Totally geodesic surfaces in twist knot complements

We would like to thank our advisor Matthew Stover for asking us 1.1, for multiple helpful suggestions in proving the main theorems, and for the careful reading of the first draft of this paper. We also thank David Futer and Rose Kaplan-Kelly for pointing out the application of 3.5 towards right-angled knots. Finally, we would like to thank the referee for their comments and their suggestions to use Sage, an open source, and the initial Sage codes.

Let $K_{j}$ be the twist knot with $j$ half-twists, $M_{j}:=S^{3}\smallsetminus K_{j}$, and $\Gamma_{j}$ be its fundamental group. The group $\Gamma_{j}$ has a presentation $\Gamma_{j}=\langle a,b\mid b=w_{j}aw_{j}^{-1}\rangle$ where $w_{j}$ is given by a word of length $2j$ in $a^{\pm 1}$ and $b^{\pm 1}$, namely:

$$w_{j}=\begin{cases}(a^{-1}b^{-1}ab)^{(j-1)/2}a^{-1}b^{-1}&j\equiv 1\mod 2\\ (a^{-1}bab^{-1})^{j/2}&j\equiv 0\mod 2\end{cases}$$

(1)

The two generators $a$, $b$ correspond to the two meridians of the twist knot chosen to have the same orientation. The longitude of the twist knots that commute with $a$ can be computed diagrammatically and is given by:

$$\ell_{j}=\begin{cases}a\overline{w}_{j}b^{2}w_{j}a&j\equiv 1\mod 2\\ \overline{w}_{j}w_{j}&j\equiv 0\mod 2\\ \end{cases}$$

(2)

where $\overline{w}_{j}$ is $w_{j}$ spelled backwards. Finally, we note that the above presentation is different from that in [10, Equation 2.1] by an isomorphism interchanging $a$ and $b$.

For $j\geq 2$, $M_{j}$ admits a complete hyperbolic structure of finite volume. The discrete faithful representation $\rho_{j}:\Gamma_{j}\to\operatorname{SL}_{2}({\mathbb{C}})$ sends all conjugates of the meridians of the knot to parabolic isometries; i.e. conjugate in $\operatorname{SL}_{2}({\mathbb{C}})$ to:

$$\begin{pmatrix}1&1\\ 0&1\end{pmatrix}$$

The fact that $z_{j}$ must be a root of an integral polynomial was first observed by Riley for the class of two-bridge knots [18, Theorem 2]. For the twist knots, a recursion for the polynomials $\Lambda_{j}(z)$ was given by Hoste and Shanahan [10, Theorem 2]. We make the following observation which will be used in the proof of 1.3:

Since the presentation of $\Gamma_{j}$ that we are using is slightly different from that in [10, Equation 2.1], we give a proof of the above theorem for completeness.

• Proof of 2.1.

  Consider the free group $\operatorname{F}_{2}$ on two generators $A$ and $B$ along with the surjective homomorphism $\operatorname{F}_{2}\to\Gamma_{j}$ sending $A$, $B$ to $a$, $b$ respectively. Let $W_{j}$ be the lift of $w_{j}$ given by Equation 1. Consider the homomorphism $P:\operatorname{F}_{2}\to\operatorname{SL}_{2}({\mathbb{Z}}[z])$ defined by

  $$P(A)=\begin{pmatrix}1&1\\ 0&1\\ \end{pmatrix}\quad\text{and}\quad P(B)=\begin{pmatrix}1&0\\ z&1\end{pmatrix}$$

  We claim that

  $$P(W_{j})=\begin{pmatrix}\Lambda_{j}(z)&\mu_{j}(z)\\ z\mu_{j}(z)&\Lambda_{j-1}(z)\end{pmatrix}$$

  (4)

  for some $\mu_{j}(z)\in{\mathbb{Z}}[z]$.

  For convenience of notation, we identify elements of $\operatorname{F}_{2}$ with their image under $P$. We consider two cases according to whether $j$ is even or odd. When $j$ is even, we have $W_{j}=([A^{-1},B])^{j/2}$. Applying Cayley–Hamilton to $[A^{-1},B]$, we get

  $$([A^{-1},B])^{2}-\operatorname{tr}([A^{-1},B])[A^{-1},B]+I=0$$

  Note that $\displaystyle\operatorname{tr}([A^{-1},B])=z^{2}+2$. Therefore, we have

  $$W_{j+2}=(z^{2}+2)W_{j}-W_{j-2}$$

  (5)

  This gives us the recursion in Equation 3. Now we observe that the upper diagonal (resp. lower diagonal) entries of $W_{2}=[A^{-1},B]$ and of $W_{0}=I$ give us the initial conditions for $\Lambda_{j}(z)$ (resp. $\Lambda_{j-1}(z)$). For the off-diagonal entries, we observe that they satisfy the same recursion as in Equation 3 and their initial conditions differ by a factor of $z$. When $j$ is odd, we proceed similarly to compute $P(W_{j})$ where the initial conditions are given by $W_{-1}=B^{-1}A^{-1}$ and $W_{1}=A^{-1}B^{-1}$.

  To finish the proof, we note that $\rho_{j}$ is obtained by factoring $\operatorname{ev}_{j}\circ P$ through the canonical projection $\operatorname{F}_{2}\to\Gamma_{j}$ where $\operatorname{ev}_{j}:\operatorname{SL}_{2}({\mathbb{Z}}[z])\to\operatorname{SL}_{2}({\mathbb{Z}}[z_{j}])$ is the homomorphism induced by the evaluation map ${\mathbb{Z}}[z]\to{\mathbb{Z}}[z_{j}]$. Using Equation 4, we see that the relation $w_{j}a-bw_{j}=0$ is satisfied if and only if $\Lambda_{j}(z_{j})=0$. ∎

Given a finitely-generated non-elementary subgroup $\Gamma$ of $\operatorname{PSL}_{2}({\mathbb{C}})$, we can define the trace field of $\Gamma$ to be ${\mathbb{Q}}(\operatorname{tr}\Gamma):={\mathbb{Q}}(\operatorname{tr}\gamma\mid\gamma\in\Gamma)$ and the invariant trace field of $\Gamma$ to be $k\Gamma:={\mathbb{Q}}(\operatorname{tr}\gamma^{2}\mid\gamma\in\Gamma)$. If $\Gamma$ additionally has finite covolume, then ${\mathbb{Q}}(\operatorname{tr}\Gamma)$ and $k\Gamma$ are number fields [13, Theorem 3.1.2]. In general, the trace field and invariant trace field are different. The trace field is not an invariant of the commensurability class of $\Gamma$, but the invariant trace field is. These fields may coincide, such as link complements in a ${\mathbb{Z}}/2$-homology sphere. [13, Corollary 4.2.2]. Finally, suppose that $\Gamma$ is generated by $\gamma_{1}$ and $\gamma_{2}$; then ${\mathbb{Q}}(\operatorname{tr}\Gamma)={\mathbb{Q}}(\operatorname{tr}\gamma_{1},\operatorname{tr}\gamma_{2},\operatorname{tr}\gamma_{1}\gamma_{2})$ [13, Equation 3.25].

It follows from the discussion above and 2.1 that the trace field of twist knot complements is ${\mathbb{Q}}(z_{j})$. Hoste and Shanahan proved that $\Lambda_{j}(z)$ is the monic minimal polynomial of $z_{j}$ over ${\mathbb{Z}}$ of degree $j$ [10, Theorem 1]. Therefore, $[{\mathbb{Q}}(z_{j}):{\mathbb{Q}}]=j$ and so twist knots are pairwise non-commensurable.

• Proof.

  Let us consider $\overline{W}_{j}$ to be the lift of $\overline{w}_{j}$ in $\operatorname{F}_{2}$ given by spelling $W_{j}$ backwards. We first claim that:

  $$P(\overline{W}_{j})=\begin{pmatrix}\Lambda_{j-1}(z)&\mu_{j}(z)\\ z\mu_{j}(z)&\Lambda_{j}(z)\end{pmatrix}$$

  Observe that $\overline{W}_{j}$ satisfy the same recurrence in Equation 5, where $\overline{W}_{0}$, $\overline{W}_{1}$ provide the base case.

  We next claim that:

  $$(-1)^{j+1}\Lambda_{j}(z)+\Lambda_{j-1}(z)=[(-1)^{j}z-2]\mu_{j}(z)$$

  We proceed by induction on $j$. The base case is provided by $W_{0}=I$ and $W_{1}=A^{-1}B^{-1}$. Assume the statement for $W_{k}$ for all $k<j+2$. We have

  	$\displaystyle(-1)^{j+3}\Lambda_{j+2}(z)+\Lambda_{j+1}(z)$	$\displaystyle=(z^{2}+2)[(-1)^{j+3}\Lambda_{j}(z)+\Lambda_{j-1}(z)])-[(-1)^{j+3}\Lambda_{j-2}(z)+\Lambda_{j-3}(z)]$
  		$\displaystyle=(z^{2}+2)[(-1)^{j}z-2]\mu_{j}(z)-[(-1)^{j}z-2]\mu_{j-2}(z)$
  		$\displaystyle=[(-1)^{j+2}z-2]\mu_{j+2}(z)$

  where the last equality follows from the recurrence of $\mu_{j}(z)$ given by Equation 5. Therefore, we have

  $$\Lambda_{j-1}(z_{j})=[(-1)^{j}z_{j}-2]\mu_{j}$$

  This identity gives us the image of $w_{j}$ and $\overline{w}_{j}$ under $\rho_{j}$. We note that by the determinant of $P(\overline{W}_{j})$,

  $$z_{j}\mu_{j}^{2}=-1$$

  (6)

  Applying $\rho_{j}$ to Equation 2 and using Equation 6, we get the image of $\ell_{j}$ under $\rho_{j}$ where

  $$\tau_{j}=(4\mu_{j}^{2}+2)=-\frac{4}{z_{j}}+2$$

  (7)

  ∎

We end this subsection by defining boundary slopes for cusped totally geodesic surfaces in $M_{j}$. Let us identify the universal cover ${\mathbb{H}}^{3}$ of $M_{j}$ with the upper-half space model and the ideal boundary of ${\mathbb{H}}^{3}$ with ${\mathbb{C}}\cup\{\infty\}$. The action of $\Gamma_{j}$ on ${\mathbb{H}}^{3}$ is given by $\rho_{j}$. We say that $P\in{\mathbb{C}}\cup\{\infty\}$ is a cusp point of $M_{j}$ if its stabilizer in $\Gamma_{j}$ is conjugate to $\langle a,\ell_{j}\rangle$. In particular, the point $\infty$ is the cusp point of $M_{j}$ whose stabilizer in $\Gamma_{j}$ is the ${\mathbb{Z}}^{2}$-subgroup generated by $a$ and $\ell_{j}$ acting as translations by $1$ and $\tau_{j}$ respectively. At every cusp point in ${\mathbb{H}}^{3}$, we choose a sufficiently small horoball neighborhood such that the stabilizer is a conjugate of $\langle a,\ell_{j}\rangle$ and such that these horoball neighborhoods are invariant under the action of $\Gamma_{j}$. We shall refer to the image of these horoball neighborhood as the cusp of $M_{j}$. Similarly given any hyperplane $H$ in ${\mathbb{H}}^{3}$, a point $P$ in the circle/line at infinity of $H$ is called a cusp point of $H$ if it is a fixed point of a parabolic element in $\operatorname{Stab}_{\Gamma_{j}}(H)$.

Let $\Sigma$ be a properly immersed cusped totally geodesic surface in $M_{j}$. Since $M_{j}$ has one cusp and $\Sigma$ is complete, all cusps of $\Sigma$ must be contained in the cusp of $M$. Given any cusp $c$ of $\Sigma$, we consider a lift $\widetilde{\Sigma}$ of $\Sigma$ to ${\mathbb{H}}^{3}$ by putting the cusp point corresponding to $c$ at $\infty$. The cusp cross-section of $c$ in $\Sigma$ is obtained by intersecting $\Sigma$ with the cusp cross-section of $M$. This intersection lifts to a horocycle at $\infty$ in $\widetilde{\Sigma}$. The stabilizer of this horocycle is an element of the form $a^{p}\ell_{j}^{q}$ in $\langle a,\ell_{j}\rangle$. In this case, we say that $p/q$ is a boundary slope of $\Sigma$.

More generally, suppose that $\zeta$ is a cusp point of a hyperplane $H$ in ${\mathbb{H}}^{3}$ such that the parabolic element in the stabilizer of $H$ in $\Gamma_{j}$ fixing $\zeta$ is conjugate to $a^{p}\ell_{j}^{q}$ in $\Gamma_{j}$. Then we say that $p/q$ is the boundary slope of $H$ at $\zeta$.

For all $j\geq 2$, $M_{j}$ contains an immersed thrice-punctured sphere $N$ as shown in Figure 1. By Adams [1, Theorem 3.1], $N$ is a totally geodesic surface. Moreover:

• Proof.

  From the diagram, we see that the fundamental group of $N$ is generated by $b$ and:

  $$\begin{cases}a^{-1}(bab^{-1}a^{-1})^{(j-1)/2}bab^{-1}(aba^{-1}b^{-1})^{(j-1)/2}a&j\equiv 1\mod 2\\ (b^{-1}aba^{-1})^{j}b^{-1}ab(ab^{-1}a^{-1}b)^{j}&j\equiv 0\mod 2\end{cases}$$

  Thus, we see that $\pi_{1}(N)$ is conjugate to the subgroup $\Delta_{j}$ generated by $x:=a$ and $y_{j}$ as stated. Under the discrete faithful representation, we have

  $$\rho_{j}(\Delta_{j})=\left\langle\begin{pmatrix}1&1\\ 0&1\end{pmatrix},\begin{pmatrix}-1&1\\ -4&3\\ \end{pmatrix}\right\rangle$$

  which is conjugate to the principal congruence subgroup of level 2 of $\operatorname{PSL}_{2}({\mathbb{Z}})$.

  Note that since both $x$ and $y_{j}$ are conjugate to $a$, the surface $N$ has two slopes 1/0. The parabolic element corresponding to the remaining cusp is given by

  $$x^{-1}y_{j}^{-1}=\begin{pmatrix}-1&0\\ 4&-1\end{pmatrix}=b^{-2}w_{j}\ell_{j}w_{j}^{-1}$$

  Since $b^{-2}w_{j}\ell_{j}$ and $a^{-2}\ell_{j}$ are conjugate to each other in $\Gamma_{j}$, the slope of the remaining cusp is $-2/1$. ∎

We will also need the following fact about the action of $\rho_{j}(\Delta_{j})$ on ${\mathbb{H}}_{\mathbb{R}}^{2}$.

• Proof.

  Let $C$ denote the set of cusp points in ${\mathbb{H}}^{2}_{\mathbb{R}}$ with respect to the action of $\rho_{j}(\Delta_{j})$. Since $a\in\rho_{j}(\Delta_{j})$, we see that $\infty\in C$. The fundamental domain for the action is an ideal quadrilateral with vertices at $\infty,0,\frac{1}{2}$ and $1$. There are three distinct orbits corresponding to $0$, $\frac{1}{2}$ and $\infty$. To see the description of these orbits, we first note that the integer matrices

  $$\begin{pmatrix}v^{\prime}&u\\ -4u^{\prime}&v\end{pmatrix},\quad\begin{pmatrix}u-2v^{\prime}&v^{\prime}\\ v-2u^{\prime}&u^{\prime}\end{pmatrix},\text{ and}\begin{pmatrix}u&v^{\prime}\\ v&u^{\prime}\end{pmatrix}$$

  map $0$, $\frac{1}{2}$, and $\infty$ to $u/v$ when $v$ is odd, $2\mod 4$, or $0\mod 4$, respectively. The determinants of these matrices can be chosen to be 1 because $4u$ is relatively prime to $v$ when both $v$ is odd and $u$ is relatively prime to $v$. By Remark 2.6, to be in $\rho_{j}(\Delta_{j})$ (up to sign), it is sufficient to show that the lower diagonal entry of each matrix is congruent to $0\mod 4$. This is immediate for the first and the third matrices. For the second matrix, note that $u^{\prime}$ has to be an odd integer since $v$ is an even integer relatively prime to $u^{\prime}$. Since $v$ is congruent to $2\mod 4$, the difference $v-2u^{\prime}$ is congruent to $0\mod 4$. ∎

We conclude this section by describing certain lifts of $N$ in ${\mathbb{H}}^{3}$ explicitly. The parametrization of these lifts play an important role in the proof of 1.3. We call a totally geodesic hyperplane in ${\mathbb{H}}^{3}$ vertical if it contains $\infty$ in its circle/line at infinity. For convenience, we let $v_{j}:=w_{j}^{-1}a^{(-1)^{j}}w_{j}^{-1}a^{-1}$ and note that $v_{j}y_{j}v_{j}^{-1}=a$. Observe that up to the action of $\operatorname{Stab}_{\Gamma_{j}}(\infty)$, the surface $N$ has three distinct vertical lifts in ${\mathbb{H}}^{3}$.

A direct computation shows that

$$w_{j}^{-1}(0)=\infty\quad\text{and}\quad v_{j}(\tfrac{1}{2})=\infty$$

Therefore, all vertical lifts of $N$ are orbits of ${\mathbb{H}}^{2}_{\mathbb{R}}$, $w^{-1}_{j}({\mathbb{H}}^{2}_{\mathbb{R}})$, and $v_{j}({\mathbb{H}}^{2}_{\mathbb{R}})$ under the action of $\operatorname{Stab}_{\Gamma_{j}}(\infty)=\langle a,\ell_{j}\rangle$. Since the slope at $0$ (resp. $\frac{1}{2}$) on ${\mathbb{H}}^{2}_{\mathbb{R}}$ is $-2/1$ (resp. $1/0$), the vertical hyperplane $w^{-1}_{j}({\mathbb{H}}^{2}_{\mathbb{R}})$ (resp. $v_{j}({\mathbb{H}}^{2}_{\mathbb{R}})$) has slope $-2/1$ (resp. $1/0$) at infinity,. Furthermore, we have

$$w_{j}^{-1}(\infty)=(-1)^{j+1}+\frac{2}{z_{j}}=-\frac{1}{2}\tau_{j}+1+(-1)^{j+1}$$

and

$$v_{j}(\infty)=-2\mu_{j}^{2}+\frac{1}{2}+(-1)^{j+1}=-\frac{1}{2}\tau_{j}+\frac{3}{2}+(-1)^{j+1}$$

Thus when $j$ is odd, the line at infinity of $w_{j}^{-1}({\mathbb{H}}^{2}_{\mathbb{R}})$ goes through the point $1$ and has slope $-2/1$. Similarly, when $j$ is odd, the line at infinity of $v_{j}({\mathbb{H}}^{2}_{\mathbb{R}})$ goes through the point $-\tau_{j}/2+5/2$ and has slope $1/0$. Figure 2 depicts the lines at infinity of the vertical lifts of $N$.

Note that in Figure 2, the element $a$ acts by translating upward and $\ell_{j}$ acts by translating to the right. In other words, the real axis in ${\mathbb{C}}$ is put in the vertical direction. Under this convention, the boundary slopes at infinity of a vertical hyperplane is precisely the slope of its line at infinity.

The following proposition of Reid [16, Proposition 2] gives an arithmetic constraint on the existence of totally geodesic surfaces inside hyperbolic $3$-manifolds.

This proposition gives a convenient way to rule out closed totally geodesic surfaces, as remarked in [16, Lemma 5].

• Proof.

  Firstly, since $[{\mathbb{Q}}(\operatorname{tr}\Gamma):{\mathbb{Q}}]=j$ is an odd prime [10, Theorem 1], the field ${\mathbb{Q}}(\operatorname{tr}\Gamma)$ has odd degree over $\mathbb{Q}$ and contains no proper subfield except ${\mathbb{Q}}$. Secondly, the fact that $\text{Im}(\rho)\leq\operatorname{SL}_{2}({\mathbb{Z}}[z])$ implies that $\Gamma$ has integral traces, see Remark 2.2. ∎

To prove the main theorems, we need to rule out cusped totally geodesic surfaces that are not freely homotopic to $N$. To this end, we examine general behavior of the intersection of two totally geodesic surfaces, and then we show that any cusped totally geodesic surface in $M$ not freely homotopic to $N$ must intersect $N$ because this thrice-punctured sphere has two distinct boundary slopes.

We have the following lemma of Fisher, Lafont, Miller and Stover [9, Lemma 3.1] which describes the intersection of totally geodesic hypersurface and immersed totally geodesic submanifolds in finite-volume hyperbolic $n$-manifold. We restate their lemma for dimension $3$. A geodesic in a cusped finite-volume hyperbolic $3$-manifold ${\mathbb{H}}^{3}/\Gamma$ is called a cusp-to-cusp geodesic if it is the image of a geodesic in ${\mathbb{H}}^{3}$ connecting two cusp points under the action of $\Gamma$ on ${\mathbb{H}}^{3}$.

Before we prove 1.3 in this section, we make the following observations as well as restate that we drop the index $j$ from all notations. Let $\Sigma\subset M$ be a cusped totally geodesic surface that is not freely homotopic to $N$. Let $\widetilde{\Sigma}$ be a hyperplane lift of $\Sigma$ to ${\mathbb{H}}^{3}$. Since $\Sigma$ is a cusped surface, the line/circle at infinity of $\widetilde{\Sigma}$ contains a cusp point of $M$. If $\widetilde{\Sigma}$ is not a vertical hyperplane, we replace $\widetilde{\Sigma}$ by a $\Gamma$-translate that moves a cusp point of $\widetilde{\Sigma}$ to infinity. Without loss of generality, we assume that $\widetilde{\Sigma}$ is a vertical hyperplane.

Since $\Sigma$ is distinct from $N$, we can see from Figure 2 that the line at infinity of $\widetilde{\Sigma}$ must intersect the line at infinity of either ${\mathbb{H}}^{2}_{\mathbb{R}}$ or $w^{-1}({\mathbb{H}}^{2}_{\mathbb{R}})$ transversely. Therefore, $\widetilde{\Sigma}$ intersects some vertical lift of $N$ along a geodesic $(P,\infty)$ for some $P\in{\mathbb{C}}$. Since $\infty$ is a cusp point of $M$, 3.3 implies that $(P,\infty)$ projects onto a cusp-to-cusp geodesic in $M$. In particular, $P$ is a cusp point of $M$.

Using the geodesic $(P,\infty)$, we construct parabolic elements in $\operatorname{Stab}_{\Gamma}(\widetilde{\Sigma})$ fixing each end point and analyze the trace of the product of these elements. Since $P$ is a cusp point of $M$, there exists $\gamma\in\Gamma$ such that $\gamma(\infty)=P$. Since $P$ and $\infty$ are cusp points of $\widetilde{\Sigma}$, $\operatorname{Stab}_{\Gamma}(\widetilde{\Sigma})$ contains the nontrivial elements $a^{p}\ell^{q}$ and $\gamma a^{m}\ell^{n}\gamma^{-1}$ fixing $\infty$ and $P$, respectively, where $\gcd(p,q)=\gcd(m,n)=1$. In other words, $p/q$ (resp. $m/n$) is the boundary slope of $\widetilde{\Sigma}$ at $\infty$ (resp. $P$).

The following is a critical preliminary calculation in our analysis. We note that $\operatorname{tr}(\rho(\pi_{1}\Sigma))$ must contain only real algebraic integers in ${\mathbb{Q}}(z)$. The condition that ${\mathbb{Q}}(z)$ does not contain any proper real subfield other than ${\mathbb{Q}}$ implies that $\operatorname{tr}(\rho(\pi_{1}\Sigma))\subseteq{\mathbb{Z}}$. Suppose that

$$\rho(\gamma)=\begin{pmatrix}\alpha&\beta\\ \delta&\eta\end{pmatrix}$$

Then, dropping the $\rho$ for convenience of notation, we must have

$$\operatorname{tr}(\gamma a^{m}\ell^{n}\gamma^{-1}a^{p}\ell^{q})=(-1)^{n+q+1}[-2+(m+n\tau)(p+q\tau)\delta^{2}]\in{\mathbb{Z}}$$

which is true if and only if

$$\delta^{2}(nq\tau^{2}+(mq+np)\tau+mp)\in{\mathbb{Z}}$$

(8)

We shall refer to this as the trace condition, which gives us the following:

• Proof.

  It follows from the previous discussion that $\Sigma$ has a vertical lift $\widetilde{\Sigma}$ to ${\mathbb{H}}^{3}$. The lift $\widetilde{\Sigma}$ must intersect either ${\mathbb{H}}^{2}_{\mathbb{R}}$ or $w^{-1}({\mathbb{H}}^{2}_{\mathbb{R}})$, whose respective slopes are $1/0$ and $-2/1$, along some vertical geodesic $\theta$. One end point of $\theta$ is $\infty$, so by 3.3, $\theta$ covers a cusp-to-cusp geodesic. This implies that the other end point of $\theta$ must also be a cusp point in ${\mathbb{H}}^{2}_{\mathbb{R}}$ or $w^{-1}_{j}({\mathbb{H}}^{2}_{\mathbb{R}})$. Therefore, either $\theta=(s/t,\infty)$ for $s/t\in{\mathbb{Q}}$ or $\theta=(w^{-1}(s/t),\infty)$ for $s/t\in{\mathbb{Q}}\cup\{\infty\}$.

  Suppose that the boundary slopes on $\widetilde{\Sigma}$ are $p/q$ at $\infty$ and $m/n$ at the other end point of $\theta$. As boundary slopes, neither $(m,n)$ nor $(p,q)$ is $(0,0)$. We have nontrivial elements $a^{p}\ell^{q}$ and $\gamma a^{m}\ell^{n}\gamma^{-1}$ in $\operatorname{Stab}_{\Gamma}(\widetilde{\Sigma})$ where $\gamma\in\Gamma$ has the property that $\gamma^{-1}(s/t)=\infty$ or $\gamma^{-1}w^{-1}(s/t)=\infty$. We consider these two cases separately, and use the general form

  $$\gamma=\begin{pmatrix}\alpha&\beta\\ \delta&\eta\end{pmatrix}$$

  Note that the end points of $\theta$ are distinct since $\theta$ is a geodesic in ${\mathbb{H}}^{3}$. Consequently, $\gamma$ cannot fix $\infty$, and so $\delta\neq 0$.

  Case 1:

  Suppose that $\widetilde{\Sigma}$ intersects ${\mathbb{H}}_{\mathbb{R}}^{2}$ nontrivially along a geodesic $\theta=(s/t,\infty)$ for some $s/t\in{\mathbb{Q}}$, as seen in Figure 3. In view of 2.7, we may choose $\gamma$ to be either

  $$\begin{pmatrix}t^{\prime}&s\\ -4s^{\prime}&t\end{pmatrix}w,\quad\begin{pmatrix}s-2t^{\prime}&t^{\prime}\\ t-2s^{\prime}&s^{\prime}\end{pmatrix}v^{-1},\text{ or }\begin{pmatrix}s&t^{\prime}\\ t&s^{\prime}\end{pmatrix}$$

  according to whether $s/t$ belongs to $[0]$, $[\frac{1}{2}]$ or $[\infty]$, respectively. It suffices to compute $\delta$ in these three circumstances by using the trace condition Equation 8.

  Since $v^{-1}(\infty)=\frac{1}{2}$, we see that $v^{-1}$ is of the form

  $$\pm\begin{pmatrix}1&*\\ 2&*\end{pmatrix}$$

  This implies that $\delta=t\in{\mathbb{Z}}\smallsetminus\{0\}$ in the last two situations. We show that this contradicts the assumption that $(p,q)\neq(0,0)\neq(m,n)$. It follows from (7) that the trace field is the same as the cusp field. That is, ${\mathbb{Q}}(z)={\mathbb{Q}}(\tau)$. Then

  $$[{\mathbb{Q}}(\tau):{\mathbb{Q}}]=[{\mathbb{Q}}(z):{\mathbb{Q}}]\geq 3,$$

  so the minimal polynomial of $\tau$ over $\mathbb{Q}$ has degree $\geq 3$. In particular, $\tau$ cannot satisfy the quadratic polynomial over ${\mathbb{Z}}$ given by the trace condition. Therefore, we have $nq=0$ and $mq+np=0$ since $\delta^{2}\neq 0$. Since $\widetilde{\Sigma}$ intersects ${\mathbb{H}}^{2}_{\mathbb{R}}$ nontrivially, the boundary slope at $\infty$ on $\widetilde{\Sigma}$ is not meridional; in other words, $q\neq 0$. We thus have $n=0$, which implies that $mq=0$, and so $m=0$. This contradicts the assumption that $(m,n)\neq(0,0)$.

  In the first situation, we get $\delta=t\mu z$ and so $\delta^{2}=-t^{2}z$. The trace condition becomes

  	$\displaystyle\delta^{2}(nq\tau^{2}+(mq+np)\tau+mp)$	$\displaystyle\in{\mathbb{Z}}$
  	$\displaystyle z[nq(-4z^{-1}+2)^{2}+(mq+np)(-4z^{-1}+2)+mp]$	$\displaystyle\in{\mathbb{Q}}$
  	$\displaystyle z[16nqz^{-2}-(16nq+4mq+4np)z^{-1}+4nq+2mq+2np+mp]$	$\displaystyle\in{\mathbb{Q}}$
  	$\displaystyle 16nqz^{-1}+(2n+m)(2q+p)z$	$\displaystyle\in{\mathbb{Q}}$
  	$\displaystyle 16nq+(2n+m)(2q+p)z^{2}$	$\displaystyle\in{\mathbb{Q}}z$

  Since the minimal polynomial of $z$ has degree at least $3$, satisfying the above is equivalent to satisfying

  $$16nq=(2n+m)(2q+p)=0$$

  Again, since $\widetilde{\Sigma}$ intersects ${\mathbb{H}}^{2}_{\mathbb{R}}$ nontrivially, we have $q\neq 0$, so $m/n=1/0$ and $p/q=-2/1$.

  

  Case 2:

  Suppose that $\widetilde{\Sigma}$ is parallel to ${\mathbb{H}}^{2}_{\mathbb{R}}$, or equivalently $\theta=(w^{-1}(s/t),\infty)$ for some $s/t\in{\mathbb{Q}}$. We may assume that $p/q=1/0$. In view of 2.7, we may choose $\gamma=\begin{pmatrix}\alpha&\beta\\ \delta&\eta\end{pmatrix}$ to be either

  $$w^{-1}\begin{pmatrix}-t^{\prime}&s\\ -4s^{\prime}&t\end{pmatrix}w,\text{ or }w^{-1}\begin{pmatrix}s-2t^{\prime}&t^{\prime}\\ t-2s^{\prime}&s^{\prime}\end{pmatrix}v^{-1},\text{ or }w^{-1}\begin{pmatrix}s&t^{\prime}\\ t&s^{\prime}\end{pmatrix}$$

  When $s/t\in[0]$, we get $\delta=sz$, so

  	$\displaystyle\delta^{2}(m+n\tau)$	$\displaystyle\in{\mathbb{Z}}$
  	$\displaystyle s^{2}z^{2}(m+n(-4z^{-1}+2))$	$\displaystyle\in{\mathbb{Z}}$

  This is satisfied if and only if $s^{2}n=s^{2}(2n+m)=0$ because the minimal polynomial of $z$ is at least cubic over ${\mathbb{Q}}$. If $s=0$, then $w^{-1}(0)=\infty$ since endpoints of $\theta$ must be distinct, therefore $s\neq 0$. When $s\neq 0$, we must have $m=n=0$, which is a contradiction.

  When $s/t\in[\frac{1}{2}]$ or $[\infty]$, we see that $\delta=-s\mu z$ and use an equivalent calculation to Case 1 to conclude that the boundary slopes are $p/q=1/0$ and $m/n=-2/1$.

  ∎