We now leave the topic of Riemann surfaces, and turn now to the (loosely related) topic of conformal mapping (and quasiconformal mapping). Recall that a conformal map from an open subset of the complex plane to another open set is a map that is holomorphic and bijective, which (by Rouché’s theorem) also forces the derivative of to be nowhere vanishing. We then say that the two open sets are conformally equivalent. From the Cauchy-Riemann equations we see that conformal maps are orientation-preserving and angle-preserving; from the Newton approximation we see that they almost preserve small circles, indeed for small the circle will approximately map to .

In previous quarters, we proved a fundamental theorem about this concept, the Riemann mapping theorem:

Theorem 1 (Riemann mapping theorem)Let be a simply connected open subset of that is not all of . Then is conformally equivalent to the unit disk .

This theorem was proven in these 246A lecture notes, using an argument of Koebe. At a very high level, one can sketch Koebe’s proof of the Riemann mapping theorem as follows: among all the injective holomorphic maps from to that map some fixed point to , pick one that maximises the magnitude of the derivative (ignoring for this discussion the issue of proving that a maximiser exists). If avoids some point in , one can compose with various holomorphic maps and use Schwarz’s lemma and the chain rule to increase without destroying injectivity; see the previous lecture notes for details. The conformal map is unique up to Möbius automorphisms of the disk; one can fix the map by picking two distinct points in , and requiring to be zero and to be positive real.

It is a beautiful observation of Thurston that the concept of a conformal mapping has a discrete counterpart, namely the mapping of one circle packing to another. Furthermore, one can run a version of Koebe’s argument (using now a discrete version of Perron’s method) to prove the Riemann mapping theorem through circle packings. In principle, this leads to a mostly elementary approach to conformal geometry, based on extremely classical mathematics that goes all the way back to Apollonius. However, in order to *prove* the basic existence and uniqueness theorems of circle packing, as well as the convergence to conformal maps in the continuous limit, it seems to be necessary (or at least highly convenient) to use much more modern machinery, including the theory of quasiconformal mapping, and also the Riemann mapping theorem itself (so in particular we are not structuring these notes to provide a completely independent proof of that theorem, though this may well be possible).

To make the above discussion more precise we need some notation.

Definition 2 (Circle packing)A (finite)circle packingis a finite collection of circles in the complex numbers indexed by some finite set , whose interiors are all disjoint (but which are allowed to be tangent to each other), and whose union is connected. Thenerveof a circle packing is the finite graph whose vertices are the centres of the circle packing, with two such centres connected by an edge if the circles are tangent. (In these notes all graphs are undirected, finite and simple, unless otherwise specified.)

It is clear that the nerve of a circle packing is connected and planar, since one can draw the nerve by placing each vertex (tautologically) in its location in the complex plane, and drawing each edge by the line segment between the centres of the circles it connects (this line segment will pass through the point of tangency of the two circles). Later in these notes we will also have to consider some infinite circle packings, most notably the infinite regular hexagonal circle packing.

The first basic theorem in the subject is the following converse statement:

Theorem 3 (Circle packing theorem)Every connected planar graph is the nerve of a circle packing.

Of course, there can be multiple circle packings associated to a given connected planar graph; indeed, since reflections across a line and Möbius transformations map circles to circles (or lines), they will map circle packings to circle packings (unless one or more of the circles is sent to a line). It turns out that once one adds enough edges to the planar graph, the circle packing is otherwise rigid:

Theorem 4 (Koebe-Andreev-Thurston theorem)If a connected planar graph is maximal (i.e., no further edge can be added to it without destroying planarity), then the circle packing given by the above theorem is unique up to reflections and Möbius transformations.

Exercise 5Let be a connected planar graph with vertices. Show that the following are equivalent:

- (i) is a maximal planar graph.
- (ii) has edges.
- (iii) Every drawing of divides the plane into faces that have three edges each. (This includes one unbounded face.)
- (iv) At least one drawing of divides the plane into faces that have three edges each.
(

Hint:use Euler’s formula , where is the number of faces including the unbounded face.)

Thurston conjectured that circle packings can be used to approximate the conformal map arising in the Riemann mapping theorem. Here is an informal statement:

Conjecture 6 (Informal Thurston conjecture)Let be a simply connected domain, with two distinct points . Let be the conformal map from to that maps to the origin and to a positive real. For any small , let be the portion of the regular hexagonal circle packing by circles of radius that are contained in , and let be an circle packing of with all “boundary circles” tangent to , giving rise to an “approximate map” defined on the subset of consisting of the circles of , their interiors, and the interstitial regions between triples of mutually tangent circles. Normalise this map so that is zero and is a positive real. Then converges to as .

A rigorous version of this conjecture was proven by Rodin and Sullivan. Besides some elementary geometric lemmas (regarding the relative sizes of various configurations of tangent circles), the main ingredients are a rigidity result for the regular hexagonal circle packing, and the theory of quasiconformal maps. Quasiconformal maps are what seem on the surface to be a very broad generalisation of the notion of a conformal map. Informally, conformal maps take infinitesimal circles to infinitesimal circles, whereas quasiconformal maps take infinitesimal circles to infinitesimal ellipses of bounded eccentricity. In terms of Wirtinger derivatives, conformal maps obey the Cauchy-Riemann equation , while (sufficiently smooth) quasiconformal maps only obey an inequality . As such, quasiconformal maps are considerably more plentiful than conformal maps, and in particular it is possible to create piecewise smooth quasiconformal maps by gluing together various simple maps such as affine maps or Möbius transformations; such piecewise maps will naturally arise when trying to rigorously build the map alluded to in the above conjecture. On the other hand, it turns out that quasiconformal maps still have many vestiges of the rigidity properties enjoyed by conformal maps; for instance, there are quasiconformal analogues of fundamental theorems in conformal mapping such as the Schwarz reflection principle, Liouville’s theorem, or Hurwitz’s theorem. Among other things, these quasiconformal rigidity theorems allow one to create conformal maps from the limit of quasiconformal maps in many circumstances, and this will be how the Thurston conjecture will be proven. A key technical tool in establishing these sorts of rigidity theorems will be the theory of an important quasiconformal (quasi-)invariant, the *conformal modulus* (or, equivalently, the extremal length, which is the reciprocal of the modulus).

** — 1. Proof of the circle packing theorem — **

We loosely follow the treatment of Beardon and Stephenson. It is slightly more convenient to temporarily work in the Riemann sphere rather than the complex plane , in order to more easily use Möbius transformations. (Later we will make another change of venue, working in the Poincaré disk instead of the Riemann sphere.)

Define a *Riemann sphere circle* to be either a circle in or a line in together with , together with one of the two components of the complement of this circle or line designated as the “interior”. In the case of a line, this “interior” is just one of the two half-planes on either side of the line; in the case of the circle, this is either the usual interior or the usual exterior plus the point at infinity; in the last case, we refer to the Riemann sphere circle as an *exterior circle*. (One could also equivalently work with an orientation on the circle rather than assigning an interior, since the interior could then be described as the region to (say) the left of the circle as one traverses the circle along the indicated orientation.) Note that Möbius transforms map Riemann sphere circles to Riemann sphere circles. If one views the Riemann sphere as a geometric sphere in Euclidean space , then Riemann sphere circles are just circles on this geometric sphere, which then have a centre on this sphere that lies in the region designated as the interior of the circle. We caution though that this “Riemann sphere” centre does not always correspond to the Euclidean notion of the centre of a circle. For instance, the real line, with the upper half-plane designated as interior, will have as its Riemann sphere centre; if instead one designates the lower half-plane as the interior, the Riemann sphere centre will now be . We can then define a Riemann sphere circle packing in exact analogy with circle packings in , namely finite collections of Riemann sphere circles whose interiors are disjoint and whose union is connected; we also define the nerve as before. This is now a graph that can be drawn in the Riemann sphere, using great circle arcs in the Riemann sphere rather than line segments; it is also planar, since one can apply a Möbius transformation to move all the points and edges of the drawing away from infinity.

By Exercise 5, a maximal planar graph with at least three vertices can be drawn as a triangulation of the Riemann sphere. If there are at least four vertices, then it is easy to see that each vertex has degree at least three (a vertex of degree zero, one or two in a triangulation with simple edges will lead to a connected component of at most three vertices). It is a topological fact, not established here, that any two triangulations of such a graph are homotopic up to reflection (to reverse the orientation). If a Riemann sphere circle packing has the nerve of a maximal planar graph of at least four vertices, then we see that this nerve induces an explicit triangulation of the Riemann sphere by connecting the centres of any pair of tangent circles with the great circle arc that passes through the point of tangency. If was not maximal, one no longer gets a triangulation this way, but one still obtains a partition of the Riemann sphere into spherical polygons.

We remark that the triangles in this triangulation can also be described purely from the abstract graph . Define a *triangle* in to be a triple of vertices in which are all adjacent to each other, and such that the removal of these three vertices from does not disconnect the graph. One can check that there is a one-to-one correspondence between such triangles in a maximal planar graph and the triangles in any Riemann sphere triangulation of this graph.

Theorems 3, 4 are then a consequence of

Theorem 7 (Riemann sphere circle packing theorem)Let be a maximal planar graph with at least four vertices, drawn as a triangulation of the Riemann sphere. Then there exists a Riemann sphere circle packing with nerve whose triangulation is homotopic to the given triangulation. Furthermore, this packing is unique up to Möbius transformations.

Exercise 8Deduce Theorems 3, 4 from Theorem 7. (Hint:If one has a non-maximal planar graph for Theorem 3, add a vertex at the interior of each non-triangular face of a drawing of that graph, and connect that vertex to the vertices of the face, to create a maximal planar graph to which Theorem 4 or Theorem 7 can be applied. Then delete these “helper vertices” to create a packing of the original planar graph that does not contain any “unwanted” tangencies. You may use without proof the above assertion that any two triangulations of a maximal planar graph are homotopic up to reflection.)

Exercise 9Verify Theorem 7 when has exactly four vertices. (Hint:for the uniqueness, one can use Möbius transformations to move two of the circles to become parallel lines.)

To prove this theorem, we will make a reduction with regards to the existence component of Theorem 7. For technical reasons we will need to introduce a notion of non-degeneracy. Let be a maximal planar graph with at least four vertices, and let be a vertex in . As discussed above, the degree of is at least three. Writing the neighbours of in clockwise or counterclockwise order (with respect to a triangulation) as (starting from some arbitrary neighbour), we see that each is adjacent to and (with the conventions and ). We say that is *non-degenerate* if there are no further adjacencies between the , and if there is at least one further vertex in besides . Here is another characterisation:

Exercise 10Let be a maximal planar graph with at least four vertices, let be a vertex in , and let be the neighbours of . Show that the following are equivalent:

- (i) is non-degenerate.
- (ii) The graph is connected and non-empty, and every vertex in is adjacent to at least one vertex in .

We will then derive Theorem 7 from

Theorem 11 (Inductive step)Let be a maximal planar graph with at least four vertices , drawn as a triangulation of the Riemann sphere. Let be a non-degenerate vertex of , and let be the graph formed by deleting (and edges emenating from ) from . Suppose that there exists a Riemann sphere circle packing whose nerve isat least(that is, and are tangent whenever are adjacent in , although we also allow additional tangencies), and whose associated subdivision of the Riemann sphere into spherical polygons is homotopic to the given triangulation with removed. Then there is a Riemann sphere circle packing with nerve whose triangulation is homotopic to the given triangulation. Furthermore this circle packing is unique up to Möbius transformations.

Let us now see how Theorem 7 follows from Theorem 11. Fix as in Theorem 7. By Exercise 9 and induction we may assume that has at least five vertices, and that the claim has been proven for any smaller number of vertices.

First suppose that contains a non-degenerate vertex . Let be the the neighbours of . One can then form a new graph with one fewer vertex by deleting , and then connecting to (one can think of this operation as contracting the edge to a point). One can check that this is still a maximal planar graph that can triangulate the Riemann sphere in a fashion compatible with the original triangulation of (in that all the common vertices, edges, and faces are unchanged). By induction hypothesis, is the nerve of a circle packing that is compatible with this triangulation, and hence this circle packing has nerve at least . Applying Theorem 11, we then obtain the required claim for .

Now suppose that contains a degenerate vertex . Let be the neighbours of traversed in order. By hypothesis, there is an additional adjacency between the ; by relabeling we may assume that is adjacent to for some . The vertices in can then be partitioned as

where denotes those vertices in that lie in the region enclosed by the loop that does not contain , and denotes those vertices in that lie in the region enclosed by the loop that does not contain . One can then form two graphs , formed by restricting to the vertices and respectively; furthermore, these graphs are also maximal planar (with triangulations that are compatible with those of ). By induction hypothesis, we can find a circle packing with nerve , and a circle packing with nerve . Note that the circles are mutually tangent, as are . By applying a Möbius transformation one may assume that these circles agree, thus (cf. Exercise 9) , . The complement of the these three circles (and their interiors) determine two connected “interstitial” regions (that are in the shape of an arbelos, up to Möbius transformation); one can check that the remaining circles in will lie in one of these regions, and the remaining circles in lie in the other. Hence one can glue these circle packings together to form a single circle packing with nerve , which is homotopic to the given triangulation. Also, since a Möbius transformation that fixes three mutually tangent circles has to be the identity, the uniqueness of this circle packing up to Möbius transformations follows from the uniqueness for the two component circle packings , .

It remains to prove Theorem 11. To help fix the freedom to apply Möbius transformations, we can normalise the target circle packing so that is the exterior circle , thus all the other circles in the packing will lie in the closed unit disk . Similarly, by applying a suitable Möbius transformation one can assume that lies outside of the interior of all the circles in the original packing, and after a scaling one may then assume that all the circles lie in the unit disk .

At this point it becomes convenient to switch from the “elliptic” conformal geometry of the Riemann sphere to the “hyperbolic” conformal geometry of the unit disk . Recall that the Möbius transformations that preserve the disk are given by the maps

for real and (see Theorem 19 of these notes). It comes with a natural metric that interacts well with circles:

Exercise 12Define the Poincaré distance between two points of by the formulaGiven a measurable subset of , define the

hyperbolic areaof to be the quantitywhere is the Euclidean area element on .

- (i) Show that the Poincaré distance is invariant with respect to Möbius automorphisms of , thus whenever is a transformation of the form (1). Similarly show that the hyperbolic area is invariant with respect to such transformations.
- (ii) Show that the Poincaré distance defines a metric on . Furthermore, show that any two distinct points are connected by a unique geodesic, which is a portion of either a line or a circle that meets the unit circle orthogonally at two points. (
Hint:use the symmetries of (i) to normalise the points one is studying.)- (iii) If is a circle in the interior of , show that there exists a point in and a positive real number (which we call the
hyperbolic centerandhyperbolic radiusrespectively) such that . (In general, the hyperbolic center and radius will not quite agree with their familiar Euclidean counterparts.) Conversely, show that for any and , the set is a circle in .- (iv) If two circles in are externally tangent, show that the geodesic connecting the hyperbolic centers passes through the point of tangency, orthogonally to the two tangent circles.

Exercise 13 (Schwarz-Pick theorem)Let be a holomorphic map. Show that for all . If , show that equality occurs if and only if is a Möbius automorphism (1) of . (This result is known as the Schwarz-Pick theorem.)

We will refer to circles that lie in the closure of the unit disk as *hyperbolic circles*. These can be divided into the finite radius hyperbolic circles, which lie in the interior of the unit disk (as per part (iii) of the above exercise), and the horocycles, which are internally tangent to the unit circle. By convention, we view horocycles as having infinite radius, and having center at their point of tangency to the unit circle; they can be viewed as the limiting case of finite radius hyperbolic circles when the radius goes to infinity and the center goes off to the boundary of the disk (at the same rate as the radius, as measured with respect to the Poincaré distance). We write for the hyperbolic circle with hyperbolic centre and hyperbolic radius (thus either and , or and is on the unit circle); there is an annoying caveat that when there is more than one horocycle with hyperbolic centre , but we will tolerate this breakdown of functional dependence of on and in order to simplify the notation. A *hyperbolic circle packing* is a circle packing in which all circles are hyperbolic circles.

We also observe that the geodesic structure extends to the boundary of the unit disk: for any two distinct points in , there is a unique geodesic that connects them.

In view of the above discussion, Theorem 7 may now be formulated as follows:

Theorem 14 (Inductive step, hyperbolic formulation)Let be a maximal planar graph with at least four vertices , let be a non-degenerate vertex of , and let be the vertices adjacent to . Suppose that there exists a hyperbolic circle packing whose nerve is at least . Then there is a hyperbolic circle packing homotopic to such that the boundary circles , are all horocycles. Furthermore, this packing is unique up to Möbius automorphisms (1) of the disk .

Indeed, once one adjoints the exterior unit circle to , one obtains a Riemann sphere circle packing whose nerve is at least , and hence equal to since is maximal.

To prove this theorem, the intuition is to “inflate” the hyperbolic radius of the circles of until the boundary circles all become infinite radius (i.e., horocycles). The difficulty is that one cannot just arbitrarily increase the radius of any given circle without destroying the required tangency properties. The resolution to this difficulty given in the work of Beardon and Stephenson that we are following here was inspired by Perron’s method of subharmonic functions, in which one faced an analogous difficulty that one could not easily manipulate a harmonic function without destroying its harmonicity. There, the solution was to work instead with the more flexible class of subharmonic functions; here we similarly work with the concept of a *subpacking*.

We will need some preliminaries to define this concept precisely. We first need some hyperbolic trigonometry. We define a hyperbolic triangle to be the solid (and closed) region in enclosed by three distinct points in and the geodesic arcs connecting them. (Note that we allow one or more of the vertices to be on the boundary of the disk, so that the sides of the triangle could have infinite length.) Let be the space of triples with and not all of infinite. We say that a hyperbolic triangle with vertices is a *-triangle* if there are hyperbolic circles with the indicated hyperbolic centres and hyperbolic radii that are externally tangent to each other; note that this implies that the sidelengths opposite have length respectively (see Figure 3 of Beardon and Stephenson). It is easy to see that for any , there exists a unique -triangle in up to reflections and Möbius automorphisms (use Möbius transforms to fix two of the hyperbolic circles, and consider all the circles externally tangent to both of these circles; the case when one or two of the are infinite may need to be treated separately.). As a consequence, there is a well defined angle for subtended by the vertex of an triangle. We need some basic facts from hyperbolic geometry:

Exercise 15 (Hyperbolic trigonometry)

- (i) (Hyperbolic cosine rule) For any , show that the quantity is equal to the ratio
Furthermore, establish the limiting angles

(Hint: to facilitate computations, use a Möbius transform to move the vertex to the origin when the radius there is finite.) Conclude in particular that is continuous (using the topology of the extended real line for each component of ). Discuss how this rule relates to the Euclidean cosine rule in the limit as go to zero. Of course, by relabeling one obtains similar formulae for and .

- (ii) (Area rule) Show that the area of a hyperbolic triangle is given by , where are the angles of the hyperbolic triangle. (Hint: there are several ways to proceed. For instance, one can prove this for small hyperbolic triangles (of diameter ) up to errors of size after normalising as in (ii), and then establish the general case by subdividing a large hyperbolic triangle into many small hyperbolic triangles. This rule is also a special case of the Gauss-Bonnet theorem in Riemannian geometry. One can also first establish the case when several of the radii are infinite, and use that to derive finite cases.) In particular, the area of a -triangle is given by the formula
- (iii) Show that the area of the interior of a hyperbolic circle with is equal to .

Henceforth we fix as in Theorem 14. We refer to the vertices as *boundary vertices* of and the remaining vertices as *interior vertices*; edges between boundary vertices are *boundary edges*, all other edges will be called *interior edges* (including edges that have one vertex on the boundary). Triangles in that involve two boundary vertices (and thus necessarily one interior vertex) will be called *boundary triangles*; all other triangles (including ones that involve one boundary vertex) will be called *interior triangles*. To any triangle of , we can form the hyperbolic triangle with vertices ; this is an -triangle. Let denote the collection of such hyperbolic triangles; because is a packing, we see that these triangles have disjoint interiors. They also fit together in the following way: if is a side of a hyperbolic triangle in , then there will be another hyperbolic triangle in that shares that side precisely when is associated to an interior edge of . The union of all these triangles is homeomorphic to the region formed by starting with a triangulation of the Riemann sphere by and removing the triangles containing as a vertex, and is therefore homeomorphic to a disk. One can think of the collection of hyperbolic triangles, together with the vertices and edges shared by these triangles, as a two-dimensional (hyperbolic) simplicial complex, though we will not develop the full machinery of such complexes here.

Our objective is to find another hyperbolic circle packing homotopic to the existing circle packing , such at all the boundary circles (circles centred at boundary vertices) are horocycles. We observe that such a hyperbolic circle packing is completely described (up to Möbius transformations) by the hyperbolic radii of these circles. Indeed, suppose one knows the values of these hyperbolic radii. Then each hyperbolic triangle in is associated to a hyperbolic triangle whose sides and angles are known from Exercise 15. As the orientation of each hyperbolic triangle is fixed, each hyperbolic triangle is determined up to a Möbius automorphism of . Once one fixes one hyperbolic triangle, the adjacent hyperbolic triangles (that share a common side with the first triangle) are then also fixed; continuing in this fashion we see that the entire hyperbolic circle packing is determined.

On the other hand, not every choice of radii will lead to a hyperbolic circle packing with the required properties. There are two obvious constraints that need to be satisfied:

- (i) (Local constraint) The angles of all the hyperbolic triangles around any given interior vertex must sum to exactly .
- (ii) (Boundary constraint) The radii associated to boundary vertices must be infinite.

There could potentially also be a global constraint, in that one requires the circles of the packing to be disjoint – including circles that are not necessarily adjacent to each other. In general, one can easily create configurations of circles that are local circle packings but not global ones (see e.g., Figure 7 of Beardon-Stephenson). However, it turns out that one can use the boundary constraint and topological arguments to prevent this from happening. We first need a topological lemma:

Lemma 16 (Topological lemma)Let be bounded connected open subsets of with simply connected, and let be a continuous map such that and . Suppose furthermore that the restriction of to is a local homeomorphism. Then is in fact a global homeomorphism.

The requirement that the restriction of to be a local homeomorphism can in fact be relaxed to local injectivity thanks to the invariance of domain theorem. The complex numbers can be replaced here by any finite-dimensional vector space.

*Proof:* The preimage of any point in the interior of is closed, discrete, and disjoint from , and is hence finite. Around each point in the preimage, there is a neighbourhood on which is a homeomorphism onto a neighbourhood of . If one deletes the closure of these neighbourhoods, the image under is compact and avoids , and thus avoids a neighbourhood of . From this we can show that is a covering map from to . As the base is simply connected, it is its own universal cover, and hence (by the connectedness of ) must be a homeomorphism as claimed.

Proposition 17Suppose we assign a radius to each that obeys the local constraint (i) and the boundary constraint (ii). Then there is a hyperbolic circle packing with nerve and the indicated radii.

*Proof:* We first create the hyperbolic triangles associated with the required hyperbolic circle packing, and then verify that this indeed arises from a circle packing.

Start with a single triangle in , and arbitrarily select a -triangle with the same orientation as . By Exercise 15(i), such a triangle exists (and is unique up to Möbius automorphisms of the disk). If a hyperbolic triangle has been fixed, and (say) is an adjacent triangle in , we can select to be the unique -triangle with the same orientation as that shares the side in common with (with the and vertices agreeing). Similarly for other permutations of the labels. As is a maximal planar graph with non-degenerate (so in particular the set of internal vertices is connected), we can continue this construction to eventually fix every triangle in . There is the potential issue that a given triangle may depend on the order in which one arrives at that triangle starting from , but one can check from a monodromy argument (in the spirit of the monodromy theorem) using the local constraint (i) and the simply connected nature of the triangulation associated to that there is in fact no dependence on the order. (The process resembles that of laying down jigsaw pieces in the shape of hyperbolic triangles together, with the local constraint ensuring that there is always a flush fit locally.)

Now we show that the hyperbolic triangles have disjoint interiors inside the disk . Let denote the topological space formed by taking the disjoint union of the hyperbolic triangles (now viewed as abstract topological spaces rather than subsets of the disk) and then gluing together all common edges, e.g. identifying the edge of with the same edge of if and are adjacent triangles in . This space is homeomorphic to the union of the original hyperbolic triangles , and is thus homeomorphic to the closed unit disk. There is an obvious projection map from to the union of the , which maps the abstract copy in of a given hyperbolic triangle to its concrete counterpart in in the obvious fashion. This map is continuous. It does not quite cover the full closed disk, mainly because (by the boundary condition (ii)) the boundary hyperbolic triangles touch the boundary of the disk at the vertices associated to and but do not follow the boundary arc connecting these vertices, being bounded instead by the geodesic from the vertex to the vertex; the missing region is a lens-shaped region bounded by two circular arcs. However, by applying another homeomorphism (that does not alter the edges from to or to ), one can “push out” the edge of this hyperbolic triangle across the lens to become the boundary arc from to . If one performs this modification for each boundary triangle, one arrives at a modified continuous map from to , which now has the property that the boundary of maps to the boundary of the disk, and the interior of maps to the interior of the disk. Also one can check that this map is a local homeomorphism. By Lemma 16, is injective; undoing the boundary modifications we conclude that is injective. Thus the hyperbolic triangles have disjoint interiors. Furthermore, the arguments show that for each boundary triangle , the lens-shaped regions between the boundary arc between the vertices associated to and the corresponding edge of the boundary triangle are also disjoint from the hyperbolic triangles and from each other. On the other hand, all of the hyperbolic circles and in and their interiors are contained in the union of the hyperbolic triangles and the lens-shaped regions, with each hyperbolic triangle containing portions only of the hyperbolic circles with hyperbolic centres at the vertices of the triangle, and similarly for the lens-shaped regions. From this one can verify that the interiors of the hyperbolic circles are all disjoint from each other, and give a hyperbolic circle packing with the required properties.

In view of the above proposition, the only remaining task is to find an assignment of radii obeying both the local condition (i) and the boundary condition (ii). This is analogous to finding a harmonic function with specified boundary data. To do this, we perform the following analogue of Perron’s method. Define a *subpacking* to be an assignment of radii obeying the following

- (i’) (Local sub-condition) The angles around any given interior vertex sum to at least .

This can be compared with the definition of a (smooth) subharmonic function as one where the Laplacian is always at least zero. Note that we always have at least one subpacking, namely the one provided by the radii of the original hyperbolic circle packing . Intuitively, in each subpacking, the radius at an interior vertex is either “too small” or “just right”.

We now need a key monotonicity property, analogous to how the maximum of two subharmonic functions is again subharmonic:

- (i) Show that the angle (as defined in Exercise 15(i)) is strictly decreasing in and strictly increasing in or (if one holds the other two radii fixed). Do these claims agree with your geometric intuition?
- (ii) Conclude that whenever and are subpackings, that is also a subpacking.
- (iii) Let be such that for . Show that , with equality if and only if for all . (
Hint:increase just one of the radii . One can either use calculus (after first disposing of various infinite radii cases) or one can argue geometrically.)

As with Perron’s method, we can now try to construct a hyperbolic circle packing by taking the supremum of all the subpackings. To avoid degeneracies we need an upper bound:

Proposition 19 (Upper bound)Let be a subpacking. Then for any interior vertex of degree , one has .

The precise value of is not so important for our arguments, but the fact that it is finite will be. This boundedness of interior circles in a circle packing is a key feature of hyperbolic geometry that is not present in Euclidean geometry, and is one of the reasons why we moved to a hyperbolic perspective in the first place.

*Proof:* By the subpacking property and pigeonhole principle, there is a triangle in such that . The hyperbolic triangle associated to has area at most by (2); on the other hand, it contains a sector of a hyperbolic circle of radius and angle , and hence has area at least , thanks to Exercise 15(iv). Comparing the two bounds gives the claim.

Now define to be the (pointwise) supremum of all the subpackings. By the above proposition, is finite at every interior vertex. By Exercise 18, one can view as a monotone increasing limit of subpackings, and is thus again a subpacking (due to the continuity properties of as long as at least one of the radii stays bounded); thus is the maximal subpacking. On the other hand, if is finite at some boundary vertex, then by Exercise 18(i) one could replace that radius by a larger quantity without destroying the subpacking property, contradicting the maximality of . Thus all the boundary radii are infinite, that is to say the boundary condition (ii) holds. Finally, if the sum of the angles at an interior vertex is strictly greater than , then by Exercise 18 we could increase the radius at this vertex slightly without destroying the subpacking property at or at any other of the interior vertices, again contradicting the maximality of . Thus obeys the local condition (i), and we have demonstrated existence of the required hyperbolic circle packing.

Finally we establish uniqueness. It suffices to establish that is the unique tuple that obeys the local condition (i) and the boundary condition (ii). Suppose we had another tuple other than that obeyed these two conditions. Then by the maximality of , we have for all . By Exercise 18(iii), this implies that

for any triangle in . Summing over all triangles and using (2), we conclude that

where the inner sum is over the pairs such that forms a triangle in . But by the local condition (i) and the boundary condition (ii), the inner sum on either side is equal to for an interior vertex and for a boundary vertex. Thus the two sides agree, which by Exercise 18(iii) implies that for all . This proves Theorem 14 and thus Theorems 7, 3, 4.

** — 2. Quasiconformal maps — **

In this section we set up some of the foundational theory of quasiconformal mapping, which are generalisations of the conformal mapping concept that can tolerate some deviations from perfect conformality, while still retaining many of the good properties of conformal maps (such as being preserved under uniform limits), though with the notable caveat that in contrast to conformal maps, quasiconformal maps need not be smooth. As such, this theory will come in handy when proving convergence of circle packings to the Riemann map. The material here is largely drawn from the text of Lehto and Virtanen.

We first need the following refinement of the Riemann mapping theorem, known as Carathéodory’s theorem:

Theorem 20 (Carathéodory’s theorem)Let be a bounded simply connected domain in whose boundary is a Jordan curve, and let be a conformal map between and (as given by the Riemann mapping theorem). Then extends to a continuous homeomorphism from to .

The condition that be a Jordan curve is clearly necessary, since if is not simple then there are paths in that end up at different points in but have the same endpoint in after applying , which prevents being continuously extended to a homeomorphism. If one relaxes the requirement that be a Jordan curve to the claim that is locally connected, then it is possible to modify this argument to still obtain a continuous extension of to , although the extension will no longer be necessarily a homeomorphism, but we will not prove this fact here.

*Proof:* We first prove continuous extension to the boundary. It suffices to show that for every point on the boundary of the unit circle, the diameters of the sets go to zero for some sequence of radii .

First observe from the change of variables formula that the area of is given by , where denotes Lebesgue measure (or the area element). In particular, this integral is finite. Expanding in polar coordinates around , we conclude that

Since diverges near , we conclude from the pigeonhole principle that there exists a sequence of radii decreasing to zero such that

and hence by Cauchy-Schwarz

If we let denote the circular arc , we conclude from this and the triangle inequality (and chain rule) that is a rectifiable curve with length going to zero as . Let denote the endpoints of this curve. Clearly they lie in . If (say) was in , then as is a homeomorphism from to , would have one endpoint in rather than , which is absurd. Thus lies in , and similarly for . Since the length of goes to zero, the distance between and goes to zero. Since is a Jordan curve, it can be parameterised homeomorphically by , and so by compactness we also see that the distance between the parameterisations of and in must also go to zero, hence (by uniform continuity of the inverse parameterisation) and are connected along by an arc whose diameter goes to zero. Combining this arc with , we obtain a Jordan curve of diameter going to zero which separates from the rest of . Sending to infinity, we see that (which decreases with ) must eventually map in the interior of this curve rather than the exterior, and so the diameter goes to zero as claimed.

The above construction shows that extends to a continuous map (which by abuse of notation we continue to call ) from to , and the proof also shows that maps to . As is a compact subset of that contains , it must surject onto . As both and are compact Hausdorff spaces, we will now be done if we can show injectivity. The only way injectivity can fail is if there are two distinct points on that map to the same point. Let be the line segment connecting with , then is a Jordan curve in that meets only at . divides into two regions; one of which must map to the interior of , which implies that there is an entire arc of which maps to the single point . But then by the Schwarz reflection principle, extends conformally across this arc and is constant in a non-isolated set, thus is constant everywhere by analytic continuation, which is absurd. This establishes the required injectivity.

This has the following consequence. Define a *Jordan quadrilateral* to be the open region enclosed by a Jordan curve with four distinct marked points on it in counterclockwise order, which we call the *vertices* of the quadrilateral. The arcs in connecting to or to will be called the *-sides*; the arcs connecting to or to will be called *-sides*. (Thus for instance each cyclic permutation of the vertices will swap the -sides and -sides, while keeping the interior region unchanged.) A key example of a Jordan quadrilateral are the (Euclidean) rectangles, in which the vertices are the usual corners of the rectangle, traversed counterclockwise. The -sides then are line segments of some length , and the -sides are line segments of some length that are orthogonal to the -sides. A *vertex-preserving conformal map* from one Jordan quadrilateral to another will be a conformal map that extends to a homeomorphism from to that maps the corners of to the respective corners of (in particular, -sides get mapped to -sides, and similarly for -sides).

Exercise 21Let be a Jordan quadrilateral with vertices .

- (i) Show that there exists and a conformal map to the upper half-plane (viewed as a subset of the Riemann sphere) that extends continuously to a homeomorphism and which maps to respectively. (Hint: first map to increasing elements of the real line, then use the intermediate value theorem to enforce .)
- (ii) Show that there is a vertex-preserving conformal map from to a rectangle (Hint: use Schwarz-Christoffel mapping.)
- (iii) Show that the rectangle in part (ii) is unique up to affine transformations. (
Hint:if one has a conformal map between rectangles that preserves the vertices, extend it via repeated use of the Schwarz reflection principle to an entire map.)

This allows for the following definition: the *conformal modulus* (or *modulus* for short, also called *module* in older literature) of a Jordan quadrilateral with vertices is the ratio , where are the lengths of the -sides and -sides of a rectangle that is conformal to in a vertex-preserving vashion.. This is a number between and ; each cyclic permutation of the vertices replaces the modulus with its reciprocal. It is clear from construction that the modulus of a Jordan quadrilateral is unaffected by vertex-preserving conformal transformations.

Now we define quasiconformal maps. Informally, conformal maps are homeomorphisms that map infinitesimal circles to infinitesimal circles; quasiconformal maps are homeomorphisms that map infinitesimal circles to curves that differ from an infinitesimal circle by “bounded distortion”. However, for the purpose of setting up the foundations of the theory, it is slightly more convenient to work with rectangles instead of circles (it is easier to partition rectangles into subrectangles than disks into subdisks). We therefore introduce

Definition 22Let . An orientation-preserving homeomorphism between two domains in is said to be-quasiconformalif one has for every Jordan quadrilateral in . (In these notes, we do not consider orientation-reversing homeomorphisms to be quasiconformal.)

Note that by cyclically permuting the vertices of , we automatically also obtain the inequality

or equivalently

for any Jordan quadrilateral. Thus it is not possible to have any -quasiconformal maps for (excluding the degenerate case when are empty), and a map is -conformal if and only if it preserves the modulus. In particular, conformal maps are -conformal; we will shortly establish that the converse claim is also true. It is also clear from the definition that the inverse of a -quasiconformal map is also -quasiconformal, and the composition of a -quasiconformal map and a -quasiconformal map is a -quasiconformal map.

It is helpful to have an alternate characterisation of the modulus that does not explicitly mention conformal mapping:

Proposition 23 (Alternate definition of modulus)Let be a Jordan quadrilateral with vertices . Then is the smallest quantity with the following property: for any Borel measurable one can find a curve in connecting one -side of to another, and which is locally rectifiable away from endpoints, such thatwhere denotes integration using the length element of (not to be confused with the contour integral ).

The reciprocal of this notion of modulus generalises to the concept of extremal length, which we will not develop further here.

*Proof:* Observe from the change of variables formula that if is a vertex-preserving conformal mapping between Jordan quadrilaterals , and is a locally rectifiable curve connecting one -side of to another, then is a locally rectifiable curve connecting one -side of to another, with

and

As a consequence, if the proposition holds for it also holds for . Thus we may assume without loss of generality that is a rectangle, which we may normalise to be with vertices , so that the modulus is . For any measurable , we have from Cauchy-Schwarz and Fubini’s theorem that

and hence by the pigeonhole principle there exists such that

On the other hand, if we set , then , and for any curve connecting the -side from to to the -side from to , we have

Thus is the best constant with the required property, proving the claim.

Here are some quick and useful consequences of this characterisation:

Exercise 24 (Rengel’s inequality)Let be a Jordan quadrilateral of area , let be the shortest (Euclidean) distance between a point on one -side and a point on the other -side, and similarly let be the shortest (Euclidean) distance between a point on one -side and a point on the other -side. Show thatand that equality in either case occurs if and only if is a rectangle.

- (i) If are disjoint Jordan quadrilaterals that share a common -side, and which can be glued together along this side to form a new Jordan quadrilateral , show that . If equality occurs, show that after conformally mapping to a rectangle (in a vertex preserving fashion), , are mapped to subrectangles (formed by cutting the original parallel to the -side).
- (ii) If are disjoint Jordan quadrilaterals that share a common -side, and which can be glued together along this side to form a new Jordan quadrilateral , show that . If equality occurs, show that after conformally mapping to a rectangle (in a vertex preserving fashion), , are mapped to subrectangles (formed by cutting the original parallel to the -side).

Exercise 26 (Continuity from below)Suppose is a sequence of Jordan quadrilaterals which converge to another Jordan quadrilateral , in the sense that the vertices of converge to their respective counterparts in , each -side in converges (in the Hausdorff sense) to the -side of , and the similarly for -sides. Suppose also that for all . Show that converges to . (Hint:map to a rectangle and use Rengel’s inequality.)

Proposition 27 (Local quasiconformality implies quasiconformality)Let , and let be an orientation-preserving homeomorphism between complex domains which is locally -quasiconformal in the sense that for every there is a neighbourhood of in such that is -quasiconformal from to . Then is -quasiconformal.

*Proof:* We need to show that for any Jordan quadrilateral in . The hypothesis gives this claim for all quadrilaterals in the sufficiently small neighbourhood of any point in . For any natural number , we can subdivide into quadrilaterals with modulus with adjacent -sides, by first conformally mapping to a rectangle and then doing an equally spaced vertical subdivision. Similarly, each quadrilateral can be subdivided into quadrilaterals of modulus by mapping to a rectangle and performing horizontal subdivision. By the local -quasiconformality of , we will have

for all , if is large enough. By superadditivity this implies that

for each , and hence

Applying superadditivity again we obtain

giving the claim.

We can now reverse the implication that conformal maps are -conformal:

*Proof:* By covering by quadrilaterals we may assume without loss of generality that (and hence also ) is a Jordan quadrilateral; by composing on left and right with conformal maps we may assume that and are rectangles. As is -conformal, the rectangles have the same modulus, so after a further affine transformation we may assume that is the rectangle with vertices for some modulus . If one subdivides into two rectangles along an intermediate vertical line segment connecting say to for some , the moduli of these rectangles are and . Applying the -conformal map and the converse portion of Exercise 25, we conclude that these rectangles must be preserved by , thus preserves the coordinate. Similarly preserves the coordinate, and is therefore the identity map, which is of course conformal.

Next, we can give a simple criterion for quasiconformality in the continuously differentiable case:

Theorem 29Let , and let be an orientation-preserving diffeomorphism (a continuously (real) differentiable homeomorphism whose derivative is always nondegenerate) between complex domains . Then the following are equivalent:

- (i) is -quasiconformal.
- (ii) For any point and phases , one has
where denotes the directional derivative.

*Proof:* Let us first show that (ii) implies (i). Let be a Jordan quadrilateral in ; we have to show that . From the chain rule one can check that condition (ii) is unchanged by composing with conformal maps on the left or right, so we may assume without loss of generality that and are rectangles; in fact we may normalise to have vertices and to have vertices where and . From the change of variables formula (and the singular value decomposition), followed by Fubini’s theorem and Cauchy-Schwarz, we have

3 and hence , giving the claim.

Now suppose that (ii) failed, then by the singular value decomposition we can find and a phase such that

for some real with . After translations and rotations we may normalise so that

But then from Rengel’s inequality and Taylor expansion one sees that will map a unit square with vertices to a quadrilateral of modulus converging to as , contradicting (i).

Exercise 30Show that the conditions (i), (ii) in the above theorem are also equivalent to the boundfor all , where

are the Wirtinger derivatives.

We now prove a technical regularity result on quasiconformal maps.

Proposition 31 (Absolute continuity on lines)Let be a -quasiconformal map between two complex domains for some . Suppose that contains the closed rectangle with endpoints . Then for almost every , the map is absolutely continuous on .

*Proof:* For each , let denote the area of the image of the rectangle with endpoints . This is a bounded monotone function on and is hence differentiable almost everywhere. It will thus suffice to show that the map is absolutely continuous on whenever is a point of differentiability of .

Let , and let be disjoint intervals in of total length . To show absolute continuity, we need a bound on that goes to zero as uniformly in the choice of intervals. Let be a small number (that can depend on the intervals), and for each let be the rectangle with vertices , , , This rectangle has modulus , and hence has modulus at most . On the other hand, by Rengel’s inequality this modulus is at least , where is a quantity that goes to zero as (holding the intervals fixed). We conclude that

On the other hand, we have

By Cauchy-Schwarz, we thus have

sending , we conclude

giving the claim.

Exercise 32Let be a -quasiconformal map between two complex domains for some . Suppose that there is a closed set of Lebesgue measure zero such that is conformal on . Show that is -conformal (and hence conformal, by Proposition 28). (Hint:Arguing as in the proof of Theorem 29, it suffices to show that of maps the rectangle with endpoints to the rectangle with endpoints , then . Repeat the proof of that theorem, using the absolute continuity of lines at a crucial juncture to justify using the fundamental theorem of calculus.)

Recall Hurwitz’s theorem that the locally uniform limit of conformal maps is either conformal or constant. It turns out there is a similar result for quasiconformal maps. We will just prove a weak version of the result (see Theorem II.5.5 of Lehto-Virtanen for the full statement):

Theorem 33Let , and let be a sequence of -quasiconformal maps that converge locally uniformly to an orientation-preserving homeomorphism . Then is also -quasiconformal.

It is important for this theorem that we do not insist that quasiconformal maps are necessarily differentiable. Indeed for applications to circle packing we will be working with maps that are only piecewise smooth, or possibly even worse, even though at the end of the day we will recover a smooth conformal map in the limit.

*Proof:* Let be a Jordan quadrilateral in . We need to show that . By restricting we may assume . By composing with a conformal map we may assume that is a rectangle. We can write as the increasing limit of rectangles of the same modulus, then for any we have . By choosing going to infinity sufficiently rapidly, stays inside and converges to in the sense of Exercise 26, and the claim then follows from that exercise.

Another basic property of conformal mappings (a consequence of Morera’s theorem) is that they can be glued along a common edge as long as the combined map is also a homeomorphism; this fact underlies for instance the Schwarz reflection principle. We have a quasiconformal analogue:

Theorem 34Let , and let be an orientation-preserving homeomorphism. Let be a real analytic (and topologically closed) contour that lies in except possibly at the endpoints. If is -quasiconformal, then is -quasiconformal.

We will generally apply this theorem in the case when disconnects into two components, in which case can be viewed as the gluing of the restrictions of this map to the two components.

*Proof:* As in the proof of the previous theorem, we may take to be a rectangle , and it suffices to show that . We may normalise to have vertices where , and similarly normalise to be a rectangle of vertices , so we now need to show that . The real analytic contour meets in a finite number of curves, which can be broken up further into a finite horizontal line segments and graphs for various closed intervals and real analytic . For any , we can then use the uniform continuity of the to subdivide into a finite number of rectangles where on each such rectangle, meets the interior of in a bounded number of graphs whose horizontal variation is . This subdivides into a bounded number of Jordan quadrilaterals . If we let denote the distance between the -sides of , then by uniform continuity of and the triangle inequality we have

as . By Rengel’s inequality, we have

since , we conclude using superadditivity that

and hence by Cauchy-Schwarz

and thus

Summing in , we obtain

giving the desired bound after sending .

It will be convenient to study analogues of the modulus when quadrilaterals are replaced by generalisations of annuli. We define a *ring domain* to be a region bounded between two Jordan curves , where (the inner boundary) is contained inside the interior of (the outer boundary). For instance, the annulus is a ring domain for any and . In the spirit of Proposition 23, define the *modulus* of a ring domain to be the supremum of all the quantities with the following property: for any Borel measurable one can find a rectifable curve in winding once around the inner boundary , such that

We record some basic properties of this modulus:

Exercise 35Show that every ring domain is conformal to an annulus. (There are several ways to proceed here. One is to start by using Perron’s method to construct a harmonic function that is on one of the boundaries of the annulus and on the other. Another is to apply a logarithm map to transform the annulus to a simply connected domain with a “parabolic” group of discrete translation symmetries, use the Riemann mapping theorem to map this to a disc, and use the uniqueness aspect of the Riemann mapping theorem to figure out what happens to the symmetry.)

- (i) Show that the modulus of an annulus is given by .
- (ii) Show that if is -quasiconformal and is an ring domain in , then . In particular, the modulus is a conformal invariant. (There is also a converse to this statement that allows for a definition of -quasiconformality in terms of the modulus of ring domains; see e.g. Theorem 7.2 of Lehto-Virtanen.) Use this to extend the definition of modulus of a ring domain to other domains conformal to an annulus, but whose boundaries need not be Jordan curves.
- (iii) Show that if one ring domain is contained inside another (with the inner boundary of in the interior of the inner boundary of ), then .

As a basic application of this concept we have the fact that the complex plane cannot be quasiconformal to any proper subset:

Proposition 37Let be a -quasiconformal map for some ; then .

*Proof:* As is homeomorphic to , it is simply connected. Thus, if we assume for contradiction that , then by the Riemann mapping theorem is conformal to , so we may assume without loss of generality that .

By Exercise 36(i), the moduli of the annuli goes to infinity as , and hence (by Exercise 36(ii) (applied to ) the moduli of the ring domains must also go to infinity. However, as the inner boundary of this domain is fixed and the outer one is bounded, all these ring domains can be contained inside a common annulus, contradicting Exercise 36(iii).

For some further applications of the modulus of ring domains, we need the following result of Grötzsch:

Theorem 38 (Grötzsch modulus theorem)Let , and let be the ring domain formed from by deleting the line segment from to . [Technically, is not quite a ring domain as defined above, but as mentioned in Exercise 36(ii), the definition of modulus remains valid in this case.] Let be another ring domain contained in whose inner boundary encloses both and . Then .

*Proof:* Let , then by Exercise 35 we can find a conformal map from to the annulus . As is symmetric around the real axis, and the only conformal automorphisms of the annulus that preserve the inner and outer boundaries are rotations (as can be seen for instance by using the Schwarz reflection principle repeatedly to extend such automorphisms to an entire function of linear growth), we may assume that obeys the symmetry . Let be the function , then is symmetric around the real axis. One can view as a measurable function on ; from the change of variables formula we have

so in particular is square-integrable. Our task is to show that ; by the definition of modulus, it suffices to show that

for any rectifiable curve that goes once around , and thus once around and in . By a limiting argument we may assume that is polygonal. By repeatedly reflecting around the real axis whenever crosses the line segment between and , we may assume that does not actually cross this segment, and then by perturbation we may assume it is contained in . But then by change of variables we have

by the Cauchy integral formula, and the claim follows.

Exercise 39Let be a sequence of -quasiconformal maps for some , such that all the are uniformly bounded. Show that the are a normal family, that is to say every sequence in contains a subsequence that converges locally uniformly. (Hint:use an argument similar to that in the proof of Proposition 37, combined with Theorem 38, to establish some equicontinuity of the .)

There are many further basic properties of the conformal modulus for both quadrilaterals and annuli; we refer the interested reader to Lehto-Virtanen for details.

** — 3. Rigidity of the hexagonal circle packing — **

We return now to circle packings. In order to understand finite circle packings, it is convenient (in order to use some limiting arguments) to consider some infinite circle packings. A basic example of an infinite circle packing is the *regular hexagonal circle packing*

where is the hexagonal lattice

and is the unit circle centred at . This is clearly an (infinite) circle packing, with two circles in this packing (externally) tangent if and only if they differ by twice a sixth root of unity. Between any three mutually tangent circles in this packing is an open region that we will call an *interstice*. It is inscribed in a *dual circle* that meets the three original circles orthogonally and can be computed to have radius ; the interstice can then be viewed as a hyperbolic triangle in this dual circle in which all three sides have infinite length.

Next we need two simple geometric lemmas, due to Rodin and Sullivan.

Lemma 40 (Ring lemma)Let be a circle that is externally tangent to a chain of circles with disjoint interiors, with each externally tangent to (with the convention ). Then there is a constant depending only on , such that the radii of each of the is at least times the radius of .

*Proof:* Without loss of generality we may assume that has radius and that the radius of is maximal among the radii of the . As the polygon connecting the centers of the has to contain , we see that . This forces , for if was too small then would be so deep in the cuspidal region between and that it would not be possible for to escape this cusp and go around . A similar argument then gives , and so forth, giving the claim.

Lemma 41 (Length-area lemma)Let , and let consist of those circles in that can be connected to the circle by a path of length at most (going through consecutively tangent circles in ). Let be circle packing with the same nerve as that is contained in a disk of radius . Then the circle in associated to the circle in has radius .

The point of this bound is that when is bounded and , the radius of is forced to go to zero.

*Proof:* We can surround by disjoint chains of consecutively tangent circles , in . Each circle is associated to a corresponding circle in of some radius . The total area of these circles is at most the area of the disk of radius . Since , this implies from the pigeonhole principle that there exists for which

and hence by Cauchy-Schwarz

Connecting the centers of these circles, we obtain a polygonal path of length that goes around , and the claim follows.

For every circle in the circle packing , we can form the inversion map across this circle on the Riemann sphere, defined by setting

for and , with the convention that maps to and vice versa. These are conjugates of Möbius transformations; they preserve the circle and swap the interior with the exterior. Let be the group of transformations of generated by these inversions ; this is essentially a Schottky group (except for the fact that we are are allowing for conjugate Möbius transformations in addition to ordinary Möbius transformations). Let denote the union of all the interstices in , and let be the union of the images of the interstitial regions under all of these transformations. We have the following basic fact:

Proposition 42has Lebesgue measure zero.

*Proof:* (Sketch) I thank Mario Bonk for this argument. Let denote all the circles formed by applying an element of to the circles in . If lies in , then it lies inside one of the circles in , and then after inverting through that circle it lies in another circle in , and so forth; undoing the inversions, we conclude that lies in infinite number of nested circles. Let be one of these circles. contains a union of six interstices bounded by and a cycle of six circles internally tangent to and consecutively externally tangent to each other. Applying the same argument used to establish the ring lemma (Lemma 40), we see that the six internal circles have radii comparable to that of , and hence has density in the disk enclosed by , which also contains . The ring lemma also shows that the radius of each circle in the nested sequence is at most times the one enclosing it for some absolute constant , so in particular the disks shrink to zero in size. Thus cannot be a point of density of , and hence by the Lebesgue density theorem this set has measure zero.

We also need another simple geometric observation:

Exercise 43Let be mutually externally tangent circles, and let be another triple of mutually external circles, with the same orientation (e.g. and both go counterclockwise around their interstitial region). Show that there exists a Möbius transformation that maps each to and which maps the interstice of conformally onto the interstice of .

Now we can give a rigidity result for the hexagonal circle packing, somewhat in the spirit of Theorem 4 (though it does not immediately follow from that theorem), and also due to Rodin and Sullivan:

Proposition 44 (Rigidity of infinite hexagonal packing)Let be an infinite circle packing in with the same nerve as the hexagonal circle packing . Then is in fact equal to the hexagonl circle packing up to affine transformations and reflections.

*Proof:* By applying a reflection we may assume that and have the same orientation. For each interstitial region of there is an associated interstitial region of , and by Exercise 43 there is a Möbius transformation . These can be glued together to form a map that is initially defined (and conformal) on the interstitial regions ; we would like to extend it to the entire complex plane by defining it also inside the circles .

Now consider a circle in . It is bounded by six interstitial regions , which map to six interstitial regions that lie between the circle corresponding to and six tangent circles . By the ring lemma, all of the circles have radii comparable to the radius of . As a consequence, the map , which is defined (and piecewise Möbius) on the boundary of as a map to the boundary of , has derivative comparable in magnitude to also. By extending this map radially (in the sense of defining for and , where is the centre of , we see from Theorem 29 that we can extend to be -quasiconformal in the interior of except possibly at for some , and to a homeomorphism from to the region consisting of the union of the disks in and their interstitial regions. By many applications of Theorem 34, is now -quasiconformal on all of , and conformal in the interstitial regions . By Proposition 37, surjects onto , thus the circle packing and all of its interstitial regions cover the entire complex plane.

Next, we use a version of the Schwarz reflection principle to replace by another -quasiconformal map that is conformal on a larger region than . Namely, pick a circle in , and let be the corresponding circle in . Let and be the inversions across and respectively. Note that maps the circle to , with the interior mapping to the interior and exterior mapping to the exterior. We can then define a modified map by setting equal to on or outside , and equal to inside (with the convention that maps to ). This is still an orientation-preserving function ; by Theorem 34 it is still -quasiconformal. It remains conformal on the interstitial region , but is now also conformal on the additional interstitial region . Repeating this construction one can find a sequence of -quasiconformal maps that map each circle to their counterparts , and which are conformal on a sequence of sets that increase up to . By Exercise 39, the restriction of to any compact set forms a normal family (the fact that the circles map to the circles will give the required uniform boundedness for topological reasons), and hence (by the usual diagonalisation argument) the themselves are a normal family; similarly for . Thus, by passing to a subsequence, we may assume that the converge locally uniformly to a limit , and that also converge locally uniformly to a limit which must then invert . Thus is a homeomorphism, and thus -quasiconformal by Theorem 33. It is conformal on , and hence by Proposition 32 it is conformal. But the only conformal maps of the complex plane are the affine maps (see Proposition 15 of this previous blog post), and hence is an affine copy of as required.

By a standard limiting argument, the perfect rigidity of the infinite circle packing can be used to give approximate rigidity of finite circle packings:

Corollary 45 (Approximate rigidity of finite hexagonal packings)Let , and suppose that is sufficiently large depending on . Let and be as in Lemma 41. Let be the radius of the circle in associated to , and let be the radius of an adjacent circle . Then .

*Proof:* We may normalise and . Suppose for contradiction that the claim failed, then one can find a sequence tending to infinity, and circle packings with nerve with , such that the radius of the adjacent circle stays bounded away from . By many applications of the ring lemma, for each circle of , the corresponding circle in has radius bounded above and below by zero. Passing to a subsequence using Bolzano-Weierstrass and using the Arzela-Ascoli diagonalisation argument, we may assume that the radii of these circles converge to a positive finite limit . Applying a rotation we may also assume that the circles converge to a limit circle (using the obvious topology on the space of circles); we can also assume that the orientation of the does not depend on . A simple induction then shows that converges to a limit circle , giving a circle packing with the same nerve as . But then by Lemma 44, is an affine copy of , which among other things implies that . Thus converges to , giving the required contradiction.

A more quantitative version of this corollary was worked out by He. There is also a purely topological proof of the rigidity of the infinite hexagonal circle packing due to Schramm.

** — 4. Approximating a conformal map by circle packing — **

Let be a simply connected bounded region in with two distinct distinguished points . By the Riemann mapping theorem, there is a unique conformal map that maps to and to a positive real. However, many proofs of this theorem are rather nonconstructive, and do not come with an effective algorithm to locate, or at least approximate, this map .

It was conjectured by Thurston, and later proven by Rodin and Sullivan, that one could achieve this by applying the circle packing theorem (Theorem 3) to a circle packing in by small circles. To formalise this, we need some more notation. Let be a small number, and let be the infinite hexagonal packing scaled by . For every circle in , define the *flower* to be the union of this circle, its interior, the six interstices bounding it, and the six circles tangent to the circle (together with their interiors). Let be a circle in such that lies in its flower. For small enough, this flower is contained in . Let denote all circles in that can be reached from by a finite chain of consecutively tangent circles in , whose flowers all lie in . Elements of will be called *inner circles*, and circles in that are not an inner circle but are tangent to it will be called *border circles*. Because is simply connected, the union of all the flowers of inner circles is also simply connected. As a consequence, one can traverse the border circles by a cycle of consecutively tangent circles, with the inner circles enclosed by this cycle. Let be the circle packing consisting of the inner circles and border circles. Applying Theorem 3 followed by a Möbius transformation, one can then find a circle packing in with the same nerve and orientation as , such that all the circles in associated to border circles of are internally tangent to . Applying a Möbius transformation, we may assume that the flower containing in is mapped to the flower containing , and the flower containing is mapped to a flower containing a positive real. (From the exercise below will lie in such a flower for small enough.)

Let be the union of all the solid closed equilateral triangles formed by the centres of mutually tangent circles in , and let be the corresponding union of the solid closed triangles from . Let be the piecewise affine map from to that maps each triangle in to the associated triangle in .

Exercise 46Show that converges to as in the Hausdorff sense. In particular, lies in for sufficiently small .

Exercise 47By modifying the proof of the length-area lemma, show that all the circles in have radius that goes uniformly to zero as . (Hint: for circles deep in the interior, the length-area lemma works as is; for circles near the boundary, one has to encircle by a sequence of chains that need not be closed, but may instead terminate on the boundary of . The argument may be viewed as a discrete version of the one used to prove Theorem 20.) Using this and the previous exercise, show that converges to in the Hausdorff sense.

From Corollary 45 we see that as , the circles in corresponding to adjacent circles of in a fixed compact subset of have radii differing by a ratio of . We conclude that in any compact subset of , adjacent circles in in also have radii differing by a ratio of , which implies by trigonometry that the triangles of in are approximately equilateral in the sense that their angles are . By Theorem 29 is -quasiconformal on each such triangle, and hence by Theorem 34 it is -quasiconformal on . By Exercise 39 every sequence of has a subsequence which converges locally uniformly on , and whose inverses converge locally uniformly on ; the limit is then a homeomorphism from to that maps to and to a positive real. By Theorem 33 the limit is locally -conformal and hence conformal, hence by uniqueness of the Riemann mapping it must equal . As is the unique limit point of all subsequences of the , this implies (by the Urysohn subsequence principle) that converges locally uniformly to , thus making precise the sense in which the circle packings converge to the Riemann map.

## 24 comments

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12 April, 2018 at 11:44 am

Tom HutchcroftThanks for the great post, Terry. I would like to use this comment to direct your readers towards some further reading about the beautiful theory of circle packing and its applications, and in particular to its applications in probability with which I am most familiar.

Good places for more background are Ken Stephenson’s book, and Steffen Rhode’s survey of Oded Schramm’s contributions to the subject https://arxiv.org/pdf/1007.2007.pdf.

First, regarding the material covered in the post:

There is an entirely combinatorial proof of the existence part of the Circle Packing Theorem due to Brightwell and Scheinerman https://epubs.siam.org/doi/abs/10.1137/0406017.

A very short proof of uniqueness for finite circle packings was found by Schramm, who (as I understand it) anonymously added it to this wikipedia article https://en.wikipedia.org/wiki/Circle_packing_theorem. It does not appear elsewhere as far as I know.

The question of existence and uniqueness of circle packings of infinite triangulations in prescribed domains was studied by He and Schramm

https://link.springer.com/article/10.1007/BF02570699 https://www.jstor.org/stable/2946541, who in particular proved the natural analogue of the uniformization theorem in this context, namely that every triangulation of the plane can be circle packed in either the Euclidean plane or the hyperbolic plane, but not both. The earlier paper of Schramm that you mentioned (http://www.ams.org/journals/jams/1991-04-01/S0894-0347-1991-1076089-9/home.html) implies that these packings are unique up to reflections and Mobius transformations. They also proved that, if the triangulation has bounded degrees, then it gets packed in the Euclidean plane if and only if simple random walk on it is recurrent (i.e., visits every vertex infinitely often almost surely). They also studied circle packings of non-simply connected domains, a topic which is closely related to the Koebe conjecture.

The existence and uniqueness problems for infinite packings were later revisited by He in the wonderful paper https://arxiv.org/abs/math/9901148, which in my opinion has the best proofs of both results. His proofs also apply to more general “disk patterns” in which circles are required to intersect with specified angles rather than necessarily being tangent.

Now, for applications in probability:

As I mentioned already, He and Schramm proved that a bounded degree triangulation of the plane can be circle packed in the hyperbolic plane if and only it is transient for simple random walk. This result was extended by Gurel-Gurevich, Nachmias, and Suoto (https://projecteuclid.org/euclid.ecp/1483671681), who proved that the random walk on a bounded degree triangulation circle packed in a domain D is transient if and only if Brownian motion on D is transient, i.e. leaves D in finite time almost surely.

Benjamini and Schramm (https://link.springer.com/article/10.1007/s002220050109) proved that in this case, the simple random walk on the circle packed triangulation converges to a point in the ideal boundary of the hyperbolic plane, and that the law of the limit point is non-atomic and has full support. They used this result to deduce that a bounded degree planar graph admits non-constant bounded harmonic functions if and only if it is transient (equivalently, the invariant sigma-algebra of the random walk on the triangulation is non-trivial if and only if the walk is transient), and in this case it also admits non-constant bounded harmonic functions of finite Dirichlet energy. They also proved the same result using square tiling instead of circle packing in https://projecteuclid.org/euclid.aop/1065725179. I should mention here that the regularity of the harmonic measures on the boundary is not very well understood. For example, it is an open problem to determine whether the harmonic measure of the random walk on the 7-regular hyperbolic triangulation is absolutely continuous with respect to Lebesgue (it is believed not to be, and should have a fractal structure).

The result of Benjamini and Schramm just mentioned allows one to construct harmonic functions on the graph by taking functions on the boundary and integrating them against the harmonic measure. Angel, Barlow, Gurel-Gurevich, and Nachmias (https://projecteuclid.org/euclid.aop/1463410036) proved that, in fact, every bounded harmonic function on a bounded degree triangulation arises in this way. In other words, the boundary of the hyperbolic plane can be identified with the Poisson boundary of the triangulation. They also proved the stronger result that every positive harmonic function on the triangulation admits a representation the harmonic extension of a measure on the boundary of the hyperbolic plane, i.e. that the boundary of the hyperbolic plane can be identified with the Martin boundary of the triangulation. In my paper https://arxiv.org/pdf/1707.07751.pdf, I prove a related representation theorem for harmonic Dirichlet functions on bounded degree triangulations. Angel, Barlow, Gurel-Gurevich and Nachmias also proved various quite strong estimates for the random walk on the triangulation, to the effect that the random walk on the circle packing behaves similarly to a quasi-conformal image of Brownian motion. Their results regarding the Poisson boundary followed earlier work by Georgakopoulos (https://arxiv.org/abs/1301.1506), which established a corresponding result for square tilings. Both results were revisited in my work with Peres (https://projecteuclid.org/euclid.ejp/1511578855), which gave a simplified and unified proof that works for both embeddings.

Circle packing has been very important in the study of random planar maps. In particular, Benjamini and Schramm (https://projecteuclid.org/euclid.ejp/1461097653) used circle packing to prove that every “distributional limit” of finite, bounded degree planar graphs is almost surely recurrent. (These limits are now commonly known as Benjamini-Schramm limits.) Alternative proofs of their result have recently been given by Angel, Nachmias, Ray, and myself (https://arxiv.org/abs/1501.04677) and by Lee (https://arxiv.org/abs/1701.01598).

Gurel-Gurevich and Nachmias (https://arxiv.org/abs/1206.0707) later improved upon the methods of Benjamini and Schramm to prove that the Uniform Infinite Planar Triangulation (UIPT) is recurrent. The UIPT is defined to be the distributional limit of uniform triangulations of the sphere with n vertices as n goes to infinity. The circle packing of the UIPT is conjectured to converge under rescaling to so-called Liouville Quantum Gravity (LQG). Despite the recent and great advances in the understanding of LQG (which includes too many papers and authors to list here), this particular conjecture still seems quite far off.

Some more papers using circle packing to study probabilistic questions on planar graphs include https://projecteuclid.org/euclid.ecp/1456943503, https://arxiv.org/abs/1603.07320, https://www.math.ubc.ca/~mathav/index_files/qs.pdf.

15 April, 2018 at 8:00 am

j.c.Thanks very much for this helpful overview! Your paper with Peres on boundaries of planar graphs for both square tilings and circle packings is particularly interesting.

One non-mathematical comment: Schramm was not anonymous on wikipedia (his username was “OdedSchramm” https://en.wikipedia.org/wiki/User:OdedSchramm ) and his contributions were quite extensive. For instance, the history of the wikipedia page on the circle packing theorem shows that he created the page and wrote the majority of it from 28 – 31 March 2008; for instance, his elementary proof of uniqueness was added in this edit https://en.wikipedia.org/w/index.php?title=Circle_packing_theorem&diff=202165964&oldid=202158966

14 April, 2018 at 2:10 pm

Asaf NachmiasHere’s a neat elementary argument due to Ohad Feldheim and Ori Gurel-Gurevich that avoids the topology in the proof of Prop 17. It’s stated in the Euclidean rather than the hyperbolic setting, but it seems to imply Prop 17 cause one can calculate the Euclidean centers and radii from the hyperbolic ones.

Assume you have an assignment of positive numbers (lengths) to the edges of a maximal planar graph so that for each internal vertex the angles formed at by triangles with edge lengths corresponding to the edges of the faces incident to sum to . Then there is a proper drawing of the graph where an edge is drawn as a straight line segment of length .

Proof: By induction on the number of vertices. For 3 vertices it is obvious. Otherwise take an internal vertex and draw the triangles corresponding to the faces incident . Since the sum of angles is everything “fits” and you get a polygon where is somewhere inside. Triangulate the polygon (two ears theorem) and record the lengths of the diagonals you added. You now have a new maximal planar graph with one vertex less and the new assignment of lengths also give angle sum at each internal vertex. Use the induction hypothesis to draw the new graph with straight lines, and now the location of and the edges emanating from it are determined since the polygon above is drawn exactly the same, so you can add them.

16 April, 2018 at 6:30 am

Terence TaoThis is a nice use of the two ears theorem! It doesn’t fit well pedagogically with the arrangement I have in my blog, but it is certainly worth noting.

15 April, 2018 at 5:04 am

John MangualIt seems quite appealing to build a complex analysis out of arrangements of circles or squares. There are some excellent graphics on the internet (that may not have been available before). E.g. on Wikipedia https://en.wikipedia.org/wiki/Circle_packing#Uniform_packings

My best guess is that holomorphic functions arise naturally in trying to solve differential equations such as Laplace equation or heat equation . However, there is no free lunch since Riemann existence theorem requires Green’s functions to solve.

15 April, 2018 at 5:24 pm

geoishardThis is lovely! Unfortunately I’m confused by Exercise 31. If I’m not being silly, it contradicts Theorem 5.14 (and the sentence following 5.14) in this removability survey: https://arxiv.org/pdf/1503.02582.pdf. But it seems to be important, so maybe I am being silly. Any ideas/help?

15 April, 2018 at 5:34 pm

Terence TaoOops, I had omitted the rather important hypothesis of quasiconformality, now added.

15 April, 2018 at 6:12 pm

geoishardAh, thanks! Very nice.

15 April, 2018 at 9:00 pm

Lexing YingThank you! In Exercise (30), should it be (K-1)/(K+1)?

[Corrected, thanks – T.]17 April, 2018 at 7:47 pm

Dan AsimovSo now I’m wondering what the derivative of a 1-parameter family of circle-packings is …

17 April, 2018 at 11:15 pm

Coloring Problems for Arrangements of Circles (and Pseudocircles) | Combinatorics and more[…] Koebe’s circle packing theorem (see this recent post) this is precisely the four color theorem so the answer is […]

20 April, 2018 at 4:27 am

John MangualHave you seen the notes of Stanislav Smirnov “Discrete Complex Analysis and Probability” https://arxiv.org/abs/1009.6077 these are from 8 years ago.

Actually these are totally different from the circle packings approach outlined by Thurston, Rodin, Sullivan, Stephenson and others. There’s also “Discrete Differential Geometry” by Bobenko-Suris. http://page.math.tu-berlin.de/~bobenko/ddg-book.html

I have trouble memorizing the Cauchy-Riemann equations, so I start with the Taylor approximation and I observe it could not matter if . I’m not 100% sure I can even do it still. Now that I’ve dropped out of school, I rarely use this material. however .

So I will obtain or so that looks right. Good.

For the longest time, I’ve wondered… what’s so special about the Cauchy-Riemann equations? I haven’t even used this result in any intelligent way yet. For now, I content myself to know this is connected to circle packings and the ising model.

20 April, 2018 at 10:19 am

AnonymousIn Exercise 10, should the first sentence read “Let G be a maximal planar graph with at least four vertices” rather than “at most”?

[Corrected, thanks – T.]24 April, 2018 at 3:37 pm

Anonymoussame exercise (10): should the last set be G\{v,v1,…,vd} rather than {v,v1,…,vd}?

[Corrected, thanks -T.]30 April, 2018 at 11:28 am

AnonymousIn Exercise 15(i), should the last formula be

cos alpha_1(r_1,infty,infty) = 1 – 2exp(-2r_1)?

[Corrected, thanks – T.]2 May, 2018 at 2:18 pm

246C notes 3: Univalent functions, the Loewner equation, and the Bieberbach conjecture | What's new[…] To prove (i), it suffices from (11) and the Schwarz-Pick lemma (Exercise 13 from Notes 2) to establish this claim when . We can also assume that since the claim is trivial when . From the […]

6 May, 2018 at 4:00 am

John MangualIs are discrete conformal mappings such as (square tilings and circle packings) quasiconformal? We have that and

Looks like sections 1 on Circle Packings and Section 2 on Quasiconformal maps combine in into sections 3 and 4 on rigidity and approximation.

You can draw a map of the United States out of squares using discrete conformal mappings. I wonder how they handled Utah-Arizona-New Mexico-Colorado?

These notes of Kenyon and Parry offer six models of the hyperbolic plane. At the moment I can only think of and .

http://library.msri.org/books/Book31/files/cannon.pdf

10 May, 2018 at 2:19 pm

AnonymousIn Exercise 35, part (ii), the phi and mod should be interchanged.

[Corrected, thanks – T.]29 May, 2018 at 9:30 am

246C notes 4: Brownian motion, conformal invariance, and SLE | What's new[…] This gives an inequality similar in spirit to the Grötzsch modulus estimate from Notes 2: […]

16 July, 2018 at 7:25 am

Nick LohrIn the two paragraphs following right after the statement in Theorem 11, you mention Theorem 14 without stating it. Do you mean to say Theorem 11? In the last paragraph before Exercise 12, you say, “It remains to show Theorem 7.” Do you mean Theorem 11?

Also, isn’t it true that there exists a unique hyperbolic triangle with all sides infinite in $\overline{D(0,1)}$ up to reflections and Möbius automorphisms? If so, is there a reason why its omitted from this discussion?

Lastly, a pedagogical question: Is it possible to reformulate Theorem 7 in the hyperbolic setting and prove Theorem 3,4 from it? This way one can get rid of Theorem 11 entirely and just deal with Theorem 14, right?

17 July, 2018 at 3:04 pm

Terence TaoThanks for the corrections. While a hyperbolic triangle with all infinite sides is indeed unique up to isometries, a hyperbolic triangle with three tangent circles centred at the vertices is not unique in this fashion, and this is really the geometric object of relevance in circle packings. Also some of the continuity statements fail if one tries to add this endpoint to the configuration space of triangles.

One can rearrange the presentation of the various theorems slightly, and one can certainly phrase Theorem 7 in the hyperbolic setting by specialising to the case when one of the circles is the exterior of the unit disk, but the inductive step still requires the use of Mobius transformations that preserve the Riemann sphere but not the hyperbolic disk, so I don’t see how to easily move the proof to entirely lie within the hyperbolic setting.

15 February, 2019 at 9:04 am

Nick LohrThank you. A few small typos: Before and after you use Cauchy-Schwarz in the proof of Caratheodory’s theorem the in the integrals should be . In the proof of Proposition 23, the first equality should be an integral over and not . Also, in this proof, you use but isn’t it ?

[Corrected, thanks – T.]1 April, 2019 at 7:08 am

Nick LohrSome more typos. In the statement and proof of Proposition 31, the should be , the ‘s should be just ‘s, and there is a that should be . Right before Exercise 35 when you are defining modulus for the annulus, the should be just to avoid complex numbers in the inequality.

[Corrected, thanks – T.]9 April, 2019 at 9:43 am

Nick LohrSome more (important) typos (I’m doing this for my senior thesis which is why I am being so pesky). In the first indented line in Theorem 38, I think we only know in the first equality. In the second indented line in this theorem, the should be and the should be flipped to (with underneath the area form). In the last indented line of the theorem, the ‘s should be ‘s and the needs to be flipped to .

[Corrected, thanks – T.]