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 is an orientable hyperbolic 3–orbifold of finite volume, then has finitely many ends (i.e. cusp neighborhoods) and each end is isometric to , where is some quotient of the 2–torus . Moreover, the stabilizer subgroup of each cusp , generated by a pair of parabolics.

Before the correction, I had written that the cusp neighborhoods are isometric to . As the proof of the above Lemma implies, this is true if and only if the covering group does not contain torsion elements that fix the point at . 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 , 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 ). 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 is a Bianchi group. Then the number of cusps on is equal to the class number of , so there is always at least one cusp. In what follows we will show that the cusp cross sections of are tori unless . Moreover, unless , all cusp cross sections are non–rigid.

Lemma 2: If has finite order, then is an elliptic Möbius transformation.

Proof: Suppose for some positive integer . Observe that is not conjugate to a matrix of the form for some complex number since otherwise would be parabolic, hence have infinite order. So has two distinct fixed points on , therefore is conjugate to a marix of the form . It is an elementary group–theoretic fact that conjugates of torsion elements are also torsion, which implies that . It follows that is a root of unity, so is real and is less than . Since the square trace map is invariant under conjugation. and the trace is invariant up to multiplication by , we conclude that is elliptic.

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

Lemma 3: contains a non-real root of unity if and only if .

Proof: The “if” direction is clear given the element and in . Conversely, suppose contains an –root of unity . Then contains the subfield , which is Galois of degree . Recall the fact that implies . In particular, one has for all primes dividing . It follows that (otherwise contains a subfield of strictly larger degree), and so . Now note if , then . We conclude that . When , then ; and when , . Any imaginary quadratic field containing (resp. or ) must also contain (resp. ), hence be equal to (resp. ) by comparing degrees.

Lemma 4: contains a non–identity element fixing if and only if .

Proof: Let and suppose is an element fixing . Observe that we can identify the point with in for any . Consider . It is easy to see that if and only if , so . Then since , hence is a unit. By Dirichlet’s unit Theorem, is necessarily a root of unity; and since by hypothesis, Lemma~3 implies . For future reference we note that these elements are and , respectively, where .

Theorem 5: The cusp cross sections of the Bianchi orbifold are tori unless . When (resp. ), the cusp cross section is a pillowcase (resp. ).

Proof: When , Lemma~4 shows that there are no torsion elements in that fix the cusp cross sections, hence each such cross section is a torus by Lemma~1. When , the unique tosion element fixing the cusp at is , which acts as on . In particular the action of on induces a degree two quotient map . So is a pillowcase, which has orbifold structure . This last statement can be seen directly by noting that stabilizes the group of –roots of unity in , and the successive pairwise dihedral angles of the corresponding lines in have cone angle . If , then fixes , where . Note that has order and that it permutes the third roots of unity in . Passing to we see that permutes the lines , which clearly have successive pairwise cone–angle . It follows that the cusp cross section in is topologically with three singular points of order , i.e .