has at least two connected components. A second component either contains
$n_{j}$ for some $j$ or, for some $k\neq i,$ $\gamma_{k}$ passes under or over
some tie $\kappa$ of $\eta_{i}.$ However, the latter case violates Lemma 4.33
as either $\eta$ is blocked by $\gamma_{b}$ and $\kappa$ or $\eta_{i}$ is
blocked by $\gamma_{i}$ and $\gamma_{k}.$ Thus, $\eta\Cap\eta_{i}$ is non-
empty and contains some horoball of $\eta.$ ∎

######  Lemma 4.38 (No accidental true tangencies).

$B_{k}\cap N_{j}\notin\mathcal{T}$ for all $k=1,2$ and $j=1,\ldots,7.$

###### Proof..

Assume otherwise. Without loss of generality, let $B_{k}=B_{2}$ and
$N_{j}=N_{1}$ with $B_{2}\cap N_{1}\in\mathcal{T}$. Send $B_{2}$ to
$H_{\infty}$ and let $f$ acting on $(\mathcal{H},\mathcal{T})$ take
$\gamma(B_{1},H_{\infty})$ to $\gamma(N_{1},H_{\infty}).$ Then $f(\eta)$ is a
necklace of at most full-sized horoballs that winds around
$\gamma(N_{1},H_{\infty}).$ Projecting to the plane, we obtain piecewise
linear curves $\mu=\pi(\operatorname{frame}(\eta))$ and
$\mu_{f}=\pi(\operatorname{frame}(f(\eta)))$, which are embedded by Lemma
4.34. Each curve then divides the plane into two open regions: the “inside”
containing $b_{1}$ or $n_{1}$ accordingly, and the “outside”.

Notice that if $\gamma$ is a tie of $\operatorname{frame}(f(\eta))$ with one
endpoint inside $\mu$, the other outside $\mu$, and passes over
$\operatorname{frame}(\eta)$, then $\eta$ has non-trivial winding count around
$\gamma$. However, this would contradict the Three Eyes Lemma 4.33 as $\eta$
also has non-trivial winding around $\gamma(B_{1},H_{\infty})$. By symmetry,
the same applies to any tie of $\operatorname{frame}(\eta)$ that has endpoints
on either side of $\mu_{f}$ and goes over $\operatorname{frame}(f(\eta))$.

Since $\mu$ and $\mu_{f}$ are both simple, there are three possible isotopy
types, which we now consider.

Case 1: $\mu_{f}$ winds around $b_{1}$. This is impossible by the the Two Eyes
Lemma 4.32.

Case 2: $\mu_{f}$ does not wind around $b_{1}$ or $\mu_{f}$ does goes through
$b_{1}$. For this to hold, $\mu_{f}$ must intersect the half-open line segment
$[b_{1},n_{1})$ between $b_{1}$ and $n_{1}$. In particular, $\mu_{f}$ must
cross from outside of $\mu$ to the inside, and then again from inside to
outside. Looking at the ties, we have seen that these crossing cannot arise
from $\operatorname{frame}(f(\eta))$ passing over
$\operatorname{frame}(\eta)$. Every underpass of
$\operatorname{frame}(f(\eta))$ under $\operatorname{frame}(\eta)$ must be
cancelled out by another underpass, because a tie of
$\operatorname{frame}(\eta)$ lying over such an underpass must have endpoints
on the same side of $\mu_{f}$. Thus, by parity, we must have at least two
crossings via shared horoballs $A_{1}$ and $A_{2}$. We further assume that a
neighborhood of $a_{i}$ in $\mu_{f}$ has a point inside $\mu$ for $i=1,2$.
This implies that that $A_{1}$ and $A_{2}$ cannot be neighbors in both $\eta$
and $f(\eta)$, as otherwise $\pi(\gamma(A_{1},A_{2}))$ is an edge of both
$\mu$ and $\mu_{f}$, so $\mu_{f}$ would not inside $\mu$. In particular, we
get that the shortest paths between $A_{1}$ and $A_{2}$ in $\eta$ and
$f(\eta)$ are distinct. Gluing them together would give a $\leq 6$-necklace,
contradicting minimality.

We have exhausted all possible avenues, so our proof by contradiction is
complete. ∎

######  Lemma 4.39.

$\eta\Cap\eta_{i}$ cannot contain $N_{i+2}$ or $N_{i-1}$

###### Proof.

Assume otherwise. Notice that
$f_{i}^{-1}\left(\left\\{N_{i},N_{i+1}\right\\}\right)=\left\\{B_{1},B_{2}\right\\}$
and $f_{i}^{-1}(\eta_{i})=\eta$ by definition. If $N_{i+2}$ or $N_{i-1}$ are
in $\eta_{i}$ then $f_{i}^{-1}(N_{i+2})$ or $f_{i}^{-1}(N_{i-1})$ are in
$\eta.$ Then, $f_{i}^{-1}(\gamma_{i+1})$ or $f_{i}^{-1}(\gamma_{i-1})$ is a
tie between $B_{k}$ and $\eta,$ which contradicts Lemma 4.38. ∎

######  Lemma 4.40.

If $\eta\Cap\eta_{i}\neq\emptyset$ then $\eta_{i}$ contains both $B_{1}$ and
$B_{2}.$

###### Proof.

Normalize so that $B_{2}=H_{\infty}$ as before. Looking at the 7-necklace
$\eta$ wrapping around $B_{1},$ we heuristically see that there isn’t very
much room for the 7-necklace $\eta_{i}$ to snake around $N_{i}$ and $N_{i+1}.$
By Lemma 4.39, $\eta$ and $\eta_{i}$ can only share $N_{i+3},N_{i-2},$ or
$N_{i+4}.$ By symmetry, the cases of $N_{i+3}$ and $N_{i-2}$ are identical. We
will assume that $\eta\Cap\eta_{i}$ includes $N_{i+3}.$ The case of $N_{i+4}$
only requires minor modifications to the arguments below.

We claim that $B_{1}$ and $B_{2}$ must both be in the necklace $\eta_{i}.$
First, we prove that if we assume that neither $B_{1}$ nor $B_{2}$ is in
$\eta_{i},$ then we are led to a contradiction. Second, we will prove that if
exactly one of $B_{1}$ or $B_{2}$ is in $\eta_{i},$ then we are also led to a
contradiction.

Case 1: Assume neither $B_{1}$ nor $B_{2}$ is in $\eta_{i}.$

Normalize via a Möbius action $g$ on $(\mathcal{H},\mathcal{T})$ sending
$N_{i}$ to $H_{\infty}$ and $N_{i+1}$ to a full-sized horoball. We will denote
$g(N_{i+3})$ as $C_{3}$; it is preceded and succeeded in the necklace
$g(\eta_{i})$ by $C_{2}$ and $C_{4};$ see Figure 16. Note that we have drawn
parts of the two necklaces $g(\eta)$ and $g(\eta_{i}),$ and these agree at
$g(N_{i+3})=C_{3}.$ Note also that $C_{2}$ or $C_{3}$ could possibly be
$g(N_{i+4}),$ but no other necklace horoball nor $B_{1},B_{2}$.

The necklace $g(\eta_{i})$ consists of horoballs that are less than or equal
to full-size, and is blocked by $g(\gamma_{i}).$ In particular, $g(\eta_{i})$
must encircle the full-sized horoball $g(N_{i+1}).$ If $g(\eta_{i})$ were a
6-necklace, then all the horoballs in it would have to be full-sized and form
a hexagonal pattern with $g(N_{i+1}).$ Because $g(\eta_{i})$ is a 7-necklace
we have more room to maneuver, but not much more room as seen in Lemma 4.34.

\begin{overpic}[scale={1.0}]{figures/necklace_sharing.pdf}
\put(45.0,43.0){$B_{1}$} \put(17.0,61.0){$N_{i+3}$} \put(78.0,41.0){$N_{i}$}
\put(64.0,73.5){$N_{i+1}$} \end{overpic}

\begin{overpic}[scale={1.4}]{figures/necklace_N_view_sharing.pdf}
\put(2.0,31.0){$g(N_{i+1})$} \put(63.0,60.0){$g(B_{1})$}
\put(81.0,15.0){$g(B_{2})$} \put(30.0,14.0){$g(N_{i+2})$}
\put(72.0,30.0){$g(N_{i+4})$} \put(42.5,42.5){$C_{2}$}
\put(55.0,30.0){$C_{3}$} \put(51.0,15.0){$C_{4}$} \end{overpic}

Figure 16: (Left) $Z$ horoball and view from $B_{2}$. (Right) View from
$g(N_{i})=H_{\infty}$ in Case 1.

Our main focus will be on the five horoballs $C_{2},$ $C_{3},$ $C_{4},$
$g(B_{1}),$ $g(B_{2}).$ The horoball centers run around $C_{3}$ in a pattern
either of the form $g(B_{\ast}),g(B_{\ast}),C_{\ast},C_{\ast}$ (“BBCC”) or of
the form $g(B_{\ast}),C_{\ast},g(B_{\ast}),C_{\ast}$ (“BCBC”). (Degeneracies
of collinearity can be construed in either of these categories, since the
inequalities of Lemma 4.34 are not strict.) We will first treat the former,
more difficult case.

Case 1A: $B_{1},B_{2}\notin\eta_{i},$ $BBCC$ pattern.

We will analyze four visual angles based at $C_{3}$: $\angle(C_{2},\ C_{3},\
C_{4})$; $\angle(C_{4},\ C_{3},\ g(B_{2}))$; $\angle(g(B_{2}),\ C_{3},\
g(B_{1}))$; and $\angle(g(B_{1}),\ C_{3},\ C_{2}).$ Of course, these angles
can also be interpreted as dihedral angles at the geodesic running from
$C_{3}$ to $\infty.$ The four angles should add up to $2\pi$, but because of
various constraints we will show that the calculated sum is in excess of
$2\pi$, thereby obtaining a contradiction.

Claim I: $\angle(g(B_{2}),\ C_{3},\ g(B_{1}))\geq 2\pi/3.$

First note that this is the dihedral angle of the edge from $C_{3}$ to
$\infty$ in the ideal tetrahedron formed by the centers of $g(B_{2}),\ C_{3},\
g(B_{1})$ and $\infty.$ Consider the isometric tetrahedron formed by the
centers of $B_{2},\ N_{i+3},\ B_{1},\ N_{i}$ (that is, apply $g^{-1}$ to the
previous ideal tetrahedron). The dihedral angle at the edge from $C_{3}$ to
infinity in the first tetrahedron corresponds to the dihedral angle at the
edge from $N_{i+3}$ to $N_{i}$ in the second tetrahedron, but this dihedral
angle is equal to the dihedral angle at the edge from $B_{1}$ to $B_{2}$ (note
that the center of $B_{2}$ is infinity). The dihedral angle at
$\gamma(B_{1},B_{2})$ is equal to the visual angle from the center of $N_{i}$
to the center of $N_{i+3}$ measured at the center of $B_{1}.$ This visual
angle must be at least $2\pi/3$ by Lemma 4.34.

This proves Claim I.

Claim II: $\angle(C_{2},\ C_{3},\ C_{4})\geq 2\pi/3.$

First, we contend that the Euclidean height of $C_{3}$ is between $1/4$ and
$1/3.$ The $\leq 1/3$ constraint is equivalent to showing
$d_{\mathbb{H}}(N_{i},N_{i+3})\geq\log 3$. To prove this, consider the picture
with $B_{2}$ at infinity and refer to Lemma 4.34. Since
$\exp(d_{\mathbb{H}}(N_{i},N_{i+1}))=d_{\mathbb{E}}(n_{i},n_{i+3})^{2}/(h_{i}h_{i+3})$,
we just need to show $d_{\mathbb{E}}(n_{i},n_{i+3})^{2}\geq 3h_{i}h_{i+3}$.
The law of cosines gives

$d_{\mathbb{E}}(n_{i},n_{i+3})^{2}=x_{i}^{2}+x_{i+3}^{2}-2x_{i}x_{i+3}\cos\beta\geq
x_{i}^{2}+x_{i+3}^{2}+x_{i}x_{i+3}\geq h_{i}+h_{i+3}+\sqrt{h_{i}h_{i+3}}\geq
3h_{i}h_{i+3},$

where $\beta\geq 2\pi/3$ is the visual angle measured at $B_{1}$ and we use
the fact that $\sqrt{h_{i}}\leq x_{i}$ for all $i$. The last inequality
follows from basic calculus. Thus, the height of $C_{3}$ is at most $1/3$.

The lower bound of $1/4$ also comes from Lemma 4.34 (8), where a 7-chain of at
most full-sized horoballs surrounding a full-sized horoball (here $N_{i+1}$)
must have all horoballs of Euclidean height at least $1/4$.

Using these height bounds and the fact that $g(\eta_{i})$ winds around
$\gamma(g(N_{i}),g(N_{i+1}))$, we now show that the visual angle
$\angle(C_{2},\ C_{3},\ C_{4})$ is at least $2\pi/3$.

\begin{overpic}[scale={1.1}]{figures/local_nekl} \put(18.0,47.0){$\alpha_{1}$}
\put(18.0,61.0){$\alpha_{2}$} \put(-55.0,56.0){$g(N_{i+1})$ center}
\put(65.0,0.0){$C_{4}$ center, height $h_{1}$} \put(90.0,47.0){$C_{3}$ center,
height $h$} \put(70.0,96.0){$C_{2}$ center, height $h_{2}$}
\put(67.0,38.0){$\beta_{1}$} \put(67.0,62.0){$\beta_{2}$}
\put(77.0,23.0){$d_{1}$} \put(77.0,75.0){$d_{2}$} \put(26.0,20.0){$y_{1}$}
\put(26.0,84.0){$y_{2}$} \end{overpic} Figure 17: Diagram for Case II of Lemma
4.40.

The argument is similar to the arguments proving Lemma 4.34. With labels as in
Figure 17, the law of cosines gives

$\cos(\beta_{1}+\beta_{2})=\frac{d_{1}^{2}-y_{1}^{2}+d_{2}^{2}-y_{2}^{2}+2\cos(\alpha_{1}+\alpha_{2})y_{1}y_{2}}{2d_{1}d_{2}}$

One can then see that the maximum value of $\cos(\beta_{1}+\beta_{2})$ must be
on the diagonal $(y_{1},\alpha_{1})=(y_{2},\alpha_{2})$ as a quick gradient
analysis shows that we can always flow towards this diagonal while increasing
$\cos(\beta_{1}+\beta_{2})$. Since necklace horoballs are tangent, we know
$d_{i}^{2}=hh_{i}$. With $y=y_{i}$ and $\alpha=\alpha_{i}$, we have

$\cos(\beta_{1}+\beta_{2})\leq\frac{d_{1}^{2}+d_{2}^{2}-4\sin(\alpha)y^{2}}{2d_{1}d_{2}}=\frac{hh_{1}+hh_{2}-4\sin(\alpha)y^{2}}{2h\sqrt{h_{1}h_{2}}}=\frac{h_{1}+h_{2}}{2\sqrt{h_{1}h_{2}}}+\frac{2}{h}\frac{y}{\sqrt{h_{1}}}\frac{y}{\sqrt{h_{2}}}\sin^{2}(\alpha)$

Since we only need to consider realizable parameters on the diagonal, Lemma
4.34 gives that $y/\sqrt{h_{i}}\geq 1$ and
$\alpha=(\alpha_{1}+\alpha_{2})/2\geq\pi/6$, so we must have

$\cos(\beta_{1}+\beta_{2})\leq\frac{h_{1}+h_{2}}{2\sqrt{h_{1}h_{2}}}-\frac{2}{h}\frac{1}{4}\leq
1-\frac{3}{2}=-\frac{1}{2}$

where we use the fact that $1/4\leq h\leq 1/3$ and a straightforward upper
bound on the first term as $1/4\leq h_{i}\leq 1$. Thus, the visual angle
$\angle(C_{2},\ C_{3},\ C_{4})=\beta_{1}+\beta_{2}$ is at least $2\pi/3$.

This proves Claim II.

If we can show that the other two angles $\angle(C_{4},\ C_{3},\ g(B_{2}))$
and $\angle(g(B_{1}),\ C_{3},\ C_{2})$ are each greater than $\pi/3$, then we
will have our contradiction. To get control of these angles we first need
lower bounds for the sizes of $C_{2}$ and $C_{4}.$ To minimize $C_{2}$ we must
first maximize $C_{3},$ so we assume that the height of $C_{3}$ is $1/3.$ We
start with a warm-up example. Assume $C_{2}$ and $C_{4}$ are of the same
height, $h^{2}.$ Then the distance between the centers of $C_{2}$ and $C_{3}$
is $h/(\sqrt{3}),$ as is the $C_{3}$ to $C_{4}$ distance. The sum of these two
distances must be greater than or equal to the distance between the centers of
$C_{2}$ and $C_{4},$ which, as we saw above, is at least 1. So the smallest
allowable height for $C_{2}=C_{4}$ satisfies $2h/\sqrt{3}=1,$ and we see that
the smallest height for $C_{2}=C_{4}$ is $3/4.$ We have finished the warm-up
example.

If we let one of the two horoballs, say $C_{4},$ increase then the smallest
height for the other horoball decreases. The worst case is when $C_{4}$ has
height 1, in which case the distance between the centers of $C_{3}$ and
$C_{4}$ is $1/\sqrt{3},$ and the sum of the distances between the centers of
$C_{2},C_{3}$ and $C_{3},C_{4}$ is $(h+1)/\sqrt{3}.$ We compare that with the
distance between the centers of $C_{2}$ and $C_{4}.$ We can assume the center
of $C_{2}$ is $(-1/2,\sqrt{h^{2}-1/4})$ and the center of $C_{4}$ is
$(1/2,\sqrt{3}/2).$ We can then calculate that the minimum height is
$0.62127.$444 One might wonder whether relaxing the condition that the
horoball $C_{2}$ abuts the horoball $g(N_{i+1})$ in our setup might allow for
a smaller $C_{2}$? (Similarly for $C_{4}.$) However, a straightforward
calculation shows that rolling $C_{2}$ along $C_{1}$ does not allow for a
decrease in the size of $C_{2}.$ With these height bounds established, we make
the following claim.

Claim III: $\angle(C_{4},\ C_{3},\ g(B_{2}))>\pi/3.$

We argue by contradiction and start by assuming that the angle is less than
$\pi/3$. We know that $C_{4}$ abuts $C_{3},$ and although $C_{4}$ and
$g(B_{2})$ might not abut, the smallest angle at $C_{3}$ will occur when
$C_{4}$ abuts $g(B_{2}),$ so we will make this assumption. Note, we have
freedom to rotate $g(B_{2})$ around $C_{3}$ to make this happen since it is
not a necklace horoball and since the height of $C_{4}$ is at least $1/4$ by
Lemma 4.34 (i.e under the rotation we cannot not miss $C_{4}$ because it is
big enough).

We begin by getting some control over the height of $g(B_{2})$; this is the
same as the height of $N_{i}$ in the original picture (both depend on the
hyperbolic distance from $B_{2}$ to $N_{i}$). Consider the triangle formed by
the centers of the 3 horoballs $N_{i+3},\ B_{1},\ N_{i}$ in the original
horoball diagram. It will be convenient to use some notation from [GMM]: We
will work with the so-called Euclidean lengths between horoballs; given two
horoballs separated by hyperbolic distance $o(n),$ the associated Euclidean
length is $\exp(o(n)/2).$ We will often associate to a single horoball $X$ its
Euclidean distance denoted $e_{X}$ from $H_{\infty}$; we can compute that the
Euclidean height of this horoball is then $1/e_{X}^{2}.$ As computed before,
if $e_{p},e_{q}$ are Euclidean lengths from two horoballs to $H_{\infty},$ and
$e_{r}$ is the Euclidean length between those two horoballs, then the
Euclidean distance between the centers of these two horoballs is ${e_{r}\over
e_{p}e_{q}}$ as given by Lemma 4.28.

In our set up, let $e_{p}$ and $e_{q}$ be the Euclidean lengths associated to
the horoballs $N_{i}$ and $N_{i+3}$ respectively. Note that $e_{p}$ is the
Euclidean length from horoball $N_{i}$ to horoball $B_{2},$ which is also the
Euclidean length from $g(N_{i})$–which is $H_{\infty}$–to $g(B_{2}).$ Let
$e_{r}$ be the Euclidean length from horoball $N_{i}$ to horoball $N_{i+3},$
and $d$ the Euclidean distance between their centers. Note that the Euclidean
length of the (full-sized) horoball $B_{1}$ is 1.

Consider the triangle formed by the centers of $N_{i},\ B_{1},\ N_{i+3}$ with
edges of length $y_{1},y_{2}$ and $d$, which was defined above. Since $B_{1}$
has height $1$, we know that $y_{1}\geq 1/e_{p}$ and $y_{2}\geq 1/e_{q}$. Let
$\alpha$ be the angle at the center of $B_{1}.$ By the Euclidean Law of
Cosines, we have $d^{2}=y_{1}^{2}+y_{2}^{2}-2\cos(\alpha)y_{1}y_{2}.$ As
observed above, $\alpha$ is between $2\pi/3$ and $\pi$. Hence,
$1\leq-2\cos(\alpha)\leq 2$ and $d^{2}\geq
1/e_{p}^{2}+1/e_{q}^{2}+1/(e_{p}e_{q}).$ So,

$(e_{r}/(e_{p}e_{q}))^{2}\geq 1/e_{p}^{2}+1/e_{q}^{2}+1/(e_{p}e_{q})$
$e_{r}^{2}\geq e_{q}^{2}+e_{p}^{2}+e_{p}e_{q}$ $0\geq
e_{p}^{2}+(e_{q})e_{p}+(e_{q}^{2}-e_{r}^{2})$

So, we have a quadratic inequality in $e_{p}$ and the solution is

${-e_{q}-\sqrt{4e_{r}^{2}-3e_{q}^{2}}\over 2}\leq
e_{p}\leq{-e_{q}+\sqrt{4e_{r}^{2}-3e_{q}^{2}}\over 2}$

The broadest range of values is when $e_{r}$ is maximized and $e_{q}$ is
minimized. Because $e_{r}$ is the Euclidean length from $N_{i}$ to $N_{i+3},$
i.e., the Euclidean length from $g(N_{i})$ to $C_{3},$ we see that
$\sqrt{3}\leq e_{r}\leq 2$ (corresponding to the height of the horoball
$C_{3}$ being in the range from $1/3$ to $1/4$). So the maximum value for
$e_{r}$ is 2. The minimum value for $e_{q}$ is 1 and this occurs when
$N_{i+3}$ is a full-sized horoball. Plugging into the formula for $e_{p}$ we
see that the range of possible values for $e_{p}$ is from 1 to
$(-1+\sqrt{13})/2=1.302775\dots.$ This is a crude bound and we can improve it
if we have better control over $e_{q}.$

$e_{q}$ is the Euclidean length from $N_{i+3}$ to $B_{2},$ which is equal to
the Euclidean length from $C_{3}$ to $g(B_{2}).$ The Euclidean distance $t$
between the centers of the horoballs $g(B_{2})$ and $C_{3}$ is then
$t=e_{q}/(e_{p}e_{C_{3}}).$ We will analyze $t$ values (so that we can control
$e_{q}$ values) by studying the angle $\angle(C_{4},\ C_{3},\ g(B_{2}))$; we
want to eliminate $t$ values that result in an angle (at $C_{3}$) of more than
$\pi/3$. $C_{3}$ and $C_{4}$ abut, and although $C_{4}$ and $g(B_{2})$ do not
necessarily abut, we can swing $C_{4}$ towards $g(B_{2})$ until they do abut
and in so doing, decrease the angle at $C_{3}.$ As such, we will assume that
$C_{4}$ and $g(B_{2})$ abut.

Thus we have a Euclidean triangle formed by the centers of $C_{4},\ C_{3},\
g(B_{2}).$ We should denote the associated Euclidean lengths by
$e_{C_{4}},e_{C_{3}},e_{g(B_{2})},$ but in the interest of simplifying some
formulas we will abuse notation and denote them respectively as $C_{4},\
C_{3},\ B_{2}.$ Note that in this notation $B_{2}=e_{p}.$ Now, the edges of
the triangle are $t,\ 1/(B_{2}C_{4}),\ 1/(C_{3}C_{4})$; note that
$1/(B_{2}C_{4})$ is the edge opposite the angle at $C_{3}$ which we will call
$\gamma.$ Using the Euclidean Law of Cosines, we have

$\displaystyle 1/(B_{2}C_{4})^{2}$
$\displaystyle=t^{2}+1/(C_{3}C_{4})^{2}-(2t/(C_{3}C_{4}))\cos(\gamma)$
$\displaystyle 0$
$\displaystyle=t^{2}+(-2\cos(\gamma)/(C_{3}C_{4}))t+1/(C_{3}C_{4})^{2}-1/(B_{2}C_{4})^{2}$

Hence,

$t=\frac{1}{2}\left(2\cos(\gamma)/(C_{3}C_{4})\pm\sqrt{(2\cos(\gamma)/(C_{3}C_{4}))^{2}-4((1/(C_{3}C_{4})^{2}-(1/(B_{2}C_{4})^{2})}\right)$
$t=\frac{1}{C_{3}C_{4}B_{2}}\left(B_{2}\cos(\gamma)\pm\sqrt{B_{2}^{2}\cos^{2}(\gamma)-B_{2}^{2}+C_{3}^{2}}\right)$

So,

$e_{q}=C_{3}e_{p}t=C_{3}B_{2}t=\frac{1}{C_{4}}\left(B_{2}\cos(\gamma)\pm\sqrt{B_{2}^{2}\cos^{2}(\gamma)-B_{2}^{2}+C_{3}^{2}}\right)$

We want to resolve the choice of $\pm$ above. Recall that we have the bounds
$1\leq C_{2},C_{3}\leq 2$, since the horoballs have heights at least $1/4$,
$1\leq B_{2}=e_{p}\leq 1.302775\dots$ and $0\leq\gamma\leq\pi/3$, by above.
Since $e_{q}\geq 1$ by definition, it is easy to see that over these
parameters the must choose $+$ in the $\pm$ above.

What’s the smallest allowable $e_{q}$ (where the angle $\gamma$ is between 0
and $\pi/3$)? Previously, we only needed the trivial bound $e_{q}\geq 1,$ but
here we can do better. Now, first, we take the smallest $\cos(\gamma),$ i.e.
$1/2,$ to obtain

$e_{q}=\frac{1}{2C_{4}}\left(B_{2}+\sqrt{B_{2}^{2}-4B_{2}^{2}+4C_{3}^{2}}\right)=\frac{1}{2C_{4}}\left(B_{2}+\sqrt{-3B_{2}^{2}+4C_{3}^{2}}\right)$

Next, we take the smallest $C_{3},$ which is $\sqrt{3}$ to obtain

$e_{q}=\frac{1}{2C_{4}}\left(B_{2}+\sqrt{12-3B_{2}^{2}}\right)$

The minimum occurs at the largest allowable value of $B_{2}=e_{p},$ which is
$1.302775\dots.$ Combining this with our above upper bound on
$C_{4}=\sqrt{1/0.62127}=1.2687,$ we get that $e_{q}\geq 1.54928.$

We now have better control over $e_{q}$ and we can go back up to our $e_{p}$
control calculation and improve it.

$\displaystyle e_{p}$
$\displaystyle\leq\left(-e_{q}+\sqrt{4e_{r}^{2}-3e_{q}^{2}}\right)/2$
$\displaystyle\leq-1.5492+\sqrt{16-3(1.5492)^{2}})/2$ $\displaystyle\leq
0.708532$

This is a contradiction, and we conclude that our original assumption that the
angle at $C_{3}$ is $\pi/3$ or less is incorrect.

This proves Claim III.

Claim IV: $\angle(g(B_{1}),\ C_{3},\ C_{2})>\pi/3$.

Taking our original set-up we can invert in the totally geodesic plane
equidistant from $B_{1}$ and $B_{2}.$ We then redo the analysis with $B_{1}$
replacing $B_{2},$ and $C_{2}$ replacing $C_{4}.$ We see that the angle at
$C_{3}$ must be greater than $\pi/3$.

This proves Claim IV, and thus concludes the proof of Case 1A.

Case 1B: $B_{1},B_{2}\notin\eta_{i},$ $BCBC$ pattern.

We operated under the assumption that the cyclic positioning of the horoballs
around $C_{3}$ was (clockwise) $B_{1},B_{2},C_{4},C_{2}.$ We need to rule out
$B_{1},C_{4},B_{2},C_{2}$ (the other case follows similarly). In particular,
we need to rule out that $C_{4}$ lies in the angle $B_{1},C_{3},B_{2}.$ We
argue by contradiction and assume that $C_{4}$ is within that angle. We
construct the triangle with centers at $B_{1},B_{2},C_{4}$ and see that its
angle at $C_{4}$ is maximized when the Euclidean distance between the centers
of $B_{1}$ and $B_{2}$ is 1. Denote the distance between the centers of
$B_{1}$ and $C_{4}$ as $a$ and the distance between the centers of $B_{2}$ and
$C_{4}$ as $b.$ Then
$a^{2}+b^{2}\geq(\sqrt{0.62127})^{2}+(\sqrt{0.62127})^{2}>1.$ Now, using the
Euclidean Law of Cosines, we see that the angle at $C_{4}$ must be less than
$\pi/2$. Consequently, the angle $B_{1},C_{3},B_{2}$ must be less than $\pi/2$
as well, which is a contradiction.

This proves Case 1B, and finishes the proof of Case 1.

Case 2: Assume $B_{1}$ or $B_{2},$ but not both, in $\eta_{i}$.

The previous argument is still valid as long as $g(B_{1})$ is not $C_{2}$
(similarly for $g(B_{2})$ and $C_{4}$). But, if $g(B_{1})$ were $C_{2}$ then
there would be a tangency in $\mathcal{T}$ between $g(B_{1})$ and $C_{3},$
hence $B_{1}$ and $N_{i+3}.$ But the latter is not allowed by Lemma 4.38,
above.

This proves Case 2.

Therefore, $B_{1},B_{2}\in\eta_{i}.$ ∎

We can now combine all the pieces to prove part of the main result of of this
section.

######  Proposition 4.41.

If $\eta$ is a globally minimal $\leq 7$-necklace in an oriented geometric
horoball system $(\mathcal{H},\mathcal{T}),$ then $\eta$ is unknotted,
unblocked, and unlinked.

###### Proof.

By Theorem 4.17, we know that $\eta$ is unknotted. Assume, by way of
contradiction that $\eta$ is blocked by some $\gamma_{b}=\gamma(B_{1},B_{2})$
for $b\in\mathcal{T}.$ We can further assume that
$b\in\mathcal{T}^{\prime\prime}$ since $\eta$ is minimal and we can make the
unknotting disk disjoint form $\mathcal{T}^{\prime}$ ties by Lemma 4.26. Thus,
the $N_{i}$ are at most full size. It was shown in [AK13, Proposition 2.8]
that all $\leq 6$-necklaces in an oriented geometric horoball system are
unblocked. Note, however, [AK13] gives an example of a non-orientable
3-manifold with a blocked 6-necklace.

We now turn to the case where $k=7.$

By Lemmas 4.37 and 4.40, $\eta_{i}$ contains both $B_{1}$ and
$B_{2}=H_{\infty}$ for all $i=1,\ldots,7.$ By Lemma 4.35, each $\eta_{i}$
contains the tie $\gamma_{b}.$ In particular, this implies that $\eta_{i}$
must be one of the seven conjugates of $\eta$ that pass though $\gamma_{b}.$
Further, $\eta_{i}\neq\eta_{j}$ for $i\neq j$ as $f_{i}\neq f_{j}.$ Thus
$\\{\eta_{i}\\}$ is exactly the set of seven conjugates of $\eta$ passing
through $\gamma_{b}.$

Let $\zeta_{i}=f_{i}^{-1}(\eta),$ which are again all distinct. Both
$\eta_{i}$ and $\zeta_{i}$ contain the tie $\gamma_{b}$ and are conjugates of
$\eta.$ Thus, $\eta_{i}=\zeta_{j}$ for some pair $(i,j).$ Since
$i,j=1,\ldots,7,$ by parity we have that $\eta_{k}=\zeta_{k}$ for some $k.$
Then $f_{k}(\eta)=f_{k}^{-1}(\eta)$ and $f_{k}^{2}(\eta)=\eta.$ This is a
contradiction as $f_{k}^{2}$ cannot have finite order.

Therefore, $\eta$ must be unblocked. By Proposition 4.20, $\eta$ is therefore
unlinked. ∎

###  4.5 Full Necklace Structures

We will need an additional condition on our compressing disks. This condition
is called “fullness” and will allow us to enumerate manifolds combinatorially
in the next section.

Throughout this section, we use the notation of [GMM11].

######  Definition 4.42.

Let $M$ be a compact 3-manifold with $\partial M$ a union of tori and
$B\subset\partial M$ is either a torus, a disc, or one or two annuli. A
general-based necklace-$n$ structure $(M,B,\Delta)$ is a handle structure on
$M$ of the form $B\times I$ with a single 1-handle and a single valence-n
2-handle attached to $B\times 1$, where we require that if $I_{1},I_{2}\subset
B\times 1$ are the islands, then the core of each bridge is an essential arc
in $B\times 1\smallsetminus\operatorname{int}(I_{1}\cup I_{2}).$

######  Definition 4.43.

A necklace-$n$ structure is a general based necklace-$n$ structure where $B$
is a torus.

######  Definition 4.44.

We say that $\Delta$ is full if every lake of $\Delta$ is a disc.

Following the convention of [GMM11], a compact hyperbolic 3-manifold $(N,T),$
is a manifold $N$ whose interior supports a complete hyperbolic structure of
finite volume, where $T$ is a component of $\partial N,$ i.e. $T$ identifies a
preferred cusp.

######  Definition 4.45.

Let $N$ be a compact hyperbolic 3-manifold. An internal necklace-$n$ structure
on $N$ consists of a non-elementary embedding $f:M\to N,$ where $(M,B,\Delta)$
is a necklace-$n$ structure and each component of $\partial M$ is either
boundary parallel in $N$ or bounds a solid torus complementary to $M.$

######  Remarks 4.46.

  1. (i)

$f$ as above is $\pi_{1}$-surjective.

  2. (ii)

By adding 1-handles we can subdivide the 2-handle into valence-3 2-handles.
Thus a necklace-n structure gives rise to a Mom-$(n-2)$ structure. On the
other hand a Mom-2 structure gives rise to a necklace$\leq 5$ structure with
the same underlying manifold.

  3. (iii)

Since an internal Mom-$n$ structure on a hyperbolic 3-manifold must have
$n\geq 2$ it follows that an internal necklace structure must have $n\geq 4.$

The following is the necklace analogue of Lemma 1.13 [GMM11].

######  Lemma 4.47.

A non-elementary embedding of the necklace$n$ manifold $M$ into the compact
hyperbolic 3-manifold $N$ will fail to give an internal necklacen structure on
N if and only if some component of $\partial M$ maps to a convolutube. In that
case, a reimbedding of $M,$ supported in a neighborhood of the convolutubes
gives rise to an internal necklace$n$ structure on $N.$ Components of
$\partial M$ that are boundary parallel in $N$ are fixed under the
reimbedding.∎

######  Proposition 4.48.

Let $(N,T)$ be a 1-cusped hyperbolic 3-manifold with the internal necklace-$k$
structure $(M,T,\Delta).$ If $\Delta$ is not full, then there exists a full
necklace-$k^{\prime}$ structure $(M^{\prime},B^{\prime},\Delta^{\prime})$ on
$N$ with $k^{\prime}<k.$ Furthermore, if $k\leq 7$ or $k$ is odd and
$k^{\prime}=k-1,$ then this structure can be chosen so that $B^{\prime}=T.$

######  Remark 4.49.

Note that the key conclusion here is fullness. It is an open question if the
structure can be made hyperbolic. In this paper, we will be shown this to be
true for $k\leq 7.$

###### Proof.

To start with, if $\Delta$ is not full, then each non-disc lake is an annulus.
Indeed, a lake with smaller Euler characteristic can only be attained if all
bridges are parallel (i.e. a punctured torus or a punctured annulus). In that
case, we see that $M$ contains a boundary connect sum of $B\times I$ and a
lens space
$\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\displaystyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\textstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{$\scriptstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{$\scriptscriptstyle\smallsetminus$}}\mspace{-4.0mu}}}\operatorname{int}(N(B^{3})),$
which is impossible Since $\pi_{1}(N)$ is torsion free it follows that the
inclusion map $B\to N$ induces a $\pi_{1}$-surjection, which is a
contradiction.

Through repeated application of Lemmas 4.51 and 4.52 below we will produce
internal necklace structures of smaller and smaller necklace number, hence
will eventually obtain the full necklace-$k^{\prime}$ structure
$(M^{\prime},B^{\prime},\Delta^{\prime}).$ Repeat applications may be
necessary since the result at a given step may have annular lakes. If
$|\partial M^{\prime}|=2$ and $B^{\prime}\neq T,$ then by replacing $\Delta$
by its dual handle structure we can assume that $B^{\prime}=T.$ If $k$ is odd
and $k^{\prime}=k-1,$ then we shall see that only Step 1 will apply, it will
be invoked exactly once and $|\partial M^{\prime}|=2.$ Thus, if $k\leq 7$ we
can either arrange for $B^{\prime}=T$ or $k^{\prime}\leq 5$ and $|\partial
M^{\prime}|=3.$

Assume $k^{\prime}\leq 5$ and $|\partial M^{\prime}|=3.$ We now show that
either $M^{\prime}$ is hyperbolic or $N$ has a full internal necklace$\leq 5$
structure $(M^{\prime\prime},B^{\prime\prime},\Delta^{\prime\prime})$ with
$M^{\prime\prime}$ hyperbolic and $|\partial M^{\prime\prime}|=2$ in which
case we are finished as before. By Remark 4.46, it follows that
$\Delta^{\prime}$ subdivides to a full Mom-$\leq 3$ structure on $N.$ By
Theorem 4.1 [GMM11] and its proof we can assume that either $M^{\prime}$ is
hyperbolic or $N$ has an internal Mom-2 structure whose underlying manifold is
hyperbolic. In the latter case, using Remark 4.46, it follows that there
exists a full internal necklace-$\leq 5$ structure
$(M^{\prime\prime},B^{\prime\prime},\Delta^{\prime\prime})$ on $N$ such that
$M^{\prime\prime}$ is hyperbolic and is the underlying manifold to a Mom-2
structure. By [GMM11] $|\partial M^{\prime\prime}|=2,$ so we can arrange that
$B^{\prime}=T.$

Finally, if $M^{\prime}$ is hyperbolic, then since $|\partial M^{\prime}|=3,$
it follows from [GMM11] that $M^{\prime}=s776,$ also known as the magic
manifold. Since $\operatorname{Diff}(s776)$ acts transitively on its ends, we
can arrange that $B^{\prime}=T.$ ∎

######  Remark 4.50.

In what follows we make repeated use of the following standard topological
facts about hyperbolic 3-manifolds. Suppose that $N$ is a compact hyperbolic
3-manifold and
$M=N\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\displaystyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\textstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{$\scriptstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{$\scriptscriptstyle\smallsetminus$}}\mspace{-4.0mu}}}\operatorname{int}(N(L))$
where $L$ is a link with components $L_{1},\cdots,L_{n}.$ Suppose that
$A\subset M$ is a properly embedded annulus whose boundary components are
essential in $\partial M.$

  1. 1.

If $\partial A\subset T,$ where $T$ is a component of $\partial N,$ then $A$
is boundary parallel in $N.$

  2. 2.

If $A$ connects $T$ to $\partial N(L_{i}),$ then $|A\cap D_{i}|=1,$ where
$D_{i}$ is a meridian of $N(L_{i}).$

  3. 3.

If $A$ intersects the distinct components $N(L_{i})$ and $N(L_{j}),$ then
$|A\cap D_{m}|=1$ for some $m\in\\{i,j\\}.$

  4. 4.

If $\partial A\subset N(L_{i}),$ then $A$ separates $M.$

###### Proof.

This is obvious possibly except for the case that $A$ hits two tubes. In that
case the tubes and $N(A)$ give a Seifert-fibered space $X$ with two singular
fibers of index $p,q.$ If $\partial X$ is incompressible in $N,$ then this
gives a $\pi_{1}$ contradiction. If $N$ compresses, then there are three
cases. In each case $N$ is a non-trivial connect sum, which is a
contradiction. If the boundary of the compressing disc is parallel to a fiber,
then $N$ is a non-trivial connect sum with two lens spaces. If the boundary
hits once, then it is a connect sum with one lens space. If it hits more than
once, then $N$ is a connect sum with a Seifert-fibered space with three
singular fibers. ∎

######  Lemma 4.51 (Step 1).

If $\Delta$ has a single annular lake and is necklace$n,$ then N has an
internal necklace$<n$ structure.

###### Proof.

First observe that $\partial M$ has two or three components where the latter
occurs if and only if the attaching curve of the 2-handle is separating. Thus,
if $k$ is odd, then $|\partial M|=2.$ Let $A$ be an annulus in $B\times I$
that cuts open the lake. By Remark 4.50, the resulting manifold
$M_{1}=M\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\displaystyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\textstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{$\scriptstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{$\scriptscriptstyle\smallsetminus$}}\mspace{-4.0mu}}}\operatorname{int}(N(A))$
is isotopic to $N$ or
$N\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\displaystyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\textstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{$\scriptstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{$\scriptscriptstyle\smallsetminus$}}\mspace{-4.0mu}}}\operatorname{int}N(K)$
depending if $|\partial M|=2$ or $3.$ $M_{1}$ has a general based necklace
structure $(M_{1},B_{1},\Delta_{1}),$ where $B_{1}$ is an annulus. Note that
$\Delta_{1}$ has two islands $I_{1},I_{2}$ and $k$ bridges. Also $x$ parallel
bridges connect $I_{1}$ to $I_{1},$ $x$ parallel bridges connect $I_{2}$ to
$I_{2},$ and the bridges that connect $I_{1}$ to $I_{2}$ fall into two
parallel families containing $a$ and $b$ bridges respectively. Assume that
$a\geq b.$ Note that $\Delta$ has a single annular lake if and only if $a>0.$
Also that $k=2x+a+b,$ hence $\Delta$ has a single annular lake if $k$ is odd.

Now $M_{1}=B_{1}\times I\cup\eta\cup\sigma,$ where $\eta$ is the 1-handle and
$\sigma$ is the 2-handle. Let $\sigma^{\prime}$ be the core of the 2-handle
and $J=B_{1}\times I\cup\eta.$ Let $E$ be a properly embedded disc in
$B_{1}\times I$ such that $|E\cap\partial\sigma^{\prime}|=2x+b.$ Let
$\alpha\subset\operatorname{int}(J)$ be a simple closed curve disjoint from
$E,$ that goes through $\eta$ once and then parallels part of $B_{1}$ along
the $a$-bridges. Similar to the operation in Lemma 4.6 [GMM11] we can hollow
out $N(\alpha)$ from $M_{1}$ to obtain a new internal necklace-$(2x+b)$
structure $(M_{2},B_{2},\Delta_{2})$ on $N$ where
$M_{2}=M_{1}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\displaystyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\textstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{$\scriptstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{$\scriptscriptstyle\smallsetminus$}}\mspace{-4.0mu}}}\operatorname{int}(N(\alpha)),B_{2}=\partial
N(\alpha),N(E)$ becomes the 1-handle and $\sigma$ continues as the 2-handle. ∎

######  Lemma 4.52 (Step 2).

If $\Delta$ has two annular lakes and is necklace$m,$ then $m=2n$ and $N$ has
an internal necklace$\leq n$-structure.

###### Proof.

Let $M_{1}$ be obtained by splitting $B\times I$ along vertical annuli which
intersect $B\times 1$ in the cores of the annular lakes. Using Remark 4.50
this gives rise to a general based internal necklace structure
$(M_{1},B_{1},\Delta_{1})$ where $B_{1}$ consists of two annuli and $M_{1}$ is
obtained by starting with $B_{1}\times I,$ attaching a 1-handle $\eta$ to the
components of $B_{1}\times 1$ and then attaching a valence-$k$ 2-handle
$\sigma.$ With notation as in Lemma 4.51, we have $a=b=0$ and $2x=k.$ Let $C$
and $C^{\prime}$ be the components of $B_{1}$ and let $E$ be a compressing
disc for $C^{\prime}\times I$ disjoint from $\eta.$ Let
$(M_{2},B_{2},\Delta_{2})$ be the necklace-$k$ 3-manifold constructed as
follows. First let $M_{2}=M_{1}.$ Next observe that $B_{1}\times I\cup\eta$ is
isotopic to $C\times I\cup\eta^{\prime}$ where $\eta^{\prime}$ is a 1-handle
with cocore $E.$ Here $\eta\cup(C^{\prime}\times
I\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\displaystyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{$\textstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{$\scriptstyle\smallsetminus$}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{$\scriptscriptstyle\smallsetminus$}}\mspace{-4.0mu}}}N(E))$
have been absorbed into $C\times I.$ With respect to this structure, $\sigma$
is now of valence $x=k/2.$ The $\Delta_{2}$ necklace structure is based on the
annulus $B_{2}.$ The proof of Step 1 now produces an internal necklace-$\leq
x$ structure $(M_{3},B_{3},\Delta_{3})$ on $N.$ ∎

##  5 Enumerating internal necklace structures

We are almost ready to prove Theorem 1.1. So far, by Sections 2 and 4, we have
shown that if $N$ is a complete hyperbolic 3-manifold with at least one cusp
of maximal cusp volume 2.62, then $N$ has a full, internal necklace-$k$
structure with $k\leq 7$—that is, a nonelementary embedding of a full handle
structure based on a torus, with one 1-handle and one 2-handle of valence at
most 7. In the present section, we enumerate full necklace-$n$ structures
("Problem Bead"), and classify their possible nonelementary embeddings
("Problem NE") to show that our original cusped hyperbolic 3-manifold $N$ is
gotten by Dehn filling of at least one of the 16 cusped hyperbolic 3-manifolds
listed in Table 1. This is all put together in the proof of Theorem 1.1 in
subsection 5.4. We start off by codifying some common notions in the
literature of hyperbolic 3-manifolds.

In the theory of Jørgensen and Thurston, one gets around the infinity of
certain classes of hyperbolic 3-manifolds by showing these classes admit
finite sets of ancestors where every element of the given class is a Dehn
filling of one of these ancestral manifolds.

######  Definition 5.1.

Suppose $\mathcal{S}$ is a set of 3-manifolds. An ancestral set for
$\mathcal{S}$ is a set $\mathcal{A}$ of 3-manifolds such that for every
$s\in\mathcal{S}$, there exists $A\in\mathcal{A}$ and a list $c$ of Dehn-
filling coefficients for the cusps of $A$ such that $s$ is homeomorphic to
$A[c]$.

Every set of 3-manifolds is, trivially, an ancestral set for itself. Usually,
however, the objective is to obtain a finite ancestral set for an infinite
class of 3-manifolds. For instance,

######  Theorem 5.2 (Jørgensen-Thurston).

For all $V>0$ the class $VOL(V)$ of hyperbolic 3-manifolds with volume at most
$V$ admits a finite ancestral set.

Another example of this is the theorem of Gabai, Meyerhoff, and Milley.

######  Theorem 5.3.

The set $\\{m125,m129,\cdots,s959\\}$ is an ancestral set for the class of
1-cusped orientable hyperbolic 3-manifolds with volume at most 2.848.

Our main result, Theorem 1.1, likewise gives a finite ancestral set for the
class of orientable hyperbolic 3-manifolds with a maximal cusp of volume at
most 2.62. Seeing as how we have proven that every such 3-manifold admits a
nonelementary embedding from a necklace manifold of at most 7 beads, it
behooves us to solve the following computational problems.

######  Definition 5.4.

A generalized link exterior is the interior of a compact connected orientable
3-manifold555Some authors require that the given closed 3-manifold be
irreducible. We do not make this requirement. whose boundary is a disjoint
union of at least one torus.

######  Definition 5.5.

Suppose $M$ is a compact 3-manifold. $NE^{+}(M)$ is the class of all finite-
volume hyperbolic manifolds into which $M$ admits a nonelementary embedding.

To make the problem of finding $NE^{+}(M)$ computationally accessible and
algorithmic, we replace $M$ with an oriented finite triangulation666This in
not an ideal triangulation, but a triangulation of the compact manifold $M$
with boundary..

######  Problem 5.6 (NE).

Given an oriented finite triangulation $\mathscr{T}$ of $M$, find an algorithm
to compute a finite ancestral set for $NE^{+}(M)$ consisting of hyperbolic
3-manifolds.

######  Problem 5.7 (Bead).

Given a natural number $n$, to enumerate the necklace manifolds with $n$
beads.

The reason for the restriction in problem NE to hyperbolic 3-manifolds is that
they admit good algorithms for determining hyperbolicity of Dehn fillings.

The main results of this section are solutions to problems NE and Bead. The
solution to problem NE relies upon algorithms in normal surface theory and a
reformulation of results in [GMM11]. The solution to problem Bead is a simple
exercise in recursion.

###  5.1 Problem NE

Before beginning the solution to problem NE, we note in passing the following
fact: if $A$ and $B$ are ancestral sets, respectively, for sets $U$ and $V$ of
3-manifolds, and $W\subset U\cup V$, then $A\cup B$ is an ancestral set for
$W$.

####  5.1.1 Nonelementary embeddings

We first show that Definition 5.1 gives just the set of hyperbolic Dehn
fillings of $M$, though prima facie it could give a proper superset of these
fillings.

######  Lemma 5.8.

If $M$ is a generalized link exterior, then $NE^{+}(M)$ is the class of cusped
hyperbolic Dehn fillings of $M$.

One could prove this as an elementary consequence of Proposition 1.7 of
[GMM11]. The idea there is to embed convolutubes (tori that only compress
towards the generalized link exterior boundary) in disjoint balls in $N$. But
we provide a “hands-on” proof in the Appendix C.

######  Definition 5.9.

A fault in a generalized link exterior $M$ is a properly embedded surface with
nonnegative Euler characteristic that is one of the following:

  * •

nonorientable;

  * •

a sphere not bounding a ball (an essential sphere);

  * •

a disc not separating a ball from $M$ (a compressing disc);

  * •

an incompressible torus not parallel to $\partial M$ (an essential torus); or

  * •

an incompressible, $\partial$-incompressible annulus (an essential annulus).

A generalized link exterior is irreducible when it admits no essential sphere;
is boundary-irreducible when it also admits no compressing disc; is
geometrically atoroidal when it also admits no essential torus; and, finally,
is anannular when it also admits no essential annulus.

The following lemma, which relies to a large extent on Lemma 1.13 of [GMM11],
enables the computation of a finite ancestral set for $NE^{+}(M)$ for any
given a triangulation $\mathcal{T}$ of $M$, ultimately by recursion on the
natural decomposition of $M$ along spheres and tori.

######  Lemma 5.10.

Let $M$ be a generalized link exterior.

  1. 1.

If $M$ admits a nonseparating fault, then $NE^{+}(M)=\emptyset$.

  2. 2.

If $M=L\ \\#\ R$, then $NE^{+}(M)\subset NE^{+}(L)\cup NE^{+}(R)$.

  3. 3.

If $M=L\cup_{\phi}R$, where $\phi:T_{L}\to T_{R}$ is a homeomorphism between
two torus boundary components $T_{L}$ and $T_{R}$ of $L$ and $R$, then
$NE^{+}(M)\subset NE^{+}(L)\cup NE^{+}(R)$.

  4. 4.

If $M$ is irreducible, boundary-irreducible, and is geometrically atoroidal,
but has an annulus separating it into two solid tori, then
$NE^{+}(M)=\emptyset$.

We note that if $M$ is irreducible, boundary-irreducible, and geometrically
atoroidal, and does not admit a nonseparating fault, but has an essential
annulus, then that annulus cuts $M$ into two solid tori. It is markedly easier
to show an annulus cuts $M$ into two solid tori than to show further that that
annulus is essential, so the former is the condition we use in the algorithm
for computing $NE^{+}(M)$.

###### Proof.

  1. 1.

If $i:M\hookrightarrow N$ were a nonelementary embedding, and $\Sigma$ were a
nonseparating fault in $M$, then $i(\Sigma)\subset N$ would be a nonseparating
fault in $N$, contrary to hyperbolicity of $N$.

  2. 2.

Suppose $M=L\ \\#\ R$. Let $S\subset M$ be a reducing sphere with the surgery
of $M$ along $S$ being $L\sqcup R$, i.e. such that $M\setminus S\approx
L^{\ast}\sqcup R^{\ast}$, $L^{\ast}$ and $R^{\ast}$ being $L$ and $R$
punctured. Suppose further that $i:M\hookrightarrow N$ is a nonelementary
embedding of $M$ into an orientable hyperbolic 3-manifold $N$. Then $i(S)$ is
a sphere in $N$. Thus, since $N$ is hyperbolic, $N\setminus S=N^{\ast}\cup
D^{3}$. Then exactly one of $L^{\ast}$ and $R^{\ast}$ has image in $D^{3}$.
Say it’s $L^{\ast}$. Then $\pi_{1}(M)=\pi_{1}(L^{\ast})\ast\pi_{1}(R^{\ast})$
and

$i_{\ast}(\pi_{1}(M))\simeq i_{\ast}(\pi_{1}(L^{\ast}))\ast
i_{\ast}(\pi_{1}(R^{\ast}))\simeq i_{\ast}(\pi_{1}(R^{\ast})).$

Hence $i|_{R^{\ast}}$ is a nonelementary embedding. In this case, capping
$i(R^{\ast})$ off with the $D^{3}$ above yields a nonelementary embedding on
$R$ into $N$. Similarly, if $R^{\ast}$ has image in $D^{3}$, then $i$ can be
modified to a nonelementary embedding of $L$ into $N$. Hence,
$NE^{+}(M)\subset NE^{+}(L)\cup NE^{+}(R).$

  3. 3.

Suppose $i:M\hookrightarrow N$ is a nonelementary embedding. Then by item 1,
the identified torus $T=T_{L}/\phi=T_{R}/\phi$ splits $M$ into two components.
By item (iv) of Lemma 1.10 of [GMM11], on one of those components, $i$
restricts to a nonelementary embedding.

  4. 4.

Suppose $M$ is irreducible, $\partial$-irreducible, and geometrically
atoroidal, but admits an annulus splitting it into two solid tori. Then
$\pi_{1}(M)\simeq\mathbb{Z}\ast_{\mathbb{Z}}\mathbb{Z}$. Such a group does not
admit a nonelementary map into $\operatorname{PSL}_{2}\mathbb{C}$. Therefore
$M$ does not admit a nonelementary embedding into an orientable hyperbolic
3-manifold.

∎

######  Theorem 5.11.

Algorithm 1 is a solution to problem NE for nontrivial generalized link
exteriors.

Algorithm 1 Problem NE

procedure probNE($\mathcal{T}$)

Assume $\mathcal{T}$ is a generalized link exterior.

if $\pi_{1}(\mathcal{T})$ already has a common axis presentation then

return $\emptyset$

else if $\mathcal{T}$ admits a strict angle structure then

return $\\{\mathcal{T}\\}$

Let $\mathcal{F}$ be the set of fundamental normal surfaces in $\mathcal{T}.$

if there is a nonseparating fault in $\mathcal{F}$ then

return $\emptyset$

else if there is a reducing sphere $S$ in $\mathcal{F}$ then

Calculate $\mathcal{T}_{L},\mathcal{T}_{R}$ such that
$\mathcal{T}=\mathcal{T}_{L}\ \\#\ \mathcal{T}_{R}.$

return
$\mbox{{prob}}\mbox{{NE}}(\mathcal{T}_{L})\cup\mbox{{prob}}\mbox{{NE}}(\mathcal{T}_{R})$

else if there is an essential torus $T$ in $\mathcal{F}$ then

Calculate $\mathcal{T}_{L},\mathcal{T}_{R}$ such that
$\mathcal{T}-T=\mathcal{T}_{L}\sqcup\mathcal{T}_{R}.$

return
$\mbox{{prob}}\mbox{{NE}}(\mathcal{T}_{L})\cup\mbox{{prob}}\mbox{{NE}}(\mathcal{T}_{R})$

else if there is an annulus $A$ in $\mathcal{F}$ splitting $\mathcal{T}$ into
two solid tori then

return $\emptyset$

else

return $\\{\mathcal{T}\\}$

The initial two sanity checks are discussed below in $\S$5.2. We turn now to
the bulk of the algorithm.

###### Proof.

Suppose $M$ is a nontrivial generalized link exterior.

If $M$ admits a nonseparating fault, then by Lemma 5.10.1,
$NE^{+}(M)=\emptyset$, so the algorithm returns a correct ancestral set.

If $M$ admits a reducing sphere $S$, then, letting $L\ \\#\ R$ be the connect-
sum decomposition of $M$ along $S$, Lemma 5.10.2 shows that $NE^{+}(M)\subset
NE^{+}(L)\cup NE^{+}(R)$. By induction on the size of the prime and JSJ
decompositions of $M$, we may let $U=\bigcup_{Y\in NE^{+}(L)}NE^{+}(Y)$ and
$V=\bigcup_{Y\in NE^{+}(R)}NE^{+}(Y)$ be the ancestral sets of $NE^{+}(L)$ and
$NE^{+}(R)$, respectively. Thus $U\cup V$ is an ancestral set for $NE^{+}(M)$,
so the algorithm returns a correct ancestral set.

If $M$ is a solid torus, then $\pi_{1}(M)\simeq\mathbb{Z}$, and so $M$ admits
no nonelementary embeddings. Thus the algorithm returns a correct ancestral
set.

Otherwise, $M$ is irreducible and $\partial$-irreducible. If in this case $M$
admits a separating essential torus $T$, then, letting $L\sqcup R\approx M-T$
be the components of the exterior of $T$ in $M$, Lemma 5.10.3 shows that
$NE^{+}(M)\subset NE^{+}(L)\cup NE^{+}(R)$. As with the case of a reducing
sphere, we recursively compute the ancestral sets $U$ and $V$ of $NE^{+}(L)$
and $NE^{+}(R)$, respectively. It follows that $U\cup V$ is an ancestral set
for $NE^{+}(M)$, so that the algorithm returns a correct ancestral set.

Otherwise, $M$ is irreducible, $\partial$-irreducible, geometrically
atoroidal, and admits no nonseparating faults. Thus, since $M$ is a nontrivial
generalized link exterior, either $M$ is Seifert-fibered over a disc with two
exceptional fibers, or $M$ is hyperbolic. In the former case, $M$ admits an
essential annulus $A$ separating it into two solid tori, and so by Lemma
5.10.4, $NE^{+}(M)=\emptyset$, so the algorithm returns a correct ancestral
set in this case, again by induction on the complexity of the decomposition of
$M$.

Otherwise, and finally, if $M$ admits no faults in $\mathcal{F},$ then by
Lemma 5.13 in section 5.1.2, $M$ admits no faults at all, and therefore is
hyperbolic by Thurston’s Haken hyperbolization theorem. So by Lemma 5.8,
$\\{M\\}$ is an ancestral set for $NE^{+}(M)$. Thus the algorithm returns a
correct ancestral set in this case. ∎

Python listings implementing this code using Regina are available under the
name problemNE.py at [Gita].

####  5.1.2 Fault finding

To turn the above lemmas into implementable code requires algorithms for
various tasks, all to do with finding faults. To determine the existence of a
fault and, if one exists, to find one, we may appeal to the theory of normal
surfaces as implemented in Regina, which we now briefly review.

######  Definition 5.12.

Fix a triangulation $\mathcal{T}$ of a 3-manifold. A normal isotopy is an
isotopy through isomorphisms of $\mathcal{T}$. A normal disc in a tetrahedron
$\tau$ is a properly embedded disc either separating one vertex of $\tau$ from
the others, or separating two pairs of vertices.777That is, regarding $\tau$
as an affine simplex, a normal disc is a properly embedded disc normally
isotopic to an affine disc transverse to $\tau$. A normal surface is a
properly embedded surface transverse to $\mathcal{T}$ that is a disjoint union
of normal discs.

The following theorem is the theoretical backbone of our use of normal surface
theory. It is a corollary of well-known results (cf. [Mat07]).

######  Proposition 5.13.

Suppose $\mathcal{T}$ is a triangulation whose underlying space is a
3-manifold $M$. There is a finite set $\mathcal{F}$ of normal surfaces in
$\mathcal{T}$, called fundamental normal surfaces, that is computable from
$T$, such that

  * •

if $M$ has an embedded projective plane, then there is one in $\mathcal{F}$;

  * •

else, if $M$ has an essential $S^{2}$, then there is one in $\mathcal{F}$;

  * •

else, if $M$ has a compressing $D^{2}$, then there is one in $\mathcal{F}$;

  * •

else, if $M$ has an embedded Klein bottle, then there is one in $\mathcal{F}$;

  * •

else, if $M$ has an essential torus, then there is one in $\mathcal{F}$;

  * •

else, if $M$ has an embedded Möbius band, then there is one in $\mathcal{F}$;

  * •

else, if $M$ has an essential annulus, then there is one in $\mathcal{F}$.

In particular, $M$ admits a fault if and only if it admits a fault in
$\mathcal{F}$.

###### Proof.

This is essentially a concatenation of Theorems 3.3.30, 4.1.12, 4.1.13,
4.1.36, 6.4.7, and 6.4.8 of [Mat07]. ∎

######  Remark 5.14.

In fact our implementation only enumerates vertex normal surfaces. This
smaller collection of surfaces sufficed for our needs.

######  Proposition 5.15.

Algorithm 2 determines whether or not a given triangulation is homeomorphic to
$T^{2}\times I$.

Algorithm 2 Homeomorphism to $T^{2}\times I$

procedure is $T^{2}\times I$($\mathcal{T}$)

if $\mathcal{T}$ has different homology from $T^{2}\times I$ then

return False

Simplify $\mathcal{T}$ so that each boundary component has a one-vertex
triangulation.

Pick a boundary component $k$; it has three edges.

for edge $e$ of $k$ do

Let $\mathcal{T}_{e}$ be gotten from $\mathcal{T}$ by closing-the-book along
$e$.

if $\mathcal{T}_{e}$ is not a solid torus then

return False

return true

###### Proof.

We give a brief outline. For more details, see [Har20]. Closing-the-book along
an edge in a one-vertex triangulation of the torus accomplishes a Dehn filling
along the flip of the edge. Dehn filling $T^{2}\times I$ always yields a solid
torus. Hence, if closing-the-book doe not yield a solid torus, then certainly
$\mathcal{T}$ is not $T^{2}\times I$. Suppose, on the other hand, that each
filling of $\mathcal{T}$ along a flip $\alpha_{e}$ of an edge of the boundary
triangulation yields a solid torus. Then $\mathcal{T}$ admits three Dehn
slopes distance 3 apart that fill to solid tori. By [Gab89] and [Gab90], it is
not difficult to show that $\mathcal{T}$ must be homeomorphic to $T^{2}\times
I$. ∎

######  Remark 5.16.

The above algorithm, though very short, relies crucially upon solid torus
recognition. The lowest known upper bounds on worst-case running time for
solid torus recognition are exponential in the number of tetrahedra. So the
same is true of the above algorithm.

Algorithms 3, 4, 5, and 6 therefore accomplish the tasks required from
Algorithm 1, assuming the ambient triangulation in which $S$ lies is
connected, and, in Algorithm 5, assuming that triangulation is irreducible. In
Algorithm 1, this assumption is satisfied at the point where Algorithm 5 is
needed.

Algorithm 3 Is a normal surface a non-separating fault?

procedure isNonSeparatingFault(S)

if $\chi(S)<0$ then

return False

Let $\mathcal{T}$ be the triangulation in which $S$ lies.

Let $\mathcal{T}^{\prime}$ be $\mathcal{T}$ cut along $S$.

if $\mathcal{T}^{\prime}$ is connected then

return True

else

return False

Algorithm 4 Is a normal surface an essential sphere?

procedure isEssentialSphere(S)

if $S$ is not a sphere then

return False

Let $\mathcal{T}$ be the triangulation in which $S$ lies.

Let $\mathcal{T}^{\prime}$ be $\mathcal{T}$ cut along $S$.

if $\mathcal{T}^{\prime}$ is connected then

return True

$\mathcal{T}^{\prime}$ has two components; call them $L$ and $R$.

if $L$ or $R$ is a ball then

return False

else

return True

Algorithm 5 Is a normal surface an essential torus?

precondition The ambient triangulation is irreducible.

procedure isEssentialTorus(S)

if $S$ is not a torus then

return False

Let $\mathcal{T}$ be the triangulation in which $S$ lies.

Let $\mathcal{T}^{\prime}$ be $\mathcal{T}$ cut along $S$.

if $\mathcal{T}^{\prime}$ is connected then

return True

$\mathcal{T}^{\prime}$ has two components; call them $L$ and $R$.

if $L$ or $R$ is $\partial$-compressible or $T^{2}\times I$ then

return False

else

return True

Algorithm 6 Is a normal surface an annulus separating $\mathcal{T}$ into two
solid tori?

procedure isSolidTorusAnnulus(S)

if $S$ is not an annulus then

return False

Let $\mathcal{T}$ be the triangulation in which $S$ lies.

Let $\mathcal{T}^{\prime}$ be $\mathcal{T}$ cut along $S$.

if $\mathcal{T}^{\prime}$ is connected then

return False

$\mathcal{T}^{\prime}$ has two components; call them $L$ and $R$.

if $L$ and $R$ are both solid tori then

return True

else

return False

Python listings implementing the above algorithms using Regina are available
under the names faultFinding.py and problemNE.py at [Gita].

###  5.2 Sanity checks

The previous section technically completes our solution to problem NE.
However, a few additional considerations speed up our computations. We turn to
these now.

First, all necklace gluings are orientable. Likewise, they are all connected.
If we were testing arbitrary MOM gluings, then we would add a connectedness
test. One other requirement, though, is that the links of all the vertices
must be tori. Many necklace gluings are not generalized link exteriors, and
hence are of no concern to us in our enumeration. So, we must implement a test
for being a generalized link exterior.

Another fast, extremely useful check is a fundamental group check. Regina has
routines for calculating and simplifying presentations of fundamental groups
of triangulations (and ideal triangulations). A significant plurality of
necklace gluings of bead number at most 7 admit fundamental group
presentations of one of the following forms, and these presentations admit no
nonelementary embeddings into $\operatorname{PSL}_{2}\mathbb{C}$.

######  Definition 5.17.

Suppose $P=\langle a,b\ |\ \ldots\,\rangle$ is a two-generator finite
presentation of some group. A common axis commutator is a relation of the form
$[a^{p},b^{q}]$ for some $p,q\in\Z$. A common axis equation is a relation of
the form $a^{p}b^{q}$ for some $p,q\in\Z$. We call these common axis
relations.

######  Lemma 5.18.

Presentations with common axis relators do not present fundamental groups of
3-manifolds admitting nonelementary embeddings into hyperbolic 3-manifolds.

###### Proof.

Suppose such a presentation $P$ presents $G<\operatorname{PSL}_{2}\mathbb{C}$.

If $P$ has a common axis commutator or equation, then $a^{p}$ and $b^{q}$
commute. Since $G<\operatorname{PSL}_{2}\mathbb{C}$, $a$ and $b$ must commute
as well, since commuting elements have the same axis, and $a^{p},b^{q}$ have
the same axes as $a,b$, respectively. So $G$ is a quotient of $\Z\oplus\Z$. ∎

Python listings implementing these sanity checks using Regina are available
under the name sanity.py at [Gita].

Yet another sanity check one can do on a triangulation is to try to find a
strict angle structure on it.

######  Definition 5.19.

Suppose $Q_{\tau}$ is the set of all normal isotopy types of quadrilaterals in
a tetrahedron $\tau$, and that $Q_{T}=\bigcup_{\tau\in T}Q_{\tau}$. A strict
angle structure is a function $\theta:Q_{T}\to(0,\pi)$ such that

$\sum_{q\in Q_{\tau}}\theta(q)=\pi$

for all $\tau\in T$, and such that

$\sum_{q\ opposite\ e}\theta(q)=2\pi$

for all edges $e$ of $T$.

######  Theorem 5.20 ([Lac00],[FG09]).

If an ideal triangulation of a generalized link exterior $M$ admits a strict
angle structure, then $M$ is hyperbolic.

Regina already provides a routine that can usually find such structures in
practice. Note that this result implies that any oriented finite triangulation
of $M$ has no faults.

The final sanity check one may perform on a triangulation is to determine if
it lies in a census of known manifolds. Regina provides such censuses, and our
computations in the next section rely overwhelmingly on the pre-computed
properties of census manifolds. We describe this reliance in more detail
there.

###  5.3 Problem Bead

We now enumerate all necklace manifolds with a given number of beads, solving
problem Bead. The representations of the manifolds are the same as in [GMM11].
The actual enumeration is also done similarly, but with an appeal to
isomorphism signatures instead of to symmetries of dipyramids.

We review the representations briefly. A full handle structure without
3-handles deformation retracts to a “core” 2-skeleton of the handle structure.
This 2-skeleton is a spine for the 3-manifold. Taking the dual of this complex
gives one an ideal cellulation $C$ of the 3-manifold by dipyramids, an
$n$-dipyramid being the suspension of an $n$-gon. The face-pairing maps
preserve the distinction between the polygons’ vertices and the suspension
points. We call such a gluing of dipyramids a MOM gluing. It is important to
note that this construction required the fullness condition.

We triangulate an $n$-dipyramid with $n$ tetrahedra by identifying $n$
tetrahedra “around an edge,” which edge runs between the suspension points of
the dipyramid. The suspension base is not a subcomplex of this triangulation.
A MOM gluing thus yields a natural ideal triangulation of the corresponding
handle structure. We call such triangulations and face-pairings split, as they
are precisely those triangulations that admit splitting surfaces, normal
surfaces with no triangles and one quadrilateral in every tetrahedron. A full
necklace manifold on $n$ beads is, among other things, a full handle structure
with one 1-handle of valence $n$. The split triangulation of such a handle
structure is a split face-pairing of a single $n$-dipyramid. For any pair of
faces, there is only one face-pairing between them preserving the distinction
between $n$-gon vertices and suspension points and reversing the orientation
on the faces. A full necklace manifold on $n$ beads is thus determined merely
by which pairs of faces are glued. This puts these structures in bijection
with partitions of $D_{n}=\\{i:0\leq i<2n\\}$ into pairs, or perfect matchings
on $D_{n}.$ These are our representations of these manifolds.

Lines 1309–1401 of enum_gluings.cpp and the function IncrementMatching
starting at line 297 of enum_utils.cpp from [GMM11] give an explicit,
“unrolled” iteration over perfect matchings. Algorithm 7 iterates over these
perfect matchings recursively instead. We found the latter more transparent
and easier to implement, though its call stack presumably makes it less space-
efficient.

Algorithm 7 Do $f(\mu)$ for all perfect matchings $\mu$ on $l\cup m$ with
subpartition $p$ that match the first element of $l$ to something else in $l$.

procedure pfm($p,l,m,f$)

if $l=nil$ then

if $m=\emptyset$ then

do $f(p)$

else if $tail(l)\neq nil$ then

let $x_{0},\ x_{1},\ l_{2}=head(l),\ head(tail(l)),\ tail(tail(l))$

do pfm$(cons((x_{0},x_{1}),p),l_{2}\cup m,\emptyset,f)$

do pfm$(p,cons(x_{0},l_{2}),\\{x_{1}\\}\cup m,f)$

Notice that our ultimate goal in enumerating these gluings is to get
triangulations of the underlying 3-manifolds. Therefore, it would suffice to
enumerate the isomorphism signatures of all such triangulations as defined in
[Bur11]. We thank Neil Hoffman for this suggestion. This enables a simpler and
more efficient enumeration than was done in [GMM11]. We maintain a hash table
sigs of the isomorphism signatures seen so far, and then for every perfect
matching, construct the associated triangulation and put its isomorphism
signature into sigs if it isn’t there already. That is, define $f(\mu)$ to be
the operation of putting the isomorphism signature of $\mu$’s gluing into sigs
if it isn’t there already, and run
$\mbox{{pfm}}(nil,[0,\ldots,2n-1],\emptyset,f)$. This solves problem Bead.

######  Remark 5.21.

There are 242 cusped necklace manifolds with 4 to 7 beads. Among these, 35
admit a strict angle structure; 63 admit a common axis group presentation; and
the remaining 144 admit a nonseparating fault. Census checks play a small role
in this calculation. The opposite was true for the calculation in section 6.

###  5.4 An ancestral set for manifolds of low cusp volume

We conclude this section with a proof of the main technical result of the
paper.

See 1.1

###### Proof.

Suppose $N$ is a complete nonelementary hyperbolic 3-manifold with at least
one maximal cusp of volume at most 2.62. Since $N$ is complete and hyperbolic,
we may identify $N$ with $\mathbb{H}^{3}/\Gamma$ for some discrete, torsion-
free, nonelementary $\Gamma<PSL_{2}\mathbb{C}.$ Since $N$ has a maximal cusp
of volume at most 2.62, there is a bicuspid triple $(P,S,L)\in\mathscr{P}$
whose associated bicuspid marking’s group $G$ is conjugate into $\Gamma.$
Without loss of generality, we may thus assume $G<\Gamma.$ Let $m,n,g$ be the
generating set for $G$ associated with $(P,S,L)$. Because
$(P,S,L)\in\mathscr{P},$ by Theorem 1.8 there is a reduced non-identity word
$w$ in $m,n,g$ with $g$-length at most 7 such that $w$ is trivial in $G,$ and
hence trivial in $\Gamma.$

Let $\mathscr{H}$ be the associated horoball system upstairs, the preimage of
the chosen maximal cusp $\kappa$ of $N$ under the universal cover
$\mathbb{H}^{3}\to\mathbb{H}^{3}/\Gamma=N$. Letting $\ell$ be the $g$-length
of $w$, by Lemma 4.2 there is an $\ell$-necklace in $\mathscr{H}$. Letting
$\eta$ be a minimal necklace in $\mathscr{H}$, $\eta$ must be a $k$-necklace
with $k\leq\ell\leq 7$. By Proposition 4.41, $\eta$ is unknotted, unblocked,
and unlinked. Moreover, by Lemma 4.11, we may assume $\eta$ is simple. Hence,
there is a compressing disk for $\eta$ which descends to an embedded disk $D$
in $N$. In fact, in $N$ we have a handlebody $M$ based on the maximal cusp
neighborhood (actually, based on $T\times[0,1]$ where $T$ is the torus
boundary of the maximal cusp neighborhood), with one-handle $\sigma$
associated to ${\cal O}(1)$ and 2-handle $\tau$ gotten by thickening our
embedded disk $D$. By construction, $M$ is non-elementarily embedded in $N$.
Lemma 4.47 and Proposition 4.48 tell us that we can assume this nonelementary
embedding produces an internal necklace-$n$ structure on $N$ with $n\leq 7$,
and that this structure is full.

The solution of problem Bead in section 5.3 enumerates all full necklace
structures with given bead number. Let $FNS_{\leq 7}$ be this set of
triangulations so enumerated. Then $N\in NE^{+}(M^{\prime})$ for some element
$M^{\prime}\in FNS_{\leq 7}$. The solution of problem NE in Theorem 5.11
computes a finite ancestral set $A(M^{\prime})$ for $NE^{+}(M^{\prime})$ for
all $M^{\prime}\in FNS_{\leq 7}.$ So by Lemma 5.8 $N$ is a Dehn filling of
some element $M\in\bigcup_{M^{\prime}\in FNS_{\leq 7}}A(M^{\prime})$. This
union has the 28 elements given in Table 2, which are all hyperbolic as one
would expect from Theorem 5.11. Since they are hyperbolic, with SnapPy one may
easily check Table 8, which asserts that the associated manifolds from Table 2
are Dehn fillings of the magic manifold s776. Thus, $N$ is a Dehn filling of
some element of Table 1. ∎

m125 | m129 | m202 | m203 | m292 | m295 | m328
---|---|---|---|---|---|---
m359 | m367 | s441 | s443 | s596 | s647 | s774
s776 | s780 | s782 | s785 | v1060 | v2124 | v2355
v2533 | v2644 | v2731 | v3108 | v3127 | v3211 | v3376
Table 2: Initial ancestral set with Dehn fillings of s776 in bold. m$125:(2,1)$ | m$129:(3,-1)$ | m$202:(1,2)$
---|---|---
m$203:(3,-2)$ | m$292:(3,1)$ | m$295:(4,-1)$
m$328:(2,-3)$ | m$359:(4,-3)$ | m$367:(5,-2)$
s$441:(4,1)$ | s$443:(5,-1)$ | v$1060:(-6,1)$
Table 8: Dehn fillings of s776 in the shortest-next-shortest basis of any
cusp.

The Python listing provegordon.py available at [Gita] performs the above
calculations in the course of proving Theorem 1.7.

######  Remark 5.22.

The paper [GMM11] showed that every MOM-$\leq 4$ manifold that embedded
nonelementarily into a hyperbolic 3-manifold was itself hyperbolic. Our
calculations showed likewise that every necklace-$\leq 7$ manifold that embeds
nonelementarily into a hyperbolic 3-manifold is itself hyperbolic. Both show
this by cutting along surfaces. However, [GMM11] performed the cutting in the
text of the paper, showing that a calculation could skip over non-hyperbolic
MOM structures. We did not skip over them, cutting along the surfaces during
the calculation.

##  6 Applications

###  6.1 Gordon’s conjecture

In this section we apply our previous work, together with some more
computations, to prove the following conjecture of Cameron Gordon from 1995.
To state the theorem, let $e(M)$ denote the number of exceptional Dehn
fillings of $M$. See 1.7

Our proof of this theorem depends upon the following important result, known
as the 6-Theorem:

######  Theorem ([Ago00], [Lac00]).

Suppose $N$ is a hyperbolic generalized link exterior and $s$ is a slope on
$\partial N$. If $\ell(s)>6$ on the maximal cusp torus, then $N(s)$ is
hyperbolic.

We note their original conclusion was that $N$ is irreducible and has infinite
word-hyperbolic fundamental group, a condition they called hyperbolike.
However, by Perelman’s affirmative resolution of Thurston’s Geometrization
Conjecture, this implies that $N$ is in fact hyperbolic.

######  Lemma 6.1 ([Ago10]).

A 1-cusped orientable hyperbolic 3-manifold $N$ with more than 8 exceptional
Dehn fillings admits a cusp of maximal area at most 36/7.

###### Proof.

Suppose the set $S$ of $N$’s exceptional Dehn fillings has more than 8
members. Then by Lemma 8.2 of [Ago00], the pairwise intersection numbers of
elements of $S$ are not all bounded above by 6. So some two distinct
exceptional Dehn fillings $s,s^{\prime}$ must have $\Delta(s,s^{\prime})\geq
7$. Now,

$\Delta(s,s^{\prime})=\frac{\ell(s)\ell(s^{\prime})\sin(\theta)}{A},$

where $\ell$ is geodesic length and $\theta$ is the angle between $s$ and
$s^{\prime}$ on the maximal cusp torus, and $A$ is its area. By the 6-Theorem,
the lengths of $s$ and $s^{\prime}$ are at most 6. Therefore, $A\leq 36/7$. ∎

###### Proof of Thm. 1.7..

Just for this proof, let us call 1-cusped orientable hyperbolic 3-manifolds
admitting at least 9 exceptional Dehn fillings superexceptional. Suppose $M$
is a superexceptional manifold. By Lemma 6.1 below, $M$ has a cusp of volume
at most $18/7=2.\overline{571428}$, which is less than $2.62$. Thus, by
Theorem 1.1, $M$ is a Dehn filling either of s776 or of one of the 15 other
manifolds detailed in Table 1. The Dehn fillings of s776 were classified in
detail in [MP06]; in particular, their Corollary A.6 shows that the only
superexceptional Dehn filling of s776 is $m004$. So either $M$ is m004, or $M$
is a Dehn filling of a manifold in Table 1. An simple special case of the
method of isolated slopes used in [MP06] can be implemented as discussed in
the following subsections. Running this method on each manifold of Table 1
only ever yields m004. Therefore, m004 is the only superexceptional manifold.
That is, m004 is the only 1-cusped hyperbolic 3-manifold admitting at least 9
exceptional Dehn fillings. ∎

As we remarked in the introduction, Crawford [Cra18] carried out an analysis
of these Dehn filling families using an alternate approach.

####  6.1.1 Determining short slopes

The methods of [MP06] depended critically on being able to enumerate all
slopes on a maximal cusp with lengths at most a given bound $B$, which, fixing
$B$, we will call short slopes. (In [MP06] and [MPR14] $B=2\pi;$ for us,
$B=6.$) The software from the more recent paper [MPR14] follows the methods of
[MP06]. Instead of enumerating just the short slopes, this software in general
enumerates a set containing those slopes, but that is sufficient for our
needs. Moreover, for manifolds of small complexity like those in Table 1, this
larger set includes few extraneous long slopes, if any at all. It is also now
possible to rigorously enumerate small slopes in SnapPy when imported as a
module in Sage; this is the approach we take.

####  6.1.2 Heuristics for determining hyperbolicity of closed 3-manifolds

We now turn to the question of determining the hyperbolicity of closed
3-manifolds. With Regina we can at least determine whether or not a closed
3-manifold admits any faults, using the algorithms from section 5.1.2. If a
closed, orientable 3-manifold admits a fault, then it is not hyperbolic. If,
on the other hand, it admits no fault, then all we can conclude is that it is
geometric. If it is not hyperbolic, then the 3-manifold is small-Seifert-
fibered. There are algorithms to recognized small-Seifert-fibered manifolds,
given by Tao Li in [Li06] and Hyam Rubinstein in [Rub04]. Unfortunately their
algorithms are not yet feasible to implement in code. Fortunately, Regina is
able to recognize some triangulations of many such 3-manifolds. In particular,
it recognizes all the nonhyperbolic faultless closed triangulations we
encountered as small-Seifert-fibered spaces, after some randomizations and
simplifications.

Thus we turn to figuring out how one can verify the hyperbolicity of a closed
3-manifold. In many cases SnapPy in SAGE can already do so, when the given
triangulation is a geometric triangulation—that is, when the triangulation
supports a solution to Thurston’s gluing equations each tetrahedron of which
is positively oriented. Our methods then amounted to coaxing out various
different triangulations of the same 3-manifold, until finding one that SnapPy
can prove geometric. Previous methods incorporated the taking of covers. But
we found this did not help much. The techniques we found most useful for
proving hyperbolicity were partial filling and dual curve drilling.

We first remind the reader briefly of how SnapPy works with closed
3-manifolds. At the core of SnapPy is a modification of Newton’s method,
applied to Thurston’s gluing equations for an ideal triangulation. If the
triangulation is not ideal in the usual sense, then the solution space for the
equations might not be 0-dimensional, regardless of the conditions one places
on the cusps. SnapPy’s Newton’s method is not designed for this situation.
Instead, one represents a closed 3-manifold as a Dehn filling of a generalized
link exterior. With an ideal triangulation of the generalized link exterior,
one gets a 0-dimensional solution space after imposing the completeness
equations as usual. But also, one gets a 0-dimensional solution space by
imposing, instead of the completeness conditions, a modified condition on the
cusp depending on the choice of Dehn filling coefficient. If SnapPy’s Newton’s
method settles on a putative solution to these modified equations, and SAGE
verifies this solution, and if, of course, the tetrahedra are positively
oriented, then the incompleteness in the structure amounts to a missing
geodesic. Filling in this geodesic recovers the original closed 3-manifold,
and proves it admits a hyperbolic structure.

In the course of our work, we are given a generalized link exterior $M$ and a
Dehn filling coefficient $c$ on its boundary, and we wish to determine whether
or not $M(c)$ is hyperbolic. One has several options at this point. The most
obvious option is to attach to Thurston’s gluing equations the cusp conditions
corresponding to $c$, and hope SnapPy finds a geometric solution to those
equations. If this does not work, one has the following two other options. If
$M$ is multiply-cusped, then one may pick a cusp $\tau$, take the slope $s$ of
$c$ on $\tau$, construct an ideal triangulation of $M(s)$, and consider the
generalized link exterior $M(s)$ with the Dehn filling coefficient $c-s$.
$M(s)(c-s)$ is homeomorphic to $M(c)$. But SnapPy very well may be able to
find a positive solution for the equations coming from $M(s)(c-s)$ and not on
those coming from $M(c)$. This was a useful technique, which decreased the
number of cusps. The opposite technique, that of drilling out curves instead
of filling them in, also proved useful. SnapPy can drill out a curve in the
dual 1-skeleton of a triangulation, putting an ideal triangulation on its
complement, and remembering which slope on the new boundary component is the
meridian $\mu$. (We thank Neil Hoffman for pointing out this feature to us.)
Again, SnapPy might be able to find a positive solution for the equations on
$(M-\gamma)(c+\mu)$ and not on $M(c)$. These heuristics sufficed for us to
verify the hyperbolicity of the hyperbolic closed 3-manifolds we encountered.

####  6.1.3 Resolving Gordon’s conjecture

To resolve Gordon’s conjecture, we now only require an accounting of
exceptional slopes. Suppose $M$ is a hyperbolic 3-manifold with 2 cusps. We
wish to determine the superexceptional Dehn fillings of $M$. To that end,
recall that for us a slope $s$ is short when it is of length at most 6 on a
maximal torus boundary of its associated cusp. We may first classify the short
slopes of $M$ on each cusp into hyperbolic and exceptional. It is routine to
accomplish such a classification. To enumerate short slopes on $\tau$, just
have SnapPy determine an area form estimate on $\tau$, then use the methods of
section 6.1.1. To determine which among them are hyperbolic and which are
exceptional, use the methods of sections 5.1.2 and 6.1.2. For each cusp
$\tau$, let $hyp(\tau)$ be the hyperbolic short slopes on $\tau$ and
$exc(\tau)$ the exceptional slopes, and let $short(\tau)$ be all the short
slopes.

Pick a cusp $\tau$. For each hyperbolic short slope $s$ on $\tau$, we may
determine the exceptional fillings of $M(s)$ as above. This directly decides
whether or not $M(s)$ is superexceptional. So our problem comes down to
determining whether or not there are any superexceptional long fillings on
$\tau$. There are infinitely many such slopes, so a simple enumeration is out
of the question. We use the following lemma to treat these slopes. Here, we
let $hyp(s)$ and $exc(s)$ be the short hyperbolic and exceptional slopes on
the unique cusp of $M(s)$.

######  Lemma 6.2.

Suppose $M$ is an orientable hyperbolic 2-cusped 3-manifold. Suppose $\tau$ is
a cusp of $M$ and $s$ is a long slope on $\tau$. Let $\tau^{\prime}$ be the
other cusp of $M$ . Then

(6.1) $|exc(s)|\leq|exc(\tau^{\prime})|+|\\{s^{\prime}\in hyp(\tau^{\prime})\
|\ s\in exc(s^{\prime})\\}|.$

###### Proof.

Suppose that $s$ is a long slope on $\tau$. Then by the 6-Theorem, $M(s)$ is
hyperbolic. Suppose $s^{\prime}$ is an exceptional filling of $M(s)$. We may
interpret this as the exceptional filling $M(s,s^{\prime})$ of $M$. Since this
is an exceptional filling and $s$ is long, by the 6-Theorem, $s^{\prime}$ must
be short on its cusp $\tau^{\prime}$ in $M$. So $s^{\prime}\in
short(\tau^{\prime})$. This already gives us the bound
$|exc(s)|\leq|short(\tau^{\prime})|=|hyp(\tau^{\prime})|+|exc(\tau^{\prime})|$.
However, we may do even better. Suppose $s^{\prime}\in hyp(\tau^{\prime})$ and
$M(s,s^{\prime})$ is exceptional. Then $s$ is an exceptional slope on
$M(s^{\prime})$. So $s\in exc(s^{\prime})$. We thereby get the better bound

$|exc(s)|\leq|exc(\tau^{\prime})|+|\\{s^{\prime}\in hyp(\tau^{\prime})\ |\
s\in exc(s^{\prime})\\}|.$

∎

Python listings implementing the above bound are available in gordon.py at
[Gita].

###  6.2 Results on hyperbolic volume

The arguments in this subsection largely follow those in Section 9 of [GMM09].
As in [GMM09] we will use the following theorem of Futer, Kalfagianni, and
Purcell:

######  Theorem ([FKP08], 1.1).

Let $M$ be a complete, finite-volume hyperbolic manifold with cusps. Suppose
$C_{1},\ldots,C_{k}$ are disjoint horoball neighborhoods of some subset of the
cusps. Let $s_{1},\ldots,s_{k}$ be slopes on $\partial C_{1},\ldots,\partial
C_{k},$ each with length greater than $2\pi.$ Denote the minimal slope length
by $\ell_{\min}$. If $M(s_{1},\ldots,s_{k})$ satisfies the geometrization
conjecture, then it is a hyperbolic manifold, and

$\operatorname{vol}(M(s_{1},\ldots,s_{k}))\geq\left(1-\left(\frac{2\pi}{\ell_{\min}}\right)^{2}\right)^{3/2}\operatorname{vol}(M).$

######  Corollary 6.3.

Every one-cusped orientable hyperbolic three-manifold of volume at most 3.07
is a Dehn filling of a manifold in Table 1.

###### Proof.

Suppose $N$ is a 1-cusped orientable hyperbolic 3-manifold. Let $V_{C}(N)$ be
the volume of its maximal cusp. As in [Mey86], by horoball packing results of
Böröczky, $V_{C}(N)\leq\sqrt{3}\cdot vol(N)/(2\mathcal{V}),$ where
$\mathcal{V}$ is the volume of a regular ideal hyperbolic tetrahedron. If
$vol(N)<3.07<2.62\cdot 2\mathcal{V}/\sqrt{3},$ then $V_{C}(N)<2.62.$ Thus by
Theorem 1.1 $N$ is a Dehn filling of a manifold in Table 1. ∎

See 1.5

###### Proof.

By Corollary 6.3 we just need to enumerate all the one-cusped fillings of
elements of Table 1 with volume at most 3.07. Suppose $M(\textbf{s})$ is a
Dehn filling of a manifold $M$ in Table 1. Suppose the volume of the Dehn
filling is at most 3.07. Then by [FKP08, 1.1] above,

$\ell_{\min}(\textbf{s})\leq
2\pi\cdot(1-(3.07/\operatorname{vol}(M))^{2/3})^{-1/2}=FKP(M).$

Let s be a Dehn filling coefficient of s776 with $\ell_{\min}(\textbf{s})\leq
FKP(\mbox{{s776}}).$ Let $s_{i}$ be the slope of s with minimum length. Then
$\mbox{{s776}}(\textbf{s})$ is also a Dehn filling of $\mbox{{s776}}(s_{i}).$
So a one-cusped Dehn filling of s776 with volume at most 3.07 is also a Dehn
filling of a two-cusped Dehn filling of s776 along a slope of length at most
$FKP(\mbox{{s776}}).$ There are finitely many slopes on s776 of length at most
$FKP(\mbox{{s776}}).$ The two-cusped fillings of s776 along such slopes that
are not in Table 1 are as follows:

m125 | m129 | m202 | m203 | m292 | m295 | m328 | m329 | m357 | m359
---|---|---|---|---|---|---|---|---|---
m366 | m367 | m388 | m391 | s441 | s443 | s503 | s506 | v1060 | v1061
Table 9: Two-cusped fillings of s776 along slopes of length less than
$FKP(\mbox{{s776}}).$

For each $M$ in Tables 1 and 9, for each cusp of $M,$ we have SnapPy in Sage
determine a maximal horoball neighborhood $C$ of the cusp, then enumerate the
slopes on $\partial C$ of length at most $FKP(M).$ The resulting one-cusped
manifolds with volume less than 3.07 are in Table 2. ∎

See 1.6

###### Proof.

As in [GMM09] we use the following lemma of Agol, Culler, and Shalen.

######  Lemma 6.4 ([ACS06, 3.1]).

Suppose that $M$ is a closed, orientable hyperbolic 3-manifold, and that $C$
is a shortest geodesic in $M$ such that $\mbox{tuberad}(C)\geq(\log 3)/2.$ Set
$N=\mbox{drill}_{C}(M).$ Then
$\operatorname{vol}(N)<3.02\cdot\operatorname{vol}(M).$

The proof of the lemma above relies on [AST07, Prop. 10.1] and [Prz06, Cor.
4.4]. The proofs of these results only assume the given geodesic is embedded,
not shortest. The same is true of the proof of this lemma. So the word
“shortest” above can be omitted. Suppose then that $N$ is a closed orientable
hyperbolic 3-manifold of volume at most $3.07/3.02.$ Then either $N$ is a Dehn
filling of a one-cusped hyperbolic 3-manifold of volume at most 3.07, or else
no embedded geodesic in $N$ admits a tube of radius at least $(\log 3)/2.$ For
instance, no embedded geodesic in the manifold Vol3 admits such a tube. But in
fact, it is the unique orientable hyperbolic 3-manifold with this property
[GT15, Cor. 2]. Thus either $N$ is a Dehn filling of a one-cusped hyperbolic
3-manifold of volume at most 3.07, or else $N$ is Vol3. By Theorem 1.5 either
$N$ is a Dehn filling of an element of Table 2 or $N$ is Vol3. In
volume_bounds.py we use SnapPy, the lemma of Futer, Kalfagianni, and Purcell,
and our hyperbolicity tests to enumerate the hyperbolic Dehn fillings of
elements of Table 2 with volume at most $3.07/3.02.$ The only such fillings
are m003(2,1) m004(5,1), and Vol3. ∎

##  Appendix A Group isomorphisms

Here we list the group isomorphisms necessary for completing the proof of
Theorem 1.2. As these are isomorphisms between presentations, they can easily
be verified by hand.

######  Lemma A.1.

For each pair for relators $r_{1}$ and $r_{2}$ in Table 4, there is an
isomorphism between $\Gamma(r_{1},r_{2})=\langle m,n,g\mid
mnMN,r_{1},r_{2}\rangle$ and $\pi_{1}(m003)$. Similarly, we have isomorphisms
for each pair in Table 5 and $\pi_{1}(m004)$.

###### Proof.

From SnapPy, we have $\pi_{1}(m003)=\langle a,b\mid abAAbabbb\rangle$. Using
the quotpic package by Holt and Rees [HR99], we list the isomorphisms $\phi$
in the Table 10.

Group | $\phi(a)$ | $\phi(b)$ | $\phi^{-1}(m)$ | $\phi^{-1}(n)$ | $\phi^{-1}(g)$
---|---|---|---|---|---
$\Gamma(GNgMgNG,mGnGmGmnGmnG)$ | $G$ | $mGn$ | $BBABA$ | $aaBA$ | $A$
$\Gamma(MnGmGMngg,MgggMgNg)$ | $gMgN$ | $gMg$ | $BABBA$ | $Ab$ | $BA$
$\Gamma(mGGmgMNg,mmnGGmGmGG)$ | $G$ | $Gm$ | $Ab$ | $BAbabA$ | $A$
$\Gamma(mgNgmGG,mGmGGmnGG)$ | $G$ | $Gm$ | $Ab$ | $BABBA$ | $A$
$\Gamma(mnGGmngMg,NGmGGGmG)$ | $MGmGN$ | $GGm$ | $ABBAB$ | $aB$ | $ABB$
$\Gamma(mnGNggNG,GGmnGmGmnG)$ | $Gnm$ | $Gm$ | $aBABab$ | $Ba$ | $aBABa$
$\Gamma(nGNmggNmG,NgMgNggg)$ | $gNgM$ | $gNg$ | $Ab$ | $BABBA$ | $BA$
$\Gamma(ngMgnGG,mNGmGGGmG)$ | $mGM$ | $NmG$ | $ABABB$ | $AAba$ | $BABAbab$
Table 10: Isomorphisms with $\pi_{1}(m003)=\langle a,b\mid abAAbabbb\rangle$.

Similarly, $\pi_{1}(m004)=\langle a,b\mid aaabABBAb\rangle$. We list the
isomorphisms $\psi$ in the Table 11.

Group | $\psi(a)$ | $\psi(b)$ | $\psi^{-1}(m)$ | $\psi^{-1}(n)$ | $\psi^{-1}(g)$
---|---|---|---|---|---
$\Gamma(MNgmGMGmg,MgMGmgNgmG)$ | $gN$ | $nGnGM$ | $BAA$ | $ABabaBAb$ | $BabaBAb$
$\Gamma(MnGmgMgmG,gmGMgMGmg)$ | $g$ | $mG$ | $ba$ | $baBAbABa$ | $a$
$\Gamma(MnGmgMgmG,mgmGMgNgMG)$ | $Ng$ | $mGnGn$ | $baa$ | $bABabaBA$ | $bABabaB$
$\Gamma(mGMnGmgMg,MGMgmGnGmg)$ | $Gn$ | $NgNgM$ | $BAA$ | $BabABAba$ | $BabABAb$
$\Gamma(mnGMgmgMG,MgmGGmgMG)$ | $G$ | $mg$ | $ba$ | $AbaBabAB$ | $A$
Table 11: Isomorphisms with $\pi_{1}(m004)=\langle a,b\mid aaabABBAb\rangle$.

∎

##  Appendix B Technical necklace theory

The following lemma is used in the proof of Lemma 4.26. As the details are a
bit technical, we chose to leave it out of the main exposition and place it in
the appendix.

######  Lemma B.1.

Let $B_{0}$ be a full-sized horoball centered at $(0,0)$ and let $B_{1},B_{2}$
be a tangent pair of at most full-sized horoballs. Let $\mathcal{D}$ be the
disk though $b_{1}=\partial_{\infty}B_{1}$ and $b_{2}=\partial_{\infty}B_{2}$
that makes an angle of $\pi/3$ with the oriented ray from $b_{2}$ though
$b_{1}$. If all the horoballs are disjoint, then the visual angle of
$\mathcal{D}$ from $(0,0)$ is at most $\pi/3$.

\begin{overpic}[scale={1.5}]{figures/escape_plane_five_lemma.pdf}
\put(12.0,48.0){$B_{0}$} \put(51.0,57.0){$\pi-\theta$}
\put(69.0,55.0){$B_{1}$} \put(47.0,12.0){$B_{2}$}
\put(51.3,26.0){${\frac{\pi}{3}}$} \put(75.0,22.0){$D$} \end{overpic} Figure
18: .

###### Proof.

Let $A$, $B$, and $C$ be, respectively, the centers of $B_{0}$, $B_{1}$, and
$B_{2}$. Let $h_{A}$, $h_{B}$, and $h_{C}$ likewise be their heights. Then
$h_{A}=1$ since $A$ is full-sized, and $h_{B},h_{C}\leq 1$ since $B,C$ are at
most full-sized. Likewise, by Lemma 4.28, $AB^{2}\geq h_{A}h_{B}=h_{B}$ and
$AC^{2}\geq h_{A}h_{C}=h_{C}$ since the interiors of $B_{0}$, $B_{1}$, and
$B_{2}$ are all disjoint. Moreover, $BC^{2}=h_{B}h_{C}$ since $B_{1}$ and
$B_{2}$ are tangent. From these inequalities, we can conclude $BC\leq AB\cdot
AC$ and $BC\leq 1$. Conversely, if $BC\leq AB\cdot AC$ and $BC\leq 1$, then
there is some choice of heights $h_{B}$ and $h_{C}$ yielding a collection of
horoballs as given above. So the conditions $BC\leq AB\cdot AC$ and $BC\leq 1$
together are equivalent to the antecedent of the lemma.

On the other hand, we can also rewrite the consequent. Let $r$ be the radius
of $\mathcal{D}$, $D$ its center, and $d=AD$. Then the visual angle of
$\mathcal{D}$ at $A$ is at most $\gamma$ if and only if $\sin(\gamma/2)\leq
r/d$. So $2r\leq d$ is equivalent to the consequent. Thus the lemma is
equivalent to the following: if $BC\leq AB\cdot AC$ and $BC\leq 1$, then
$2r\leq d$.

We now want to rewrite this in terms of $r$ and $d$ and one other parameter.
First, by construction of $\mathcal{D}$, $BC^{2}=3r^{2}$. So $BC\leq 1$ is
equivalent to $3r^{2}\leq 1$.

Next, consider all the possible configurations with a given pair of $d$ and
$r$. These are parametrized by the angle $\angle ADB$. However, we find it
more convenient to use instead the parameter $\phi=\angle ADB-\pi/3$, which
measures the deviation from the symmetric position where $AB=AC$ and $A$ and
$D$ lie on opposite sides of the supporting line of $\overline{BC}$. By the
law of cosines,

$\displaystyle AB^{2}$ $\displaystyle=d^{2}+r^{2}-2dr\cos(\pi/3+\phi)$
$\displaystyle=x-y(\cos(\phi)/2-\sqrt{3}\sin(\phi)/2)\mbox{ and}$
$\displaystyle AC^{2}$ $\displaystyle=d^{2}+r^{2}-2dr\cos(\pi/3-\phi)$
$\displaystyle=x-y(\cos(\phi)/2+\sqrt{3}\sin(\phi)/2),$

letting $x=d^{2}+r^{2}$ and $y=2dr.$ Thus

$AB^{2}\cdot AC^{2}=y^{2}\cos^{2}(\phi)-xy\cos(\phi)+x^{2}-3y^{2}/4.$

Hence the other part of the antecedent we may rewrite as

(B.1) $3r^{2}\leq
4d^{2}r^{2}\cos^{2}(\phi)-2\cos(\phi)(dr^{3}+d^{3}r)+d^{4}-d^{2}r^{2}+r^{4}.$

Finally for this setup, we note first that, by assumption,
$\phi\in[-\pi/3,\pi/3]$, for otherwise $A$ and $D$ would lie on the same side
of the supporting line of $\overline{BC}$, contrary to how we drew
$\mathcal{D}$; and second, that, of course, we assume $d>0$ and $r>0$. Thus
the lemma is equivalent to the following statement:

> Suppose $d,r>0$ and $\phi\in[-\pi/3,\pi/3]$. Suppose $3r^{2}\leq 1$.
> Finally, suppose B.1 holds. Then $d\geq 2r$.

To show this, let us first bound the extreme values of the right-hand side of
B.1 for $\phi\in[-\pi/3,\pi/3]$. To that end, we enumerate the critical points
in $[-\pi/3,\pi/3]$. Using $x$ and $y$ as above,

$\frac{d}{d\phi}\left(AB^{2}\cdot AC^{2}\right)=y\sin(\phi)(x-2y\cos(\phi))$

First, $y=2dr=0$ is impossible. Next, $\sin(\phi)=0$ when $\phi=0$, i.e. when
$\cos(\phi)=1$, since $\phi\in[-\pi/3,\pi/3]$. The last critical point
condition yields $\cos(\phi)=x/(2y)$. Finally, the edge cases $\phi=\pm\pi/3$
both yield $\cos(\phi)=1/2$. Thus the extreme values of $AB^{2}\cdot AC^{2}$
for $\phi\in[-\pi/3,\pi/3]$ are in the set

$\displaystyle S$ $\displaystyle=\left\\{(y-2x)^{2}/4,\quad(x-y)(y+2x)/2,\quad
3(x-y)(x+y)/4\right\\}$
$\displaystyle=\left\\{(d^{2}-dr+r^{2})^{2},\quad(d-r)^{2}(d^{2}+dr+r^{2}),\quad
3(d-r)^{2}(d+r)^{2}/4\right\\}.$

Therefore, under our assumptions, $3r^{2}\leq s$ for some $s\in S$. These are
all quartic forms. Dividing them all by $r^{4}$ and setting $z=d/r$ yields the
set

$T=\\{(z^{2}-z+1)^{2},\quad(z-1)^{2}(z^{2}+z+1),\quad
3(z-1)^{2}(z+1)^{2}/4\\}.$

Assuming $0<r\leq\sqrt{1/3}$, we have $3r^{2}/r^{4}=3/r^{2}\geq 9$. So it will
suffice to show that if any element of $T$ is at least 9, then $z\geq 2$. We
finish by case analysis.

Suppose $(z^{2}-z+1)^{2}\geq 9$. Since $z^{2}-z+1\leq-3$ is impossible, we
have $z^{2}-z+1\geq 3$. Now, $z^{2}-z-2=(z-2)(z+1)$. Since $z>0$,
$z^{2}-z+1\geq 3$ implies $z\geq 2$.

Suppose $(z-1)^{2}(z^{2}+z-1)\geq 9$. Let $f(z)=(z-1)^{2}(z^{2}+z-1)$. Then
$f^{\prime}(z)=(z-1)(4z^{2}+z+1)$. Now, $4z^{2}+z+1$ is positive for all
$z\in\mathbb{R}$. So $f$ is decreasing on $(-\infty,1]$ and increasing on
$[1,\infty)$. Calculating, $f(0)=1$, $f(1)=0$, and $f(2)=7$. Hence if $z\geq
0$ and $f(z)\geq 9$, then $z>2$.

Finally, suppose $3(z-1)^{2}(z+1)^{2}/4\geq 9$. Then $(z^{2}-1)^{2}\geq 12$.
Since $z^{2}-1\leq-\sqrt{12}$ is impossible, $z^{2}-1\geq\sqrt{12}$. Since
$z>0$, $z\geq\sqrt{1+\sqrt{12}}>2$.

∎

##  Appendix C Convolotubes

The following is a “hands-on” proof of Lemma 5.8. Since this Lemma is a
consequence of Proposition 1.7 of [GMM11], we only include the proof here for
completeness. Recall that we want to show that if $M$ is a generalized link
exterior, then $NE^{+}(M)$ is the class of cusped hyperbolic Dehn fillings of
$M$.

###### Proof of Lemma 5.8.

Abusing notation, we realize an embedding in $NE^{+}(M)$ as $M\subset N$ where
$N$ is an orientable hyperbolic finite-volume $N$. Suppose $T$ is a torus
boundary component of $M$. It therefore is either $\partial$-parallel in $N$
or compressible in $N,$ since $N$ is atoroidal. Suppose $T$ is compressible.
Let $\Delta$ be a compressing disk for $T$ in $N$. Up to isotopy,
$\Delta\cap\partial M$ is a multicurve on $\Delta$ with $\partial\Delta$ as a
component. There may or may not be other components.

  * •

Suppose $\Delta\cap\partial M=\partial\Delta$. Then either $\Delta\subset M$,
or $\Delta$ lies in a component of $N\setminus int(M)$.

    * –

Suppose $\Delta\subset M$. Then $\Delta$ is a compressing disc for $T$ in $M$.
Let $U$ be a regular neighborhood of $\Delta\cup T$ in $M$, so that $\partial
U=T\sqcup\Sigma$ for some sphere $\Sigma\subset M$. Thus $M$ is a (possibly
trivial) connect-sum with a solid torus, along $\Sigma$. Now, since $N$ is
irreducible, $\Sigma$ bounds a ball $B$ in $N$. Either $T\subset B$, or not.

      * *

If $T\subset B$, then $U\subset B$. Hence $M=M^{\prime}\cup_{\Sigma}U$, where
$M^{\prime}=M\setminus int(B)$. Let $i:M\hookrightarrow N$ be the inclusion of
$M$ into $N$. We may modify $i$ on $U$ to a different embedding so that $i(U)$
is the complement of an unknot in $B$. Thus after so modifying $i$, $T$ bounds
a solid torus in $N$ away from $M$, and no other boundary tori of $M$ are
modified, thus reducing the number of convolutubes of $i(\partial M)$.

      * *

Otherwise, $U\cup B$ is a solid torus $\mathcal{T}$ in $N$, and
$M\subset\mathcal{T}$. Thus the embedding of $M$ in $N$ is elementary.

    * –

Suppose $\Delta$ lies in a component of $N\setminus M$. Now let $U$ be a
regular neighborhood of $\Delta\cup T$ in $N\setminus int(M)$, so that
$\partial U=T\sqcup\Sigma$ for some sphere $\Sigma\subset N\setminus int(M)$.
Since $N$ is irreducible, $\Sigma$ bounds a ball $B$ in $N$. Either $T\subset
B$ or not.

      * *

Suppose $T\subset B$. Then $M\subset B$, for $M$ is connected and lies to that
side of $T$ away from $\Sigma$. Hence the embedding of $M$ in $N$ is
elementary.

      * *

Otherwise, $T$ bounds the solid torus $U\cup B$ in $N$ to the side of $T$ away
from $M$.

Therefore, if $\Delta\cap\partial M=\partial\Delta$, then either the embedding
is elementary, or it admits a (possibly trivial) redefinition such that $T$
bounds a solid torus in $N$ away from $M$.

  * •

Suppose instead that $int(\Delta)\cap\partial M\neq\emptyset$. Then it has an
innermost component $\gamma$. It bounds a unique disc $\Delta^{\prime}$ in
$\Delta$ such that $\Delta^{\prime}\cap\partial M=\gamma$. Then
$\partial\Delta^{\prime}$ lies on some other boundary component $T^{\prime}$
of $M$. On this component $T^{\prime}$, $\partial\Delta^{\prime}$ is either
essential or not.

    * –

If $\partial\Delta^{\prime}$ is inessential on $T^{\prime}$, then it bounds a
disc $\delta$ on $T^{\prime}$. Then $\partial\Delta\cup_{\gamma}\delta$ is a
sphere $\Sigma$ in $N$, and thus bounds a ball $B$ in $N$. Hence we may
isotope $\Delta$, fixing $\partial\Delta$, through $B$, and then slightly
pushing off $\delta$, to a new disc $D$ such that $int(D)\cap\partial
M=(int(\Delta)\cap\partial(M))\setminus\gamma$, thus reducing the number of
components in $int(\Delta)\cap\partial(M)$, and leaving $M$ invariant.

    * –

If $\partial\Delta^{\prime}$ is essential on $T^{\prime}$, then
$\partial\Delta^{\prime}$ is a compressing disc for $T^{\prime}$ with
$\Delta^{\prime}\cap\partial M=\partial\Delta^{\prime}$.

Thus by induction on $int(\Delta)\cap\partial M$, there is some compressing
disc $\Delta$ for some boundary component of $M$ such that $\Delta\cap\partial
M=\partial\Delta$, reducing to the previous case.

Therefore, if $M\subset N$ is nonelementary, then there is some nonelementary
embedding $i:M\hookrightarrow N$ such that $N\setminus i(M)$ consists of
boundary collars and solid tori—that is, such that $i$ manifests $N$ as a Dehn
filling of $M$.

Conversely, if $N$ is a Dehn filling of $M$, then the Dehn filling embedding
$i:M\hookrightarrow N$ has $i_{\ast}(\pi_{1}(M))=i_{\ast}(\pi_{1}(N))$. In
particular, if $N$ is hyperbolic, then $i_{\ast}(\pi_{1}(N))$ is nonabelian,
so the embedding is nonelementary.

Thus the class of hyperbolic 3-manifolds $N$ admitting nonelementary
embeddings from $M$ is the same as the class of hyperbolic Dehn fillings of
$M$. ∎

##  Appendix D Computational aspects

In this appendix, we will go over the technical details needed to validate
Propositions 2.15 and 3.5. We organize the discussion around the theoretical
aspects and the arithmetic of complex affine 1-jets, followed by a description
of the files, code, and data provided.

As described in Section 2, the proofs are provided as binary trees where each
terminal node corresponds a (sub)box of the parameter space $\mathscr{B}$ with
a label that encodes an inequality which holds for all points in that box.
This inequality is rigorously verified by the programs verify and identify to
hold over the box. Each inequality is encoded either as a boundary condition
or a word. In the following sections, we prove the validity of how each
inequality is constructed and the validity of the arithmetic used by verify.

Recall that the main parameter space objects are the parent box $\mathscr{B}$,
the compact subset $\mathscr{P}$ that is guaranteed to contain a bicuspid
triple for every cusped hyperbolic 3-manifolds with cusp volume $\leq 2.62$,
and $\mathfrak{D}$, the set of all geometric bicuspid triples in
$\mathscr{P}$.

We have four types of terminal conditions at a (sub)box
$\mathscr{B}_{\mathfrak{b}}$ of $\mathscr{B}$. These are:

  * •

Boundary conditions — these prove that $\mathscr{B}_{\mathfrak{b}}$ lies
outside of $\mathscr{P}$ by showing that $\mathscr{B}_{\mathfrak{b}}$ that the
inverse of one of the inequalities in Proposition 2.5 holds over the entire
box.

  * •

Killed words — a killer word $w$ associated to $\mathscr{B}_{\mathfrak{b}}$ is
used to prove that $\mathscr{B}_{\mathfrak{b}}\subset\mathcal{K}_{w}$, which
by Corollary 2.10 means that
$\mathscr{B}_{\mathfrak{b}}\cap\mathfrak{D}=\emptyset$. This is equivalent to
saying that the bicuspid triples in $\mathscr{B}_{\mathfrak{b}}$ are either
indiscrete or are incorrectly marked, meaning that a correct marking is found
elsewhere in $\mathscr{P}\mathscr{B}$.

  * •

Necklace words — a necklace word $w$ associated to
$\mathscr{B}_{\mathfrak{b}}$ is used to prove
$\mathscr{B}_{\mathfrak{b}}\subset\mathcal{U}_{w}$, which by Lemma 2.8 means
that any triple in $\mathscr{B}_{\mathfrak{b}}\cap\mathfrak{D}$ is a relator
of $g$-length at most that of $w$. In verify, we allow this $g$-length to be
at most $7$, while in identify we limit this to $3$.

  * •

Variety words — a variety word $w$ associated to $\mathscr{B}_{\mathfrak{b}}$
is used to prove $\mathscr{B}_{\mathfrak{b}}\subset\mathcal{V}_{w}$, which by
Lemma 2.13 means that $w$ is a relator for any bicuspid group associated to a
triple in $\mathscr{B}_{\mathfrak{b}}\cap\mathfrak{D}$. This condition is only
used in identify.

The inequalities in the definitions of $\mathcal{K}_{w}$, $\mathcal{U}_{w}$,
and $\mathcal{V}_{w}$ are straightforward to check after constructing the
matrix corresponding to $w$ at points of the box $\mathscr{B}_{\mathfrak{b}}$.
To prove that these conditions hold over the entire box, we use 1-jets with
error and round-off error for computations as is done in [GMT03]. If one
wanted to use interval arithmetic instead, we expect that further subdivision
would be necessary.

###  D.1 1-jets and error control

Our computational tool for verifying the above mentioned inequalities are
1-jets. Let

$\mathscr{A}=\\{(z_{0},z_{1},z_{2})\in\mathbb{C}^{3}:|z_{i}|\leq 1\\}$

and consider a (holomorphic) function $g:\mathscr{A}\to\mathbb{C}.$ A $1$-jet
approximation with error $\epsilon$ of $g$ is a linear map
$(z_{0},z_{1},z_{2})\mapsto c+a_{0}z_{0}+a_{1}z_{1}+a_{0}z_{2}$ such that

$|g(z_{0},z_{1},z_{2})-c+a_{0}z_{0}+a_{1}z_{1}+a_{0}z_{2})|\leq\epsilon.$

Considering all maps approximated by a given 1-jet, one defines the jet-set

$S(c,a_{0},a_{1},a_{2};\epsilon)=\\{g:\mathscr{A}\to\mathbb{C}:|g(z_{0},z_{1},z_{2})-c+a_{0}z_{0}+a_{1}z_{1}+a_{0}z_{2})|\leq\epsilon\\}.$

In [GMT03], the authors derive the arithmetic for jets. For example, given
jet-sets $S_{1}$ and $S_{2}$, they show to to compute the parameters for a
jet-set $S_{3}$ where $f\cdot g\in S_{3}$ for all $f\in S_{1}$ and $g\in
S_{2}$. They do this for all basic arithmetic operations $\pm,\times,/,$ and
$\sqrt{\,}$. In code, we call these ACJs for “affine complex jets.”

For our computations, we use this exact same arithmetic. Note, to make this
arithmetic fast, the parameters of $S(c,a_{0},a_{1},a_{2};\epsilon)$ are all
given as floating-point doubles. Recall that floating-point numbers are a
finite subset of the reals represented via a list of bits on a computer. As
such, any arithmetic operation between two floating-point numbers must make a
choice about which floating-point result to give, as the real value of this
result may not be representable as floating-point number. On a machine
conforming to the IEEE-754 standard [iee85], all operations performed with
floating points numbers are guaranteed to round in a consistent way to a
closest floating-point representative, as long as overflow and underflow have
not occurred. The code supplied as part of verify and identify checks that
underflow and overflow does not happen during validation.

Just as in Section 7 of [GMT03], each boxcode corresponds to a (sub)box of our
parameter space that has a floating-point center
$(c_{0},c_{1},c_{2},c_{3},c_{4},c_{5})$ and a floating-point size
$(s_{0},s_{1},s_{2},s_{3},s_{4},s_{5})$. Using the IEEE-754 standard, the
floating-point size is large enough so that the floating-point version of the
box, i.e.
$\\{(x_{0},x_{1},x_{2},x_{3},x_{4},x_{5})\in\mathbb{R}^{6}:|x_{i}-c_{i}|\leq
s_{i}\\}$, contains the true box as a proper subset. Recall that over a box we
associate the coordinate functions $L=x_{3}+\mathfrak{i}x_{0}$,
$S=x_{4}+\mathfrak{i}x_{1}$, and $P=x_{5}+\mathfrak{i}x_{2}$ for our
parameters. We can replace these coordinate functions with linear maps
$g_{L},g_{L},g_{P}:\mathscr{A}\to\mathbb{C}$ given by:

$g_{L}(z_{0},z_{1},z_{2})=c_{3}+\mathfrak{i}c_{0}+(s_{3}+\mathfrak{i}s_{0})z_{0},\quad
g_{S}(z_{0},z_{1},z_{2})=c_{4}+\mathfrak{i}c_{1}+(s_{4}+\mathfrak{i}s_{1})z_{1},\quad
g_{L}(z_{0},z_{1},z_{2})=c_{5}+\mathfrak{i}c_{2}+(s_{5}+\mathfrak{i}s_{2})z_{2}.$

Notice that, by construction, the linear maps see all the $L,S,P$ values over
the given box. We can now think of $g_{L},g_{S},g_{P}$ as living in jet-sets.
For example, $g_{P}\in
S(c_{5}+\mathfrak{i}c_{2},0,0,s_{5}+\mathfrak{i}s_{2};0)$ and similarly for
others. It follows that any evaluations with jet-arithmetic will include the
true values over the box up to the error that accumulates. See [GMT03],
Sections 7 and 8 for concrete details.

###  D.2 Elimination conditions

Recall that the computational proofs of our results are encoded by binary
trees where each terminal leaf contains a string that encodes the necessary
list of inequalities that we must hold true over the box. The code in verify.c
and identify.c is responsible for reading the tree starting from the root
node, building the box ACJ parameters for each boxcode, and checking each
condition. The evaluations of each condition occurs in elimination.c. The code
in these files is commented and we believe the reader will be best served by
following the comments there as well as the README.md.

We want to stress a few main points. First, the jet arithmetic code is
entirely borrowed from [GMT03], aside from a tiny patch of a clear typo in the
published code, see the patch file in the scripts folder of [Gitb]. In
addition, we make use of the floating-point rounding estimates from [GMT03] in
a few places in our code. In particular, we apply this error control to
evaluate boundary conditions such as cusp area and bounds on real and
imaginary parts of parameters. Since most of these quantities are monotonic in
the parameters, we simply construct validated over and under estimates at the
vertices of our box. These are the nearer, further, and greater points, which
are nearer than all box points to $\vec{0}$, further than all box points for
$\vec{0}$, and greater than all box points, respectively. This simple
constructions allows us to easily check the boundary conditions of Proposition
2.5. For a detailed explanation of how these points are chosen, see the
comments in box.c.

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