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

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