Characterizing slopes for the $(-2,3,7)$-pretzel knot

In this note we exhibit concrete examples of characterizing slopes for the knot $12n242$, aka the $(-2,3,7)$-pretzel knot. Although it was shown by Lackenby that every knot admits infinitely many characterizing slopes, the non-constructive nature of the proof means that there are very few hyperbolic knots for which explicit examples of characterizing slopes are known.


Introduction
Given a knot K ⊆ S 3 , we say that p/q ∈ Q is a characterizing slope for K if the oriented homeomorphism type of the manifold obtained by p/q-surgery on K determines K uniquely. That is, p/q is a characterizing slope for K there does not exist any knot K = K such that S 3 p/q (K) ∼ = S 3 p/q (K ). It was shown by Lackenby that every knot admits infinitely many characterizing slopes and for a hyperbolic knot any slope p/q with q sufficiently large is characterizing [Lac19]. Although these results show the existence of characterizing slopes, the proofs are non-constructive and so there are very few hyperbolic knots for which explicit examples of characterizing slopes are known. Ozsváth and Szabó have shown every slope is characterizing for the figure-eight knot 4 1 [OS19] and recent work of Baldwin and Sivek implies that every non-integer slope is characterizing for 5 2 [BS22]. The aim of this article is to exhibit explicit examples of characterizing slopes for the knot 12n242, also known as the (−2, 3, 7)-pretzel knot. Since 12n242 is a hyperbolic L-space knot -Fintushel and Stern showed that it admits two lens space surgeries [FS80]-it has only finitely many non-characterizing slopes that are not negative integers [McC19]. The following theorem is a quantitative version of this fact. As far as the author is aware, these are the first known explicit examples of characterizing slopes on a hyperbolic knot with genus greater than one.
The key input allowing us to prove Theorem 1 is the fact that 12n242 is one of the knots with smallest volume (up to reflection it one of only three hyperbolic knots with volume smaller than 3.07) [GHM + 21]. A result of Futer, Kalfagianni and Purcell on the change in volume of a hyperbolic manifold under Dehn filling [FKP08] can then be used to restrict potential non-characterizing slopes coming from surgeries on hyperbolic knots with large volume (and satellites thereof). A miscellany of invariants can then be used to rule out non-characterizing slopes coming from hyperbolic knots with small volume. In principle, one could use a similar approach to derive information about the characterizing slopes of the other small volume knots: 4 1 and 5 2 . However, much better results have already been obtained by other means for both of these knots [OS19,BS22], so we restrict our analysis to 12n242.
We note that Theorem 1 says nothing about negative integer characterizing slope. Although there are knots which possess infinitely many integer non-characterizing slopes [BM18], all known examples admit infinitely many non-characterizing slopes of both sign. This suggests that 12n242 (and L-space knots more generally) should admit only finitely many integer non-characterizing slopes. However establishing such a result remains an interesting and challenging problem.
Non-characterizing slopes. Lackenby has shown for a hyperbolic knot K any slope p/q with q sufficiently large is characterizing for K [Lac19]. For example, Theorem 1 shows that q ≥ 49 is sufficiently large for 12n242. However the "sufficiently large" here is inherently dependant on the specific knot in question. To illustrate this dependence, we exhibit a family of hyperbolic two-bridge knots {K q } q≥1 such that for each q, the slope 1 q is non-characterizing for K q . This family is shown in Figure 2 with the details of the construction discussed in Section 4.
Conventions. The following notational conventions will be in force throughout the paper: • When considering a rational number p/q ∈ Q, we will always assume this to be written with p and q coprime and q ≥ 1. • Given two oriented 3-manifolds Y and Y , we will use Y ∼ = Y to denote the existence of an orientation-preserving homeomorphism between them. • For a knot K, we will denote its Alexander polynomial by ∆ K (t). We will always assume this is normalized so that ∆ K (1) = 1 and ∆ K (t) = ∆ K (t −1 ). • Given a knot K in S 3 , we will use mK to denote its mirror.
• An L-space knot is one which admits positive L-space surgeries.
Acknowledgements. The author would like to thank Steve Boyer and Patricia Sorya for interesting conversations. He would also like to acknowledge the support of NSERC and FRQNT.
We will informally refer to the five knots in Theorem 2 as the "low volume knots" and the remaining hyperbolic knots as the "large volume knots". For our purposes it will be useful to note that the volume of 4 1 satisfies (2.1) vol(4 1 ) ≈ 2.0988 ≤ 2.1 and the volume of 12n242 satisfies 2.2. Slope lengths. Let K be a hyperbolic knot in S 3 , that is, S 3 \ K admits a complete finitevolume hyperbolic structure with one cusp. Given a slope σ on K and horoball neighbourhood N of the cusp we can assign a length to σ by considering the minimal length of a curve representing σ on ∂N (measured in the natural Euclidean metric on ∂N ). Since S 3 \ K has a unique cusp, there is a unique maximal horoball neighbourhood of this cusp. We will use K (σ) to denote the length of σ with respect this maximal horoball neighbourhood.
Lemma 3. Let K and K be hyperbolic knots in S 3 with vol(K ) < vol(K). If r and r are slopes such that S 3 Proof. Futer, Kalfagianni and Purcell have shown that if = K (r) > 2π, then we have the following volume bound [FKP08, Theorem 1.1]: Furthermore, since Thurston showed that volume strictly decreases under Dehn filling [Thu82], we have that vol(S 3 r (K)) = vol(S 3 r (K )) < vol(K ). Together these bounds give which can be easily rearranged to give the desired bound on K (p/q).
Next we need a mechanism for converting bound on K (p/q) into bounds on p and q.
2.3. Hyperbolic surgeries on satellite knots. We will use the following result to understand noncharacterizing slopes coming from satellite knots.
Lemma 5. Let K be a satellite knot such that S 3 p/q (K ) hyperbolic for some p/q ∈ Q. Then there is a hyperbolic knot J with g(J) < g(K ) and an integer w > 1 such that S 3 p/q (K) ∼ = S 3 p/(qw 2 ) (J). Moreover, if q ≥ 2, then K is a cable of J with winding number w.
Proof. Let T be an incompressible torus in S 3 \ K . We can consider K as a knot in the solid torus V bounded by T . Thus we can consider K as a satellite with companion given by the core J of V . By choosing T to be "innermost", we can ensure that S 3 \ J contains no further incompressible tori. That is, we can assume that J is not a satellite knot. By Thurston's trichotomy for knots in S 3 , this implies that J is a torus knot or a hyperbolic knot [Thu82]. Since S 3 p/q (K ) is hyperbolic, it is atoroidal and irreducible. Consequently, after surgery the solid torus V must become another solid torus. However, Gabai has classified knots in a solid torus with non-trivial solid torus surgeries, showing that K is either a torus knot or a 1-bridge braid in V [Gab89]. Moreover since solid torus fillings on 1-bridge braids only occur for integer surgery slopes, K is a cable of J unless q = 1. In either event, we have that S 3 p/q (K ) ∼ = S 3 p/q (J), where the slope p/q is determined by the curve bounding a disk after surgering V . Using a homological argument one can show that q = qw 2 , where w > 1 is the winding number of K in V [Gor83, Lemma 3.3]. Since S 3 p/q (K) is a hyperbolic manifold, J cannot be a torus knot. It follows that J must be a hyperbolic knot. The only remaining statement is the inequality g(J) < g(K ). This follows from Schubert's formula for the genus of a satellite knot [Sch53], which asserts that for a knot K = P (J) with pattern P of winding number w ≥ 0, there is a constant g(P ) ≥ 0 such that g(K ) = g(P ) + wg(J).
We obtain the necessary inequality since w ≥ 2.
2.4. The Casson-Walker invariant. It will also be convenient to use surgery obstructions derived from the Casson-Walker invariant [Wal92]. For any rational homology sphere Y , this is a rationalvalued invariant λ(Y ) ∈ Q. Boyer and Lines showed that this satisfies the following surgery formula [BL90]: where ∆ K (1) denotes the second derivative of the Alexander polynomial ∆ K (t) evaluated at t = 1. This formula immediately yields the following observation.
Lemma 6 can be used to obstruct non-characterizing slopes coming from cables.
Lemma 7. Let K and K be knots. If there is K a non trivial cable of K and a non-zero slope p/q ∈ Q such that S 3 p/q (K) ∼ = S 3 p/q (K ), then there are coprime integers r, s, with s ≥ 2 such that ∆ K (1) = (r 2 − 1)(s 2 − 1) 12 + s 2 ∆ K (1) Proof. Suppose that K is the (r, s)-cable of K , where s ≥ 2 is the winding number. By the usual formula for the Alexander polynomial of a satellite knot, we have that where T r,s denotes the (r, s)-torus knot. Taking second derivatives we obtain 1 (2.3) ∆ K (1) = ∆ Tr,s (1) + s 2 ∆ K (1).
1 The reader should note that since ∆ K (t) = ∆ K (t −1 ) we have that ∆ K (1) = 0 2 Since the direct calculation is somewhat involved, we include a derivation for completeness, but relegate it to the appendix.
2.5. An obstruction from ν + . Here we take some input from knot Floer homology. Recall that for a knot K in S 3 , Ni and Wu derived a non-increasing sequence of non-negative integers V 0 (K), V 1 (K), . . . from the knot Floer chain complex which can be used to calculate the d-invariants of surgeries on K. For p/q > 0 and an appropriate identification of Spin c (S 3 p/q (K)) and Spin c (S 3 p/q (U )) with {0, 1, . . . , p − 1}, we have [NW15, Proposition 1.6] Hom and Wu defined the invariant ν + (K) to be the smallest index i for which V i = 0 [HW16]. In particular we have ν + (K) = 0 if and only if V 0 = 0.
Proof. Since −S 3 p/q (K) ∼ = S 3 −p/q (mK), we can assume that p/q > 0. Summing the formula (2.6) over all spin c -structures on S 3 p/q (mK) and S 3 p/q (K) we see that Which implies that S 3 p/q (mK) and S 3 p/q (mK) cannot be homeomorphic.

Proof of Theorem 1
Throughout this section we take K = 12n242. Suppose that p/q = 0 is a non-characterizing slope for K satisfying . The length bound (3.1) implies that the manifold S 3 p/q (K) is hyperbolic and, using Lemma 3, that S 3 p/q (K) cannot be obtained by any surgery on the figure-eight knot 4 1 . By Thurston's trichotomy for knots in S 3 , the knot K is either a torus knot, a hyperbolic knot or a satellite knot. Since torus knots never yield a hyperbolic manifold by surgery, we may ignore the first possibility and restrict our attention to the latter two options. Claim 1. If K is a hyperbolic knot, then q < 49 and |p| < 49(2g(K ) − 1).
Proof. Suppose that K is a hyperbolic knot. Condition (3.1) eliminates the possibility that K is 4 1 . By consideration of the Casson-Walker invariant as in Lemma 6, we see that K is not 5 2 or m5 2 . Using the ν + invariant as in Lemma 8, we see that K is not m12n242. Thus having exhausted all the low volume knots in Theorem 2, we may conclude that vol(K ) > 3.07. Thus by Lemma 3 we have the bound Using Lemma 4, this yields the required bound.
Proof. Suppose that K is a satellite knot and that q ≥ 2. By (3.1), the manifold S 3 p/q (K) is hyperbolic and Lemma 5 applies to show that K is a cable of a hyperbolic knot J such that g(J) < g(K ) and S 3 p/q (J) ∼ = S 3 p/q (K) for some q > q. By the assumption (3.1) we see that J is not 4 1 . Furthermore, applying the Casson-Walker invariant as in Lemma 7, we see that J cannot be 5 2 , m5 2 , 12n242 or m12n242. This is because there are no non-trivial integer solutions with s ≥ 2 to the equations: 24 = (r 2 − 1)(s 2 − 1) 12 + 4s 2 and 24 = (r 2 − 1)(s 2 − 1) 12 + 24s 2 .
Thus, having ruled out all the knots of small volume in Theorem 2, the only remaining possibility is that J must be a knot with vol(J) > 3.07. Thus by Lemma 3 we have the bound Applying Lemma 4 along with the inequalities q < q and g(J) < g(K ) give the required bounds.
Claim 3. If K is a satellite knot and p/q ≥ 9, then |p| < 49(2g(K ) − 1), Proof. Suppose that K is a satellite knot and p/q ≥ 2g(K) − 1 = 9. Since K is an L-space knot, this implies that S 3 p/q (K) is a hyperbolic L-space. By Lemma 5 there is a hyperbolic knot J such that g(J) < g(K ) and S 3 p/q (J) ∼ = S 3 p/q (K) for some q > q. Since ∆ K (1) = 0, [BL90, Proposition 5.1] shows that J is not 12n242. Furthermore, since S 3 p/q (J) is an L-space and none of 4 1 , 5 2 , m5 2 or m12n242 are L-space knots, Theorem 2 allows us to conclude that vol(J) > 3.07. Thus as before we arrive at the bounds Applying Lemma 4(b) and g(J) < g(K ) gives the required bounds.
We now convert these statements into results on characterizing slopes. The bound q ≥ 49 is straight forward.
Claim 4. The slope p/q is a characterizing slope for K whenever q ≥ 49.
Proof. Together Claim 1 and Claim 2 show that p/q is a charactering slope for K whenever K (p/q) ≥ 14.17 and q ≥ 49. However, Lemma 4(a) shows that K (p/q) ≥ 14.17 is automatically satisfied whenever q ≥ 49.
In order to obtain the other conditions on charactering slopes, we need to invoke results linking the genera of K and K Claim 5. The slope p/q is a characterizing slope for K whenever p ≥ max{24q, 441}.
Proof. Since S 3 18 (K) is a lens space, it bounds a sharp 4-manifold. Thus [McC21, Theorem 1.2] applies to show that S 3 p/q (K) bounds a sharp 4-manifold for all p/q ≥ 18. In particular, we may apply [McC21, Theorem 1.1] to show that if p/q ≥ 4g(K) + 4 = 24, then g(K ) = g(K) = 5. Thus Claim 1 and Claim 3 imply that p/q is a characterizing slope for K whenever the conditions p ≥ 24q, p ≥ 49(2g(K) − 1) = 441 and K (p/q) ≥ 14.17 are all satisfied. Lemma 4(b) shows that the bound K (p/q) ≥ 14.17 is redundant, being implied by p ≥ 441. Thus we have a characterizing slope for K if p ≥ 24q and p ≥ 441.
This completes the proof of all bounds in Theorem 1.

Constructing some non-characterizing slopes
In this section we construct some examples of knots with non-characterizing slopes with arbitrarily large denominator. The generic construction is the following. Let L = C ∪ K be a link with two unknotted components and linking number link(C , K ) = ω. Let Y be the manifold obtained by performing 1/n-surgery on both components on L for some non-zero integer n ∈ Z. Since C and K are both unknotted, performing 1/n surgery on one or other of them individually again results in S 3 . Performing such a surgery shows that Y arises by (nω 2 + 1 n )-surgery on the knots K and C, where K is the image of K in the copy of S 3 obtained by surgering C and C is the image of C after surgering K . If one chose L wisely, then the knots K and C will be distinct and thus the slope nω 2 + 1 n will be non-characterizing for K and C.
Using this idea, we can prove the following.
Proposition 10. Let K be a knot with g(K) ≥ 2 which can be unknotted by adding q positive full twists along two oppositely oriented strands. Then 1 q is a non-characterizing slope for K. Proof. The hypothesis on unknotting implies that we can take a link L = C ∪ K with unknotted components such that (a) K can be obtained from K by performing 1/q-surgery on C and (b) C bounds a disk D that intersects K in two oppositely oriented points. If we take the disk D and add a tube that follows an arc of K , we obtain an embedded genus one surface Σ with boundary C which is disjoint from K . Since Σ is disjoint from K , it is preserved under surgery on K and hence shows that the knot C obtained by performing 1/q surgery on K has genus at most one. Since K is assumed to have genus at least two, this implies that C is not isotopic to K and hence that 1/q is a non-characterizing slope for K.
Example 11. Using the preceding proposition, we can show that for every q ≥ 1, there is a hyperbolic 2-bridge knot K q for which 1 q is a non-characterizing slope. Figure 2 depicts a two-bridge knot K q of genus two that can be unknotted by adding q positive full twists along two oppositely-oriented strands. The genera of these knots can be easily verified, since Seifert's algorithm always yields a minimal genus Seifert surface when applied to an alternating diagram [Cro59,Mur58]. Thus Proposition 10 applies to K q .
We also note that sufficiently complicated knots with unknotting number one must always have an non-characterizing slope. Since every slope is characterizing for the trefoil and the figure-eight knot [OS19], we see that the condition on the genus cannot be relaxed.
• If K can be unknotted by changing a positive crossing, then +1 is non-characterizing for K.
• If K can be unknotted by changing a negative crossing, then −1 is non-characterizing for K.
2q − 1 crossings . . . Figure 2. A link K ∪ C , such that twisting along C yields the two-bridge link K q . Proposition 10 implies that K q has 1 q as a non-characterizing slope.