HHIDDEN SYMMETRIES VIA HIDDEN EXTENSIONS
ERIC CHESEBRO AND JASON DEBLOIS
Abstract.
This paper introduces a new approach to finding knots and linkswith hidden symmetries using “hidden extensions”, a class of hidden symme-tries defined here. We exhibit a family of tangle complements in the ball whoseboundaries have symmetries with hidden extensions, then we further extendthese to hidden symmetries of some hyperbolic link complements. A hidden symmetry of a manifold M is a homeomorphism of finite-degree coversof M that does not descend to an automorphism of M . By deep work of Mar-gulis, hidden symmetries characterize the arithmetic manifolds among all locallysymmetric ones: a locally symmetric manifold is arithmetic if and only if it hasinfinitely many “non-equivalent” hidden symmetries (see [13, Ch. 6]; cf. [9]).Among hyperbolic knot complements in S only that of the figure-eight is arith-metic [10], and the only other knot complements known to possess hidden sym-metries are the two “dodecahedral knots” constructed by Aitchison–Rubinstein [1].Whether there exist others has been an open question for over two decades [9,Question 1]. Its answer has important consequences for commensurability classesof knot complements, see [11] and [2].The partial answers that we know are all negative. Aside from the figure-eight,there are no knots with hidden symmetries with at most fifteen crossings [6] andno two-bridge knots with hidden symmetries [11]. Macasieb–Mattman showed thatno hyperbolic ( − , , n ) pretzel knot, n ∈ Z , has hidden symmetries [8]. Hoffmanshowed the dodecahedral knots are commensurable with no others [7].Here we offer some positive results with potential relevance to this question. Ourfirst main result exhibits hidden symmetries with the following curious feature. Definition 0.1.
For a manifold M (possibly with boundary) and a submanifold S of M , a hidden extension of a self-homeomorphism φ of S is a hidden symmetryΦ : M → M of M , where p i : M i → M are connected, finite-sheeted covers for i = 1 ,
2, that lifts φ on a component of p − ( S ).We use a family { L n } of two-component links constructed in previous work [3].For each n , L n is assembled from a tangle S in B , n copies of a tangle T in S × I ,and the mirror image S of S . Figure 1 depicts L , with light gray lines indicatingthe spheres that divide it into copies of S and T . For n ∈ N and m ≥
0, we willalso use a tangle T n ⊂ L m + n : the connected union of S with n copies of T . Forinstance, L contains a copy of T (which is pictured in Figure 2 below) and of T .Upon numbering the endpoints of T n as indicated in Figure 2, order-two evenpermutations determine mutations : mapping classes of ∂ ( B − T n ) induced by180-degree rotations of the sphere obtained by filling the punctures. Theorem 1.8.
For n ∈ N , the mutation of ∂ ( B − T n ) determined by (1 3)(2 4) hasa hidden extension over a cover of B − T n and for any m ∈ N , taking T n ⊂ L m + n ,a hidden extension over a cover of S − L m + n . a r X i v : . [ m a t h . G T ] S e p ERIC CHESEBRO AND JASON DEBLOIS TS Figure 1.
The link L In particular, this gives the first proof that the S − L m + n have hidden symme-tries. Its heart is the fact that though (1 3)(2 4) does not extend over S − L m + n ,it is represented by an isometry of the totally geodesic ∂ ( B − T n ) that is inducedby an isometry m ( n )1 of H in the commensurator (see eg. [9, p. 274]) of the groupΓ m + n uniformizing S − L m + n . Lemma 1.6 asserts the analogous fact for the group∆ n uniformizing B − T n , which implies the other assertion of Theorem 1.8.In Section 2 we attack the same problem on the same examples, but from adifferent direction. The idea in this section is to produce hidden symmetries withoutprior knowledge of an orbifold cover such as was used in Theorem 1.8. Insteadwe leverage the decomposition of L n into tangle complements, producing explicithidden extensions of the mutation over covers of these and solving a gluing problemto piece them together to produce a hidden symmetry of L n . One nice byproductof this approach is an explicit description of the hidden symmetry. We show: Theorem 2.9.
For each n ∈ N there is an -sheeted cover N n → B − T n anda hidden extension Ψ : N n → N n of the mutation (1 3)(2 4) acting on S ( n ) − T n .Moreover, for each m ∈ N , Ψ extends to a hidden symmetry of an -sheeted coverof S − L m + n that contains N n . Given that we are motivated by hidden symmetries of knot complements, thefollowing question is natural:
Question.
Is there a knot K in S and a hidden symmetry of S − K that is ahidden extension of a symmetry of some surface in S − K ? In fact as the referee has pointed out, one might ask this about the knownexamples with hidden symmetries. While it seems unlikely that the figure-eightknot complement has hidden extensions, given the classification of incompressiblesurfaces there (see [12, § T n also lies in many knots in S which are distinct from the 3 known examples ofknots with hidden symetries. If an analog of Theorem 2.9 could be proved for anysuch 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 ( √− T n cannot lie in knots whose complements have hiddensymmetries, and intend to study this further.We conclude the introduction with two related problems. IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 3 S (0) S (1) S T
Figure 2.
The tangle T ⊂ B can be decomposed along a sphere S (0) into a tangle S ⊂ B and a tangle T ⊂ S × I . Problem 1.
Classify tangles in the ball with complements whose boundary has asymmetry with hidden extension.
Problem 2.
Given a tangle T in the ball B , and a symmetry φ of ∂B − T witha hidden extension across a cover of B − T , classify the links L containing T suchthat the hidden extension of φ extends to a cover of S − L .1. Existence of hidden extensions
The goal of Section 1.1 is to describe the tangle complements B − T n fromboth the topological and geometric perspectives, by collecting relevant definitionsand results scattered throughout [3] and re-assembling them here in a more helpfulorder. In this sub-section we merely summarize geometric details, referring theinterested reader to [3] for proofs. In Section 1.2 we prove Theorem 1.8.1.1. The topology and geometry of T n . The solid lines in Figure 2 describe atwo string tangle T ⊂ B . ∂B is shown as a dotted line labeled S (1) . There is anadditional sphere S (0) shown in the figure. If we cut ( B , T ) along S (0) we obtaina pair of tangles ( B , S ) and ( S × I, T ). Orienting I so that S (0) = S × { } , welet ∂ − T = T ∩ S (0) and ∂ + T = T ∩ S (1) .Let r T : ( S × I, T ) → ( S × I, T ) be the reflection homeomorphism visible inFigure 1, and let T be the subtangle of T that lies to the left of the fixed point setof r T . That is, T = T ∩ S × [0 , / T as a tangle in S × I .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 .There and henceforth, we work with the upper half space model for H and use thestandard representation of Isom( H ) as a Z extension of PSL ( C ). If d ∈ PSL ( C ),we write d for the matrix whose entries are the complex conjugates of the entriesof d . When we apply this operation to each element of a subgroup Γ < PSL ( C )we obtain a subgroup denoted by Γ.For a Kleinian group Γ, we denote the convex core of H / Γ as C (Γ). We willuse the term natural map as in [3] (see below Definition 3.1 there) to refer to therestriction to C (Λ) of the orbifold covering map H / Λ → H / Γ, for Λ < Γ. Becausethe limit set of Γ contains that of Λ, the natural map takes C (Λ) into C (Γ). ERIC CHESEBRO AND JASON DEBLOIS
The geometric models for B − S and ( S × I ) − T described in parts (1) and(2) of Proposition 1.1 are hyperbolic 3-manifolds with totally geodesic boundaryproduced by pairing certain faces of the right-angled ideal octahedron and cuboc-tahedron, respectively, but they are described in the Proposition as convex coresof the quotients of H by the groups generated by the face-pairing isometries. Theequivalence of these two forms of description is proved in Lemma 2.1 of [3]. Proposition 1.1. (1)
For s = (cid:0) − (cid:1) and t = (cid:0) i − ii − i (cid:1) , ∆ = (cid:104) s , t (cid:105) is aKleinian group, and there is a homeomorphism f S : M S . = B − S → C (∆ ) . (2) For f , g and h below, Γ T = (cid:104) f , g , h (cid:105) is a Kleinian group, and there is ahomeomorphism f T : M T . = ( S × I ) − T → C (Γ T ) . f = (cid:0) − (cid:1) g = (cid:16) − i √ − i √ − − i √ (cid:17) h = (cid:16) i √ − − i √ − i √ − i √ (cid:17) (3) For c = (cid:16) i √
20 1 (cid:17) , Γ T = (cid:10) Γ T , c − Γ T c (cid:11) is a Kleinian group, and there isa homeomorphism f T : M T . = ( S × I ) − T → C (Γ T ) satisfying: • composing the inclusion M T → M T with f T yields f T ; and • for r T as above, f T ◦ r T ◦ f − T is induced by x (cid:55)→ c − ¯ x c . (4) The intersection ∆ ∩ Γ T is a Fuchsian group Λ stabilizing the hyperplane H = R × (0 , ∞ ) of H . This is the intersection of the convex hulls of thelimit sets of ∆ and Γ T , and the natural maps from H / Λ to C (∆ ) and C (Γ T ) map to totally geodesic boundary components. (5) The image of the natural map H / Λ → C (Γ T ) is the image of ∂ − M T . =( S × { } ) − T under f T . The same holds with each instance of T herereplaced by T .For the homeomorphism j : ( ∂B , ∂S ) → ( S × { } , ∂ − T ) such that ( B , S ) ∪ j ( S × I, T ) ∼ = ( B , T ) , f T ◦ j ◦ f − S : ∂C (∆ ) → C (Γ T ) fac-tors through H / Λ as the composition of a natural map with the inverse ofanother. We now turn back to topology and give an inductive definition of the tangles T n , assembling ( B , T n ) from a single copy of ( B , S ) and n of ( S × I, T ) foreach n ∈ N , using T as pictured in Figure 2 as the base case. Numbering thepoints of ( S (0) , ∂T ) and ( S (1) , ∂T ) as shown in the figure, let ( S × { } , ∂ + T ),( S × { } , ∂ − T ), and ( ∂B , ∂S ) inherit numberings from their inclusions to thesespheres. Note that the resulting numbering of ( S × ∂I, ∂T ) is r T -invariant.Now for n >
1, assume for 1 ≤ k < n that tangles T k ⊂ B with labeled endpointsare defined, and, for k >
1, inclusions ( B , T k − ) (cid:44) → ( B , T k ) and ι k : ( S × I, T ) → ( B , T k ), such that: • ( B , T k ) = ( B ∪ ι k ( S × I ) , T k − ∪ ι k ( T )); • ι k preserves labels on ∂ + T ; and • the included image of B intersects i k ( S × I ) in a sphere S ( k ) , with( S ( k ) , S ( k ) ∩ T k ) = ( ∂B , ∂T k − ) = ι k ( S × { } , ∂ + T ).Define T n ⊂ B as the quotient of the disjoint union ( B , T n ) (cid:116) ( S × I, T ) byidentifying ι n − ( x,
1) to ( x,
0) for each x ∈ S ; let the inclusion of ( B , T n − ) and ι n : ( S × I, T ) → ( B , T n ) be induced by the respective inclusions into the disjointunion; and label the endpoints of T n coherently with T ∩ ( S × { } ) using ι n . It isclear by construction that the inductive hypothesis applies to ( B , T n ). IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 5
Having topologically described the T n , our next order of business is to givegeometric models for their complements; that is to describe hyperbolic manifoldswith totally geodesic boundary homeomorphic to the B − T n . In parallel withour topological description of T n , these are assembled from copies of the geometricmodels described in Proposition 1.1. To this end, we define:Γ ( j ) T = c − j − Γ T c j − Λ ( j ) = c − j Λ c j F ( j ) = c − j ( H ) / Λ ( j ) Note for each j that C (Γ ( j ) T ) is isometric to C (Γ T ), so it is just a copy of M T , and F ( j ) is isometric to H / Λ. Now with ∆ as in Proposition 1.1, for n ≥ n = (cid:68) ∆ , Γ (1) T , . . . , Γ ( n ) T (cid:69) . (In [3], ∆ is denoted as Γ S and ∆ n as Γ ( n ) − .) The consequence of Propositions 3.10and 3.12 of [3] below shows that C (∆ n ) is a geometric model for B − T n . Proposition 1.2.
For each n ∈ N there is a homeomorphism f n : B − T n → C (∆ n ) . Moreover, the natural map C (∆ n − ) → C (∆ n ) is an isometric embedding,and there is another, ι n : C (Γ T ) → C (∆ n ) factoring through an isometry C (Γ T ) → C (Γ ( i ) T ) , such that for n > the following diagrams commute. B − T n − f n − (cid:47) (cid:47) (cid:15) (cid:15) C (∆ n − ) (cid:15) (cid:15) M T f T (cid:47) (cid:47) ι n (cid:15) (cid:15) C (Γ T ) ι n (cid:15) (cid:15) B − T n f n (cid:47) (cid:47) C (∆ n ) B − T n f n (cid:47) (cid:47) C (∆ n ) This also holds for n = 1 , taking f . = f S : M S → C (∆ ) at the top left.The natural map F ( j ) → f n ( S ( j ) − ∂T j ) is an isometry onto a totally geodesicsurface in C (∆ n ) when ≤ j ≤ n ; in particular, F ( n ) is isometric to ∂C (∆ n ) . Our final task in comprehensively describing the T n is to translate the tangleendpoint 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 listingrepresentatives for the parabolic conjugacy classes in Λ as words in Γ S and Γ T . p = s − = f − p = stst − = fg − f − h − gp = ( tst ) s − ( tst ) − = ( h − fg ) − g − ( h − fg ) , p = p p p − A calculation shows that p = ( ) p = (cid:0) − − (cid:1) p = (cid:0) −
14 25 − (cid:1) p = (cid:0) − − (cid:1) From Lemma 2.4 of [3] we obtain the next proposition.
Proposition 1.3.
For any n ∈ N , j ≤ n , and k ∈ { , , , } , the parabolic conju-gacy class in Λ ( j ) which corresponds to the point labeled k in S ( j ) is represented by p ( j ) k = c − j p k c j . Also Λ ( j ) is generated by any three of the p ( j ) k ’s. We finish by giving a geometric model for the mutation with a hidden extension.The result below follows from Proposition 1.3 above and Lemma 5.5 of [3].
ERIC CHESEBRO AND JASON DEBLOIS
Lemma 1.4.
Let m = (cid:0) − − (cid:1) . For each n ≥ , m ( n )1 . = c − n m c n normalizes Λ ( n ) and induces a cycle representation (1 3)(2 4) on the the four cusps of F ( n ) ,where each cusp is numbered according to its corresponding parabolic isometry p ( n ) j . The proof of existence.
The key new tool we need to prove Theorem 1.8 isa discrete group containing both ∆ n , with finite index, and also the isometry m that induces the mutation (1 3)(2 4). (The group G m + n of [3, Lemma 6.2] playsthis role for the group Γ m + n uniformizing S − L m + n , by [3, Prop. 6.3].) We willuse 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-spacemodel of H bounded by the Euclidean hemisphere of unit radius centered at 0 ∈ C and, for k ∈ N , let B k be the Euclidean translate of B centered at k ( − i √ i is the imaginary unit. For complex numbers z and w , refer by z H + w to thegeodesic plane ( z R + w ) × (0 , ∞ ). Definition 1.5.
For an integer n ≥
0, define Q n to be the polyhedron of H bounded by H + i/ i H , i H + 1 /
2, and ∂ B k for k ∈ { , , . . . , n } . Further define:(1) f by first reflecting in i H and then in i H + 1 / b by first reflecting in H + i/ ∂ B ; and(3) for k ≥ a k by reflecting in i H + 1 / ∂ B k . B B B H + i/ i H i H + 1 / Figure 3.
Bounding hyperplanes for Q viewed from above.As defined, we have f = ( ) b = ( ii ) a = (cid:0) − − (cid:1) a = (cid:16) − i √ i √ − i √ (cid:17) . In particular, we see that (cid:104) f , a (cid:105) = PSL ( Z ) contains m . Lemma 1.6.
For each integer n ≥ , the orientation-preserving subgroup H n ofthe group generated by reflections in the faces of Q n satisfies: (1) H n is a Kleinian group generated by { f , b , a , . . . , a n } . (2) ∆ n < H n with finite index. IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 7 (3)
The projection c − n ( H ) → H /H n factors through an isometric embeddingof H / PSL ( Z ) onto ∂C ( H n ) .Proof. Clearly f ∈ H n , b ∈ H n , and a i ∈ H n for 0 ≤ i ≤ n . Let r (cid:48) denotethe reflection across i H + 1 /
2. It is not hard to see that Q n ∪ r (cid:48) ( Q n ) is a convexpolyhedron in H with one face in each of H + i √ i H , i H + 1, ∂ B k , and ∂ r (cid:48) ( B k )for k ∈ { , , . . . , n } . The following facts can be explicitly verified: • The face in H + i √ i H and i H + 1 at right angles, thosein ∂ B and r (cid:48) ( ∂ B ) at an angle of π/
3, and no others. The product a b rotates by π about an axis that bisects this face, preserving it. • The face in i H meets each of those in ∂ B k at an angle of π/
2, and none ofthose in r (cid:48) ( ∂ B j ). The element f takes this face to the one in i H + 1. • The face in ∂ B k shares an edge with the face in r (cid:48) ( ∂ B k (cid:48) ) if and only if k = k (cid:48) ; in this case at an angle of 2 π/
3. The element a k takes the latterto the former. The faces in ∂ B k and ∂ B k − meet at an angle of π/ k >
0; likewise those in ∂ B k and ∂ B k +1 for k < n ; and ∂ B k ∩ ∂ B k (cid:48) = ∅ for k (cid:48) / ∈ { k − , k, k + 1 } .Hence, { f , a b , a , . . . , a n } is a face-pairing for Q n ∪ r (cid:48) ( Q n ). Poincare’s polyhe-dron theorem implies that this set of isometries generates a discrete group whosefundamental domain is Q n ∪ r (cid:48) ( Q n ). By construction, this group is contained in H n . It is equal to H n because their fundamental domains have the same volume.The numbered formulas (8) and (9) above Proposition 6.3 of [3] express thegenerators of ∆ in terms of a , b , and f and they express Γ T in terms of a , a ,and f . Therefore, ∆ < H n and Γ T < H n . It can be verified directly that, forevery k , c − a k c = a k +1 and that c commutes with f . It follows that c − k Γ T c k < H n for all k ≤ n −
1. Moreover, the second paragraph of the proof of [3, Proposition6.3] expresses the generators of c − Γ T c in terms of a , a , and f . So, if n ≥ H n . Now, by definition, we have ∆ n < H n .The polyhedron P n of [3, Lemma 6.2] consists of points ( z, t ) ∈ Q n such thatthe imaginary coordinate of z is at least − n √
2. Every face of P n is a face of Q n except the unique face F of P n contained in H − n · i √
2. The face F is orthogonalto ∂ B n , i H , and i H + 1 / Q n . This means that a single face of P n ∪ r (cid:48) ( P n ) contains F , meeting only thebounding hyperplanes i H , i H + 1, ∂ B n , and r (cid:48) ( ∂ B n ). Moreover, these intersectionsare all orthogonal, so F projects to the sole totally geodesic boundary componentof the orbifold ( P n ∪ r (cid:48) ( P n )) /H n .We claim that ( P n ∪ r (cid:48) ( P n )) /H n = C ( H n ). We first show that P n ∪ r (cid:48) ( P n )is contained in the convex hull of the limit set of H n , which implies that ( P n ∪ r (cid:48) ( P n )) /H n is contained in the convex core. Inspecting Figures 3 and 4 in [3], oneobserves that P n ∪ r (cid:48) ( P n ) is contained in the union P ∪ (cid:83) n − k =0 c − k ( P ), where P and P are the regular ideal octahedron and right angled ideal cuboctahedron describedin Corollaries 2.2 and 2.3 of [3]. Both P and P are the convex hulls of their idealpoints, and each of these is a parabolic fixed point of ∆ or Γ T , respectively. (Onecan show this directly, or appeal to the third-from-last paragraph of the proof of [3,Lemma 2.1].) Since each parabolic fixed point of a Kleinian group lies in its limitset, it follows that P ∪ (cid:83) n − k =0 c − k ( P ) is in the convex hull of the limit set of H n .As a subset, P n ∪ r (cid:48) ( P n ) shares this property. On the other hand, the penultimate ERIC CHESEBRO AND JASON DEBLOIS paragraph of [3, proof of Lemma 2.1] shows that ( P n ∪ r (cid:48) ( P n )) /H n contains C ( H n )and this proves our claim.By the above, c − n ( H ) projects to ∂C ( H n ) under the quotient map H → H /H n . By Proposition 1.2, the same plane projects to ∂C (∆ n ) under H → H / ∆ n . It follows that the orbifold covering map H / ∆ n → H /H n restricts toone C (∆ n ) → C ( H n ). Since these both have finite volume, the map is finite-to-one,and hence ∆ n has finite index in H n .Among all bounding hyperplanes of Q n ∪ r (cid:48) ( Q n ), only i H , i H + 1, ∂ B n , and r (cid:48) ( ∂ B n ) meet the hyperplane H − n · i √
2. Each of these intersections is a rightangle. Thus, F is a quadrilateral and { a n , f } is an edge pairing for F . This impliesthat F / (cid:104) a n , f (cid:105) is the boundary of ( P n ∪ r (cid:48) ( P n )) /H n .We mentioned above that c − n a c n = a n and f c = cf , so (cid:104) a n , f (cid:105) = c − n (cid:104) a , f (cid:105) c n .Therefore, the projection H − n · i √ c − n ( H ) → H /H n factors through anisometric embedding of H / PSL ( Z ). (cid:3) 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 thelimit sets of G and H are equal, so the natural map C ( H ) → C ( G ) is an orbifoldcover. Theorem 1.8.
For n ∈ N , the mutation of ∂ ( B − T n ) determined by (1 3)(2 4) hasa hidden extension over a cover of B − T n and for any m ∈ N , taking T n ⊂ L m + n ,a hidden extension over a cover of S − L m + n .Proof. As we mentioned in the introduction to this paper, we may view ( B , T n ) asa subset of ( S , L n + m ). (For more rigor, compare the definitions at the beginningof this section with [3, Definitions 3.8].) Here it is bounded by the sphere S ( n ) ,with the mirror image ( B , T m ) of T m on the other side. If the mutation (1 3)(2 4)extended over B − T n , then S − L n + m would be homeomorphic to its mutantby (1 3)(2 4) along S ( n ) . By Mostow-Prasad rigidity, these two links would beisometric, but by Theorem 2 of [3] they are not. (In the notation of that result, L m + n = L (0 ,..., and its mutant is L (0 ,..., ,..., with the sole “1” the ( m + 1)thentry.) This also implies it does not extend over S − L m + n .However, because m ( n )1 lies in the finite extension H n of ∆ n it normalizes thenormal core Ω n of ∆ n in H n and determines a self-isometry ˜Ψ of H / Ω n . This is afinite cover of H / ∆ n which by Fact 1.7 above restricts to a cover C (Ω n ) → C (∆ n ).In particular, the boundary of C (Ω n ) is totally geodesic. One component of ∂C (Ω n ) is the quotient of c − n ( H ) by its stabilizer ˜Λ ( n ) = Ω n ∩ Λ ( n ) in Ω n . Since m ( n )1 normalizes both Λ ( n ) and Ω n , it normalizes ˜Λ ( n ) and determines an isometryof c − n ( H ) / ˜Λ ( n ) lifting the one determined by m ( n )1 on ∂C (∆ n ).A completely analogous argument applies to S − L m + n , replacing ∆ n by Γ n from Prop. 3.12 of [3] and H n by G n from Prop. 6.3 there. (cid:3) Matching covers
In this section, we build an explicit hidden extension of the mutation (13)(24) of ∂ ( B − T n ). To find an appropriate cover, we use the decomposition of B − T n alongthe spheres S ( j ) into one copy of M S and n copies of M T and find appropriate coversof these pieces which glue together to give a cover of B − T n with the necessaryproperties. Figure 4 is a schematic depiction of how this will be done. IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 9 (cid:102) M S (cid:102) M (1) T C (Ω T ) (cid:102) M (2) TJ −→ ←→ R T R T −→ M S M (1) T M (2) Tj −→ r T −→ Figure 4.
Assembling a cover of ( S × I ) − T that has a hiddenextension of the mutation (1 3)(2 4).For convenience, in this section we will supress the homeomorphisms f S , f T ,and f T of Proposition 1.1 and simply make the identifications: M S = C (∆ ) M T = C (Γ T ) M T = C (Γ T ) . Further, for n ∈ N and 1 ≤ i ≤ n we will identify the i th copy M ( i ) T of ( S × I, T )in ( B , T n ) with C (Γ ( i ) T ) (compare Proposition 1.2).To produce the covers (cid:102) M S and the (cid:102) M ( i ) T of Figure 4 we will divide the orbifold O n = H /H n branched-covered by B − T n into pieces covered by M S and the M ( i ) T , then analyze the corresponding subgroups of H n . For M S we use Q fromDefinition 1.5: the polyhedron bounded by H + i/ i H , i H + 1 /
2, and ∂ B . Ourfirst lemma shows that the orientation-preserving subgroup H of the reflectiongroup generated by Q contains the group ∆ = Γ S uniformizing M S . Our seconduses H and some elementary number theory to find a cover of M S with abundantsymmetry.We follow a similar strategy for M T , producing a polyhedron P T and a group H T < H n , which Lemma 2.3 shows contains the group Γ T uniformizing M T . Wewill use the permutation representation of H T given by acting on left cosets of Γ T to find a cover of M T with a hidden extension of (1 3)(2 4). Doubling this coveracross a boundary component yields the model (cid:102) M T for the (cid:102) M ( i ) T . Lemma 2.1.
The reflection group H (recall Lemma 1.6) has the following addi-tional properties: (1) H is a Kleinian group which contains ∆ as a subgroup of index 12. (2) H = (cid:104) a , b , f | a = b = ( b − a ) = ( a f ) = 1 (cid:105) . (3) PSL ( Z ) = Stab H ( H ) . (4) The projection
H → H /H factors through an isometric embedding of H / PSL ( Z ) onto ∂C ( H ) .Proof. As in Lemma 1.6, r (cid:48) is the reflection through i H + 1 /
2. From the proof ofLemma 1.6, we know that the collection { a , b a , f } is a face-pairing for Q ∪ r (cid:48) ( Q ). So the Poincar´e polyhedron theorem gives the presentation above. Asidefrom the fact that [ H , ∆ ] = 12, the rest of the lemma follows from the special case n = 0 in Lemma 1.6. But, if we compare the volume of a regular ideal octahedronin H with the volume of P , we see that [ H , ∆ ] = 12. (cid:3) Lemma 2.2.
There is an index five subgroup Ω < ∆ which is normal in H .Define Λ = Ω ∩ Λ . (1) [Λ : Λ ] = 5 , (2) p , p ∈ Λ , and (3) p and p project to generators of Λ / Λ .Proof. In the ring of Gaussian integers 5 = (1 + 2 i )(1 − i ). So, restricting themap Z [ i ] → Z [ i ] / (1 + 2 i ) to Z gives a ring epimorphism Z → Z [ i ] / (1 + 2 i ). Thequotient ring Z [ i ] / (1 + 2 i ) is isomorphic to Z / Z and we obtain a group epimor-phism PSL ( Z [ i ]) → PSL ( Z / Z ) which restricts to an epimorphism PSL ( Z ) → PSL ( Z / Z ). Since PSL ( Z ) < H < PSL ( Z [ i ]), the restriction to H is also ontoand the kernel Ω of this map has index | PSL ( Z / Z ) | = 60 in H .Using the explicit descriptions of s and t from Section 1, we see that ∆ maps ontothe parabolic subgroup { ( ∗ ) } of PSL ( Z [ i ] / (1 + 2 i )) which has order 5. Hence,∆ ∩ Ω has index five in ∆ . Since [ H : ∆ ] = 12, it follows that [ H : ∆ ∩ Ω ] =60. Therefore, ∆ contains Ω .Similarly, the explicit descriptions of the p j ’s from Section 1 show that Λ mapsonto this same parabolic subgroup and [Λ : Λ ] = 5. The final assertion is alsoimmediate from these descriptions. (cid:3) Lemma 2.3.
Let P T be the polyhedron bounded by ∂ B , ∂ B , i H and i H + 1 / .The orientation-preserving subgroup H T of the group generated by reflections inthe sides of P T is a Kleinian group such that (1) H T = (cid:104) a , a , f | a = a = 1 , ( a a − ) = ( a f ) = ( a f ) = 1 (cid:105) , (2) ∂C ( H T ) consists of a pair of totally geodesic surfaces, (3) PSL ( Z ) and Γ T are subgroups of H T , and (4) [ H T : Γ T ] = [Stab H T ( H ) : Λ] = 12 .Proof. First, recall from just before Lemma 1.6 and in the proof of Lemma 1.6 thatClaim (3) has already been established.Now, for visual intuition, compare P T with the right-angled ideal cuboctahedron P . The intersection P T ∩ P is the portion of P T which lies between H and H − i √
2. The intersection P T ∩P has two additional faces contained in H and H− i √ ∞ . These additional faces are either perpendicular toor disjoint from those of P T .Let F be the face of P T ∩ P contained in i H , A its face in ∂ B , and A the face in ∂ B . Let r (cid:48) denote reflection across i H + 1 /
2. A fundamental domainfor H T is the union P T ∪ r (cid:48) ( P T ). By construction, a , a , f ∈ H T and theseisometries 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 T is generated by a , a , and f . Upon noting that, for each i ∈ { , } , A i intersects F and A − i IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 11 perpendicularly and r (cid:48) ( A i ) at an angle of 2 π/
3, the above presentation comes fromthe usual edge-cycle relations.We claim that C ( H T ) is the quotient of Q T . = P ∩ ( P T ∪ r ( P T )) by the face-pairing isometries, with totally geodesic boundary. As remarked above, the faces Q T ∩ H and Q T ∩ ( H − i √
2) of Q T intersect the others perpendicularly, so Q T projects under the quotient map H /H T to a suborbifold with totally geodesicboundary. This is contained in C ( H T ), as Q T ⊂ P , which by Corollary 2.3 of[3] is contained in the convex hull of the limit set of Γ T which is contained in theconvex hull of the limit set of H T . Arguing as in the proof of Lemma 2.1 of [3]gives the reverse inclusion and hence the claim.That [ H T : Γ T ] = 12 follows from volume considerations. Note that a and a each preserve P and fix the point ( A ∩ A ) ∩ ( r ( A ) ∩ r ( A )); thus the group theygenerate has these properties as well. Since Q T contains a neighborhood in P ofits ideal vertex ∞ , (cid:104) a , a (cid:105) acts freely on the set ideal vertices of P . It is not hardto show directly that this action is transitive. Since P has twelve ideal vertices, itsvolume is twelve times that of Q T . Since these are fundamental domains for M T and C ( H T ) the associated cover has degree twelve.Recall from above that the faces Q T ∩ H and Q T ∩ ( H − i √
2) of Q T projectto ∂C ( H T ). The face-pairings of Q T induce edge-pairings on these faces, andone checks directly that each face has its edges identified to each other. It followsthat ∂C ( H T ) has two components. Since ∂M T also has two components, eachcomponent of ∂M T covers twelve-to-one. Since ∂M T has a component isometricto F (0) , [Stab H T ( H ) : Λ] = 12. (cid:3) Lemma 2.4.
There is a homomorphism φ : H T → S determined by φ ( a ) = (1 5 9)(2 6 10)(3 7 11)(4 8 12) φ ( a ) = (1 8 10)(2 7 9)(3 6 12)(4 5 11) φ ( f ) = (1 5 11 10 3)(2 7 6 8 12) It has the following properties. (1) | φ ( H T ) | = 660 , (2) φ (Γ T ) = (cid:104) φ ( h ) , φ ( f ) (cid:105) ∼ = Z (cid:111) Z , and (3) φ (Λ) = (cid:104) φ ( f ) (cid:105) = φ (Γ T ) ∩ φ ( m Γ T m − ) is the largest subgroup of φ (Γ T ) normalized by φ ( m ) .Remark . The homomorphism φ above is the permutation representation of H T given by its action on the left cosets of Γ T . This fact is not needed in the proofbelow or the rest of the paper. Proof.
That φ is a homomorphism follows from the presentation for H T given inLemma 2.3. Our expressions for a , a , f , f , g , h , and m as matrices make it easyto verify the equalities f = a f a − g = (cid:0) a − a (cid:1) f − (cid:0) a − a (cid:1) − h = a a f − a m = (cid:0) f a − (cid:1) f − . This gives φ ( f ) = (2 7 5 9 3)(4 6 11 10 12) φ ( g ) = (2 9 12 7 4)(3 11 6 8 5) φ ( h ) = (2 12 7 8 6 3 4 11 10 9 5) φ ( m ) = (1 8)(2 12)(3 4)(5 11)(6 9)(7 10) . We see that φ (Γ T ) = (cid:104) φ ( h ) , φ ( f ) (cid:105) ∼ = Z (cid:111) Z , because φ ( g ) = φ ( fh − ) and φ ( fhf − ) = φ ( h ).Under the action of H T on Z given by φ , Γ T is a subgroup of Stab H T (1).We claim that, in fact, these groups are equal. Let C = { , , } D = { , , } E = { , , } F = { , , } and observe that φ (cid:104) a , a (cid:105) preserves the triples C, D, E , and F . This gives a homo-morphism ψ : (cid:104) a , a (cid:105) → S with ψ ( a ) = ( D E F ) ψ ( a ) = ( C D F ) . Since these two elements generate A we have that the image of ψ is the order12 group A . The group (cid:104) a , a (cid:105) also acts by isometry on the polyhedron P andacts freely and transitively on its set of ideal vertices. Hence |(cid:104) a , a (cid:105)| = 12 and ψ : (cid:104) a , a (cid:105) → A is an isomorphism.Since |(cid:104) a , a (cid:105)| = 12 and Γ T is torsion-free, the elements of (cid:104) a , a (cid:105) make up acomplete set of representatives for the left cosets of Γ T . If k ∈ Stab H T (1) then k = an , where a ∈ (cid:104) a , a (cid:105) and n ∈ Γ T . Then φ ( a ) · φ ( a ) φ ( n ) · φ ( k ) · ψ ( a ) · C = C . If we list the elements of A , we see thatthe only possibilities for a are the identity or a ± . Since a ± do not fix 1, we musthave k ∈ Γ T as claimed.We know now that ker φ < Γ T , which implies that the images of distinct leftcosets of Γ T in H T have empty intersection in φ ( H T ). Therefore, | φ ( H T ) | =55 ·
12 = 660.Our formulas for the p j ’s in terms of f , g , and h give φ ( p ) = φ ( p ) = φ ( f − ) and φ ( p ) = id, so φ (Λ) = (cid:104) φ ( f − ) (cid:105) . Since m normalizes Λ, φ ( m ) normalizes φ (Λ);and indeed we have that φ ( m fm − ) = φ ( f − ). On the other hand, φ ( m hm − ) doesnot stabilize 1 while φ (Γ T ) does, so φ (Γ T ) ∩ φ ( m Γ T m − ) is properly containedin φ (Γ T ). This intersection contains φ (Λ), so since | φ (Γ T ) | = 55 the intersec-tion equals φ (Λ). It moreover follows that φ (Λ) is the largest subgroup of φ (Γ T )normalized by φ ( m ). (cid:3) Define Ω T = φ − ( φ Λ), and let p T : H / Ω T → H / Γ T be the correspondingcover. If we let θ : H → PSL ( Z [ i ] / (1 + 2 i )) be the map used in the proof ofLemma 2.2 and identify θ (Λ) and φ (Λ) with the isomorphism φ f (cid:55)→ ( ), we seethat θ = φ . In particular, Λ = Λ ∩ ker φ . Lemma 2.6.
The preimage of p − T ( F (0) ) has three components. One is the inclusion-induced image of F (0) in H / Ω T , which projects isometrically to H / Γ T . Theother two are respectively isometric to g ( H ) / ( g Λ g − ) and g − ( H ) / ( g − Λ g ) , eachof which projects five-to-one into H / Γ T .Proof. Lemma 2.4 implies that [Γ T : Ω T ] = 11. Fact 1.7 implies that p T restrictsto a covering map C (Ω T ) → M T . By definition of Ω T , Λ is a subgroup of Ω T and corresponds to a component of ∂C (Ω T ) that covers ∂ − M T one-to-one.To understand the entire preimage p − T ( ∂ − M T ) we pass to the cover (cid:101) N → M T corresponding to ker φ . Let f = φ ( f ) and h = φ ( h ), so the deck group of the cover IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 13 is φ (Γ T ) = (cid:104) f, h (cid:105) . Λ is the stabilizer of H in ker φ , so (cid:101) S = H / Λ is a boundarycomponent of (cid:101) N . Since [Λ : Λ ] = 5, (cid:101) S is a 5-fold cover of ∂ − M T .Locate a base point x ∈ ∂ − M T so that elements of Λ = π ( ∂ − M T , x ) arerepresented by loops in ∂ − M T , and let ˜ x ∈ (cid:101) S be in the preimage of x . As usual,our choice of the point ˜ x gives an action of π ( M T , x ) on (cid:101) N . By Lemma 2.4, f ∈ φ (Λ), so f k . ˜ x ∈ (cid:101) S for every k . Moreover, for each fixed j , ( h j f k ) . ˜ x and( h j f k (cid:48) ) . ˜ x = ( h j f k ) . ( f k (cid:48) − k . ˜ x ) occupy the same component of ∂ (cid:101) N for any k and k (cid:48) .This means that ∂ (cid:101) N has eleven components covering ∂ − M T . Each component isof the form h j ( (cid:101) S ) for some j ∈ { , . . . , } and contains the orbit of ˜ x under theleft coset h j (cid:104) f (cid:105) . Recall from the proof of 2.4 that f h = h f . Therefore, (cid:104) f (cid:105) actson the components of the preimage of ∂ − M T in ∂ (cid:101) N . Its three orbits are { (cid:101) S } { h ( (cid:101) S ) , h ( (cid:101) S ) , h ( (cid:101) S ) , h ( (cid:101) S ) , h ( (cid:101) S ) } { h ( (cid:101) S ) , h ( (cid:101) S ) , h ( (cid:101) S ) , h ( (cid:101) S ) , h ( (cid:101) S ) } . Because C (Ω T ) is the quotient of (cid:101) N by the action of (cid:104) f (cid:105) , any component of thelatter two orbits projects injectively to p − ( ∂ − M T ). From the proof of Lemma 2.4,we know that φ ( g ) = f h − = h f and hence φ ( g − ) = hf − . Since h ( (cid:101) S ) and h ( (cid:101) S )lie in different (cid:104) f (cid:105) -orbits the lemma’s final claim follows. (cid:3) Let (cid:102) M T be the double of C (Ω T ) across p − T (( S × { } ) − T ) and let R T : (cid:102) M T → (cid:102) M T be the doubling involution. It is straightforward to show that there is an 11-foldcover p T : (cid:102) M T → M T that restricts on C (Ω T ) to p T such that p T ◦ R T = r T ◦ p T .Let ∂ − (cid:102) M T = p − T ( ∂ − M T ). Corollary 2.7.
The mutation of ∂ − M T determined by (1 3)(2 4) has a hidden ex-tension Ψ : (cid:102) M T → (cid:102) M T that preserves each component of p − T ( ∂ − M T ) and commuteswith R T .Proof. Since φ ( m ) normalizes (cid:104) φ ( f ) (cid:105) in S , m normalizes Ω T . So m induces aself-isometry Ψ of H / Ω T which, by Fact 1.7, preserves M T . We claim that Ψ preserves p − T (( S × { j } ) − T ) for j ∈ { , } .From Lemma 2.3, we know that ∂C ( H T ) has two components. So, under thebranched cover M T → C ( H T ), the images of the two components of ∂M T are dis-tinct. Recall that Proposition 1.1(5) implies ∂ − M T = H / Λ. Since each componentof p − T ( ∂ − M T ) is the quotient of a Γ T -translate of H by its stabilizer in Ω T , thismeans that the H T - and Γ T -orbits of H are identical. Since m ∈ PSL ( Z ) < H T ,the Γ T -orbit of H is perserved by m . This proves the claim.Now, using the claim, we obtain an isometry Ψ of (cid:102) M T that commutes with R T and agrees with Ψ on C (Ω T ). In the last part of the proof of Proposition 6.6 from[3], we show that m does not normalize Γ T . Hence, Ψ is a hidden symmetry of M T . Since m normalizes Λ, Ψ restricts to a lift of (1 3)(2 4) on the component of p − T ( ∂ − M T ) which is the image of H . Therefore, Ψ is a hidden extension of thismutation over C (Ω T ) and Ψ is a hidden extension over (cid:102) M T . (cid:3) Corollary 2.8.
There is a covering space p S : (cid:102) M S → M S with degree 11 and anisometry J : ∂ (cid:102) M S → ∂ − (cid:102) M T , which lifts the map j : ∂M S → ∂ − M T , such that J − Ψ J extends across (cid:102) M S .Proof. Let (cid:102) M S be the disjoint union M S (cid:116) (cid:101) N a (cid:116) (cid:101) N b , where (cid:101) N a and (cid:101) N b are copies of C (Ω ). Define J component-wise as follows. • Recall the isometric embedding ι (0) − : F (0) → ∂M S and let ι + : F (0) → ∂ (cid:102) M T be the isometric embedding given by Lemma 2.6. Define J | ∂M S as ι + ◦ ( ι (0) − ) − . • Let ι a : H / Λ → ∂ (cid:101) N a be the isometric embedding guaranteed by Lemma2.2 and let ι g : H / Λ → ∂ (cid:102) M T be the composition of the isometric em-bedding g ( H ) / g Λ g − → ∂ − (cid:102) M T given by Lemma 2.6 with the naturalisometry H / Λ → g ( H ) / g Λ g − . Define J | ∂ (cid:101) N a as ι g ◦ ι − a . • Define J | ∂ (cid:101) N b as ι g − ◦ ι − b in analogy with the case above.To see that J lifts j , notice first that Proposition 1.2 implies that j = ι (0)+ ( ι (0) − ) − .Now, for x ∈ H , the covering map H / Ω → H / ∆ sends Ω ( x ) to Γ ( x ) and H / Λ → F (0) sends Λ ( x ) to Λ( x ). So ι a and ι b lift ι . Similarly, ι g and ι g − lift ι (0)+ , since ι (0)+ factors as the natural composition F (0) → x ( H ) / x Λ x − → ∂ − M T whenever x ∈ Γ T .Lemma 5.8 of [3] implies that m normalizes ∆ . So, the restriction of J − Ψ J to ∂M S extends over M S . A calculation shows that gm g − = (cid:0) − − (cid:1) , so gm g − preserves H and m preserves g − ( H ). For x ∈ H , the map J − Ψ J | ∂ (cid:101) N a takesΩ ( x ) to Ω ( g − m g ( x )). This map extends over (cid:101) N a since g − m g ( x ) ∈ H and,by Lemma 2.2, Ω is normal in H .Since Ψ takes ∂ − (cid:102) M T to itself and preserves the components covered by H and g − ( H ), it also preserves the component covered by g ( H ). For x ∈ H , the map J − Ψ J | ∂ (cid:101) N b takes Ω ( x ) to Ω ( gm g − ( x )). As before, this map extends over (cid:101) N b because Ω is normal in H . (cid:3) Theorem 2.9.
For each n ∈ N there is an -sheeted cover N n → B − T n anda hidden extension Ψ : N n → N n of the mutation (1 3)(2 4) acting on S ( n ) − T n .Moreover, for each m ∈ N , Ψ extends to a hidden symmetry of an -sheeted coverof S − L m + n that contains N n .Proof. Let ∂ + (cid:102) M ( j ) T = ∂ (cid:102) M ( j ) T − ∂ − (cid:102) M ( j ) T and compose R T : ∂ + (cid:102) M T → ∂ − (cid:102) M T withmarking maps to obtain isomorphisms R T : ∂ + (cid:102) M ( j ) T → ∂ − (cid:102) M ( j +1) T . Define N n to bethe adjunction space (cid:102) M S ∪ J (cid:102) M (1) T ∪ R T · · · ∪ R T (cid:102) M ( n ) T . Because J lifts j and R T lifts r T , the covering maps p S and p T determine coveringspaces p n : N n → B − T n , which restrict to p S and p T on the factors of theadjunction space.The hidden extension Ψ is given on each of the (cid:102) M ( i ) T by the eponymous symmetryfrom Corollary 2.7 and on (cid:102) M S by the extension of J − Ψ J described in Corollary2.8. Corollary 2.7 implies that Ψ : N n → N n does not descend to B − T n .Let p n : N m → B − T m be the mirror image of the cover p n and Ψ : N m → N m the mirror image of Ψ. Using the mirror map ∂N n → ∂N m to glue, weform an adjunction space (cid:102) M m + n = N n ∪ N m . The covering maps p n and p n determine a covering space (cid:102) M m + n → S − L m + n and Ψ and Ψ determine anisometry (cid:102) M m + n → (cid:102) M m + n . As defined, the covering space and isometry restrictto p n and Ψ on N n ⊂ (cid:102) M m + n . (cid:3) IDDEN SYMMETRIES VIA HIDDEN EXTENSIONS 15
Remark . The referee has asked whether the covers of B − T n and S − L m + n described in Theorem 2.9 have minimal degree among those admitting hiddenextensions of (1 3)(2 4). We suspect this is so but cannot quite prove it. Below weprove a related but weaker assertion. Suppose Φ : M → M is a hidden symmetryof M T that restricts to a lift of (1 3)(2 4) on a component S of p − ( ∂ − M T ), where p : M → M T and p : M → M T are finite-degree connected covers of M T . IfΦ is induced by some n ∈ H T from Lemma 2.3 then the p i have degree at least 11.Let Γ and Γ be the finite-index subgroups of Γ T respectively correspondingto the M i . There exist finite-index subgroups Λ i of Λ, and g i ∈ Γ T − Γ i so that S is represented in Γ by g Λ g − and S = Φ( S ) by g Λ g − . For each i , therestriction of p i to S i has p i ∗ ( g i λg − i ) = λ for λ ∈ Λ i . That Φ : S → S lifts(1 3)(2 4) translates at the level of induced maps to m λ m − = g − Φ ∗ ( g λg − ) g = g − n g λg − n − g for each λ ∈ Λ , since m : Λ → Λ is the induced map of (1 3)(2 4). But thecentralizer of Λ in PSL ( C ) is trivial, so we have n = g m g − . Lemma 2.4(3) nowimplies that Γ has index at least 11 in Γ T . Acknowledgement.
We are grateful to the referee for asking interesting ques-tions, and for suggesting changes that have improved the quality of this paper.
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Department of Mathematical Sciences, University of Montana
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