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
.