Geometry of planar surfaces and exceptional fillings

We build the circle packing in four steps, beginning with the Fuchsian manifold associated to $R$ given by face pairings of blue hemispheres.

Step 1: Select circles tangent to ideal vertices. Select an ideal vertex $v$ of $P$ and choose a small circle on $S^{2}_{\infty}$ tangent to the equatorial plane at $v$. Also take its image under reflection in the equatorial plane. The face pairings of $P$ take these circles to a finite collection of circles, all tangent to the equatorial plane at ideal vertices of $P$. By choosing the initial radius small enough, these circles will be mapped to disjoint circles under $\Gamma$. Add such circles for each ideal vertex of $P$, and color them green. Note the green circles are preserved under the action of $\Gamma$.

Step 2: Select circles on edges. Next, select an ideal edge of $P$. This corresponds to half of a blue circle in the northern hemisphere of $S^{2}_{\infty}$. Choose a finite collection of pairwise tangent circles orthogonal to this blue circle with the initial and final circles tangent to the green circles at the endpoints of the corresponding ideal edge. One of the face pairings of $\Gamma$ takes this collection of circles to a collection of pairwise tangent circles on another blue half-circle, identified to the original. See for example Figure 1(a). Again we may ensure the circles have disjoint interiors, and again reflect through the equatorial plane. Repeat this process for each ideal edge, and color the resulting circles green. When finished, we have a collection of pairwise tangent green circles with disjoint interiors, orthogonal to blue circles. The collection of blue circles, green circles, and the equatorial plane is preserved under the action of $\Gamma$.

Step 3: Select circles in interior. Now in the region of the northern hemisphere of $S^{2}_{\infty}$ disjoint from all interiors of blue circles, add finitely many circles with disjoint interiors, tangent to existing green circles. The result will not necessarily be a circle packing, with all triangular interstices. However, we may arrange such that all interstices are triangles and quadrilaterals, with only a finite number of quadrilateral interstices. Continue to color these circles green, and reflect them across the equator to obtain a collection of circles in $S^{2}_{\infty}$, colored green and blue, that meet in triangular interstices and a finite number of quadrilateral interstices. As in [5], the quadrilaterals are parameterized by ${\mathbb{R}}$, with rational numbers corresponding to quadrilaterals which admit an actual circle packing.

Step 4: Deform to a circle packing. Again following Brooks, for each green circle, adjoin to $\Gamma$ the reflection in that green circle. Also add reflection in the equator. This forms a new group $\Gamma^{\prime}$, which is still discrete by the Poincaré polyhedron theorem. Then as in [6], there exists a quasi-conformal deformation of $\Gamma^{\prime}$ by an arbitrarily small amount $\delta$ to a new group $\Gamma_{\delta}^{\prime}$, taking reflections in circles to reflections in circles, for which all quadrilaterals have rational parameters. That is, the result admits a finite circle packing, by adding finitely many circles to the quadrilateral interstices. Let $\Gamma_{\delta}$ denote the image of $\Gamma$ under this deformation. It is still a Fuchsian group, because $\Gamma_{\delta}^{\prime}$ preserves the plane corresponding to the equator, by a reflection. Then $R^{\prime}={\mathbb{H}}^{2}/\Gamma_{\delta}$ is a quasi-conformal deformation of $R$ that is $\delta$-close to $R$, and $R^{\prime}$ satisfies the conclusions of the lemma. ∎

Given $R$ and $H$, the cusp neighborhood $H$ lifts to a disjoint collection of horoballs in the Fuchsian manifold ${\mathbb{H}}^{3}/\Gamma$. We let $\widetilde{H}$ denote finitely many horoballs, one at each ideal vertex of $P$. We prove the result by stepping back through the proof of Lemma 2.3, ensuring that we may choose circles at each step to be disjoint from the collection $\widetilde{H}$, or to meet them only in noncompact regions if they share a point on the boundary at infinity.

In the first step of the proof of Lemma 2.3, we choose green circles tangent to vertices of $P$. To see that these can be chosen as in the statement of the lemma, consider the (finitely many) vertices of $P$ identified by face pairings. As we shrink any circle tangent to such a vertex, its images under face pairings also shrink. Thus we may shrink any such circle until the hyperplane bounded by it meets only the horoball at the ideal vertex at which it is tangent, and similarly for its translates under face pairings.

Now view the ball model of ${\mathbb{H}}^{3}$ as the unit ball in ${\mathbb{R}}^{3}$. Let $C$ denote the complement on $S^{2}$ of the interiors of disks bounded by blue circles and the green circles tangent to the equator, selected in the last paragraph. Then $C$ is compact, as is the collection of finite (closed) horoballs $\widetilde{H}$, and these two sets are disjoint. Thus the Euclidean distance $d(C,\widetilde{H})$ is some positive value $d_{0}>0$. Let $r_{0}=d_{0}/2$. Then any Euclidean hemisphere with center in $C$ of radius $r_{0}$ will have Euclidean distance at least $d_{0}/2$ from $\widetilde{H}$, hence its hyperbolic distance will also be bounded away from zero.

Use this to complete the second and third steps of selecting the circle packing. We may ensure that all circles selected in these steps have radius less than $r_{0}$; since their centers lie in $C$, they are disjoint $\widetilde{H}$. By compactness, we may also ensure that the triangle and quadrilateral interstices are shaped such that if we were to add any additional circles to the collection, those circles must have radius less than $r_{0}/2$, so adding any finite collection of circles to the existing collection yields new hemispheres disjoint from $\widetilde{H}$.

Followng the fourth step, we next adjust the hyperbolic structure to $R^{\prime}$ admitting a circle packing. Further, for any neighborhood of $(R^{\prime},H^{\prime})$ in decorated moduli space, we may obtain an embedded cusp neighborhood $H^{\prime}$ of $R^{\prime}$ such that $(R^{\prime},H^{\prime})$ lies in that neighborhood. Let $\widetilde{H}^{\prime}$ denote the finitely many horoballs that are lifts of the horoball neighborhood $H^{\prime}$ with centers on points corresponding to deformed points of $P$.

Before deforming $R$ to $R^{\prime}$, the distance between horoballs $\widetilde{H}$ and those green hemispheres that are not tangent to the equator is bounded away from $0$, say by $d>0$, so we can ensure our deformation is small enough that the distance from new hemispheres to $\widetilde{H}^{\prime}$ is still at least $d/2>0$. Moreover, we can ensure that after the deformation, interstices are still sufficiently small that any added circles correspond to hemispheres disjoint $\widetilde{H}^{\prime}$. Thus when we complete to a circle packing, all green circles satisfy the conclusion of the lemma. ∎

Suppose by way of contradiction that some red circle meets a horoball $H_{0}$ of $\widetilde{H}^{\prime}$. The red circle is determined by a triangular interstice, with three sides determined by green circles. At most one of these circles is tangent to the equator, so at most one meets the center of a horoball. Thus two green circles disjoint from $\widetilde{H}^{\prime}$ meet in a vertex $v$ of the interstice. Apply an isometry taking ${\mathbb{H}}^{3}$ to the upper half space model, taking $v$ to infinity. Then we have two green hemispheres that are disjoint from $\widetilde{H}^{\prime}$ mapping to parallel vertical planes. Ensure one maps to the vertical plane parallel to the real line through the point $-i$ in ${\mathbb{C}}$, and one maps to the parallel vertical plane through $i$, and the third has boundary mapping to the unit circle in ${\mathbb{C}}$. The horoball $H_{0}$ maps to a horoball disjoint from all three green circles, except possibly centered at a point on the unit circle. In any case, its center will be on ${\mathbb{C}}$ in the infinite strip between the two parallel vertical planes. Finally, the red circle maps to the unique hyperplane meeting the three points of tangency of the green; this is the vertical plane with boundary on the imaginary axis of ${\mathbb{C}}$.

By assumption, the image of $H_{0}$ intersects the red vertical plane. However, it cannot meet any of the green planes. Thus its Euclidean radius is strictly less than the imaginary coordinate of its center. Since its center lies no closer to the imaginary axis than the unit circle, it is impossible that it intersects the red plane while remaining disjoint from the two green vertical planes. This is a contradiction. ∎

Starting with $(R,H)$, obtain $(R^{\prime},H^{\prime})$ and its circle packing as in Lemma 2.4, with $\Gamma^{\prime}$ denoting the group of isometries of ${\mathbb{H}}^{3}$ giving face pairings of the blue hemispheres associated with $R^{\prime}$. Adjoin to $\Gamma^{\prime}$ reflections in the green circles and reflections in the equatorial plane. Finally, adjoin reflections in the red planes. The quotient of ${\mathbb{H}}^{3}$ by this group gives a finite volume hyperbolic 3–manifold with fundamental domain cut out by red, blue, and green hemispheres in ${\mathbb{H}}^{3}$. It contains an embedded surface isometric to $R^{\prime}$, namely the image of the equatorial plane.

By construction, each cusp of $R^{\prime}$ is embedded in a distinct cusp of $M$. To finish the proof it suffices to show that the embedded cusp neighborhood $H^{\prime}$, which has lengths within $\epsilon$ of those corresponding to $H$, lifts to an embedded neighborhood of (some of) the cusps of $M$.

For suppose lifts of $H^{\prime}$ are not embedded. Since $H^{\prime}$ is embedded in $R^{\prime}$, horoballs $\widetilde{H}^{\prime}$ with centers on the equator are all disjoint. Thus some horoball $H_{1}$ must intersect a horoball $H_{2}$ in the fundamental domain for $M$, and at least one of those horoballs, say $H_{2}$, does not have its center on the equator. But then $H_{2}$ cannot have its center at any of the ideal vertices of the fundamental domain. Hence $H_{2}$ must intersect one of the red, green, or blue faces of the fundamental domain in a compact region. Because $H_{2}$ is a translate of some horoball of $\widetilde{H}^{\prime}$ under the fundamental group of $M$, it follows that one of the horoballs of $\widetilde{H}^{\prime}$ meets a red, green, or blue hemisphere in a compact region. This contradicts our choice of circle packings. ∎

Take the manifold $M$ of Theorem 2.1, with embedded surface $R_{0}:=R^{\prime}$ in the neighborhood of $R$ in the moduli space, and cusp neighborhood satisfying the requirements of the theorem. When we Dehn fill along boundary slopes of $R^{\prime}$, we attach a collection of disks to $R^{\prime}$, capping off the punctures to obtain a sphere $S$. Dehn filling along a single cusp attaches a disk to $R^{\prime}$ that meets the core $c$ of the Dehn filling solid torus exactly once. Because each cusp of $M$ meets at most one cusp of $R^{\prime}$, it follows that after all Dehn fillings, the core $c$ meets the sphere $S$ exactly once. Hence $S$ cannot bound a ball to either side, and so it is a reducing sphere for the Dehn filling. ∎