The signature and cusp geometry of hyperbolic knots

We introduce a new real-valued invariant called the natural slope of a hyperbolic knot in the 3-sphere, which is defined in terms of its cusp geometry. We show that twice the knot signature and the natural slope differ by at most a constant times the hyperbolic volume divided by the cube of the injectivity radius. This inequality was discovered using machine learning to detect relationships between various knot invariants. It has applications to Dehn surgery and to 4-ball genus. We also show a refined version of the inequality where the upper bound is a linear function of the volume, and the slope is corrected by terms corresponding to short geodesics that link the knot an odd number of times.


Introduction
In low-dimensional topology, there are two very different types of invariant: those derived from hyperbolic structures on 3-manifolds, and those invariants with connections to 4-dimensional manifolds.Of the latter type, one of the most fundamental invariants is the signature of a knot.Our main goal in this paper is to establish a new and unexpected connection between these two fields.We will show that the cusp geometry of a hyperbolic knot in the 3-sphere encodes information about the signature of the knot.
One of the most important geometric features of a hyperbolic knot K is its maximal cusp.The boundary of this cusp is a Euclidean torus that forms the boundary of a regular neighbourhood of K.This torus is isometric to C/Λ for a lattice Λ in C. The meridian and longitude of the knot give generators µ and λ for Λ.The parallelogram in C spanned by 0, µ, λ, and µ + λ forms a fundamental domain for the action of Λ on C. We introduce a new geometric quantity called the natural slope that measures how far this parallelogram is from being right-angled.It can be defined by the following formula: slope(K) = Re(λ/µ).
Alternatively, natural slope can be defined as follows.Pick a geodesic on the torus C/Λ that represents a meridian.Choose any point on such a geodesic and send off a geodesic orthogonally from this point.It runs along the knot and eventually it comes back to the initial meridian; see Figure 1.In doing so, it has gone along a longitude minus some number s of meridians.This number s is not necessarily an integer because the geodesic may return to a different point along the meridian from where it started.This real number s is the natural slope of K.
We remark that quantities with a resemblance to the natural slope have been defined by other authors [2,9].However, these other quantities do not seem to μ μ Figure 1.A geodesic running in the direction µ ⊥ that is perpendicular to the meridian µ.By the time it returns to the meridian, it has travelled one longitude minus some multiple s of the meridian.This real number s is the natural slope of K. be directly related to natural slope, and none of these previous articles seems to provide a connection between hyperbolic geometry and signature.
Experimentally, starting from the plot in Figure 2, we have observed that the natural slope of K is very highly correlated with 2σ(K), where σ(K) is the signature.See Figure 3 for plots of signature versus slope for knots up to 16 crossings in the Regina census [5] and for random knots generated by SnapPy [8] having 10 to 80 crossings in their SnapPy-simplified forms.Our goal in this paper is to prove that such a surprising connection holds and to explore its consequences.Our first main result, which we prove in Section 4, establishes that slope(K) is approximately equal to 2σ(K), but with an additive error that can be bounded by geometric quantities.
In the above formula, N is a maximal cusp and inj x (S 3 \ K) denotes the injectivity radius of a point x in S 3 \K.Note that although inj(K) −3 appears in the inequality in Theorem 1.1, in practice inj(K) tends not to be particularly small.(See Figure 12 for example.)Experimental evidence, which we provide in Section 7, suggests that c 1 should be quite small: perhaps c 1 = 0.3 suffices.This is based on the largest value 0.234 of |2σ(K) − slope(K)| inj(K) 3 / vol(K) that we managed to obtain by studying a class of knots that are closures of certain braids.
One might wonder whether there is a constant c 2 such that for every hyperbolic knot K.However, we show in Corollary 5.1 that there cannot exist such a constant.We achieve this by exhibiting a sequence of examples that are obtained by twisting 3 strands of a hyperbolic knot.Nevertheless, we can estimate σ(K) in terms of geometric quantities, with an error that is at most a linear function of vol(K).The main term in this estimate is slope(K)/2, but there are also correction terms that are defined using the complex length of short geodesics.From the complex lengths, the following parameters are computed.
Definition 1.2.Let γ be a geodesic in a hyperbolic 3-manifold with complex length cl(γ).Here, cl(γ) is chosen so that Im(cl(γ)) ∈ (−π, π].The twisting parameter tw(γ) = (tw p (γ), tw q (γ)) is the pair (p, q) of coprime integers satisfying the following: (1) p is even and q is odd and non-negative; (2) subject to this condition, the quantity |cl(γ)p + 2πiq| is minimised; (3) if there are several values of (p, q) for which this quantity is minimised, then choose the one that is minimal with respect to lexicographical ordering.
Consider a hyperbolic knot K in S 3 .For any ε ∈ R + less than the Margulis constant ε 3 , let OddGeo(ε/2) denote the set of geodesics with length less than ε/2 and having odd linking number with K.For p, q ∈ Z + , the signature correction term κ(p, q) is given by Definition 4.2 and satisfies κ(p, q) = −σ(T (p, q)) − pq/2, where T (p, q) is the (p, q) torus knot.Then we have the following refinement of Theorem 1.1, which we prove in Section 6, that does not depend on the injectivity radius: Theorem 1.3.Let ε 3 be the Margulis constant, and let ε ∈ (0, ε 3 ).Then there is a constant c 4 (depending on ε) such that for any hyperbolic knot K, the quantities σ(K) and slope(K)/2 − γ∈OddGeo(ε/2) κ(tw p (γ), tw q (γ)) differ by at most c 4 vol(K).Right: Its cusp torus, as provided by SnapPy [8].The longitude is 3.9279 and the meridian is 0.7237 + 1.0160i.Its natural slope is 1.8267 and its signature is 0. Theorem 1.1 has applications in low-dimensional topology.On the one hand, the signature of K controls the cusp shape, which in turn has consequences for the possible exceptional surgeries on K. On the other hand, the cusp shape controls the signature, which has consequence for the 4-ball genus of K. We now provide these applications.
1.1.An application to Dehn surgery.Cusp geometry is well known to control the exceptional surgeries on a knot K. Recall that a slope s on ∂N (K) is said to be exceptional if the manifold K(s) obtained by Dehn filling along s does not admit a hyperbolic structure.
The length of a slope s = q/p ∈ Q, denoted (s), is defined to be the length of any geodesic representative of s = pλ + qµ in the boundary of the maximal cusp.A theorem of Agol [1] and Lackenby [16] states that if (s) > 6, then s is not exceptional.
We relate slope length to natural slope, using the following simple geometric lemma, which we will prove in Section 2.
Lemma 1.4.If K is a hyperbolic knot, then the length of the slope q/p satisfies (q/p) ≥ |p slope(K) + q|.
Given that slope(K) and 2σ(K) are highly correlated, one would therefore expect that any exceptional slope q/p should lie within a short interval around −2σ(K).It is also known that |p| ≤ 8, by a theorem of Lackenby and Meyerhoff [17].Hence, we obtain a bounded set of slopes that contains all the exceptional ones, and that is defined in terms of the signature.
An interesting case is the (−2, 3, 7)-pretzel knot 12n242.This has signature −8 and slope approximately −18.215.It has 7 exceptional slopes: 16, 17, 18, 37/2, 19, and 20.Observe that these slopes are concentrated in a short interval [16,20] that contains both − slope(K) and −2σ(K).This close correlation between the exceptional slopes and −2σ(K) seems to be a phenomenon that had not previously been observed.Specifically, we have the following consequence of our main theorem.
Corollary 1.5.If K is a hyperbolic knot and q/p is a slope satisfying then the manifold K(q/p) obtained by q/p Dehn surgery along K is hyperbolic.
Theorem 1.3 gives a similar bound on slopes resulting in hyperbolic surgeries that does not involve inj(K).1.2.An application to 4-ball genus.One of the most important 4-dimensional quantities associated to a knot K is its 4-ball genus g 4 (K).This is defined to be the minimal possible genus of a smoothly embedded compact orientable surface in the 4-ball B 4 with boundary K.One can also define the topological 4-ball genus g top 4 (K) by considering locally-flat topologically embedded compact orientable surfaces with boundary K.The inequality g 4 (K) ≥ g top 4 (K) is immediate.The following result provides a lower bound on g top 4 (K) in terms of purely hyperbolic data.This follows immediately from our main theorem together with the well-known inequality g top 4 (K) ≥ |σ(K)|/2.
Corollary 1.6.The topological 4-ball genus g top 4 (K) of a hyperbolic knot K satisfies This corollary seems to be the first time that information about the 4-ball genus has been obtained in terms of hyperbolic geometry.Again, Theorem 1.3 gives a similar lower bound on g top 4 (K) that does not involve inj(K).1.3.Spanning surfaces.Theorem 1.1 is proved using a new construction of spanning surfaces with a specified slope.It is of independent interest.Theorem 1.7.There is a constant c 3 such that every hyperbolic knot K in S 3 has an unoriented spanning surface F satisfying Moreover, the boundary slope of this surface is n/1, where n is an even integer that is closest to slope(K).
We prove this in Section 3. The crosscap number of a knot K is the minimum of b 1 (F ) for F an unoriented spanning surface of K.When K is hyperbolic, the above theorem gives an upper bound on a version of the crosscap number where ∂F has slope n/1.
Theorem 1.1 is proved by combining this result with a theorem of Gordon and Litherland [12], which asserts that one can compute the signature of a knot K using any spanning surface F for K; see Theorem 4.1.
Note that slope also gives a lower bound on the Seifert genus: see Proposition 2.5.
1.4.Highly twisted knots.In Section 5, we show the following result for highly twisted knots: Theorem 1.8.Let K be a knot in the 3-sphere, and let C 1 , . . ., C n be a collection of disjoint simple closed curves in the complement of K that bound disjoint discs.Suppose that . ., q n ) be the knot obtained from K by adding q i full twists to the strings going through C i , for each i ∈ {1, . . ., n}.Let i be the linking number between C i and K, when they are both given some orientation.Suppose that 1 , . . ., m are even and m+1 , . . ., n are odd.Then there is a constant k, depending on K and C 1 , . . ., C n , such that the following hold, provided each |q i | is sufficiently large: The slight difference between the behaviour of σ(K(q 1 , . . ., q n )) and the behaviour of slope(K(q 1 , . . ., q n ))/2 as the q i tend to infinity enables us to construct families of knots that show the injectivity radius cannot be dropped from Theorem 1.1.1.5.Methodology.One of the novel aspects of this work was the use of machine learning.We embarked with the aim of discovering new relationships between various 3-dimensional invariants.By using machine learning, we observed an unexpected non-linear relationship between σ(K) and Re(µ), the real part of the meridional translation µ.This led us to define the natural slope, which we observed to have a strong linear correlation with σ(K).Theorems 1.1 and 1.3 are the results of our attempts to prove this correlation.

Hyperbolic knots and natural slope
A knot K is hyperbolic if its complement S 3 \ K admits a complete finitevolume hyperbolic metric.By the Mostow rigidity theorem [21], the hyperbolic structure is unique up to isometry, hence every geometric invariant of the hyperbolic structure on S 3 \ K is a topological invariant of the knot.For example, the volume vol(K) := vol(S 3 \ K) and the injectivity radius inj(K) defined in the introduction are such invariants.
For a pair of coprime integers p, q, the torus knot T (p, q) is one that can be drawn on the surface of the standard torus in the 3-sphere, and winds p times in the longitude direction and q times along the meridian.Given a knot K in S 3 and a knot K in the solid torus S 1 × D 2 , one can form the satellite of K with pattern K by mapping the solid torus in a neighbourhood of K, and considering the image of K .By the work of Thurston [20], a knot is hyperbolic if and only if it is not a torus knot or a satellite knot.In particular, every hyperbolic knot is prime; i.e., not the connected sum of two non-trivial knots.In other words, one can build all knots from hyperbolic knots and torus knots using satellite operations.Definition 2.1.For any hyperbolic knot K, the end of S 3 \K has a neighbourhood called a cusp.The boundary ∂N of a maximal cusp neighbourhood N ⊂ S 3 \ K is a Euclidean torus.Identify ∂N with C/Λ, where C is the complex plane and Λ is a lattice in C. We arrange this identification so that the longitude lifts to a straight line in C starting at 0 and ending at some λ ∈ R >0 .This is the knot's longitudinal translation.Given this normalisation, the meridian lifts to a straight line starting at 0 and ending at some complex number µ with Im(µ) > 0. This is the meridional translation of K.
We remark that the real part of meridional translation Re(µ) in the KnotInfo [19] data set for knots with at most 12 crossings is listed without signs.However, SnapPy [8] does compute the sign for hyperbolic knots.
Note that |µ| ≤ 6, where |µ| denotes the length of the meridian.Indeed, by work of Agol [1] and Lackenby [16], Dehn filling along a slope longer than 6 gives a hyperbolic 3-manifold, while Dehn filling along the meridian is S 3 , which is not hyperbolic.Furthermore, any curve on the cusp torus ∂N has length at least 1.In particular, |µ| ≥ 1.
For the maximal cusp neighbourhood N , we have vol(∂N ) = 2 vol(N ) ≤ 2 vol(K), and vol(∂N ) ≤ |λ||µ|.On the other hand, by a result of Lackenby and Purcell [18], there is a constant C such that, for K alternating, Based on experimental data, one might ask if this also holds for random knots.
Definition 2.3.The natural slope slope(K) of a hyperbolic knot K is defined as follows.Let µ ⊥ be a unit vector at the origin of C orthogonal to µ.Then some multiple of µ ⊥ is equal to λ − sµ for some s ∈ R. Then slope(K) := s.
Lemma 2.4.We have Proof. Figure 6 shows a lift of the cusp torus to the complex plane C. The point λ − sµ is shown (which is a multiple of µ ⊥ ).If we apply the transformation to C that is multiplication by 1/µ, then µ ⊥ becomes purely imaginary.So λ/µ − s is purely imaginary.Hence, s = Re(λ/µ).This is also equal to λ Re(µ)/|µ| 2 .We are now ready to prove Lemma 1.4 from the introduction: On the other hand, by Lemma 2.4, we have slope(K) = λ Re(µ)/|µ| 2 .Hence Slope gives a lower bound on the Seifert genus: Proof.By equation (2.2), we have |λ| ≤ 4πg(K) − 2π.Furthermore, |µ| ≥ 1. Together with Lemma 2.4, we obtain that and the result follows.
3. Proof of Theorem 1.7 The key to proving Theorem 1.7 is the construction of a nice triangulation of a hyperbolic knot complement: Proposition 3.1.There is a constant c 1 such that, for every hyperbolic knot K in S 3 with embedded cusp neighbourhood N , there is a triangulation T of M := S 3 \ (K ∪ int(N )) with the following properties: (1) The number t of tetrahedra of T is at most 3) is a normal curve in ∂M that intersects each edge at most once.
Proof.We remark that the validity of the conclusion in the proposition does not depend on the choice of embedded cusp neighbourhood N .We will pick N as follows.Let N max be the maximal cusp neighbourhood.Retract this to form the embedded cusp neighbourhood N , so each point of ∂N has distance 0.5 from ∂N max .
Note that the Euclidean metric on ∂N is obtained from that of ∂N max by scaling by the factor e −0.5 = 1/ √ e.Let ε := inj(K)/2.We use a variation of Jørgensen's and Thurston's method [22, §5.11] to build the triangulation T .(See also the work of Breslin [4] and Kobayashi-Rieck [15].) We pick a maximal collection of points in ∂M that are all at least ε/8 from each other.We will extend this to a collection of points P in M without adding any new points in ∂M .Our aim is to ensure that the Voronoi diagram for P in M restricts to the Voronoi diagram for P ∩ ∂M in ∂M , where the latter is given its Euclidean metric.Recall that the Voronoi diagram [23][24] corresponding to P is a cell structure of M where the interior of every 3-cell consists of the set of points in M that are closer to a specific point of P than any other point of P .Similarly, the Voronoi diagram for P ∩ ∂M is a cell structure of M where the interior of every 2-cell consists of the set of points in ∂M that are closer (in the Euclidean metric) to a specific point of P ∩ ∂M than any other point of P ∩ ∂M .
The Voronoi diagram for M can be constructed as follows.The universal cover H 3 → S 3 \ K restricts to the universal cover M → M .This set M is obtained from H 3 by removing the interior of the inverse image of N .We may arrange that one component of this inverse image is a horoball N ∞ = {(x, y, z) : z ≥ k} in the upper half-space model for H 3 , for some k > 0. Let P denote the inverse image of P in M .Each cell of the Voronoi diagram for M is the image of a cell for the Voronoi diagram for P in M .Each 2-cell that does not lie in ∂ M is equidistant from two points of P .Hence, it is totally geodesic.Our aim is to ensure that each such 2-cell that intersects the horosphere ∂N ∞ is equidistant between two points of P ∩ ∂N ∞ .This will imply that the 2-cell intersects ∂N ∞ in a Euclidean geodesic arc.The union of these arcs forms the 1-skeleton of the Voronoi diagram for P ∩ ∂N ∞ in ∂N ∞ .Thus, we can deduce that the Voronoi diagram for P in M restricts to the Voronoi diagram for P ∩ ∂M in ∂M .
We now describe how the set P is chosen.We have already picked a maximal collection of points in ∂M that are all at least ε/8 from each other.This set will be P ∩ ∂M .We then add points to this set that lie in the interior of M , but subject to the condition that each of these points in the interior of M has distance at least ε/4 from the other points in the set.We stop when it is no longer possible to add any further points with this property.Let P be the resulting set of points.
By our choice of P , each point in ∂M has distance less than ε/8 from some point of P ∩ ∂M .It also has distance at least ε/8 from each point of P ∩ int(M ).Thus, for each point of ∂M , each of its closest points in P also lies in ∂M .Now consider a 2-cell of the Voronoi diagram for M that intersects ∂N ∞ but does not lie in ∂N ∞ .This is equidistant between two points p 1 and p 2 of P .The intersection between this 2-cell and ∂N ∞ is an arc.Let x be any point in the interior of this arc.Then x is equidistant between p 1 and p 2 , and these are the closest two points of P to x.As argued above, any point of P that is closest to x must lie in ∂ M .We will show that, in fact, p 1 and p 2 lie in ∂N ∞ .Suppose not.Then one of these points lies in ∂ M − ∂N ∞ .The shortest arc from x to ∂ M − ∂N ∞ must run through the inverse image of ∂N max .One component of this inverse image is a horosphere about the point at infinity, with distance 0.5 from ∂N ∞ .Hence, the length of this arc is at least 0.5.On the other hand, each point in ∂ M has distance less than ε/8 from some point of P ∩ ∂ M .We will show below that ε/8 < 0.12 < 0.5, and hence this is a contradiction.
Thus, we have indeed guaranteed that the restriction to ∂M of the Voronoi diagram for P in M is the Voronoi diagram for P ∩ ∂M in ∂M , as claimed.We now subdivide each 2-cell of the Voronoi diagram for M into triangles without introducing any new vertices, and subdivide each 3-cell into tetrahedra by coning off from the point of P lying in it, obtaining the triangulation T of M .Since the restriction of the Voronoi diagram to ∂M agrees with that arising from its Euclidean metric, this implies that each triangle of T in ∂M is straight.
Since the open balls of radius ε/16 about the points of P are pairwise disjoint, where B(ε/16) is a ball in H 3 of radius ε/16.We claim that the number t p of tetrahedra of T incident to a point p ∈ P is at most a universal constant k.Indeed, when p lies in the interior of M , t p is exactly the number of triangles in the boundary 2-sphere S of the 3-cell of the Voronoi diagram containing p.When p lies in the boundary of M , t p is the number of triangles in this sphere that are not incident to p.When a vertex of one of these triangles lies in the interior of M , it is equidistant from at least four points of P , one of which is p.When a vertex of the triangles lies on the boundary of M , it is equidistant from at least three points of P , one of which is p.So, a vertex in S is specified by choosing two or three other points of P , each of which is at most ε/2 from p.The ball B(p, ε/2) is embedded in S 3 \ K, since ε/2 = inj(K)/4, and hence lifts to a ball B in H 3 .The balls of radius ε/16 about the inverse image of P in B are disjoint, and lie within B(9ε/16).So, the number of points of P at most ε/2 from p is bounded above by It follows that t p ≤ k := k0 3 .Then the total number of tetrahedra We may pick the Euclidean geodesic representative for the slope ν so that it misses the vertices of T .Hence ν is a normal curve, because it is a Euclidean geodesic and each triangle of T in ∂M is straight.We now show ν does not intersect any triangle in ∂M more than once.Let D be a fundamental domain in ∂N max with sides µ and ν. (See Figure 7.) We will show that the perpendicular distance h between the sides of D that are parallel to ν is at least 0.55.Hence, the perpendicular distance between sides of the corresponding fundamental domain in ∂N ∞ is at least 0.55/ √ e > 0.33.On the other hand, we will show that the length of each edge of T in ∂M is at most 0.23.This will imply that in the triangulation of ∂M , no triangle can run in D between these opposite sides, and hence that T satisfies property (2).This will complete the proof.
According to a theorem of Cao and Meyerhoff [6], the area A of the boundary of the maximal cusp is at least 3.35.Let θ be the angle of two of the four corners of D satisfying 0 < θ ≤ π/2.Say that this angle is at the vertex v 1 of D, and label the remaining vertices v 2 , v 3 , v 4 , so that the line joining v 1 to v 2 has slope µ.
Let b be the perpendicular projection of v 4 onto the line joining v 1 and v 2 .We claim that b lies between v 1 and v 2 , or possibly equals one of these vertices.Place v 1 at the origin in the complex plane.Then v 2 = ±µ and v 4 = λ − nµ.Now, by the definition of s = slope(K), the perpendicular projection of λ − sµ onto the line through v 1 and v 2 is v 1 .Hence, the perpendicular projection b of λ − nµ onto this line has distance |n − s| |µ| from v 1 .But n is a closest even integer to s, and so |n − s| ≤ 1.Therefore, b lies between v 1 and v 2 , or is equal to one of these points, as claimed.and so Therefore, The square of the reciprocal of this expression is It is easy to check that this is a convex function of |µ| and hence its maximal value over the interval 1 ≤ |µ| ≤ 6 occurs when |µ| = 1 or 6.It also is maximised by taking A as small as possible, in other words A = 3.35.We deduce that h is at least 6 × (3.35) Hence, the perpendicular distance between sides of the corresponding fundamental domain in ∂N ∞ is at least 0.55/ √ e > 0.33.We now compare this to the maximal length of an edge of T in ∂M .Each triangle of T in ∂M lies within a disc centred at a point of P ∩ ∂M with radius at most ε/8.Hence each triangle has side length at most ε/4 = inj(K)/8.Now the length L of the shortest slope s on ∂N max is at most |µ| ≤ 6.This gives an upper bound on inj(K), as follows.By applying an isometry to hyperbolic space, we may arrange that a component of the inverse image of N max in upper half space is {(x, y, z) : z ≥ 1}.We may also arrange that a covering transformation corresponding to s is (x, y, z) → (x + L, y, z).It therefore sends (0, 0, 1) to (L, 0, 1).The hyperbolic distance between these points is at most where e ∂ is the number of edges of F in ∂M .As v ≥ e ∂ , we obtain that where the second inequality is property (1) of T in Proposition 3.1.

The knot signature
Another fundamental knot invariant is the signature σ(K).Given a Seifert surface S for K; i.e., a compact, oriented, and connected surface with boundary K, one can define the Seifert form The signature is a 4-dimensional invariant, in the sense that it gives a lower bound on the topological 4-ball genus g top 4 (K), which is the minimal genus of a compact, oriented, locally-flat, connected surface bounded by K in the 4-ball B 4 .
One can also compute the signature of a knot from unoriented surfaces using the work of Gordon and Litherland [12].Let F be an unoriented surface bounding a knot K in S 3 .Let {b 1 , . . ., b n } be a basis of H 1 (F ), and let b i be the double push-off of b i into S 3 \ F .Then the Goeritz matrix G F is an n × n symmetric matrix with (i, j)-th entry lk(b i , b j ) for i, j ∈ {1, . . ., n}.Furthermore, the normal Euler number e(F ) of F is defined to be −lk(K, K ), where K is the framing of K given by F . Gordon and Litherland proved the following: Theorem 4.1.Let F be an unoriented surface bounding the knot K in S 3 .Then where σ(G F ) is the signature of the Goeritz matrix.
We are now ready to show how Theorem 1.1 follows from Theorem 1.7.
Proof of Theorem 1.1.Let F be the surface provided by Theorem 1.7, with boundary slope ν = λ − nµ, where n is a closest even integer to slope(K).Let G F be the Goeritz matrix of F .Since for the absolute constant Indeed, for any hyperbolic knot K, we have inj(K) ≤ 1.82 as shown in the proof of Proposition 3.1, and vol(K) > 2.0298, with the figure eight knot having the smallest volume, by Cao and Meyerhoff [6].
In the following definition, we introduce the signature correction κ(p, q) for integers p and q, which is related to the signature of the (p, q)-torus knot.The correction terms in Theorem 1.3 are defined in terms of κ(p, q).Definition 4.2.For any pair of positive integers (p, q), we define the signature correction κ(p, q) recursively as follows.
It is reasonably clear that this gives a well-defined value of κ(p, q).This is because it defines κ(p, q) uniquely when p = q, and when p = q, it defines κ(p, q) in terms of some κ(p , q ) where either q < q, or q = q and p < p.However, the rationale for the definition comes from the following fact due to Gordon, Litherland, and Murasugi [13]: Theorem 4.3.The signature of the (p, q)-torus link T (p, q) satisfies σ(T (p, q)) = −pq/2 − κ(p, q).The signature correction κ(p, q) arises naturally as the signature of the Goeritz form of a surface bounding the (p, q)-torus knot, as follows.
Lemma 4.4.Let V be the standard solid torus in S 3 , and let T (p, q) be the curve on ∂V that is the (p, q)-torus knot, where p is even and q is odd.Thus, p is the winding number of T (p, q) in V .Then there is a compact unoriented surface F in V with boundary T (p, q), and σ(G F ) = −κ(p, q) for any such F .Proof.Since p is even, T (p, q) is trivial in H 1 (V ; Z 2 ).It therefore bounds an unoriented surface F in V .Applying Gordon and Litherland's signature formula (Theorem 4.1) to F , we deduce that The push-off K of ∂F into F has linking number pq with ∂F .To see this, observe that K is homologous in V to p times a core curve γ of V .Similarly, ∂F is homologous in the solid torus cl(S 3 \ V ) to q times its core curve γ, which is a meridian of γ .Thus lk(∂F, K ) = pq lk(γ, γ ) = pq, where the final equality is Theorem 4.3.
Lemma 4.5.Let A be a non-singular square matrix with real entries.Let A + be a non-singular matrix obtained from A by adding a final row and final column.Then σ(A + ) is either σ(A) − 1 or σ(A) + 1.
Proof.Let λ 1 ≤ • • • ≤ λ n be the eigenvalues of A, and let λ + 1 ≤ • • • ≤ λ + n+1 be the eigenvalues of A + .Cauchy's interlacing theorem states that Hence, the number of negative eigenvalues of A + is at least the number of negative eigenvalues of A, and similarly the number of positive eigenvalues of A + is at least the number of positive eigenvalues of A.
Lemma 4.6.Let V be a solid torus embedded in S 3 .Pick a slope λ on ∂V that has winding number 1 in V .Let K be the knot on ∂V that has slope pλ + qµ, where µ is the meridian of V , and where p is even and q is odd.Then K bounds a compact unoriented surface F in V with the property that the Goeritz form Proof.Because p is even, K bounds a compact surface F in V .We may pick a basis e 1 , . . ., e n for H 1 (F ) so that e 1 , . . ., e n−1 have zero winding number around V .Let V be an embedding of V in S 3 such that K is sent to T (p, q).Let F be the image of F .
We claim that the Goeritz forms G F and G F agree on the first n − 1 rows and columns.To prove this, we view V as the regular neighbourhood of a standard unknot embedded in the horizontal plane.Then, up to isotopy, V can be obtained from V by applying Reidemeister moves and crossing changes to this unknot.None of these moves affects the first n − 1 rows and columns of the Goeritz form, for the following reason.Any given entry of the Goeritz form is lk(b i , b j ) for a suitable curve b i in the surface and b j the double push-off of another curve in the surface.When the entry of the Goeritz form lies in the first n − 1 rows and columns, these curves b i and b j have zero winding number around the solid torus.Hence, geometrically, b i winds an equal number of times around the solid torus in opposite directions, as does b j .So, when we perform a Reidemeister move or a crossing change to the solid torus, and we compare the resulting projections of b i ∪ b j to the horizontal plane, the sum of the signs of the crossings between b i and b j remains unchanged.This sum is 2 lk(b i , b j ).This proves the claim.

Highly twisted knots
The following is Theorem 1.8 from the introduction: Theorem 1.8.Let K be a knot in the 3-sphere, and let C 1 , . . ., C n be a collection of disjoint simple closed curves in the complement of K that bound disjoint discs.Suppose that . . ., q n ) be the knot obtained from K by adding q i full twists to the strings going through C i , for each i ∈ {1, . . ., n}.Let i be the linking number between C i and K, when they are both given some orientation.Suppose that 1 , . . ., m are even and m+1 , . . ., n are odd.Then there is a constant k, depending on K and C 1 , . . ., C n , such that the following hold, provided each |q i | is sufficiently large: One can use this to show that the factor inj(K) −3 cannot simply be dropped from Theorem 1.1 (cf.Conjecture 7.4 for what we expect for random knots): Corollary 5.1.There does not exist a constant c 2 such that for every hyperbolic knot K.
Proof of Theorem 1.8.The knot K(q 1 , . . ., q n ) is obtained by performing −1/q i surgery on C i , for each i ∈ {1, . . ., n}.Let L denote the link Thurston's Hyperbolic Dehn Surgery theorem, as all the |q i | tend to infinity, the hyperbolic structures on S 3 \ K(q 1 , . . ., q n ) tend in the geometric topology to the hyperbolic structure on S 3 \ L. In fact, more it true.Fix a horoball neighbourhood N of the cusps of S 3 \ L that is small enough so that the cusp torus T surrounding K lies in the complement of N .Then, if all the |q i | are sufficiently large, the inclusion (S 3 \ L) \ N → S 3 \ K(q 1 , . . ., q m ) is a bi-Lipschitz homeomorphism onto its image, with bi-Lipschitz constants that tend to 1 as all the |q i | tend to infinity.(See [3] for instance.) Let λ(K) be the longitude and µ(K) the meridian of K.These form a basis of the lattice Λ(K), where the cusp torus of K in S 3 \ L is C/Λ(K).Let γ be the image of λ(K) and µ the image of µ(K) on the cusp torus C/Λ(K(q 1 , . . ., q n )) of K(q 1 , . . ., q n ).The curves γ and µ form a basis of the lattice Λ(K(q 1 , . . ., q n )).So, we may assume that γ and µ are approximately constant complex numbers when |q i | are large.However, we have not normalised the lattice so that γ is real.We know that there is some for some s ∈ R. Here, N , µ ⊥ , γ, s , and µ all depend on q 1 , . . ., q n .But N and s tend to fixed real numbers as the |q i | go to infinity.
The key observation is that γ is not necessarily the longitude λ for K(q 1 , . . ., q n ).In fact, the linking number between γ and K(q 1 , . . ., q n ) is i 2 i q i ; see Figure 8.For suppose that the disc bounded by C i intersects K in p − points of negative sign and p + points of positive sign.So, i = p + −p − .Then, when we perform a full twist about C i , we introduce 2(p + + p − ) 2 new crossings between γ and K(q 1 , . . ., q n ).Of these, 2(p 2 + + p 2 − ) have positive sign and 4p + p − have negative sign.So the linking number between γ and K(q 1 , . . ., q n ) changes by . Each full twist about C i changes the linking number between γ and K(q 1 , . . ., q n ) by 2  i .
It follows that and hence We conclude that slope(K(q 1 , . . ., q n )) = s − n i=1 i q 2 i .On the other hand, there is a constant k such that |s | ≤ k if |q 1 |, . . ., |q n | are sufficiently large, which implies the first inequality of the theorem.
Recall that m+1 , . . ., n are odd.Suppose that q m+1 , . . ., q r are even and that q r+1 , . . ., q n are odd.Let µ m+1 , . . ., µ r be meridians for C m+1 , . . ., C r , respectively.Let F be a spanning surface for Since this link has even linking number with each component of C 1 ∪• • •∪C r , we may choose this spanning surface to be disjoint from these components.We can view this surface as properly embedded in the exterior of For i ∈ {r + 1, . . ., n}, the curve F ∩ ∂N (C i ) has slope equal to a longitude plus an odd number of meridians.By choosing the surface appropriately, we can ensure that this odd number is 1.Now perform surgery along C 1 , . . ., C n .The surface becomes a surface in the exterior of the new link.On ∂N (C i ) for i ∈ {m + 1, . . ., r}, it now has slope equal to a meridian plus q i longitudes.On ∂N (C i ) for i ∈ {r + 1, . . ., n}, it is a meridian plus q i + 1 longitudes.Since we are assuming that |q i | is sufficiently large, we can suppose that q i = 0, −1 and hence that this slope is not meridional.Within each solid torus N (C m+1 ), . . ., N (C n ), we can now insert a surface, as shown in Figure 9.Let F denote the resulting spanning surface of K(q 1 , . . ., q n ).Here, q i = 5 or 6.Also shown in Figure 9 is a collection of generators for H 1 (F ∩ N (C i )) for i ≥ m + 1.Note that H 1 (F ∩ N (C i )), for i ≥ m + 1, form direct summands of H 1 (F ).So we can extend this set of generators to a basis of H 1 (F ), by adding further elements of H 1 (F ).The associated Goeritz form G F is diagonal when restricted to the rows and columns corresponding to Each C i gives rise to |q i |/2 diagonal entries when m + 1 ≤ i ≤ r and |q i + 1|/2 entries when r + 1 ≤ i ≤ n.These entries are +1 when q i is positive and −1 when q i is negative.Hence, the signature of this matrix differs from n i=m+1 q i /2 by at most (n − r)/2.So, applying Lemma 4.5, Theorem 4.1 due to Gordon and Litherland states that . Here The second inequality of the theorem follows immediately.
Note that if we set ε = ε 3 /2, then c 4 then becomes a universal constant.However, given the present uncertainty about the precise value of ε 3 , we do not specify ε definitively.Definition 6.1.Let γ be an embedded closed geodesic in the hyperbolic 3-manifold M , and let N (γ) be a regular neighbourhood of γ consisting of points at most a certain distance r from γ.Let γ be a component of the inverse image of γ in H 3 , which we can take to be {(0, 0, z) : z > 0} in the upper half-space model.Let N (γ) be the component of the inverse image of N (γ) containing γ.We let λ be the slope on ∂N (γ) that has winding number one around N (γ) and that lifts to a path in N (γ) starting on the half-plane {(x, y, z) : y = 0, x ≥ 0} and with interior that is disjoint from the half plane {(x, y, z) : y = 0, x ≤ 0}.In the event that this path ends precisely on the half plane {(x, y, z) : y = 0, x ≤ 0}, λ is chosen so that it avoids {(x, y, z) : y ≤ 0, x = 0}.Then λ is called the canonical longitude of γ.Note that it does not necessarily have zero linking number with γ.
There is the following alternative interpretation of the canonical longitude in terms of the complex length of γ.We give T = ∂N (γ) its inherited Riemannian metric.This is homogeneous, since any two points of T differ by an isometry of T .The metric on T therefore has constant curvature, which must be zero by the Gauss-Bonnet theorem.It is therefore Euclidean.We can represent it as the quotient of the Euclidean plane E 2 by a lattice L. Each slope on T corresponds to a lattice point.We can assume that the lattice point corresponding to the meridian is a purely imaginary number µ.As the circumference of a radius r circle in the hyperbolic plane is 2π sinh(r), we have µ = 2π sinh(r)i, where r is the radius of the tube around γ. Let ν be a geodesic in T that is perpendicular to a meridian and that starts and ends on the meridian (but not necessarily at the same point).Then (ν) = cosh(r) Re(cl(γ)), where cl(γ) is the complex length of the geodesic γ and Re(cl The significance of the twisting parameter arises from the following lemma.Lemma 6.2.Let M be a hyperbolic 3-manifold and ε ∈ (0, ε 3 ).Let γ be a geodesic in M with (γ) < ε/2.Let T be the toral boundary component of M (0,3ε/4] that encloses γ, let µ ⊂ T be a meridian of γ and let λ be the canonical longitude.If (p, q) = (tw p (γ), tw q (γ)), then for some constant c 5 depending only on ε.
We now give an upper bound on L. Let s be a slope on T whose lift s to H 3 satisfies d H 3 (s(0), s(1)) = 3ε/4.This is again possible since T ⊂ ∂M (0,3ε/4] .If r ≤ 2ε, then L ≤ |µ| ≤ 2π sinh(2ε).Now suppose that r > 2ε.Let N (γ) ⊂ V be a regular neighbourhood of γ of radius r − ε.Let β be a geodesic in H 3 connecting s(0) and s(1), and let β be its projection to M .Then β is a geodesic homotopic to s of length 3ε/4, which hence lies in V \ N (γ).The nearest point projection ϕ : V \ N (γ) → T satisfies (ϕ(β)) ≤ l 0 (β) = l 0 (3ε/4) for a constant l 0 depending only on ε.Hence L ≤ k 0 := max(2π sinh(2ε), 3l 0 ε/4), as claimed.A consequence of L ≥ k 0 is that Area(T ) ≥ a 0 for a constant a 0 depending only on ε.Indeed, a disc D of radius L/2 on T about an arbitrary point of T is embedded, so We claim the length of the shortest curve in any nontrivial class in H 1 (T ; Z 2 ) is at most k 1 Area(T ) for a constant k 1 depending on ε.Indeed, let γ ⊥ 0 : I → T be a geodesic arc starting and ending on the shortest geodesic γ 0 and orthogonal to it.Then (γ ⊥ 0 ) = Area(T )/L.The points γ ⊥ 0 (0) and γ ⊥ 0 (1) divide γ 0 into two arcs, one of which has length at most L/2.Let γ 1 be a geodesic representative of the closed curve that runs along γ ⊥ 0 and then along the shorter of the two arcs in γ 0 .We obtain that (γ 1 ) ≤ L/2 + Area(T )/L.
The curves γ 0 and γ 1 give a basis for H 1 (T ; Z 2 ).Hence, the shortest representative of every nontrivial class in H 1 (T ; Z 2 ) is at most L + (L/2 + Area(T )/L).As L ∈ [k 0 , k 0 ] and Area(T ) ≥ a 0 , we have So there is some slope (a, b) on T with a even and b odd such that for some constant k 1 depending on ε.
Let T be the torus obtained from T by scaling by tanh(r) in the ν-direction.As r > ε/8, we have tanh(r) ∈ (tanh(ε/8), 1).Since tanh(r) < 1, the shortest slope (p, q) on T with p even and q odd has length at most k 1 Area(T ).The lattice that specifies T is generated by Hence, by Definition 1.2, the slope pλ + qµ on T is the shortest among slopes for which p is even and q is odd.Therefore, its length on T is at most k 1 Area(T ).So So we can set c 5 := k 1 / tanh(ε/8), which concludes the proof of the lemma.
Proof of Theorem 1.3.We claim that we can build a triangulation T of M [3ε/4,∞) with the following properties: (1) The number of tetrahedra of T is at most c vol(K), where c depends on ε.
(2) If n is a closest even integer to slope(K), then some Euclidean geodesic with slope λ − nµ on ∂N (K) is a normal curve in ∂M [3ε/4,∞) that intersects each edge of T at most once.(3) On the component T of ∂M [3ε/4,∞) corresponding to ∂N (K), the edges of T are Euclidean geodesics with length at most ε/15.We follow the construction in the proof of Proposition 3.1, but with different constants.We pick a maximal collection of points in ∂M [3ε/4,∞) that are at least ε/30 apart.We then add points to this collection in the interior of M [3ε/4,∞) that have distance at least ε/15 from each other and from the earlier points.We stop when it is not possible to add any further points, and denote the resulting collection by P .We then form the associated Voronoi diagram, subdivide the 2-cells of this cell structure into triangles without adding any new vertices, and then triangulate each 3-cell by coning from the relevant point of P .Let T be the resulting triangulation of M [3ε/4,∞) .
Exactly the same argument as in the proof of Proposition 3.1 gives that the number of tetrahedra of T is at most c vol(K), where c depends on ε.The length of each edge in ∂M [3ε/4,∞) is now at most ε/15, because we took points that were at least ε/30 apart, rather than at least ε/8 apart.Thus, all that needs to be proved are that the edges of T in T are Euclidean geodesics and that there is a Euclidean geodesic with slope λ − nµ on ∂N (K) which is a normal curve in ∂M [3ε/4,∞) that intersects each edge of T at most once.
We start by showing that the edges of T in T are Euclidean geodesics.Following the proof of Proposition 3.1, we need to show that, for each point x on T , its closest points in P all lie in T and have distance at most ε/30 from x.We also need to show that the shortest geodesic joining x to any of these points remains within the cusp.The first of these statements holds by our choice of P .
Note that T lies within M (0,ε] .By definition of the Margulis constant, M (0,ε] consists of a cusp and some regular neighbourhoods of geodesics with length at most ε.The Euclidean metrics on T and the cusp component of ∂M (0,ε] differ by a Euclidean scale factor of 4/3, and hence are hyperbolic distance ln(4/3) > 0.287 from each other.On the other hand, the 3-dimensional Margulis constant satisfies ε 3 < 0.775.(See the discussion in [11,Section 1.1].)Hence, ε/30 < ln(4/3).We deduce that for each point x in T , any shortest geodesic to a closest point in P must lie in the cusp.This implies that the restriction to T of the Voronoi diagram for P in M is equal to the Voronoi diagram for P ∩ T in T with its Euclidean metric.In particular, the edges of T in T are Euclidean geodesics, as claimed in (3).
Let N max be a maximal cusp neighbourhood around K. Then N max contains T .This torus T is a scaled copy of ∂N max .It is scaled so that for each point on T , two lifts of this point in H 3 are exactly 3ε/4 apart and no two lifts of this point are any closer than this.Say that d is the hyperbolic distance between T and ∂N max .Then the scale factor taking ∂N max to T is e −d .Now the meridian slope on ∂N max has length at most 6.Hence, the meridian slope on T has length at most 6e −d .So any point on T has two lifts to H 3 that are less than 6e −d apart, and therefore, 3ε/4 ≤ 6e −d .As in the proof of Proposition 3.1, let h be the length in ∂N max of a Euclidean geodesic that starts and ends on a geodesic with slope λ − nµ and that is orthogonal to this geodesic.It was shown there that h ≥ 0.55.Hence, the length of the corresponding geodesic on T is at least 0.55e −d ≥ (0.55/6)(3ε/4).On the other hand, the length of each edge of T on T is at most ε/15, and ε/15 < (0.55/6)(3ε/4).Hence, each such edge can intersect any geodesic with slope λ − nµ at most once.This establishes the claimed properties of T .
Let T 1 , . . ., T m be the components of ∂M [3ε/4,∞) , where T i encircles a geodesic γ i ∈ OddGeo(ε/2).Let tw(γ i ) = pλ i + qµ i , where λ i is the canonical longitude on T i and µ i is the meridian, and let C i be a curve on T i with this slope.Then Realise each C i as a Euclidean geodesic in T i missing the vertices of T i , and hence as a normal curve in T i .Since (C i ) ≤ c 5 Area(T i ) and by property (3) of the triangulation T , the normal representative of C i intersects each edge of T at most c 5 Area(T i ) times for a constant c 5 depending only on ε.
We claim that there is a connected normal curve C i in T i for i ∈ {1, . . ., m} with the following properties: (1) C i and C i are equal in H 1 (T i ; Z 2 ); (2) C i intersects each edge of T at most once.This is constructed as follows.For each edge of T that intersects C i an odd number of times, replace this intersection by a single point of intersection.These will be the points of intersection between C i and the 1-skeleton of T .Since |C i ∩ ∂t| is even for each triangle t of T , we have |C i ∩ ∂t| ∈ {0, 2}.If |C i ∩ ∂t| = 2, join the two points of C i ∩ ∂t by a normal arc of C i .The result is a collection of simple closed curves in T i that are mod 2 homologous to C i .If any of these curves are inessential in T i , remove them.The resulting curves are essential in T i .Since they are non-trivial in mod 2 homology, they consist of an odd number of parallel copies of a curve.If this odd number is greater than one, remove all but one of these curves.The result is C i , and we write Let C be the union of C and a normal curve where n is a closest even integer to slope(K).We claim that C bounds an unoriented surface in M [3ε/4,∞) .As n is even, there is a compact surface properly embedded in the exterior of K with boundary slope (1, −n).It intersects each geodesic with length at most ε/2 in a collection of meridians.For a geodesic with odd linking number with K, the number of these meridians is odd.For the others, it is even.As C i is homologous to the meridian of T i over Z 2 , we may modify the surface so that its boundary is precisely C .This proves the claim.
As C intersects each edge of T at most once, we can find a surface F in for some constant c 6 , just like in the proof of Theorem 1.7.Now C i and C i are equal in H 1 (T i ; Z 2 ).Hence, we may insert a compact connected surface F i into a regular neighbourhood N (T i ) of T i with ∂F i = C i ∪ C i .Since C i and C i intersect each edge of T at most c 5 Area(T i ) times, this surface may be chosen so that −χ(F i ) ≤ c 5 Area(T i ) for a constant c 5 depending only on ε.Hence, the surface for a constant c 7 that depends only on ε.Here, the last inequality follows from the observation that Area(T i ) ≤ c 8 vol(N (T i )) for some constant c 8 , where and V i is the solid toral component of M (0,3ε/4] of tube radius r i that encloses the geodesic γ i ∈ OddGeo(ε/2).
In each V i , we construct the surface provided by Lemma 4.6 with boundary C i = C ∩ V i .We attach these surfaces to F to form a surface F + .We now specify a basis for H 1 (F + ).We start by picking a basis for H 1 (V 1 ∩ F + ).We arrange that all but one of these basis elements have zero winding number around V 1 .We then continue to V 2 , and so on.We then extend this to a basis for H 1 (F + ) by adding some oriented curves in F .We order this basis as follows into n + 1 blocks.In the first block, we place all the basis elements of H 1 (V 1 ∩ F + ) that have zero winding number around V 1 .In the second block, we do the same for V 2 , and so on.In the final block, we place all the remaining basis elements.We saw in the proof of This, together with equation (6.3), imply that the number of elements in this final block is bounded above by a linear function of vol(K).Let G be the submatrix of the Goeritz form G F+ consisting of the first n blocks.By Lemma 4.5, σ(G) and σ(G F ) differ by at most the number of elements in the final block.Note that G is block diagonal.For the block corresponding to V i , the signature differs from σ(G F+∩Vi ) by at most one by Lemma 4.5.On the other hand, |σ(G F+∩Vi ) + κ(tw p (γ i ), tw q (γ i ))| ≤ 2 by Lemma 4.6.Hence, for some constant c 9 .By Gordon and Litherland's theorem (Theorem 4.1), The result follows as n is a closest even integer to slope(K).

Experimental data and some conjectures about random knots
We set out to find links between hyperbolic and 4-dimensional knot invariants.Initial scatter plots compared some 4-dimensional invariants (the signature and Heegaard Floer invariants τ , ν, and ε), the crossing number, and several hyperbolic invariants (volume, meridional and longitudinal translations, and the Chern-Simons invariant).As σ is strongly correlated to τ , ν, and ε, we decided to only focus on σ, which is more classical and easier to compute.
The strongest and most surprising correlation was between the signature and the real part of the meridional translation; see Figure 2.There were some more predictable relationships among the hyperbolic invariants.which indicates that the constant c 1 appearing in Theorem 1.1 is typically quite small.The largest value of this quantity we managed to obtain is less than 0.234, and we conjecture it is always at most 0.3.The left of Figure 11 shows the maximum and the right the mean of c 1 (K) by crossing number for the Regina census of knots of at most 16 crossings.See Figure 12 for a scatter plot of injectivity radius versus volume for random hyperbolic knots of 10-80 crossings.This suggests that the injectivity radius is typically not too small as the volume increases.
We will say that a property P holds asymptotically almost surely, or a.a.s., in short, if the probability that P holds for knots of n crossings tends to 1 as n → ∞.
It is known that there is a constant A such that vol(K) ≤ A c(K), where c(K) is the crossing number of K. From scatter plots, one might conjecture that there is a constant a such that a c(K) ≤ vol(K) a.a.s.Such an inequality cannot hold for all hyperbolic knots K.For example, consider twist knots.More generally, the highly twisted knots considered in Section 5 have bounded volume but unbounded crossing number.
We now consider the behaviour of the signature σ(K) for random knots K.By Theorem 4.1, σ(K) can be computed from the black surface of a checkerboard colouring of a diagram of K. Hence, it is the signature of a c(K) × c(K) matrix.If the signs of the eigenvalues of this matrix were independently distributed, then the expected value of |σ(K)| would be C c(K) for some constant C. From computational evidence, it appears the constant is about 2. Based on this heuristic, we introduce the following definition: We use the volume instead of the crossing number as it is easier to compute using SnapPy and is more regular.

Figure 2 .
Figure 2. A plot of signature versus the real part of the meridional translation, Re(µ), coloured by longitudinal translation, for a dataset of knots randomly generated by SnapPy.

Figure 3 .
Figure 3.A plot of signature versus slope for knots up to 16 crossings in the Regina census (left) and for a dataset of knots randomly generated by SnapPy having 10 to 80 crossings in their SnapPy-simplified form (right).

Figures 4
Figures 4 and 5 illustrate the relationship between signature and slope in Theorem 1.1 for the knots 6 1 and 12a52, respectively.

Figure 4 .
Figure 4. Left: The stevedore knot 6 1 , which is a slice knot.Right: Its cusp torus, as provided by SnapPy[8].The longitude is 3.9279 and the meridian is 0.7237 + 1.0160i.Its natural slope is 1.8267 and its signature is 0.

Figure 5 .
Figure 5. Left: The knot 12a52.Right: Its cusp torus.The longitude is 27.7228 and the meridian is −1.2838 + 0.5145i.Its natural slope is −18.6064 and its signature is −8.Note how far the parallelogram is from being right-angled; this is the defining feature of having very positive or very negative slope.

Figure 6 .
Figure 6.The calculation of natural slope

Figure 7 .
Figure 7.A fundamental domain D in ∂N max with sides µ and ν (K) is at most 1.82 and ε/4 is at most 0.23.This completes the proof.Proof of Theorem 1.7.Let the triangulation T and the curve ν be as in Propositions 3.1.Since ν = λ − nµ for n even, [ν] = [λ] ∈ H 1 (∂M ; Z 2 ), so ν bounds an unoriented surface S in M .If we make S transverse to the 1-skeleton of T , it defines a simplicial 1-cocycle c ∈ C 1 (M ; Z 2 ) via c(e) = |S ∩ e| mod 2 for each edge e of T .If we connect the midpoints of the edges e of T such that c(e) = 1, we obtain a surface F that intersects each tetrahedron T of the triangulation T in at most one triangle or square.In particular, F is a normal surface.Furthermore, ∂F = ν as ν is a normal curve that intersects each triangle in ∂M at most once.Discard any closed components of F .Let t be the number of tetrahedra of T .Furthermore, write v, e, and f for the number of vertices, edges, and faces of F , respectively.By the above, f ≤ t.Then χ(F ) = v − e + f , and since F is not a disk, |χ(F )| = e − f − v. Since every face of F is a triangle or a quadrilateral,

Figure 9 .
Figure 9.The part of the spanning surface in N (C i ) for i ≥ m+1.Here, q i = 5 or 6.

Figure 11 .
Figure 11.The maximum (left) and the mean (right) of c 1 (K) as functions of the crossing number for knots up to 16 crossings in the Regina census.

Figure 12 .
Figure 12.A scatter plot of injectivity radius versus volume for random knots of 10-80 crossings.