Parabolic and Hyperbolic Packings

Abstract

In this chapter we discuss countably infinite connected simple graphs that are locally finite , that is, the vertex degrees are finite. In a similar fashion to the previous chapter, an infinite planar graph is a connected infinite graph such that there exists a drawing of it in the plane. We recall that a drawing is a correspondence sending vertices to points of $\mathbb {R}^2$ and edges to continuous curves between the corresponding vertices such that no two edges cross. An infinite planar map is an infinite planar graph equipped with a set of cyclic permutations {σ _v : v ∈ V } of the neighbors of each vertex v, such that there exists a drawing of the graph which respects these permutations, that is, the clockwise order of edges emanating from a vertex v coincides with σ _v.

4.1 Infinite Planar Maps

In this chapter we discuss countably infinite connected simple graphs that are locally finite , that is, the vertex degrees are finite. In a similar fashion to the previous chapter, an infinite planar graph is a connected infinite graph such that there exists a drawing of it in the plane. We recall that a drawing is a correspondence sending vertices to points of $\mathbb {R}^2$ and edges to continuous curves between the corresponding vertices such that no two edges cross. An infinite planar map is an infinite planar graph equipped with a set of cyclic permutations {σ _v : v ∈ V } of the neighbors of each vertex v, such that there exists a drawing of the graph which respects these permutations, that is, the clockwise order of edges emanating from a vertex v coincides with σ _v.

Unlike the finite case, one cannot define faces as the connected components of the plane with the edges removed since the drawing may have a complicated set of accumulation points. This is the reason that we have defined faces in Sect. 3.1 combinatorially, that is, based solely on the edge set and the cyclic permutation structure. This definition makes sense in both the finite and infinite case. In the latter case we may have infinite faces.

A (finite or infinite) planar map is a triangulation if each of its faces has exactly 3 edges. Given a drawing of a triangulation, the Jordan curve theorem implies that the edges of each face bound a connected component of the plane minus the edges. We will often refer to the faces as these connected components. A triangulation is called a plane triangulation if there exists a drawing of it such that every point of the plane is contained in either a face or an edge and any compact subset of the plane intersects at most finitely many edges and vertices. The term disk triangulation is also used in the literature and means the same with the unit disk taking the place of the plane in the previous definition. Of course these two definitions are equivalent since the plane and the open disk are homeomorphic. For example, take the product of the complete graph K ₃ on 3 vertices with an infinite ray $\mathbb {N}$ and add a diagonal edge in each face that has 4 edges; this is a plane triangulation. However, the product of K ₃ with a bi-infinite ray $\mathbb {Z}$ together with the same diagonals is a triangulation but not a plane triangulation, since it cannot be drawn in the plane without an accumulation point.

It turns out that there is a combinatorial criterion for a triangulation to be a plane/disk triangulation. We say that an infinite graph is one-ended if the removal of any finite set of its vertices leaves exactly one infinite connected component.

Lemma 4.1

An infinite triangulation is a plane triangulation if and only if it is one-ended.

Proof

Suppose G = (V, E) is a plane triangulation and consider a drawing of the graph with no accumulation points in the plane such that every point of the plane belongs to either an edge or a face. Let A ⊆ V be a finite set of vertices and take $B \subset \mathbb {R}^2$ to be a ball around the origin which contains every vertex of A, every edge touching a vertex of A and every face incident to such an edge. Let u ≠ v be two vertices drawn outside of B and take a continuous curve γ between them in $\mathbb {R}^2 \setminus B$. By definition of B, this path only touches faces and edges that are not incident to the vertices of A and hence one can trace a discrete path from u to v in the graph that “follows” γ and avoids A. Since B intersects only finitely many edges and vertices, we learn that G ∖ A has a unique infinite component.

Conversely, assume now that G is one-ended and consider a drawing of G in the plane. By the stereographic projection we project the drawing to the unit sphere $\mathbb {S}^2$ in $\mathbb {R}^3$. Denote by $\mathcal {I}$ the complement in $\mathbb {S}^2$ of the union of all faces and edges. Since G is an infinite triangulation this union is an open set, hence $\mathcal {I}$ is a closed set and its boundary $\partial \mathcal {I}$ is precisely the set of accumulation points of the drawing. Since $\mathcal {I}$ is closed, each connected component of $\mathcal {I}$ must be closed as well and hence contain at least one accumulation point. Since G is one-ended $\mathcal {I}$ cannot have more than one connected component, since otherwise we would be able to separate the two components by a finite set of edges and obtain two infinite connected components. Now choose a point $p \in \mathcal {I}$ and rotate the sphere so that p is the north pole. Project back the rotated sphere to the plane and consider the drawing in the plane. In this drawing the union of all faces and edges must be a simply connected set. By the Riemann mapping theorem this set is homeomorphic to the whole plane, and we deduce that the triangulation is a plane triangulation. □

4.2 The Ring Lemma and Infinite Circle Packings

The circle packing theorem Theorem 3.5 is stated for finite planar maps. However, it is not hard to argue that any infinite map also has a circle packing. To this aim we will prove what is known as Rodin and Sullivan’s Ring Lemma [70]; we will use it many times throughout this book. Given circles C ₀, C ₁, …, C _M with disjoint interiors, we say that C ₁, …, C _M completely surround C ₀ if they are all tangent to C ₀ and C _i is tangent to C _i+1 for i = 1, …, M (where C _M+1 is set to be C ₁).

Lemma 4.2 (Ring Lemma, Rodin and Sullivan [70])

For every integer M > 0 there exists A > 0 such that if C ₀ is a circle completely surrounded by M circles C ₁, …, C _M , and r _i is the radius of C _i for every i = 0, 1, …, M, then r ₀∕r _i ≤ A for every i = 1, …, M.

Proof

We may scale the picture so that r ₀ = 1. Assume that the radius of C ₂ is small and consider the circles C ₁ and C ₃ to its left and right. It cannot be that both C ₁ and C ₃ have large radii compared to C ₂ since in this case they will intersect; see Fig. 4.1. Hence, one of them has to be small as well. Assume without loss of generality that it is C ₃. By similar reasoning, one of C ₁ and C ₄ has to be small. We continue this argument this way and get a path of circles of small radii; thus, for the circles C ₁, …, C _M to completely surround C ₀ we learn that M must be large. □

Fig. 4.1

C ₂ is small, but both C ₁ and C ₃ are large

For a circle packing P and a vertex v, denote by C _v the circle corresponding to v, by the center of that circle, and by $\operatorname {\mathrm {rad}}(v)$ its radius. We write G(P) for the tangency graph of the packing P, that is, the graph in which each vertex is a circle of P and two such circles form an edge when they are tangent.

Claim 4.3

Let G be an infinite simple planar map. Then there exists a circle packing P such that G(P) is isomorphic to G as planar maps.

Proof

If G is not a triangulation, then it is always possible to add in each face new vertices and edges touching them so the resulting graph is a planar triangulation (in an infinite face we have to put infinitely many vertices). After circle packing this new graph, we can remove all the circles corresponding to the added vertices and remain with a circle packing of G. Thus, we may assume without loss of generality that G is a triangulation.

Fix a vertex x, and let G _n be the graph distance ball of radius n around x. Apply the circle packing theorem to G _n to obtain a packing P _n, and scale and translate it so that $\operatorname {\mathrm {rad}}(x)=1$ and is the origin.

Consider a neighbor y of x. By the Ring Lemma (Lemma 4.2), there exists a constant A = A(x, y) > 0 such that $A^{-1}\le \operatorname {\mathrm {rad}}(y) \le A$. By compactness there exists a subsequence of packings $P_{n_k}$ for which $\operatorname {\mathrm {rad}}_{n_k}(y)$ and both converge. By taking further subsequences for the rest of x’s neighbors, and then for the rest of the graph’s vertices, it follows by a diagonalization argument that there exists a subsequence such that the radii and centers of all vertices converge. The limiting packing P _∞ satisfies that G(P _∞) is isomorphic to G. □

4.3 Statement of the He–Schramm Theorem

Given a circle packing P of a triangulation G, we define the carrier of P, denoted $\operatorname {\mathrm {Carrier}}(P)$, to be the union of the closed discs bounded by the circles of P together with the spaces bounded between any three circles that form a face (i.e., the interstices). When P is a circle packing of an infinite one-ended triangulation, the argument in Lemma 4.1 shows that $\operatorname {\mathrm {Carrier}}(P)$ is simply connected.

We say that G is circle packed in $\mathbb {R}^2$ when $\operatorname {\mathrm {Carrier}}(P)=\mathbb {R}^2$. Denote by $\mathbb {U}$ the disk $\{z \in \mathbb {R}^2 : |z|<1\}$; we say that G is circle packed in $\mathbb {U}$ when $\operatorname {\mathrm {Carrier}}(P)=\mathbb {U}$. See Fig. 4.2.

Fig. 4.2

Two circle packings with carriers $\mathbb {R}^2$ (left) and $\mathbb {U}$ (right)

Let G be a plane triangulation. Then G can be drawn in the plane $\mathbb {R}^2$ or alternatively in the disk $\mathbb {U}$ (since they are homeomorphic), but can it be circle packed both in $\mathbb {R}^2$ and in $\mathbb {U}$? A celebrated theorem of He and Schramm [40] states that this cannot be done: each plane triangulation can be circle packed in either the plane or the disk, but not both. In fact, the combinatorial property of G that determines on which side of the dichotomy we are is the recurrence or transience of the simple random walk on G (assuming also that G has bounded degrees, that is, sup_{x ∈ V (G)}deg(x) < ∞). This is the content of the He–Schramm theorem, which we are now ready to state.

Theorem 4.4 (He and Schramm [40])

Let G be an infinite simple plane triangulation with bounded degrees.

• 1. If G is recurrent, then there exists a circle packing P of G such that $\operatorname {\mathrm {Carrier}}(P) =~\mathbb {R}^2$.

• 2. If G is transient, then there exists a circle packing P of G such that $\operatorname {\mathrm {Carrier}}(P) =~\mathbb {U}$.

• 3. If P is a circle packing of G with $\operatorname {\mathrm {Carrier}}(P) = \mathbb {R}^2$ , then G is recurrent.

• 4. If P is a circle packing of G with $\operatorname {\mathrm {Carrier}}(P) = \mathbb {U}$ , then G is transient.

Remark 4.5

Schramm [73] proved that a circle packing P of a triangulation G(P) with $\operatorname {\mathrm {Carrier}}(P)=\mathbb {R}^2$ is uniquely determined up to dilations, rotations and translations. If $\operatorname {\mathrm {Carrier}}(P) = \mathbb {U}$ the same holds up to Möbius transformations of $\mathbb {U}$ onto itself (see also [37]). Hence the packings guaranteed to exist in Theorem 4.4 (1) and Theorem 4.4 (2) are unique in this sense.

Corollary 4.6

Any bounded degree plane triangulation can be circle-packed in $\mathbb {R}^2$ or $\mathbb {U}$ , but not both.

Remark 4.7

In fact, it is proved in [40] that the corollary above holds without the assumption of bounded degree. Furthermore, in [40] Theorem 4.4 (1) and Theorem 4.4 (4) are proved without the bounded degrees assumption, but the other two statements require this assumption.

The following example demonstrates why the bounded degree condition is necessary for Theorem 4.4 (2) and Theorem 4.4 (3).

Example 4.8

Let P be a triangular lattice circle packing (as in Fig. 4.3), and let C ₀, C ₁, C ₂, … be an infinite horizontal path of circles in P going (say) to the right. In the upper face shared by C _n and C _n+1, draw 2 ⁿ circles which form a vertical path and each of them tangent both to C _n and C _n+1; the last circle of these is also tangent to the upper neighbor of C _n and C _n+1. See Fig. 4.3.

The resulting graph is a plane triangulation and the carrier of the packing is $\mathbb {R}^2$. However, it is an easy exercise to verify that the tangency graph of this circle packing is transient.

Fig. 4.3

Unbounded degree transient triangulation circle packed in $\mathbb {R}^2$

In the rest of this chapter we prove Theorem 4.4. We begin by proving parts 3 and 4, in which a circle packing is given and we use its geometry to estimate certain effective resistances. Afterwards we prove parts 1 and 2, in which we use the electrical estimates to deduce facts about the geometry of the circle packing.

4.4 Proof of the He–Schramm Theorem

4.4.1 Proof of Theorem 4.4 (3)

Denote the circle packing P = {C _v}_{v ∈ V} where V is the vertex set of G and C _v denotes the circle corresponding to the vertex v. Write Δ for the maximum degree of G and fix a vertex v ₀. By scaling and translating we may assume that $C_{v_0}$ is a radius 1 circle around the origin. For a real number R > 0, let V _R = V _B(0,R) denote set of vertices v for which is in the Euclidean ball of radius R around the origin.

Lemma 4.9

There exist C = C(Δ) > 1 and c = c(Δ) > 0 such that for every R ≥ 1 we have

• (i) There are no edges between V _R and V ∖ V _CR , and

• (ii) $\mathcal {R}_{\mathrm {eff}}\left (V_R\leftrightarrow V \setminus V_{CR} \right )\ge c$.

Proof

We begin with part (i). For every v ∈ V _R it holds that $\operatorname {\mathrm {rad}}(v)\le R$ since $C_{v_0}$ is centered at the origin. By the Ring Lemma (Lemma 4.2), there exists A = A(Δ) such that $\operatorname {\mathrm {rad}}(u)\le AR$ for every u ∼ v , and therefore . Hence (i) holds with C = A + 2.

To prove part (ii) we define

Recall from Lemma 2.32 that $\mathcal {R}_{\mathrm {eff}}\left (V_R\leftrightarrow V \setminus V_{CR} \right )\ge \mathcal {E}(h)^{-1}$. By the triangle inequality, for an edge {x, y} with both endpoints in V _CR ∖ V _R we have

and it is straightforward to check that the same bound holds also when one of the edge’s endpoints is in V _R or V ∖ V _CR. Thus, using the Ring Lemma’s (Lemma 4.2) constant A = A(Δ) from part (i),

$$\displaystyle \begin{aligned} \mathcal{E}(h) \le \sum_{x \in V_{CR} \setminus V_R} \sum_{y: y \sim x} \frac{( (A+1)\operatorname{\mathrm{rad}}(x) )^2}{(C-1)^2R^2} \le \frac{\varDelta(A+1)^2}{\pi(C-1)^2R^2}\cdot \sum_{x\in V_{CR \setminus V_R}} \operatorname{\mathrm{area}}(C_x), \end{aligned}$$

where $\operatorname {\mathrm {area}}(C_x)$ is the area that C _x encloses (that is, $\pi \operatorname {\mathrm {rad}}(x)^2$). We have that $\sum _x \operatorname {\mathrm {area}}(C_x)\le \operatorname {\mathrm {area}}(B(\mathbf {0},2CR)) = 4\pi C^2R^2$, hence if C = A + 2, then

$$\displaystyle \begin{aligned} \mathcal{E}(h) \le 4\varDelta C^2, \end{aligned}$$

and the result follows for c = (4ΔC ²)⁻¹. □

Proof of Theorem 4.4 (3)

Consider the unit current flow I from v ₀ to ∞ and fix any R ≥ 1. Restricting this flow to the edges which have at least one endpoint in the annulus V _CR ∖ V _R gives a unit flow from V _R to V ∖ V _CR, by part (i) of Lemma 4.9. Hence, by part (ii) of that lemma and by Thomson’s principle (Theorem 2.28), the energy contributed to $\mathcal {E}(I)$ from these edges is at least c. In the same manner, the edges which have at least one endpoint in the annulus $V_{C^{2k+1}R} \setminus V_{C^{2k}R}$ contribute at least c to $\mathcal {E}(I)$. Part (i) of Lemma 4.9 implies that all these edge sets are disjoint, hence $\mathcal {E}(I)=\infty$ and we learn that G is recurrent (Corollary 2.39). □

4.4.2 Proof of Theorem 4.4 (4)

We will use the given circle packing of G to create a random path to infinity with finite energy. This gives transience by Claim 2.46. This proof strategy is similar to that of Theorem 2.47.

Proof of Theorem 4.4 (4)

Let v ₀ be a fixed vertex of the graph, and apply a Möbius transformation to make the circle of P corresponding to v ₀ be centered at the origin 0. We now use Claim 2.46 to construct a flow θ from v ₀ to ∞ by choosing a uniform random point p on $\partial \mathbb {U}$, taking the straight line from 0 to p and considering the set of all circles in the packing P that intersect this line in the order that they are visited; this set forms an infinite simple path in the graph which starts at v ₀.

To bound the energy of the flow, we claim that there exists some constant C (which may depend on the graph G and the packing P) such that the probability that the random path uses the vertex v is bounded above by $C \operatorname {\mathrm {rad}}(v)$. Indeed, since there are only finitely many vertices with centers at distance at most 1∕2 from 0, we may assume that the center of v is of distance at least 1∕2 from 0. In this case, in order for v to be included in the random path the circle of v must intersect the line between 0 and p. By the Ring Lemma (Lemma 4.2) the neighbors of v have circles of radii comparable to $\operatorname {\mathrm {rad}}(v)$ and so the probability of the line touching them is at most $C \operatorname {\mathrm {rad}}(v)$. Since the vertex degree is bounded by Δ and $\sum _{v \in V} \pi \operatorname {\mathrm {rad}}(v)^2$ is at most the area of $\mathbb {U}$, we find that

$$\displaystyle \begin{aligned} \mathcal{E}(\theta) \le C \varDelta \sum_{v \in V} \operatorname{\mathrm{rad}}(v)^2 \leq C \varDelta \, . \end{aligned}$$

Hence G is transient by Corollary 2.39 □

4.4.3 Proof of Theorem 4.4 (1)

We apply Claim 4.3 to obtain a circle packing P of G and prove that $\operatorname {\mathrm {Carrier}}(P)=\mathbb {R}^2$. Fix some vertex v and rescale and translate so that P(v) is the unit circle $\partial \mathbb {U}$. Assume by contradiction that $\operatorname {\mathrm {Carrier}}(P) \neq \mathbb {R}^2$ and let $p \in \mathbb {R}^2 \setminus \operatorname {\mathrm {Carrier}}(P)$ be a point not in the carrier. Rotate the packing so that p = R for some real number R > 1. Let U ∈ [−1, 1] and consider the circle C _U = {z : |z − p| = R − U}. We traverse C _U from the point U counterclockwise and consider all the circles of P which intersect C _U. These circles form a simple path in the graph G starting from v. Since $\operatorname {\mathrm {Carrier}}(P)$ is simply connected by Lemma 4.1 and $p\not \in \operatorname {\mathrm {Carrier}}(P)$ it cannot be that $C_U\subset \operatorname {\mathrm {Carrier}}(P)$. Thus, as we traverse C _U counterclockwise we must hit the boundary of $\operatorname {\mathrm {Carrier}}(P)$. We conclude that the path in G we obtained in this manner is an infinite simple path starting at v.

We now let U be a uniform random variable in [−1, 1] and let μ denote the probability measure on random infinite paths starting at v we obtained as described above. Let θ be the flow induced by μ as in Claim 2.46. We wish to bound the energy $\mathcal {E}(\theta )$. Consider a vertex w ∈ G and its corresponding circle C _w and let B be the Euclidean ball of radius R + 1 around p. If C _w does not intersect B, it cannot be included in the random path by our construction. If it does intersect this ball, then the probability that the random path intersects it is bounded above by its radius. Thus as in the proof of Theorem 4.4 (4),

$$\displaystyle \begin{aligned}\mathcal{E}(\theta) \leq C \varDelta \sum_{w : C_w \cap B \neq \emptyset} \operatorname{\mathrm{rad}}(w)^2 \, ,\end{aligned}$$

where Δ is the maximal degree of G and we have used the Ring Lemma (Lemma 4.2). We learn that $\mathcal {E}(\theta )$ is bounded above by a constant multiple of the area of all circles of P that intersect B. Since $p \not \in \operatorname {\mathrm {Carrier}}(P)$, by the Ring Lemma (Lemma 4.2), any circle of P that intersects B cannot have radius more than AR for some large A ≥ R (since otherwise, all the circles surrounding this vertex will have radius more than R + 1, contradicting the fact that $p \not \in \operatorname {\mathrm {Carrier}}(P)$). We learn that all the circles counted in the sum above are contained in the Euclidean ball of radius (A + 1)R + 1 around p. Since these circles has disjoint interiors, the sum of their area is bounded above by the area of the Euclidean ball above. We conclude that $\mathcal {E}(\theta )<\infty$, hence G is transient by Corollary 2.39 and we have reached a contradiction. □

4.4.4 Proof of Theorem 4.4 (2)

We will use the following simple corollary of the circle packing theorem, Theorem 3.5. A finite triangulation with boundary is a finite connected simple planar map in which all faces are triangles except for a distinguished outer face whose boundary is a simple cycle.

Claim 4.10

Let G be a finite triangulation with boundary. Then, there is a circle packing P of G such that all circles of the outer face are internally tangent to $\partial \mathbb {U}$ and all other circles of P are contained in $\mathbb {U}$.

Proof

Denote by v ₁, …, v _m the vertices of the outer face ordered according to the cycle they form. Add a new vertex v ^∗ to the graph and connect it to v ₁, …, v _m according to their order. We obtain a finite triangulation G ^∗. Apply Theorem 3.5 to obtain a circle packing $P=\{C_v\}_{v \in V(G^*)}$. By translating and dilating we may assume that $C_{v^*}$ is centered at the origin and has radius 1. Apply the map $z \mapsto \frac {1}{z}$ on this packing. Since this map preserves circles, the image of the circles $\{C_v\}_{v \in V(G^*)\setminus \{v^*\}}$ under this map is precisely the desired circle packing. □

Furthermore, we will require an auxiliary general estimate. Given a circle packing P and a set of vertices A, we write $\operatorname {\mathrm {diam}}_P(A)$ for the Euclidean diameter of the union of all circles in P corresponding to the vertices of A.

Lemma 4.11

Let P be a circle packing contained in $\mathbb {U}$ of a finite triangulation with boundary with maximum degree Δ, such that the circle of a chosen non-boundary vertex v ₀ is centered at the origin and has radius r ₀ . Assume that $r_0 \geq r_{\min }$ for some constant $r_{\min }>0$ . Then there exists a constant $c=c(r_{\min }, \varDelta )>0$ such that for any connected set A of vertices,

$$\displaystyle \begin{aligned} \mathcal{R}_{\mathrm{eff}}(v_0 \leftrightarrow A) \geq c \log \frac{1}{\operatorname{\mathrm{diam}}_P(A)} \, .\end{aligned} $$

(4.1)

If in addition all circles of the outer face are tangent to $\partial \mathbb {U}$ and A contains a vertex of the outer face, then

$$\displaystyle \begin{aligned} \mathcal{R}_{\mathrm{eff}}(v_0 \leftrightarrow A) \leq c^{-1} \log \frac{1}{\operatorname{\mathrm{diam}}_P(A)\wedge{1 \over 2}} \, .\end{aligned} $$

(4.2)

Proof

Write $\varepsilon = \operatorname {\mathrm {diam}}_P(A)$ and let z(A) denote the union of all circles corresponding to the vertices of A. We begin with the proof of (4.1), which goes along similar lines to the proof of Lemma 4.9. Let $z_0\in \mathbb {R}^2$ be such that z(A) ⊂{|z − z ₀|≤ ε}. For any r > 0 denote by V _r the set of vertices whose corresponding circles have centers inside {|z − z ₀|≤ r}, so that A ⊂ V _ε. Repeating the proof of Lemma 4.9 shows that there exists a constant C = C(Δ) > 0 such that

• (i) There are no edges between V _r and V ∖ V _Cr, and

• (ii) $\mathcal {R}_{\mathrm {eff}}(V_r \leftrightarrow V \setminus V_{Cr}) \geq C^{-1}$, as long as V _r and V ∖ V _Cr are non-empty.

Regarding this proof, we note that it is possible that the set {|z − z ₀|≤ r} is not contained in $\mathbb {U}$ (unlike the proof of Lemma 4.9 when the carrier is all of $\mathbb {R}^2$), however, this only works in our favor. The proof of (4.1) now proceeds similarly to the proof of Theorem 4.4 (3). When ε is small enough (depending only on $r_{\min }$ and Δ), by the Ring Lemma (Lemma 4.2), the Euclidean distance between the circle corresponding to v ₀ and A is at least some constant (which again depends only on $r_{\min }$ and Δ) so that $v_0 \not \in V_{C^K \varepsilon }$ for some $K = \varOmega (\log (1/\varepsilon ))$. For each k = 0, 2, 4, …, K the sets of edges which have at least one endpoint in the annulus $V_{C^{k+1} \varepsilon } \setminus V_{C^k \varepsilon }$ are disjoint by (i). By (ii), each of these sets of edges contribute at least C ⁻¹ to the energy of the unit current flow from A to v ₀, concluding the proof of (4.1) using Thomson’s principle (Theorem 2.28).

For the proof of (4.2) we construct a unit flow from v ₀ to A that has energy $O(\log (1/\varepsilon ))$. The construction is in the same spirit as the proof of Theorem 4.4 (4), but there are some technical difficulties to overcome. Since A contains a vertex that is tangent to $\partial \mathbb {U}$, we choose $z_0\in \partial \mathbb {U}$ that belongs to a circle of A. By rotating the packing we may assume that z ₀ = e ^iε∕4.

We now treat two cases separately. In the first case we assume that there exists z ₁ in z(A) such that $\arg (z_1)\in [0,\varepsilon /2]$ and |z ₁|≤ 1 − ε∕2 such that the path in z(A) from z ₀ to z ₁ remains in the sector $\arg (z) \in [0,\varepsilon /2]$. Consider the points

$$\displaystyle \begin{aligned}x_0=-r_0 \qquad x_1=r_0 \qquad y_1 = 1-\varepsilon/3 \qquad y_0=1 \, ,\end{aligned}$$

and note that x ₀, x ₁ are the leftmost and rightmost points on the circle of v ₀. Let C ₀ and C ₁ be the upper half plane semi-circles in which x ₀, y ₀ and x ₁, y ₁ are antipodal points, respectively. The choice of y ₀, y ₁ is made so that the path between z ₀ to z ₁ in z(A) must cross the region bounded by C ₀, C ₁ and the intervals [x ₀, x ₁], [y ₁, y ₀], by our assumption on z ₁ as long as ε is small enough. See Fig. 4.4, left.

Fig. 4.4

Left: for any t ∈ [0, 1] the semi-circle C _t must intersect the path in A between z ₀ and z ₁. Right: the quadrilateral Q _i is bounded between $\ell _{\theta _i}$, $\ell _{\theta _{i+1}}$, C ₀ and C ₁

For each t ∈ [0, 1] write C _t for the upper half plane semi-circle in which ty ₁ + (1 − t)y ₀ and tx ₁ + (1 − t)x ₀ are antipodal points, so that C _t continuously interpolates between C ₀ and C ₁. See Fig. 4.4, left. Choose t ∈ [0, 1] uniformly at random and consider the random path γ which traces C _t from left to right. This random path starts at the circle of v ₀ and must hit the path between z ₀ and z ₁ by our previous discussion. Hence, the circles of P that intersect γ must contain a path in the graph from v ₀ to A. By Claim 2.46 we obtain a flow I from v ₀ to A whose energy $\mathcal {E}(I)$ we now bound.

For an angle θ ∈ [0, π] we denote by w _θ(t) the point at angle θ, seen from the center of C _t, on the semi-circle C _t . It is an exercise to see that the set of points {w _θ(t) : t ∈ [0, 1]} form a straight line interval ℓ _θ. Furthermore, when t is chosen uniform in [0, 1], the intersection of C _t and ℓ _θ is a uniformly chosen point on ℓ _θ. Set θ ₀ = 0 and θ _i = 2ⁱ⁻¹ ε for i = 1, …, K − 1 where $K=O(\log (1/\varepsilon ))$ such that θ _K−1 ∈ [π∕4, π∕2] and set θ _K = π. We will obtain the bound $\mathcal {E}(I)=O(K)$ by bounding from above by a constant the contribution to $\mathcal {E}(I)$ coming from edges which intersect the quadrilateral Q _i of $\mathbb {R}^2$ bounded by $\ell _{\theta _i}, \ell _{\theta _{i+1}}, C_0,C_1$; see Fig. 4.4, right. The random path γ restricted to Q _i can be sampled by choosing a uniform random point on $\ell _{\theta _i}$, setting t ∈ [0, 1] to be the unique number such that C _t intersects $\ell _{\theta _i}$ at the chosen point, and tracing the part of C _t from $\ell _{\theta _i}$ to $\ell _{\theta _{i+1}}$. The lengths of the four curves bounding Q _i are all of order 2ⁱ ε and so we deduce that if v corresponds to a circle of radius O(2ⁱ ε) which intersects Q _i, then the probability that it is visited by γ is $O( \operatorname {\mathrm {rad}}(v)/2^i \varepsilon )$. Since the sum of $\operatorname {\mathrm {rad}}(v)^2$ over such v’s is at most the area of Q _i up to a multiplicative constant (note that some of these circles need not be contained in Q _i) it is at most O(2²ⁱ ε ²). Since the degrees are bounded we deduce that the contribution to the energy from edges touching such v’s is O(1). Lastly, if v corresponds to a larger circle, then we bound its probability of being visited by γ by 1 and note that there can only be O(1) many such v’s whose circles intersects Q _i. Thus the contribution from these is another O(1). Since there are $O(\log (1/\varepsilon ))$ such i’s we learn that $\mathcal {E}(I) = O(\log (1/\varepsilon ))$ finishing our proof in this case using Thomson’s principle (Theorem 2.28).

In the second case, we assume that there exists z ₁ ∈ z(A) such that $\arg (z_1) \not \in [0,\varepsilon /2]$ and |z ₁|≥ 1 − ε. It is clear that since $\operatorname {\mathrm {diam}}_P(A)=\varepsilon$ either the first or the second case must occur. Denote $z_0^{\prime }=|z_1|e^{i\varepsilon /4}$ and let x ₀, x ₁ be antipodal points on the circle of v ₀ such that the straight line between x ₀ and x ₁ is parallel to the straight line between $z_0^{\prime }$ and z ₁. The vertices $z_0^{\prime }, z_1, x_0, x_1$ form a trapezoid, see Fig. 4.5. We choose a uniform random point t ∈ [0, 1] and stretch a straight line from tx ₀ + (1 − t)x ₁ to $t z_0 + (1-t)z_1^{\prime }$. We then continue it by a straight line from $t z_0 + (1-t)z_1^{\prime }$ to $w \in \partial \mathbb {U}$ where $\arg (w) = \arg (t z_0 + (1-t)z_1^{\prime })$. Denote the resulting path by γ _t and note that it starts inside the circle of v ₀ and must hit the path between z ₀ and z ₁ in z(A). Thus, the set of all circles which intersect γ _t form a path in the graph that starts at v ₀ and ends at A; this random choice of γ _t gives us as usual a unit flow from v ₀ to A by Claim 2.46. By repeating the same argument as in the previous case (that is, splitting the trapezoid into $O(\log (1/\varepsilon ))$ many trapezoids of constant aspect ratio), we see that the contribution to the energy of the flow induced by the random path γ _t of the edges in the trapezoid is $O(\log (1/\varepsilon ))$. Furthermore, the same argument gives that the edges in the quadrilateral formed by the vertices $z_0, z_0^{\prime }, z_1$ and $e^{i \arg (z_1)}$ also contribute at most a constant to the energy, concluding our proof for the second case by Thomson’s principle again (Theorem 2.28). □

Fig. 4.5

The resistance across the trapezoid on vertices $x_0, x_1, z_0^{\prime }, z_1$ is $O(\log (1/\varepsilon ))$ when $|z_0^{\prime }-z_1|=\varTheta (\varepsilon )$

Proof of Theorem 4.4 (2)

Denote by d _G(u, v) the graph distance between the vertices u, v of G. Fix some v ₀ ∈ V and let

$$\displaystyle \begin{aligned} B_j &= \{v: d_G(v_0,v) \leq j\}, \\ V_j &= B_j \cup \{\text{finite components of }V\setminus B_j\}, \\ E_j &= \{\text{edges induced by }V_j\}. \end{aligned} $$

The graph G _j = (V _j, E _j) with the map structure inherited from G is a finite triangulation with boundary. Indeed, it is straightforward to check that it is 2-connected (i.e., the removal of a single vertex does not disconnect the graph) which implies that the outer face forms a simple cycle, see [20, Proposition 4.2.5]. Furthermore, since G is one-ended and we have added all the finite components in V ∖ B _j there cannot be a face with more than 3 edges except for the outer face which we denote by ∂G _j.

Thus G _j is an increasing sequence of finite triangulations with boundary such that ∪_j G _j = G. We apply Claim 4.10 to pack G _j inside the unit disk $\mathbb {U}$ such that the circles of ∂G _j are tangent to $\partial \mathbb {U}$. By applying a Möbius transformation from $\mathbb {U}$ onto $\mathbb {U}$, we may assume that the circle corresponding to v ₀ is centered at the origin. We denote this packing by P _j and let $r_0^j$ be the radius of v ₀ in P _j.

Since G is transient it follows that there exists some c = c(Δ) > 0 such that $r_0^j \geq c$ for all j by Corollary 2.39. Indeed, if $r_0^j \leq \varepsilon$, we learn by Lemma 4.9 and the proof of Theorem 4.4 (3) that $\mathcal {R}_{\mathrm {eff}}(v_0 \leftrightarrow \infty ) \geq c' \log (\varepsilon ^{-1})$ for some c′ = c′(Δ) > 0.

As we did in Claim 4.3, we now take a subsequence in which the centers and radii of all vertices converge. Denote the resulting limiting packing by P _∞. This packing has all circles inside $\mathbb {U}$ and we therefore deduce that $\operatorname {\mathrm {Carrier}}(P_\infty ) \subseteq \mathbb {U}$. It is a priori possible that $\operatorname {\mathrm {Carrier}}(P_\infty )$ is some strict subset of $\mathbb {U}$, i.e., that all the circles stabilize inside some strict subset of $\mathbb {U}$. We now argue that this is not possible.

Let Z be the set of accumulation points of $\operatorname {\mathrm {Carrier}}(P_\infty )$; it suffices to show that $Z\subset \partial \mathbb {U}$ since any simply connected domain $\varOmega \subset \mathbb {U}$ for which $\partial \varOmega \subset \partial \mathbb {U}$ must equal $\mathbb {U}$. Since Z is a compact set, let z ∈ Z minimize |z| among all z ∈ Z; it suffices to show that $z\in \partial \mathbb {U}$. Fix ε > 0 and put

The graph spanned on the vertices U _ε(z) may be disconnected, yet by our choice of z it is clear that U _ε(z) contains an infinite connected component. Indeed, one can draw a straight line from the origin to z without intersecting Z and consider the set of all circles intersecting this line; from some point onwards the vertices corresponding to these circles will reside in U _ε(z).

Therefore, let W _ε(z) be an infinite connected component of the graph spanned on U _ε(z). Let J = J(z, ε) be the first integer such that V _J ∩ W _ε(z) ≠ ∅. Since the G _j’s are increasing finite sets and W _ε(z) is an infinite connected set, we have that ∂G _j ∩ W _ε(z) ≠ ∅ for all j ≥ J. Consider now any connected component A _j of the graph spanned on the vertices V _j ∩ W _ε(z).

Denote by $P_\infty ^{j}$ the finite circle packing obtained from P _∞ by taking only the circles of V _j. It has the same adjacenty graph as P _j but it is a different packing. Since A _j ⊂ W _ε(z), it follows that $\operatorname {\mathrm {diam}}_{P_\infty ^{j}}(A_j) \leq 4\varepsilon$. By Lemma 4.11, Eq. (4.1), applied to the set A _j in the packing $P_\infty ^{j}$, we deduce that $\mathcal {R}_{\mathrm {eff}}(v_0 \leftrightarrow A_j; G_j) \geq c \log (1/\varepsilon )$. Since A _j is a connected component of V _j ∩ W _ε(z) and since W _ε(z) is an infinite connected set of vertices in G, it follows that A _j must contain a vertex of ∂V _j. Thus, we may apply Lemma 4.11, Eq. 4.2, to the set A _j, this time in the packing P _j, to get that there exists some c > 0 such that

$$\displaystyle \begin{aligned} \operatorname{\mathrm{diam}}_{P_j}(A_j) \leq \varepsilon^c \, .\end{aligned} $$

(4.3)

Choose some v _J ∈ ∂G _J ∩ W _ε(z) so that . For each j ≥ J choose v _j ∈ ∂G _j ∩ W _ε(z) so that v _j and v _J are in the same connected component A _j of the graph spanned on V _j ∩ W _ε(z). Since the circle of v _j in P _j touches $\partial \mathbb {U}$ we learn by (4.3) that the distance of the circle of v _J in P _j from $\partial \mathbb {U}$ is at most ε ^c for all j ≥ J. Since the circle corresponding to v _J in P _∞ is the limit of its circles in P _j we deduce that the distance of from $\partial \mathbb {U}$ is at most ε ^c. Hence the distance of z from $\partial \mathbb {U}$ is at most ε + ε ^c. Since ε was arbitrary we obtain that $z\in \partial \mathbb {U}$, as required. □

4.5 Exercises

• 1. Let G be a triangulation of the plane with maximal degree at most 6. Prove that the simple random walk on G is recurrent.

• 2. Let G be a plane triangulation that can be circle packed in the unit disc {z : |z| < 1}. Show that the simple random walk on G is transient. (Note that G may have unbounded degrees)

• 3.(*) Let P be a circle packing of a finite simple planar map with degree bounded by D such that all of its faces are triangles except for the outerface. Assume that the carrier of P is contained in [−11, 11]², contains [−10, 10]² and that all circles have radius at most 1. Let h be the harmonic function taking the value 1 on all vertices with centers left of the line $\{-10\}\times \mathbb {R}$, taking the value 0 on all vertices with centers right of the line $\{10\}\times \mathbb {R}$, and is harmonic anywhere else. Assume x and y are two vertices such that their centers are contained in [−1, 1]² and that the Euclidean distance between these centers is at most 𝜖 > 0. Show that

  $$\displaystyle \begin{aligned}|h(x) - h(y)| \leq {C \over \log(1/\epsilon)} \, ,\end{aligned}$$

  for some constant C = C(D) > 0 independent of 𝜖. [Hint: assume h(x) < h(y) and consider the sets A = {v : h(v) ≤ h(x)} and B = {v : h(v) ≥ h(y)}].