text:
\section{Introduction}
\label{sec.intro}

\subsection{Tetrahedral manifold s}
\label{sub.goal}

We call a cusped hyperbolic 3-manifold {\em tetrahedral} if it can be
decomposed into regular ideal tetrahedra. The combinatorial data of this
decomposition is captured in the {\em combinatorial tetrahedral tessellation} which can be defined simply as an
ideal triangulation where all edges have order 6. By Mostow rigidity, a
combinatorial tetrahedral tessellation \myRemoved{removed uniquely}
determines a tetrahedral manifold. However, there might be several non-isomorphic
\myChanged{(i.e., not related by just relabeling tetrahedra and vertices)}
combinatorial tetrahedral tessellation s yielding the same tetrahedral
manifold. That is why we introduce the two terms tetrahedral manifold{} and combinatorial tetrahedral tessellation{} to
distinguish whether we regard isometric or combinatorially isomorphic objects
as equivalent.

The tetrahedral manifold{} were also called {\em maximum volume} in~\cite{A,VMF,VTF1, VTF2}
because they are precisely the ones with maximal volume among all hyperbolic
manifolds with a fixed number of tetrahedra. Thus, they also appear at the
trailing ends of the \texttt{SnapPy} \cite{SnapPy} census manifolds sharing
the same
letter\footnote{The case of the letter \texttt{m} is exceptional because it
spans several number of tetrahedra for purely historic reasons.} (e.g.,
\texttt{m405} to \texttt{m412}, \texttt{s955} to
\texttt{s961}, \texttt{v3551}, \texttt{t12833} to \texttt{t12845},
\texttt{o9\_44249}). Moreover, the number of tetrahedra and the Matveev
complexity~\cite{Matveev} also coincides for these manifolds.

The census of tetrahedral manifold s illustrates a number of phenomena of arithmetic
hyperbolic manifolds including symmetries visible
in the canonical cell decomposition \myChanged{but hidden by} the combinatorial tetrahedral tessellation.
In particular, the
canonical cell decomposition might have non-tetrahedral cells.

Several manifolds that have played a key role in the development of hyperbolic
geometry are tetrahedral, e.g., the complements of the figure-eight knot, the
minimally twisted 5-chain link (which conjecturally is also the minimum
volume orientable hyperbolic manifold with 5 cusps) and the Thurston
congruence link.
The last two have the special property that their combinatorial tetrahedral tessellation{} is maximally
symmetric, i.e., any tetrahedron can be taken to any other tetrahedron in
every orientation-preserving configuration via a combinatorial isomorphism.
One of the authors has classified link complements with this special
property in previous work \cite{Goerner}.

We also construct several new links with tetrahedral complement.

\subsection{Our results and methods}
\label{sub.results}

Our main goals   (see \cite{Goerner:tetcensus} for the data) are the
creation of
\begin{enumerate}
\renewcommand{\theenumi}{\alph{enumi}}
\item \label{goal:ctt}
The census of combinatorial tetrahedral tessellation s up to 25 (orientable case), respectively, 21
(non-orientable case) tetrahedra.
\item \label{goal:homeoCtt}
The grouping by isometry type and the corresponding canonical cell
decompositions.\\
We ship this as a \texttt{Regina} \cite{Regina} file containing
triangulations in a
hierarchy reflecting the grouping.
\item \label{goal:tman}
The corresponding census of tetrahedral manifold s.\\
We ship this as a \texttt{SnapPy} census containing a representative
triangulation for each isometry type.
This census can be used just like any other \texttt{SnapPy} census.
\item \label{goal:morphisms}
The list of covering maps between the combinatorial tetrahedral tessellation s.
\end{enumerate}

For~(\ref{goal:ctt}), we use a new approach differing from the traditional
one that starts by enumerating 4-valent graphs used first by
Callahan-Hildebrand-Weeks~\cite{CHW} \myChanged{or variations of the
traditional approach such as by Burton and Pettersson \cite{Burton:edges}}.
The advantage of our new approach is
that it scales to a substantially higher number of tetrahedra because it
allows for early pruning of triangulations with edges of wrong order. We also
deploy isomorphism signatures to avoid recounting combinatorially isomorphic
triangulations. Recall that the isomorphism signature is an improvement
by Burton~\cite{burton:encode}  of the (non-canonical) dehydration sequences.
It is a complete invariant of the combinatorial isomorphism type of a
triangulation. Algorithms~~1 and~2 used for the enumeration of combinatorial tetrahedral tessellation s are
described  in Section~\ref{section2}.
Isomorphism signatures of orientable  combinatorial tetrahedral tessellation s with at most seven tetrahedra
are presented in Table~\ref{table:isomorph}.

For~(\ref{goal:homeoCtt}), we use a new invariant we call the
{\em isometry signature} (see Section~\ref{section3}). It is a complete
invariant of the isometry type
of a cusped hyperbolic 3-manifold. It is defined as the isomorphism
signature of the canonical retriangulation of the canonical cell
decomposition ~\cite{EP}. To compute it, we use exact arithmetic to certify
the canonical cell decomposition even when the cells are not tetrahedral,
expanding on work by Dunfield, Hoffman, Licata~\cite{DHL}.

For~(\ref{goal:morphisms}), we wrote a script that finds combinatorial
homomorphisms from a triangulation to another triangulation.

Several of the techniques here are new and can be generalized: The
{\em isometry
signature} is an invariant that is defined for any finite-volume cusped
hyperbolic
3-manifolds. It is a complete isometry invariant (and thus by Mostow
rigidity a complete homotopy
invariant) that can be effectively computed and,
in general, be certified whenever the manifold is orientable and the
canonical cell
decomposition contains only tetrahedral cells using
{\tt hikmot}~\cite{hikmot}.
We also provide an improvement of the code provided in \cite{DHL}
to certify
canonical triangulations that is simpler and generalizes to any number
of cusps.

\myRemoved{After reading Burton/Petterson's paper more, I realized that
their approach shares more with the traditional one than with the one
taken here.}

Applying the above discussed methods we obtain the following result.

\begin{theorem}
The number of  combinatorial tetrahedral tessellation s and tetrahedral manifold s
up to 25 tetrahedra for orientable manifolds and up to 21 tetrahedra for
non-orientable manifolds are listed in Table~\ref{table:censusNumbers}.
\end{theorem}

\begin{table}
\caption{Number of triangulations in the census.\label{table:censusNumbers}}
\bigskip
\begin{tabular}{r||r|r||r|r||r}
& \multicolumn{2}{|c||}{combinatorial} & \multicolumn{2}{|c||}{tetrahedral}
& \multicolumn{1}{c}{homology} \\
& \multicolumn{2}{|c||}{tet. tessellations} & \multicolumn{2}{|c||}{manifolds}
& \multicolumn{1}{c}{links} \\
Tetrahedra & orientable & non-or. & orientable & non-or. & \\ \hline \hline
1 & 0 & 1 & 0 & 1 & 0 \\ \hline
2 & 2 & 2 & 2 & 1 & 1 \\ \hline
3 & 0 & 1 & 0 & 1 & 0 \\ \hline
4 & 4 & 4 & 4 & 2 & 2 \\ \hline
5 & 2 & 12 & 2 & 8 & 0 \\ \hline
6 & 7 & 14 & 7 & 10 & 0 \\ \hline
7 & 1 & 1 & 1 & 1 & 0 \\ \hline
8 & 14 & 10 & 13 & 6 & 5 \\ \hline
9 & 1 & 6 & 1 & 6 & 0 \\ \hline
10 & 57 & 286 & 47 & 197 & 12 \\ \hline
11 & 0 & 17 & 0 & 17 & 0 \\ \hline
12 & 50 & 117 & 47 & 80 & 7 \\ \hline
13 & 3 &8 & 3 & 8 & 0 \\ \hline
14 & 58 &134 & 58 & 113 & 25 \\ \hline
15 & 91 & 975 & 81 & 822 & 0 \\ \hline
16 & 102 & 175 & 96 & 142 & 32 \\ \hline
17 & 8 & 52 & 8 & 52 & 0 \\ \hline
18 & 213 & 1118 & 199 & 810 & 66 \\ \hline
19 & 25 & 326 & 25 & 326 & 0 \\ \hline
20 & 1886 & 26320 & 1684 & 22340 & 209 \\ \hline
21 & 31 & 251 & 31 & 251 & 0 \\ \hline
22 & 390 &- & 381 & - & 148 \\ \hline
23 & 58 &- & 58 & - & 0 \\ \hline
24 & 1544 & - & 1465 & - & 378 \\ \hline
25 & 7563 & - & 7367 & - & 0
\end{tabular}
\end{table}

All  combinatorial tetrahedral tessellation s and tetrahedral manifold s indicated in Table~\ref{table:censusNumbers} are
enumerated in supplement files available in~\cite{Goerner:tetcensus}.

Knots and links with tetrahedral complement are shown in
Figures~\ref{f.tetlinks}, \ref{f.tetlinks12} and \ref{fig.remarkable}.

\subsection{Features of the tetrahedral census}

Properties of tetrahedral manifolds that make them
interesting to study include:

\begin{itemize}
\item
\myChanged{The tetrahedral manifolds are arithmetic as they are a proper
subset of the commensurability class of figure-eight knot complement,
closed under finite coverings, see Section~\ref{sec:Margulis}.}

\myComment{
All tetrahedral manifolds are arithmetic and commensurable with the
figure-eight
knot complement, see Lemma~\ref{lem.arithmetic}. The converse is not true:
In Remark~\ref{rem.nottrue}, we list examples of
manifolds commensurable with the figure-eight knot which are not tetrahedral.

\item
The tetrahedral census is closed under finite (possibly irregular) coverings.
We investigate the resulting category in Section~\ref{section:cttCat}.
}

\item
The tetrahedral manifolds are exactly those with maximal volume among all
cusped hyperbolic manifolds with a fixed number of tetrahedra.
\item
Their Matveev complexity equals the number of regular ideal tetrahedra.

\item
Many combinatorial tetrahedral tessellation s \myChanged{hide symmetries, i.e, there are isometries of the
corresponding tetrahedral manifold{} that are not induced from a combinatorial
isomorphism of the combinatorial tetrahedral tessellation.}

\item
A substantial fraction of tetrahedral manifolds are link complements.
\end{itemize}

\section{The enumeration of combinatorial tetrahedral tessellation s}
\label{section2}

\begin{algorithm}
\SetKwProg{Proc}{Procedure}{}{end}%
\SetKwProg{Fn}{Function}{}{end}%
\SetKwFunction{Search}{RecursiveFind}%
\SetKwFunction{Main}{FindAllTetrahedralTessellations}%
\Fn{\Main{integer max, bool orientable}}{
\KwResult{Returns all (non-)orientable tetrahedral tessellations up to
combinatorial isomorphism with at most \ArgSty{max} tetrahedra.}
~

result $\leftarrow \{\}$ \tcc*{resulting triangulations}
already\_seen $\leftarrow \{\}$ \tcc*{isomorphism signatures encountered
earlier}
~

\Proc{\Search{Triangulation $t$}}{
\KwResult{Searches all triangulations obtained from $t$ by gluing faces
or adding tetrahedra.}
~

\tcc{Close order 6 edges and reject unsuitable triangulations}
\If{ \FuncSty{FixEdges(}$t$\FuncSty{)} = ``valid'' }{
\tcc{Skip triangulations already seen earlier}
\If{ \FuncSty{isomorphismSignature(}$t$\FuncSty{)} $\not\in$ already\_seen}{
already\_seen $\leftarrow$ already\_seen $\cup$
\{\FuncSty{isomorphismSignature(}$t$\FuncSty{)}\}\;

\If{ t has no open faces }{
\tcc{t orientable by construction if orientable = true}
\If{t is non-orientable or orientable = true}{
result $\leftarrow$ result $\cup ~\{t\}$\;
}
}
\Else{
\tcc{This choice results in faster enumeration}
choose an open face $F_1$=(tetrahedron, $f_1$) of $t$ adjacent to an
open edge of highest order\;
\If{t has less than max tetrahedra}{
\FuncSty{ResursiveFind(}\ArgSty{t} with a new tetrahedron glued to $F_1$
via an odd permutation\FuncSty{)}
}

\For{each open face $F_2 \not=F_1$ of t}{
\For{each $p \in S_4$}{
\If{$p(f_1)=f_2$}{
\If{p is odd or orientable = false}{
\FuncSty{RecursiveFind(}$t$ with $F_1$ glued to $F_2$ via $p$\FuncSty{)}\;
}
}
}
}
}
}
}
}
\FuncSty{RecursiveFind(}triangulation with one unglued tetrahedron\FuncSty{)}\;
\KwRet{result}
}
\caption{The main function to enumerate all tetrahedral tessellations.}
\label{algo:main}
\end{algorithm}

\begin{algorithm}
\SetKwProg{Fn}{Function}{}{end}
\SetKwFunction{FRecurs}{FixEdges}%
\Fn{\FRecurs{Triangulation t}}{
\KwResult{t is modified in place. Returns``valid'' or ``invalid''.}
\While{t has open edge e of order 6}{
close edge $e$\;
}
\KwRet{``valid'' if every edge e
\begin{itemize}\vspace{-1mm}\setlength{\itemsep}{0pt}\item has order $<6$
(if open) or $=6$ (if closed) and\item has no projective plane as vertex
link.\end{itemize}}
}
\caption{A helper function closing order 6 edges and rejecting
triangulations which cannot result in tetrahedral tessellations.}
\label{algo:fixEdges}
\end{algorithm}

We use Algorithm~\ref{algo:main} to enumerate the combinatorial tetrahedral tessellation s. The input is the
maximal number of tetrahedra to be considered and a flag indicating whether we
wish to enumerate the orientable or the non-orientable tessellations. The
result is a set of ideal triangulations where each edge has order 6
resulting in manifolds of the desired orientability.

As pointed out in the introduction our algorithm differs from the
traditional approach:
we recursively try all possible ways open faces can be face-paired without
enumerating 4-valent graphs first. This will, of course, result in many
duplicates, so
we keep a set of isomorphism signatures (see \cite{burton:encode}) of
previously encountered triangulations
around to prevent recounting. Recall that an isomorphism signature is,
unlike a dehydration sequence, a complete invariant of the combinatorial
isomorphism type of a triangulation.

The advantage of this approach is that we can insert a procedure that can
prune the
search space early on. In our case, this procedure is given in
Algorithm~\ref{algo:fixEdges} and rejects ideal triangulations where edges
have the wrong
order. It also rejects ideal triangulations with non-manifold topology.
These can occur
when the tetrahedra around an edge cannot be oriented consistently and the
vertex link of
the center of the edge becomes a projective plane $\mathbb{RP}^2$.

The algorithm has been implemented using \texttt{Regina} and we briefly
recall how a triangulation is presented. The vertices of each tetrahedron
are indexed 0, 1, 2, 3 and the faces are indexed by the number of the vertex
opposite  to it. Triangulations in intermediate stages will have unpaired
faces. We call a face open if it is unpaired, otherwise closed.
A triangulation consists of a number of tetrahedra and for each
tetrahedron $T_1$ and each face index $f_1=0, ..., 3$, we store two pieces
of data to encode whether and how the face $F_1=(T_1,f_1)$ is glued to
another face $F_2=(T_2,f_2)$ with face index $f_2$ of another
(not necessarily distinct) tetrahedron $T_2$:
\begin{enumerate}
\item A pointer to $T_2$. If $F_1$ is an open face, this pointer is null.
\item An element $p\in S_4$ such that $p(f_1) = f_2$ and the vertex
$i\not =f_1$ of $T_1$ is glued to $p(i)$ of $T_2$.
\end{enumerate}

The face pairings implicitly determine edge classes. We call such an edge
open if it is adjacent to an open face (necessarily so exactly two) and
otherwise closed. Closing an open edge means gluing the two open adjacent
faces by the suitable permutation.

The source for the implementation is in
\texttt{src/genIsomoSigsOfTetrahedralTessellations.cpp}. See
\cite[\texttt{data/}]{Goerner:tetcensus} for isomorphism signatures of all
combinatorial tetrahedral tessellation s from Table~\ref{table:censusNumbers} and Table~\ref{table:isomorph}
for orientable tessellations with $n \leq 7$ tetrahedra. Also names of
manifolds in the census are presented (see Section~\ref{section4}).

\begin{table}[h]
\caption{Isomorphism signatures for all orientable combinatorial tetrahedral tessellation s with $n \leq 7$
tetrahedra. \label{table:isomorph}}
\bigskip
\begin{tabular}{r|l|l||r|l|l}
n & Signatures &Name & n & Signatures & Name \cr \hline
2 & \texttt{cPcbbbdxm} & $\otet02_{0000}$ & 6 & \texttt{gLLPQccdfeefqjsqqjj} & $\otet06_{0000}$ \cr \hline
2 & \texttt{cPcbbbiht} & $\otet02_{0001}$& 6 & \texttt{gLLPQccdfeffqjsqqsj} & $\otet06_{0001}$ \cr \hline
4 & \texttt{eLMkbbdddemdxi} & $\otet04_{0000}$ &  6 & \texttt{gLLPQceefeffpupuupa} & $\otet06_{0002}$ \cr \hline
4 & \texttt{eLMkbcddddedde} & $\otet04_{0001}$ & 6 & \texttt{gLMzQbcdefffhxqqxha} & $\otet06_{0003}$ \cr \hline
4 & \texttt{eLMkbcdddhxqdu} & $\otet04_{0002}$ & 6 & \texttt{gLMzQbcdefffhxqqxxq} & $\otet06_{0004}$ \cr \hline
4 & \texttt{eLMkbcdddhxqlm} & $\otet04_{0003}$ & 6 & \texttt{gLvQQadfedefjqqasjj} & $\otet06_{0005}$ \cr \hline
5 & \texttt{fLLQcbcedeeloxset} & $\otet05_{0000}$ & 6 & \texttt{gLvQQbefeeffedimipt} & $\otet06_{0006}$ \cr \hline
5 & \texttt{fLLQcbdeedemnamjp} & $\otet05_{0001}$ & 7 & \texttt{hLvAQkadfdgggfjxqnjnbw} & $\otet07_{0000}$ \cr
\end{tabular}
\end{table}

\section{The isometry signature}
\label{section3}

In the previous section, we enumerated all combinatorial tetrahedral tessellation s with a given maximal
number of tetrahedra up to combinatorial isomorphism. In the next step,
we want to find the equivalence classes of those combinatorial tetrahedral tessellation s yielding the same
tetrahedral manifold{} up to isometry.

We do this by grouping combinatorial tetrahedral tessellation s by their {\em isometry signature} which we
define, compute and certify in this section. To summarize, the isometry
signature is the isomorphism signature of the canonical retriangulation
of the canonical cell decomposition. If, however, the canonical cell
decomposition has simplices as cells, we short-circuit and just use the
isomorphism signature of the canonical cell decomposition itself.
We can certify the isometry signature by using exact computations to determine
which faces in the proto-canonical triangulation are transparent.

The code implementing the certified canonical retriangulation can be found in
\texttt{src/canonical\_o3.py}. The code to group (and name) the combinatorial tetrahedral tessellation s by
isometry signature
is in
\texttt{src/identifyAndNameIsometricIsomoSigsOfTetrahedralTessellations.py}.

\subsection{Definition} \label{sec:defCanonicalCell}

Recall that the hyperboloid model of 3-dimensional hyperbolic space $\mathbb H^3$
in (3+1)-Minkowski space (with inner product defined by
$\langle x, y\rangle = x_0y_0 + x_1y_1+x_2y_2-x_3y_3$) is given by
$$
S^+=\left\{x=(x_0,...,x_3) \,\, | \,\,
x_3>0, \quad \langle x,x\rangle = -1\right\}.
$$

For a cusped hyperbolic manifold $M$, choose a horotorus cusp neighborhood
of the same volume for each cusp. Lift $M$ and the cusp neighborhoods to
$\mathbb H^3\cong S^+$. The cusp neighborhoods lift to a $\pi_1(M)$-invariant set
of horoballs. For each horoball $B\subset S^+$, there is a dual
vector $v_B$ that is light-like (i.e., $\langle v_B, v_B \rangle =0$) and
such that $w\in B \Leftrightarrow \langle v_B, w\rangle > -1$.
The boundary of the convex hull of all $v_B$ has polygonal faces.

\begin{definition}
The {\em canonical cell decomposition} of $M$ is given by the radial
projection of the polygonal faces of the boundary of the convex hull of
all $v_B$ onto $S^+$.
\end{definition}

The canonical cell decomposition was introduced by Epstein and
Penner~\cite{EP}. It does not depend on a particular choice of cusp
neighborhoods as long as they all have the same volume, or equivalently, same
area.

\begin{definition}
A triangulation which is obtained by subdividing the cells of the canonical
cell decomposition and inserting (if necessary) flat tetrahedra is called
a {\em proto-canonical
triangulation}. If it contains no flat tetrahedra, i.e., all tetrahedra are
positively
oriented, it is called a {\em geometric proto-canonical triangulation}.
\end{definition}

The result of calling \texttt{canonize} on a \texttt{SnapPy} manifold is a
proto-canonical triangulation. If the canonical cell decomposition has
cells which are not ideal tetrahedra (non-regular or regular), there might
be more than
one proto-canonical triangulation of the same manifold. A face of a
proto-canonical triangulation which is part of a 2-cell of the canonical
cell decomposition is called {\em opaque}. Otherwise, a face is called
{\em transparent}.

\begin{definition}
Consider a 2-cell in the canonical cell decomposition which is an $n$-gon.
Pick
the suspension of such an $n$-gon by the centers of the two neighboring
3-cells.
These suspensions over all 2-cells form a decomposition of $M$ into
topological diamonds.
Each diamond
can be split into $n$ tetrahedra along its central axis. The result is
called the {\em canonical retriangulation}.
\end{definition}

The canonical retriangulation carries exactly the same information as the
canonical cell decomposition (just packaged as a triangulation) and thus
only depends on (and uniquely determines)
the isometry type of the manifold. \texttt{SnapPy} uses it internally to
compute, for example, the symmetry group of a hyperbolic manifold $M$ by
enumerating the combinatorial isomorphisms of the canonical retriangulation
of $M$. Similarly, \texttt{SnapPy} uses it to check whether two manifolds
are isometric.

\begin{definition}
The {\em isometry signature} of $M$ is the isomorphism signature of the
canonical retriangulation if the canonical cell decomposition has
non-simplicial cells. Otherwise, it is the isomorphism signature of the
canonical cell decomposition itself.
\end{definition}

\begin{example}
The triangulation of \texttt{m004} given in the \texttt{SnapPy} census
already is the canonical cell decomposition. Thus, the isometry signature
of the manifold \texttt{m004} is the isomorphism signature of the census
triangulation,
namely \texttt{cPcbbbiht} presented in Table~\ref{table:isomorph}. In the
census of tetrahedral hyperbolic manifolds \texttt{m004} named
$\otet02_{0001}$. Recall that this manifold is the figure-eight knot
complement.\\
The cell decomposition for  \texttt{m202} given in the \texttt{SnapPy}
census is not canonical. The isomorphism signature of its \texttt{SnapPy}
triangulation is \texttt{eLMkbbdddemdxi} presented in
Table~\ref{table:isomorph}. In the census of tetrahedral hyperbolic
manifolds  \texttt{m202} named $\otet04_{0000}$. Observe, that
$\otet04_{0000}$ is the complement of a 2-component link presented in
Figure~\ref{f.tetlinks}. The isometry signature of \texttt{m202} is
\texttt{jLLzzQQccdffihhiiqffofafoaa} that is realized by a triangulation
with ten tetrahedra.
\end{example}

\subsection{Computation of the tilt}

Consider an ideal triangulation ${\mathcal T} = \cup_{i} T_{i}$ of a cusped
manifold $M$ with a shape
assignment for each tetrahedron, i.e., a $z_i\in\mathbb C\setminus\{0,1\}$
determining an embedding of the tetrahedron $T_{i}$ as ideal tetrahedron in
$\mathbb H^3$ up to isometry. If the shapes fulfill the consistency equations
(also known as gluing equations) in
logarithmic form and have positive imaginary parts, we call the triangulation
together with the shape assignment a {\em geometric ideal triangulation}.
Thurston shows that a geometric ideal triangulation glues up to a complete
hyperbolic structure on $M$. Given a geometric ideal triangulation and a
face $F$ of it, the
tilt $\mathrm{Tilt}(F)$ is a real number defined by Weeks~\cite{Weeks}
which determines whether a given triangulation is proto-canonical and
which faces are transparent.

We now describe how to compute $\mathrm{Tilt}(F)$ following
the notation in \cite{DHL} and use it to determine the
canonical retriangulation.

\subsubsection{Computation of a cusp cross section}

The ideal tetrahedra intersect the boundary of a neighborhood of a cusp
in Euclidean triangles and we call the resulting assignment of lengths to
edges a {\em cusp cross
section}. We first compute
a cusp cross section $C_c$ for some neighborhood of each cusp $c$ by picking an
edge $e_j$ for each cusp and assigning length $e_j=1$ to it. We recursively
assign lengths to the other edges by using that the ratio of two edge
lengths
is given by the respective $|z_i^*|$ where $z_i^*$ is one of the edge
parameters $z_i, z'_i=\frac{1}{1-z_i}, z''_i=1-\frac{1}{z_i}$:
\begin{equation*}
e_l =e_k \cdot |z_i^*|. \label{eqn:EdgeRatioCusp}
\end{equation*}

\subsubsection{Computation of  the cusp area}

We can compute the area of each Euclidean triangle $t$ as
\begin{equation*}
A(t)=\frac{1}{2} e_k^{\myChanged{2}} \cdot \mathrm{Im}(z_i^*)
\end{equation*}
where $e_k$ and $z_i^*$ are as above. The cusp area $A(C_c)$ of the cusp cross
section $C_c$ is simply the sum of the areas $A(t)$ over all its Euclidean
triangles $t$.

\subsubsection{Normalization of the cusp area}

\myChanged{We need to scale} each cusp cross section \myChanged{to have the
same target area $A$.
The new edge lengths and areas are given by
\begin{equation*}
e'_l = e_l \cdot\sqrt{\frac{A}{A(C_c)}}\quad\mbox{and}
\quad A'(t)=A(t)\frac{A}{A(C_c)}.
\end{equation*}}

\subsubsection{Computation of the circumradius for each Euclidean triangle}

Let $R^i_v$ denote the circumradius of the Euclidean triangle $t$ that is
the cross section of the tetrahedron $i$ near vertex $v\in\{0,1,2,3\}$. If
$e'_j$, $e'_k$, and
$e'_l$ are the edge lengths of $t$, elementary trigonometry implies
\begin{equation*}
R^i_v = \frac{e'_j e'_k e'_l}{4 A'(t)}.
\end{equation*}

\subsubsection{Computation of the tilt of a vertex}

Compute
\begin{equation} \label{eqn:Tilt}
\mathrm{Tilt}(i, v) = R^i_v - \sum_{u\not=v} R^i_u\
\frac{\mathrm{Re}(z_i^*)}{|z_i^*|}
\end{equation}
where $z_i^*$ is the edge parameter for the edge from $u$ to $v$.

\subsubsection{Computation of the tilt of a face}

If the face $F$ opposite to vertex $v$ of tetrahedron $i$ is glued to that
opposite of
$v'$ of tetrahedron $i'$, the tilt of the face is defined as
\begin{equation*}
\mathrm{Tilt}(F) = \mathrm{Tilt}(i,v) + \mathrm{Tilt}(i',v').
\end{equation*}

\subsubsection{Determination of transparent faces and canonical
retriangulation}

Weeks proves that \cite{Weeks}
a geometric ideal triangulation is a geometric proto-canonical
triangulation if all
$\mathrm{Tilt}(F)\leq 0$. In that case, a face $F$ is transparent if and
only if
$\mathrm{Tilt}(F)=0$.

\texttt{SnapPy} implements an algorithm to compute the canonical
retriangulation.
It can be refactored so that it takes as input the opacities of the faces
and is
purely combinatorial. In case of a geometric (!) proto-canonical
triangulation,
Weeks' arguments in the \texttt{SnapPy} code prove that this
algorithm works correctly.

For all manifolds we encountered, several randomization trials were always
sufficient to ensure that the ideal triangulation returned by \texttt{SnapPy}'s
\texttt{canonize} is always geometric proto-canonical. Thus, the result of the
purely combinatorial canonical retriangulation algorithm is known to be correct
as long as we certify the input to be a geometric proto-canonical triangulation
with certified opacities of its faces.

\begin{remark}
Even though we can certify the results for all listed manifolds in the
tetrahedral census, it is not known if
\begin{itemize}
\item
every cusped hyperbolic manifold has a geometric proto-canonical triangulation,
\item
every cusped hyperbolic manifold has a geometric ideal triangulation.
\end{itemize}
Moreover, it is known that \texttt{SnapPy}'s
implementation can give the wrong canonical retriangulation if we use as
input a non-geometric (!)
proto-canonical triangulation. As pointed out by Burton, the triangulation
\texttt{x101} in the non-orientable cusped \texttt{SnapPy} census is such
an example
where flat tetrahedra cause \texttt{SnapPy} to give an incorrect canonical
retriangulation.\\
It is unclear to the authors which of the following factors contribute to
the incorrect result:
\begin{itemize}
\item Numerical precision issues.
\item \texttt{SnapPy}'s extension of the above definition of
$\mathrm{Tilt}(F)$ to flat tetrahedra (where some
$A(t)=0$ and thus $R^i_v=\infty$) using \texttt{CIRCUMRADIUS\_EPSILON}.
\item Week's arguments for the purely combinatorial part of the canonical
retriangulation algorithm seem to implicitly assume that there are no
flat-tetrahedra.
\end{itemize}
\myChanged{
The existence of geometric triangulations of a hyperbolic manifold can be
proven when some tetrahedra are allowed to be flat
 \cite{PetronioWeeks:partiallyFlatTrig}. It can also be proven virtually
\cite{LuoSchleimerTillmann:virtualGeometricTrig}.}
\end{remark}

\begin{figure}
\begin{center}
\includegraphics[height=3.8cm]{draws/tetInCube.pdf}
\end{center}
\caption{Subdivision of a cube into 5 tetrahedra. For a regular ideal
hyperbolic cube, all tetrahedra are again regular ideal. The subdivision
introduced additional diagonals on the faces.\label{fig:tetInCube}}
\end{figure}

\begin{remark} \label{remark:burtonPair}
Call a manifold that can be decomposed into regular ideal cubes
{\em cubical}. Recall that a regular ideal cube can be subdivided into 5
regular ideal tetrahedra, see Figure~\ref{fig:tetInCube}. However, this
does not imply that a cubical manifold is tetrahedral.\\
A counter-example is the manifold appearing in the census as
$\texttt{x101}$ and $\texttt{x103}$. Its canonical cell decomposition
consists of one regular ideal cube. As Burton explained~\cite{burton:x101},
$\texttt{x101}$ subdivides the cube into 5 regular ideal tetrahedra but
needs to insert a flat tetrahedron to match the diagonals on the cube.
Thus, it is not a tetrahedral manifold (but still has a tetrahedral
double-cover $\mathrm{ntet}{10}_{0093}).$\\
\texttt{x103} splits the same cube into 6 non-regular tetrahedra and is
a geometric proto-canonical triangulation.
\end{remark}

\subsection{Certification for tetrahedral manifolds}

Let $\sqrt{\mathbb Q^+}$ denote the multiplicative group of all square roots of
positive rational numbers and let $\mathbb Q(\sqrt{\mathbb Q^+})\subset \mathbb C$ be the
field generated by $\sqrt{\mathbb Q^+}$.

\begin{lemma}
If we pick as target area $A=\sqrt{3}$, we have
for a geometric proto-canonical triangulation of a tetrahedral manifold $M$:
$$
z_i^* \in \mathbb Q(\sqrt{-3});\qquad A(C_c)\in \mathbb Q^+\sqrt{3};
$$
$$
|z_i^*|, e_l, A(t), e'_l, A'(t), R^i_v \in \sqrt{\mathbb Q^+};\qquad
\mathrm{Tilt}(F)\in \mathbb Q(\sqrt{\mathbb Q^+}).
$$
\end{lemma}

\begin{proof}
$M$ and thus its universal cover can be decomposed into regular ideal
tetrahedra. The resulting regular tessellation in $\mathbb H^3$ can be chosen
to have vertices at $\mathbb Q(\sqrt{-3})$ (also see Section~\ref{sec.properties}),
thus the shapes of any ideal triangulation of $M$ are in $\mathbb Q(\sqrt{-3})$.\\
Develop a cusp cross section constructed above in $\mathbb C$ such that the
vertices of the edge set to length 1 are at 0 and 1. Then all vertices
have complex coordinates in $\mathbb Q(\sqrt{-3})$ and a fundamental domain in
$\mathbb C$ for the cusp is a parallelogram spanned by two complex numbers in
$\mathbb Q(\sqrt{-3})$.
The area $A(C_c)$ of such a parallelogram is in $\mathbb Q^+\sqrt{3}$.\\
The rest follows from the above formulas.
\end{proof}

We can represent a $z_i^*$ exactly by $r_1+r_2\sqrt{\myChanged{-}3}$ with
$r_1, r_2\in\mathbb Q$. We can represent the other quantities exactly using
Corollary \ref{cor.Galois}
below.

\begin{lemma}
\label{lem.Galois}
Let $p_1,\dots,p_r$ denote a list of distinct prime numbers and
$K=\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_r})$ denote the corresponding number field.
Then,
\begin{itemize}
\item[(a)]
$K/\mathbb Q$ is Galois with Galois group $G(K/\mathbb Q)=(\mathbb Z/2\mathbb Z)^r$.
\item[(b)]
If $\mathbb Q \subset L \subset K$ is a subfield of $K$ such that $[L:\mathbb Q]=2$
then $L=\mathbb Q(\sqrt{p_I})$ where $p_I=\prod_{i \in I} p_i$ for some nonempty
$I \subset \{1,\dots,r\}$.
\item[(c)]
The $\mathbb Q$-linear map
$$
\mathbb Q(\sqrt{p_1}) \otimes_\mathbb Q \mathbb Q(\sqrt{p_2}) \otimes_\mathbb Q \dots
\otimes_\mathbb Q \mathbb Q(\sqrt{p_r}) \to K, \qquad x_1 \otimes \dots \otimes x_r \mapsto
\prod_{i=1}^r x_i
$$
is an isomorphism of $\mathbb Q$-vector spaces.
\end{itemize}
\end{lemma}

\begin{proof}
We will prove this by induction on $r$. When $r=1$ (a) is obvious and (b)
follows from the fundamental theorem of Galois theory~\cite[VI,Thm.1.2]{Lang}.

Assume that the lemma is true for $r-1$, and let
$K_1=\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_{r-1}})$ and $K_2=\mathbb Q(\sqrt{p_r})$.
Then, we claim that $K_1 \cap K_2 =\mathbb Q$. Indeed, otherwise we have $K_2 \subset
K_1$ and by part (b) it follows that $\mathbb Q(\sqrt{p_r})=\mathbb Q(\sqrt{p_I})$
for some nonempty subset $I \subset \{1,\dots,r-1\}$. So,
$\sqrt{p_r}=a + b \sqrt{p_I}$ for $a,b \in \mathbb Q$. Squaring, we get
$$
p_r = a^2 + b^2 p_I, \qquad a b = 0.
$$
If $a=0$ then $p_r=b^2 p_I$ and since $I$ is nonempty, it follows that
$p_i^2$ divides $p_r$ where $p_i, p_r$ are distinct primes, a contradiction.
If $b=0$ then $p_r=a^2$ and $p_r$ is a prime number, also a contradiction.
This shows that $K_1 \cap K_2 =\mathbb Q$. Let
$K=K_1 K_2 =\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_{r}})$ denote the composite field.
It follows by \cite[VI,Thm.1.14]{Lang} that $K$ is a Galois extension
with Galois group $G(K/\mathbb Q)=G(K_1/\mathbb Q) \times G(K_2/\mathbb Q) = (\mathbb Z/2\mathbb Z)^r$.
This proves part (a) of the inductive part. Part (b) follows from part (a)
by the fundamental theorem of Galois theory \cite[VI,Thm.1.2]{Lang} and the
classification of all index 2 subgroups of $(\mathbb Z/2\mathbb Z)^r$.
Part (c) follows from part (a) and the induction hypothesis.
\end{proof}

Part (c) of Lemma \ref{lem.Galois} implies the following corollary.

\begin{corollary}
\label{cor.Galois}
Every element in $\mathbb Q(\sqrt{\mathbb Q^+})$ has a unique representative of the form
\begin{equation}
\label{eq.ri}
r_1\sqrt{n_1}+\dots+r_k \sqrt{n_k}
\end{equation}
where $r_i\in\mathbb Q\setminus 0$ and $n_1<\dots <n_k$ are square-free positive
integers.
\end{corollary}

\begin{remark}
\label{rem.Galois.effective}
For the purpose of effective exact computation, we need an explicit way of
adding, subtracting, multiplying and dividing
 expressions of the form ~\eqref{eq.ri}.
This is obvious except for
division where we give the following algorithm: To compute $n/d$ where $n$
and $d$ are two such forms and $d$ contains a non-rational term
$r_j\sqrt{n_j}$, pick a prime $p$ dividing $n_j$. We can write $d$ as
$d_0 + \sqrt{p} d_1$ such that $d_0$ contains no term $r_i\sqrt{n_i}$ with
$p | n_i$. We now have
$$
\frac{n}{d} = \frac{n (d_0-\sqrt{p} d_1)}{(d_0+\sqrt{p}d_1)(d_0-\sqrt{p} d_1)}
= \frac{n (d_0-\sqrt{p} d_1)}{d_0^2 - p d_1^2}.
$$
The new denominator is simpler because it contains no more terms
$r_i\sqrt{n_i}$ with $p | n_i$. Thus, by repeating this process we can
eliminate all primes in the terms of the denominator.
\end{remark}

When we say {\em using interval arithmetics}, we mean:
\begin{enumerate}
\item We convert the exact representation of each quantity in
$\mathbb Q(\sqrt{\mathbb Q^+})$, respectively, $\mathbb Q(\sqrt{-3})$ to an interval $[a,b]$,
respectively,  a complex interval $[a,b] + [a',b'] i$. These intervals have
interval semantics: the true value of the quantity is guaranteed to be
contained in the interval.
\item Any operations such as + or $\log$ are carried out such that interval
semantics is preserved, i.e., the resulting interval is again guaranteed to
contain the true value of the computed quantity.
\item An inequality involving an interval is considered certified only if
it is true for all values in the interval. E.g., if the interval given for
$x$ is $[a,b]$, then $x<0$ is certified only if $b<0$.
\end{enumerate}

We can now certify the geometric proto-canonical triangulation and the
opacities of its faces. Our input is a candidate geometric proto-canonical
triangulation obtained by calling \texttt{SnapPy}'s \texttt{canonize} on a
tetrahedral manifold. We first guess exact values $z_i$ from the
approximated shapes reported by \texttt{SnapPy}. Using those guesses, we verify
\begin{enumerate}
\item the rectangular form of the edge equations exactly,
\item $\mathrm{Im}(z_i)>0$ for each tetrahedron (using interval arithmetics),
\item $|e|<10^{-7}$ for each edge where $e$ is the error of the logarithmic
form of the edge equation (using interval arithmetics),
\item all the equations~(\ref{eqn:EdgeRatioCusp}) exactly,
\item $\mathrm{Tilt}(F)<0$ (using interval arithmetics) for an opaque face,
respectively, $\mathrm{Tilt}(F)=0$ (using exact arithmetics) for a
transparent face.
\end{enumerate}

(1) implies that the error in (3) will be a multiple of $2\pi i$ so a
small enough error implies that the logarithmic form of the edge equations
is fulfilled exactly. Together with (2), this means that the tetrahedra
yield a (not necessarily complete) hyperbolic structure. Completeness is
ensured by (4) which checks that the cusp cross section is Euclidean.
Checking (4) really means verifying that the recursion process to obtain
the edge lengths could construct a consistent result. (5) certifies the
geometric proto-canonical triangulation and the opacities of the faces.

\begin{remark}
Note that in the process, we actually produce complex intervals for the
shapes from \texttt{SnapPy}'s approximations certified to contain the true
values. We can do this because we know that the shapes are in the field
$\mathbb Q(\sqrt{-3})$ and thus can guess exact solutions and verify them exactly.
An alternative method to obtain certified intervals from approximated shapes
is the Krawczyk test implemented in \texttt{hikmot} \cite{hikmot}. We could
not use it here though, because it cannot deal with non-orientable
manifolds. The edge equations for a non-orientable manifold are polynomials
in $z_i^*$ and ${1}/{\bar{z}_i^*}$.
\end{remark}

\begin{remark}
We could have also avoided guessing by tracking \texttt{SnapPy}'s
algorithm to obtain a proto-canonical triangulation. We know that the
shapes of the tetrahedral tessellation are all exactly represented by
$\frac{1}{2}+\frac{1}{2}\sqrt{\myChanged{-}3} $ and that \texttt{SnapPy}
is performing 2-3 and 3-2 moves during the algorithm. However, this would
require changes to the \texttt{SnapPea} kernel since it does not report the
sequence of moves it performed.
\end{remark}

For guessing a rational representation from an approximation, we use the
\texttt{fractions} module shipped with \texttt{python}. It essentially
computes the continued fraction for a given real number and evaluates it
at a stage where the resulting denominator is less than a given bound
(10000 in our case). For the (complex) interval arithmetics, we use
\texttt{sage}. Our implementation in \texttt{python} is based on the script
given in \cite{DHL}.

\subsection{Certification in the generic case}

Dunfield, Hoffman, L\myChanged{icat}a give an implementation in \cite{DHL}
to certify a triangulation to be the canonical cell decomposition (which
cannot contain non-tetrahedral cells). Though not needed here, we want to
point out that their implementation can be both simplified and generalized
to any number of cusps.

They start with certified complex intervals for the shapes returned by
\texttt{hikmot} \cite{hikmot}. But instead of following the complicated
procedure in \cite[Section 3.7]{DHL}, one can simply apply interval
arithmetics to the above equations to compute $\mathrm{Tilt}(F)$. The
result is an interval $[a,b]$ for each $\mathrm{Tilt}(F)$ that is
guaranteed to contain the true value of $\mathrm{Tilt}(F)$. If $b<0$ for
each interval, then the $\mathrm{Tilt}(F)$ are certified to be less than 0,
thus the given ideal triangulation is the canonical cell decomposition.

We provide a version of \texttt{canonical.py} here that implements this.

\section{Results of the implementation of algorithms}
\label{section4}

We implemented the algorithms described in the previous section, see
\cite{Goerner:tetcensus} for the resulting data. The longest algorithm
to run was the enumeration of the combinatorial tetrahedral tessellation s: the orientable case up to 25
tetrahedra and the non-orientable one up to 21 tetrahedra each took about
$\approx 6$ weeks CPU time and $\approx  70$Gb on a Xeon E5-2630, 2.3Ghz.
The number of resulting combinatorial tetrahedral tessellation s and tetrahedral manifold s are listed in
Table~\ref{table:censusNumbers}.

\subsection{Names of tetrahedral manifolds}

We give the tetrahedral manifold s names such as ``$\otet08_{0002}$'' (orientable),
respectively, ``$\ntet02_{0000}$'' (non-orientable) with ``tet'' followed
by the number of tetrahedra and an index.
The different combinatorial tetrahedral tessellation s corresponding to the same tetrahedral manifold{} are named with an
additional index, e.g., ``$\otet08_{0002}\#0$'', ``$\otet08_{0002}\#1$''. We
choose as canonical representative for an isometry class the first combinatorial tetrahedral tessellation{},
e.g., $\otet08_{0002}\#0$ for the tetrahedral manifold{} $\otet08_{0002}$.\\
The indices are canonical: before indexing the combinatorial tetrahedral tessellation s and tetrahedral manifold s, we
first sort the combinatorial tetrahedral tessellation s within an isometry class lexicographically by
isomorphism signature and then sort the tetrahedral manifold s lexicographically by the
isomorphism signature of their canonical representative.

\subsection{SnapPy census}

Our census of tetrahedral manifold s can be easily accessed from \texttt{SnapPy}. Simply
change to the directory \texttt{snappy} accompanying this article and type
\texttt{from tetrahedralCuspedCensus import *}. The two censuses
\texttt{TetrahedralOrientableCusped\-Census} and
\texttt{TetrahedralNonorientableCuspedCensus} have the same methods as any
other census such as \texttt{OrientableCuspedCensus}. Here are examples of
how to use them:

\begin{verbatim}
>>> from tetrahedralCuspedCensus import *
>>> M=TetrahedralOrientableCuspedCensus['otet02_0000'] # also m003
>>> TetrahedralOrientableCuspedCensus.identify(Manifold('m004'))
otet02_0001(0,0)
>>> len(TetrahedralOrientableCuspedCensus(tets=5)) # Number with 5 tets
2
>>> for M in TetrahedralOrientableCuspedCensus(tets=5):
...     print OrientableCuspedCensus.identify(M)
m410(0,0)
m412(0,0)(0,0)
>>> TetrahedralOrientableCuspedCensus.identify(Manifold("m208"))
>>>
\end{verbatim}

\myChanged{The last example shows that \texttt{m208} is not a tetrahedral
manifold since it has only 5 tetrahedra and thus would be in the
tetrahedral census. Note that \texttt{SnapPy}'s \texttt{is\_isometric\_to}
is using numerical methods and can fail to find an isomorphism. To verify
that \texttt{m208} is not tetrahedral, one can certify its isometry
signature\footnote{\myChanged{
We plan a future publication describing how to generalize the techniques
for certifying isometry signatures to all cusped hyperbolic manifold.
The third named author has already incorporated this into SnapPy, beginning
with version 2.3.2, see SnapPy documentation.}} and check that it is not
in the data files \cite{Goerner:tetcensus} provided with this paper.}

\subsection{Regina files}

We also provide the census of combinatorial tetrahedral tessellation s as two \texttt{Regina} files (for
orientable and non-orientable) in the \texttt{Regina} directory
accompanying this article. Each file groups the combinatorial tetrahedral tessellation s first by number of
tetrahedra and then by isometry class. The container for each isometry class
contains the different combinatorial tetrahedral tessellation s as well as the canonical retriangulation.

The \texttt{Regina} files can be inspected using the \texttt{Regina} GUI or
the \texttt{Regina} python API. An example of how to traverse the tree
structure in the file is given in \texttt{regina/example.py}.

\subsection{Morphisms}
\label{section:morphisms}

Similarly to combinatorial isomorphism, we can define a
{\em combinatorial homomorphism} between combinatorial tetrahedral tessellation s, but without the
requirement that different tetrahedra in the source go to the different
tetrahedra in the destination. It assigns to each tetrahedron in the
source a tetrahedron in the destination and a permutation in $S_4$
indicating which vertex of the source tetrahedron is mapped to which
vertex of the destination tetrahedron. These permutations have to be
compatible with the gluings of the source and destination tetrahedra.
If the tessellations are connected and have no open faces, the source
triangulation needs to have the same number of or a multiple of the number
of tetrahedra as the destination. Topologically, a combinatorial
homomorphism is a covering map that preserves the triangulation. We have
implemented a procedure to list all combinatorial homomorphisms for a pair
of triangulations in python.

We give a list of all pairs $(M, N)$ of combinatorial tetrahedral tessellation s such that there is a
combinatorial homomorphism from $M$ to $N$ as a text file
\texttt{data/morphisms.txt}. We do not include the trivial pairs $(M,M)$
or pairs $(M,N)$ which factor through another combinatorial tetrahedral tessellation{} as those can be
recovered trivially through the reflexive and transitive closure. We also
give some of the resulting graphs in \texttt{misc/graphs}. We discuss an
example in more detail later in Section~\ref{section:cttCat}.

\section{Properties of tetrahedral manifold s}
\label{sec.properties}

\myChanged{
\subsection{Tetrahedral manifolds are arithmetic}
\label{sub.arithmetic}

Recall that two manifolds (or orbifolds) are commensurable if they have
a common finite cover. Commensurability is an equivalence relation.
The commensurability class of the figure-eight knot complement {\tt m004}
consists exactly of the cusped
hyperbolic orbifolds and manifolds with invariant trace field $\mathbb Q(\sqrt{-3})$ that
are arithmetic or, equivalently, that have integral traces
\cite[Theorem 8.2.3 and 8.3.2]{MR}. Thus, tetrahedral manifolds are
also arithmetic with the same invariant trace field since
\begin{lemma}
Tetrahedral manifolds are commensurable to {\tt m004}.
\end{lemma}

More precisely, the commensurability class of {\tt m004} also contains the orbifold
$\mathfrak{R}=\mathbb H^3/\mathrm{Isom}(\{3,3,6\})$ where the Coxeter group
$\mathrm{Isom}(\{3,3,6\})$ is the symmetry group of the regular
tessellation $\{3,3,6\}$ by regular ideal tetrahedra. This orbifold can be used
to characterize the tetrahedral manifolds in this commensurability class:

\begin{lemma}
A manifold $M$ is a covering space of $\mathfrak{R}$ if and only if it is
tetrahedral.
\end{lemma}

\begin{proof}
A combinatorial tetrahedral tessellation{} of a manifold $M$ lifts to the tessellation $\{3,3,6\}$ in its
universal cover $\mathbb H^3$. Thus, $\pi_1(M)$ is a subgroup of the symmetry group
$\mathrm{Isom}(\{3,3,6\})$. Consequently, $M$ is a cover of $\mathfrak{R}$.

Conversely, a covering map $M\to\mathfrak{R}$ induces a combinatorial tetrahedral tessellation{} on the manifold
$M$ with the standard fundamental domain of $\mathfrak{R}$ lifting to the
barycentric subdivision of the combinatorial tetrahedral tessellation.
\end{proof}

\subsection{Implications of the Margulis Theorem}
\label{sec:Margulis}

Since {\tt m004} is arithmetic, Margulis Theorem implies that its
commensurator is not discrete and thus the commensurability class of
{\tt m004} contains no minimal element
\cite{neumannReid:Topology90Arithmetic,MR,Walsh}. In particular,
$\mathfrak{R}$ is not the minimal element of the commensurability class.
We thus expect to see the following phenomena in the commensurability
class containing the tetrahedral manifolds:
\begin{itemize}
\item Non-tetrahedral manifolds that are still commensurable with
{\tt m004}. For example, the following manifolds in SnapPy's
{\tt OrientableCuspedCensus} up to 8 simplices have this property:
$${\tt m208, s118, s119, s594, s595, s596, v2873, v2874}$$
\item Tetrahedral manifolds $M$ with different covering maps
$M\to\mathfrak{R}$ inducing non-isomorphic combinatorial tetrahedral tessellation s of the same manifold $M$.
\item Combinatorial tetrahedral tessellations ``hiding symmetries'',
defined as follows.
\end{itemize}

\begin{definition}
A combinatorial tetrahedral tessellation{} $T$ hides symmetries if the corresponding tetrahedral manifold{} $M$ has an
isometry that is not induced from a combinatorial automorphism of  $T$.
In other words, if there is an isometry $M\to M$ that does not commute
with the covering map $M\to\mathfrak{R}$ corresponding to $T$.
\label{def:hidesSymmetries}
\end{definition}

In this section, we will illustrate these phenomena using the tetrahedral
census.

\begin{remark}
 By definition, the canonical cell decomposition and thus the canonical
retriangulation sees all isometries, so we can detect this by checking
that the number of combinatorial automorphisms of the canonical
retriangulation is higher than those of the combinatorial tetrahedral tessellation. To enable the reader to
do this, the Regina file containing the tetrahedral census
\cite{Goerner:tetcensus} includes the canonical retriangulation as well.
The combinatorial automorphisms can be found using the method
{\tt findAllIsomorphisms} of a Regina triangulation or {\tt find\_morphisms}
in {\tt src/morphismMethods.py}.
 \end{remark}
}

\myChanged{

\begin{remark}
The minimum volume orientable cusped hyperbolic orbifold
$\mathfrak{M}=\mathbb H^3/\mathrm{PGL}(2,\mathbb Z[\zeta])$ and the Bianchi orbifold
$\mathfrak{B}=\mathbb H^3/\mathrm{PSL}(2,\mathbb Z[\zeta])$ of discriminant $D=-3$ where
$\zeta=\frac{1+\sqrt{-3}}{2}$ are related to $\mathfrak{R}$ as follows
with each map being a 2-fold covering \cite{neumannReid:SmallVolOrbs}:
$$
\mathfrak{B}\to\mathfrak{M}\to\mathfrak{R}.
$$
Similarly to $\mathfrak{R}$ being the quotient of $\mathbb H^3$ by the symmetry
group of the regular tessellation $\{3,3,6\}$, $\mathfrak{M}$ corresponds
to orientation-preserving symmetries, and $\mathfrak{B}$ corresponds to the
symmetry group of the regular tessellation $\{3,3,6\}$ after two-coloring
the regular ideal tetrahedra.

Thus, the manifold covering spaces of $\mathfrak{M}$ correspond to the
orientable combinatorial tetrahedral tessellation s, and the manifold covering spaces of $\mathfrak{B}$
correspond to orientable combinatorial tetrahedral tessellation s whose tetrahedra can be two-colored. Regina
displays the dual 1-skeleton of a triangulation in its UI under
``Skeleton: face pairing graph'', so we can check whether a combinatorial tetrahedral tessellation{} is a
cover of $\mathfrak{B}$ by testing whether the graph Regina shows is
two-colorable.  For example,
all orientable combinatorial tetrahedral tessellation s with \myChanged{fewer} than 5 tetrahedra are covers of
$\mathfrak{B}$. But $\otet05_{0000}$ and $\otet06_{0000}$ are not.
\end{remark}

\begin{remark}
Related results include: \cite{BMR95} show that all once-punctured torus
bundles in the commensurability class of the figure eight-knot complement
{\tt m004} are actually cyclic covers of the tetrahedral manifolds
{\tt m003} and {\tt m004} and thus tetrahedral. The non-arithmetic
hyperbolic once-punctured torus bundles are studied in
\cite{goodmanHeardHodgson:CommensuratorsOfCuspedHyperbolicManifolds}
where an algorithm is given to compute the commensurator of a cusped
non-arithmetic hyperbolic manifold. \cite{RV} study symmetries of or
hidden by cyclic branched coverings of 2-bridge knots.
\end{remark}

}

\myComment{
\subsubsection{Tetrahedral manifold s are arithmetic}
\label{sub.arithmetic}

Two manifolds (or orbifolds) are commensurable if they have a common finite
cover. Commensurability is an equivalence relation. In this section we
use nations from \texttt{SnapPea} census of manifolds. Recall that
{\tt m004} is the figure-eight knot complement and its volume is
$v_{4} = 2.02988321282\ldots$.
The next lemma follows from combining Theorems 3.3.4, 8.2.3 and 8.3.2
of \cite{MR}:

\begin{lemma}
\label{lem.1}
For a cusped hyperbolic manifold $M$, the following are
equivalent:
\begin{itemize}
\renewcommand{\theenumi}{\alph{enumi}}
\item[(a)]
$M$ is commensurable to {\tt m004}.
\item[(b)]
$M$ is arithmetic with invariant trace field $\mathbb Q(\sqrt{-3})$.
\item[(c)]
the invariant trace field of $M$ is $\mathbb Q(\sqrt{-3})$ and $M$ has
integral traces.
\end{itemize}
\end{lemma}

\begin{lemma}
\label{lem.arithmetic}
All tetrahedral manifolds are arithmetic with invariant trace field
$\mathbb Q(\sqrt{-3})$ and commensurable to each other.
\end{lemma}

We will prove this together with Lemma~\ref{lem.RMB} later.

\begin{remark}
\label{rem.nottrue}
The converse to Lemma~\ref{lem.arithmetic} does not hold.
There are 25 manifolds in the {\tt OrientableCensus} with volume
an integer multiple of $v_4$ that are not tetrahedral.
Of those,
$$
{\tt m208, s118, s119, s594, s595, s596, v2873, v2874}
$$
are arithmetic, and commensurable to {\tt m004}.
\end{remark}

\begin{remark}
The converse to Lemma~\ref{lem.arithmetic} holds for
once-punctured torus bundles. Indeed, in~\cite{BMR95} it is shown that a
hyperbolic once-punctured torus bundle is arithmetic and commensurable to
{\tt m004}
if and only if it is a cyclic cover of it or its sister {\tt m003}, both
of which are
tetrahedral manifolds ($\otet02_{0001}$ and $\otet02_{0000}$, respectively).
Hence such a torus bundle is tetrahedral.
\end{remark}

\subsubsection{Relationship to the Bianchi orbifold}
\label{subsection:bianchiRelation}

We call a triangulation two-colorable if the tetrahedra can be assigned
two colors such that no two neighboring tetrahedra have the same color. Let
us now define three orbifolds $\mathfrak{B}$, $\mathfrak{M}$ and
$\mathfrak{R}$. Recall that $\{3,3,6\}$ is the regular tessellation of
$\mathbb H^3$ by regular ideal tetrahedra. Let $\{3,3,6\}^+$ denote the
two-colored regular tessellation. Symmetries of $\{3,3,6\}^+$ are supposed
to preserve the coloring.

If we use the upper half space model of $\mathbb H^3$ which has boundary
$\mathbb C\cup\{\infty\}$ such that
$\mathrm{PSL}(2,\mathbb C)\cong\mathrm{PGL}(2,\mathbb C)\cong\mathrm{Isom}^+(\mathbb H^3)$, we can move
$\{3,3,6\}$ such that one tetrahedron $T$ has vertices at
$\{0,1,\zeta,\infty\}$ where $\zeta=\frac{1+\sqrt{-3}}{2}$. We then have
the following relationship between groups (each inclusion has index 2):
$$
\xymatrix@1{\mathrm{Isom}^+(\{3,3,6\}^+)\,
\ar@{}[r]|-*[@]{\subset} \ar@{=}[d]  & \mathrm{Isom}^+(\{3,3,6\})\,
\ar@{}[r]|-*[@]{\subset}  \ar@{=}[d]  & \mathrm{Isom}(\{3,3,6\})\\
 \mathrm{PSL}(2,\mathbb Z[\zeta]) \, \ar@{}[r]|-*[@]{\subset} & \mathrm{PGL}(2,\mathbb Z[\zeta]). }
$$

We denote the corresponding quotients by
$$
\xymatrix@1{\mathfrak{B} \qquad \ar[r] & \qquad \mathfrak{M}\qquad
\ar[r] & \qquad \mathfrak{R}.}
$$

Note that $\mathfrak{R}$ is a reflection orbifold, $\mathfrak{M}$ is also
the orientable cusped minimum volume orbifold and $\mathfrak{B}$ is  the
Bianchi orbifold for discriminant $D=-3$

\begin{remark}
\label{remark:fundDom}
We can explicitly give fundamental domains for these orbifolds.
Consider the barycentric subdivision of $\{3,3,6\}$ which divides each
tetrahedron into 24 simplices. Let $T'$ denote a tetrahedron adjacent to
$T$, $F$ a face shared by $T$ and $T'$ and $E$ an adjacent edge.
\smallskip
\begin{center}
\begin{tabular}{|c|c|}
\hline
Orbifold & Fundamental Domain\\ \hline \hline
$\mathfrak{R}$ & 1 simplex \\ \hline
$\mathfrak{M}$ & 2 simplices in $T$ touching $E$ and $F$\\ \hline
$\mathfrak{B}$ & 4 simplices in $T$ and $T'$ touching $E$ and $F$\\ \hline
\end{tabular}
\end{center}
\end{remark}

We can now characterize tetrahedral manifolds as follows:

\begin{lemma}
\label{lem.RMB}
\rm{(a)} A manifold $M$ is a covering space of $\mathfrak{R}$ if and only
if it is tetrahedral.\\
\rm{(b)} A manifold $M$ is a covering space of $\mathfrak{M}$ if and only
if it is orientable and tetrahedral.\\
\rm{(c)} A manifold $M$ is a covering space of $\mathfrak{B}$ if and only
if it is orientable and has a two-colorable combinatorial tetrahedral tessellation{}.
\end{lemma}

\begin{remark}
\label{remark:baryOrientability}
Part (c) of Lemma \ref{lem.RMB} is similar to the fact that a triangulation
is orientable if and only if its barycentric subdivision is two-colorable.
\end{remark}

\begin{corollary}
\label{cor.2color}
We can efficiently determine in \texttt{Regina} whether a given combinatorial tetrahedral tessellation{} is a
cover of the Bianchi orbifold $\mathfrak{B}$ by checking whether its dual
1-skeleton is a
two-colorable graph. This graph can be found for a triangulation in the
\texttt{Regina} UI under ``Skeleton: face pairing graph''. For example,
all orientable combinatorial tetrahedral tessellation s with \myChanged{fewer} than 5 tetrahedra are covers of
$\mathfrak{B}$. But $\otet05_{0000}$ and $\otet06_{0000}$ are not.
\end{corollary}

\begin{proof}[Proof of Lemma~\ref{lem.arithmetic} and \ref{lem.RMB}]
First, we show (a) of Lemma~\ref{lem.RMB}.
The combinatorial tetrahedral tessellation{} of $M$ lifts to the tessellation $\{3,3,6\}$ in its universal
cover $\mathbb H^3$.
Thus, $\pi_1(M)$ is a subgroup of the symmetry group
$\mathrm{Isom}(\{3,3,6\})$, and thus $M$ is a cover of $\mathfrak{R}$.

Now, assume that $M$ is a cover of $\mathfrak{R}$. Pick a fundamental
domain for $\mathfrak{R}$ as in Remark~\ref{remark:fundDom}. Let $v$ be
the vertex of this simplex corresponding to a center of a tetrahedron in
$\{3,3,6\}$ and $F$ be the opposite face. As $M$ is a cover of
$\mathfrak{R}$, $F$ lifts to $M$ where it subdivides $M$ into tetrahedra.
Note that this requires $M$ to be a manifold cover so that the link of a
lift of $v$ is indeed a tetrahedron.

(b) follows similarly since $\mathfrak{M}$ is the quotient by those
symmetries in $\mathrm{Isom}(\{3,3,6\})$ that are orientation-preserving.

(c) follows from the fact that a two-coloring of a combinatorial tetrahedral tessellation{} lifts to a
two-coloring of $\{3,3,6\}$, respectively, vice versa descends to a combinatorial tetrahedral tessellation{}
for a subgroup of $\mathrm{Isom}^+(\{3,3,6\})$.

As {\tt m004} is in the same commensurability class as the above three
orbifolds, Lemma~\ref{lem.arithmetic} follows.
\end{proof}

\subsubsection{Hidden symmetries}

Here, we say that a {\em combinatorial tetrahedral tessellation{} realizes a hidden symmetry} if the
corresponding tetrahedral manifold{} has an isometry that does not descend to a
combinatorial isomorphism of the combinatorial tetrahedral tessellation{}. In other words, we are applying
the following definition with $O=\mathfrak{R}$:

\begin{definition}
Let $M\to O$ be a covering map of hyperbolic orbifolds. A
{\em hidden symmetry} of $M\to O$ is a symmetry of $M$ that does not
descend to $O$, i.e., there is no symmetry of $O$ making the following
diagram commute:
$$
\xymatrix{M \ar[d] \ar[r] & M \ar[d] \\ O\ar[r] & O.}
$$
\end{definition}

\begin{remark}
Since the combinatorial tetrahedral tessellation{} as well as the associated canonical retriangulation
are available in the \texttt{Regina} files shipped with this paper, the
reader can easily check whether a combinatorial tetrahedral tessellation{} has hidden symmetries: this is
the case if and only if the number of combinatorial automorphisms of the
canonical retriangulaton is larger than that of the combinatorial tetrahedral tessellation. The combinatorial
automorphisms can be found with \texttt{Regina}'s
\texttt{findAllIsomorphisms} or \texttt{find\_morphisms} in
\texttt{src/morphismMethods.py}.
\end{remark}

Given an orbifold $O$, there is a related notion of hidden symmetry
that can be applied to just a group $\Gamma$ with $O=\mathbb H^3/\Gamma$ and
that is defined in terms of the commensurator and normalizer. We refer
the reader to the existing literature
\cite{Walsh,MR,neumannReid:Topology90Arithmetic}, especially
\cite{goodmanHeardHodgson:CommensuratorsOfCuspedHyperbolicManifolds}
which gives an algorithm for constructing the commensurator. For our
case, where $O=\mathfrak{R}$ and $\Gamma=\mathrm{Isom}(\{3,3,6\})$ is
arithmetic, we only remark that Margulis Theorem (see
\cite[Thm.10.3.5]{MR}) implies that we expect many combinatorial tetrahedral tessellation{} with hidden
symmetries. For hidden symmetries of cyclic branched coverings of
2-bridge knots we mention \cite{RV}.
}

\myComment{
We expect that many combinatorial tetrahedral tessellation s with hidden symmetries exists because
$\mathrm{Comm}(\Gamma)$

We expect many combinatorial tetrahedral tessellation s with hidden

Recall that the commensurator of a subgroup
$\Gamma\subset\mathrm{Isom}(\mathbb H^3)$ (see \cite{Walsh})
$$
\mathrm{Comm}(\Gamma) = \{g\in\mathrm{Isom}(\mathbb H^3)\ |\
\Gamma\mbox{ and }g\Gamma g^{-1}\mbox{ are commensurable}\},
$$
is a complete  commensurability invariant of the group $\Gamma$ and
contains the normalizer
$$
\mathrm{N}(\Gamma)=\{g\in\mathrm{Isom}(\mathbb H^3)\ |\ \Gamma = g\Gamma g^{-1}\}.$$

\begin{definition}
Let $\Gamma\subset\mathrm{Isom}(\mathbb H^3)$. A {\em virtual symmetry} of
$\Gamma$ is a coset $g\mathrm{N}(\Gamma)\in \mathrm{Virt}(\Gamma)=
\mathrm{Comm}(\Gamma)/\mathrm{N}(\Gamma)$. A non-trivial virtual symmetry
is called a {\em hidden symmetry}.
\end{definition}

Given an orbifold $O$, we first need to fix a group $\Gamma$ such that
$O\cong \mathbb H^3/\Gamma$ before
we can write down an element $g\mathrm{N}(\Gamma)\in\mathrm{Virt}(\Gamma)$.
However, virtual symmetry can still be defined to be intrinsic to $O$:

\begin{lemma}
Given an orbifold $O$, there is a canonical set $\mathrm{Virt}(O)$
associated to $O$ such that $\mathrm{Virt}(O)$ is canonically isomorphic
to $\mathrm{Virt}(\Gamma)$ once a particular $\Gamma$ with
$O\cong\mathbb H^3/\Gamma$ is chosen.
\end{lemma}

\begin{proof}
Assume we have two groups $\Gamma$ and $\Gamma'$ with
$O\cong \mathbb H^3/\Gamma$. There is a correspondence of
$\mathrm{Virt}(\Gamma)$ and $\mathrm{Virt}(\Gamma')$: if
$h\in\mathrm{Isom}(\mathbb H^3)$ conjugates $\Gamma$ to $\Gamma'$, then an
element $g\mathrm{N}(\Gamma)\in\mathrm{Virt}(\Gamma)$ is send to an
element $gh\mathrm{N}(\Gamma)\in\mathrm{Virt}(\Gamma')$. This element
is well-defined since $h$ is determined up to an element in
$\mathrm{N}(\Gamma)$. Hence, we can obtain a canonical object by
thinking of a {\em virtual symmetry of $O$} as an equivalence class in
$$\mathrm{Virt}(O)=\frac{\{(\Gamma, g\mathrm{N}(\Gamma))\ |\
O\cong\mathbb H^3/\Gamma, g\in\mathrm{Comm}(\Gamma)\}} {(\Gamma,
g\mathrm{N}(\Gamma)) \sim (h\Gamma h^{-1}, gh\mathrm{N}(\Gamma))}.$$
\end{proof}

\begin{lemma}
\label{lemma:hiddenSyms}
Let $O$ be a hyperbolic orbifold. Let $M\to O$ be a covering map. There is
a well-defined map
$$
\mathrm{Isom}(M)\to \mathrm{Virt}(O)
$$
such that (non-)hidden symmetries of $M\to O$ become (non-)hidden
symmetries of $O$.\\
Furthermore, each virtual symmetry of $O$ is realized by a finite covering
$M\to O$, i.e., for every element in $\mathrm{Virt}(O)$, there is an
orbifold $M$ with a covering map $M\to O$ such that there is a symmetry
in $\mathrm{Isom}(M)$ corresponding to the element in $\mathrm{Virt}(O)$.
\end{lemma}

\begin{proof}
The manifold $M$ is a quotient by a subgroup $\Gamma'\subset \Gamma$. A
symmetry of $M$ can be represented by an element $g$ such that
$g\Gamma' g^-1 = \Gamma'$.
The covering map $M\to O$ determines $\Gamma'$ up to conjugation by an
element in $N(\Gamma)$. A symmetry of $M$ is an element $n$ with
$\Gamma$... {\bf \textcolor{red}{$\diamondsuit$} need to finish proof, wonder whether I can reference
literature, but, to my frustration, no article explains it that clearly.}
\end{proof}

In this paper, we are only interested in the case where $O=\mathfrak{R}$
and fix $\Gamma=\mathrm{Isom}(\{3,3,6\})$, so $\mathrm{N}(\Gamma)=\Gamma$.
Margulis Theorem (see \cite[Thm.10.3.5]{MR}) states that
$\mathrm{Comm}(\Gamma)$ is dense, and thus
$\mathrm{Comm}(\Gamma)/\mathrm{N}(\Gamma)$ infinite, if and only if $O$
is arithmetic. This implies the following lemma (where the partial
ordering is given by $M'<M$ if $M'\leftarrow M$):

\begin{lemma}
\label{lemma:infMinEl}
The poset of combinatorial tetrahedral tessellation s has infinitely many minimal elements.
\end{lemma}

\begin{proof}
The orbifold $O=\mathfrak{R}$ is arithmetic and thus has infinitely many
hidden symmetries. For each hidden symmetry, consider all combinatorial tetrahedral tessellation{} $M\to O$
that realize it. Among those, pick one with the minimal number of
tetrahedra. Such a combinatorial tetrahedral tessellation{} is necessarily minimal in the above poset. As
a combinatorial tetrahedral tessellation{} can realize only finitely many hidden symmetries, there must be
infinitely many such minimal combinatorial tetrahedral tessellation s.
\end{proof}

\begin{remark}
If $O$ is non-arithmetic, $\mathbb H^3/\mathrm{Comm}(\Gamma)$ is an orbifold and
the minimal element in the commensurability class of $O$. An algorithm for
computing $\mathrm{Comm}(\Gamma)$ for cusped non-arithmetic manifolds is
given in \cite{goodmanHeardHodgson:CommensuratorsOfCuspedHyperbolicManifolds}.
\end{remark}

}

\subsection{The category of combinatorial tetrahedral tessellation s}
\label{section:cttCat}

\myChanged{ To study the commensurability class containing the tetrahedral
manifolds, we think of it as a category. For this,} recall
 the notion of a combinatorial homomorphism from
Section~\ref{section:morphisms}. On the underlying topological space, a
combinatorial homomorphism is a covering map. We thus get two categories
with a forgetful functor \myChanged{$\mathcal{T}\to\mathcal{M}$}:

\begin{definition}
\label{def:categories}
The {\em category \myChanged{$\mathcal{M}$} of \myChanged{manifolds
commensurable with tetrahedral manifold s}} has as objects manifolds commensurable
\myChanged{with} {\tt m004} and as morphisms covering maps.\\
The {\em category \myChanged{$\mathcal{T}$} of combinatorial tetrahedral tessellation s} has as objects
combinatorial tetrahedral tessellation s and as morphisms combinatorial homomorphisms.
\myRemoved{changed order of $\mathcal{T}$
and $\mathcal{M}$}
\end{definition}

\myRemoved{Removed Remark}
\myComment{
\begin{remark}
The first category is equivalent to the category whose objects are
covering maps $M\to \mathfrak{R}$ (with $M$ begin a manifold) and
whose morphisms are commutative diagrams of covering maps
$$
\xymatrix{M \ar[rd] \ar[rr] & & M' \ar[ld]\\ & \mathfrak{R}}
$$
and the forgetful functor takes $M\to \mathfrak{R}$ to $M$.
\end{remark}}

We show a small part of these categories in Figure~\ref{fig:catCtt}
and observe:
\begin{itemize}
\item $\otet04_{0001}\#0$ has two 2-covers (indicated by the solid arrows)
giving two different triangulations $\otet08_{0002}\#0$ and
$\otet08_{0002}\#1$. These triangulations are not combinatorially
isomorphic but yield isometric manifolds (indicated by the dashed line).
\item The figure-eight knot complement, $\otet02_{0001}\#0$, and its
sister, $\otet02_{0000}\#0$, have a common cover $\otet04_{0002}\#0$. More
general, any two combinatorial tetrahedral tessellation s have a common cover combinatorial tetrahedral tessellation {} \myChanged{as they}
are in the same commensurability class.
\item $\otet02_{0001}\#0$ and $\otet02_{0000}\#0$ show that the graph is a
poset with more than one minimal element. In fact, most combinatorial tetrahedral tessellation s in our
census are minimal elements and we conjecture that there are infinitely
many such minimal elements.
\item The figure also shows a manifold \texttt{m208}, \myChanged{which is
non-tetrahedral}. \myChanged{However, as with any manifold in this
commensurability class, it still has a tetrahedral covering space, here}
$\otet08_{0010}\#0$ (the arrow has to be dashed because \texttt{m208} is not
tetrahedral so the map is not a combinatorial homomorphism).
\end{itemize}

\begin{remark}
\label{remark:ExampleHiddenSym}
The last example shows that the \myChanged{combinatorial tetrahedral tessellation{}} $\otet08_{0010}\#0$
\myChanged{hides symmetries as in Definition~\ref{def:hidesSymmetries}}.
To see this, notice that the covering space $\otet08_{0010} \to
\texttt{m208}$ is 2-fold, thus regular and \texttt{m208} is the quotient
of $\otet08_{0010}$ by the group $G=\mathbb Z/2\mathbb Z$ of deck transformations. If
$G$ preserved the combinatorial tetrahedral tessellation{} $\otet08_{0010}\#0$, the quotient \texttt{m208} would
have an induced combinatorial tetrahedral tessellation. But \texttt{m208} is not tetrahedral, thus the
nontrivial
element of $G$ is a symmetry of $\otet08_{0010}\#0$ which is not a
combinatorial homomorphism.
\end{remark}

\begin{figure}
\begin{center}
\scalebox{0.39}{\includegraphics{draws/Orientable8.pdf}}
\end{center}
\caption{A small part of the category
\myChanged{$\mathcal{T}$} of combinatorial tetrahedral tessellation s (solid arrows)
and the larger category \myChanged{$\mathcal{M}$} of \myChanged{manifolds
commensurable with {\tt m004}} (dashed arrows). Multiple morphisms
between two objects are collapsed to just one arrow, automorphisms and
morphisms factoring through another object are dropped.
\label{fig:catCtt}}
\end{figure}

\myRemoved{Small improvements to Figure 2.}

\subsection{Canonical cell decompositions}

\subsubsection{Examples}

The canonical cell decomposition of a tetrahedral manifold{} can:
\begin{itemize}
\item Be a {\bf combinatorial tetrahedral tessellation}.\\
{\bf Examples:} $\otet02_{0000}$ and $\otet10_{0010}$. The latter one has
two combinatorial tetrahedral tessellation s, $\otet10_{0010}\#0$ being the canonical cell decomposition.
\item Be a {\bf coarsening} of a combinatorial tetrahedral tessellation{}.\\ (i.e., the combinatorial tetrahedral tessellation{} is a
subdivision of the canonical cell decomposition.)\\
{\bf Example:} $\otet05_{0001}$. The canonical cell decomposition consist
of single regular ideal cube that can be subdivided into 5 tetrahedra
(see Figure~\ref{fig:tetInCube}) such that the diagonals introduced on
the faces are compatible. This yields the unique (up to combinatorial
isomorphism) combinatorial tetrahedral tessellation{} for this manifold. We elaborate on the relationships
to cubes below.
\item {\bf Neither} of the above.\\
 In which case, the canonical cell decomposition can still
\begin{itemize}
\item Consists of (non-regular) tetrahedra.\\
{\bf Example:} $\otet08_{0010}$.
\item Contain cells which are not tetrahedra\\
{\bf Example:} $\otet08_{0001}$. Its canonical cell decomposition contains
some hexahedra obtained by gluing two non-regular tetrahedra.
\end{itemize}
\end{itemize}

\subsubsection{Cubical manifolds}

\myChanged{
Recall from Remark~\ref{remark:burtonPair} that a manifold was called
cubical if it can be decomposed into regular ideal cubes.
Figure~\ref{fig:tetInCube} showed that there are two choices of picking
alternating vertices of a cube, which span a tetrahedron and thus yield a
subdivision of a regular ideal cube into 5 regular ideal tetrahedra. Even
though each cube of a combinatorial cubical tessellation can be subdivided
into regular ideal tetrahedra individually, this only yields a combinatorial tetrahedral tessellation{} if
the choices made are compatible with the face-pairings of the combinatorial
cubical tessellation. We saw $\otet05_{0001}$ above as an example where this
was possible and {\tt x103} in Remark~\ref{remark:burtonPair} as an example
where this was impossible.

}

\myComment{
Even though each ideal cube can be divided into 5 regular ideal tetrahedra
individually, the remark showed an example (\texttt{x103}) where this did
not yield a combinatorial tetrahedral tessellation{}. An example of the opposite where this yields a combinatorial tetrahedral tessellation{}
is the above manifold $\otet05_{0001}$.\\
}

If a manifold is both tetrahedral and cubical, the canonical cell
decomposition can actually consist of regular cubes or regular ideal
tetrahedra (or neither). This is illustrated by the two cubical links given
by Aitchison and Rubinstein \cite{AR:dodecahedral}:
\begin{itemize}
\item The canonical cell decomposition of the complement $\otet10_{0011}$
of the alternating 4-chain link \texttt{L8a21} (see Figure~\ref{f.tetlinks})
consists of two regular ideal cubes.
\item The complement $\otet10_{0006}$ of the other cubical link
\texttt{L8a20} (see Figure~\ref{f.tetlinks}) admits two combinatorial tetrahedral tessellation s up to
combinatorial isomorphism, one of which is equal to the canonical cell
decomposition.
\end{itemize}

\begin{remark}
\myChanged{Figure~\ref{fig:tetInCube} also shows that the choice of 5
regular ideal tetrahedra to subdivide a cube hides symmetries of the cube,
namely, the rotation by $\pi/2$ of the cube that takes one choice to the
other. This rotation is an element in the commensurator but not in the
normalizer of $\mathrm{Isom}(\{3,3,6\})$ and thus a hidden symmetry of
$\mathfrak{R}$. A combinatorial tetrahedral tessellation{} arising as subdivision of a combinatorial
cubical tessellation can hide the symmetries of the combinatorial cubical
tessellation corresponding to this rotation, i.e., there can be symmetries
of the combinatorial cubical tessellation that are not symmetries of the combinatorial tetrahedral tessellation.
}

\myComment{
Cubical manifolds can actually have symmetries visible in the cubical
structure but not the tetrahedral one. To describe this hidden symmetry,
imagine a regular ideal cube and pick a set of alternating vertices.
These alternating vertices span a regular ideal tetrahedron, see
Figure~\ref{fig:tetInCube}. A rotation by $\pi/2$ takes the tetrahedron
to the tetrahedron spanned by the complementary set of vertices. This
rotation is a hidden symmetry of the tetrahedron, and thus $\mathfrak{R}$.\\
The combinatorial tetrahedral tessellation{} of $\otet10_{0011}$ actually realizes this hidden symmetry.
In other words, its two regular ideal cubes can be divided each into 5
regular ideal tetrahedra in a way such that the diagonals introduced on
the faces of each cube match. If we flip all the diagonals on all the
faces, we obtain a different subdivision. Thus, when subdividing to obtain
the combinatorial tetrahedral tessellation, a choice was made and the manifold $\otet10_{0011}$ has a symmetry
taking one of these choices to the other. This symmetry is hidden in the
combinatorial tetrahedral tessellation{}.\\
}
\myChanged{An example of this is $\otet10_{0011}$. }
Other examples are obtained by subdividing the cubical regular
tessellation link complements $\mathcal{U}_{1+\zeta}^{\{4,3,6\}}$,
$\mathcal{U}_2^{\{4,3,6\}}$\myChanged{,} and $\mathcal{U}_{2+\zeta}^{\{4,3,6\}}$
classified in \cite{Goerner}. By definition, each of these three manifolds
can be decomposed into ideal regular cubes such that each flag of a cube,
an adjacent face and an edge adjacent to the face can be taken to any other
flag by a symmetry. In particular, these manifolds contain a symmetry
flipping the diagonals of the faces of the cubes.
\end{remark}

\subsubsection{Canonical combinatorial tetrahedral tessellation s}

\myChanged{We call a combinatorial tetrahedral tessellation{} a regular tessellation if it corresponds} to a regular
covering space of $\mathfrak{R}$ or $\mathfrak{M}$. This is equivalent to
saying that the combinatorial automorphisms act transitively on flags
consisting of a tetrahedron, an adjacent face and an adjacent edge (we drop
the vertex in the flag to allow chiral combinatorial tetrahedral tessellation s) \cite{Goerner}.

\begin{lemma}
Consider a combinatorial tetrahedral tessellation{} \myChanged{$T$}. \myChanged{$T$ is equal to the
canonical cell decomposition of the corresponding tetrahedral manifold{} $M$ if $T$ is a
regular tessellation
or if $M$ has only one cusp. In particular, a tetrahedral manifold{}
with only one cusp has a unique combinatorial tetrahedral tessellation. If $T$ is equal to the canonical cell
decomposition, then $T$ hides no symmetries.}
\myComment{
\begin{center}
\includegraphics{draws/cttImplications.pdf}
\end{center}}
\end{lemma}

\begin{proof}
\myChanged{
Recall from Section~\ref{sec:defCanonicalCell} that the canonical cell
decomposition relies on choosing cusp neighborhoods of the same volume
for each cusp. If $T$ is regular, then each cusp neighborhood intersects
$T$ in the same triangulation. This is also true if $M$ has only one cusp
and there is only one cusp neighborhood to choose. Thus, each end of a
tetrahedron intersects the cusp neighborhoods in the same volume. $T$ lifts
to the regular tessellation $\{3,3,6\}$ of $\mathbb H^3$ and the cusp
neighborhoods lift to horoballs with the same symmetry. Hence, the canonical
cell decomposition is equal to $T$. The other statement follows from the
canonical cell decomposition not hiding any symmetries by definition.}
\myComment{
When computing the tilts in Equation~\ref{eqn:Tilt} for a 1-cusped combinatorial tetrahedral tessellation{},
there is no rescaling of cusp cross sections necessary and all tetrahedra
are regular. Thus, all $R^i_v$ are equal and all
$\mathrm{Im}(z_i^*)=\sqrt{3}/2$ making all tilts negative. Hence, 1-cusped
tetrahedral manifold has a unique combinatorial tetrahedral tessellation{} which is always equal to the
canonical cell decomposition.

Since all symmetries of a manifold are visible in the canonical cell
decomposition, all symmetries are combinatorial isomorphisms if the combinatorial tetrahedral tessellation{}
is equal to the canonical cell decomposition. Thus, such a combinatorial tetrahedral tessellation{} realizes
no hidden symmetries.}
\end{proof}

\myChanged{
\begin{remark}
For some cubical tessellations such as $\mathcal{U}_{1+\zeta}^{\{4,3,6\}}$,
$\mathcal{U}_2^{\{4,3,6\}}$, and $\mathcal{U}_{2+\zeta}^{\{4,3,6\}}$, we can
partition the cusps into two disjoint sets such that no edge connects two
cusps of the same set. If, in the construction of the canonical cell
decomposition, we now pick for cusps in one set cusp neighborhoods of a
volume slightly different from those for cusps in the other set, we no
longer obtain the cubical tessellation but one of the two subdivided combinatorial tetrahedral tessellation s
depending on which set of cusps we favored.
\end{remark}
}

\section{Tetrahedral links}
\label{sec.links}

\subsection{Some facts about tetrahedral links}

Consider a cusped 3-manifold $M$, i.e., the interior of a compact
3-manifold $\bar{M}$ with boundary $\partial\bar{M}$ a disjoint union of
tori. We say that $M$ is a \emph{homology link complement} if the long
exact sequence in homology associated to $(\bar{M},\partial \bar{M})$ is
isomorphic to that of the complement of a link in $S^3$. Let
$i:\partial\bar{M}\to \bar{M}$ denote the inclusion of the boundary. We
thank C. Gord\myChanged{o}n for pointing out to us that (b) implies (d).

\begin{lemma}
\label{lem.homology.link}
Let $M$ be a cusped 3-manifold. The following are equivalent:
\begin{enumerate}
\renewcommand{\theenumi}{\alph{enumi}}
\item $M$ is a homology link complement.
\item $H_1(M;\mathbb Z)=\mathbb Z^c$ where $c$ is the number of cusps.
\item The cuspidal homology
$H^{\mathrm{cusp}}_1(M)=H_1(\bar{M};\mathbb Z)/\mathrm{Im}(i_*)$ vanishes.
\item $M$ is the complement of a link in an integral homology sphere.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) implies (b) since $H_1(\partial \bar{M})\cong\mathbb Z^{2c}$ determines $c$
and $H_1(M)=\mathbb Z^c$ for a link complement in $S^3$. The equivalence of (b)
and (c) was shown in \cite[Lem.6.9]{Goerner}. To prove that (b) implies (d),
we work
by induction on $c$. For $c=0$, $M$ is a homology sphere and thus the
complement of the empty link.
Assuming it is true for $c-1$, pick a component $T$ of $\partial \bar{M}$
and let $H$ be
the image of $H_1(T;\mathbb Z)$ in $H_1(\bar{M};\mathbb Z)$ under the map induced by
inclusion. By Poincare duality,
$H$ has rank $1$ or $2$ (apply
\cite[Chapter VI, Theorem 10. 4]{bredon:top_and_geo} to $\bar{M}$ with
all boundary components but $T$ Dehn-filled). Now we claim that $H$
contains a rank $1$ direct summand of $H_1(\bar{M};\mathbb Z)$ (so one can now do a
Dehn filling on $T$ to reduce $c$ by 1). For if not, then $H$ is contained in
$p H_1(\bar{M};\mathbb Z)$ for some prime $p$. Then
$H_1(T;\mathbb Z_p)$ maps trivially in $H_1(\bar{M};\mathbb Z_p)$, contradicting duality.

It is left to show that (d) implies (a). This follows easily from
Alexander duality \cite{BZ}.
\end{proof}

A homology link $M$ is the complement of a link in the 3-sphere if
and only if there is a Dehn-filling of it with trivial fundamental group.
In that
case, the filling is a homotopy 3-sphere, hence a standard 3-sphere (by
Perelman's Theorem), and the
link is the complement of the core of the filling. \texttt{SnapPy} can
compute the homology of a hyperbolic manifold as well as a presentation of
its fundamental group, before or after filling. Note that links are in
general not determined by their complement, i.e., there are 3-manifolds
that arise as the complement of infinitely many different links~\cite{Gordon}.
On the other hand,
the only tetrahedral knot is the figure-eight knot. This follows from the
fact that
tetrahedral manifolds are arithmetic, and the only arithmetic knot is the
figure-eight
knot~\myChanged{\cite[Theorem 2]{reid:arithmeticity}}. \myRemoved{Changed
reference}

\subsection{A list of tetrahedral links}

Of the 124 orientable tetrahedral manifolds with at most 12 tetrahedra,
27 are homology links and \texttt{SnapPy} identified 13 of them with link
exteriors in its census. Of the remaining 14 homology links,
\begin{itemize}
\item
$\otet04_{0000}$ is the Berge \myChanged{manifold, the complement of a link
in \cite{MP:magicmfdfilling}}\myRemoved{Changed reference},
\item
11 are link complements, with corresponding links shown in
Figure~\ref{f.tetlinks} and~\ref{f.tetlinks12}.\\
(These links were found by drilling some curves until the manifold could
be identified as a complement of a link in SnapPy's \texttt{HTLinkExteriors}.
We then found a framing of some components of the link
such that Dehn-filling gives back the tetrahedral manifold. This gives us a Kirby diagram
of the tetrahedral manifold. Using the Kirby Calculator \cite{kirbycalculator}, we
successfully removed all Dehn-surgeries and obtained a link.)
\item $\otet08_{0003}$ and $\otet10_{0023}$
(with 2 and 1 cusps respectively) are not link complements.\\
(This can be shown
using {\tt fef\_gen.py} \myChanged{based on} \cite{MPR:fivechainfilling}
and available from \cite{fefgen}\myRemoved{Changed reference} to list all
exceptional slopes.\myRemoved{No footnote as this bug has been resolved in
version 1.3 of {\texttt fef\_gen}}
and then compute homologies for those.)
\end{itemize}

The data in Table~\ref{table:censusNumbers} also suggest:

\begin{conjecture}
Every tetrahedral link complement has an even number of tetrahedra (i.e.,
a corresponding combinatorial tetrahedral tessellation{} has an even number of tetrahedra).
\end{conjecture}

\subsection{A remarkable tetrahedral link}
\label{sub.remarkable}

Of the 11580 orientable tetrahedral manifold s with at most 25 tetrahedra, 885
are homology links, and have at most 7 cusps. There is a unique tetrahedral manifold \
with 7 cusps, $\otet20_{0570}$, which is a link complement, and a
2-fold cover of the minimally twisted 5-chain link $L10n113=\otet10_{0027}$.
This remarkable link is shown in Figure~\ref{fig.remarkable}.

\begin{figure}[!hptb]
\centering{
\includegraphics[height=0.12\textheight]{draws/K4a1.pdf}\hspace{5mm}
\includegraphics[height=0.12\textheight]{draws/Berge.pdf}\hspace{5mm}
\includegraphics[height=0.12\textheight]{draws/L6a2.pdf}\hspace{5mm}
\includegraphics[height=0.12\textheight]{draws/L10n46.pdf}
\\
$$
\otet02_{0001}(K4a1) \qquad \quad
\otet04_{0000} \qquad \qquad \quad
\otet04_{0001}(L6a2) \quad
\otet08_{0002}(L10n46)
$$
\\
\includegraphics[height=0.14\textheight]{draws/L8a20.pdf}\hspace{2mm}
\includegraphics[height=0.14\textheight]{draws/L10n88.pdf}\hspace{2mm}
\includegraphics[height=0.14\textheight]{draws/L11n354.pdf}\hspace{2mm}
\includegraphics[height=0.14\textheight]{draws/L8a21.pdf}
\\
$$
\otet10_{0006}(L8a20) \qquad
\otet10_{0042}(L10n88) \qquad
\otet10_{0008}(L11n354) \qquad
\otet10_{0011}(L8a21)
$$
\includegraphics[height=0.14\textheight]{draws/L10n101.pdf}\hspace{5mm}
\includegraphics[height=0.14\textheight]{draws/L12n2201.pdf}\hspace{5mm}
\includegraphics[height=0.14\textheight]{draws/L10n113.pdf}
\\
$$
\otet10_{0014}(L10n101) \qquad
\otet10_{0028}(L12n2201) \qquad
\otet10_{0027}(L10n113)
$$
\includegraphics[height=0.15\textheight]{draws/L12n1739.pdf}\hspace{5mm}
\includegraphics[height=0.15\textheight]{draws/L14n38547.pdf} \hspace{5mm}
\includegraphics[height=0.15\textheight]{draws/L14n24613.pdf}
\\
$$
\otet10_{0043}(L12n1739) \qquad
\otet08_{0009}(L14n38547) \qquad
\otet08_{0001}(L14n24613)
$$
\includegraphics[height=0.15\textheight]{draws/mv8_11.pdf}\hspace{5mm}
\includegraphics[height=0.15\textheight]{draws/mv10_16.pdf} \hspace{5mm}
\includegraphics[height=0.15\textheight]{draws/mv10_17.pdf}\hspace{5mm}
\includegraphics[height=0.15\textheight]{draws/mv10_37.pdf}
\\
$$
\otet08_{0005} \qquad \qquad \quad
\otet10_{0007} \qquad \qquad \qquad \quad
\otet10_{0003} \qquad \qquad \quad
\otet10_{0025}
$$
\caption{The tetrahedral links with at most 10 tetrahedra.}
\label{f.tetlinks}
}
\end{figure}

\begin{figure}[!hptb]
\centering{
\includegraphics[height=0.14\textheight]{draws/otet12_0001.pdf}\hspace{5mm}
\includegraphics[height=0.14\textheight]{draws/otet12_0005.pdf}\hspace{5mm}
\\
$$
\otet12_{0001} \qquad \qquad \qquad \quad \otet12_{0005}
$$
\includegraphics[height=0.14\textheight]{draws/otet12_0006.pdf}\hspace{5mm}
\includegraphics[height=0.14\textheight]{draws/otet12_0010.pdf}\hspace{5mm}
\\
$$
\otet12_{0006} \qquad \qquad \qquad \quad \otet12_{0010}
$$
\\
\includegraphics[height=0.14\textheight]{draws/L10a157.pdf}\hspace{12mm}
\includegraphics[height=0.14\textheight]{draws/L12n2208.pdf}\hspace{12mm}
\includegraphics[height=0.14\textheight]{draws/L13n9382.pdf}\hspace{2mm}
\\
$$
\otet12_{0007}(L10a157) \qquad
\otet12_{0009}(L12n2208) \qquad
\otet12_{0018}(L13n9382)
$$
\caption{The tetrahedral links with 12 tetrahedra.}
\label{f.tetlinks12}
}
\end{figure}

\begin{figure}[!hptb]
\centering{
\includegraphics[height=0.20\textheight]{draws/otet20_0570.pdf}
\caption{The remarkable link $\otet20_{0570}$.}
\label{fig.remarkable}
}
\end{figure}

\subsection*{Acknowledgment}
E.F., V.T. and A.V. were supported in part by the Ministry of Education and
Science of the Russia (the state task number 1.1260.2014/K) and RFBR grant
16-01-00414. S.G. was
supported in part by a National Science Foundation grant DMS-14-06419. M. G.
was supported in part by a National Science Foundation grant DMS-11-07452.

We would like to thank Frank Swenton for adding the features to the
Kirby calculator \cite{kirbycalculator} necessary to draw the new
tetrahedral links, and Cameron Gordon for completing the proof of Lemma
\ref{lem.homology.link}.

\bibliographystyle{hamsalpha}

redpajama_set_name: RedPajamaArXiv


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text:
Three reasons why Russia can beat Portugal
Corners could work in the hosts' favour tomorrow
Russia's focal point in attack Fyodor Smolov has been unstoppable recently
The crowd at the Spartak Stadium will be the 12th man
By Igor Borunov with Russia
European champions Portugal are viewed as the favourites going into Wednesday's game against the hosts, especially following Cristiano Ronaldo's superb performance against Mexico on Sunday. However, there are a few reasons why Russia can cause an upset against the eighth-best side in the world according to the FIFA/Coca-Cola Ranking.
Russia's supremacy in the air Before the tournament, head coach Stanislav Cherchesov identified set pieces as one of the Sbornaya's biggest threats. Russia scored from a dead-ball situation in three consecutive warm-up matches ahead of the FIFA Confederations Cup 2017 – against Belgium, Hungary and Chile – and centre-back Viktor Vasin is proving especially dangerous in the air, with two goals in his last four appearances.
Vasin came inches away from improving his goalscoring record in the seventh minute of the Opening Match against New Zealand on Saturday, when he nodded Aleksandr Samedov's corner onto the post. As for Portugal, Fernando Santos' players actually conceded from a corner late on against Mexico, Hector Moreno leaping highest in the box to salvage a draw.
Smolov is on fire Striker Fyodor Smolov wholly repaid the faith shown in him against the All Whites, coming away with the award for Man of the Match. He started and finished off the move that made it 2-0 in the second half and also completed more successful dribbles (nine) than anyone else in the first round of fixtures.
This follows back-to-back seasons in which Smolov has finished top of the goalscoring charts in the Russian Premier League. Over the past two years, he has scored 55 goals in all competitions, including six for the national team. Nevertheless, he is modest when it comes to his achievements. "I don't consider myself a hero at all and we haven't achieved anything just yet," he told FIFA.com. "We've beaten New Zealand; now let's get ready for a tough encounter with Portugal."
Phenomenal support at the Spartak Stadium More than 50,000 fans cheered Russia on during the Opening Match in Saint Petersburg and, while the capacity at the Spartak Stadium is a little lower, the support is expected to be even more passionate.
The excitement around the Portugal game has been simmering in Russia for a long time and a sell-out was announced well before the start of the tournament. The Spartak Stadium is sure to be bouncing on Wednesday, especially after the positive start against New Zealand.
It is also worth remembering this arena is home to champions, as Spartak Moscow won the Russian title here for the first time in 16 years just a few months ago. The local crowd is therefore used to the winning feeling. The Red-Whites have the largest contingent of players in the national team (five) and the likes of Denis Glushakov and Aleksandr Samedov are no strangers to scoring great goals on their home turf.

redpajama_set_name: RedPajamaCommonCrawl


The oddsmakers have posted odds for seven NFL teams to make or miss the playoffs.
By comparing our projections to betting odds, we found four teams offering value.
There's only two weeks left in the regular season and the number of playoff spots available are dwindling.
The Kansas City Chiefs and Los Angeles Chargers have punched their tickets to the postseason in the AFC while the New Orleans Saints, Los Angeles Rams and Chicago Bears have clinched playoff berths in the NFC.
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The Colts don't control their own destiny, which is why Indy is plus-money to make the playoffs. In addition to winning out, Andrew Luck & Co. need a loss by either the Pittsburgh Steelers or Baltimore Ravens.
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The Action Network NFL Power Rankings would make the Colts 1.1-point favorites over the Titans in Tennessee.