# A maximum volume conjecture for hyperbolic polyhedra

Giulio Belletti

###### Abstract

We propose a volume conjecture for hyperbolic polyhedra that is similar in
spirit to the recent volume conjecture by Chen and Yang on the growth of the
Turaev-Viro invariants. Using Barrett’s Fourier transform we are able to prove
this conjecture in a large family of examples. As a consequence of this
result, we prove the Turaev-Viro volume conjecture for a new infinite family
of hyperbolic manifolds.

###### Contents

  1. 1 Introduction
  2. 2 The Kauffman bracket and the Yokota invariant
    1. 2.1 The Kauffman bracket
    2. 2.2 The definition of the Yokota invariant from the Kauffman bracket
  3. 3 Volume conjecture for polyhedra
    1. 3.1 The volume conjecture for polyhedra
    2. 3.2 The maximum volume conjecture
  4. 4 The Fourier Transform
  5. 5 The Turaev-Viro volume conjecture
  6. A Appendix: Numerical evidence for Conjecture 2

##  1 Introduction

In [CY18] Chen and Yang proposed and provided extensive computations for the
following conjecture:

######  Conjecture 1 (The Turaev-Viro volume conjecture).

Let $M$ be a hyperbolic $3$-manifold, either closed, with cusps, or compact
with geodesic boundary. Then as $r$ varies along the odd natural numbers,

$\lim_{r\rightarrow\infty}\frac{2\pi}{r}\log\left(TV\left(M,e^{\frac{2\pi
i}{r}}\right)\right)=\mathrm{Vol}(M)$ (1)

This conjecture has been verified for the complements of the Borromean rings
[DKY18], of the figure eight knot [DKY18], all the hyperbolic integral Dehn
surgeries on the figure eight knot [Oht18], and all complements of fundamental
shadow links [BDKY18].

In this paper, we propose a version of the Chen-Yang volume conjecture for
hyperbolic polyhedra, for a newly defined Turaev-Viro type invariant of graphs
that we denote with $TV_{r}(\Gamma,q)$ where $q$ is a root of unity.

######  The Maximum Volume Conjecture.

Let $\Gamma\subseteq S^{3}$ be a $3$-connected planar graph. Then

$\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left(TV_{r}(\Gamma,e^{\frac{2\pi
i}{r}})\right)=\sup_{P}\mathrm{Vol}(P)$

where $P$ varies among all proper generalized hyperbolic polyhedra (see
Definition 3.2) with $\Gamma$ as a $1$-skeleton, and $r$ ranges across all odd
natural numbers.

The relationship between Conjecture 1 and Kashaev’s volume conjecture is
similar to the relationship between the Maximum Volume Conjecture and the
volume conjecture for polyhedra of [CGvdV15].

We are able to prove the Maximum Volume Conjecture for a large family of
examples:

######  Theorem 4.8.

The Maximum Volume Conjecture is verified for any planar graph obtained from
the tetrahedron by applying any sequence of the following two moves:

  * •

blowing up a trivalent vertex (see Figure 1) or

  * •

triangulating a triangular face (see Figure 2).

$\longrightarrow$

Figure 1: Truncating a vertex

$\longrightarrow$

Figure 2: Triangulating a face

The Maximum Volume Conjecture naturally leads to the question of what is the
supremum of all volumes of polyhedra sharing the same $1$-skeleton. This is
answered in [Bel20, Theorem 4.2] by the following

######  Theorem 1.1.

For any $3$-connected planar graph $\Gamma$,

$\sup_{P}\mathrm{Vol}(P)=\mathrm{Vol}(\overline{\Gamma})$

where $P$ varies among all proper generalized hyperbolic polyhedra with
$1$-skeleton $\Gamma$ and $\overline{\Gamma}$ is the rectification of
$\Gamma$.

The rectification of a graph is defined in [Bel20, Section 3.4]; for the
purpose of this paper it suffices to say that $\overline{\Gamma}$ is the
polyhedron with $1$-skeleton $\Gamma$ with every edge tangent to
$\partial\mathbb{H}^{3}$ in the Klein model of hyperbolic space (hence, which
has dihedral angle $0$ at each edge). This polyhedron can be canonically
truncated to give an ideal right-angled hyperbolic polyhedron, hence it makes
sense to speak of $\mathrm{Vol}(\overline{\Gamma})$ as the volume of the
truncation.

Theorem 4.8 leads to the proof of the Turaev-Viro volume conjecture for a new
infinite family of cusped manifolds. These are complements of certain links in
$S^{3}\\#^{g}(S^{1}\times S^{2})$; their hyperbolic structure is obtained by
gluing right-angled octahedra.

In Section 2 we set the notation, give the basic properties of the Kauffman
bracket and define the Yokota invariant. In Section 3 we discuss previous
volume conjectures for polyhedra and introduce the Turaev-Viro invariant of a
graph and the Maximum Volume Conjecture. In Section 4 we introduce the Fourier
transform of Barrett, and use it to prove Theorem 4.8. Section 5 contains the
proof of the Turaev-Viro volume conjecture for a new family of manifolds.
Finally in an appendix we propose numerical evidence for a related volume
conjecture for polyhedra.

Acknowledgments. I wish to thank my advisors for their guidance and support.
Furthermore I wish to thank Renaud Detcherry, Efstratia Kalfagianni and Tian
Yang for their comments on a preliminary version of this paper.

##  2 The Kauffman bracket and the Yokota invariant

###  2.1 The Kauffman bracket

Throughout the rest of the paper $r\geq 3$ is an odd integer and
$q=e^{\frac{2\pi i}{r}}$. All the definition we give in this section are
standard; the only notable difference is that in some papers (e.g. [Bar03])
the graphs are colored with half-integer colors, while here we use integers.

The _quantum integer_ $[n]$ is defined as $\frac{q^{n}-q^{-n}}{q-q^{-1}}$, and
the quantum factorial $[n]!$ is $\prod_{i=1}^{n}[i]$. Furthermore, we denote
with $I_{r}$ the set of all even natural numbers $\leq r-2$.

######  Remark 2.1.

Because of the choice of root of unity $q$, we need to work with the $SO(3)$
version of the quantum invariants, rather than the $SU(2)$ version. This
essentially amounts to using only even numbers as colors.

######  Definition 2.2.

We say that a triple $(a,b,c)$ of natural numbers is _$r$ -admissible_ if

  * •

$a,b,c\leq r-2$;

  * •

$a+b+c$ is even and $\leq 2r-4$;

  * •

$a\leq b+c$, $b\leq a+c$ and $c\leq a+b$.

We say that a $6$-tuple $(n_{1},n_{2},n_{3},n_{4},n_{5},n_{6})$ is
$r$-admissible if the $4$ triples $(n_{1},n_{2},n_{3})$,
$(n_{1},n_{5},n_{6})$, $(n_{2},n_{4},n_{6})$ and $(n_{3},n_{4},n_{5})$ are
$r$-admissible.

For $n\in I_{r}$ define

$\Delta_{n}=(-1)^{n+1}[n+1]$ (2)

For an $r$-admissible triple $(a,b,c)$ we can define

$\Theta(a,b,c)=(-1)^{\frac{a+b+c}{2}}\frac{[\frac{a+b+c}{2}+1]!}{[\frac{a+b-c}{2}]![\frac{a-b+c}{2}]![\frac{-a+b+c}{2}]!}$
(3)

and $\Delta(a,b,c):=\Theta(a,b,c)^{-\frac{1}{2}}$. Notice that the number
inside the square root is real; by convention we take the positive square root
of a positive number, and the square root with positive imaginary part of a
negative number.

If $v$ is a trivalent vertex of a graph whose incident edges are colored by an
admissible triple $a,b,c$ we write for short $\Theta(v)$ and $\Delta(v)$
instead of $\Theta(a,b,c)$ and $\Delta(a,b,c)$.

Moreover, for an $r$-admissible $6$-tuple
$(n_{1},n_{2},n_{3},n_{4},n_{5},n_{6})$ we can define its $6j$-symbol as usual
as

$\displaystyle\begin{vmatrix}n_{1}&n_{2}&n_{3}\\\
n_{4}&n_{5}&n_{6}\end{vmatrix}=\prod_{i=1}^{4}\Delta(v_{i})\sum_{z=\max
T_{i}}^{\min
Q_{j}}\frac{(-1)^{z}[z+1]!}{\prod_{i=1}^{4}[z-T_{i}]!\prod_{j=1}^{3}[Q_{j}-z]!}$
(4)

where:

  * •

$v_{1}=(n_{1},n_{2},n_{3})$, $v_{2}=(n_{1},n_{5},n_{6})$,
$v_{3}=(n_{2},n_{4},n_{6})$, $v_{4}=(n_{3},n_{4},n_{5})$;

  * •

$T_{1}=\frac{n_{1}+n_{2}+n_{3}}{2}$, $T_{2}=\frac{n_{1}+n_{5}+n_{6}}{2}$,
$T_{3}=\frac{n_{2}+n_{4}+n_{6}}{2}$ and $T_{4}=\frac{n_{3}+n_{4}+n_{5}}{2}$;

  * •

$Q_{1}=\frac{n_{1}+n_{2}+n_{4}+n_{5}}{2}$,
$Q_{2}=\frac{n_{1}+n_{3}+n_{4}+n_{6}}{2}$ and
$Q_{3}=\frac{n_{2}+n_{3}+n_{5}+n_{6}}{2}$.

######  Remark 2.3.

Notice that if $z\geq r-1$ the summand in (4) corresponding to $z$ is equal to
$0$.

######  Definition 2.4.

An _$r$ -admissible coloring_ for a tetrahedron $T$ is an assignment of an
$r$-admissible $6$-tuple $(n_{1},n_{2},n_{3},n_{4},n_{5},n_{6})\in I_{r}^{6}$
to the set of edges of $T$, as shown in figure 3. More in general, we say that
an $r$-admissible coloring for a $3$-valent graph $\Gamma\subseteq S^{3}$ is
an assignment of elements of $I_{r}$ to the edges of $\Gamma$ such that the
colors at each vertex form an admissible triple. Even more generally we say
that an assignment of elements of $I_{r}$ to edges of a (not necessarily
$3$-valent) graph is a _coloring_ , and a graph $\Gamma$ together with its
coloring $col$ is a _colored graph_ $(\Gamma,col)$.

$n_{4}$$n_{6}$$n_{5}$$n_{2}$$n_{1}$$n_{3}$ Figure 3: An admissible coloring
for a tetrahedron Figure 4: A Theta graph

######  Definition 2.5.

The _Kauffman bracket_ is the unique map

$\langle\cdot\rangle:\\{\textrm{colored trivalent framed graphs in
}S^{3}\\}\rightarrow\mathbb{C}$

satisfying the following properties:

  1. 1.

If $\Gamma$ is the planar circle colored with $n$ then
$\langle\Gamma\rangle=\Delta_{n}$;

  2. 2.

If $\Gamma$ is a Theta graph (see Figure 4) colored with the $r$-admissible
triple $a,b,c$ then $\langle\Gamma\rangle=1$;

  3. 3.

If $\Gamma$ is a tetrahedron colored with the $r$-admissible $6$-tuple
$(n_{1},\dots,n_{6})$ then $\langle\Gamma\rangle$ is the $6j$-symbol;

  4. 4.

The _fusion rule_ :

$\left\langle\vbox{\hbox{\includegraphics[width=56.9055pt]{fusion_rule.pdf}}}\right\rangle=\sum_{i\in
I_{r}}\Delta_{i}\left\langle\vbox{\hbox{\includegraphics[width=85.35826pt]{fusion_rule2.pdf}}}\right\rangle$
(5)

  5. 5.

If $\Gamma$ has a bridge (that is to say, an edge that disconnects the graph
if removed) colored with $i\neq 0$, then $\langle\Gamma\rangle=0$;

  6. 6.

If at some vertex of $\Gamma$ the colors do not form an $r$-admissible triple,
$\langle\Gamma\rangle=0$;

  7. 7.

If $\Gamma$ is colored with an $r$-admissible coloring such that the color of
an edge $e$ is equal to $0$, then
$\langle\Gamma\rangle=\frac{1}{\sqrt{\Delta_{a}\Delta_{b}}}\langle\Gamma^{\prime}\rangle$
where $\Gamma^{\prime}$ is $\Gamma$ with $e$ removed, and $a,b$ are the colors
of the edges that share a vertex with $e$ (notice that since the coloring is
$r$-admissible, two edges sharing the same vertex with $e$ will have the same
color);

  8. 8.

The framing change:

$\left\langle\vbox{\hbox{\includegraphics[height=56.9055pt]{frame_rule1.pdf}}}\right\rangle=(-1)^{\frac{b+c-a}{2}}q^{\frac{b(b+2)+c(c+2)-a(a+2)}{4}}\left\langle\vbox{\hbox{\includegraphics{frame_rule2.pdf}}}\right\rangle$

  9. 9.

If $\Gamma$ is the disjoint union of $\Gamma_{1}$ and $\Gamma_{2}$, then
$\langle\Gamma\rangle=\langle\Gamma_{1}\rangle\langle\Gamma_{2}\rangle$.

It is absolutely not clear from the definition that such a map exists; a proof
is in [KL94, Chapter 9]. However, it is straightforward to see that Properties
1-9 are enough to calculate $\langle\Gamma\rangle$. Taking any planar diagram
of $\Gamma$, apply a fusion rule near each crossing, and then undo the
crossing using Property 8; therefore we only need to calculate
$\langle\cdot\rangle$ on planar graphs. Here repeated applications of the
Fusion rule create a bridge, and rules 5, 7 and 9 allow to compute
$\langle\Gamma\rangle$ from the Kauffman bracket of two graphs with fewer
vertices.

######  Remark 2.6.

There are a few different normalizations of the Kauffman bracket in the
literature. Here we use the _unitary normalization_ ; it should be noted that
[KL94] uses a different one, however the results there apply to the unitary
normalization with little modification.

In what follows sometimes we will color the edges of $\Gamma$ with linear
combinations of colors; the Kauffman bracket can be extended linearly to this
context. In particular, we will use _Kirby’s color_ $\Omega:=\sum_{i\in
I_{r}}\Delta_{i}i$.

###  2.2 The definition of the Yokota invariant from the Kauffman bracket

In this subsection we give an overview of the Yokota invariant, which
generalizes the Kauffman bracket invariant of trivalent graphs to graphs with
vertices of any valence; it was first introduced in [Yok96].

Suppose $\Gamma\subseteq S^{3}$ is a framed graph with vertices of valence
$\geq 3$; as before $r>2$ is odd and $q=e^{2\pi i/r}$.

For a vertex $v$ of $\Gamma$, we can take a small ball $B$ containing $v$, and
replace $\Gamma\cap B$ with a trivalent planar tree in $B$ having the same
endpoints in $\partial B\cap\Gamma$ (see figure 5). We call this procedure a
_desingularization_ of $\Gamma$ at $v$. Notice that if $v$ has valence greater
than $3$, then this procedure is not unique; however, any desingularization is
related to any other via a sequence of Whitehead moves (see figure 6).

$\xrightarrow{\hskip 28.45274pt}$

Figure 5: Desingularization of a vertex of valence $6$

We say that the trivalent graph $\Gamma^{\prime}$ is a desingularization of
$\Gamma$ if it is obtained from $\Gamma$ by desingularization of each vertex
of valence $>3$.

$\xrightarrow{\hskip 28.45274pt}$

Figure 6: A Whitehead move

######  Definition 2.7.

Let $(\Gamma,col)$ be a framed graph in $S^{3}$ colored with elements of
$I_{r}$. Let $\Gamma^{\prime}$ be a desingularization of $\Gamma$. Call
$e_{1}^{\prime},\dots,e_{k}^{\prime}$ the edges of $\Gamma^{\prime}$ that were
added by the desingularization. If $k>0$, the _Yokota invariant_ of
$(\Gamma,col)$ is

$Y_{r}(\Gamma,col):=\sum_{col^{\prime}\in
I_{r}^{k}}\left(\prod_{i=1}^{k}\Delta_{col^{\prime}(e_{i}^{\prime})}\right)|\langle\Gamma^{\prime},col\cup
col^{\prime}\rangle|^{2}$

with $col^{\prime}$ coloring the edges $e_{1}^{\prime},\dots,e_{k}^{\prime}$.
If instead $k=0$ (i.e. $\Gamma=\Gamma^{\prime}$ i.e. $\Gamma$ is trivalent)
then $Y_{r}(\Gamma,col)=|\langle\Gamma,col\rangle|^{2}$.

As we did with the Kauffman bracket, we extend linearly the Yokota invariant
to linear combinations of colors. Notice that in this case, even if $\Gamma$
is trivalent, we may get
$Y_{r}(\Gamma,col)\neq\lvert\langle\Gamma,col\rangle\rvert^{2}$.

######  Remark 2.8.

We stress the fact that we are using the unitary normalization for the
Kauffman bracket. If we instead used the Kauffman normalization
$\langle\cdot\rangle_{K}$ of [KL94], the definition of the Yokota invariant of
$\Gamma,col$ would be

$Y_{r}(\Gamma,col):=\sum_{col^{\prime}\in
I_{r}^{k}}\frac{\prod_{i=1}^{k}\Delta_{col^{\prime}(e^{\prime}_{i})}}{\prod_{v\textrm{
vertex of }\Gamma}\Theta(v)}|\langle\Gamma^{\prime},col\cup
col^{\prime}\rangle_{K}|^{2}$

######  Proposition 2.9.

[Yok96] The Yokota invariant does not depend on the choice of
desingularization.

We can easily extend the Yokota invariant to graphs with $1$-valent and
$2$-valent vertices as well via the following formulas.

$Y\left(\includegraphics[width=56.9055pt]{lefteq2valence.pdf}\right)=\frac{\delta_{i,j}}{\Delta_{i}}Y\left(\includegraphics[width=85.35826pt]{righteq2valence.pdf}\right);$

$Y\left(\vbox{\hbox{\includegraphics{lefteq1valence.pdf}}}\right)=\delta_{i,0}Y\left(\vbox{\hbox{\includegraphics{righteq1valence.pdf}}}\right);$

we normalize the graph with a single vertex and no edges to have invariant
equal to $1$.

$a$$b$$c$

$\longrightarrow$ $a$$b$$c$$a$$b$$c$

Figure 7: A vertex sum of two trivalent vertices

Now we give four important properties of the Yokota invariant.

######  Proposition 2.10.

The following hold:

  1. 1.

The Yokota invariant does not depend on the framing of $\Gamma$.

  2. 2.

If an edge $e$ of $\Gamma$ is colored with the Kirby color $\Omega$, and
$\Gamma^{\prime}$ is obtained from $\Gamma$ via a Whitehead move on the edge
$e$ (keeping every color the same) then
$Y_{r}(\Gamma,col)=Y_{r}(\Gamma^{\prime},col)$.

  3. 3.

If $\Gamma$ is a vertex sum of $\Gamma_{1},\Gamma_{2}$ along trivalent
vertices $v_{1}\in\Gamma_{1}$ and $v_{2}\in\Gamma_{2}$ (see Figure 7), then
$Y_{r}(\Gamma,col)=Y_{r}(\Gamma_{1},col_{1})Y_{r}(\Gamma_{2},col_{2})$ where
$col_{1},col_{2}$ are the restrictions of $col$ to $\Gamma_{1},\Gamma_{2}$
respectively.

  4. 4.

If a link $L\subseteq S^{3}$ is colored with $col\in I_{r}^{\lvert L\rvert}$,
then $Y_{r}(L,col)=\lvert J^{L}_{col}(q)\rvert^{2}$ where $J^{L}_{col}$ is the
colored Jones polynomial of $L$ colored with $col$.

###### Proof.

Part 1 holds because $\langle\Gamma\rangle$ depends on the framing of $\Gamma$
only up to a factor of $q^{a}$, thus when taking squared norms this becomes
$1$. Part 2 is essentialy the well definition of the Yokota invariant: both
sides of the equality are equal to the Yokota invariant of the graph obtained
by collapsing $e$ to a point.

Part 3 follows from the analogous property for the Kauffman bracket; this is
obtained via two applications of the fusion rule and one application of the
bridge rule 5 (see Figure 8). Part 4 is just the fact that in the case of $L$,
seen as a trivalent graph with no vertices, $Y_{r}(L,col)=\lvert\langle
L,col\rangle\rvert^{2}=\lvert J_{col}^{L}(q)\rvert^{2}$. ∎

$\xrightarrow{\hskip 42.67912pt}$

Figure 8: Applying the fusion rule to three edges arising from a vertex sum.

It is very important that the vertex sum in Proposition 2.10.3 is done between
trivalent vertices; the assertion is false in general. However, a particular
case still holds.

######  Definition 2.11.

Let $\Gamma,col$ be a colored planar graph and $v$ one of its vertices. We
define the _double_ of $\Gamma$ at the vertex $v$ to be the graph
$\Gamma^{2},col^{2}$ obtained from a vertex sum of two copies, both colored
with $col$, of $\Gamma$ at the vertex $v$.

######  Lemma 2.12.

$Y_{r}(\Gamma^{2},col^{2})=Y_{r}(\Gamma,col)^{2}$

###### Proof.

We give the proof in the case of $v$ having valence $4$; the general case is
identical, except that the notation is heavier.

By linearity, we can assume that $\Gamma^{2}$ only has trivalent vertices.

Apply the fusion rule to the edges arising from the vertex sum until you
obtain a bridge which is eliminated.

We have

$\displaystyle
Y_{r}\left(\vbox{\hbox{\includegraphics[width=56.9055pt]{double1.pdf}}}\right)=\left(\sum_{i,j,k\in
I_{r}}\Delta_{i}\Delta_{j}\Delta_{k}\left\langle\vbox{\hbox{\includegraphics[height=42.67912pt]{vertexsum3.pdf}}}\right\rangle\right)^{2}=$
$\displaystyle\left(\sum_{i\in
I_{r}}\Delta_{i}\left\langle\vbox{\hbox{\includegraphics[height=42.67912pt]{double2.pdf}}}\right\rangle\right)^{2}=\sum_{i,j\in
I_{r}}\Delta_{i}\Delta_{j}\left\langle\vbox{\hbox{\includegraphics[height=42.67912pt]{double2.pdf}}}\right\rangle\left\langle\vbox{\hbox{\includegraphics[height=42.67912pt]{double3.pdf}}}\right\rangle$
$\displaystyle=\sum_{i,j\in
I_{r}}\Delta_{i}\Delta_{j}\left\langle\vbox{\hbox{\includegraphics[height=42.67912pt]{double4.pdf}}}\right\rangle^{2}=Y_{r}(\Gamma,col)^{2}$

∎

######  Remark 2.13.

The Kauffman bracket (hence, the Yokota invariant) can also be defined in the
much larger setting of framed trivalent graphs in closed oriented
$3$-manifolds; since we will not need such a generality that carries some more
technical details, we will restrict ourselves to the $S^{3}$ case.

##  3 Volume conjecture for polyhedra

###  3.1 The volume conjecture for polyhedra

Costantino first conjectured in [Cos07] that the growth of the $6j$-symbol is
given by the volume of a hyperbolic tetrahedron. A Volume Conjecture for
trivalent graphs (and their Kauffman bracket invariant) was proposed in
[vdV09] and later refined in [CGvdV15] to the case of planar trivalent graphs
and simple hyperbolic polyhedra. The conjecture of [CGvdV15] evaluates the
invariant at the first root of unity $q=e^{\pi i/r}$; the downside of this
choice is that they have to consider poles of the Kauffman bracket, instead of
its values directly. Recently, Murakami and Kolpakov [KM18] proposed a volume
conjecture for polyhedra at the second root of unity $q=e^{2\pi i/r}$, but
only stated it for simple polyhedra without hyperideal vertices; remarkably
this conjecture directly involves the value of the Kauffman bracket. Here we
propose an extension of Kolpakov-Murakami’s volume conjecture to a very
general setting, and then propose a volume conjecture for polyhedra that is
similar in spirit to Chen-Yang’s volume conjecture.

Geometric background.

Recall the projective model for hyperbolic space
$\mathbb{H}^{3}\subseteq\mathbb{R}^{3}\subseteq\mathbb{RP}^{3}$ where
$\mathbb{H}^{3}$ is the unit ball of $\mathbb{R}^{3}$ (for the basic
definitions see for example [BB02]). Notice that for convenience we have
picked an affine chart $\mathbb{R}^{3}\subseteq\mathbb{RP}^{3}$, so that it
always make sense to speak of segments between two points, half spaces,
etcetera; this choice is inconsequential, up to isometry. It should be
mentioned that isometries, in this model, correspond to projective
transformations that preserve the unit sphere.

We can associate to a point $p$ lying in
$\mathbb{R}^{3}\backslash\overline{\mathbb{H}^{3}}$ a plane
$\Pi_{p}\subseteq\mathbb{H}^{3}$, called the _polar plane_ of $p$, such that
all lines passing through $\mathbb{H}^{3}$ and $p$ are orthogonal to
$\Pi_{p}$. If $p\in\mathbb{R}^{3}\backslash\overline{\mathbb{H}^{3}}$, denote
with $H_{p}\subseteq\mathbb{H}^{3}$ the half space delimited by the polar
plane $\Pi_{p}$ on the other side of $p$; in other words, $H_{p}$ contains
$0\in\mathbb{R}^{3}$. If the line from $p$ to $p^{\prime}$ passes through
$\mathbb{H}^{3}$, then $\Pi_{p}$ and $\Pi_{p^{\prime}}$ are disjoint [BB02,
Lemma 4]. In particular, if the segment from $p$ to $p^{\prime}$ intersects
$\mathbb{H}^{3}$, then $\Pi_{p}\subseteq H_{p^{\prime}}$ and
$\Pi_{p^{\prime}}\subseteq H_{p}$; if however the segment does not intersect
$\mathbb{H}^{3}$, but the half line from $p$ to $p^{\prime}$ does, then
$H_{p}\subseteq H_{p^{\prime}}$. If $p$ gets pushed away from
$\mathbb{H}^{3}$, then $\Pi_{p}$ gets pushed closer to the origin of
$\mathbb{R}^{3}$.

######  Definition 3.1.

A _projective polyhedron_ in $\mathbb{RP}^{3}$ is a convex polyhedron in some
affine chart of $\mathbb{RP}^{3}$. Alternatively, it is the closure of a
connected component of the complement of finitely many planes in
$\mathbb{RP}^{3}$ that does not contain any projective line.

######  Definition 3.2.

Following [RH93, Definition 4.7]

  * •

We say that a projective polyhedron
$P\subseteq\mathbb{R}^{3}\subseteq\mathbb{RP}^{3}$ is a _generalized
hyperbolic polyhedron_ if each edge of $P$ intersects $\mathbb{H}^{3}$.

  * •

A vertex of a generalized hyperbolic polyhedron is _real_ if it lies in
$\mathbb{H}^{3}$, _ideal_ if it lies in $\partial\mathbb{H}^{3}$ and
_hyperideal_ otherwise.

  * •

a generalized hyperbolic polyhedron $P$ is _proper_ if for each hyperideal
vertex $v$ of $P$ _the interior_ of the polar half space $H_{v}$ contains all
the other real vertices of $P$ (see Figure 9, left).

  * •

We define the _truncation_ of a generalized hyperbolic polyhedron $P$ at a
hyperideal vertex $v$ to be the intersection of $P$ with $H_{v}$; similarly
the _truncation_ of $P$ is the truncation at every hyperideal vertex, that is
to say $P\cap\left(\cap_{v\textrm{ hyperideal}}H_{v}\right)$. We say that the
_volume_ of $P$ is the volume of its truncation. Notice that the volume of a
non-empty generalized hyperbolic polyhedron could be $0$ if the truncation is
empty.

$\Pi_{v}$$v$ Figure 9: A proper truncation

In the remainder of the paper we simply say _proper polyhedra_ for proper
generalized hyperbolic polyhedra.

When it has positive volume, the truncation of a generalized hyperbolic
polyhedron $P$ is itself a polyhedron; some of its faces are the truncation of
the faces of $P$, while the others are the intersection of $P$ with some
truncating plane; we call such faces _truncation faces_. If an edge of the
truncation of $P$ lies in a truncation face we say that the edge is arising
from the truncation.

######  Remark 3.3.

For proper polyhedra the dihedral angles at the edges arising from the
truncation are $\frac{\pi}{2}$.

######  Remark 3.4.

An important feature of the truncation of a proper polyhedron $P$ is that it
determines $P$ (once we know which faces of $P$ are truncation faces), since
it is enough to remove the truncation faces to undo the truncation (see Figure
10).

$\longrightarrow$

Figure 10: Removing the truncation faces recovers the original polyhedron.

We are always going to consider _face marked_ polyhedra; this means that each
face of a polyhedron is uniquely determined, and therefore they never have any
symmetry.

######  Remark 3.5.

If $\Gamma$ is the $1$-skeleton of a projective polyhedron, then it is
$3$-connected (that is to say, it cannot be disconnected by removing two non-
adiacent vertices). Furthermore, any $3$-connected planar graph is the
$1$-skeleton of a proper polyhedron [Ste22]. If a planar graph is
$3$-connected, then it admits a unique embedding in $S^{2}$ (up to isotopies
of $S^{2}$ and mirror symmetry) [Fle73, Corollary 3.4]. Hence when in the
following we consider a planar graph, it is always going to be $3$-connected
and embedded in $S^{2}$. In particular, it will make sense to talk about the
dual of $\Gamma$, denoted with $\Gamma^{*}$. The graph $\Gamma^{*}$ is the
$1$-skeleton of the cellular decomposition of $S^{2}$ dual to that of
$\Gamma$.

######  Definition 3.6.

Let $\Gamma$ be a planar $3$-connected graph; the space of all the face-marked
proper polyhedra with $1$-skeleton $\Gamma$ considered up to isometry (i.e.
projective transformations preserving the unit sphere) is denoted as
$\mathcal{A}_{\Gamma}$.

######  Remark 3.7.

It is important not to mix up the $1$-skeleton of a projective polyhedron with
the $1$-skeleton of its truncation. In what follows, whenever we refer to
$1$-skeletons we always refer to those of projective polyhedra (and not their
truncation) unless specified.

We propose the following formulation of the volume conjecture, generalizing
the previously mentioned versions.

######  Conjecture 2 (The volume conjecture for polyhedra).

Let $P$ be a proper polyhedron with dihedral angles
$\alpha_{1},\dots,\alpha_{m}$ at the edges $e_{1},\dots,e_{m}$, and
$1$-skeleton $\Gamma$. Let $col_{r}$ be a sequence of $r$-admissible colorings
of the edges $e_{1},\dots,e_{m}$ of $\Gamma$ such that

$2\pi\lim_{r\rightarrow+\infty}\frac{col_{r}(e_{i})}{r}=\pi-\alpha_{i}.$

Then

$\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left\lvert
Y_{r}(\Gamma,col_{r})\right\rvert=\mathrm{Vol}(P).$

######  Remark 3.8.

In the case where $P$ is a simple polyhedron in $\mathbb{H}^{3}$ (i.e. a
compact polyhedron with only trivalent vertices) this conjecture is the same
as the volume conjecture of Kolpakov-Murakami [KM18].

Conjecture 2 was verified in [CM] for tetrahedra with at least one hyperideal
vertex; we provide some further supporting numerical evidence for Conjecture 2
for some pyramids in the Appendix, and prove it for a large family of examples
in Proposition 4.9 and the subsequent remark (however, only for a single
sequence of colors).

###  3.2 The maximum volume conjecture

The Turaev-Viro invariant of the complement of a link $L\subseteq S^{3}$ is
related to the Jones polynomial of $L$ via a simple formula.

######  Proposition 3.9.

[DKY18, Theorem 1.1] For any link $L=L_{1}\sqcup\cdots\sqcup L_{k}\subseteq
S^{3}$,

$TV_{r}(S^{3}\setminus
L)=\frac{2^{k+1}\sin\left(2\pi/r\right)^{2}}{r}\sum_{col\in
I_{r}^{k}}\left|J^{L}_{col}(q)\right|^{2}$ (6)

This motivates us to give the following definition of the Turaev-Viro
invariant of a graph.

######  Definition 3.10.

Let $\Gamma\subseteq S^{3}$ be a planar graph with $e$ edges. We define the
Turaev-Viro invariant of $\Gamma$, in analogy with Propositions 3.9 and
2.10.4, as

$TV_{r}(\Gamma):=\sum_{col\in I_{r}^{e}}\lvert Y_{r}(\Gamma,col)\rvert.$

We now state a volume conjecture for polyhedra in the vein of Chen-Yang’s
volume conjecture.

######  Conjecture 3 (The Maximum Volume Conjecture).

Let $\Gamma\subseteq S^{3}$ be a $3$-connected planar graph. Then

$\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left(TV_{r}(\Gamma)\right)=\sup_{P\in\mathcal{A}_{\Gamma}}\mathrm{Vol}(P)$

where $r$ ranges across all odd natural numbers.

######  Remark 3.11.

If Conjecture 2 is true, then of course

$\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left(TV_{r}(\Gamma)\right)\geq\sup_{P\in\mathcal{A}_{\Gamma}}\mathrm{Vol}(P).$

However, there could be a sequence of colorings $col_{r}$ such that
$Y_{r}\left(\Gamma,col_{r}\right)$ grows faster than for any sequence
satisfying the hypotheses of Conjecture 2; therefore Conjecture 2 does not
imply the Maximum Volume Conjecture. Nevertheless we believe the Maximum
Volume Conjecture to be easier to prove than Conjecture 2, as it only concerns
the largest values for the volume and the Turaev-Viro invariants, and not
those of any particular geometric structure.

######  Remark 3.12.

It is proven in [Bel20] that
$\sup_{P\in\mathcal{A}_{\Gamma}}\mathrm{Vol}(P)=\mathrm{Vol}(\overline{\Gamma})$
where $\overline{\Gamma}$ is the rectification of $\Gamma$. The rectification
of $\Gamma$ is defined as the unique projective polyhedron with $1$-skeleton
$\Gamma$ and with every edge tangent to $\partial\mathbb{H}^{3}$. While
$\overline{\Gamma}$ is not a proper (or even generalized hyperbolic)
polyhedron, we can still speak of its truncation and its volume; for more
details see [Bel20, Section 3.4].

######  Theorem 4.8.

The Maximum Volume Conjecture is verified for any planar graph obtained from
the tetrahedron by applying any sequence of the following two moves:

  * •

blowing up a trivalent vertex (see Figure 1) or

  * •

triangulating a triangular face (see Figure 2).

This theorem will be proven in Section 4.

##  4 The Fourier Transform

In this section we prove Theorem 4.8. The first main tool used is a sharp
upper bound on the asymptotic growth of a $6j$-symbol [BDKY18].

######  Theorem 4.1.

For any $r$ and any $r$-admissible $6$-tuple
$n_{1},n_{2},n_{3},n_{4},n_{5},n_{6}$, we have

$\frac{2\pi}{r}\log\left|\Bigg{|}\begin{matrix}n_{1}&n_{2}&n_{3}\\\
n_{4}&n_{5}&n_{6}\end{matrix}\Bigg{|}_{q=e^{\frac{2\pi i}{r}}}\right|\leq
v_{8}+O\left(\frac{\log(r)}{r}\right).$

where $v_{8}\sim 3.66$ is the volume of the regular ideal right-angled
octahedron. Furthermore, this inequality is sharp, with the upper bound
achieved at the $6$-tuple $\frac{r-2\pm 1}{2},\dots,\frac{r-2\pm 1}{2}$ with
the signs chosen so that $\frac{r-2\pm 1}{2}$ is even.

The second main tool used to prove Theorem 4.8 is the Fourier Transform
introduced in [Bar03] by Barrett. We describe it here in a slightly different
context and notation.

Figure 11: The $0$-framed Hopf link

Let $H\subseteq S^{3}$ be the $0$-framed Hopf link as in figure 11. If $i,j\in
I_{r}$ we denote with $H(i,j)\in\mathbb{C}$ the value of the Kauffman bracket
of the Hopf link colored with $i,j$; an easy induction on $j$ shows that

$\displaystyle
H(i,j)=(-1)^{i+j}[(i+1)(j+1)]=(-1)^{i+j}\frac{\sin\left(\frac{2\pi}{r}\left(i+1\right)\left(j+1\right)\right)}{\sin\left(\frac{2\pi}{r}\right)}.$

Furthermore denote with

$N:=\frac{r}{4\sin^{2}\left(\frac{2\pi}{r}\right)}=\langle
U,\Omega\rangle^{2}=\left(\sum_{i\in I_{r}}\Delta_{i}^{2}\right)^{2}$

where $U$ is the $0$-framed unknot in $S^{3}$ colored with the color
$\Omega:=\sum\limits_{i\in I_{r}}\Delta_{i}i$.

######  Remark 4.2.

Once again we remark that we are using the $SO(3)$ version of the invariants
evaluated at $q=e^{2\pi i/r}$. However, the Fourier transform and its
properties hold with any choice of primitive $2r$-th root of unity, or any
choice of primitive $4r$-th root of unity for the $SU(2)$ case; the proofs
work verbatim in every other case.

The following proposition was first noticed by Barrett in [Bar03]; a concise
proof was later given in [BFMGI07]. For the sake of completeness, we include a
detailed proof of this result.

######  Proposition 4.3.

If $\Gamma$ is a planar framed graph and $\Gamma^{*}$ is its planar dual, then

$Y_{r}(\Gamma^{*},col^{\prime})=N^{-g}\sum_{col\textrm{coloring of
}\Gamma}Y_{r}(\Gamma,col)H(col,col^{\prime})$

where

$H(col,col^{\prime}):=\prod_{e\textrm{ edge of
}\Gamma}H(col(e),col^{\prime}(e^{*})),$

and $g$ is the genus of a regular neighborhood of $\Gamma$.

###### Proof.

The proof is entirely diagrammatic; when we display an equality between
(linear combinations of) diagrams, we mean that they have the same Kauffman
bracket. Throughout the proof we will liberally add $\Omega$-colored,
$0$-framed unknots that are unlinked from anything else; this will generate an
ambiguity of a power of $N$ that we will account for at the end.

First we show that $Y_{r}(\Gamma,col)$ is equal to the Kauffman bracket of the
link $L$ obtained from $\Gamma$ as in Figure 12. Every vertex is replaced by a
circle colored with $\Omega$, and every edge is replaced by a circle colored
with the same color as the edge, wrapping around once each of the two circles
corresponding to its vertices.

$\xrightarrow{\hskip 22.76228pt}$ $\Omega$

Figure 12: The _Chainmail Rule._ Each circle has the same color as its
corresponding edge.

This can be shown by using the definition of $Y$ after applying the following
identity to $L$:

$\vbox{\hbox{\includegraphics{fusionrule.pdf}}}=\sum_{i\in
I_{r}}\Delta_{i}\vbox{\hbox{\includegraphics[height=85.35826pt]{fusionrule2.pdf}}}$
(7)

This holds for any number of strands; it is obtained by repeated application
of the fusion rule followed by the well known fact (see [Lic93, Lemma 6]) that
if a diagram contains the skein element in Figure 13 it is equal to $0$ unless
$i=0$.

$\Omega$$i$ Figure 13:

When passing from $\Gamma$ to $L$ we still speak of edges and vertices of $L$:
we mean the circles corresponding to edges and vertices of $\Gamma$
respectively. Slightly more improperly we speak of faces of $L$, by which we
mean the portions of the plane delimited by edges of $L$.

Now we apply the following relation (obtained from the connected sum formula
for the Kauffman bracket):

$\sum_{i\in
I_{r}}H(i,j)\vbox{\hbox{\includegraphics{fouriertransf2.pdf}}}=\vbox{\hbox{\includegraphics{fouriertransf1.pdf}}}$
(8)

to each component of $L$ that corresponds to an edge of $\Gamma$: this gives
us the Fourier transform of $Y_{r}(\Gamma,col)$. We call the meridional
circles added via this process the _transverse_ circles; they will correspond
to edges of $\Gamma^{*}$.

Figure 14: Stretching edges towards the center and adding an extra component.

Take a face $F$ of $L$ and stretch the circles transverse to its edges so that
they are close to the center of $F$ (see Figure 14). Add an unknot $U$ colored
with $\Omega$ at the center of $F$ and handleslide it along all the edges of
$F$; the result is that $U$ gets linked to each transversal circle and remains
unlinked from any edge or vertex of $\Gamma$ as in the left part of Figure 15.
The circle $U$ will correspond to a vertex in $\Gamma^{*}$. Repeat this
procedure for every face of $L$.

Now apply the inverse of the chainmail relation in Figure 12 to each circle
corresponding to a vertex of $\Gamma$ and each circle corresponding to a
vertex of $\Gamma^{*}$. The result is going to be $4$ unlinked graphs (and
several unlinked unknots that for now we ignore), $2$ of which give
$Y_{r}(\Gamma^{*},col^{\prime})$ and two of which give $Y_{r}(\Gamma,\Omega)$
(where we still denote with $\Omega$ the coloring of $\Gamma$ with color
$\Omega$ on each edge).

Figure 15: The central component gets linked by handleslides.

######  Lemma 4.4.

$Y_{r}(\Gamma,\Omega)=N^{g}$

###### Proof.

Figure 16: Bicycle with $3$ wheels

The Yokota invariant does not change when performing a Whitehead move on an
edge colored with $\Omega$ (Proposition 2.10.2). Therefore, we can change
$\Gamma$ to be a “bicycle” graph as in 16, with some circles connected
linearly by segments; since a Whitehead move does not change the genus of the
regular neighborhood, there are $g$ circles. Because of the bridge rule 5, the
Kauffman bracket is $0$ unless the colors of every connecting edge is $0$, and
therefore

$Y_{r}(\Gamma,\Omega)=\left(\sum_{i_{1},\dots,i_{g}\in
I_{r}}\Delta_{i_{1}}^{2}\cdots\Delta_{i_{g}}^{2}\right)=N^{g}$

∎

It only remains to check how many factors of $N$ are added or lost through
this procedure. At the beginning we added an unknot for each vertex of
$\Gamma$, and then for each face. However when we applied the inverse of the
chainmail relation we removed the exact same number of components; therefore
there is no additional $N$ factor.

∎

######  Proposition 4.5.

$\lim_{r\rightarrow\infty}\frac{TV_{r}(\Gamma)}{TV_{r}(\Gamma^{*})}=O(r^{n})$

for some $n\in\mathbb{N}$.

###### Proof.

Let $col_{max}$ be a coloring of $\Gamma$ such that $\lvert
Y_{r}(\Gamma,col)\rvert$ is maximum.

Thanks to Proposition 4.3

$\displaystyle\frac{TV_{r}(\Gamma^{*})}{TV_{r}(\Gamma)}\leq\frac{\sum_{col}\left\lvert
Y_{r}(\Gamma^{*},col)\right\rvert}{\left\lvert
Y_{r}(\Gamma,col_{max})\right\rvert}=N^{-g}\sum_{col}\left\lvert\sum_{col^{\prime}}H(col,col^{\prime})\frac{Y_{r}(\Gamma,col^{\prime})}{\left\lvert
Y_{r}(\Gamma,col_{max})\right\rvert}\right\rvert\leq$ $\displaystyle
N^{-g}\sum_{col,col^{\prime}}\lvert H(col,col^{\prime})\rvert$

and the latter is a sum of polynomially many polynomial terms. ∎

######  Corollary 4.6.

If $\lim\limits_{r\to+\infty}\frac{\pi}{r}\log\left(TV_{r}(\Gamma)\right)$
exists, then
$\lim\limits_{r\to+\infty}\frac{\pi}{r}\log\left(TV_{r}(\Gamma^{*})\right)$
also exists; moreover the two quantities are equal.

######  Corollary 4.7.

The Maximum Volume Conjecture is true for $\Gamma$ if and only if it is true
for $\Gamma^{*}$.

###### Proof.

Corollary 4.6 of [Bel20] states that the maximum volume of $\Gamma$ is the
same as the maximum volume of $\Gamma^{*}$; this and Corollary 4.6 imply the
thesis.

∎

Before we go on to prove Theorem 4.8, we propose a slightly sharper
conjecture.

######  Conjecture 4.

If $\Gamma$ is a planar $3$-connected graph and $col$ is any $r$-admissible
coloring of its edges, then

$\frac{\pi}{r}\log\left\lvert
Y_{r}(\Gamma,col)\right\rvert\leq\mathrm{Vol}(\overline{\Gamma})+O\left(\frac{\log(r)}{r}\right)$

Moreover, the inequality is sharp, with equality attained by the sequence of
colorings giving the color $\frac{r-2\pm 1}{2}$ to each edge (the sign is
chosen so that the colors are even).

This is clearly a direct generalization of Theorem 4.1. It is straightforward
to show that Conjecture 4 implies the Maximum volume conjecture, since
$TV_{r}$ is a sum of polynomially many terms of the type $Y_{r}(\Gamma,col)$.
However it is slightly more precise since it specifies the error term and
which colorings give the maximum growth.

######  Theorem 4.8.

If $\Gamma$ is obtained from the tetrahedron by a sequence of blow-ups of
vertices or triangulations of triangular faces, then Conjecture 4 is verified.

###### Proof.

The proof is by induction on the number $g$ of blow-ups or triangulations
needed to construct $\Gamma$. Notice that if $\Gamma$ is obtained from
$\Gamma^{\prime}$ via a blow-up, $\Gamma^{*}$ is obtained from
$(\Gamma^{\prime})^{*}$ by a triangulation of a triangular face, and vice-
versa.

We first prove that $\mathrm{Vol}(\overline{\Gamma})=(g+1)v_{8}$. The case of
$g=0$ is well known and appears in [Ush06]. Take now any $\Gamma$ obtained
from $\Gamma^{\prime}$ by a blow-up of a vertex $v$; we can take the
rectification $\overline{\Gamma^{\prime}}$ and glue a right-angled ideal
octahedron to the face corresponding to $v$. Notice that the gluing is done
along an ideal triangular face, and along right dihedral angles. It is
immediate to see that this gluing gives the truncation of $\overline{\Gamma}$:
the $1$-skeleton is the same and there are only right angles. Therefore, by
blowing up a vertex the maximum volume grows by $v_{8}$. Dually, triangulating
a triangular face makes the maximum volume grow by $v_{8}$ as well.

We now prove that

$\frac{\pi}{r}\log\left\lvert
Y_{r}(\Gamma,col)\right\rvert\leq(g+1)v_{8}+O\left(\frac{\log(r)}{r}\right).$

The base case $g=0$ is Theorem 4.1.

If $\Gamma$ is obtained from $\Gamma^{\prime}$ as a blow-up of a single
vertex, then

$Y_{r}(\Gamma,col)=Y_{r}(\Gamma^{\prime},col_{1})Y_{r}(T,col_{2})$

where $T$ is a tetrahedron, and $col_{1},col_{2}$ are the colorings induced by
$col$ on $\Gamma^{\prime}$ and $T$ respectively. Therefore,
$Y_{r}(\Gamma,col)\leq Y_{r}(\Gamma^{\prime},col_{1})Y_{r}(T,col_{2})$ and by
induction

$\frac{\pi}{r}\log\left\lvert
Y_{r}(\Gamma,col)\right\rvert\leq(g+1)v_{8}+O\left(\frac{\log(r)}{r}\right).$

By duality, this inequality also holds if $\Gamma$ is obtained from
$\Gamma^{\prime}$ by triangulating a single triangular face.

The sharpness of the upper bound is proven in the following proposition.

######  Proposition 4.9.

If $\Gamma$ is as above and $col=(\frac{r-2\pm 1}{2},\dots,\frac{r-2\pm
1}{2})$ (where the signs are chosen so that $r-2\pm 1$ is a multiple of $4$),
then

$\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left(Y_{r}(\Gamma,col)\right)=(g+1)v_{8}.$

###### Proof.

The proof is once again by induction; the base case is Theorem 4.1. Suppose
$\Gamma$ is obtained from the tetrahedron by $g$ blow-ups and triangulations,
and at least $1$ blow-up. Then, $\Gamma$ is a vertex sum of $\Gamma_{1}$ and
$\Gamma_{2}$, with both graphs obtained from the tetrahedron via $g_{1}$ and
$g_{2}$ blow-ups or triangulations respectively, and $g_{1}+g_{2}=g-1$. Since
$Y_{r}(\Gamma,col)=Y_{r}(\Gamma_{1},col_{1})Y_{r}(\Gamma_{2},col_{2})$ (with
$col_{1},col_{2}$ the colorings induced by $col$ on $\Gamma_{1},\Gamma_{2}$
respectively), we have

$\displaystyle\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left(Y_{r}(\Gamma,col)\right)$
$\displaystyle=\lim_{r\rightarrow+\infty}\frac{\pi}{r}\log\left(Y_{r}(\Gamma_{1},col_{1})Y_{r}(\Gamma_{2},col_{2})\right)=$
$\displaystyle=(g_{1}+1+g_{2}+1)v_{8}=(g+1)v_{8}.$

We need to deal with the case of $\Gamma$ being obtained via $g$
triangulations. In this case, $\Gamma^{*}$ is obtained from the tetrahedron
via $g$ blow-ups. Apply the Fourier transform to $Y_{r}(\Gamma,col)$:

$Y_{r}(\Gamma,col)=\sum_{col^{\prime}}H(col,col^{\prime})Y_{r}(\Gamma^{*},col^{\prime});$

however, since $col$ is constantly $\frac{r-2\pm 1}{2}$ and even,

$\displaystyle H\left(\frac{r-2\pm
1}{2},j\right)=(-1)^{j}\frac{\sin\left(\frac{2\pi}{r}\frac{r\pm
1}{2}(j+1)\right)}{\sin(2\pi/r)}=$
$\displaystyle(-1)^{j}\frac{\sin\left(\pi(j+1)\pm\frac{\pi}{r}(j+1)\right)}{\sin(2\pi/r)}=-\frac{\sin(\pm\frac{\pi}{r}(j+1))}{\sin(2\pi/r)}$

which has $\mp$ sign since $0\leq j\leq r-1$. Moreover, since $\Gamma^{*}$ is
a trivalent graph,
$Y_{r}(\Gamma^{*},col^{\prime})=\lvert\langle\Gamma^{*},col^{\prime}\rangle\rvert^{2}$
is non-negative for every coloring; therefore, $Y_{r}(\Gamma,col)$ is a sum
with constant sign of $Y_{r}(\Gamma^{*},col^{\prime})$ over all possible
colorings. This shows that $Y_{r}(\Gamma,col)$ grows as the maximum growth of
$Y_{r}(\Gamma^{*},col^{\prime})$ over all colorings, which is $(g+1)v_{8}$. ∎

∎

######  Remark 4.10.

Proposition 4.9 actually proves the Conjecture 3 for a large family of
polyhedra (albeit for a single sequence of colors each) since the volume of a
polyhedron with internal angles $0$ is the volume of its rectification (notice
how $2\pi\frac{r\pm 1-2}{2}\rightarrow\pi$). Moreover, because of 2.12,
Conjecture 3 is verified (for the sequence above) for any graph obtained from
the tetrahedron via blow-ups, triangulations and doubles.

##  5 The Turaev-Viro volume conjecture

In this section we apply Theorem 4.8 to prove the Turaev-Viro volume
conjecture for an infinite family of examples.

######  Proposition 5.1.

Let $\Gamma\subseteq S^{3}$ be a graph obtained from the tetrahedron by a
sequence of $g-1$ blow-ups of vertices or triangulations of triangular faces
as in the hypothesis of Theorem 4.8; let $e_{1},\dots,e_{k}$ be its edges, and
denote with $h$ the number of vertices of $\Gamma$. Then there is a
$k$-component link $L=L_{1}\sqcup\cdots\sqcup L_{k}$ in
$S^{3}\\#^{h-1}\left(S^{1}\times S^{2}\right)$ such that for any $col\in
I_{r}^{k}$ coloring (seen both as a coloring of $\Gamma$ and as a coloring of
$L$) we have

$Y_{r}(\Gamma,col)=\left(\frac{2}{\sqrt{r}}\sin(2\pi/r)\right)^{h}RT_{r}(S^{3}\\#^{h-1}\left(S^{1}\times
S^{2}\right),L,col)$

###### Proof.

We have seen in the proof of Proposition 4.3 that there is a way to associate
to any $\left(\Gamma,col\right)$ a skein element $\mathcal{L}$ in
$\mathcal{S}(S^{3})$ such that $Y_{r}(\Gamma,col)=\langle\mathcal{L}\rangle$.
The skein $\mathcal{L}$ is a link with $k+h$ components; $k$ of these
components are in bijection with the edges of $\Gamma$ and are colored with
the corresponding color of $col$. The other $h$ are unknotted components in
bijection with the vertices of $\Gamma$ and are colored with $\Omega$. Pick a
component of $\mathcal{L}$ colored with $\Omega$: it is possible to
handleslide it along each other $\Omega$-colored component without modifying
the Kauffman bracket. After it is handleslid along each component, it becomes
unlinked from everything, therefore $\langle\mathcal{L}\rangle=\langle
U\rangle\langle\mathcal{L}^{\prime}\rangle=\left(\frac{2}{\sqrt{r}}\sin(2\pi/r)\right)\langle\mathcal{L}^{\prime}\rangle$
where $U$ is an unknotted, unlinked component colored with $\Omega$ and
$\mathcal{L}^{\prime}$ is the remaining part of the skein. By the definition
of the Reshetikhin-Turaev invariant of links

$\langle\mathcal{L}^{\prime}\rangle=\left(\frac{2}{\sqrt{r}}\sin(2\pi/r)\right)^{h-1}RT_{r}(S^{3}\\#^{h-1}\left(S^{1}\times
S^{2}\right),L,col)$

where $L$ is the link obtained from $\mathcal{L}^{\prime}$ by doing a
$0$-framed Dehn surgery on the components of $\mathcal{L}$ colored with
$\Omega$. Notice that $L$ only depends on $\Gamma$ and not on the coloring. ∎

As we did previously, when writing the formulas we drop the factor
$\left(\frac{2}{\sqrt{r}}\sin(2\pi/r)\right)^{h-1}$ to simplify the notation;
since this is a factor that grows polynomially in $r$, dropping it is
inconsequential when proving the volume conjecture.

Figure 17: The building block: a ball with $4$ disks in its boundary, and $6$
arcs connecting them.

######  Proposition 5.2.

The link $L$ obtained from $\Gamma$ the construction in the proof of
Proposition 5.1 is hyperbolic, and its hyperbolic structure is obtained by
gluing $4g$ right-angled hyperbolic ideal octahedra.

###### Proof.

Let $\overline{\Gamma}$ be the rectification of $\Gamma$, and let $P$ be its
truncation. We have seen in the proof of Theorem 4.8 that $P$ can be obtained
by gluing $g$ right-angled hyperbolic octahedra. Take two copies of $P$ and
glue them along each corresponding truncation face. This gives a manifold
homeomorphic to a handlebody of genus $h-1$ with some annuli removed from the
boundary; the decomposition into octahedra makes it into a finite volume
manifold $M$ with geodesic boundary. Take the double of $M$ along the geodesic
boundary: this gives a manifold $N$ which is homeomorphic to
$S^{3}\\#^{h-1}\left(S^{1}\times S^{2}\right)\setminus L$. To see this, take
the octahedron $O$ and truncate a small link of each of its vertices. This
truncation can be seen as the basic building block of the fundamental shadow
links (see Figure 17): each truncated vertex corresponds to an arc, four of
the faces of the octahedron correspond to the discs and the remaining four
faces correspond to the regions of the spheres delimited by the arcs. The
polyhedron $P$ is obtained by gluing octahedra together; glue the building
blocks in the same pattern to obtain a ball with $h$ discs on its boundary and
some arcs connecting the discs. If we take the double of this ball along the
discs we obtain a genus $h-1$ handlebody with a link in its boundary. Doubling
the handlebody gives $S^{3}\\#^{h}(S^{1}\times S^{2})$ and the link in the
boundary is exactly $L$. ∎

######  Proposition 5.3.

Let $\Gamma$ be a graph as in the hypotheses of Proposition 5.1; let $t$ be
the maximal number of disjoint triangular faces in the truncation of
$\overline{\Gamma}$. Let $L$ be the link associated to $\Gamma$ by the
construction of Proposition 5.1, and $E_{L}$ be its exterior. Then $E_{L}$
contains at most $t+2g-2$ disjoint thrice-punctured spheres.

###### Proof.

The reasoning in this proof is similar to the proof of [CFMP07, Proposition
3.4].

Let $P$ be the truncation of $\overline{\Gamma}$; we have seen that $E_{L}$ is
obtained by doubling $P$ along the truncation faces (to obtain a hyperbolic
manifold with geodesic boundary $H$) and doubling again along the geodesic
boundary.

The truncation faces of $P$ can be colored with black and the remaining with
white; this way two faces of the same color never share an edge.

As we have seen $E_{L}$ decomposes into octahedra; take $O$ an octahedron in
this decomposition, and let $S$ be any thrice-punctured sphere.

Figure 18: The $6$ geodesic in a thrice punctured sphere cutting it into
triangles.

Claim: $S\cap O$ is either the empty set or a facet of $O$.

Figure 19: A square arising as the intersection of a thrice-punctured sphere
and an octahedron of $E_{L}$.

We first look at $S\cap O$ as a subset of $S$. It must be a convex region of
$S$ delimited by geodesics. Since $S$ contains exactly $6$ closed geodesics
the possible configurations are easy to list. Figure 18 shows the $6$
geodesics cutting $S$ into triangles; the possibilities for $S\cap O$ can be
obtained by looking at all the possible ways to glue these triangles to obtain
a convex set. The convex subsets of $S$ obtained by gluing triangle regions
are:

  1. 1.

a triangle with $1$ ideal vertex (obtained by taking a single triangle
region);

  2. 2.

a triangle with $2$ ideal vertices (obtained by gluing two triangle regions
without an ideal vertex in common);

  3. 3.

a square with $1$ ideal vertex and $2$ right angles (obtained by gluing two
triangle regions with an ideal vertex in common);

  4. 4.

a triangle with $2$ ideal vertices and a right angle (obtained by gluing a
triangle region to the triangle in $2$);

  5. 5.

a square with $2$ ideal vertices (obtained by gluing two triangles in $2$
along a common geodesic side);

  6. 6.

a bigon with $1$ ideal point in its interior (obtained by gluing all triangle
regions sharing an ideal vertex);

  7. 7.

a bigon with $1$ ideal point in its boundary (obtained by gluing two triangle
regions that have all the edges on the same geodesics);

  8. 8.

a region with $3$ ideal points (obtained in several possible ways).

Every other possible way of gluing together the triangle regions of Figure 18
does not give a convex subset.

On the other hand, $S\cap O$ as a subset of $O$ must coincide with the
intersection of $O$ with a plane $\Pi\subseteq\mathbb{H}^{3}$; therefore it
cannot be either a bigon with an ideal point. Moreover, $\Pi\cap O$ cannot be
a triangle with one or two ideal vertices, nor can it be a square with one
ideal vertex and two right angles. The remaining possibilities are that it is
a vertex, an edge, a region with $3$ ideal points, or a square with two ideal
vertices (see Figure 19). However by construction $O$ is glued to at least
three octahedra which are different from $O$ and each other; therefore the
case of a square with two ideal vertices is impossible since the intersection
of $S$ with these octahedra must also be a square with $2$ ideal vertices,
which would contradict the fact that $S$ is a thrice-punctured sphere. Finally
the only properly embedded, totally geodesic surface with exactly $3$ ideal
points in $O$ must be a triangular face.

Let $\mathcal{S}$ be a set of disjoint thrice-punctured spheres. This
determines a set of disjoint ideal triangles in each of the four copies of $P$
that make up $E_{L}$; some of them are in the boundary of a polyhedron while
some of them are properly embedded. Each polyhedron contains exactly $g-1$
disjoint properly embedded geodesic triangles (the ones that decompose $P$
into octahedra). These glue up to give $2g-2$ disjoint thrice-punctured
spheres in $E_{L}$. Furthermore, a disjoint collection $T_{1},\dots,T_{t}$ of
triangles in $\partial P$ induces a set of disjoint thrice-punctured spheres.
Therefore, there are at most $2g-2+t$ disjoint thrice-punctured spheres in
$E_{L}$.

∎

######  Remark 5.4.

If $M$ is the exterior of a fundamental shadow link with volume $2nv_{8}$,
then it contains exactly $2n$ disjoint thrice-punctured spheres. This can be
used to show that some of the exterior of the links provided by Proposition
5.1 are not fundamental shadow links; the simplest such example is the link
associated to the graph shown in Figure 20. An easy check shows that the
truncation of $\overline{\Gamma}$ contains at most $6$ disjoint triangular
faces, which means that $E_{L}$ contains at most $10$ thrice-punctured
spheres; on the other hand a fundamental shadow link with the same volume as
$E_{L}$ must contain $12$ such spheres.

More in general, if $\Gamma$ is obtained from the tetrahedron through at least
one triangulation and at least one blow-up, then the associated manifold is
not diffeomorphic to the exterior of a fundamental shadow link (and there is
at least one such manifold of volume $4nv_{8}$ for each $n>1$).

Figure 20: A graph whose link is not a fundamental shadow link.

######  Theorem 5.5.

Let $\Gamma$ be as in Proposition 5.1 and let $L\subseteq
S^{3}\\#^{h-1}\left(S^{1}\times S^{2}\right)$ obtained from $\Gamma$ by the
construction of Proposition 5.1. Then the Turaev-Viro volume conjecture holds
for the exterior of $L$.

###### Proof.

Theorem 4.8 implies that

$\frac{\pi}{r}\log\lvert RT_{r}(S^{3}\\#^{h-1}\left(S^{1}\times
S^{2}\right),L,col)\rvert=\frac{\pi}{r}\log\lvert Y_{r}(\Gamma,col)\rvert\leq
gv_{8}+O\left(\frac{\log(r)}{r}\right).$

Furthermore if we denote with $c$ the coloring $\left(\frac{r\pm
1}{2},\dots,\frac{r\pm 1}{2}\right)$ (where the sign is chosen so that the
color is always even), we have

$\frac{\pi}{r}\log\lvert RT_{r}(S^{3}\\#^{h-1}\left(S^{1}\times
S^{2}\right),L,c)\rvert=\frac{\pi}{r}\log\lvert
Y_{r}(\Gamma,c)\rvert=gv_{8}+O(\log(r)/r).$

If $E_{L}$ is the exterior of $L$,

$TV_{r}(E_{L})=\sum_{col\in I_{r}^{k}}\lvert
RT_{r}(S^{3}\\#^{h-1}\left(S^{1}\times S^{2}\right),L,col)\rvert^{2}$

by [BDKY18, Proposition 5.3], and $\mathrm{Vol}(E_{L})=4gv_{8}$ by Proposition
5.2, which implies the thesis since

$\lim_{r\rightarrow\infty}\frac{2\pi}{r}\log\left(TV_{r}(E_{L})\right)=4gv_{8}$

because the sum in the formula for $TV_{r}(E_{L})$ has polynomially many terms
all with the same sign. ∎

######  Remark 5.6.

There is an overlap between Theorem 5.5 and Theorem 1.1 of [BDKY18]. Some
links of Theorem 5.5 are also Fundamental Shadow Links (FSL); namely, those
links corresponding to graphs obtained from the tetrahedron by blow-ups.
However as we have seen many others are not.

##  Appendix A Appendix: Numerical evidence for Conjecture 2

Supporting evidence for Conjecture 2 in the case of simple polyhedra can be
found in [KM18]. In this appendix we show numerical computations supporting
the conjecture for the square and pentagonal pyramids; all the calculations
are performed with the Mathematica software. The notebook is available at
https://sites.google.com/sns.it/giulio-belletti-homepage/research; all
calculations were performed on a Dell XPS 13 laptop.

_The ideal regular square pyramid._

By Bao-Bonahon ([BB02, Theorem 1]) there is a unique square pyramid such that
the angles at the base are $\frac{\pi}{4}$ and the vertical angles are
$\frac{\pi}{2}$. Such a pyramid is ideal and is maximally symmetric; it is
decomposed into two ideal tetrahedra with angles
$\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{2}$ hence its hyperbolic volume is
equal to $4\Lambda\left(\frac{\pi}{4}\right)=\frac{v_{8}}{2}\sim 1.83193$
(where $\Lambda$ is the Lobachevski function). Consider the coloring of Figure
21; it converges to the angles of the ideal pyramid in the sense of Conjecture
2.

$\left[\frac{r}{4}\right]$$\left[\frac{r}{4}\right]$$\left[\frac{r}{4}\right]$$\left[\frac{r}{4}\right]$$\left[\frac{3r}{16}\right]$$\left[\frac{3r}{16}\right]$$\left[\frac{3r}{16}\right]$$\left[\frac{3r}{16}\right]$
Figure 21: The coloring of a square pyramid associated to the ideal regular
pyramid

Its Yokota invariant is given by

$\sum_{i\in
I_{r}}\Delta_{i}\begin{vmatrix}\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&i\\\
\left[\frac{3r}{16}\right]&\left[\frac{3r}{16}\right]&\left[\frac{3r}{16}\right]\end{vmatrix}^{4}$

where $\left[x\right]$ is the rounding of $x$ to the nearest integer.

_The $0$-angled squared pyramid_

Because of the arguments of [Bel20], the square pyramid with every dihedral
angle equal to $0$ exists and attains the maximum volume of any square pyramid
(it is in fact the rectified pyramid). Its truncation is the right-angled
ideal square antiprism. The volume of a right-angled ideal antiprism with
$n$-gonal face is given by

$2n\left(\Lambda\left(\frac{\pi}{4}+\frac{\pi}{2n}\right)+\Lambda\left(\frac{\pi}{4}-\frac{\pi}{2n}\right)\right)$

and for $n=4$ this gives $\sim 6.02305$.

Color the pyramid with the color $\left[\frac{r}{4}\right]$ at every vertex;
this coloring converges to the angles of the rectified pyramid.

Its Yokota invariant is given by

$\sum_{i\in
I_{r}}\Delta_{i}\begin{vmatrix}\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&i\\\
\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]\end{vmatrix}^{4}$

and its growth is shown in the following graph.

_The ideal regular pentagonal pyramid._

As before there is a unique ideal pentagonal pyramid with vertical angles
$\frac{3\pi}{5}$ and base angles $\frac{\pi}{5}$; this pyramid is maximally
symmetric. We can decompose it into $3$ ideal tetrahedra, two with dihedral
angles $\frac{\pi}{5},\frac{\pi}{5},\frac{3\pi}{5}$ and the remaining with
dihedral angles $\frac{\pi}{5},\frac{2\pi}{5},\frac{2\pi}{5}$. Its volume then
is

$5\Lambda\left(\frac{\pi}{5}\right)+2\Lambda\left(\frac{2\pi}{5}\right)+\Lambda\left(\frac{3\pi}{5}\right)\sim
2.49339.$

$\left[\frac{r}{5}\right]$$\left[\frac{r}{5}\right]$$\left[\frac{r}{5}\right]$$\left[\frac{r}{5}\right]$$\left[\frac{r}{5}\right]$$\left[\frac{r}{10}\right]$$\left[\frac{r}{10}\right]$$\left[\frac{r}{10}\right]$$\left[\frac{r}{10}\right]$$\left[\frac{r}{10}\right]$
Figure 22: The coloring of the pentagonal pyramid corresponding to an ideal
regular pyramid

Consider the coloring in Figure 22, converging to the angles of the ideal
pyramid. Its Yokota invariant is

$\sum_{i,j\in
I_{r}}\Delta_{i}\Delta_{j}\left|\left(\begin{vmatrix}\left[\frac{r}{10}\right]&\left[\frac{r}{10}\right]&i\\\
\left[\frac{r}{5}\right]&\left[\frac{r}{5}\right]&\left[\frac{r}{5}\right]\end{vmatrix}\begin{vmatrix}\left[\frac{r}{10}\right]&\left[\frac{r}{10}\right]&j\\\
\left[\frac{r}{5}\right]&\left[\frac{r}{5}\right]&\left[\frac{r}{5}\right]\end{vmatrix}\begin{vmatrix}\left[\frac{r}{10}\right]&i&j\\\
\left[\frac{r}{5}\right]&\left[\frac{r}{5}\right]&\left[\frac{r}{5}\right]\end{vmatrix}\right)\right|^{2}.$

_The $0$-angled pentagonal pyramid_

The volume of the rectified pyramid is $\sim 8.13789$, and the corresponding
Yokota invariant is

$\sum_{i,j\in
I_{r}}\Delta_{i}\Delta_{j}\left|\left(\begin{vmatrix}\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&i\\\
\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]\end{vmatrix}\begin{vmatrix}\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&j\\\
\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]\end{vmatrix}\begin{vmatrix}\left[\frac{r}{4}\right]&i&j\\\
\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]&\left[\frac{r}{4}\right]\end{vmatrix}\right)\right|^{2}.$

Because of the greater range of the sum, it is considerably slower to compute
than the other examples; we were only able to arrive to level $r=321$, and the
Yokota invariant is within $4\%$ of the volume. However this is similar to the
error (at level $321$) in the previous examples.

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Giulio Belletti, Scuola Normale Superiore, Pisa, Italy

E-mail address<EMAIL_ADDRESS>