Exceptional surgeries in 3-manifolds
EEXCEPTIONAL SURGERIES IN –MANIFOLDS KENNETH L. BAKER AND NEIL R. HOFFMAN
Abstract.
Myers shows that every compact, connected, orientable 3–manifold with no 2–sphere boundarycomponents contains a hyperbolic knot. We use work of Ikeda with an observation of Adams-Reid to showthat every 3–manifold subject to the above conditions contains a hyperbolic knot which admits a non-trivialnon-hyperbolic surgery, a toroidal surgery in particular. We conclude with a question and a conjectureabout reducible surgeries.
Myers shows that there are hyperbolic knots in every compact, connected, orientable 3–manifold whoseboundary contains no 2–spheres [Mye93]. Might there be such a 3–manifold for which every hyperbolic knothas no non-trivial exceptional surgeries?One approach to showing the answer is
Yes would be to prove that there exists a 3–manifold in whichevery hyperbolic knot has cusp volume larger than 18 so that the 6–Theorem [Ago00, Lac00] would obstructany exceptional surgery. However, [ACF +
06, Corollary 5.2] implies that every closed, connected, orientable3–manifold contains infinitely many hyperbolic knots with cusp volume at most 9. So this approach will notwork. Furthermore, the knots constructed in [ACF +
06, Corollary 5.2] do not necessarily have any exceptionalsurgery, so that work does not address our question.In this short note we demonstrate the answer to the question is actually No by constructing hyperbolicknots with a non-trivial toroidal surgery in any 3–manifold. Theorem 1.
Let M be a compact, connected, orientable –manifold such that ∂M contains no –spheres.There exist infinitely many hyperbolic knots in M that admit a toroidal surgery.Proof. Let M be a compact, connected orientable 3–manifold whose boundary contains no 2–spheres. In[Ike12], Ikeda shows that M contains an infinite family of embedded genus 2 handlebodies in M , each withhyperbolic and anannular complement of its interior where its genus 2 boundary is totally geodesic. Let H be any one of these handlebodies.In Lemma 2 we find a knot K in H that bounds an embedded once-punctured Klein bottle Σ such that H − K is a one-cusped anannular hyperbolic manifold in which ∂H is totally geodesic. Therefore M − K decomposes along ∂H into two anannular hyperbolic manifolds. Thus, following an observation of Adams andReid [AR93, Observation 2.1], M − K is a hyperbolic manifold containing a quasi-Fuchsian surface isotopicto ∂H , and K is a hyperbolic knot in M . (Note that while ∂H is totally geodesic in both M − int ( H ) and H − K , its hyperbolic structure may not be the same in these two manifolds. Hence ∂H is not necessarilytotally geodesic in M − K .)Since K bounds the once-punctured Klein bottle Σ, surgery on K along the slope σ of ∂ Σ produces amanifold M K ( σ ) containing an embedded Klein bottle (cid:98) Σ. The manifold M K ( σ ) will be toroidal unless thetorus ∂ N ( (cid:98) Σ) compresses. However, Lemma 2 shows that K may be further chosen in M so that H K ( σ ) − (cid:98) Σis also a one-cusped anannular hyperbolic manifold in which ∂H is totally geodesic boundary. Therefore, as M K ( σ ) − (cid:98) Σ decomposes along ∂H into the hyperbolic manifolds M − int ( H ) and H K ( σ ) − (cid:98) Σ, it follows that ∂ N ( (cid:98) Σ) must be incompressible in M K ( σ ). (cid:3) Lemma 2.
There is a knot K in a genus handlebody H that bounds a once-punctured Klein bottle Σ sothat H − K is a one-cusped anannular hyperbolic manifold in which ∂H is totally geodesic. Hence surgery on K along the slope σ of ∂ Σ produces a manifold H K ( σ ) containing an embedded Klein bottle (cid:98) Σ . Furthermore, K may be chosen so that H K ( σ ) − (cid:98) Σ is also a one-cusped anannular hyperbolic manifold in which ∂H istotally geodesic.Proof. Figure 2(a) shows a surgery description of a trivial 3–strand tangle in the ball, along with an arc k that has its endpoints on the tangle strands. Figure 2(b) shows the result of an isotopy in which the tangle a r X i v : . [ m a t h . G T ] J a n s more obviously trivial at the expense of elongating the arc k . The double branched cover of this trivial3–strand tangle is a handlebody H in which the arc k lifts to a knot K . Figure 2(c),(d), and (e) illustratethe construction of the knot K in the handlebody H . In (c), two caps with red dual arcs are attached tothe 3–strand tangle to form a trivial 1–strand tangle in the ball. After straightening the strand in (d), thedouble branched cover is taken in (e). The two caps each lift to 2–handles attached to H . The two red arcsin (e) are the co-cores of these two 2–handles, so H is obtained by drilling them out. For lifting the surgerydescription, note that a curve linking the branch locus once with surgery coefficient 1 / a lifts to a singlecurve with surgery coefficient 1 /a . Thus for each pair of integers n, m , we obtain a knot K in a genus 2handlebody H .Figure 2(f) shows a surgery description of the double of ( H, K ) across ∂H , the link K ∪ K in H ∪ H = S × S S × S , obtained by mirroring Figure 2(e) and performing 0 surgery on the components formedfrom the co-cores of the 2–handles and their mirrors. A verified computation in SnapPy [CDGW] confirmsthat the complement of the link K ∪ K in S × S S × S is hyperbolic for choices of n, m ∈ Z with | n | and | m | suitably large. (More specifically, a verified computation in SnapPy shows that after doing the two0-surgeries on the two red components in Figure 2(f), the resulting 6–component link in S × S S × S hasa hyperbolic complement. Then there is a constant N such that the 2–component link complement resultingfrom the surgeries on the green and purple components will be hyperbolic if both | n | > N and | m | > N ;see [Koj88, Lemma 5] or [BHL19, Theorem 3.1].) Since the double has the reflective symmetry in which ∂H is the fixed set, it must be a totally geodesic surface. Hence H − K must be a one-cusped anannularhyperbolic manifold in which ∂H is totally geodesic.In Figure 2(e) one observes that K bounds a once-punctured Klein bottle Σ in H that is disjoint from thetwo curves of the surgery description. As such, Dehn surgery on K in H along the boundary slope σ = ∂ Σproduces the manifold H K ( σ ) which contains the Klein bottle (cid:98) Σ obtained by capping off Σ with a meridionaldisk of the surgery.All that remains is to show that (cid:98)
Σ is essential in the filling. First, we may understand the complementof (cid:98)
Σ through tangles. As apparent in Figure 2(e), the surface Σ may be taken to be invariant under theinvolution of H from the branched covering so that the fixed set intersects Σ in two points and an arc. ThenΣ descends to a disk D containing the arc k in its boundary and meeting the branch locus in the remainderof its boundary and two points in its interior. This disk D may be tracked from its initial quotient of Σ inFigure 2(d) back to Figure 2(a). Now Figure 1(a) shows the exterior of the arc k while Figure 1(b) showsthe rational tangle filling associated to σ –framed surgery on K . In particular, the disk D − k is completedto a disk (cid:98) D containing the closed component of the branch locus as its boundary and meeting the strandsof the branch locus in two interior points. Indeed, the double branched cover of the tangle Figure 1(b) isthe manifold H K ( σ ) in which (cid:98) D lifts to (cid:98) Σ. Finally, Figure 1(c) shows the tangle that is the complement ofa small regular neighborhood of (cid:98) D .Figure 3(a) shows a rational tangle filling of Figure 1(c) with the arc k (cid:48) that is the core of the rationaltangle. This 3–strand tangle is a trivial tangle as made more apparent in Figures 3(b), (c), and (d) whichisotop the tangle while elongating arc k (cid:48) . As before, (c) shows the attachment of two caps with dual arcs and(d) straightens the resulting 1–strand tangle. Figure 3(e) shows the double branched cover which illustratesthe lift of the arc k (cid:48) as the knot K (cid:48) in another genus 2 handlebody H (cid:48) . Again, the two caps each lift to 2–handles attached to H (cid:48) , the two red arcs in (e) are the co-cores of these two 2–handles, and so H (cid:48) is obtainedby drilling them out. Note that the knot K (cid:48) in H (cid:48) depends on the previously chosen pair of integers n, m ofthe surgery description.It now follows that, by construction, H K ( σ ) − (cid:98) Σ is homeomorphic to H (cid:48) − K (cid:48) . We show that H (cid:48) − K (cid:48) is a one-cusped anannular hyperbolic manifold in which ∂H (cid:48) is totally geodesic just as we did for H − K .Figure 3(f) shows a surgery description of the double of ( H (cid:48) , K (cid:48) ) across ∂H (cid:48) , the link K (cid:48) ∪ K (cid:48) in H (cid:48) ∪ H (cid:48) = S × S S × S , obtained by mirroring Figure 3(e) and performing 0 surgery on the components formedfrom the co-cores of the 2–handles and their mirrors. A verified computation in SnapPy [CDGW] confirmsthat the complement of the link K (cid:48) ∪ K (cid:48) in S × S S × S is hyperbolic if both | n | and | m | are suitablylarge. Since the double has the reflective symmetry in which ∂H (cid:48) is the fixed set, it must be a totally geodesicsurface. Hence H (cid:48) − K (cid:48) is a one-cusped anannular hyperbolic manifold in which ∂H (cid:48) is totally geodesic. /2n 1/2m 1/2n 1/2m 1/2n 1/2m (a) (b) (c) Figure 1. (b) A surgery description of a 3–strand tangle in the ball with an unknot compo-nent that bounds a disk intersected twice by the strands. (a) The complement of a rationaltangle in this tangle. (c) The complement of a small neighborhood of the disk bounded bythe unknot component.Since H K ( σ ) − (cid:98) Σ ∼ = H (cid:48) − K (cid:48) , we obtain the desired results whenever | n | and | m | are large enough to besuitably large in both situations. (cid:3) Remark 3.
To give a concrete example, taking n = m = 1 is sufficient for the knots K ⊂ H and K (cid:48) ⊂ H (cid:48) tobe hyperbolic. Certainly, one could verify the hyperbolicity of these knots by hand in the spirit of what wasdone in [AR93], but the argument would take longer. Hence we content ourselves with verified computationsin SnapPy [CDGW].What can be said about other kinds of exceptional surgeries? Considerations of Betti numbers show thatmany closed, compact, orientable 3–manifolds cannot contain a knot with a Dehn surgery to a lens space ora small Seifert fibered space. In light of the Cabling Conjecture [GAnS86] whose proof would imply that nohyperbolic knot in S has a reducible surgery, it is reasonable to expect that there are 3–manifolds in whichno hyperbolic knot admits a reducible surgery. However, we are presently unaware of any 3–manifold knownto not have a hyperbolic knot with a non-trivial reducible surgery. Question 4.
Which compact, connected, orientable –manifolds do not contain a hyperbolic knot with anon-trivial reducible surgery? While non-trivial reducible surgeries on hyperbolic knots in reducible manifolds do exist, see e.g. [HM03],we suspect that manifolds whose prime decompositions have at least 3 summands are candidates.
Conjecture 5.
A closed orientable –manifold with at least summands does not contain a hyperbolic knotwith a non-trivial reducible surgery. Towards the conjecture, suppose K is a hyperbolic knot in a closed orientable 3–manifold M with atleast 3 summands. One may hope that each planar meridional surfaces in the knot complement M − K arising from K intersecting multiple reducing spheres would contribute a certain amount to the length ofthe shortest longitude of K . From this, at least if M had sufficiently many summands, one would be able touse the 6-Theorem to obstruct a non-trivial reducible surgery. However this would also obstruct a toroidalsurgery contrary to Theorem 1. Indeed, it would also contradict [ACF +
06, Corollary 5.2] which shows thatthe topology of M cannot force all longitudes of hyperbolic knots in M to be long.On the other hand, combinatorial structures in knot complements can induce obstructions. For instance,hyperbolic alternating knots in S that have at least 9 twist regions (in twist-reduced diagrams) provide anobstruction the existence of non-trivial exceptional fillings; see [Lac00, Theorem 5.1]. Acknowledgments 6.
KB thanks Jacob Caudell for conversations related to [Cau21, Conjecture 5] thatprompted this note. This work was partially supported by Simons Foundation grant
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Topology Appl. , 49(2):115–127, 1993. /2n 1/2m 1/2n 1/2m1/2n 1/2m1/n -1/n1/m -1/m k’ k’k’ K’K’ K’_ k’ Figure 3. (a) A rational tangle filling of Figure 1(c) with its core arc k (cid:48) . (b) & (c) Anisotopy showing the filled tangle is a rational 3–strand tangle. The arc k (cid:48) is carried along.(c) Attached to the rational 3–strand tangle are two caps with their dual arcs to form a 1–strand tangle in the ball. (d) The 1–strand tangle is straightened. (e) The double branchedcover is formed. Drilling out the red arcs leaves a genus 2 handlebody H (cid:48) containing theknot K (cid:48) that covers k (cid:48) . (f) A surgery description of the double of ( H (cid:48) , K (cid:48) ) is formed. Department of Mathematics, University of Miami, Coral Gables, FL 33146, USA
Email address : [email protected] Department of Mathematics, Oklahoma State University, Stillwater, OK 74078, USA
Email address : [email protected]@okstate.edu