# Cusps on Bianchi Orbifolds II

I need to clarify a minor mistake (now corrected) that was made out of haste in my previous post. Thankfully this leads nicely into a discussion that I was planning on writing about anyways. Recall the following result from part I (Lemma 8):

Lemma 1: If ${M}$ is an orientable hyperbolic 3–orbifold of finite volume, then ${M}$ has finitely many ends (i.e. cusp neighborhoods) and each end is isometric to ${\widetilde{T}\times [0,\infty)}$, where ${\widetilde{T}}$ is some quotient of the 2–torus ${T^2}$. Moreover, the stabilizer subgroup of each cusp ${{\mathbb Z}\oplus {\mathbb Z}}$, generated by a pair of parabolics.

Before the correction, I had written that the cusp neighborhoods are isometric to ${T^2\times [0,\infty)}$. As the proof of the above Lemma implies, this is true if and only if the covering group ${\Gamma}$ does not contain torsion elements that fix the point at ${\infty}$. The existence of such elements introduces interesting delicacies that are relevant to some of the problems I have been working on. In general, if there are elliptic elements that fix a cusp at ${\infty}$, then the cusp is called rigid. This means that it cannot be deformed; and in particular, Dehn surgery cannot be performed (unless the cusp is ${S^2(2,2,2,2)}$). We will explain and investigate these last statements in a later post when we discuss arithmetic knot complements (Cusps on Bianchi Orbifolds III).

Assume from here-on-out that ${\Gamma = \Gamma_d = \mathrm{PSL}_2({\cal O}_d)}$ is a Bianchi group. Then the number of cusps on ${\mathop{\mathbb H}^3/\Gamma_d}$ is equal to the class number of ${K_d}$, so there is always at least one cusp. In what follows we will show that the cusp cross sections of ${\mathop{\mathbb H}^3/\Gamma_d}$ are tori unless ${d \in \{1,3\}}$. Moreover, unless ${d = 3}$, all cusp cross sections are non–rigid.

Lemma 2: If ${\gamma \in \mathrm{PSL}_2({\mathbb C})}$ has finite order, then ${\gamma}$ is an elliptic Möbius transformation.

Proof: Suppose ${\gamma^k = \left[\begin{smallmatrix} 1 & \\ & 1 \end{smallmatrix}\right]}$ for some positive integer ${k}$. Observe that ${\gamma}$ is not conjugate to a matrix of the form ${\left[\begin{smallmatrix} 1 & \alpha \\ & 1 \end{smallmatrix}\right]}$ for some complex number ${\alpha}$ since otherwise ${\gamma}$ would be parabolic, hence have infinite order. So ${\gamma}$ has two distinct fixed points on ${{\mathbb C}}$, therefore is conjugate to a marix of the form ${\left[\begin{smallmatrix} \lambda & \\ & \lambda^{-1} \end{smallmatrix}\right]}$. It is an elementary group–theoretic fact that conjugates of torsion elements are also torsion, which implies that ${\lambda^k = 1}$. It follows that ${\lambda}$ is a root of unity, so ${\lambda + \lambda^{-1}}$ is real and ${|\lambda + \lambda^{-1}|}$ is less than ${2}$. Since the square trace map is invariant under conjugation. and the trace is invariant up to multiplication by ${-1}$, we conclude that ${\gamma}$ is elliptic. $\Box$

The Dirichlet Unit Theorem implies that the group of units in any imaginary quadratic field has rank ${0}$. Hence any unit in ${{\cal O}_d}$ is a root of unity. It turns out that the existence of such elements is rare, as the following Lemma shows.

Lemma 3: ${{\cal O}_d}$ contains a non-real root of unity if and only if ${d \in \{1,3\}}$.

Proof: The “if” direction is clear given the element ${i \in {\cal O}_1}$ and ${\zeta_3 = \frac{1+\sqrt{-3}}{2}}$ in ${{\cal O}_3}$. Conversely, suppose ${{\cal O}_d}$ contains an ${n^{th}}$–root of unity ${\zeta_n\neq \pm 1}$. Then ${K_d}$ contains the subfield ${{\mathbb Q}(\zeta_n)}$, which is Galois of degree ${\phi(n)}$. Recall the fact that ${b\mid a}$ implies ${\phi(b)\mid \phi(a)}$. In particular, one has ${p-1\mid \phi(n)}$ for all primes ${p}$ dividing ${n}$. It follows that ${p-1 \leq 3}$ (otherwise ${K_d}$ contains a subfield of strictly larger degree), and so ${n = 2^{e_1}3^{e_2}}$. Now note if ${n > 6}$, then ${\phi(n) > 2}$. We conclude that ${n \in \{2,3,4,6\}}$. When ${n \in \{2,4\}}$, then ${\zeta_n = \pm 4}$; and when ${n\in \{3,6\}}$, ${\zeta_n \in \{\frac{\pm 1 \pm \sqrt{-3}}{2}\}}$. Any imaginary quadratic field containing ${\pm i}$ (resp. ${\pm \zeta_3}$ or ${\pm \overline{\zeta_3}}$) must also contain ${K_1}$ (resp. ${K_3}$), hence be equal to ${K_1}$ (resp. ${K_3}$) by comparing degrees. $\Box$

Lemma 4: ${\Gamma_d}$ contains a non–identity element fixing ${\infty}$ if and only if ${d \in \{1,3\}}$.

Proof: Let ${\gamma = \left[\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right] \in \mathrm{PSL}_2({\cal O}_d)}$ and suppose ${\gamma}$ is an element fixing ${\infty}$. Observe that we can identify the point ${\infty}$ with ${\frac{z}{0}}$ in ${\mathop{\mathbb P} {\mathbb C}}$ for any ${z\in {\mathbb C}}$. Consider ${\gamma \cdot \infty = \frac{a \frac{z}{0} + b}{c \frac{z}{0} + d}}$. It is easy to see that ${\gamma \cdot \frac{z}{0} = \frac{z'}{0}}$ if and only if ${b = c = 0}$, so ${\gamma = \left[\begin{smallmatrix} a & 0 \\ 0 & d \end{smallmatrix}\right]}$. Then ${d = a^{-1}}$ since ${\mathrm{det }\gamma = 1}$, hence ${a \in {\cal O}_d}$ is a unit. By Dirichlet’s unit Theorem, ${a}$ is necessarily a root of unity; and since ${\gamma \neq \left[\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix}\right]}$ by hypothesis, Lemma~3 implies ${d \in \{1,3\}}$. For future reference we note that these elements are ${\left[\begin{smallmatrix} i & 0 \\ 0 & -i \end{smallmatrix}\right]}$ and ${\left[\begin{smallmatrix} \omega & 0 \\ 0 & \omega^2 \end{smallmatrix}\right]}$, respectively, where ${\omega = \frac{-1 + \sqrt{-3}}{2}}$. $\Box$

Theorem 5: The cusp cross sections of the Bianchi orbifold ${\mathop{\mathbb H}^3/\Gamma_d}$ are tori unless ${d\in \{1,3\}}$. When ${d = 1}$ (resp. ${d = 3}$), the cusp cross section is a pillowcase (resp. ${S^3(3,3,3)}$).

Proof: When ${d \not \in \{1,3\}}$, Lemma~4 shows that there are no torsion elements in ${\Gamma_d}$ that fix the cusp cross sections, hence each such cross section is a torus by Lemma~1. When ${d = 1}$, the unique tosion element fixing the cusp at ${\infty}$ is ${\gamma = \left[\begin{smallmatrix} i & 0 \\ 0 & -i \end{smallmatrix}\right]}$, which acts as ${-1}$ on ${H_1(T^2,{\mathbb Z})}$. In particular the action of ${\gamma}$ on ${T^2}$ induces a degree two quotient map ${g: T^2 \rightarrow P = T^2/\langle \gamma \rangle}$. So ${P}$ is a pillowcase, which has orbifold structure ${S^2(2,2,2,2)}$. This last statement can be seen directly by noting that ${\gamma}$ stabilizes the group of ${4^{th}}$–roots of unity in ${{\mathbb C}}$, and the successive pairwise dihedral angles of the corresponding lines in ${\mathop{\mathbb P} {\mathbb C}}$ have cone angle ${\frac{\pi}{2}}$. If ${d = 3}$, then ${\gamma = \left[\begin{smallmatrix} \omega & 0 \\ 0 & \omega^2 \end{smallmatrix}\right]}$ fixes ${\infty}$, where ${\omega = \frac{-1 + \sqrt{-3}}{2}}$. Note that ${\gamma}$ has order ${3}$ and that it permutes the third roots of unity in ${{\mathbb C}}$. Passing to ${\mathop{\mathbb P} {\mathbb C}}$ we see that ${\gamma}$ permutes the lines ${\{{\mathbb R} e^{\frac{i\pi}{3}}, {\mathbb R} e^{\frac{i\pi}{3}}, {\mathbb R} e^{\frac{i\pi}{3}}}$, which clearly have successive pairwise cone–angle ${\frac{\pi}{3}}$. It follows that the cusp cross section in ${\mathop{\mathbb H}^3/\Gamma_d}$ is topologically ${S^2}$ with three singular points of order ${3}$, i.e ${S^2(3,3,3)}$. $\Box$