Bounds on Volume Increase Under Dehn Drilling Operations

Martin Bridgeman ^{∗}

June 14, 1995

Abstract

In this paper we investigate how the volume of hyperbolic manifolds increases under the process of removing a curve, that is, Dehn drilling. If the curve we remove is a geodesic we are able to show that for a certain family of manifolds the volume increase is bounded above by π ·l where l is the length of the geodesic drilled. Also we construct examples to show that there is no lower bound to the volume increase in terms of a linear function of a positive power of length and in particular volume increase is not bounded linearly in length.

1 Introduction

Let M and M′ be three-dimensional manifolds such that M′ is homeomorphic to M − α, where α is a simple closed curve in M. Then we say that M′ is obtained by Dehn drilling on M and α is the curve drilled.

If M and M′ are complete hyperbolic manifolds of finite volume then Gromov ([Th]) showed that the volume of M′ is greater than the volume of M. Therefore hyperbolic Dehn drilling always increases volume, but what this increase depends on is unknown. In the following we study the case of drilling out a geodesic and show how the volume increase is related to the length of the geodesic drilled.

Conjecture: 1 Dehn drilling a geodesic increases volume by at most π · L where L is the length of the geodesic to be drilled.

In this paper we will prove this conjecture for a class of Dehn drillings as well as give motivation for why the conjecture should be true in general. We also show that in general there is no linear lower bound on the volume increase by showing that there is no linear lower bound for the set of Dehn drillings we consider.

∗Research at MSRI is supported in part by NSF grant DMS-9022140.

2 Motivation

The motivation for the conjecture and the idea behind the proof in the special case to be considered is the following. If M′ is obtained by hyperbolic Dehn drilling a geodesic α in M then we first construct a continuous one-parameter family of hyperbolic cone-manifolds M_{θ}, 0 < θ ≤ 2π interpolating between M and M′. By interpolating we mean that M_{2π} = M, lim_{θ→0} M_{θ} = M′ and M_{θ} is homeomorphic to M with cone axis α and cone angle θ.

Having constructed a one-parameter family of cone-manifolds interpolating between M and M′, we apply results of Craig Hodgson ([Ho]) which describe the derivative of volume of a smooth one-parameter family of cone manifolds.

Before we give Hodgson’s formula, we state the Sclafli formula for the variation of volume of a one-parameter family of polyhedra in a space of constant curvature, on which Hodgson’s formula is based.

Theorem 1 (Sclafli, Hodgson [Ho]) Let X_{t} be a smooth one-parameter family of polyhedra in a simply connected n-dimensional space of constant curvature K. Then the derivative of the volume of X_{t} satisfies the equation:

(n − 1)KdV ol(X_{t}) = ∑_{F} V_{n−2}(F)dϕ_{F}

where the sum is over all co-dimension two faces of X_{t}, V_{n−2} is the (n − 2)-dimensional volume and ϕ_{F} is the dihedral angle at F.

In the cone manifold case we have the following.

Theorem 2 (Hodgson[Ho]) Let C_{t} be a smooth family of (curvature K) cone-manifold structures on a manifold with fixed topological type of singular locus. Then the derivative of volume of C_{t} satisfies

(n − 1)KdV ol(C_{t}) = ∑_{Σ} V_{n−2}(Σ)dϕ_{Σ}

where the sum is over all components Σ of the singular locus of C and θ_{Σ} is the cone angle along Σ.

Therefore in our case where the singular locus is a single cone axis α we have

dV = −\frac{1}{2}l(θ)dθ
ΔV = \frac{1}{2} ∫_{0}^{2π} l(θ)dθ

where θ is the cone angle along α, l(θ) is the length of α in M_{θ} and where ΔV is the volume increase under Dehn drilling.

Figure 1: Ideal hyperbolic drum

Therefore if the function l is monotonically increasing in θ then it follows that

ΔV ≤ π · L

where L is the length of the geodesic α in M.

The conjecture and the proof of the conjecture for the class of Dehn drillings we consider, is based on this two-fold approach of interpolation and monotonicity.

Basic Polyhedra

We will now prove that for a Dehn drilling operation on basic polyhedra, the conjecture is true. This Dehn drilling operation has some interesting properties including giving a new way to enumerate all basic polyhedra and hence all link projections ([B2]). A basic polyhedron is an ideal hyperbolic polyhedron P with all dihedral angles being right angles. Therefore all vertices have valence four. Let B be the set of all basic polyhedra. Examples of basic polyhedra are the ideal hyperbolic drums {T_{n}}_{n≥3} which have two faces being regular ideal n-gons, corresponding to the top and bottom of the drum and every other face being an ideal triangle (figure 1).

As all the dihedral angles are π/2, basic polyhedra can also be thought of as orbifolds. In [B1] it was shown that if P is a basic polyhedron then it has a four-fold cover (in the orbifold sense) L_{P} which is a hyperbolic link complement in S^{3} having a boundary component for each vertex of P.

Surgery and Dehn drilling

If P is a basic polyhedron and F is a non-triangular face of P then we define surgery on P as follows. Choose two non-adjacent edges e_{1}, e_{2} of F and pinch them together to form a new combinatorial polyhedron \overline{P} (figure 2). By Andreev’s theorem ([Th]), \overline{P} has a realization P′

Figure 2: Surgery

which is an ideal hyperbolic polyhedron with all dihedral angles right angles and therefore is also a basic polyhedron. We say P′ is obtained by surgery on P and therefore surgery is an operation defined on the set B.

In the sense of orbifolds, P′ is obtained by Dehn drilling the unique geodesic perpendicular to both e_{1} and e_{2} in P and if we lift to the four-fold covers, we see that L_{P′} is obtained by Dehn drilling a simple closed geodesic in L_{P} ([B2]). Thus surgery corresponds to a Dehn drilling operation.

We define a partial ordering on B by saying P ≺ P′ if P′ is the result of repeated surgeries starting with P. This partial ordering has as initial objects the set {T_{n}}_{n≥3} mentioned above ([B2]) and therefore we can enumerate all basic polyhedra by repeated surgery on these initial objects.

Theorem 3 Surgery increases volume by at most π · L where L is the length of the geodesic drilled.

To prove this theorem we must first interpolate between P and P′ where P′ is obtained by surgery on P.

3 Interpolating Surgery

Let P be a basic polyhedron with F a non-triangular face of P and e_{1}, e_{2} be two nonadjacent edges of F. Also let g be the unique geodesic in F perpendicular to e_{1}, e_{2}. To interpolate between P and P′ we need a continuous one-parameter family of polyhedra P_{θ}, 0 < θ ≤ π which have combinatorial type of P ∪ {g}, with dihedral angle θ along g and all other edges having the same dihedral angles as the corresponding edge of P and with P_{π} = P, lim_{θ→0} P_{θ} = P′. Intuitively P_{θ} is obtained by bending P along g an amount π − θ. To find such a one-parameter family of polyhedra we use results of Hodgson and Rivin ([HR]) which characterizes compact hyperbolic polyhedra in terms of their spherical duals.

Figure 3: Cutting open S to make S_{θ}

Spherical Dual of a Polyhedron

If P is a hyperbolic polyhedron then we form its spherical dual P^{∗} which is a spherical complex combinatorially dual to the polyhedron P as follows. Each vertex corresponds to a spherical polygon with edges having lengths being the exterior dihedral angles of the edges incident to the vertex and with face angles being the complementary angles of the face angles at the vertex (complement of α = π−α). Similarly an ideal vertex corresponds to a spherical hemisphere with vertices dividing the equator into arcs of length less than π. Let S = P^{∗} and define spherical complexes S_{θ} by cutting S open at e_{1}^{∗}, e_{2}^{∗} (dual edges to e_{1}, e_{2}), and attaching a spherical bigon B with angle π − θ. B is further divided into two triangles T_{1}, T_{2} by joining the two points on the equator by an edge g^{∗}. If P_{θ} is realisable as a hyperbolic polyhedron then its spherical dual is S_{θ} with g^{∗} dual to g. Note that each of e_{1}^{∗}, e_{2}^{∗} split into two edges e_{i}^{∗}, \overline{e}_{i}^{∗}, i = 1, 2 (figure 3). Thus we have a continuous one-parameter family of spherical complexes S_{θ}, 0 ≤ θ ≤ π which interpolates between the spherical duals of P and P′. Before we can interpolate between P and P′ we need to know more about the map between hyperbolic polyhedra and spherical complexes.

The following theorem due to Craig Hodgson and Igor Rivin gives a characterization of compact hyperbolic polyhedra in terms of their spherical duals.

Theorem 4 ([HR]) A metric space (M, g) homeomorphic to S^{2} can arise as the spherical dual P^{∗} of a compact hyperbolic polyhedron P in H^{3} if an only if the following hold.

(a) The metric has constant curvature 1 away from a finite collection of cone points c_{i}.

(b) The cone angles at c_{i} are greater than 2π.

(c) The lengths of closed geodesics of (M, g) are all strictly greater than 2π.

Moreover P is uniquely determined.

Approximating by compact Polyhedra

Since the polyhedron P_{θ} is not compact, we will find a family of compact hyperbolic polyhedra P_{θ}^{t}, t > 0 approximating P_{θ} by applying the above result. We define spherical complexes S_{θ}^{t} by modifying S_{θ} so that it is the dual of a compact hyperbolic polyhedron as follows. We increase the length of every edge of S_{θ} which does not belong to B by a factor of (1 + t) and add a vertex v_{i} at the pole of each hemisphere H_{i}, and join v_{i} by an edge to each vertex on the equator of H_{i}. Note that if we let t = 0 then S_{θ}^{0} = S_{θ}. It will be shown, using the above characterization (theorem 4), that S_{θ}^{t} is the dual of a compact hyperbolic polyhedron P_{θ}^{t}. Furthermore P_{θ}^{t} has combinatorial type of P_{θ} with its ideal vertices truncated and the truncating faces are perpendicular to the faces of P_{θ}^{t} intersecting it. Thus in the Klein model, all the planes intersecting a truncating face meet in a single point outside S_{∞}^{2} corresponding to the unique point perpendicular to the truncating face. A generalized hyperbolic polyhedron is when we allow vertices to be outside S_{∞}^{2} in the Klein model and we call the vertices outside of S_{∞}^{2} hyper-ideal. In this paper we will not distinguish between a generalized polyhedron and the unique polyhedron obtained by truncating the hyper-ideal vertices. Thus P_{θ}^{t} corresponds to a generalized hyperbolic polyhedron with combinatorial type P_{θ} and with the vertices corresponding to ideal vertices of P_{θ} being hyper-ideal. Intuitively as t tends to 0 these hyper-ideal vertices tend to S_{∞}^{2} and become ideal.

Lemma: 1 Given θ, 0 < θ ≤ π then there exists δ > 0 such that the spherical complex S_{θ}^{t} is the dual of a unique compact hyperbolic polyhedron P_{θ}^{t} for 0 < t < δ.

Proof :Obviously S_{θ}^{t} is a spherical complex satisfying (a) and (b) of the characterization (theorem 4), and therefore we need only show that there are no geodesics of length ≤ 2π. Let γ^{t} be a geodesic in S_{θ}^{t}. If γ^{t} intersects the interior of a hemisphere H in a single arc then l(γ^{t}|_{H^{o}}) = π. Therefore if γ^{t} intersects more than two hemispheres in their interiors then l(γ^{t}) ≥ 3π.

Case 1: γ^{t} doesn’t intersect the interior of bigon B. Then we get a curve in S_{π}^{t} by removing the bigon and stitching back the edges. This curve we also call γ^{t} and it is a geodesic in S_{π}^{t}.

If γ^{t} intersects the interior of two hemispheres H_{1}, H_{2} and if γ^{t} doesn’t contain an edge then H_{1}, H_{2} intersect in points a distance ≥ π apart giving a contradiction. Thus γ^{t} must contain an edge and l(γ^{t}) ≥ 2π + π/2 > 2π.

If γ^{t} intersects the interior of only one hemisphere H, then it must enter and leave H along an edge. We replace the arc in H^{o} by an edge path of the same length by going along the boundary of H. This gives us a new geodesic which we still call γ^{t} satisfying

l(γ^{t}) ≥ π/2 + π/2 + π/2(1 + t) + π/2(1 + t) = 2π + tπ > 2π

Therefore all geodesics in S_{θ}^{t} not intersecting the interior of B have length bounded away from 2π.

Case 2: γ^{t} intersects interior of B. There are a number of possibilities.

If γ^{t} enters B at either the apex of T_{1} or T_{2} then it has to leave at the other apex and l(γ^{t}|_{B^{o}}) = π. We replace this arc by edge path e_{1}^{∗} ∪ e_{2}^{∗} or \overline{e}_{1}^{∗} ∪ \overline{e}_{2}^{∗} depending on which gives a geodesic as a result (at least one must give a geodesic). Thus we have a geodesic which has the same length as γ^{t} and doesn’t intersect the interior of B. Therefore γ^{t} has length > 2π by the first case.

If γ^{t} intersects the interior of two hemispheres H_{1}, H_{2} and both are not incident to B then γ^{t} must contain an edge of S_{θ}^{t} and therefore l(γ^{t}) ≥ 2π + π/2 > 2π. If either H_{1}, H_{2} are incident to B only at an apex then γ^{t} intersects B in an arc from one apex to the other and has been already shown to be > 2π. Therefore either γ^{t} contains an edge and is longer than 2π or H_{1}, H_{2} intersect at a point p along their boundaries where γ^{t} traverses from H_{1} to H_{2}. But H_{1}, H_{2} correspond to hemispheres in S_{π}^{t} whose intersection is contained in e_{1}^{∗} ∪e_{2}^{∗} and thus H_{1}, H_{2} cannot intersect at point p. Therefore any geodesic in S_{θ}^{t} intersecting the interior of two or more hemispheres has length bounded away from 2π.

If γ^{t} only intersects the interior of one hemisphere H and it enters B^{o} along an edge and leaves along an edge then it must intersect B^{o} along the equatorial edge. Therefore there is a geodesic in S_{π}^{t} of length l(γ^{t}) − (π − θ) obtained by removing the bigon as before and taking what is left of γ^{t} as the curve. This must be of length > 2π, therefore γ^{t} has length ≥ 2π + (π − θ) > 2π. If γ^{t} enters through H and leaves along an edge at vertex v of B (v is not an apex), then we first replace the interior arc in H by a boundary edge path b of H of the same length, we call this new curve γ^{t} also. Now γ^{t} contains two sides of a spherical triangle \overline{T} in B. H intersects B in an edge e and let v_{1} be the vertex of B contained in b ∩ e. Now replace the part of γ^{t} consisting of two sides of the spherical triangle \overline{T} by the edge of B joining v, v_{1} (figure 4). This gives a geodesic of smaller length that is an edge path and therefore l(γ^{t}) > 2π.

If γ^{t} intersects no hemisphere in its interior then γ^{t} is an edge geodesic containing the equatorial edge of B. Removing the bigon we get a geodesic in S_{π}^{t} of smaller length which implies l(γ^{t}) > 2π.

Since no geodesic of S_{θ}^{t} has length ≤ 2π, then S_{θ}^{t} is the spherical dual of a unique compact hyperbolic polyhedron P_{θ}^{t} (by theorem 4). ■

To find what is the combinatorial structure of P_{θ}^{t} we need to find a cell division of S_{θ}^{t} into convex spherical polygons by geodesics of length less than π and whose endpoints are cone points. This is how we described S_{θ}^{t} originally, but this may not be the only cell division possible. To show that it is, consider the apex v of one of the triangles T_{1}, T_{2}. The only other vertices a distance less than π from v are those vertices connected to v by an edge in the original cell division and therefore the edges from v must be a subset of the original edges from v. Also all edges must be included as if any are left out then there is an angle ≥ π subtended at v. Now considering a vertex v_{1} on the equator we have an angle of π inside the bigon B at v_{1}. Therefore the equatorial edge of B must also be included in the division. Using the fact that any geodesic intersecting the interior of a hemisphere must have length ≥ π and that no vertex can have an angle of ≥ π subtended at it, it is easy to show that all other edges of the original division are necessary. There cannot be any more edges as any

Figure 4: Replacing geodesic by edge geodesic

other edge from vertex to vertex would cross another edge in an interior point which can’t happen. Therefore P_{θ}^{t} has dual given by the original cell division of S_{θ}^{t}.

P_{θ}^{t} is a general hyperbolic polyhedron with combinatorial type P ∪ g and hyper-ideal vertices corresponding to the ideal vertices of P. The dihedral angles between a truncating plane and the planes it intersects are all π/2 . Also the dihedral angle along the edge corresponding to g is θ and except for the four edges incident to g which have dihedral angles π/2, all others have dihedral angle π/2(1 − t). Similarly the face angles on all truncating faces are π/2(1 − t) except where the four edges incident to g intersect. The face angles on the two faces perpendicular to g, at the vertices incident to g, are θ, and all other face angles are π/2. As t tends to zero all dihedral angles approach π/2 except along g which has fixed dihedral angle θ.

4 Monotonicity and Volume Increase

Lemma 1 shows that P_{θ}^{t} exists for 0 < t ≤ k(θ), 0 < θ ≤ π where k(θ) > 0. We can choose k(θ) = c · θ where c is a positive constant. Hodgson and Rivin showed ([HR]) that the map from the space of compact hyperbolic polyhedra to their spherical duals is a homeomorphism and hence P_{θ}^{t} is a continuous two-parameter family of polyhedra in the above domain. When all the dihedral angles are acute we have continuity by Alexandrof ([A]) and therefore lim_{t→0}P_{π}^{t} = P. Thus we extend P_{θ}^{t} by letting P_{π}^{0} = P. Also if 0 < θ ≤ π/2 then lim_{t→0}P_{θ}^{t} = P_{θ}^{0} where P_{θ}^{0} is the unique hyperbolic polyhedron with combinatorial type of P ∪ {g} and dihedral angles π/2 except along g which has dihedral angle θ. We can now take the limit as (t, θ) tends to (0, 0) and we get that lim_{(t,θ)→(0,0)}P_{θ}^{t} = P′. Therefore P_{θ}^{t} is well defined and continuous on the triangle 0 ≤ t ≤ c · θ, 0 ≤ θ ≤ π except for along the open interval of length π/2 from (0, π/2) to (0, π).

We now choose a continuous one-parameter family of polyhedra interpolating between P and P′ by taking the image of a piecewise linear path joining (0, π) and (0, 0) in the domain of P_{(·)}^{(·)} (figure 5). Thus the one-parameter family X interpolating between P and P′ is the product of four continuous one-parameter families X_{i}, i = 1, …, 4 obtained by taking the image of the path joining the four points (0, π), (0, t), (c/t, t), (c/t, 0), (0, 0) by alternately horizontal and vertical lines.

The volume change from P to P′ ΔV satisfies

ΔV = ΔV_{1} + ΔV_{2} + ΔV_{3} + ΔV_{4}

where ΔV_{i} is the volume change over X_{i}.

By continuity given ϵ > 0 we can choose a t > 0 such that

|ΔV_{i}| ≤ ϵ i ≠ 2
|l^{t}(π) − l^{0}(π)| ≤ ϵ

where l^{0}(π) = l, the length of the original geodesic in P.

If l^{t} is monotonic then

0 ≤ ΔV_{2} ≤ 1/2(π − t/c)l^{t}(π) ≤ π/2 · l^{t}(π)
⇒ 0 ≤ ΔV ≤ π/2 · l^{t}(π) + 3ϵ ≤ π/2 · l + ϵ̃

Therefore by choosing smaller values of t we have

ΔV ≤ π/2 · l

Taking the four-fold cover of P given by the hyperbolic link complement L_{P} we have

ΔV′ = 4 · ΔV ≤ 4 · (π/2 · l) = 2πl = πL

where ΔV′ is the volume increase from L_{P} to L_{P′} and L is the length of the geodesic drilled from L_{P} to obtain L_{P′}.

Monotonicity

The following theorem is based on Cauchy Rigidity and is the step needed to prove monotonicity, but first a definition.

Definition: 1 A signed polyhedron P is a polyhedron whose edges are labeled with either + , − or 0.

A vertex v of a signed polyhedron has type given by listing the signs on the edges incident to v in a clockwise or anti-clockwise manner (with the usual equivalence) and a face F is called a zero-face if all its edges are marked with zero.

Figure 5: 2-parameter domain and path from P to P′

Theorem 5 Let P be a trivalent signed polyhedron such that, except for two disjoint faces F_{1}, F_{2}, every face is either a zero-face or has at least four sign changes. Then the space obtained by collapsing all zero-edges P̃ is topologically a bouquet of spheres where either

(1) Each sphere has the trivial cell division (a point and a disc).

or

(2) There is exactly one sphere S̃ in the bouquet with a non-trivial cell division. S̃ has two cells with no sign changes and every other cell has exactly four sign changes. Also all vertices of S̃ have type (+ + −) , (− − +) or (+ − +−) and therefore valence three or four.

Proof :Let Σ be the set of edges of P and Σ = Σ_{+} ∪ Σ_{−} ∪ Σ_{0} be the partition into + , − and 0 edges. If |Σ_{0}| = 0 then construct a vector field on P as follows. In a neighborhood of Σ the vector field is transverse to Σ_{+} and parallel to Σ_{−}. The index of a vertex v can easily be calculated from its type. One way is to read the type of a vertex from left to right, count −1/4 for sign changes, −1/2 for same sign and then adding 1. This is never positive if the valence is at least three and if the valence of v is greater than four its index is always negative. In our case the vertices all have valence three but in general only vertices of type (+ + −), (− − +) and (+ − +−) (which have index zero), have non-negative indices (figure 6).

The vector field is extended over a face F as follows. As we go round the edges of F the sign changes 2n times , which means that there are n disjoint + segments joined together by n disjoint − segments. Choose a center point c_{F} in the interior of F and for each + segment we join an interior point of it to the center c_{F} by a prong and extend the vector field in a neighborhood of these n prongs in the obvious way. Cutting along the prongs divides F into n pieces over which the vector field can easily be extended. The index of the critical point c_{F} is 1 − n/2 (figure 7). Note that if the face has no sign changes then the vector field can still be extended with c_{F} having index 1, in keeping with the formula. The interior of each face has at most one critical point and all other critical points are vertices of P. Each vertex contributes a non-positive index and each face F ≠ F_{1}, F_{2} has at least four sign changes,

Figure 6: Indices of vertices

which implies c_{F} has non-positive index. In order to have the sum of the indices equal to two (the Euler number of P), the indexes of both c_{F_{1}}, c_{F_{2}} must be one and therefore they are of constant sign. Also all vertices and centers c_{F} , F ≠ F_{1}, F_{2} must have index zero. Therefore all vertices have type (+ + −), (− − +) and each F has exactly four sign changes.

If |Σ_{0}| ≠ 0 then we cannot find a vector field extending over Σ_{0}. We avoid this problem by collapsing these zero-edges. First consider \overline{Σ}_{0} ⊆ Σ_{0} to be the collection of edges belonging to loops in Σ_{0}. This decomposes P into regions {R_{i}} which have disjoint interiors and are n_{i}-connected regions whose boundaries consist of zero-edges. Some regions correspond to a zero-face which we call zero-regions. To find a cell division with no zero-edges we must collapse Σ_{0}. Each region R_{i} becomes a sphere S̃_{i} in a bouquet of spheres P̃. The zero-regions become spheres with the trivial cell division of a point and a disc and every other face F collapses to a cell F̃ of some S̃_{i}.

We collapse Σ_{0} by first collapsing \overline{Σ}_{0}. Each R_{i} becomes a sphere S_{i}. If R_{i} is not a zeroface then we have each component of ∂R_{i}, β_{ij} collapses to a vertex v_{ij} on a sphere S_{i}. As R_{i} is not a face there is an edge path γ_{ij} splitting R_{i} into two regions R_{ij}^{1}, R_{ij}^{2} with endpoints on β_{ij}. If there are no other paths from β_{ij} to the interior then either R_{i} is the union of two faces or there are two faces which intersect in two separate edges (figure 8). Both give contradictions. Therefore collapsing β_{ij} gives vertex v_{ij} with valence at least three. Thus the vertices on S_{i} after collapsing \overline{Σ}_{0}, have valence at least three. If we now collapse Σ_{0} − \overline{Σ}_{0} to get sphere S̃_{i} in P̃ the valences cannot go down as the only zero-edges left on S_{i} form a collection of disjoint trees.

Therefore if S̃_{i} does not correspond to a zero-face it has a decomposition into faces all but two of which have at least four sign changes and all vertices have valence at least three. Thus

Figure 7: Extend vector field over face Figure 8: Valence after collapse is at least three as before we can give S̃_{i} a vector field which implies that there can be at most one sphere S̃ which has a non-trivial cell division. On S̃ the vertices all have type (+ + −), (− − +) or (+ − +−) , the faces corresponding to F_{1}, F_{2} give rise to faces F̃_{1}, F̃_{2} of S̃ which have constant sign on their boundaries and all other faces have exactly four sign changes. ■

Now we apply the above theorem. If we compare the corresponding faces of P_{θ_{1}}^{t} and P_{θ_{2}}^{t} then apart from the two faces F_{1}, F_{2} which are incident and perpendicular to g , all others have identical face angles. Let Q be the marked polyhedron combinatorially equivalent to both P_{θ_{1}}^{t}, P_{θ_{2}}^{t} and marked with + , − or 0 depending on whether the length of the edge in P_{θ_{1}}^{t} is bigger, smaller or equal in length to the corresponding edge of P_{θ_{2}}^{t}. By the below result in planar hyperbolic geometry (lemma 2), any face F ≠ F_{1}, F_{2} is either a zero-face or has at least four sign changes. Therefore by the above (theorem 5), both faces F_{1}, F_{2} have no sign changes. If θ_{2} > θ_{1} then again using the below lemma, both faces must have constant sign +. If the edge g is + then we get two (+ + +) vertices and if the edge g is 0 then after collapsing we get a (+ + ++) vertex, both giving contradictions.

Therefore g must be marked with − and thus l^{t} is monotonic in θ.

Theorem 5 gives more information about the pattern on a marked polyhedron P. The region R corresponding to S̃ is obtained by removing the zero-faces of P and has no zeroloops with interior edges. When we collapse the boundary curves of R to get sphere S then any vertex of S has valence three or four. Any zero-edge path of more than one edge will give a vertex in S̃ with valence greater than four and therefore every zero-edge in S is isolated. These isolated zero-edges in S must join two trivalent vertices which have alternating signs as you go around a neighborhood of the zero-edge.

In the case of the markings on Q we obtain figure 9.

The following is the technical lemma needed above. A proof can be found in the referenced paper.

Lemma: 2 ([HR]) Let E_{1}, E_{2}, …, E_{n} be lines in H^{2} defining a convex polygonal curve C. Let A_{i} = E_{i} ∩ E_{i+1}(i = 1, …, n − 1) be the vertices of C , let ν_{i} = E_{i}^{⊥}(i = 1, …, n) be the outward unit normals to the Minkowski inner product ⟨·, ·⟩, and let l_{i} be the length of E_{i} ∩ C = A_{i−1}A_{i}(i = 1, …, n − 2). Let E′_{1}, E′_{2}, …, E′_{n} be lines in H^{2} defining another such curve C′ with A′_{i}, ν′_{i}, l′_{i} defined as above. Assume that

(i) < A_{i} =< A′_{i} for i = 2, …, n − 1 and

(ii) l′_{i} ≥ l_{i} for i = 2, …, n − 1.

Then

⟨ν′_{1}, ν′_{n}⟩ ≤ ⟨ν_{1}, ν_{n}⟩,

with equality if and only if A_{1}A_{2}…A_{n−1} is congruent to A′_{1}A′_{2}…A′_{n−1}.

An easy consequence of this lemma is the following. Let F_{a}, F_{b} be two hyperbolic n-gons with face angles equal. If we label an n-gon F with +, 0 or − by comparing lengths of corresponding sides of F_{a}, F_{b}, then either F has all sides marked with 0, in which case F_{a}, F_{b}

Figure 9: How edge lengths in P_{θ}^{t} vary with θ

are congruent, or, F has at least four sign changes as we go around the perimeter. This shows that Q satisfies the hypothesis of the theorem.

To show that F_{1}, F_{2} are marked + we need to relate the Minkowski inner product to face angles. If two edges of a compact hyperbolic polygon intersect in an angle θ < π then the inner product of their outward normals is cos(θ_{ext}) where θ_{ext} is the exterior angle of θ. Now the convex polygons p_{1}, p_{2} corresponding to F_{1} in P_{θ_{1}}^{t}, P_{θ_{2}}^{t} respectively, have the same face angles except at the vertex incident to g, where p_{i} has the angle θ_{i}. Thus if the sides of p_{1} where all smaller or equal in length to the sides of p_{2} then we would have by the above lemma that the inner product of the outward normals of the two sides intersecting at g in p_{1} is greater than or equal to the same inner product in p_{2}, that is cos(π − θ_{1}) ≥ cos(π − θ_{2}). But θ_{1} < θ_{2}, and therefore the sides of p_{1} must be greater than or equal in length to the corresponding sides in p_{2}. This gives the needed step to show that both F_{1}, F_{2} are of constant sign +.

5 Combinatorial Bound on Volume Increase

Here we will show that bending along a geodesic g on a face F which is perpendicular to two edges of F increases volume at most linearly in the number of edges of F.

The idea is that as θ, the exterior dihedral angle at g varies, F is still intrinsically an ideal n-gon. If l(θ) is long then this forces F to be narrow somewhere across g. This implies that the two faces F_{1}, F_{2} obtained by bending F along g (which meet with exterior dihedral angle θ), have faces F_{3}, F_{4} adjacent to F_{1}, F_{2} respectively which are connected by a short path on the bent face F. As F_{3}, F_{4} cannot intersect, this shows that l(θ) must decrease as we bend otherwise F_{3}, F_{4} would be forced to intersect.

We will describe the proof for when we have a continuous one-parameter family of polyhedra P_{θ} interpolating between P and P′. We obtain the result in general by approximating as before by the one-parameter family X^{t}.

Long Geodesics give Thin Polygons

An ideal hyperbolic quadrilateral has two perpendiculars and it can be shown using elementary hyperbolic geometry that the hyperbolic sine’s of the half lengths of these are reciprocal. The following lemma is a generalization of this fact.

Lemma: 3 If F is an ideal hyperbolic n-gon and g the unique perpendicular to edges e_{1}, e_{2} of F then ∃ e_{3}, e_{4} separated by g with distance d between them satisfying

sinh(d/2) · sinh(\frac{(π − 2)l}{2π(n − 3)}) ≤ 1

where l is the length of g.

Proof :Choose ϵ > 0 such that the one-sided ϵ-neighborhood of g has the same area as F.

⇒ l sinh(ϵ) = (n − 2)π

If we take N_{ϵ} the two-sided ϵ−neighborhood of g then for each x ∈ g there is an arc γ_{x} perpendicular to g centered at x of radius ϵ. We define three subsets of g as follows:

g_{i} = {x ∈ g : |γ_{x} ∩ ∂F | = i}

Therefore l = l_{0} + l_{1} + l_{2} where l_{i} = length of g_{i}, i = 0, 1, 2.

Each of the sets g_{i} is a collection of intervals. If A_{1} = Area of F outside N_{ϵ} (cusp region) and A_{2} = area of F consisting of points whose perpendicular geodesic to g has length less than ϵ (narrow region) then

Area(F) = Area(F ∩ N_{ϵ}) + A_{1}
Area(F ∩ N_{ϵ}) = 2l_{0} sinh(ϵ) + l_{1} sinh(ϵ) + A_{2}
⇒ l sinh(ϵ) = 2l_{0} sinh(ϵ) + l_{1} sinh(ϵ) + A_{1} + A_{2}
⇒ l_{2} sinh(ϵ) = l_{0} sinh(ϵ) + A

where A = A_{1} + A_{2} (cusp plus narrow region).

If e is one of the n − 2 edges of F not intersecting g, drop perpendiculars to g from the endpoints of e on S_{∞}^{1} to the two points x, y on g. This forms a quadrilateral Q_{e} with area π and base [x, y]. Let A_{e} be the area of the region in Q_{e} which is either a distance greater than ϵ from g or whose perpendicular to g has length less than ϵ. A_{e} is the contribution of

Figure 10: A_{e} is minimum

Figure 11: General picture of A_{e}

Q_{e} to A. If e doesn’t intersect N_{ϵ} then the A_{e} is minimum when e is tangent to ∂N_{ϵ} (figure 10). If L is the length of the base of the quadrilateral tangent to ∂N_{ϵ} then

sinh(L/2) = \frac{1}{sinh(ϵ)}
⇒  A_{e}  ≥ π − L sinh(ϵ) = π − \frac{L}{sinh(L/2)}
⇒  A_{e}  ≥ π − 2

If e intersects ∂N_{ϵ} at points x′, y′ then project down onto g to get points x_{1}, y_{1} (figure 11). Now if \overline{L} is the length of either [x, x_{1}] or [y_{1}, y] (both have the same length) then

\overline{L} ≤ L/2
⇒ A_{e} = π − 2\overline{L} sinh(ϵ)
⇒ A_{e} ≥ π − 2
⇒ A ≥ (n − 2)(π − 2)

Since we have that l_{2} sinh(ϵ) = l_{0} sinh(ϵ) + A then

l_{2} ≥ \frac{A}{sinh(ϵ)} ≥ \frac{(n − 2)(π − 2)}{sinh(ϵ)}
⇒ l_{2} ≥ \frac{π − 2}{π} · l

The subset g_{2} consists of at most (n − 3) intervals. Therefore by the above, one of these intervals [a, b] must have length D satisfying

D ≥ \frac{π − 2}{π(n − 3)} · l

Consider the quadrilateral Q formed by taking the two perpendicular arcs γ_{a}, γ_{b} to g, centered at a, b respectively, of length 2ϵ. Choose the other sides α, β to join the endpoints of γ_{a}, γ_{b}. There are two edges e_{3}, e_{4} of F separated by g, which enter and leave Q through the sides γ_{a}, γ_{b}. To show that e_{3}, e_{4} are close note that they are closer than α and β are. The two geodesics α, β describe an ideal quadrilateral by extending α, β to the sphere at infinity. This quadrilateral has a diagonal of length D′ which is longer than D whose other diagonal has length d′. If d is the distance between e_{3} and e_{4} then

sinh(D/2) ≤ sinh(D′/2) = \frac{1}{sinh(d′/2)} ≤ \frac{1}{sinh(d/2)}

Therefore we have

sinh(d/2) · sinh(\frac{(π − 2)l}{2π(n − 3)}) ≤ 1  (1)

■ This lemma shows that long geodesics must give thin polygons.

Thin Bent Plane Forces Intersection

In the hyperbolic plane consider a geodesic γ represented by a circle C of radius R intersecting the unit circle at right angles. The two geodesics passing through the origin o meeting γ on S_{∞}^{1} form an angle of ϕ at o (visual angle). If x is the Euclidean distance from o to γ and d is the hyperbolic distance then we have by simple planar geometry

R  = \frac{1 − x^{2}}{2x}
tan(ϕ/2) = R
2 tanh^{−1}(x) = d
⇒  x  = tanh(d/2)
⇒  R  = \frac{1}{sinh(d)}
⇒ tan(ϕ/2) = \frac{1}{sinh(d)}

Figure 12: Sideview of the planes

In P_{θ} we have planes F_{1}, F_{2} intersecting with dihedral angle ϕ and let F_{3}, F_{4} be the two planes intersecting in edges e_{3}, e_{4} respectively of lemma 3. We also have planes F′_{3}, F′_{4} defined by geodesics α, β which intersect F_{1}, F_{2} at right angles.

If F′_{3}, F′_{4} intersect then F_{3}, F_{4} must also intersect. Therefore since planes F_{3}, F_{4} cannot intersect then neither can F′_{3}, F′_{4}. Looking along the geodesic g, F′_{3}, F′_{4} are given by two geodesics γ_{3}, γ_{4} and F′_{3}, F′_{4} by two geodesics passing through the origin and perpendicular to γ_{3}, γ_{4} respectively (figure 12).

Both γ_{3}, γ_{4} have the same visual angle ϕ′. If they are not to intersect then

ϕ′/2 ≤ ϕ/2
⇒ tan(ϕ′/2) ≤ tan(ϕ/2)
⇒ \frac{1}{sinh(d′/2)} ≤ tan(ϕ/2)

where d′ is the length of the short curve on F joining F′_{3}, F′_{4}. By lemma 3 we have

sinh(\frac{(π − 2)l(ϕ)}{2π(n − 3)}) ≤ tan(ϕ/2)

Therefore

l(ϕ) ≤ \frac{2π(n − 3)}{(π − 2)} · sinh^{−1}(tan(ϕ/2))  (2)

Therefore we obtain a combinatorial bound on the volume increase by integrating the above equation. This integration is possible as although sinh^{−1}(tan(ϕ/2)) tends to infinity at π it is still integrable. Therefore we have

ΔV ≤ \frac{π(n − 3)}{(π − 2)} · ∫_{0}^{π} sinh^{−1}(tan(ϕ/2)dϕ = K · (n − 3)  (3)

where K is some fixed constant. We can derive this bound in general using the one-parameter family of approximations X^{t} in the same manner as before.

Figure 13:

6 Non-existence of lower bound for volume increase

If the volume increase under Dehn drilling is bounded below by a constant times some positive power of the length of the geodesic drilled then we have that

ΔV ≥ c · l^{a}

where c, a are fixed positive constants and l is the length of the geodesic drilled. We have shown in the previous section, that the volume increase by drilling out a geodesic belonging to an ideal n-gon of a basic polyhedron is bounded linearly in n. Therefore to prove that no lower bound exists we need to find a sequence of basic polyhedra each containing a face of bounded valence and the set of lengths of geodesics perpendicular to two sides unbounded. Let {P_{k}}_{k=0}^{∞} be a sequence of basic polyhedra and N a number such that each P_{k} has a marked face F_{k} which is an n_{k}-gon, n_{k} < N. Also let l_{k} be the length of the longest geodesic α_{k} in F_{k} perpendicular to two edges of F_{k}. Then the volume increase ΔV_{k}, by drilling out α_{k} is bounded above by some constant K depending only on N. If the sequence l_{k} is not bounded then by the lower bound for volume increase ΔV_{k} ≥ c · l_{k}^{a}. Therefore ΔV_{k} is unbounded giving a contradiction. We now construct the required family using circle packings.

Circle Patterns and Packings

If P is a hyperbolic polyhedron then in the upper half-space model each face F of P lies on a hemisphere H_{F} which intersects S_{∞}^{2} in a circle C_{F}. Furthermore the angle at which two circles C_{F_{1}}, C_{F_{2}} intersect equals the dihedral angle between the faces F_{1}, F_{2}. Thus looking for a polyhedron of prescribed dihedral angles and combinatorics is equivalent to finding a circle pattern with prescribed angles of intersection and combinatorial pattern. We will use the terminology circle packing to refer to a collection of circles with intersections being tangential.

In the case of constructing a basic polyhedron we require that the circles meet in groups of four with opposite circles tangent and neighboring circles meeting at right-angles (figure 13).

We start our circle pattern with a circle C_{0} of radius one centered at the origin O. We further take circles C_{1}, C′_{1} of radius r such that both are perpendicular to C_{0} and have centers on the positive, negative real axis respectively. Thus there are unique circles C_{2}, C′_{2} which are perpendicular to C_{0}, with centers on the positive, negative imaginary axis respectively and are tangent to both C_{1}, C′_{1}. This configuration of five circles depends only on our choice of r and therefore we have a one-parameter family C_{r} of circle patterns. By simple planar hyperbolic geometry the radius of both C_{2}, C′_{2} is 1/r. This collection of five circles describes an ideal quadrilateral on the hyperbolic plane H_{0} associated with circle C_{0}. Furthermore r is a coordinate system for the shape of the quadrilateral with it becoming narrow in the horizontal direction as r tends to infinity and becoming narrow in the vertical direction r tends to 0.

To extend C_{r} to the circle pattern of a basic polyhedron we perform a moebius transformation M on C_{r}. M consists of a rotation of π/2 about the origin followed by an expansion by a real number k_{r}. The factor k_{r} is chosen so that M(C_{1}) is tangent to the original C_{1}. For r large these four tangent points are the only intersections between C_{r} and M(C_{r}) and so we extend circle pattern C_{r} by adding M(C_{r}) to form circle pattern \overline{C_{r}} = C_{r} ∪ M(C_{r}) (figure 14).

Now take the collection of hemispheres H_{r} associated with \overline{C_{r}} and let N_{r} be the region between the two hemispheres H_{0}, M(H_{0}) cut out by H_{r}. N_{r} is an infinite volume hyperbolic orbifold with four infinite volume funnels. Each of these funnels meet S_{∞}^{2} in a quadrilateral bounded by circles which are tangent at points of intersection. It can be easily shown that all four of these quadrilaterals are equivalent up to moebius transformation to a quadrilateral Q_{r}.

To form a basic polyhedron from N_{r} we must truncate the funnels of N_{r}. We do this by first finding a packing of Q_{r} by a finite number of circles such that the interstices are all triangular. Then having packed each of the four copies of Q_{r}, we are left with a finite number of triangular interstices which we truncate by adding the unique circle passing through the three vertices of each interstice. These circles are also perpendicular to the three sides of the interstices and therefore we have truncated the four infinite volume funnels of N_{r} to get a basic polyhedron P_{r}.

Unfortunately it is not always possible to find a finite packing of an arbitrary circular quadrilateral but using work of Robert Brooks([Br]) we will show that there is a sequence {r_{n}} such that Q_{r_{n}} is packable and r_{n} monotonically tends to infinity.

We define the continued fraction c(Q) of a quadrilateral Q by using the greedy algorithm. Assume that the four sides have been labelled left, right, top, bottom. We first pack Q from left to right by starting at the left side of Q and packing in the largest circle tangent to the left, top and bottom of Q. Now Q contains a smaller quadrilateral Q_{1} which we again pack with the largest circle tangent to the left, top and bottom sides. After n_{1} circles have been added the remaining quadrilateral Q_{n_{1}} is too narrow to fit any more circles which are tangent to the left, top and bottom so we now pack Q_{n_{1}} from top to bottom with circles tangent to the top, left and right sides. Again after n_{2} circles have been packed into Q_{n_{1}} we are left with a quadrilateral Q_{n_{1},n_{2}} too wide to pack with a circle tangent to the top, left and right. The packing algorithm is defined recursively by now applying the algorithm to the quadrilateral Q_{n_{1},n_{2}} (figure 15). To each quadrilateral we get a sequence of integers {n_{i}} and we define the function c on the space of quadrilaterals to be the real number with these Figure 14: Circle pattern

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Figure 15: Greedy algorithm

as its continued fraction expansion, that is

c(Q) = n_{1} + \frac{1}{n_{2} + \frac{1}{n_{3} + \frac{1}{n_{4} + ⋯}}}

Note that the greedy algorithm gives a finite packing of Q if and only if c(Q) is rational. Now let S be the space of all configurations of four circles up to moebius transformation which have disjoint interiors and such that the complement of the interiors is two circular quadrilaterals. Then Robert Brooks showed ([Br]) that the map \overline{c} : S → ℜ^{2} defined by mapping a configuration of circles to the pair of continued fractions of its two circular quadrilateral regions, is not only continuous but a homeomorphism. This implies that c is continuous and therefore c(Q_{r}) is a continuous function of r. It can be shown that as r tends to infinity the modulus of Q_{r} tends to infinity. Therefore c(Q_{r}) tends to infinity and by continuity there exists a sequence {r_{n}} such that c(Q_{r_{n}}) = n. Therefore Q_{r_{n}} can be packed with n circles from left to right.

Now basic polyhedron P_{n} is defined by truncating the funnels of N_{r_{n}} as described above. Figure 16 is a combinatorial picture of P_{n}. The polyhedra P_{n} have the property of each containing an ideal quadrilateral face F_{n} whose modulus tends to infinity as a function in n. Thus F_{n} contains a geodesic α_{n}, perpendicular to two sides of F_{n} whose length monotonically increases to infinity as a function of n. Therefore this sequence of polyhedra is a counterexample to the existence of a lower bound to volume increase under Dehn drilling. We get a similar counter-example for the manifold case by looking at the four-fold covers of these polyhedra when viewed as orbifolds.

Conclusion

The two bounds we derive are very different. One is a bound on the volume increase by the length of the geodesic drilled and the other is a purely combinatorial bound. I believe that these techniques can be applied to a wider class than I have done and further hope to prove the conjecture at the outset by decomposing the manifold into polyhedra and extending the results in this paper.

References

[A] Alexandroff Convex Polyhedra

[B1] M. Bridgeman Volume increase under Dehn drilling operations Phd. thesis Princeton June 1994

[B2] M. Bridgeman The Structure and Enumeration of Link Projections To Appear Trans. A.M.S. (1995)

[Br] R. Brooks On the Deformation Theory of Classical Schottky Groups Duke Mathematical Journal December 1985

[Th] W. P. Thurston The Geometry and Topology of 3-manifolds

[HR] Craig Hodgson , Igor Rivin A characterization of Compact Convex Polyhedra in Hyperbolic 3-Space

[Ho] Craig Hodgson Degeneration & Regeneration of Geometric structures on Three-Manifolds Phd. thesis Princeton June 1986 Figure 16: Polyhedron with thin ideal quadrilateral face

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