Hidden symmetries via hidden extensions

0.43 MB·English

by  Eric Chesebro

HIDDEN SYMMETRIES VIA HIDDEN EXTENSIONS
ERICCHESEBROANDJASONDEBLOIS
6
1 Abstract. Thispaperintroducesanewapproachtofindingknotsandlinks
0 withhiddensymmetriesusing“hiddenextensions”,aclassofhiddensymme-
2 triesdefinedhere. Weexhibitafamilyoftanglecomplementsintheballwhose
boundaries have symmetries with hidden extensions, then we further extend
p
thesetohiddensymmetriesofsomehyperboliclinkcomplements.
e
S

9
Ahidden symmetry ofamanifoldM isahomeomorphismoffinite-degreecovers
1
of M that does not descend to an automorphism of M. By deep work of Mar-

] gulis, hidden symmetries characterize the arithmetic manifolds among all locally
T
symmetric ones: a locally symmetric manifold is arithmetic if and only if it has
G
infinitely many “non-equivalent” hidden symmetries (see [13, Ch. 6]; cf. [9]).
. AmonghyperbolicknotcomplementsinS3 onlythatofthefigure-eightisarith-
h
t metic [10], and the only other knot complements known to possess hidden sym-
a
metriesarethetwo“dodecahedralknots”constructedbyAitchison–Rubinstein[1].
m
Whether there exist others has been an open question for over two decades [9,
[
Question 1]. Its answer has important consequences for commensurability classes

2 of knot complements, see [11] and [2].
v The partial answers that we know are all negative. Aside from the figure-eight,
6
there are no knots with hidden symmetries with at most fifteen crossings [6] and
2
notwo-bridgeknotswithhiddensymmetries[11]. Macasieb–Mattmanshowedthat
7
0 no hyperbolic (−2,3,n) pretzel knot, n ∈ Z, has hidden symmetries [8]. Hoffman
0 showed the dodecahedral knots are commensurable with no others [7].
. Hereweoffersomepositiveresultswithpotentialrelevancetothisquestion. Our
1
0 first main result exhibits hidden symmetries with the following curious feature.
5
Definition 0.1. For a manifold M (possibly with boundary) and a submanifold
1
S of M, a hidden extension of a self-homeomorphism φ of S is a hidden symmetry
:
v Φ: M → M of M, where p : M → M are connected, finite-sheeted covers for
1 2 i i
Xi i=1,2, that lifts φ on a component of p−1(S).
1
r We use a family {L } of two-component links constructed in previous work [3].
a n
Foreachn,L isassembledfromatangleS inB3,ncopiesofatangleT inS2×I,
n
and the mirror image S of S. Figure 1 depicts L , with light gray lines indicating
2
the spheres that divide it into copies of S and T. For n ∈ N and m ≥ 0, we will
also use a tangle T ⊂ L : the connected union of S with n copies of T. For
n m+n
instance, L contains a copy of T (which is pictured in Figure 2 below) and of T .
2 1 2
Upon numbering the endpoints of T as indicated in Figure 2, order-two even
n
permutations determine mutations: mapping classes of ∂(B3 − T ) induced by
n
180-degree rotations of the sphere obtained by filling the punctures.
Theorem 1.8. For n∈N, the mutation of ∂(B3−T ) determined by (13)(24) has
n
a hidden extension over a cover of B3−T and for any m∈N, taking T ⊂L ,
n n m+n
a hidden extension over a cover of S3−L .
m+n
1
2 ERICCHESEBROANDJASONDEBLOIS
S
T
Figure 1. The link L
2
In particular, this gives the first proof that the S3−L have hidden symme-
m+n
tries. Its heart is the fact that though (13)(24) does not extend over S3−L ,
m+n
it is represented by an isometry of the totally geodesic ∂(B3−T ) that is induced
n
by an isometry m(n) of H3 in the commensurator (see eg. [9, p. 274]) of the group
1
Γ uniformizingS3−L . Lemma1.6assertstheanalogousfactforthegroup
m+n m+n
∆ uniformizing B3−T , which implies the other assertion of Theorem 1.8.
n n
In Section 2 we attack the same problem on the same examples, but from a
differentdirection. Theideainthissectionistoproducehiddensymmetrieswithout
prior knowledge of an orbifold cover such as was used in Theorem 1.8. Instead
we leverage the decomposition of L into tangle complements, producing explicit
n
hiddenextensionsofthemutationovercoversoftheseandsolvingagluingproblem
to piece them together to produce a hidden symmetry of L . One nice byproduct
n
of this approach is an explicit description of the hidden symmetry. We show:
Theorem 2.9. For each n ∈ N there is an 11-sheeted cover N → B3 −T and
n n
a hidden extension Ψ: N → N of the mutation (13)(24) acting on S(n) −T .
n n n
Moreover, for each m∈N, Ψ extends to a hidden symmetry of an 11-sheeted cover
of S3−L that contains N .
m+n n
Given that we are motivated by hidden symmetries of knot complements, the
following question is natural:
Question. Is there a knot K in S3 and a hidden symmetry of S3 −K that is a
hidden extension of a symmetry of some surface in S3−K?
In fact as the referee has pointed out, one might ask this about the known
examples with hidden symmetries. While it seems unlikely that the figure-eight
knot complement has hidden extensions, given the classification of incompressible
surfaces there (see [12, §4.10]), we have no corresponding conjecture about the
dodecahedral knot complements. This would be interesting to know.
Another tantalizing possibility arises from the observation that each tangle T
n
also lies in many knots in S3 which are distinct from the 3 known examples of
knots with hidden symetries. If an analog of Theorem 2.9 could be proved for any
such knot it would give a new example whose complement has hidden symmetries.
We have ruled out many possibilities using a criterion given in [11, Corollary 2.2]:
√
a knot complement with hidden symmetries has cusp field Q(i) or Q( −3). This
conditioncanbeeasilycheckedwithwithSnapPy[5]andSnap(see[4]). Wesuspect
that there is a reason the T cannot lie in knots whose complements have hidden
n
symmetries, and intend to study this further.
We conclude the introduction with two related problems.
HIDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 3
2
S 2
3
3
4 4
1 1
T
S(0) S(1)
Figure 2. ThetangleT ⊂B3 canbedecomposedalongasphere
1
S(0) into a tangle S ⊂B3 and a tangle T ⊂S2×I.
Problem 1. Classify tangles in the ball with complements whose boundary has a
symmetry with hidden extension.
Problem 2. Given a tangle T in the ball B3, and a symmetry φ of ∂B3−T with
a hidden extension across a cover of B3−T, classify the links L containing T such
that the hidden extension of φ extends to a cover of S3−L.
1. Existence of hidden extensions
The goal of Section 1.1 is to describe the tangle complements B3 − T from
n
both the topological and geometric perspectives, by collecting relevant definitions
and results scattered throughout [3] and re-assembling them here in amore helpful
order. In this sub-section we merely summarize geometric details, referring the
interested reader to [3] for proofs. In Section 1.2 we prove Theorem 1.8.
1.1. The topology and geometry of T . The solid lines in Figure 2 describe a
n
two string tangle T ⊂B3. ∂B3 is shown as a dotted line labeled S(1). There is an
1
additional sphere S(0) shown in the figure. If we cut (B3,T ) along S(0) we obtain
1
a pair of tangles (B3,S) and (S2×I,T). Orienting I so that S(0) =S2×{0}, we
let ∂ T =T ∩S(0) and ∂ T =T ∩S(1).
− +
Let r : (S2 ×I,T) → (S2 ×I,T) be the reflection homeomorphism visible in
T
Figure 1, and let T be the subtangle of T that lies to the left of the fixed point set
0
of r . That is, T =T ∩S2×[0,1/2]. Reparametrizing the underlying interval, we
T 0
may also regard T as a tangle in S2×I.
0
Proposition 1.1 below, which combines parts of Propositions 2.7, 2.8, and 3.7 of
[3], introduces geometric models for the complements of the tangles S, T and T.
0
Thereandhenceforth,weworkwiththeupperhalfspacemodelforH3 andusethe
standardrepresentationofIsom(H3)asaZ extensionofPSL (C). Ifd∈PSL (C),
2 2 2
we write d for the matrix whose entries are the complex conjugates of the entries
of d. When we apply this operation to each element of a subgroup Γ < PSL (C)
2
we obtain a subgroup denoted by Γ.
For a Kleinian group Γ, we denote the convex core of H3/Γ as C(Γ). We will
use the term natural map as in [3] (see below Definition 3.1 there) to refer to the
restrictiontoC(Λ)oftheorbifoldcoveringmapH3/Λ→H3/Γ,forΛ<Γ. Because
the limit set of Γ contains that of Λ, the natural map takes C(Λ) into C(Γ).
4 ERICCHESEBROANDJASONDEBLOIS
The geometric models for B3−S and (S2×I)−T described in parts (1) and
0
(2) of Proposition 1.1 are hyperbolic 3-manifolds with totally geodesic boundary
produced by pairing certain faces of the right-angled ideal octahedron and cuboc-
tahedron, respectively, but they are described in the Proposition as convex cores
of the quotients of H3 by the groups generated by the face-pairing isometries. The
equivalence of these two forms of description is proved in Lemma 2.1 of [3].
Proposition 1.1. (1) For s = (cid:0) 1 0(cid:1) and t = (cid:0)2i2−i(cid:1), ∆ = (cid:104)s,t(cid:105) is a
−11 i 1−i . 0
Kleiniangroup,andthereisahomeomorphismf : M =B3−S →C(∆ ).
S S 0
(2) For f, g and h below, Γ = (cid:104)f,g,h(cid:105) is a Kleinian group, and there is a
T0 .
homeomorphism f : M =(S2×I)−T →C(Γ ).
T0 T0 0 T0
f = (cid:0) 1 0(cid:1) g = (cid:16)−1+i√21−2i√√2(cid:17) h = (cid:16) 2i√√2 −3−√i√2(cid:17)
−11 −2 3−i 2 −3+i 2 −3i 2
(3) For c = (cid:16)1i√2(cid:17), Γ = (cid:10)Γ ,c−2Γ c2(cid:11) is a Kleinian group, and there is
0 1 T T0 T0
.
a homeomorphism f : M =(S2×I)−T →C(Γ ) satisfying:
T T T
• composing the inclusion M →M with f yields f ; and
T0 T T T0
• for r as above, f ◦r ◦f−1 is induced by x(cid:55)→c−2x¯c2.
T T T T
(4) The intersection ∆ ∩Γ is a Fuchsian group Λ stabilizing the hyperplane
0 T
H = R×(0,∞) of H3. This is the intersection of the convex hulls of the
limit sets of ∆ and Γ , and the natural maps from H/Λ to C(∆ ) and
0 T 0
C(Γ ) map to totally geodesic boundary components.
T .
(5) The image of the natural map H/Λ → C(Γ ) is the image of ∂ M =
T − T
(S2 ×{0})−T under f . The same holds with each instance of T here
T
replaced by T .
0
For the homeomorphism j: (∂B3,∂S) → (S2 × {0},∂ T) such that
−
(B3,S)∪ (S2 ×I,T) ∼= (B3,T ), f ◦j ◦f−1: ∂C(∆ ) → C(Γ ) fac-
j 1 T S 0 T
tors through H/Λ as the composition of a natural map with the inverse of
another.
We now turn back to topology and give an inductive definition of the tangles
T , assembling (B3,T ) from a single copy of (B3,S) and n of (S2 × I,T) for
n n
each n ∈ N, using T as pictured in Figure 2 as the base case. Numbering the
1
points of (S(0),∂T ) and (S(1),∂T ) as shown in the figure, let (S2 ×{1},∂ T),
1 1 +
(S2×{0},∂ T), and (∂B3,∂S) inherit numberings from their inclusions to these
−
spheres. Note that the resulting numbering of (S2×∂I,∂T) is r -invariant.
T
Nowforn>1,assumefor1≤k <nthattanglesT ⊂B3withlabeledendpoints
k
aredefined,and,fork >1,inclusions(B3,T )(cid:44)→(B3,T )andι : (S2×I,T)→
k−1 k k
(B3,T ), such that:
k
• (B3,T )=(B3∪ι (S2×I),T ∪ι (T));
k k k−1 k
• ι preserves labels on ∂ T; and
k +
• the included image of B3 intersects i (S2 × I) in a sphere S(k), with
k
(S(k),S(k)∩T )=(∂B3,∂T )=ι (S2×{1},∂ T).
k k−1 k +
Define T ⊂ B3 as the quotient of the disjoint union (B3,T )(cid:116)(S2 ×I,T) by
n n
identifying ι (x,1) to (x,0) for each x ∈ S2; let the inclusion of (B3,T ) and
n−1 n−1
ι : (S2×I,T)→(B3,T )beinducedbytherespectiveinclusionsintothedisjoint
n n
union; and label the endpoints of T coherently with T ∩(S2×{1}) using ι . It is
n n
clear by construction that the inductive hypothesis applies to (B3,T ).
n
HIDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 5
Having topologically described the T , our next order of business is to give
n
geometric models for their complements; that is to describe hyperbolic manifolds
with totally geodesic boundary homeomorphic to the B3 −T . In parallel with
n
our topological description of T , these are assembled from copies of the geometric
n
models described in Proposition 1.1. To this end, we define:
Γ(j) =c−2(j−1)Γ c2(j−1) Λ(j) =c−2jΛc2j F(j) =c−2j(H)/Λ(j)
T T
Note for each j that C(Γ(j)) is isometric to C(Γ ), so it is just a copy of M , and
T T T
F(j) is isometric to H/Λ. Now with ∆ as in Proposition 1.1, for n≥1 let
0
(cid:68) (cid:69)
∆ = ∆ ,Γ(1),...,Γ(n) .
n 0 T T
(In[3],∆ isdenotedasΓ and∆ asΓ(n).) TheconsequenceofPropositions3.10
0 S n −
and 3.12 of [3] below shows that C(∆ ) is a geometric model for B3−T .
n n
Proposition 1.2. For each n ∈ N there is a homeomorphism f : B3 − T →
n n
C(∆ ). Moreover, the natural map C(∆ )→C(∆ ) is an isometric embedding,
n n−1 n
and there is another, ι : C(Γ )→C(∆ ) factoring through an isometry C(Γ )→
n T n T
C(Γ(i)), such that for n>1 the following diagrams commute.
T
B3−T fn−1 (cid:47)(cid:47)C(∆ ) M fT (cid:47)(cid:47)C(Γ )
n−1 n−1 T T
ιn ιn
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47) (cid:47)(cid:47)
B3−T C(∆ ) B3−T C(∆ )
n n n n
fn fn
.
This also holds for n=1, taking f =f : M →C(∆ ) at the top left.
0 S S 0
The natural map F(j) → f (S(j) −∂T ) is an isometry onto a totally geodesic
n j
surface in C(∆ ) when 0≤j ≤n; in particular, F(n) is isometric to ∂C(∆ ).
n n
Our final task in comprehensively describing the T is to translate the tangle
n
endpoint labeling to the geometric setting, yielding a labeling of cusps of F(n),
or equivalently, of parabolic conjugacy classes in Λ(n). We begin below by listing
representatives for the parabolic conjugacy classes in Λ as words in Γ and Γ .
S T
p =s−1 =f−1
1
p =stst−2 =fg−1f−1h−1g
2
p =(tst)s−1(tst)−1 =(h−1fg)−1g−1(h−1fg),
3
p =p p p−1
4 1 2 3
A calculation shows that
p =(10) p =(cid:0)−1 5 (cid:1) p =(cid:0)−1425(cid:1) p =(cid:0)29−45(cid:1)
1 11 2 0 −1 3 −9 16 4 20−31
From Lemma 2.4 of [3] we obtain the next proposition.
Proposition 1.3. For any n∈N, j ≤n, and k ∈{1,2,3,4}, the parabolic conju-
gacy class in Λ(j) which corresponds to the point labeled k in S(j) is represented by
p(j) =c−2jp c2j. Also Λ(j) is generated by any three of the p(j)’s.
k k k
Wefinishbygivingageometricmodelforthemutationwithahiddenextension.
The result below follows from Proposition 1.3 above and Lemma 5.5 of [3].
6 ERICCHESEBROANDJASONDEBLOIS
Lemma 1.4. Let m = (cid:0)−35(cid:1). For each n ≥ 0, m(n) =. c−2nm c2n normalizes
1 −23 1 1
Λ(n) and induces a cycle representation (13)(24) on the the four cusps of F(n),
where each cusp is numbered according to its corresponding parabolic isometry p(n).
j
1.2. The proof of existence. The key new tool we need to prove Theorem 1.8 is
a discrete group containing both ∆ , with finite index, and also the isometry m
n 1
that induces the mutation (13)(24). (The group G of [3, Lemma 6.2] plays
m+n
this role for the group Γ uniformizing S3−L , by [3, Prop. 6.3].) We will
m+n m+n
use this with a standard argument to show there is a hidden extension.
As in Definitions 6.1 of [3], let B be the open half-ball in the upper half-space
0
modelofH3 boundedbytheEuclideanhemisphereofunitradiuscenteredat0∈C
√
and, for k ∈N, let B be the Euclidean translate of B centered at k(−i 2), where
k 0
i is the imaginary unit. For complex numbers z and w, refer by zH+w to the
geodesic plane (zR+w)×(0,∞).
Definition 1.5. For an integer n ≥ 0, define Q to be the polyhedron of H3
n
bounded by H+i/2, iH, iH+1/2, and ∂B for k ∈{0,1,...,n}. Further define:
k
(1) f by first reflecting in iH and then in iH+1/2;
0
(2) b by first reflecting in H+i/2 and then in ∂B ; and
0 0
(3) for k ≥0, a by reflecting in iH+1/2 and then in ∂B .
k k
H+i/2
B
0
B
1
B
2
iH iH+1/2
Figure 3. Bounding hyperplanes for Q viewed from above.
2
As defined, we have
f0 =(1011) b0 =(0i 1i) a0 =(cid:0)01−−11(cid:1) a1 =(cid:16)−i1√2−11++i√i√22(cid:17).
In particular, we see that (cid:104)f ,a (cid:105)=PSL (Z) contains m .
0 0 2 1
Lemma 1.6. For each integer n ≥ 0, the orientation-preserving subgroup H of
n
the group generated by reflections in the faces of Q satisfies:
n
(1) H is a Kleinian group generated by {f ,b ,a ,...,a }.
n 0 0 0 n
(2) ∆ <H with finite index.
n 2n
HIDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 7
(3) The projection c−n(H) → H3/H factors through an isometric embedding
n
of H/PSL (Z) onto ∂C(H ).
2 n
Proof. Clearly f ∈ H , b ∈ H , and a ∈ H for 0 ≤ i ≤ n. Let r(cid:48) denote
0 n 0 n i n
the reflection across iH+1/2. It is not hard to see that Q ∪r(cid:48)(Q ) is a convex
√ n n
polyhedron in H3 with one face in each of H+i 2, iH, iH+1, ∂B , and ∂r(cid:48)(B )
k k
for k ∈{0,1,...,n}. The following facts can be explicitly verified:
√
• The face in H+i 2 meets those in iH and iH+1 at right angles, those
in ∂B and r(cid:48)(∂B ) at an angle of π/3, and no others. The product a b
0 0 0 0
rotates by π about an axis that bisects this face, preserving it.
• The face in iH meets each of those in ∂B at an angle of π/2, and none of
k
those in r(cid:48)(∂B ). The element f takes this face to the one in iH+1.
j 0
• The face in ∂B shares an edge with the face in r(cid:48)(∂B ) if and only if
k k(cid:48)
k = k(cid:48); in this case at an angle of 2π/3. The element a takes the latter
k
to the former. The faces in ∂B and ∂B meet at an angle of π/2 for
k k−1
k >0; likewise those in ∂B and ∂B for k <n; and ∂B ∩∂B =∅ for
k k+1 k k(cid:48)
k(cid:48) ∈/ {k−1,k,k+1}.
Hence, {f ,a b ,a ,...,a } is a face-pairing for Q ∪r(cid:48)(Q ). Poincare’s polyhe-
0 0 0 0 n n n
dron theorem implies that this set of isometries generates a discrete group whose
fundamental domain is Q ∪r(cid:48)(Q ). By construction, this group is contained in
n n
H . It is equal to H because their fundamental domains have the same volume.
n n
The numbered formulas (8) and (9) above Proposition 6.3 of [3] express the
generators of ∆ in terms of a , b , and f and they express Γ in terms of a , a ,
0 0 0 0 T0 0 1
and f . Therefore, ∆ < H and Γ < H . It can be verified directly that, for
0 0 n T0 n
everyk,c−1a c=a andthatccommuteswithf . Itfollowsthatc−kΓ ck <H
k k+1 0 T0 n
for all k ≤ n−1. Moreover, the second paragraph of the proof of [3, Proposition
6.3] expresses the generators of c−2Γ c2 in terms of a , a , and f . So, if n ≥ 2,
T0 1 2 0
this group is also in H . Now, by definition, we have ∆ <H .
n n 2n
The polyhedron P of [3, Lemma 6.2] consists of points (z,t) ∈ Q such that
n √ n
the imaginary coordinate of z is at least −n 2. Every face of P is a face of Q
√ n n
except the unique face F of P contained in H−n·i 2. The face F is orthogonal
n
to ∂B , iH, and iH+1/2 and does not meet any other bounding hyperplanes of
n
Q . This means that a single face of P ∪r(cid:48)(P ) contains F, meeting only the
n n n
bounding hyperplanes iH, iH+1, ∂B , and r(cid:48)(∂B ). Moreover, these intersections
n n
are all orthogonal, so F projects to the sole totally geodesic boundary component
of the orbifold (P ∪r(cid:48)(P ))/H .
n n n
We claim that (P ∪ r(cid:48)(P ))/H = C(H ). We first show that P ∪ r(cid:48)(P )
n n n n n n
is contained in the convex hull of the limit set of H , which implies that (P ∪
n n
r(cid:48)(P ))/H is contained in the convex core. Inspecting Figures 3 and 4 in [3], one
n n
observesthatP ∪r(cid:48)(P )iscontainedintheunionP ∪(cid:83)n−1c−k(P ),whereP and
n n 1 k=0 2 1
P aretheregularidealoctahedronandrightangledidealcuboctahedrondescribed
2
in Corollaries 2.2 and 2.3 of [3]. Both P and P are the convex hulls of their ideal
1 2
points, andeachoftheseisaparabolicfixedpointof∆ orΓ , respectively. (One
0 T0
canshowthisdirectly,orappealtothethird-from-lastparagraphoftheproofof[3,
Lemma 2.1].) Since each parabolic fixed point of a Kleinian group lies in its limit
set, it follows that P ∪(cid:83)n−1c−k(P ) is in the convex hull of the limit set of H .
1 k=0 2 n
As a subset, P ∪r(cid:48)(P ) shares this property. On the other hand, the penultimate
n n
8 ERICCHESEBROANDJASONDEBLOIS
paragraph of [3, proof of Lemma 2.1] shows that (P ∪r(cid:48)(P ))/H contains C(H )
n n n n
and this proves our claim.
By the above, c−2n(H) projects to ∂C(H ) under the quotient map H3 →
2n
H3/H . By Proposition 1.2, the same plane projects to ∂C(∆ ) under H3 →
2n n
H3/∆ . It follows that the orbifold covering map H3/∆ → H3/H restricts to
n n 2n
oneC(∆ )→C(H ). Sincethesebothhavefinitevolume,themapisfinite-to-one,
n 2n
and hence ∆ has finite index in H .
n 2n
Among all bounding hyperplanes of Q ∪r(cid:48)(Q ), only iH, iH +1, ∂B , and
√ n n n
r(cid:48)(∂B ) meet the hyperplane H−n·i 2. Each of these intersections is a right
n
angle. Thus,F isaquadrilateraland{a ,f }isanedgepairingforF. Thisimplies
n 0
that F/(cid:104)a ,f (cid:105) is the boundary of (P ∪r(cid:48)(P ))/H .
n 0 n n n
Wementionedabovethatc−na cn =a andf c=cf ,so(cid:104)a ,f (cid:105)=c−n(cid:104)a ,f (cid:105)cn.
0 √ n 0 0 n 0 0 0
Therefore, the projection H − n · i 2 = c−n(H) → H3/H factors through an
n
isometric embedding of H/PSL (Z). (cid:3)
2
We will use the following simple fact below, and several more times.
Fact 1.7. If H has finite index in a non-elementary Kleinian group G then the
limit sets of G and H are equal, so the natural map C(H) → C(G) is an orbifold
cover.
Theorem 1.8. For n∈N, the mutation of ∂(B3−T ) determined by (13)(24) has
n
a hidden extension over a cover of B3−T and for any m∈N, taking T ⊂L ,
n n m+n
a hidden extension over a cover of S3−L .
m+n
Proof. Aswementionedintheintroductiontothispaper,wemayview(B3,T )as
n
a subset of (S3,L ). (For more rigor, compare the definitions at the beginning
n+m
of this section with [3, Definitions 3.8].) Here it is bounded by the sphere S(n),
with the mirror image (B3,T ) of T on the other side. If the mutation (13)(24)
m m
extended over B3 −T , then S3 −L would be homeomorphic to its mutant
n n+m
by (13)(24) along S(n). By Mostow-Prasad rigidity, these two links would be
isometric, but by Theorem 2 of [3] they are not. (In the notation of that result,
L = L and its mutant is L with the sole “1” the (m+1)th
m+n (0,...,0) (0,...,1,...,0)
entry.) This also implies it does not extend over S3−L .
m+n
However, because m(n) lies in the finite extension H of ∆ it normalizes the
1 2n n
normalcoreΩ of∆ inH anddeterminesaself-isometryΨ˜ ofH3/Ω . Thisisa
n n 2n n
finitecoverofH3/∆ whichbyFact1.7aboverestrictstoacoverC(Ω )→C(∆ ).
n n n
In particular, the boundary of C(Ω ) is totally geodesic. One component of
n
∂C(Ω ) is the quotient of c−2n(H) by its stabilizer Λ˜(n) =Ω ∩Λ(n) in Ω . Since
n n n
m(n) normalizes both Λ(n) and Ω , it normalizes Λ˜(n) and determines an isometry
1 n
of c−2n(H)/Λ˜(n) lifting the one determined by m(n) on ∂C(∆ ).
1 n
A completely analogous argument applies to S3 −L , replacing ∆ by Γ
m+n n n
from Prop. 3.12 of [3] and H by G from Prop. 6.3 there. (cid:3)
2n n
2. Matching covers
Inthissection,webuildanexplicithiddenextensionofthemutation(13)(24)of
∂(B3−T ). Tofindanappropriatecover,weusethedecompositionofB3−T along
n n
thespheresS(j)intoonecopyofM andncopiesofM andfindappropriatecovers
S T
of these pieces which glue together to give a cover of B3−T with the necessary
n
properties. Figure 4 is a schematic depiction of how this will be done.
HIDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 9
M(cid:102)S M(cid:102)T(1) M(cid:102)(2)
T
C(Ω )
T0
−J→ −R→T
R
←→T
−→j −r→T
M M(1) M(2)
S T T
Figure 4. Assembling a cover of (S2×I)−T that has a hidden
2
extension of the mutation (13)(24).
For convenience, in this section we will supress the homeomorphisms f , f ,
S T0
and f of Proposition 1.1 and simply make the identifications:
T
M =C(∆ ) M =C(Γ ) M =C(Γ ).
S 0 T0 T0 T T
Further, for n∈N and 1≤i≤n we will identify the ith copy M(i) of (S2×I,T)
T
in (B3,T ) with C(Γ(i)) (compare Proposition 1.2).
n T
To produce the covers M(cid:102)S and the M(cid:102)T(i) of Figure 4 we will divide the orbifold
O = H3/H branched-covered by B3 −T into pieces covered by M and the
n n n S
M(i), then analyze the corresponding subgroups of H . For M we use Q from
T n S 0
Definition 1.5: the polyhedron bounded by H+i/2, iH, iH+1/2, and ∂B . Our
0
first lemma shows that the orientation-preserving subgroup H of the reflection
0
group generated by Q contains the group ∆ =Γ uniformizing M . Our second
0 0 S S
uses H and some elementary number theory to find a cover of M with abundant
0 S
symmetry.
We follow a similar strategy for M , producing a polyhedron P and a group
T T0
H <H , which Lemma 2.3 shows contains the group Γ uniformizing M . We
T0 n T0 T0
willusethepermutationrepresentationofH givenbyactingonleftcosetsofΓ
T0 T0
to find a cover of M with a hidden extension of (13)(24). Doubling this cover
T0
across a boundary component yields the model M(cid:102)T for the M(cid:102)T(i).
Lemma 2.1. The reflection group H (recall Lemma 1.6) has the following addi-
0
tional properties:
(1) H is a Kleinian group which contains ∆ as a subgroup of index 12.
0 0
(2) H =(cid:104)a ,b ,f |a3 =b3 =(b−1a )2 =(a f )2 =1(cid:105).
0 0 0 0 0 0 0 0 0 0
(3) PSL (Z)=Stab (H).
2 H0
10 ERICCHESEBROANDJASONDEBLOIS
(4) The projection H → H3/H factors through an isometric embedding of
0
H/PSL (Z) onto ∂C(H ).
2 0
Proof. As in Lemma 1.6, r(cid:48) is the reflection through iH+1/2. From the proof of
Lemma 1.6, we know that the collection {a ,b a ,f } is a face-pairing for Q ∪
0 0 0 0 0
r(cid:48)(Q ). So the Poincar´e polyhedron theorem gives the presentation above. Aside
0
fromthefactthat[H ,∆ ]=12,therestofthelemmafollowsfromthespecialcase
0 0
n=0 in Lemma 1.6. But, if we compare the volume of a regular ideal octahedron
in H3 with the volume of P , we see that [H ,∆ ]=12. (cid:3)
0 0 0
Lemma 2.2. There is an index five subgroup Ω < ∆ which is normal in H .
0 0 0
Define Λ =Ω ∩Λ.
0 0
(1) [Λ:Λ ]=5,
0
(2) p ,p ∈Λ , and
2 4 0
(3) p and p project to generators of Λ/Λ .
1 3 0
Proof. In the ring of Gaussian integers 5 = (1+2i)(1−2i). So, restricting the
map Z[i] → Z[i]/(1+2i) to Z gives a ring epimorphism Z → Z[i]/(1+2i). The
quotient ring Z[i]/(1+2i) is isomorphic to Z/5Z and we obtain a group epimor-
phism PSL (Z[i]) → PSL (Z/5Z) which restricts to an epimorphism PSL (Z) →
2 2 2
PSL (Z/5Z). Since PSL (Z)<H <PSL (Z[i]), the restriction to H is also onto
2 2 0 2 0
and the kernel Ω of this map has index |PSL (Z/5Z)|=60 in H .
0 2 0
UsingtheexplicitdescriptionsofsandtfromSection1,weseethat∆ mapsonto
0
the parabolic subgroup {(10)} of PSL (Z[i]/(1+2i)) which has order 5. Hence,
∗1 2
∆ ∩Ω hasindexfivein∆ . Since[H :∆ ]=12, itfollowsthat[H :∆ ∩Ω ]=
0 0 0 0 0 0 0 0
60. Therefore, ∆ contains Ω .
0 0
Similarly, the explicit descriptions of the p ’s from Section 1 show that Λ maps
j
onto this same parabolic subgroup and [Λ : Λ ] = 5. The final assertion is also
0
immediate from these descriptions. (cid:3)
Lemma 2.3. Let P be the polyhedron bounded by ∂B , ∂B , iH and iH+1/2.
T0 0 1
The orientation-preserving subgroup H of the group generated by reflections in
T0
the sides of P is a Kleinian group such that
T0
(1) H =(cid:104)a ,a ,f |a3 =a3 =1,(a a−1)2 =(a f )2 =(a f )2 =1(cid:105),
T0 0 1 0 0 1 0 1 0 0 1 0
(2) ∂C(H ) consists of a pair of totally geodesic surfaces,
T0
(3) PSL (Z) and Γ are subgroups of H , and
2 T0 T0
(4) [H :Γ ]=[Stab (H):Λ]=12.
T0 T0 HT0
Proof. First,recallfromjustbeforeLemma1.6andintheproofofLemma1.6that
Claim (3) has already been established.
Now,forvisualintuition,compareP withtheright-angledidealcuboctahedron
T0
P . The intersection P ∩P is the portion of P which lies between H and H−
√2 T0 2 T0 √
i 2. TheintersectionP ∩P hastwoadditionalfacescontainedinHandH−i 2
T0 2
and a single ideal vertex at ∞. These additional faces are either perpendicular to
or disjoint from those of P .
T0
Let F be the face of P ∩ P contained in iH, A its face in ∂B , and A
T0 2 0 0 1
the face in ∂B . Let r(cid:48) denote reflection across iH+1/2. A fundamental domain
1
for H is the union P ∪r(cid:48)(P ). By construction, a ,a ,f ∈ H and these
T0 T0 T0 0 1 0 T0
isometries determine a face-pairing for this fundamental domain. In particular,
a (r(cid:48)(A )) = A , a (r(cid:48)(A )) = A , and f (F) = r(cid:48)(F). Thus, H is generated by
0 0 0 1 1 1 0 T0
a , a , and f . Upon noting that, for each i ∈ {0,1}, A intersects F and A
0 1 0 i 1−i