aa r X i v : . [ m a t h . G T ] M a r A SMALL INFINITELY-ENDED 2-KNOT GROUP
R. BUDNEY AND J. A. HILLMAN
Abstract.
We show that a 2-knot group discovered in the courseof a census of 4-manifolds with small triangulations is an HNNextension with finite base and proper associated subgroups, andhas the smallest base among such knot groups. introduction Nontrivial classical knot groups have one end. This is equivalent tothe asphericity of the knot complement, in the light of Dehn’s Lemmaand Poincar´e duality in the universal cover. In higher dimensions thecomplements of nontrivial knots are never aspherical. However, Ker-vaire gave a uniform algebraic characterization of n -knot groups for all n ≥
3, and a partial characterization of 2-knot groups [8]. (Artin’s spin-ning construction shows that classical knot groups are 2-knot groupsand 2-knot groups are high dimensional knot groups.) These charac-terizations have been used to provide examples of such groups withvarious properties. In particular, they have been used to find knotswhose groups have more than one end.The 2-twist spins τ K of 2-bridge knots K provide many examples of2-knot groups with two ends. The first examples of higher dimensionalknots with infinitely-ended groups were given in [5]. Their examplesare 2-knots, and the groups are HNN extensions with finite base andproper associated subgroups. The simplest of these has presentation h a, b, t | tat − = a , a = 1 , aba − = b i , with base h a, b i ∼ = Z/ Z ⋊ Z/ Z and both associated subgroups h a i ∼ = Z/ Z . (Note that the second and third relations together imply that b = 1.) It is clear from this presentation that the group is also a freeproduct with amalgamation of h a, b i with h a, t i ∼ = πτ over h a i , where3 is the trefoil knot [9]. In fact it is the group of a satellite of τ with companion Fox’s Example 10, as is clear from the analysis of thegroups of such knots in [7]. Replacing the trefoil with other 2-bridgeknots gives all of the examples of § Mathematics Subject Classification.
Key words and phrases. end, 2-knot, 4-manifold, virtually free. In § π which is an HNN extension with base the gener-alized quaternionic group Q (16) and (distinct) associated subgroups Q (8). This group is not properly the group of a satellite knot ( § § Q (16). However there are such groups with base D × Z/ Z and associated subgroups ( Z/ Z ) ( § § § Out ( π ) ∼ = ( Z/ Z ) .For us, an n -knot is a locally flat embedding of S n in S n +2 , and soorientations of the spheres determine orientations of the knot comple-ments and preferred meridians for the knots. We shall use Chapter 14of [6] as a one-stop reference for the aspects of higher dimensional knottheory that we need. All homology groups considered below shall haveintegral coefficients, and so we shall write H i ( G ) instead of H i ( G ; Z ).2. The knot exterior
The knot exterior of this paper was discovered while forming a censusof 4-manifolds triangulable with 6 or less pentachora. Precisely, BenBurton generalized the census-generation algorithm in Regina [2] toenumerate all unordered 4-dimensional delta-complexes whose vertexlinks are triangulated 3-spheres or more generally 3-manifolds. Trian-gulations having a non-spherical manifold vertex link are called cuspedtriangulated manifolds . Most of the non-trivial knot exteriors in thecensus are of ideal/cusped type. A previous paper was written on thesimplest non-trivial knot exterior in the census [1]. A future paper willdescribe the census in full.In the census there are approximately 1.4 million combinatorial classesof knot exteriors in homotopy spheres. By combinatorial class we meantriangulated manifolds, up to homeomorphism that preserve the sim-plicial subdivision from the triangulation, i.e. the homeomorphismsneed not preserve the characteristic maps of the individual simplices.Among these 1.4 million triangulations, 8521 have non-abelian fun-damental group. Most of these have finitely generated commutatorsubgroup. There are just twenty exceptional cases.Ten of these have group Φ with presentation h a, t | tat − = a i ,and so are homeomorphic to the exterior of Fox’s Example 10. Thefinal ten all have fundamental group isomorphic to the group π with SMALL INFINITELY-ENDED 2-KNOT GROUP 3 presentation h a, t | a , a − tata t − , a tata − t − , a t − a − t i . We briefly describe one of the ten triangulations giving a manifold M with fundamental group π . At present we know there are at mosttwo PL homeomorphism types represented by these triangulations.M (0123) (0124) (0134) (0234) (1234)0 4 (0123) 3 (0124) 2 (1320) 2 (0234) 1 (1234)1 5 (0123) 4 (1240) 4 (4320) 2 (1234) 0 (1234)2 0 (4031) 5 (3142) 3 (4013) 0 (0234) 1 (0234)3 5 (0421) 0 (0124) 2 (1340) 4 (1423) 5 (4031)4 0 (0123) 1 (4012) 5 (0324) 1 (4310) 3 (0342)5 1 (0123) 3 (0321) 3 (2431) 4 (0314) 2 (1402)The leftmost column lists the pentachora of the triangulation, la-belling them 0 through 5. In each row, to the right of the pentachoronindex is a collection of pairs n ( abcd ). The first number n indicates onwhich pentachoron this tetrahedral facet is glued to. The entry ( abcd )indicates the affine-linear map on the tetrahedral facet. For example,the row 1 5(0123) 4(2104) 4(4320) 2(1234) 0(1234) indicates that inthe pentachoron indexed by 1, the 4th tetrahedron is glued to the 4thtetrahedron but in pentachoron 5, with vertices (0123) glued to (0123)in that order. Similarly, the 2nd tetrahedron is glued to the 1st tetra-hedron but in pentachoron 4, with vertices (0134) sent to (4320) inthat order, etc. We leave the readers to consult the documentation for[2] for details. In summary, this triangulation has no internal vertices –after performing the above gluings, all the vertices have been identified,thus the single vertex has vertex link a 30-tetrahedron triangulation of S × S . The triangulation has 3 edges, 12 triangles, 15 tetrahedra and6 pentachora.Verification that M is a knot exterior in a homotopy sphere is similarto the argument in [1] and left to the reader. (Up to changes of ori-entation, there are at most two knots with exterior homeomorphic to M .) Budney and Burton have automated the process and it is imple-mented in the software [2]. Perhaps for some readers it would be moreappealing to read the algorithm implemented in Regina. At presentthe 4-manifolds tools are in the development repository of Regina, andthese tools will be in the general release of Regina by version 5.0. Onecan readily check that the above triangulation has a single non-trivialsymmetry, an involution that reverses orientation and acts non-triviallyon H ( M ). The involution is the simplicial map that sends pentachoron R. BUDNEY AND J. A. HILLMAN π = π ( M ), we find that the third relatorin the above presentation is a consequence of the others, and so thispresentation simplifies to h a, t | tat − = a t a − t − , a , a t = ta i . Setting b = ta t − , this becomes h a, b, t | ta t − = b, tabt − = a , a = b = ( ab ) i . (The final two relations imply that bab − = a − . Hence a = 1, and so a is a central involution.) Thus π ∼ = B ∗ H ϕ is an HNN extension, withbase B ∼ = Q (16), the generalized quaternionic group with presentation h a, b | a = b = ( ab ) i . and associated subgroups H = h a , ab i ∼ = ϕ ( H ) = h a , b i ∼ = Q (8).The commutator subgroup is an iterated generalized free product withamalgamation π ′ ∼ = . . . B ∗ H B ∗ H B . . . , and is perfect ( π ′ = π ′′ ). Since H and ϕ ( H ) are proper subgroups of B the commutator subgroup is not finitely generated. Hence no knotwith this group is fibred.3. π is not properly the group of a satellite knot If an n -knot K is a satellite of K about K relative to a simpleclosed curve γ in X ( K ) then πK ∼ = πK / hh w q ii ∗ w =[ γ ] πK , where [ γ ] ∈ πK has order q ≥ w is a meridian for K . The case q = 0 corresponds to [ γ ] having infinite order. (See [7], or page 271 of[6].) If K = τ r k is the r -twist spin of an ( n − k and ( q, r ) = 1then πK / hh w q ii ∼ = Z/qZ . Thus every 2-knot group with non-trivialtorsion is trivially the group of a satellite knot. We shall say that πK is properly the group of a satellite knot if | πK / hh w q ii| > q .Suppose that the group π of § π has a central subgroup of order 2, it is not of the form A ∗ Z B with A and B nontrivial knot groups. Hence π ∼ = G ∗ Z/qZ H ,where G is a knot group and H is the quotient of a knot group by the q th power of a meridian, for some q >
0, but is not cyclic.
SMALL INFINITELY-ENDED 2-KNOT GROUP 5
Since π is an HNN extension with finite base it is virtually free.(See Corollary IV.1.9 of [4]. The argument given there implies that π has a free subgroup of index dividing 16!.) Therefore so are G and H , and all these groups have well-defined virtual Euler characteristics.Mayer-Vietoris arguments give χ v ( π ) = χ v ( Q (16)) − χ v ( Q (8)) = 116 −
18 = − χ v ( π ) = χ v ( G ) + χ v ( H ) − q , while χ v ( G ) ≤
0, since G is an infinite virtually free group. Hence χ v ( H ) ≥ q − . Now since π ∼ = Q (16) ∗ Q (8) ϕ , any finite subgroup of π is conjugate toa subgroup of Q (16). Therefore q divides 8, and so χ v ( H ) >
0. Hence H is finite, and so it is isomorphic to a subgroup of Q (16).We then find that the only possibility is that q = 8 and H ∼ = Q (16).But then χ v ( G ) = 0, and so G ′ is finite, and is either Z/ Z or Q (16).Neither of these groups admits a meridianal automorphism, and sothere is no such knot group G . Thus we may conclude that π is notproperly the group of a satellite knot.HNN extensions arise naturally in knot theory when an n -knot K hasa minimal Seifert hypersurface V , one for which the pushoffs from V toeither side of Y = S n +2 \ V are both π -injective. The knot group πK is then an HNN extension with base π ( Y ) and associated subgroupsisomorphic to π ( V ). There are 2-knots with group Z/ Z ⋊ Z (thegroup of the 2-twist spun trefoil) which do not have minimal Seiferthypersurfaces. (See Chapter 17 of [6].) Does a 2-knot with exteriorthe manifold M of § π ∼ = Q (16) ∗ Q (8) ϕ ? (Knots related by composition withreflections of S n or S n +2 have similar Seifert hypersurfaces.)4. hnn extensions with small finite base Let G = B ∗ C φ be an HNN extension with base B and associatedsubgroups C and φ ( C ). Let j : C → B be the natural inclusion.Consideration of the Mayer-Vietoris sequence for the extension showsthat H ( G ) ∼ = Z and H ( G ) = 0 if and only if H ( j ) − H ( φ ) is anisomorphism and H ( j ) − H ( φ ) is surjective. If the H condition holdsthen B/N is perfect, where N = hh{ j ( g ) − φ ( g ) | g ∈ C }ii is the normalclosure of { j ( g ) − φ ( g ) | g ∈ C } in B . In particular, if B is solvable then R. BUDNEY AND J. A. HILLMAN N = B . The H condition holds automatically if H ( B ; Z ) = 0, inparticular, if B is a finite group of cohomological period 4.If both homological conditions hold and N = B then the stable letterof the HNN extension normally generates G , and so G is a knot group[5]. (However, such HNN extensions need not be 2-knot groups. Thegroup Z/ Z ⋊ Z is a high dimensional knot group which is an HNNextension with H = B = Z/ Z , but the Farber-Levine condition fails,since 2 mod (5). See Chapter 14 of [6].)In particular, if B is finite and C is a proper subgroup then B isnonabelian, and C/C ′ ∼ = B/B ′ . Hence H ( B ) cannot be cyclic of evenorder. Therefore B is neither a dihedral group D k with k odd, nor Z/ Z ⋊ − Z/ Z . This leaves only Q (8), D , D and A among thegroups of order less than 16.The group Q (8) has no proper subgroup with abelianization ( Z/ Z ) .If B = D k then C must be isomorphic to D l , for some l dividing k .But then H ( C ) ∼ = H ( B ) = Z/ Z , and H ( j ) and H ( φ ) are the sameisomorphism. Hence H ( j ) − H ( φ ) is not an epimorphism. Thus wemay exclude D and D . If B = A then C must be Z/ Z . Since H ( Z/ Z ) = 0 and H ( A ) = Z/ Z , this group may be excluded also.Thus the smallest possible base must have order at least 16. Thegroup Q (16) has two proper subgroups with abelianization ( Z/ Z ) .These are h a , b i and h a , ab i , and are isomorphic to Q (8). The au-tomorphism a a, b ab of Q (16) carries one onto the other. Fixgenerators x, y for Q (8). Then we may assume that j ( x ) = a and j ( y ) = b . If φ is another embedding then φ ( x ) and φ ( y ) have order4, so one must be a ± and the other a i b . If, moreover, H ( j ) − H ( φ )is an isomorphism then φ ( x ) = a i b and φ ( y ) = a ± , and i must beodd. After conjugation in Q (16) we may assume that φ ( x ) = ab and φ ( y ) = a . Since Q (16) is solvable and H ( Q (16)) = 0 the HNN exten-sion π = Q (16) ∗ Q (8) ϕ is a knot group, and it has the smallest finitebase among all such HNN extensions. Moreover, it is the unique suchgroup with HNN base Q (16).5. further examples with base of order 16 There are eight other non-abelian groups of order 16. Four aresemidirect products K ⋊ L with K and L cyclic. In three of thesefour cases H ( C ) = H ( B ) = Z/ Z , and so these may be ruled out,by the argument that excluded D and D . The fourth group M has presentation h a, x | a = x = 1 , xax = a i . The abelianization is Z/ Z ⊕ Z/ Z , and so C must be h a , x i . There is no second embedding SMALL INFINITELY-ENDED 2-KNOT GROUP 7 φ such that H ( j ) − H ( φ ) is an isomorphism, and so we may rule out M . (Note that H ( M ) = 0.)The next group to consider is ( Z/ Z ) ⋊ θ Z/ Z , the semidirect prod-uct with action generated by θ = ( ) ∈ GL (2 , F ), and with pre-sentation h a, b, x | a = b = x = 1 , ab = ba, bx = xb, axa − = bx i .(Since b = a ( ax ) , this group is generated by { a, x } .) This hasabelianization Z/ Z ⊕ Z/ Z , and so H ( C ) = Z/ Z . It follows fromthe LHS spectral sequence for B as a semidirect product that H ( B )maps onto H ( Z/ Z ; ( Z/ Z ) ) = Ker( θ − I ) = Z/ Z . Hence either H ( B ) = Z/ Z and H ( j ) − H ( φ ) = 0, or H ( B ) has order ≥
4. Inneither case is H ( j ) − H ( φ ) an epimorphism, and so we may excludethis group.The remaining three have abelianization ( Z/ Z ) , so C must be anabelian subgroup of index 2, and hence normal. These are Q (8) × Z/ Z , D × Z/ Z and the central product of D with Z/ Z , with presentation h a, c, x | a = x = 1 , a = c , ac = ca, cx = xc, xax = a − i . We may eliminate Q (8) × Z/ Z and the central product immediately,as neither has a proper subgroup with abelianization ( Z/ Z ) .The final group is B = D × Z/ Z , with presentation h a, b, x | a = b = x = 1 , ab = ba, bx = xb, xax = a − i . There are two proper subgroups isomorphic to ( Z/ Z ) . These are h a , b, x i and h a , b, ax i , and the automorphism a a , b b , x ax of B carries one onto the other Let { c , c , c } be the standard basis for( Z/ Z ) . Then we may assume that C is the image of the embedding j : ( Z/ Z ) → B determined by j ( c ) = a , j ( c ) = b and j ( c ) = x .Let V = ( Z/ Z ) , and let e i be the image of the generator of H ( V ) = Z/ Z under the inclusion of V onto the subgroup generatedby { c k | k = i } . Then { e , e , e } is a basis for H ( C ) ∼ = ( Z/ Z ) . Wealso have H ( B ) ∼ = ( Z/ Z ) , since H ( B ) = H ( D ) ⊕ ( H ( D ) ⊗ Z/ Z ),by the K¨unneth Theorem. This has a basis { f , f , f } , where f is theimage of the generator of H ( D ), f = a ⊗ b and f = x ⊗ b . Thehomomorphism H ( j ) sends e , e and e to f , f and 0, respectively.Reimbeddings satisfying the H condition must carry C to e C = h a , b, ax i . There are | GL (3 , F ) | = 168 possible isomorphisms φ . Con-jugation in B reduces this by half (since [ B : C ] = 2), but this stillleaves too many possibilities to examine easily by hand. We shall justgive one example.Let ˜ j ( c ) = ax , ˜ j ( c ) = a and ˜ j ( c ) = b . Then Im(˜ j ) = e C , and H (˜ j ) sends e , e and e to 0 , f + f and f , respectively. It followseasily that the homological conditions are satisfied. Let φ = ˜ jj − (so R. BUDNEY AND J. A. HILLMAN φ ( a ) = ax , φ ( b ) = a and φ ( x ) = b ), and let π = B ∗ C φ . Then thestable letter of the HNN extension is a normal generator for π , since B is solvable, and so π is a high-dimensional knot group. Is there a2-knot group with HNN basis B = D × Z/ Z ?6. symmetries Four of the ten (ideal) triangulations in the census which realize π have simplicial involutions which reverse orientation and acts non-trivially on H ( M ). In particular, the triangulation displayed in § M is the interior of a compact 4-manifold M ,with boundary ∂M ∼ = S × S , and the involution extends to M . It isknown that there are 13 involutions of S × S , up to conjugacy [10].In Tollefson’s classification precisely three reverse both the orientationand the meridian, and they are determined by their fixed-point sets.One has fixed-point set S ∐ S , and does not extend across D × S .The others have fixed-point set S ∐ S and S ∐ S , respectively, andextend to involutions of D × S . Computation shows that the presentinvolution fixes S ∐ S (i.e., four points) in ∂M , and thus extendsacross any homotopy 4-sphere of the form M ∪ D × S . Hence everyknot with exterior M is strongly +amphicheiral. Is any such knot alsoinvertible or reflexive?On the algebraic side, it is easy to determine the outer automorphismclasses of π , since every automorphism of an HNN extension with finitebase must carry the base to a conjugate of itself. Thus Out ( π ) isgenerated by automorphisms which fix Q (16) set-wise. If α is such anautomorphism then α ( a ) = a i and α ( b ) = a j b , for some odd i = ± j , and α ( t ) = wt ǫ , for some w ∈ π ′ = hh a ii and ǫ = ± ǫ = 1. The images of the relations give equations wb i w − = a j b and wta i + j bt − w − = a i . The first equation implies that w ∈ Q (16), by the uniqueness of normalforms for elements of an HNN extension. Hence j must be even, andso i + j = 2 k + 1, for some k . The second equation then becomes wb k a w − = a i , SMALL INFINITELY-ENDED 2-KNOT GROUP 9 and so k is also even. On following this through, we find that there isan unique such automorphism for each w ∈ Q (16). Four of these areinner automorphisms, given by conjugation by elements of the subgroup h a , ab i , and so we need only consider the automorphisms f and g , givenby f ( a ) = a , f ( b ) = b and f ( t ) = a t , and g ( a ) = a − , g ( b ) = a − b and g ( t ) = at . It is easy to see that f g = gf and f = g = id π .There is also an automorphism h such that h ( a ) = a , h ( b ) = ab and h ( t ) = ( at ) − . This automorphism induces the involution of π/π ′ ∼ = Z .We have f h = hf and ( gh ) = id π , while h is conjugation by a , andso h = id π . Thus Out ( π ) ∼ = ( Z/ Z ) , and is generated by the imagesof f , g and h .The HNN structure determines a Mayer-Vietoris sequence · · · → H ( Q (16)) → H ( π ) → H ( Q (8)) → H ( Q (16)) → . . . , where the right hand homomorphism is the difference of the homo-morphisms induced by the inclusions of the two associated subgroups.Since H ( Q (8)) ∼ = Z/ Z , H ( Q (16)) ∼ = Z/ Z and H ( Q (16)) = 0,it follows that H ( π ) is cyclic of order dividing 8. Since both of thehomomorphisms H ( Q (8)) → H ( Q (16)) induced by the inclusions areinjective, H ( π ) = 0. The automorphisms f and g preserve the associ-ated subgroups, and induce the identity on H ( Q (8)). Hence they alsoinduce the identity on H ( π ). 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Four-Manifolds, Geometries and Knots ,Geometry and Topology Monographs 5,Geometry and Topology Publications (2002). (Revision 2007).[7] Kanenobu, T. Groups of higher-dimensional satellite knots,J. Pure Appl. Alg. 28 (1983), 179–188.[8] Kervaire, M. A. Les noeuds de dimensions sup´erieures,Bull. Soc. Math. France 93 (1965), 225–271. [9] Ratcliffe, J. On the ends of higher dimensional knot groups,J. Pure Appl. Alg. 20 (1981), 317–324.[10] Tollefson, J. L. Involutions on S × S and other manifolds,Trans. Amer. Math. Soc. 183 (1973), 139–152. Mathematics and Statistics, Victoria University,Victoria, V8W 3R4, CanadaSchool of Mathematics and Statistics, University of Sydney,Sydney, NSW 2006, Australia
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