slicing with $\Pi_{2}=x_{2}x_{3}v_{2}$,
which intersects the $a_{k}x_{i}$ edges at $b_{4},\ldots,b_{k}$. We remove the
pyramid apexed at $a_{6}$ with base (in our example) of
$x_{2},x_{3},b_{4},b_{5},b_{6},v_{2}$; Fig. 3.4(f). The same methodical
technique will remove all but the last new vertex $b_{k}$, which replaces
$a_{k}$ but has degree one smaller.

Continuing the process, slicing with $\Pi_{i}=x_{i}x_{i+1}v_{2}$, up to
$i=k-1$, will lead to the complete removal of $v_{1}$, as previously
illustrated in Fig. 3.3(b), completing the inner induction. Inverse induction
on the number of vertices of the g-dome then completes the proof: each step
reduces the number of top vertices by $1$, and the degree of $v_{2}$ increases
by $k+1$, the degree lost at $v_{1}$.

With $G_{0}=G$, each $G_{j}$ is the intersection of $G_{k-1}$ with a closed
half-space containing a base edge, so it is convex for all $j=1,...,n$. Indeed
each $G_{j}$ is a g-dome, because all untouched faces continue to meet $X$ in
either an edge or a vertex, and new faces always share an edge with $X$. ∎

The partition of a g-dome into pyramids given by Theorem 3.2 is not unique.
For our example in Fig. 3.3(a), we finally get the pyramid apexed at $v_{2}$
in (b) of the figure, but we could as well have ended with a pyramid apexed at
$v_{1}$.

We will see in Chapter 4 that one g-dome of $O(n)$ vertices reduces to
$O(n^{2})$ pyramids of constant size, and $O(n)$ pyramids each of size $O(n)$.

#####  3.3.1 G-domes $\to$ Pyramids for Cube/Tetrahedron

We now return to the cube/tetrahedron example, and show that following the
proof of Theorem 3.2 partitions the two g-domes $G_{1}$ and $G_{2}$ into
pyramids. We will see that the two g-domes partition into two pyramids each.

The analysis for $G_{1}$ is illustrated in Fig. 3.5. Note $G_{1}$ has been
reoriented so that its base $X_{1}=p_{1}p_{3}p_{7}p_{5}$ is horizontal, and
relabeled $x_{1}x_{2}x_{3}x_{4}$ to match the proof.

We proceed to describe each step, making comparisons to Figs. 3.3 and 3.4.

  1. (a)

In (a) of the figure, the base of the g-dome has been oriented horizontally,
and the correspondence between the cube labels and the labels used in the
proof of Theorem 3.2 are shown. The top-canopy of $G_{1}$ is $v_{1}v_{2}$,
with $v_{1}$ of degree-$5$ and $v_{2}$ of degree-$3$. Eventually $v_{1}$ will
be removed and the degree of $v_{2}$ increased by $1$, as in Fig. 3.3.

  2. (b)

The first step is to slice off a pyramid apexed at $v_{1}$ by the plane
$x_{1}x_{2}v_{2}$. Due to the coplanar triangles, this has the effect of
“erasing” the diagonal $x_{1}v_{1}$, as would have occurred if that diagonal
had a dihedral angle different from $\pi$.

  3. (c)

Next, the slice $x_{2}x_{3}v_{2}$ cuts off a pyramid $P_{1}$ apexed at
$v_{1}$, a pyramid we will analyze in Section 4.4.1, leaving the shape in (d).

  4. (d)

A new vertex $a_{3}$ is created by the slice in (c), with degree-$4$. This is
the “replacement” for $v_{1}$ but of smaller degree. This corresponds to
$a_{3}$ in Fig. 3.4(a), although here there are no further vertices
$a_{4},\ldots$ along an $a$-chain.

  5. (e)

We again slice with $x_{2}x_{3}v_{2}$, which reduces the degree of $a_{3}$ to
$3$. This $a_{3}$ corresponds to $b_{4}$ in Fig. 3.4(f).

  6. (f)

A final slice by $x_{3}x_{4}v_{2}$ removes $a_{3}$, leaving the pyramid
$P_{2}$ shown. This completes the removal of $v_{1}$ in (a), increasing the
degree of $v_{2}$ from $3$ to $4$. Because what remains is a pyramid, the
processing stops.

So $G_{1}$ is partitioned into two pyramids, $P_{1}$ and $P_{2}$, apexed at
$p_{2}$ and $p_{6}$ respectively.

Figure 3.5: (a) g-dome $G_{1}$. (b) After slicing by $x_{1}x_{2}v_{2}$. (c,d)
Slicing by $x_{2}x_{3}v_{2}$. (e,f) Two further slices remove $a_{3}$.

We perform a similar analysis for the simpler $G_{2}$, shown in Fig. 3.6,
again reoriented so that its base $p_{8}p_{1}p_{3}$ is horizontal and
relabeled $x_{1}x_{2}x_{3}$. Here the top-canopy is $v_{1}v_{2}$, and the
first slice $x_{1}x_{2}v_{2}$ reduces $G_{2}$ to a pyramid $P_{2}$. So $G_{2}$
is also partitioned into two pyramids, $P_{3}$ and $P_{4}$, apexed at $p_{5}$
and $p_{7}$ respectively. Together $G_{1}$ and $G_{2}$ are partitioned into
four pyramids.

Figure 3.6: (a) $G_{2}$ with corresponding cube labels. (b) After slicing with
$x_{1}x_{2}v_{2}$.

We will complete this example by tailoring the four pyramids in the next
chapter.

###  Chapter 4 Tailoring via Sculpting

In this chapter we complete the proof that one slice of $P$ by plane $\Pi$ can
be tailored to the face of $Q$ lying in $\Pi$, following the sequence slice
$\to$ g-domes $\to$ pyramids $\to$ tailoring. The previous chapter established
the g-domes $\to$ pyramids link. Here we first prove the relatively
straightforward slice $\to$ g-domes process, and then concentrate on the more
complex pyramid $\to$ tailoring step.

####  4.1 Slice $\to$ G-domes

######  Lemma 4.1.

With $Q\subset P$ and $\Pi$ a plane slicing $P$ and containing a face $F$ of
$Q$, the sliced-off portion $P^{\prime}$ can be partitioned into a _fan_ of
$O(n)$ g-domes $G_{1},G_{2},\ldots$, a fan in the sense that the bases of the
g-domes all share a common edge of $F$.

Figure 4.1: Dodecahedron sliced by $\Pi$. The polyhedron between each pair
$\Pi_{i},\Pi_{i+1}$ is a g-dome.

###### Proof.

Assume $\Pi$ is horizontal, with $Q$ the portion below $\Pi$ and $P^{\prime}$
the portion above. Denote by $x_{i}$ the vertices of $Q$ in $\Pi$,
$i=1,\ldots,m$; call the top face of $Q$ with these vertices $F$. Let $e$ be
any edge of $F$, say $e=x_{1}x_{2}$, and let $F^{\prime}$ be the face of $Q$
sharing $e$ with $F$. Call the plane lying on $F^{\prime}$ $\Pi_{0}$.

Now imagine rotating $\Pi_{0}$ about $e$ toward $P^{\prime}$, noting as it
passes through each vertex $v_{1},v_{2},\ldots$ in that order. For perhaps
several consecutive vertices, the portion of $P^{\prime}$ between the previous
and the current plane is a g-dome, but rotating further takes it beyond a
g-dome. More precisely, let $\Pi_{1}$ through $v_{j_{1}}$ be the plane such
that, in the sequence

$v_{1},v_{2},\ldots,v_{j_{1}},v_{j_{1}+1},\ldots,v_{j_{2}},v_{j_{2}+1}\ldots\;,$

the portion $G_{1}$ of $P^{\prime}$ between $\Pi_{0}$ and $\Pi_{1}$, including
the vertices $v_{1},v_{2},\ldots,v_{j_{1}}$, is a g-dome based on $\Pi_{1}$,
but the portion rotating further to include one more vertex, $v_{j_{1}+1}$, is
not a g-dome. $\Pi_{2}$ through $v_{j_{2}}$ is defined similarly: the portion
$G_{2}$ between $\Pi_{1}$ and $\Pi_{2}$ including
$v_{j_{1}+1},\ldots,v_{j_{2}}$ is a g-dome based on $\Pi_{2}$, but including
$v_{j_{2}+1}$ it ceases to be a g-dome. Here for each $\Pi_{i},\Pi_{i+1}$
pair, the base $X$ of the g-dome lies in $\Pi_{i+1}$.

Fig. 4.1 illustrates the process. Here $\Pi_{1}$ is through $v_{1}=v_{j_{1}}$,
and the last g-dome lies between $\Pi_{3}$ and $\Pi$.

So we have now partitioned $P^{\prime}$ into g-domes $G_{1},G_{2},\ldots$. ∎

We will invoke Lemma 4.4 (ahead) to reduce each g-dome to its base by
tailoring, in the order $G_{1},G_{2},\ldots$. This will reduce $P^{\prime}$ to
just the top face $F$ of $Q$.

Having established the claim of the lemma for one slice, it immediately
follows that it holds for an arbitrary number of slices. This was already
illustrated in the cube/tetrahedron example in Chapter 3, and we will use it
again in Theorem 4.6.

We should note that it is at least conceivable that only a single g-dome is
needed above. See Open Problem 18.2.

####  4.2 Pyramid $\to$ Tailoring

The goal of this section is to prove that removal of a pyramid, i.e., a vertex
truncation, can be achieved by (digon-)tailoring. We reach this in Lemma 4.4:
a degree-$k$ pyramid $P$ can be removed by $k-1$ tailoring steps, each step
excising one vertex by removal of and then sealing a digon. The first $k-2$ of
these digons each have one endpoint a vertex, and so leave the total number of
vertices of $P$ at $k+1$. The $(k-1)$-st digon has both endpoints vertices,
and so its removal reduces the number of vertices to the $k$ base vertices.

We start with Lemma 4.2 which claims the result but only under the assumption
that the slice plane is close to the removed vertex. Although this lemma is
eventually superseded, it establishes the notation and the main idea.
Following that, Lemma 4.3 removes the “sufficiently small” assumption of Lemma
4.2, but in the special case of $P$ a pyramid. Finally we reach the main claim
in Lemma 4.4, which shows this special case encompasses the general case.

In the following, we use $\partial S$ to indicate the $1$-dimensional boundary
of a $2$-dimensional surface patch $S$.

#####  4.2.1 Notation

To help keep track of the notation throughout this critical section, we list
the main symbols below.

  * •

Initially $P$ and $Q$ are polyhedra with $Q\subset P$, with $P$ sliced by
plane $\Pi$ to truncate vertex $v$ of degree $k$.

  * •

Later we specialize $P$ to be the pyramid sliced off.

  * •

$X=\Pi\cap P$ is the base of the pyramid, with vertices $\partial
X=x_{1}x_{2}\ldots x_{k}$.

  * •

The lateral faces of $P$ are denoted by $L$, so $P=X\cup L$.

  * •

$D(x_{i},y_{i})$ is a digon with endpoints $x_{i}\in\partial X$ and $y_{i}$.

  * •

$P_{j}$ is the modified pyramid after $j$ digon removals.

  * •

$C(x_{i},P_{j})$ is the cut locus $C(x_{i})$ on $P_{j}$.

  * •

The first ramification point of $C(x_{i},P_{j})$ beyond $y_{i}$ is $a_{i}$.

Notation will be repeated and supplemented within each proof.

####  4.3 Small volume slices

######  Lemma 4.2.

Let $P$ be a convex polyhedron, and $Q$ the result obtained by slicing $P$
with a plane $\Pi$ at sufficiently small distance to a vertex $v$ of $P$, and
removing precisely that vertex. Then $Q$ can be obtained from $P$ by $k-1$
tailoring steps.

###### Proof.

Let the vertex $v$ to be removed have degree $k$ in the $1$-skeleton of $P$.
Let $e_{i}$, $i=1,\ldots,k$, be the edges incident to $v$, and $x_{i}$ the
intersection of the slicing plane $\Pi$ with those edges: $\\{x_{i}\\}=\Pi\cap
e_{i}$.

We will illustrate the argument with the right triangular prism shown in Fig.
4.2, where $k=3$ and $\Pi=x_{1}x_{2}x_{3}$. Note that we do not exclude the
case when some (or all) of the $x_{i}$ are vertices of $P$.

Figure 4.2: $P$ is a prism whose top face is an isosceles right triangle. The
truncated vertex $v$ has degree $k{=}3$. $x_{3}$ is the midpoint of $e_{3}$.
Digon $D_{1}\supset\\{v\\}$ is shown red; $D_{2}\supset\\{y_{1}\\}$ is purple.
$D_{1}$ is sutured closed before $D_{2}$ is excised. The last replacement
vertex $y_{k-1}$ must be identical to $x_{k}$.

Denote by ${\omega}_{P}(x_{i})$ $i=1,\ldots,k$ the curvatures of $P$ at
$x_{i}$, and by ${\omega}_{Q}(x_{i})$ the corresponding curvatures of $Q$. The
curvature ${\omega}_{P}(v)$ will be distributed to the $x_{i}$.

In the figure, ${\omega}_{P}(v)=90^{\circ}$, the curvatures of the three
$x_{i}$ are $135^{\circ},135^{\circ},0^{\circ}$ in $P$, and approximately
$156^{\circ},156^{\circ},48^{\circ}$ in $Q$. Indeed the increases sum to
$90^{\circ}$: $21^{\circ}+21^{\circ}+48^{\circ}$.

The goal now is to excise $k-1$ digons with one end at
$x_{1},x_{2},\ldots,x_{k-1}$, removing precisely the surface angle needed to
increase ${\omega}_{P}(x_{i})$ to ${\omega}_{Q}(x_{i})$. After digon removals
at $x_{1},\ldots,x_{i}$, we call the resulting polyhedron $P_{i}$.

Let a digon with endpoints $x_{i}$ and $y_{i}$ be denoted
$D_{i}=(x_{i},y_{i})$. Cut out from $P$ the digon $D_{1}=(x_{1},y_{1})$
containing only the vertex $v$ in its interior, of angle at $x_{1}$ equal to
${\omega}_{Q}(x_{1})-{\omega}_{P}(x_{1})$. By the assumption that the slice
plane $\Pi$ is sufficiently close to $v$, the curvature difference is small
enough so that $D_{1}$ includes only $v$. Again by the sufficiently-close
assumption, we may assume the digon endpoint $y_{1}$ lies on the edge of
$C(x_{1})$ incident to $v$, prior to the first ramification point of
$C(x_{1})$. After suturing closed the digon geodesics, $y_{1}$ becomes a
vertex of curvature
${\omega}_{P}(v)-({\omega}_{Q}(x_{1})-{\omega}_{P}(x_{1}))$. In the figure,
$y_{1}$ has curvature $90^{\circ}-21^{\circ}\approx 69^{\circ}$. In a sense,
$y_{1}$ “replaces” $v$.

Next cut out a digon $D_{2}=(x_{2},y_{2})$ containing only the vertex $y_{1}$
in its interior, of angle at $x_{2}$ equal to
${\omega}_{Q}(x_{2})-{\omega}_{P}(x_{2})$. The newly created vertex $y_{2}$
“replaces” $y_{1}$. Continue cutting out digons $D_{i}=(x_{i},y_{i})$ up to
$i=k-1$, each $D_{i}$ surrounding $y_{i-1}$, and replacing $y_{i-1}$ with
$y_{i}$.

Because these tailorings have sharpened the curvatures ${\omega}_{P}(x_{i})$
to match the after-slice curvatures ${\omega}_{Q}(x_{i})$, it must be that the
curvature at the last replacement vertex $y_{i-1}$ is the same as the
curvature at $x_{k}$: ${\omega}_{P}(y_{k-1})={\omega}_{Q}(x_{k})$ (to satisfy
Gauss-Bonnet). So now the tailored $P_{k-1}$ matches $Q$ in both the positions
of the vertices $x_{i}$, $i=1,\ldots,k-1$, and their curvatures; the only
possible difference is the location of $y_{i-1}$ compared to $x_{k}$. But the
rigidity result, Theorem 2.9, implies that $y_{k-1}=x_{k}$, and $P_{k-1}$ and
$Q$ are now congruent. ∎

The “sufficiently-small” assumption in the preceding proof allowed us to
assume that the digon $D_{i}=(x_{1},y_{1})$ endpoint $y_{1}$ lay on the
segment of $C(x_{i})$ incident to $v$ prior to the first ramification point
$a_{1}$ of $C(x_{1})$. Recall that
${\omega}(x_{1})+{\omega}(y_{i})={\omega}(v)$, and the further along the
segment $va_{1}$ that $y_{1}$ lies, the larger the digon angle at $x_{1}$ is.
The procedure would be problematic if the digon angle at $x_{1}$ were not
large enough even with $y_{1}$ at that ramification point $a_{1}$. The next
lemma removes the sufficiently-small assumption in the special case when $P$
is itself a pyramid, and the vertex truncation reduces $P$ its base, doubly-
covered. Following this, we will show that the case when $P$ is a pyramid is
the “worst case,” and so the general case follows.

#####  4.3.1 Pyramid case

######  Lemma 4.3.

Let $P$ be a pyramid over base $X$. Then one can tailor $P$ to reduce it to
$X$ doubly-covered, using $k-1$ digon removal steps.

###### Proof.

We continue to use the notation in the previous lemma, and introduce further
notation needed here. Let $L=P\setminus X$ be the lateral sides of the pyramid
$P$; so $P=L\cup X$. After each digon $D_{i}=(x_{i},y_{i})$ is removed and
sutured closed, the convex polyhedron guaranteed by Alexandrov’s Gluing
Theorem will be denoted by $P_{i}$. We continue to view $P_{i}$ as
$P_{i}=L_{i}\cup X_{i}$, even though already $P_{1}$, is in general no longer
a pyramid. We will see that all the digon excisions occur on $L_{i}$, while
$X_{i}$ remains isometric to the original base $X$, but no longer (in general)
planar.

We will use $C(x_{i},P_{j})$ to mean the cut locus of $x_{i}$ on $P_{j}$.
Regardless of which $P_{j}$ is under consideration, we will denote by $a_{i}$
the first ramification point of $C(x_{i})$ immediately beyond the vertex
$y_{i-1}$ surrounded by the digon $D_{i}=(x_{i},y_{i})$.

We need to establish two claims:

Claim (1):

The cut locus $C(x_{i+1},P_{i})$ is wholly contained in $L_{i}$.

Claim (2):

The digon angle ${\alpha}_{i+1}$ at $x_{i+1}$ to $a_{i+1}$ is large enough to
reduce the $L$-angle at $x_{i+1}$ to its $X$-angle on the base.

Before addressing the general case of these claims, we illustrate the
situation for $x_{1}$, referencing Fig. 4.3.

Figure 4.3: (a) A pyramid with pentagonal base $X$. Shortest paths from
$x_{1}$ to all vertices are marked green. (b) The star-unfolding with respect
to $x_{1}$. The triangles from $X$ are blue; those from $L$ are green.
$C(x_{1})\subset L$ is red. The digon $D_{1}=(x_{1},y_{1})$ is shaded.

The digon $D_{1}(x_{1},y_{1})$ surrounding $v$ places $y_{1}$ on the $va_{1}$
segment of $C(x_{1},P)=C(x_{1})$. If one imagines $y_{1}$ sliding along
$va_{1}$ from $v$ to $a_{1}$, the digon angle at $x_{1}$, call it
${\delta}_{1}$, increases. To show that $y_{1}$ can be placed so that
${\delta}_{1}$ is large enough to reduce the angle at $x_{1}$ to its angle in
$X$ will require $a_{1}$ to lie in $L$ (rather than in $X$).

It turns out that $C(x_{1})\subset L$ follows from a lemma in [AAOS97].111
Lem. 3.3: the cut locus is contained in the “kernel” of the star-unfolding
which in our case is a subset of $L$. However, after removing $D_{1}$ and
invoking Alexandrov’s Gluing Theorem, we can no longer apply this lemma. With
this background, we now proceed to the general case.

###### Claim (1): $C(x_{i+1},P_{i})\subset L$.

Assume we have removed digons at $x_{1},\ldots,x_{i}$, so that
$P_{i}=X_{i}\cup L_{i}$, and $L_{i}$ contains one vertex $y_{i}$, the endpoint
of the last digon $D_{i}$ removed, and $X_{i}$ contains no vertices. Assume to
the contrary of Claim (1) that $C(x_{i+1},P_{i})=C(x_{i+1})$ includes a point
$z$ strictly interior to $X_{i}$. Because $z\in C(x_{i+1})$, there are two
geodesic segments from $x_{i+1}$ to $z$, call them ${\gamma}^{z}_{1}$ and
${\gamma}^{z}_{2}$. Because $X_{i}$ contains no vertices, it cannot be that
both ${\gamma}^{z}_{1}$ and ${\gamma}^{z}_{2}$ are in $X_{i}$. Say that
${\gamma}^{z}_{1}$ crosses $L_{i}$. Let $p\in\partial X$ be the first point at
which ${\gamma}^{z}_{1}$ enters $X_{i}$, and let
${\gamma}_{1}\subset{\gamma}^{z}_{1}$ be the portion from $x_{i+1}$ to $p$.
See Fig. 4.4.

Figure 4.4: $X_{i}$ is flipped above $L_{i}$ in this abstract illustration.
Cauchy’s Arm Lemma ultimately shows that
$\ell({\gamma}_{1})\geq\ell({\gamma}_{2})$.

The geodesic segment ${\gamma}_{1}$ divides $L_{i}$ into two parts; let
$L^{\prime}_{i}$ be the part that does not contain the vertex $y_{i}$. Join
$x_{i+1}$ to $p$ with a geodesic ${\gamma}_{2}$ lying in $X_{i}$.
${\gamma}_{2}$ was a shortest path to $p$ in $X$, but may no longer be
shortest in $X_{i}$. ${\gamma}_{2}$ also divides $X_{i}$ into two parts; let
$X^{\prime}_{i}$ be the part sharing a portion of $\partial X$ with
$L^{\prime}_{i}$.

Now we will argue that $\ell({\gamma}_{1})\geq\ell({\gamma}_{2})$, where
$\ell({\gamma})$ is the length of ${\gamma}$. This will yield a contradiction,
for the following reasons. ${\gamma}_{1}$ is a shortest geodesic to $p$,
because it extends to ${\gamma}^{z}_{1}$, which is a shortest geodesic to $z$.
So ${\gamma}_{2}$ cannot be strictly shorter than ${\gamma}_{1}$. Therefore we
must have $\ell({\gamma}_{1})=\ell({\gamma}_{2})$, which implies that $p\in
C(x_{i+1})$. But then ${\gamma}_{1}$ cannot continue to ${\gamma}^{z}_{1}$
beyond $p$, as $p$ is a cut point.

To reach $\ell({\gamma}_{1})\geq\ell({\gamma}_{2})$, we will use the extension
of Cauchy’s Arm Lemma described in Chapter 2. Let ${\theta}^{X}_{j}$ be the
angle at $x_{j}$ in $X_{i}$, and ${\theta}^{L}_{j}$ be the angle at $x_{j}$ in
$L_{i}$. For $j>i+1$, we know that ${\theta}^{L}_{j}>{\theta}^{X}_{j}$ because
$P$ is a pyramid (see Lemma 2.8) and digon removal has not yet reached
$x_{j}$. We also know that ${\theta}^{X}_{j}\leq\pi$ because $X$ is convex.
However, ${\theta}^{L}_{j}$ could be nearly as large as
$2\pi-{\theta}^{X}_{j}$ if the pyramid $P$’s apex $v$ projects outside the
base $X$.

Let $c_{X}$ be the planar convex chain in $\partial X$ that corresponds to
$X^{\prime}_{i}$; assume $c_{X}$ is $x_{i+1},x_{i+2},\ldots,x_{j},\ldots,p$
with angles ${\theta}^{X}_{j}$. (The case where $c_{X}$ is includes the other
part of $\partial X$, $x_{i+1},x_{i},\ldots,p$ can be treated analoguously.)
Then $\ell({\gamma}_{2})$ is the length of the chord between $c_{X}$’s
endpoints. In order to apply Cauchy’s lemma, we rephrase the angles at $x_{j}$
as turn angles ${\tau}_{j}=\pi-{\theta}^{X}_{j}$. The extension of Cauchy’s
lemma guarantees that, if the chain angles are modified so that the turn
angles lie within $[-{\tau}_{j},{\tau}_{j}]$, then the endpoints chord length
cannot decrease. Roughly, opening (straightening) the angles stretches the
chord.

Define $c_{L}$ as the planar (possibly nonconvex) chain composed of the same
vertices $x_{j}$ that define $c_{X}$, but with angles ${\theta}^{L}_{j}$.
Because ${\theta}^{L}_{j}\leq 2\pi-{\theta}^{X}_{j}$, the turn angles
$\pi-{\theta}^{L}_{j}$ in $c_{L}$ satisfy

$\pi-{\theta}^{L}_{j}\geq\pi-(2\pi-{\theta}^{X}_{j})=-(\pi-{\theta}^{X}_{j})=-{\tau}_{j}\;.$

Also, because ${\theta}^{L}_{j}>{\theta}^{X}_{j}$,

$\pi-{\theta}^{L}_{j}\leq\pi-{\theta}^{X}_{j}={\tau}_{j}\;.$

So the $c_{L}$ turn angles are in $[-{\tau}_{j},{\tau}_{j}]$, and we can
conclude from Theorem 2.7 that the $c_{L}$ endpoints chord length
$\ell({\gamma}_{1})$ is at least $\ell({\gamma}_{2})$, the $c_{X}$ endpoints
chord length.

We have now reached $\ell({\gamma}_{1})\geq\ell({\gamma}_{2})$, whose
contradiction described earlier shows that indeed $C(x_{i+1})\subset L_{i}$.

###### Claim (2): ${\alpha}_{i+1}\geq{\theta}^{L}_{i+1}-{\theta}^{X}_{i+1}$.

Recall that ${\alpha}_{i+1}$ is the angle at $x_{i+1}$ of the digon from
$x_{i+1}$ to $a_{i+1}$, the first ramification point of $C(x_{i+1})$ beyond
the vertex $y_{i}$. The claim is that ${\alpha}_{i+1}$ is large enough to
reduce ${\theta}^{L}_{i+1}$ to ${\theta}^{X}_{i+1}$. We establish this by
removing a path from $C(x_{i+1})$ and tracking angles, as follows.

From Claim (1), $C(x_{i+1})\subset L_{i}$. Let ${\rho}_{i,i+2}$ be the path in
the tree $C(x_{i+1})$ from $x_{i}$ to $x_{i+2}$. See Fig. 4.5.

Figure 4.5:  Star-unfolding of a pyramid with respect to $x_{i+1}$. The
triangles from $X_{i}$ are blue; those from $L_{i}$ are green. Red points are
images of $x_{i+1}$. Digons are shaded. ${\rho}_{i,i+2}$ is purple; remainder
of $C(x_{i+1})$ is red.

Removal of ${\rho}_{i,i+2}$ from $C(x_{i+1})$ disconnects $C(x_{i+1})$ into
the edge $y_{i}a_{i+1}$, and a series of subtrees $T_{j}$. Each $T_{j}$ shares
a point $z_{j}$ with $C(x_{i+1})$. Let $D_{j}$ be the digon from $x_{i+1}$ to
$z_{j}$, and let ${\delta}_{j}$ be the angle of $D_{j}$ at $x_{i+1}$. Finally,
let ${\Delta}_{j}=\sum_{j}{\delta}_{j}$.

Note that all the ${\delta}_{j}$ angles are in $X_{i}$. In contrast, the angle
at $x_{i+1}$ in the digon $D_{i+1}=D(x_{i+1},a_{i+1})$ is in $L_{i}$, as
illustrated in Fig. 4.5. We defer justifying this claim to later.

Cut off all $D_{j}$, and also cut off $D_{i+1}$. Suture the surface closed;
call it $P^{*}=X^{*}\cup L^{*}$. By Lemma 2.6, the cut locus
$C(x_{i+1},P^{*})$ is precisely the path ${\rho}_{i,i+2}$. Therefore, by Lemma
2.2, $P^{*}$ is a doubly-covered convex polygon, so all angles at $x_{j}$ are
equal above in $L^{*}$ and below in $X^{*}$. In particular,
${\theta}^{X^{*}}_{i+1}={\theta}^{L^{*}}_{i+1}$. Now, because ${\Delta}$ angle
was removed from ${\theta}^{X}_{i+1}$,
${\theta}^{X^{*}}_{i+1}={\theta}^{X}_{i+1}-{\Delta}$. Because ${\alpha}_{i+1}$
was removed from ${\theta}^{L}_{i+1}$,
${\theta}^{L^{*}}_{i+1}={\theta}^{L}_{i+1}-{\alpha}_{i+1}$. Therefore,

$\displaystyle{\theta}^{L}_{i+1}-{\alpha}_{i+1}$ $\displaystyle=$
$\displaystyle{\theta}^{X}_{i+1}-{\Delta}\;\leq\;{\theta}^{X}_{i+1}$ (4.1)
$\displaystyle{\alpha}_{i+1}$ $\displaystyle\geq$
$\displaystyle{\theta}^{L}_{i+1}-{\theta}^{X}_{i+1}$ (4.2)

which is Claim (2).

It remains to show that $D_{i+1}$ is in $L_{i}$ rather than in $X_{i}$.
Suppose to the contrary that all the angle removal was in $X_{i}$. Then
${\theta}^{L}_{i+1}={\theta}^{X}_{i+1}-{\Delta}-{\alpha}_{i+1}$. So
${\theta}^{L}_{i+1}<{\theta}^{X}_{i+1}$, which is not possible for $P$ a
pyramid. This completes the proof of Claim (2) and the lemma. ∎

#####  4.3.2 General case

Lemma 4.3 is special in that $P$ sits over a base $X$. In the general
situation, $X$ is the intersection of $Q$ with the truncating slice plane
$\Pi$, but $X$ is not a face of $Q$. Rather in general, $X$ is inside $Q$, the
“top” of the portion $Q^{\prime}$ of $Q$ below $\Pi$.

######  Lemma 4.4.

Let $Q$ be obtained from $P$ by truncating vertex $v$. Then, if $v$ has
degree-$k$, $Q$ may be obtained from $P$ by $k-1$ tailoring steps, each the
excision of a digon surrounding one vertex.

###### Proof.

Here we argue that the general case is in some sense no different than the
special case of $P$ a pyramid just established in Lemma 4.3. In fact, the
exact same digon excisions suffice to tailor $P$ to $Q$.

First we establish additional notation. Let $\Pi$ be the plane slicing off $v$
above $\Pi$, and let $X=\Pi\cap P$. Let the “bottom” part of $P$ be
$Q^{\prime}$, with the final polyhedron $Q=Q^{\prime}\cup X$. We continue to
use $L$ to denote the portion of $P$ above $\Pi$, so $P=Q^{\prime}\cup L$.
After removal of digons at $x_{1},x_{2},\ldots,x_{i}$, we have
$P_{i}=Q^{\prime}_{i}\cup L_{i}$.

Below it will be important to distinguish between the three-dimensional
extrinsic shape of $Q^{\prime}_{i}$ and its intrinsic structure determined by
the gluings that satisfy Alexandrov’s Gluing Theorem. We will use
$\bar{Q}^{\prime}_{i}$ for the embedding in ${\mathbb{R}}^{3}$ and
$Q^{\prime}_{i}$ for the intrinsic surface, and we will similarly distinguish
between $\bar{X}_{i}$ and $X_{i}$. Note that we can no longer assume that
$C(x_{i+1},P_{i})\subset L_{i}$, for the cut locus could extend into
$Q^{\prime}$ (whereas it could not extend into $X$ in Lemma 4.3).

It suffices to show by induction that, on $P_{i}$, the following statements
hold:
(a) The shortest path ${\gamma}_{i+1}$ joining $x_{i+1}$ to $y_{i}$ is
included in $L_{i}$.
(b) The ramification point $a_{i+1}$ is still on $L_{i}$.
(c) The $x_{i+1}$ angle ${\alpha}_{i+1}$ of the digon
$D_{i+1}=D(x_{i+1},a_{i+1})$ is larger than or equal to
${\omega}_{Q}(x_{i+1})-{\omega}_{P}(x_{i+1})$ (and so sufficient to reduce the
curvature to ${\omega}_{Q}(x_{i+1})$).

To see (a), assume, on the contrary, that ${\gamma}_{i+1}$ intersects
$Q^{\prime}_{i}$. Assume, for the simplicity of the exposition, that
${\gamma}_{i+1}$ enters $Q^{\prime}_{i}$ only once, at $x_{i+1}$, and exits
$Q^{\prime}_{i}$ at $p\in\partial X$. Let ${\gamma}^{\prime}_{i+1}$ denote the
part of ${\gamma}_{i+1}$ between $x_{i+1}$ and $p$.

We now check (a) for $i=0$. $Q=Q^{\prime}\cup X$ and $X$ is planar, hence,
because the orthogonal projection of any rectifiable curve onto a plane
shortens or leaves its length the same, ${\gamma}^{\prime}_{1}$ is longer than
or has the same length as its projection ${\gamma}^{\prime\prime}_{1}$ onto
$X$. So $p$ is a cut point of $x_{1}$ along ${\gamma}_{1}$, contradicting the
extension of ${\gamma}_{1}$ as a geodesic segment beyond $p$.

By the induction assumption, all the digon excisions occur on $L_{i}$;
$Q^{\prime}_{i}$ is unchanged. Nevertheless, as part of $P_{i}$, neither
$\bar{Q}^{\prime}_{i}$ nor $\bar{X}_{i}$ is (in general) congruent to the
original $\bar{Q}$ and $\bar{X}$. However, if we consider $Q^{\prime}_{i}$ and
$X_{i}$ separate from $P_{i}$, we can reshape them so that
$\bar{Q}^{\prime}_{i}=\bar{Q}$ and $\bar{X}_{i}=\bar{X}$, precisely because
they have not changed. Then $\bar{X}$ is planar and the projection argument
used for $i=0$ works for all $i$.

Next we check (b) and (c) for $i=0$. Consider, as in Lemma 4.3, the digon
$D_{1}=(x_{1},y_{1})$ with $y_{1}\in C(x_{1})$, with again $a_{1}$ the first
ramification point of $C(x_{1})$ beyond $v$. The direction at $v$ of the edge
$va_{1}$ is only determined by the geodesic segment from $x_{1}$ to $v$, and
hence is not influenced at all by $Q^{\prime}$, because, by $i=0$ in (a), that
segment lies in $L$.

The ramification point $a_{1}$ is joined to $x_{1}$ by three geodesic
segments, two of them—say ${\gamma}_{1}$ and ${\gamma}_{2}$—included in $L$.
The third geodesic ${\gamma}_{3}$ starts from $x_{1}$ towards $Q^{\prime}$ and
finally enters $L$ to connect to $a_{1}$. See Fig. 4.6. Because these three
geodesics have the same length, the longer ${\gamma}_{3}$ is, the longer are
${\gamma}_{1}$ and ${\gamma}_{2}$, and therefore more distant is $a_{1}$ to
$v$. So $a_{1}$ is closest to $v$, and the segment $va_{1}$ shortest, when
$Q^{\prime}=X$ and $P$ is a pyramid. It is when $va_{1}$ is shortest that
there is the least “room” for $y_{1}$ on $va_{1}$ to achieve the needed digon
angle at $x_{1}$, for that angle is largest when $y_{1}$ approaches $a_{1}$.
Therefore, the case when $Q^{\prime}=X$ and $P$ is a pyramid is the worst
case, already settled in Lemma 4.3.

Figure 4.6: Three geodesics to ramification point $a_{1}$ on $C(x_{1})$.
Dashed ${\gamma}_{3}$ partially in $Q^{\prime}$. The same situation holds
under the changes: $x_{1}\to x_{i+1}$, $a_{1}\to a_{i+1}$, $v\to y_{i}$,
$Q^{\prime}\to Q^{\prime}_{i}$, $L\to L_{i}$.

Now we treat the general case for (b) and (c). Again by the induction
assumption, all changes to $P_{i}$ were made on its “upper part” $L_{i}$.

Because we ultimately need to reduce $L$ to $X$, the angle
${\omega}_{Q}(x_{i+1})-{\omega}_{P}(x_{i+1})$ necessary to be excised at
$x_{i+1}$, does not depend on $Q^{\prime}_{i}$, only on $L_{i}$. Thus the
argument used for $i=0$ carries through. The situation depicted in Fig. 4.6
remains the same, with $x_{1}$ replaced by $x_{i+1}$, $v$ replaced by $y_{i}$,
and $a_{1}$ replaced by $a_{i+1}$. The ramification point $a_{i+1}$ is closest
to $y_{i}$, and the segment $y_{i}a_{i+1}$ shortest, when $Q^{\prime}=X$ and
$P$ is a pyramid. It is when $y_{i}a_{i+1}$ is shortest that there is the
least “room” for $y_{i+1}$ on $y_{i}a_{i+1}$ to achieve the needed angle
excision at $x_{i+1}$, for that angle is largest when $y_{i+1}$ approaches
$a_{i+1}$. Therefore, the case when $Q^{\prime}=X$ and $P$ is a pyramid is the
worst case, already settled in Lemma 4.3. ∎

Note that, in the end, the digon removals in Lemma 4.2, and then in Lemma 4.3,
also work in the general case, Lemma 4.4.

####  4.4 Cube/Tetrahedron: Completion

#####  4.4.1 Pyramid Removals

We now return to the cube/tetrahedron example started in Chapter 3. We had
reduced the sliced-off portion of the original cube $P$ to four pyramids
$P_{1},P_{2},P_{3},P_{4}$. Now each of these pyramids needs to be “tailored
away” to leave the goal tetrahedron $Q$. Fig. 4.7 shows that when processed in
order, their removal reduces the portion of $P$ “above” $Q$, leaving the goal
tetrahedron $Q$.

Figure 4.7: Pyramid removals. The apex of the pyramid to be next removed is
highlighted. From (a) to (b), the pyramid $P_{1}$ with apex $p_{2}$ and base
$p_{1}p_{3}p_{6}$ is removed. Then $P_{2}$ apexed at $p_{6}$ is removed,
producing (c), then removing $P_{3}$ apexed at $p_{5}$ leads to (d). The final
removal of $P_{4}$ apexed at $p_{7}$, leaves the tetrahedron
$Q=p_{1}p_{3}p_{4}p_{8}$ previously illustrated in Fig. 3.2(a).

#####  4.4.2 Pyramid Reductions by Tailoring

Now we follow Lemma 4.4 to reduce each of the four pyramids to their bases. We
only illustrate this for the first pyramid removed, $P_{1}$, with apex $p_{2}$
and base $X=p_{1}p_{3}p_{6}$. See Fig. 4.8(a). The base is an equilateral
triangle, with edge lengths $\sqrt{2}$, with the apex is connected to the base
vertices by unit-length edges. Because the apex $p_{2}$ is a cube corner, it
has three incident $90^{\circ}$ angles, so ${\omega}(p_{2})=90^{\circ}$. The
three faces incident to $p_{2}$ are $45^{\circ}{-}45^{\circ}{-}90^{\circ}$
triangles. The base angles in $X$ are $60^{\circ}$. So each digon must reduce
the incident $90^{\circ}$ angles by $30^{\circ}$ to match $60^{\circ}$.
Starting with $p_{1}$, the digon geodesics are $\pm 15^{\circ}$ around the
$p_{1}p_{2}$ edge.

The geometry is clearest if we unfold $P_{1}$’s lateral faces into the plane,
as shown Fig. 4.8(b). Excising the first digon and sealing the cut results in
a new polyhedron, with the apex $p_{2}$ replaced by a new vertex, call it $y$,
of curvature ${\omega}(y)=60^{\circ}$ (because the two digon angles must sum
to the $90^{\circ}$ curvature at $p_{2}$). Unfortunately, we cannot display
this new polyhedron because of the difficulty of constructing what AGT
guarantees exists.

Lemma 4.4 says that just one more digon needs removal (because $p_{2}$ has
degree $k=3$), again $\pm 15^{\circ}$ this time around the $p_{3}y$ edge. This
removal reduces $P_{1}$ to its equilateral triangle base, and, despite the
nonconstructive nature of AGT, we know that the full polyhedron is exactly
what we illustrated earlier in Fig. 4.7(b).

Figure 4.8: (a) Pyramid $P_{1}$: apex $p_{2}$, base $p_{1}p_{3}p_{6}$. (b)
Excision of two digons. Each is bound by geodesics $\pm 15^{\circ}$ about the
edge to $p_{2}$. After removing $p_{2}$, a new vertex $y$ is created. The
removal of $y$ in the second excision flattens the pyramid to its base.

One further remark on the shape of $P_{1}$ after excision of the first digon.
If we imagine $P_{1}$ standing alone on its base $X$ as illustrated in Fig.
4.8(a), rather than as part of the full cube polyhedron, then it is not
difficult to reconstruct the shape of $P^{\prime}_{1}$. It is a flat doubly-
covered quadrilateral, with the triangle $p_{3}yp_{6}$ flipped over and joined
to the $p_{3}p_{6}$ edge of the equilateral triangle base $X$. However, with
this $P^{\prime}_{1}$ piece joined to the full cube, it seems much less
straightforward to determine the shape of the full polyhedron.

Each pyramid reduction proceeds in the same manner: $k-1$ digons are excised
if the apex has degree-$k$, and the lateral faces are reduced to the base. So
$P_{2}$ has apex $p_{6}$ and base $p_{1}p_{3}p_{5}p_{7}$. After removal of
three digons, the result is as illustrated in Fig. 4.7(c). $P_{3}$’s apex
$p_{5}$ has degree-$3$, so two digon removals lead to (d). The last pyramid
removed, $P_{4}$, reduces to the face $p_{1}p_{3}p_{8}$ of $Q$, completing the
tailoring of the cube $P$ to tetrahedron $Q$.

#####  4.4.3 Seals

After removal of the two digons illustrated in Fig. 4.8(b), $P_{1}$ has been
reduced to its equilateral triangle base $X$. Sealing the first digon produces
a seal ${\sigma}_{1}$, which is then clipped to a segment $s_{1}$ by the
second digon removal, which produces ${\sigma}_{2}$ along the boundary of $X$.
The seal segments then are as shown in Fig. 4.9(a).

Figure 4.9: (a) Base of $P_{1}$. (b) Base of $P_{2}$. Seal segments: red.

Fig. 4.9(b) shows the three seal segments that result by reducing $P_{2}$ to
its base. These depictions of the seal graph ${\Sigma}$ will be explored in
some detail in Chapter 5.

There is an aspect of the seals we are not tracking: As can be seen in Fig.
4.7(b,c), one of the faces of $P_{2}$ is the equilateral triangle base of
$P_{1}$. Since that base is already crossed by seal segment $s_{1}$ when
$P_{2}$ is undergoing digon removal, that segment $s_{1}$ will be reflected as
a cut in the reduced base of $P_{2}$, not depicted in Fig. 4.9(b). We have not
attempted to track this complex overlaying of seal cuts in the surface of $P$.
However, we explore seals for one pyramid tailoring in some detail in Chapter
5.

####  4.5 Hexagonal Pyramid Example

We now detail a more complex example following Lemma 4.3 to tailor a pyramid
$P$ to its base $X$. We continue to employ the notation used in the lemmas
above. The example is shown in Fig. 4.10. $X$ is a regular hexagon, and $L$
consists of $k=6$ congruent, $70^{\circ}{-}70^{\circ}{-}40^{\circ}$ isosceles
triangles. The curvature at the apex $v$ is $360^{\circ}-6\cdot
40=120^{\circ}$. The angle at each $x_{i}$ in $X$ is $120^{\circ}$ whereas the
angle in $L$ is $140^{\circ}$. So each digon excision must remove $20^{\circ}$
from $x_{i}$. As in the lemmas, we excise the digons in circular order around
$\partial X$.

Figure 4.10: Pyramid with regular hexagon base $X$; all lateral faces
congruent triangles.

We display the progress of the excisions on the layout of $L$ in Fig. 4.11(a).
Let $v=y_{0}$ for ease of notation. $D_{1}=(x_{1},y_{1})$ includes the
geodesic ${\gamma}_{1}$ from $x_{1}$ to $y_{0}$, and locates $y_{1}$ on the
cut locus segment as described in the lemmas. The digon boundary geodesics
each remove $10^{\circ}$ from the left and right neighborhood of $x_{1}$, and
meet at $y_{1}$ at an angle of $100^{\circ}$, which is then the curvature at
the new vertex: ${\omega}(y_{1})=100^{\circ}$. Notice that the digon angles
$20^{\circ}+100^{\circ}$ match the curvature ${\omega}(v)=120^{\circ}$
removed, as they must to satisfy Gauss-Bonnet.

Figure 4.11: (a) Cone $L$ flattened; ${\omega}(v)=120^{\circ}$. Digons
$D_{i}=(x_{i},y_{i-1})$ shaded. Dashed lines are geodesics ${\gamma}_{i}$ from
$x_{i}$ to the vertex $y_{i-1}$ (with $y_{0}=v$). (b) After excising all
digons $D_{1},\ldots,D_{5}$, $L_{5}$ is isometric to $X$. Seals
$\tilde{{\sigma}}_{i}$ are marked.

One should imagine that $D_{1}$ is sutured closed in Fig. 4.11(a), producing
$L_{1}$, before constructing $D_{2}(x_{2},y_{2})$. Let ${\sigma}_{i}$ be the
geodesic on $L_{i}$ that results from sealing $D_{i}$ closed; ${\sigma}_{i}$
is like a “scar” from the excision. Notice that one of the geodesics bounding
$D_{2}$ crosses ${\sigma}_{1}$.

This pattern continues as all $k{-}1=5$ digons are removed, each time
replacing vertex $y_{i-1}$ with $y_{i}$, flattening the curvature
${\omega}(y_{i})={\omega}(y_{i-1})-20^{\circ}$. Finally, after
$D_{5}(x_{5},y_{5})$ is removed, $y_{5}$ is coincident with $x_{6}$. No
further digon removal is needed, because $D_{5}$ removed $20^{\circ}$ from
$x_{6}$. So now each angle in $L_{k}=L_{5}$ at all $k{=}6$ vertices is
$120^{\circ}$, and $L_{5}$ is isometric to a flat regular hexagon, i.e., to
$X$.

This final hexagon is shown in Fig. 4.11(b). The images $\tilde{{\sigma}}_{i}$
of the seals are in general clipped versions of ${\sigma}_{i}$ on $L_{i}$,
clipped by subsequent digon removals. The particular circular order of digon
removal followed in this example and the lemmas result in a spiral pattern
formed by $\tilde{{\sigma}}_{i}$. Other excision orderings, which ultimately
would result in the same flat $L_{k-1}$ (effectively proved in Lemmas 4.2–4.4)
would create different seal patterns. As mentioned earlier, we study seals in
detail in Chapter 5.

####  4.6 Tailoring is finer than sculpting

In this section we reach one of our main results, Theorem 4.6, which says,
roughly, that any polyhedron $Q$ that can be obtained by sculpting $P$ can be
obtained by tailoring $P$. Moreover, Lemma 4.5 shows that polyhedra can be
obtained by tailoring that cannot be obtained by sculpting. So, in a sense,
tailoring is finer than sculpting.

######  Lemma 4.5.

There are shapes $P$ and sequences of tailorings of $P$ that result in
polyhedra not achievable by sculpting.

###### Proof.

We first tailor a regular tetrahedron $T$ as in Example 1.1, resulting in the
kite $K$ in Fig. 1.2(b). We now show that $K$ cannot fit inside $T$, so it
couldn’t have been sculpted from $T$. Assume $T$ has edge-length $1$. Then its
extrinsic diameter is $1$ and its intrinsic diameter is $2/\sqrt{3}$ (see,
e.g., Theorem 3.1 in [Rou03]). Moreover, the extrinsic diameter of $K$ is
precisely the intrinsic diameter of $T$, and so it cannot fit inside $T$.

Next we construct a non-degenerate example, a modification of the previous
one. Consider a non-degenerate pentahedron $F$ close enough to $K=oac_{d}b$ in
Fig. 1.2(b). For example, it could have two vertices close to the vertex $a$
of $K$. Insert into $F$ the removed digon from $T$; this is not affected by
the new vertex, because it does not interfere with the geodesic segment from
$c_{d}$ to $o$. We arrive at some surface $P$ close enough to the original
tetrahedron $T$. Therefore, the intrinsic and extrinsic diameters of $P$ and
$F$ are close enough to those of $K$ and $T$, respectively, and the above
inequality between the extrinsic diameters of $P$ and $F$ still holds, because
of the “close enough” assumption. ∎

######  Theorem 4.6.

Let $P$ be a convex polyhedron, and $Q\subset P$ a convex polyhedron resulting
from repeated slicing of $P$ with planes. Then $Q$ can also be obtained from
$P$ by tailoring. Consequently, for any given convex polyhedra $P$ and $Q$,
one can tailor $P$ “via sculpting” to obtain any homothetic copy of $Q$ inside
$P$.

###### Proof.

Lemma 4.1 established that one slice leads to domes, Theorem 3.2 showed that
each dome leads to pyramids, and Lemma 4.4 showed that each pyramid can be
reduced to its base by tailoring. Since this holds for one slice, it
immediately follows that it holds for arbitrary slicing.

Concerning the domes $\to$ pyramids step, we note that the property that each
pyramid $P_{i}$ has a common edge with $X$, established in Theorem 3.2, allows
reduction of the pyramids in the order that they are obtained in that theorem.
After each reduction, the result is still a g-dome, allowing iteration until
the original g-dome is reduced to its base.

For the homothet-copy claim of the lemma, shrink $Q$ by a dilation until it
fits inside $P$, and then apply the reductions. ∎

As we mentioned in the Preface, an informal consequence of this theorem is
that $P$ can be “whittled” to e.g., a sphere $S$:

######  Corollary 4.7.

For any convex polyhedron $P$ and any convex surface $S$, one can tailor $P$
to approximate a homothetic copy of $S$.

###### Proof.

Bring a homothetic copy of $S$ inside $P$. Perform a series of slicings of $P$
with planes tangent to $S$. Any degree of approximation desired can be
achieved by increasing the number of plane splicings. Call the result of these
slicings $Q$. Now apply Theorem 4.6. ∎

Despite this corollary, it does not seem possible to accomplish the reverse:
to start with a strictly convex surface and tailor it to a polyhedron.
However, one can of course sculpt a surface to a polyhedron.

###  Chapter 5 Pyramid Seal Graph

####  5.1 Pyramid Digon Removal

As we have seen in Theorem 4.6, tailoring by tracking sculpting ultimately
relies on digon removal reducing pyramids to their bases. We have illustrated
such reductions for a few low-degree pyramids in Figs. 4.2, 4.8, and 4.11. In
the latter two figures, we displayed the seals $\tilde{{\sigma}_{i}}$ on the
base $X$ after reduction. It however remains difficult to grasp in detail the
digon-removal process for a pyramid $P$, for at least three reasons:

  1. 1.

After removing the first digon $D_{1}$, $P_{1}$ is (in general) no longer a
pyramid. The difficulty of computationally realizing the subsequent
intermediate shapes $P_{i}$, guaranteed by AGT, makes it hard to envision the
process.

  2. 2.

The seals ${\sigma}_{i}$ that result from closing digon $D_{i}$ cross and clip
one another.

  3. 3.

The process depends on the order in which the digons are removed.

We will continue to circumvent this last difficulty by only studying the
natural order of digon removal, anchored at $x_{1},x_{2},x_{3},\ldots$ in
counterclockwise order around $\partial X$. In this section, we introduce a
different way to view digon removal that in some sense skirts the first two
difficulties.

The process is complex enough to require somewhat extensive notation, which we
list in two parts before turning to examples.

#####  5.1.1 Notation I

  * •

$P$: a pyramid, $n$ vertices around base $X$.

  * •

Base vertices $x_{1},x_{2},\ldots,x_{n}$, in counterclockwise order around
$\partial{X}$.

  * •

Apex $y_{0}$ of degree-$n$.

  * •

$y_{i}$: apex after removing digon $D_{i}$.

  * •

$D_{i}$: digon from $x_{i}$ to $y_{i}$, surrounding $y_{i-1}$.

  * •

$L_{i}$: (the remaining of the) lateral faces after removing digon $D_{i}$.
$L_{0}$: initial faces before any removals. The apex of $L_{i}$ is $y_{i}$.

  * •

$P_{i}$: The polyhedron $L_{i}\cup X$, guaranteed by AGT. $P=P_{0}$ is the
original, before any digon removal.

We should emphasize that although $P_{i}=L_{i}\cup X$, in general $X$ will not
be planar in $P_{i}$ as it is in $P_{0}$, and so $P_{i}$ is not a pyramid, as
previously mentioned.

####  5.2 Cone Viewpoint

Although we do not know the structure of $P_{i}$, except at the beginning
($i=0$) and end ($i=n{-}1$), when it is $P$ and doubly-covered $X$
respectively, we do know that the lateral faces $L_{i}$ contain only one
vertex, $y_{i}$, hence they form a subset of a cone apexed at $y_{i}$. Any
cone can be cut open along a generator (a ray on the cone from the apex) and
laid flat in the plane. Such a layout will have an angle gap of
${\omega}(y_{i})$ at the apex. It is especially useful to cut along
$x_{i}y_{i-1}$ before removing digon $D_{i}$. We will provide several
examples, after presenting more notation. We emphasize the indices $i{-1}$,
$i$, and $i{+}1$ in the following, in an attempt to avoid confusion.

#####  5.2.1 Notation II

  * •

$\bar{L}_{i-1}$: Unfolding of $L_{i-1}$ cut open along $x_{i}y_{i-1}$. So
after removing and sealing digon $D_{i-1}$, but not yet $D_{i}$.

  * •

$\bar{L}_{i}$: Unfolding of $L_{i}$ cut open along $x_{i+1}y_{i}$. So after
removing and sealing digon $D_{i}$, but not yet $D_{i+1}$. So $\bar{L}_{0}$ is
$L_{0}$ cut open along $x_{1}y_{0}$, and $\bar{L}_{1}$ is $L_{1}$ cut open
along $x_{2}y_{1}$.

  * •

${\sigma}_{i}=x_{i}y_{i}$ is the $i$-th seal after suturing closed the digon
$D_{i}$. We view the seals as directed from $x_{i}$ to $y_{i}$, so that they
have distinguished left and right sides. This direction is only used in the
proofs; the seals are illustrated as undirected segments in several figures.
When the direction plays a role, we use boldface: $\bm{{\sigma}}_{i}$.

  * •

${\Sigma}_{i}$: the seal graph after removing digon $D_{i}$.
${\Sigma}_{0}=\varnothing$, and ${\Sigma}_{1}=x_{1}y_{1}$.

  * •

$s_{j}\subseteq{\sigma}_{j}$, $1\leq j\leq i$, is the possibly truncated seal
segment in ${\Sigma}_{i}$, on the surface $P_{j}$. So, after possibly other
truncations, we reach $\tilde{{\sigma}_{j}}$, hence the informal inclusion
$\tilde{{\sigma}_{i}}\subseteq s_{j}\subseteq{\sigma}_{j}$; “informal” because
those geodesic segments live in different spaces.

  * •

$S_{i}$ is the subset of $L_{i}$ bounded by $x_{1}y_{i}$ and $x_{i}y_{i}$, the
_sealed region_ which we will later prove contains ${\Sigma}_{i}$.

####  5.3 Examples

We start with $P_{1}$, previously displayed in Fig. 4.8. $X$ is an equilateral
triangle, with the apex centered above its centroid. Fig. 5.1 shows the
removal of $n{-}1=2$ digons $D_{1},D_{2}$ that reduce $L_{0}$ to the
equilateral triangle base $X$. Images are repeated so that in one row the
transition from $L_{i-1}$ to $L_{i}$ by removal of $D_{i}$ is evident.

Figure 5.1: $P_{1}$. $X$ is an equilateral triangle. Cf. Fig. 4.9(a).

Next is the more complicated $P_{2}$ in Fig. 5.2. Here $X$ is a rectangle, and
three digons are removed, $D_{1},D_{2},D_{3}$, before reaching $X$.

Figure 5.2: $P_{2}$. $X$ is a rectangle. Cf. Fig. 4.9(b).

We should emphasize that any one of these figures could be cut out and closed
to a cone. This cone would not be rigid, and its boundary $\partial X$ would
not (in general) be planar, as we mentioned earlier.

As a last example, we extract one row illustrating removal of $D_{5}$ from a
pyramid of degree-$12$ in Fig. 5.3. We will refer to this figure subsequently.

Figure 5.3: Pyramid of $n{=}12$ base vertices $x_{i}$. (a) $\bar{L}_{4}$.
$S_{4}$ (blue). (b) Digon $D_{5}$ (yellow). (c) $\bar{L}_{5}$. ${\Sigma}_{i}$
in red.

####  5.4 Preliminary Lemmas

The viewpoint just described is simple enough to be implemented, and to allow
us to construct the seal graph ${\Sigma}$ for any pyramid $P$. See ahead to
Fig. 5.8 for examples. This leads to Theorem 5.3: ${\Sigma}$ is a tree. The
proof of this claim is somewhat intricate, and presented in Section 5.5. That
proof requires two lemmas, both involving the structure of the cut locus,
which we present first. The reader might skip these proofs until later.

######  Lemma 5.1.

After removing digons $D_{1},\ldots,D_{i}$ and closing seals
${\sigma}_{1},\ldots,{\sigma}_{i}$, $C(x_{i+1},P_{i})$ includes the path
$x_{1},\ldots,x_{i}$, with each node in that path of degree-$2$.

For example, in Fig. 5.3, $i=4$ and $C(x_{5},P_{4})$ includes
$x_{1},x_{2},x_{3},x_{4}$.

###### Proof.

We start with the leaf $x_{i}\in C(x_{i+1})$, and argue that
$(x_{i},x_{i-1},\ldots,x_{1})$ is a path ${\rho}$ in $C(x_{i+1})$, i.e., that
every point along ${\rho}$ is of degree $\leq 2$. The proof uses techniques
detailed in the proof of Lemma 4.3. In particular, Fig. 5.4 below depicts the
situation abstractly, similar to Fig. 4.4 in Lemma 4.3.

First, $x_{i}$ is of degree-$1$ in $C(x_{i+1})$: The pyramid edge
$x_{i+1}x_{i}$ is the shortest geodesic, unaffected by the digon removals up
to $D_{i}$. An edge $e_{i}$ of $C(x_{i+1})$ starts at $x_{i}$, and because of
the equal angles above on $L_{i}$ and below on $X$ at $x_{i}$, and because
$e_{i}$ is bisecting, initially it starts along the geodesic $x_{i}x_{i-1}$.
It then either continues to $x_{i-1}$, or reaches a ramification point.

Suppose the path continues $x_{i},x_{i-1},\ldots,x_{j}$, but then reaches a
ramification point $r$ on $x_{j}x_{j-1}$. Let ${\rho}_{j}$ denote this path up
to $r$. We now analyze this situation and show it is contradictory. Consult
Fig. 5.4 throughout.

Figure 5.4: Abstract depiction of $x_{i},x_{i-1},\ldots,x_{1}$ path in
$C(x_{i+1})$ (red). $X$ is flipped below; $L_{i}$ above.

The possible geodesics from $x_{i+1}$ to $r$ are: ${\gamma}_{X}$ on $X$ below,
${\gamma}_{1}$ and ${\gamma}_{2}$ on $L_{i}$ above, and possibly
${\gamma}_{3}$ both above and below. Note that there can only be the two
${\gamma}_{1}$ and ${\gamma}_{2}$ because there is just one vertex $y_{i}$ on
$L_{i}$.

First consider ${\gamma}_{1}$, which, together with ${\gamma}_{X}$, encloses
${\rho}_{j}$. The planar convex chain along ${\rho}_{j}$,
$x_{i},x_{i-1},\ldots,x_{j},r$ is congruent above and below, because the
angles above and below are equal after digon removals. Thus the chords
connecting the endpoints of the chains are equal, and so
$|{\gamma}_{1}|=|{\gamma}_{X}|$.

Next consider ${\gamma}_{2}$, which, together with ${\gamma}_{X}$, encloses
$x_{i+2},\ldots,x_{n},x_{1},\ldots,x_{j-1}$. We again compare the planar
convex chain with angles below on $X$ to the chain with angles above on
$L_{i}$. Because some of the angles above are strictly larger than their
counterparts below, we can apply Cauchy’s Arm Lemma just as we did in the
proof of Lemma 4.3(Claim (1)) to conclude that
$|{\gamma}_{2}|>|{\gamma}_{X}|$. Therefore ${\gamma}_{2}$ cannot add to the
degree of $r$ in $C(x_{i+1})$.

Finally, a geodesic ${\gamma}_{3}$ that lies on both $L_{i}$ and $X$ must have
a portion completely above on $L_{i}$, to which we may apply the same arm-
lemma argument to conclude that $|{\gamma}_{3}|>|{\gamma}_{X}|$.

Therefore, $r$ is in fact of degree $2$, it is not a ramification point, and
${\rho}$ extends from $x_{i}$ to $x_{1}$ as claimed. ∎

######  Lemma 5.2.

Assume the digons $D_{1},\ldots,D_{i}$ have been removed and $S_{i-1}$ is the
sealed region containing ${\sigma}_{1},\ldots,{\sigma}_{i-1}$ (this inclusion
will be proven later). Let $a_{i}$ be the first ramification point of
$C(x_{i},P_{i-1})$, on the segment $y_{i-1}a_{i}$. Then $a_{i}\not\in
S_{i-1}$.

First we illustrate the claim of the lemma with the example from Fig. 5.3(b),
repeated as Fig. 5.5(a). In order for the lemma to be false, the situation
instead must appear as in (b) of the figure, with $a_{5}\in S_{4}$.

###### Proof.

For the purposes of contradiction, consider the situation depicted abstractly
in Fig. 5.5(c), with $a_{i}\in S_{i-1}$. By Lemma 5.1, $x_{1},\ldots,x_{i-1}$
is a path of degree-$2$ nodes in $C(x_{i},P_{i-1})=C(x_{i})$. The edge
$y_{i-1}a_{i}$ of $C(x_{i})$ bisects the angle formed by the two images of
$x_{i}$. From $a_{i}$, $C(x_{i})$ must contain a path ${\rho}$ that connects
to $x_{i}$. Because $a_{i}\in S_{i-1}$, ${\rho}$ must start with an edge to
the right of the line containing $y_{i-1}a_{i}$.

Note that $y_{i-1}$ is the only vertex on $L_{i-1}$, so $a_{i}$ is not a
vertex, and therefore has a total surface angle ${\theta}_{a_{i}}=2\pi$. Lemma
2.3 requires a strictly leftward branch at $a_{i}$, left of the line
containing $y_{i-1}a_{i}$. Call the path continuing this left branch
$\lambda$. This path $\lambda$ is “trapped”: It cannot terminate in the
interior of $S_{i-1}$ because there are no vertices in that region. It cannot
connect to any one of $x_{1},\ldots,x_{i-1}$ because those nodes are
degree-$2$. If $\lambda$ crossed ${\rho}$ it would form a cycle. Therefore we
have reached a contradiction, and so $a_{i}\not\in S_{i-1}$. ∎

Figure 5.5: (a) Fig. 5.3(b). (b) How a counterexample might appear. (c)
Abstract counterexample. Portions of $C(x_{i})$ purple.

####  5.5 Pyramid Seal Graph is a Tree

Roughly, we prove in this section that the complete seal graph
${\Sigma}={\Sigma}_{n}$, and intermediate seal graphs ${\Sigma}_{i}$, have the
structure of a spiraling tree. Spiral trees—slit trees rather than seal
trees—will play a significant role in Part II.

We repeat some notation here for convenience. The _sealed region_ $S_{i}$ is
bounded by the segments $x_{1}y_{i}$, $x_{i}y_{i}$, and the portion of
$\partial X$ from $x_{1}$ to $x_{i}$. Also recall that each seal
$\bm{{\sigma}}_{j}=(x_{j},y_{j})$ is directed from $x_{j}$ to $y_{j}$. The
seal graph ${\Sigma}_{i}$ is composed of segments $s_{j}$, each a subsegment
of ${\sigma}_{j}$.

######  Theorem 5.3.

The seal graph ${\Sigma}_{i}$, after removal of digons $D_{1},\ldots,D_{i}$ in
counterclockwise order, has the following properties:

  1. (1)

${\Sigma}_{i}\subset S_{i}$.

  2. (2)

${\Sigma}_{i}$ is a directed tree with root $y_{i}$.

  3. (3)

Each segment $s_{j}$ of ${\Sigma}_{i}$ is a (possibly truncated) seal
${\sigma}_{j}=x_{j}y_{j}$ that remains anchored on its $x_{j}$ endpoint; i.e.,
the truncation is on the $y_{i}$-end.

  4. (4)

Each leaf $x_{j}$ is the start of a directed, convex path $\bm{\pi}_{j}$ to
the root $y_{i}$.

  5. (5)

The edges of $\bm{\pi}_{j}$ are portions of seal segments of increasing
indices.

  6. (6)

Along $\bm{\pi}_{j}$, lower-indexed edges terminate from the left on higher-
indexed edges.

  7. (7)

The last seal, $\bm{{\sigma}}_{i}=x_{i}y_{i}$, the _root segment_ of
${\Sigma}_{i}$, has no segments of ${\Sigma}_{i}$ incident to its right side.

The last segment of the complete seal graph ${\Sigma}={\Sigma}_{n}$ coincides
with the edge $x_{n-1}x_{n}$ of $\partial X$.

We again refer to Fig. 5.3(ab). When $i=4$, ${\Sigma}_{4}$ satisfies the
properties, and we seek to re-establish the properties for ${\Sigma}_{5}$ in
(c) of the figure. We enlarge (b) of that figure in Fig. 5.6 to track in this
proof. It may also help to consult the complete seal graphs in Fig. 5.8.

Figure 5.6: $i=4$, before removal of digon $D_{i+1}=D_{5}$. Detail from Fig.
5.3(b).

###### Proof.

Induction Basis. These claims are trivially true for $i=0$ because
${\Sigma}_{0}=\varnothing$. So assume $i=1$. $L_{1}$ has just had the digon
$D_{1}=x_{1}y_{1}$ excised around $y_{0}$. $\bar{L}_{1}$ is then cut open
along $x_{2}y_{1}$. ${\Sigma}_{1}$ is the single segment
${\sigma}_{1}=x_{1}y_{1}$, and all properties are easily verified.

Induction Hypothesis. Assume that all properties hold for ${\Sigma}_{i}$. The
removal of digon $D_{i+1}=x_{i+1}y_{i+1}$ leads to ${\Sigma}_{i+1}$. Now we
establish the properties for ${\Sigma}_{i+1}$.

  1. (1)

${\Sigma}_{i+1}\subset S_{i+1}$. $S_{i+1}$ grows on both sides: by the
triangle $x_{1}y_{i}y_{i+1}$ (clockwise in Fig. 5.6), and to $x_{i+1}y_{i+1}$
(counterclockwise in the figure). By Lemma 5.2, we know that $a_{i+1}\not\in
S_{i}$, and because $y_{i+1}$ is on the $y_{i}a_{i+1}$ segment of
$C(x_{i+1})$, we know that $y_{i+1}\not\in S_{i}$. Therefore the triangle
$x_{1}y_{i}y_{i+1}$ does in fact grow $S_{i}$ on the $x_{1}$-end. On the
$x_{i+1}$-end, the new seal $\bm{{\sigma}}_{i+1}=x_{i+1}y_{i+1}$ is
incorporated, and all seal segments right of $\bm{{\sigma}}_{i+1}$ are clipped
by the removal of digon $D_{i+1}$. Therefore indeed $S_{i+1}$ expands to
include all of ${\Sigma}_{i+1}$.

  2. (2)

${\Sigma}_{i+1}$ is a directed tree with root $y_{i+1}$. We know that
${\Sigma}_{i}$ is a directed tree with root $y_{i}$. We first argue that the
seal $\bm{{\sigma}}_{i+1}$ intersects the segments of ${\Sigma}_{i}$ from
right-to-left, re-establishing property (7). Let ${\gamma}_{1}$ and
${\gamma}_{2}$ be the left and right geodesics of digon
$D_{i+1}=x_{i+1}y_{i+1}$. ${\gamma}_{1}$ starts within the triangle
$\triangle=x_{i}x_{i+1}y_{i}$, at an angle $({\alpha}_{i+1}+{\beta}_{i+1})/2$
left of the $x_{i+1}y_{i}$ edge of $\triangle$. Therefore ${\gamma}_{1}$ cuts
into $S_{i}$ through the $x_{i}y_{i}$ edge of $\triangle$. The segments of
${\Sigma}_{i}$ crossed and clipped by ${\gamma}_{1}$ are crossed from right-
to-left. And since $i+1$ is the highest indexed seal, the segments of
${\Sigma}_{i}$ that meet $\bm{{\sigma}}_{i+1}$ satisfy property (7): lower-
indexed segments terminate on the left of $\bm{{\sigma}}_{i+1}$.111We should
mention that this property, that $\bm{{\sigma}}_{i+1}$ crosses segments right-
to-left, is dependent on the counterclockwise ordering of digon removal. If
after removing digon $D_{i}$, we next removed some $D_{j}$ with $j>i+1$, it
could be that it is ${\gamma}_{2}$ rather than ${\gamma}_{1}$ that clips
${\Sigma}_{i}$. This would result in seal graphs with a different structure.
Henceforth, we use ${\sigma}_{i+1}$ as determined by ${\gamma}_{1}$.

By property (8), $\bm{{\sigma}}_{i}=x_{i}y_{i}$ is the root segment of
${\Sigma}_{i}$, which we now know is crossed by $\bm{{\sigma}}_{i+1}$ right-
to-left, say crossing at point $p$. Because $y_{i+1}\not\in S_{i}$ by Lemma
5.2, $\bm{{\sigma}}_{i+1}$ has the root $y_{i}$ of ${\Sigma}_{i}$ to its
right, and all the leaves $x_{j}$ of ${\Sigma}_{i}$ to its left.

Now suppose ${\Sigma}_{i+1}$ has an undirected cycle $\Phi$; see Fig. 5.7.
Because ${\Sigma}_{i}$ has no cycle, any cycle in ${\Sigma}_{i+1}$ must have
one edge a subsegment of ${\sigma}_{i+1}$. Because all of ${\Sigma}_{i}$ right
of $\bm{{\sigma}}_{i+1}$ is removed, the cycle $\Phi$ must “rest on” the left
side of $\bm{{\sigma}}_{i+1}$. It is clear $\Phi$ cannot rest on the
$x_{i+1}p$ portion of $\bm{{\sigma}}_{i+1}$. So $\Phi$ must rest on the
portion of $\bm{{\sigma}}_{i+1}$ left of $\bm{{\sigma}}_{i}$, as illustrated.
But then, imagining removing $\bm{{\sigma}}_{i+1}$, $\Phi$ must have formed a
cycle $\Phi^{\prime}\supset\Phi$ in ${\Sigma}_{i}$, a contradiction. Therefore
${\Sigma}_{i+1}$ is indeed a tree.

The remaining properties are now easily established.

  3. (3)

Because $\bm{{\sigma}}_{i+1}$ clips segments of ${\Sigma}_{i}$ to its right,
each segment $s_{j}\in{\Sigma}_{i}$ remains anchored on $x_{j}$. And the new
segment $\bm{{\sigma}}_{i+1}$ is anchored on $x_{i+1}$.

  4. (4)

The directed, convex path $\bm{\pi}_{j}$ remains, but now may by shortened
where it joins with $\bm{{\sigma}}_{i+1}$.

  5. (5)

The segments along $\bm{\pi}_{j}$ have increasing indices, possibly now
including $i+1$.

  6. (6)

We earlier established that lower-indexed segments of ${\Sigma}_{i+1}$
terminate from the left on higher-indexed segments, possibly now including
$\bm{{\sigma}}_{i+1}$.

  7. (7)

The last and new seal $\bm{{\sigma}}_{i+1}=x_{i+1}y_{i+1}$ becomes the root
segment incident to the root $y_{i+1}$, and has no segments incident to its
right side.

Finally, it is a consequence of Lemma 4.2 and the rigidity Theorem 2.9 that
$y_{n-1}=x_{n}$ so that the $(n{-}1)$-st seal coincides with the edge
$x_{n-1}x_{n}$. ∎

Figure 5.7: Cycle $\Phi$ is not possible: removing ${\sigma}_{i+1}$ extends
$\Phi$ to $\Phi^{\prime}$. Figure 5.8: Four complete seal graphs.

#####  5.5.1 Other Digon Orderings

The proof of Theorem 5.3 depends on removing the digons in the order
$x_{1},x_{2},\ldots,x_{n-1}$ around $\partial X$. This affects Lemma 5.1’s
conclusion that $x_{1},\ldots,x_{i}$ is a path in the cut locus, which then
affects Lemma 5.2’s conclusion that the ramification point $a_{i+1}$ is
outside the sealed region $S_{i}$. In addition, which side of the removal of
digon $D_{i+1}$ clips ${\Sigma}_{i}$ is affected by knowing that $x_{i+1}$ is
adjacent to $x_{i}$ on $\partial X$. All of these consideration affect the
structure of the seal graph. We leave the question of whether the seal graph
is a tree for other orderings of digon removals to Open Problem 18.3.

Toward this open problem, we only show here, with the next result, what a
degree-$4$ vertex in ${\Sigma}$ must look like.

In the proof below, we simplify the notation of $\tilde{{\sigma}_{i}}$ on the
base to just ${\sigma}_{i}$, keeping in mind that ${\sigma}_{i}$ and
${\sigma}_{j}$ formally live in different spaces,.

######  Lemma 5.4.

If a seal graph ${\Sigma}$ for a pyramid $L$ has a degree-$4$ vertex $z$, then
there exist $i<j<k$ such that ${\sigma}_{i}$ and ${\sigma}_{j}$ end at $z$,
and ${\sigma}_{k}$ passes beyond $z$. Moreover, the digon excision order is
$D_{k}$ immediately after $D_{j}$ immediately after $D_{i}$.

###### Proof.

Consider a common point $z$ of ${\sigma}_{i}$ and ${\sigma}_{j}$, with $i<j$.
We may assume $j>i+1$ , since otherwise $\deg z=3$ in ${\Sigma}$.

Assume first that no other ${\sigma}_{k}$ passes through $z$. Assume that
$\deg z=4$ in ${\Sigma}$. This implies that the digon $D_{j}$ crosses
${\sigma}_{i}$ and, since ${\sigma}_{i}$ remains a geodesic after the excision
of $D_{j}$, ${\sigma}_{i}$ must be orthogonal to both geodesics bounding
$D_{j}$. Therefore, ${\sigma}_{i}$ creates with those two geodesics two
geodesic triangles, both of positive curvature. So each such triangle contains
a vertex inside, contradicting that $D_{j}$ itself contains only one vertex.

Assume now that ${\sigma}_{i}$ ends at $z$, as does ${\sigma}_{k}$ for some
$k\not\in\\{i,j\\}$. Notice that we cannot have four ${\sigma}$s ending at
$z$, because for the last one arriving—say ${\sigma}_{k}$—${\sigma}_{k}$ would
create a vertex at $z$, which will be excised by the digon $D_{k+1}$, breaking
that degree-$4$ configuration at $z$.

So we may assume that $z$ belongs only to ${\sigma}_{i}$, ${\sigma}_{j}$, and
${\sigma}_{k}$. We may further assume, without loss of generality, that
$i<j<k$.

Notice that both ${\sigma}_{i}$ and ${\sigma}_{j}$ end at $z$ and
${\sigma}_{k}$ does not, because otherwise ${\sigma}_{k}$ would create a
vertex on ${\sigma}_{i}$ (or on ${\sigma}_{j}$) which would be excised by the
digon $D_{k+1}$, breaking that degree-$4$ configuration at $z$.

The digon excision order follows: only $D_{j}$ could surround $y_{i}$, so its
excision was just after $D_{i}$, and similarly for $D_{j}$. ∎

For a particular digon-removal ordering, consider the inverse image of
${\Sigma}$ on $L$, and denote it by $\chi$. $\chi$ is a simple geodesic
polygon surrounding $v$. In Fig. 4.11(a), $\chi$ is the boundary of the gray
region, effectively the union of the digons (but recall that the digons
$D_{i}$ live on different surfaces $P_{i-1}$; hence “effectively”). Clearly,
excising the surface bounded by $\chi$ from $L$ all at once achieves the same
effect as excising the digons $D_{i}$ one-by-one.

The region of $L$ bound by a geodesic polygon $\chi$ is a particular instances
of what we call a _crest_ : a subset of $L$ enclosing $v$ whose removal and
suitable suturing via AGT will reduce $L$ to $X$. Note that we allow the
boundary of a crest to include portions of $\partial L$, e.g., $\chi$ in Fig.
4.11(a) includes the $x_{i}$ as well as the edge $x_{5}x_{6}$. In Chapter 7 we
will show that it is possible to construct crests directly on $L$ without
deriving them from digon removals.

###  Chapter 6 Algorithm for Tailoring via Sculpting

In this research we have also concentrated on achieving constructive proofs of
the theorems, constructive in the sense of leading to finite algorithms. In
this chapter we follow Theorem 4.6 to yield an algorithm for achieving the
tailoring of $P$ to $Q$. Throughout we measure computational complexity in
terms of $n$, where $n=\max\\{|P|,|Q|\\}$ is the number of vertices of the
larger of $P$ or $Q$; so $|P|,|Q|=O(n)$.111 A finer analysis would treat the
number of vertices of $Q$ and $P$ independently, say, $Q$ with $m$ vertices.
The algorithm steps would be the same, but the complexities would be
apportioned differently.  Our goal for all the algorithms is to achieve
polynomial-time complexity, $O(n^{k})$, but we have not worked hard to lower
$k$, through, e.g., exploitation of efficient data structures. Instead we are
content to leave improvements for future work. We will see that $k=4$ seems to
suffice.

Euler’s $V{-}E{+}F$ theorem implies that the number of edges and faces of a
polyhedron are linearly related to the number of vertices, so all components
of $P$ and $Q$ are $O(n)$. Thus for tailoring via slicing, the number of
slices of $P$ is $O(n)$: one slice per face of $Q$. Each slice leads to
g-domes, each g-dome to pyramids, and each pyramid is reduced to its base by a
series of tailoring steps: digon excisions and suturings.

We analyze the complexity in four parts, the $0$-th just the conceptual
slicing of $P$ by planes on each face of $Q$:

  * •

Algorithm 0: Slice $P$ to $Q$: $O(n)$.

  * •

Algorithm 1: slice $\to$ g-domes, following Lemma 4.1.

  * •

Algorithm 2: g-dome $\to$ pyramids, following Theorem 3.2.

  * •

Algorithm 3: pyramid $\to$ digons, following Lemma 4.4.

Algorithms $0,1,2$ operate on the extrinsic $3$-dimensional structure of the
polyhedra. Algorithm $3$ instead processes its calculations on the intrinsic
structure of the surface.

####  6.1 Algorithm 1: slice$~{}\to~{}$g-domes

Input : One slice plane $\Pi$

Output : $O(n)$ g-domes, a total of at most $O(n)$ vertices.

// Consult Fig. 4.1.

Let $F$ be the face of $Q$ lying in $\Pi$, and $e$ be an edge of $F$.

Sort vertices angularly about $e$. // $O(n\log n)$

for  _$i=0,1,2,\ldots,k$_ do

Rotate $\Pi_{i}$ about $e$ until the portion swept is not a g-dome.

// Following Lemma 4.1.

Add to g-domes list.

end for

Result: List of $O(n)$ g-domes.

Algorithm 1 From one slice, $O(n)$ g-domes.

Algorithm 1 follows the proof of Lemma 4.1, which clearly results in at most
$O(n)$ g-domes, with the sum of the complexities of the g-domes for any one
slice $O(n)$. It can be implemented to run in $O(n\log n)$ time. Therefore,
over $O(n)$ slices, each resulting in $O(n)$ g-dome vertices, we have at most
$O(n^{2}\log n)$ total g-dome complexity: $O(n\log n)$ sorting repeated $O(n)$
times.

However, it could be that a clever choice of ordering of the slices always
results in a smaller total complexity, perhaps $O(n\log n)$. Using ${\Omega}$
to indicate a lower bound,222 For example, a constant fraction of $n$ is
${\Omega}(n)$. the question is: Might each of the ${\Omega}(n)$ slices from
Algorithm $0$ lead to g-domes with a total complexity of ${\Omega}(n^{2})$,
independent of the order of slicing? Although resolving this question has only
a small effect on the overall time-complexity of the tailoring algorithm, we
take a detour to explore this issue via an example in the next subsection.

#####  6.1.1 Complexity of sculpting

######  Example 6.1.

Consider Fig. 6.1, two nested pyramids $Q\subset P$ sharing a common regular
polygon base. The combinatorial complexity of slicing $P$ with face planes of
$Q$ is order-sensitive: from ${\Omega}(n^{2})$ to ${\Omega}(n\log n)$.

Figure 6.1: $Q\subset P$. The shared base is a regular polygon of $n{=}48$
sides. (a) One slice-plane $\Pi_{0}$ lying on a face of $Q$. (b) After slicing
by $\Pi_{0}$. (c) A second, opposite slice $\Pi_{n/2}$.

###### Proof.

Let the faces of $Q$ ordered in sequence around the base be
$F_{0},F_{1},\ldots,F_{n-1}$, each $F_{i}$ determining a plane $\Pi_{i}$. Let
$P_{i}$ be the polyhedron after slicing $P$ with planes
$\Pi_{0},\Pi_{1},\ldots,\Pi_{i}$. $\Pi_{0}$ cuts $n-2$ edges of $P$, as shown
in Fig. 6.1(a,b), and effectively removes half the edges of $P$ from later
slices. If $P$ is sliced in the order $i=1,2,\ldots,n-2$, following adjacent
faces around the base, each plane cuts a diminishing number of the remaining
edges. An explicit calculation shows that plane $\Pi_{i}$ cuts
$\lfloor\frac{n+1-i}{2}\rfloor$ edges of $P_{i-1}$. And because

$\sum_{i=1}^{n-2}\left\lfloor\frac{n+1-i}{2}\right\rfloor=\Omega(n^{2})\;,$

with this plane-slice ordering, ${\Omega}(n)$ slices each have ${\Omega}(n)$
vertices, for a total quadratic complexity, ${\Omega}(n^{2})$.

However, if one instead orders the slices in a binary-search pattern, then the
total complexity is ${\Omega}(n\log n)$, as we now show. Let $n=2^{m}$ be a
power of $2$ without loss of generality. The pattern is slicing with planes
lying on faces $F_{i}$ with indices in the order

$s=\left(0,\frac{n}{2},\frac{n}{4},\frac{3n}{4},\frac{n}{8},\frac{3n}{8},\frac{5n}{8},\frac{7n}{8},\ldots\right)\;.$

We partition this sequence $s$ of indices into subsequences,
$s=(0,s_{1},s_{2},\ldots,s_{m})$, as follows:

$\displaystyle s_{1}$ $\displaystyle=$ $\displaystyle\left(\frac{n}{2}\right)$
$\displaystyle s_{2}$ $\displaystyle=$
$\displaystyle\left(\frac{n}{2^{2}},\frac{3n}{2^{2}}\right)$ $\displaystyle
s_{3}$ $\displaystyle=$
$\displaystyle\left(\frac{n}{2^{3}},\frac{3n}{2^{3}},\frac{5n}{2^{3}},\frac{7n}{2^{3}}\right)$
$\displaystyle\cdots$ $\displaystyle s_{k}$ $\displaystyle=$
$\displaystyle\left(\frac{j\,n}{2^{k}}\right),\;\textrm{where}\;j=1,3,5,\ldots,2^{k}{-}1\;.$

Notice that the number of indices in sequence $s_{k}$, $|s_{k}|=2^{k-1}$. As a
check, the total number of indices in $s$ is

$1+\sum_{k=1}^{m}|s_{k}|=1+\sum_{k=1}^{m}2^{k-1}=2^{m}=n\;.$

One can calculate that the slice at $\displaystyle{i=\frac{j\,n}{2^{k}}}$ only
cuts $\displaystyle{\frac{n}{2^{k}}}$ edges of $P_{i-1}$. Only
$\displaystyle{\frac{n}{2^{k-1}}}$ edges are “exposed” to $\Pi_{i}$: for
example, for $k=2$ and $i=1/4$, $n/2$ edges are possibly available for cutting
by $\Pi_{i}$, as can be seen in Fig. 6.1(c). However, because of the slant of
$\Pi_{i}$, only half of those, $n/4$, are in fact cut by $\Pi_{i}$.

Now we compute the total number of edges sliced by the planes following the
sequence $s$. Because $|s_{k}|=2^{k-1}$, and each slice in $s_{k}$ cuts
$\displaystyle{\frac{n}{2^{k}}}$edges, the total number of cuts over all $k$
is

$\sum_{k=1}^{m}2^{k-1}\frac{n}{2^{k}}=n\sum_{k=1}^{m}\frac{1}{2}=n\,\frac{m}{2}\;.$

And since $m=\log n$, the total complexity is ${\Omega}(n\log n)$, or an
average of ${\Omega}(\log n)$ for each of ${\Omega}(n)$ slices. ∎

In the absence of a resolution to this complexity question, we will assume
that Algorithm 0 and Algoirthm 1 together result in $O(n)$ g-domes each of
$O(n)$ vertices, produced in time $O(n^{2})$.

####  6.2 Algorithm 2: g-dome$~{}\to~{}$pyramids

Input : One g-dome $G$ of $O(n)$ vertices

Output : $O(n)$ pyramids, each of size $O(n)$; and $O(n^{2})$ pyramids, size
$O(1)$.

// Following Theorem 3.2.

for  _each of $k=O(n)$ vertex-degree reductions_ do

As in Theorem 3.2, slice with plane: $O(k)$.

Remove pyramid $P$ of $O(k)$ vertices.

“Clean-up” by removing $k$ pyramids each of size $O(1)$.

end for

Result: List of pyramids.

Algorithm 2 Partition one g-dome to $O(n)$ pyramids.

We follow Theorem 3.2 for partitioning each g-dome $G$ into pyramids. Each
vertex $v_{i}$ of the top-canopy of $G$ is removed, as in Fig. 3.3, until only
one remains. Removal of each $v_{i}$ follows the degree-removal steps
illustrated in Fig. 3.4.

Because the sum of the vertex degrees of a g-dome is $2E=O(n)$, the asymptotic
complexity of processing a g-dome with many vertices in its top-canopy is no
different than it is for just two vertices as in Fig. 3.3. Moreover, we can
assume that $v_{2}$ has degree-$3$ while $v_{1}$ has degree-$k$, with
$k=O(n)$.

First a plane slice results in a pyramid of $k$ vertices with apex $v_{1}$,
which is removed (and reduced by Algorithm 3). Next follows a “clean-up” phase
that removes $O(k)$ pyramids each of $4$ or $5$ vertices, so of constant size,
$O(1)$.

This is then repeated for the new apex of degree $k-1$: removal of a pyramid
of $k-1$ vertices, and cleanup of $O(k-1)$ pyramids of constant size. After
iterating through $k,k-1,k-2,\ldots$, the algorithm has sliced off $O(k^{2})$
pyramids of constant size, and $O(k)$ pyramids of size $O(k)$. In the worst
case $k=O(n)$, for a total complexity of $O(n^{2})$.

####  6.3 Algorithm 3: pyramid$~{}\to~{}$digons

Input : A pyramid $P$ of $O(n)$ vertices

Output : $O(n)$ digons whose removal flattens $P$ to $X$.

// Following Lemma 4.4.

// Assume apex degree-$k$, with $k=O(n)$.

for  _each of $x_{i}$, $i=1,2,\ldots,k$_ do

Construct digon $D_{i}(x_{i},y_{i})$: Locate $y_{i}$.

Locate $y_{i}$ by tracing geodesics: $O(k)$.

end for

Result: List of $O(n)$ digons

Algorithm 3 Tailor one pyramid $P$ to its base $X$.

Lastly we concentrate on the cost of removing one pyramid $P$ of $O(n)$
vertices. Following Lemma 4.4, this requires $O(n)$ digon removals. For each
digon $D_{i}(x_{i},y_{i})$, we need to calculate the location of $y_{i}$ on
$C(x_{i})$; then $y_{i}$ becomes a vertex for the removal of the next digon
$D_{i+1}$. Fortunately, there is no need to compute the cut locus $C(x_{i})$.

Let us focus on locating $y_{i}$, after the removal of
$D_{i-1}(x_{i-1},y_{i-1})$ the previous iteration. Recall that $y_{i}$ is the
only vertex on $L_{i}$, so it is immediate to find the shortest path
${\gamma}$ from $x_{i}$ to $y_{i}$. We know the angle
${\theta}={\omega}_{Q}(x_{i})-{\omega}_{P}(x_{i})$ needed to be removed by
$D_{i}$, so we know that geodesics ${\gamma}_{1}$ and ${\gamma}_{2}$ at angles
${\theta}/2$ left and right of ${\gamma}=x_{i}y_{i}$ will meet at $y_{i}$.
Tracing ${\gamma}_{1}$ and ${\gamma}_{2}$ over the surface might cross sealed
digons $D_{1},\ldots,D_{i-1}$, the blueseals ${\sigma}_{i}$ in Fig. 4.11(b).
So the cost of computing ${\gamma}_{1}\cap{\gamma}_{2}=y_{i}$ is $O(n)$.

Thus the complexity of tailoring one pyramid of $O(n)$ vertices is $O(n^{2})$.

Note that we do not need the extrinsic $3$-dimensional structure of the
intermediate polyhedra guaranteed by AGT to perform the calculations, as is
evident in the example described in Fig. 4.11.

####  6.4 Overall Tailoring Algorithm

Putting the three algorithm complexities together, and for succinctness
abbreviating $O(n)$ with $n$ and “time-complexity” with “complexity,” we have:

  1. (1)

$n$ slices $\to$ $n$ g-domes, each of size $n$: complexity $n^{2}\log n$.

  2. (2)

$1$ g-dome of size $n$ $\to$

    1. (a)

$n$ pyramids, each of size $n$: complexity $n^{2}$.

    2. (b)

$+$ $n^{2}$ constant-size pyramids: complexity $n^{2}$.

  3. (3)

$1$ pyramid of size $n$ $\to$ $n^{2}$ to reduce to base: complexity $n^{2}$.

(1) and (2) together have complexity $n^{3}$, iterating over each of the $n$
domes. (The $n\log n$ sorting need not be repeated after the g-domes are
identified.) (3) repeats $n^{2}$ over (potentially) $n^{2}$ pyramids: $n$
g-domes resulting in $n$ pyramids each. Thus the total complexity is $n^{4}$.
We summarize in a theorem:

######  Theorem 6.2.

Given convex polyhedra $P$ and $Q$ of at most $n$ vertices each, and $Q\subset
P$, $P$ can be tailored to $Q$, tracking a sculpting of $Q$ from $P$ as in
Theorem 4.6, in time $O(n^{4})$.

###  Chapter 7 Crests

In this chapter we revisit the suggestion made at the end of Chapter 5 that
the digons to reduce one pyramid to its base could be cut out all at once. As
before, let $P=L\cup X$ be a pyramid with base $X$ and lateral sides $L$.
Recall that a _crest_ is a subset of $L$ enclosing $v$ whose removal and
suitable suturing via AGT will reduce $L$ to $X$.

We derive a method for identifying a crest that does not rely on digon
removals, but rather works directly on a pyramid. This allows us to achieve in
Theorem 7.5 reshaping of $P$ to $Q$ by the removal of crests to flatten
pyramids. We call this process _crest-tailoring_ , in contrast to the _digon-
tailoring_ explored in Chapter 4. It represents a tradeoff between the
simplicity of digons and the number of excisions: for digon-tailoring, $O(n)$
per pyramid, for crests, one per pyramid. We first illustrate the process of
identifying a crest on two example pyramids before proving that it always
works.

####  7.1 Examples

Let $P=L\cup X$ with $\partial X=\partial L=(x_{1},x_{2},\ldots,x_{k})$ having
vertices $x_{i}$. The apex is $v$, which projects orthogonally to $\bar{v}$.
We will describe the procedure for identifying a crest first for $\bar{v}\in
X$ and then for $\bar{v}\not\in X$. Although the cases initially feel
different, the proofs will show that they are nearly the same.

Fig. 7.1 illustrates the case $\bar{v}\in X$. Let $T_{i}=x_{i}x_{i+1}v$ and
$\bar{T}_{i}=x_{i}x_{i+1}\bar{v}$. We proved in Lemma 2.8 that the angle
$\bar{{\theta}}_{i}$ at $x_{i}$ in $X$ is strictly smaller than than the angle
${\theta}_{i}$ on $L$, the sum of the angles in $T_{i-1}$ and $T_{i}$ incident
to $x_{i}$ (as long as $|v\bar{v}|>0$).

Figure 7.1: (a) Pyramid. (b) $X$ with projected triangles
$\bar{T}_{i}=x_{i}x_{i+1}\bar{v}$. (c) Flattening of $L$ to $\bar{L}$, in this
case by cutting the edge $x_{5}v$. The lifted triangles $T^{L}_{i}$ are shown
yellow. The crest $\chi$ is blue.

A key definition is the _lift_ of $\bar{T}_{i}$ onto $L$. Let
$\bar{{\alpha}}_{i}$ and $\bar{{\beta}}_{i}$ be the base angles of
$\bar{T}_{i}$, at $x_{i}$ and $x_{i+1}$ respectively. On $L$, extend geodesic
${\gamma}_{i}$ from $x_{i}$ at angle $\bar{{\alpha}}_{i}$, and extend geodesic
${\gamma}_{i+1}$ from $x_{i+1}$ at angle $\bar{{\beta}}_{i}$. Let $v_{i}$ be
the point on $L$ at which these geodesics meet; $v_{i}$ is the image of
$\bar{v}$. (We will not establish that indeed these geodesics meet on $L$
until Lemma 7.3.) Then $\operatorname{lift}(\bar{T}_{i})=T^{L}_{i}$ is a
geodesic triangle on $L$ isometric to $\bar{T}_{i}$. Another way to view the
lift of $\bar{T}_{i}$ is to imagine $\bar{T}_{i}$ rotating about
$x_{i}x_{i+1}$ by the dihedral angle there and pasting it on the inside of
$L$.

Yet another way to view the lift is as follows. $L$ is isometric to a cone and
can be flattened by cutting along a generator, i.e., a segment from $v$ to
$\partial L$. Let $\bar{L}$ be a particular flattening, with the cut generator
not “near” $x_{i}$ just for simplicity. Then place a copy of $\bar{T}_{i}$ on
$\bar{L}$ matching $x_{i}x_{i+1}$. Then refold $\bar{L}$ to $L$. We will
continue to reason with a flattened $\bar{L}$ but remembering that $\bar{L}$
is a representation of $L$, and so the cut edge is not relevant.

This last layout-viewpoint yields a method to construct a full crest, call it
$\chi$. The base $X$, partitioned into $\bar{T}_{i}$, can be modified by
opening the angle at $x_{i}$ from $\bar{{\theta}}_{i}$ to ${\theta}_{i}$.
After opening at all $x_{i}$, this figure can be superimposed on the flattened
$\bar{L}$, matching the boundaries $x_{1},\ldots,x_{k}$. This is illustrated
in Fig. 7.1(c). The crest is then the portions of $L$ not covered by the
lifted $\bar{T}_{i}$. It should be clear that cutting out $\chi$ and suturing