MATRIZ ESTRATEGICA Y PROGRAMATICA SECTOR DESARROLLO COMUNITARIO
6.5 SECTOR DE ATENCIÓN A GRUPOS VULNERABLES.
We claim that on the other hand, we havem≤nmF+k3, wherek3=n+k0−1.
To see this, assume that
m≥nmF+k3+ 1 =n(mF + 1) +k0.
Then we can find non-disjoint m-tilesX andY withx∈X,y∈Y. Moreover, we can pickn(mF + 1)-tilesX′ andY′ withx∈X′ andy∈Y′. By definition ofmF
we know thatX′
∩Y′=
∅; soX′ andY′ are disjointn(m
F+ 1)-tiles joined by the
connected setK=X∪Y. Hence by Lemma 5.36 the setKmust consist of at least Dm−n(mF+1)≥Dek0≥10
m-tiles; but K consists of only two m-tiles. This is a contradiction showing the desired claim.
(v) Letm′:=m′
f,C(x, y) be defined as in (8.1). Thenm′ ≥1, because the two
0-tiles have non-empty intersection. So m′−1 ≥0, and there exist (m′−1)-tiles
X and Y with x∈ X and y ∈ Y. Then X∩Y 6= ∅ by definition of m′, and so
m(x, y)≥m′−1.
Conversely, letm:=m(x, y). Supposem′< m−k
0. Then there existm′-tiles
X′ andY′ withX′
∩Y′ =
∅,m-tilesX and Y withX ∩Y 6=∅, and x∈X∩X′,
y∈Y∩Y′. HenceK=X
∪Y is a union of twom-tiles joining the disjointm′-tiles
X′ andY′; but such a union must consist of at least
Dm−m′ ≥Dek0≥10
m-tiles by Lemma 5.36. This is a contradiction showing thatm−k0≤m′. So the
claim is true with k4=k0.
8.2. Existence and basic properties of visual metrics
We are now ready to prove Propositions 8.3 and 8.4, and in particular the existence of visual metrics. In this section we will also collect various other and somewhat more technical statements related to visual metrics that will be useful later on.
Proof of Proposition 8.3. (i) Fix a Jordan curveC ⊂S2with post(f) ⊂ C. A functionq:S2×S2→[0,∞) is called aquasimetric if it has the symmetry
property q(x, y) = q(y, x), satisfies the condition q(x, y) = 0 ⇔ x = y, and the inequality
(8.4) q(x, y)≤K(q(x, z) +q(z, y)), holds for a constantK≥1 and all x, y, z∈S2.
We now define a quasimetricqonS2. For this purpose, we fix Λ>1 and set
(8.5) q(x, y) := Λ−m(x,y),
forx, y∈S2, wherem(x, y) =m
f,C(x, y)∈N0∪ {∞}is as in Definition 8.1.
Symmetry and the propertyq(x, y) = 0⇔x=y are clear. The quasi-triangle inequality (8.4) follows from Lemma 8.7 (i).
It is well known (see [He01, Proposition 14.5]) that a sufficient “snowflaking” of a quasimetric leads to a distance function that is comparable to a metric. This means there is 0< ǫ <1, and a metric̺onS2such that̺≍qǫ. Then̺is a visual
(ii) Let̺be a visual metric forf satisfying (8.2), andda fixed base metric on S2that induces the given topology ofS2. We have to show that ifx∈S2and{x
i}
is a sequence inS2, then̺(x
i, x)→0 if and only ifd(xi, x)→0 asi→ ∞.
Assume first that̺(xi, x)→0 asi→ ∞. By (8.2) this is obviously equivalent
to mi := mf,C(xi, x) → ∞. For each i there are non-disjoint mi-tiles Xi and Yi
withx∈X, xi∈Y. Thus
d(xi, x)≤diamd(Xi) + diamd(Yi)≤2 mesh(f, mi,C).
Sincef is expanding andmi → ∞, the latter expression approaches 0 asi→ ∞.
Henced(xi, x)→0 asi→ ∞.
Conversely, suppose that d(xi, x) → 0 as i → ∞. Let n ∈ N0 be arbitrary.
Thenxlies in somen-flowerWn(p) (see Lemma 5.29 (iv)). Since flowers are open
sets, we have xi ∈Wn(p) for sufficiently largei. For each of these i we can find
n-tiles X and Y with x ∈ X, xi ∈ Y, and p ∈ X ∩Y. This implies mi ≥ n.
Thereforemi→ ∞, and hence̺(xi, x)→ ∞as desired.
(iii) This follows from Lemma 8.7 (iii).
(iv) This follows from (iii) and the definition of a visual metric. (v) This follows from (iii) and Lemma 8.7 (iv).
(vi) This follows from Lemma 8.7 (ii).
If two expanding Thurston maps are topologically conjugate, then their visual metrics are closely related.
Proposition 8.8. Let f: S2
→ S2 and g:Sb2
→ Sb2 be expanding Thurston maps that are topologically conjugate. Then S2 equipped with any visual metric for f is snowflake equivalent to Sb2 equipped with any visual metric for g. Every homeomorphism h:S2→Sb2 satisfying h◦f =g◦his a snowflake equivalence.
Proof. By our assumptions there exists a topological conjugacy between f and g, i.e., a homeomorphism h: S2
→ Sb2 such that h
◦f = g◦h. Let ̺ be a visual metric on S2 for f, and̺bbe a visual metric on Sb2 forg. Let Λ >1 and b
Λ > 1 be the expansion factors of̺ and ̺, respectively. It suffices to show thatb
h: (S2, ̺)→(Sb2,b̺) is a snowflake equivalence.
To see this, pick a Jordan curve C ⊂ S2 with post(f) ⊂ C. Then Cb= h(C)
is a Jordan curve in Sb2 with post(g) =h(post(f))
⊂Cb(see Lemma 2.5). Sinceh conjugatesf andg, it follows from Proposition 5.16 (iii) and (v) or, alternatively, from the uniqueness statement in Lemma 5.12 that for each n ∈ N0 the images
of the cells in the cell decompositionDn := Dn(f,
C) ofS2 under the map h are
precisely the cells in the cell decompositionDbn :=
Dn(g,b
C) ofSb2; so we have
(8.6) Dbn ={h(c) :c∈ Dn}
for alln∈N0. This implies that b
m(h(x), h(y)) =m(x, y) for allx, y∈S2, wheremb =m
g,Cbandm=mf,C (recall Definition 8.1). Combining
this with Proposition 8.3 (iii) we see that
b
̺(h(x), h(y))≍Λb−m(h(x),h(y))b =Λb−m(x,y)= Λ−αm(x,y)≍̺(x, y)α
for allx, y∈S2, whereα= log(Λ)/b log(Λ) and the implicit multiplicative constants
8.2. EXISTENCE AND BASIC PROPERTIES OF VISUAL METRICS 177
We now prove the geometric characterization of visual metrics.
Proof of Proposition 8.4. Let f: S2 → S2 be an expanding Thurston
map, andC ⊂S2 be a Jordan curve with post(f)⊂ C.
We first show that a visual metric̺ forf has the properties (i) and (ii). By Proposition 8.3 (iii) we may assume that ̺ satisfies (8.2) for m = mf,C and a
constantC=C(≍).
(i) Let k0 ∈ N be defined as in (8.3), and let σ and τ be disjoint n-cells. If
x∈σandy∈τ are arbitrary, thenm(x, y)< n+k0. Indeed, if this were not the
case, then we could find (n+k)-tiles X and Y with x∈ X, y ∈ Y, X∩Y 6=∅, and k≥k0. ThenK =X ∪Y is a connected set meeting disjointn-cells. Hence
by Lemma 5.36 the number of (n+k)-tiles inK should be≥Dk ≥Dek0≥10. On
the other hand,K consists of two (n+k)-tiles. This is impossible. Thus ̺(x, y) ≥ (1/C)Λ−n−k0, and so we get the desired bound dist
̺(σ, τ) ≥
(1/C′)Λ−n with the constantC′ =CΛk0 that is independent ofn, σ, andτ.
(ii) Ifx, y are points in somen-tile X, thenm(x, y)≥n. Since every n-edge is contained in ann-tile, this inequality is still true ifxandy are contained in an n-edge. Hence ̺(x, y)≤CΛ−m(x,y)≤CΛ−n, and so diam
̺(τ)≤CΛ−n whenever
τ is ann-tile orn-edge, whereC=C(≍) is the constant from (8.2).
A similar lower bound for the diameter of ann-edge orn-tileτ follows from (i) and the fact that everyn-edge orn-tile contains two distinctn-vertices.
To prove the converse implication, suppose that ̺is a metric onS2 with the properties (i) and (ii) as in the statement. We want to show that ̺is visual forf. Letx, y∈S2,x
6
=y, be arbitrary, andm=mf,C(x, y).
Then we can findm-tilesX andY withx∈X,y∈Y, andX∩Y 6=∅. By (ii) we have
̺(x, y)≤diam̺(X) + diam̺(Y).Λ−m.
We can also find (m+ 1)-tilesX′ andY′ withx∈X′,y ∈Y′. By definition ofm
we then haveX′∩Y′=∅. Hence by (i)
̺(x, y)≥dist̺(X′, Y′)&Λ−m.
Since the implicit multiplicative constants in the previous inequalities are indepen- dent ofxandy, it follows that ̺is a visual metric forf.
It is possible to establish the phenomenon of “exponential shrinking” as in Proposition 8.4 (ii) also for other types of sets. For example, we have
(8.7) diam̺(Wn(p))≤CΛ−n
for every n-flower for (f,C) where the constant C is independent ofn and p. Of particular importance will be exponential shrinking for lifts of paths.
Lemma8.9. Letf:S2
→S2 be an expanding Thurston map, and̺be a visual metric for f with expansion factorΛ>1. Then for every path γ: [0,1]→S2 there exists a constantA >0 with the following property: ifn∈Nandeγ is any lift ofγ
by fn, then
diam̺(eγ)≤AΛ−n.
Proof. Pick a Jordan curveC ⊂S2with post(f)⊂ C, and let δ
0>0 be as in
(5.14) with̺as the base metric onS2. Then we can break upγinto a finite number
of pathsγi,i= 1, . . . , N, traversed in successive order such that diam̺(γi)< δ0for
i= 1, . . . , N. By Lemma 5.34 (ii) each lift of a pieceγiis contained in onen-flower,
and so the whole lifteγinN n-flowers. Hence by (8.7) we have diam̺(eγ)≤CNΛ−n,
whereC >0 is independent ofnandγ. The statement follows withA=CN.
In general, the constantA in the last lemma will depend onγ, but the proof shows that we can take the same constant A for a family of paths if there exists N ∈Nsuch that each path can be broken up into at mostN subpaths of diameter < δ0.
Let f be an expanding Thurston map, and C ⊂ S2 be a Jordan curve with
post(f)⊂ C. It is useful to define neighborhoods of points by using the cells in our decompositionsDn =
Dn(f,
C). To do this, letx∈S2, n
∈N0, and set
Un(x) =[{Y ∈Xn : there exists ann-tileX with (8.8)
x∈X andX∩Y 6=∅}.
It is convenient to define Un(x) also for negative integers n. We set Un(x) =
U0(x) =S2forn <0.
The sets Un(x) resemble metric balls defined in terms of a visual metric very
closely.
Lemma8.10. Let̺be a visual metric forf with expansion factorΛ>1. Then there are constantsK≥1 andn0∈N0 with the following properties.
(i) For allx∈S2 and all n∈Z,
B̺(x, r/K)⊂Un(x)⊂B̺(x, Kr), wherer= Λ−n.
(ii) For allx∈S2 and all r >0,
Un+n0(x)⊂B̺(x, r)⊂Un−n0(x), wheren=⌈−logr/log Λ⌉.
Proof. (i) Let m=mf,C. Ify ∈Un(x), then m(x, y)≥n, and so ̺(x, y).
Λ−n =r. This gives the inclusion Un(x)
⊂B̺(x, Kr) for a suitable constant K
independent ofxandn.
Conversely, suppose that y /∈ Un(x). Then n
≥ 1. If we pick any n-tilesX and Y with x ∈ X and y ∈ Y, then X∩Y = ∅ by definition of Un(x). So by
Proposition 8.4 (i) we have
̺(x, y)≥dist̺(X, Y)&Λ−n=r.
HenceB̺(x, r/K)⊂Un(x) ifKis suitably large independent of xandn.
(ii) Choosen0=⌈logK/log Λ⌉+1, whereKis as in (i). Then Λ−n0≤1/(ΛK).
Moreover, Λ−n
≤r≤ΛΛ−n, and so
KΛ−n−n0≤r≤(1/K)Λ−n+n0.
The desired inclusion then follows from (i).
We next show that when S2 is equipped with a visual metric, then tiles are
“quasi-round”. In particular, every tile contains points that are “deep inside” the tile.
8.2. EXISTENCE AND BASIC PROPERTIES OF VISUAL METRICS 179
Lemma 8.11. Let f:S2 → S2 be an expanding Thurston map, C ⊂ S2 be a Jordan curve withpost(f)⊂ C, and̺be a visual metric forf with expansion factor
Λ>1. Then there exists a constant C ≥1 with the following property: for every
n-tileX for (f,C)there exists a point p∈X such that
B̺(p,(1/C)Λ−n)⊂X⊂B̺(p, CΛ−n).
Proof. With a suitable constantCindependent ofn, an inclusion of the form X⊂B̺(p, CΛ−n)
holds for everyn-tileX and every pointp∈X as follows from Proposition 8.4 (ii). The main difficulty for an inclusion in the opposite direction is to find an appropriate point p. For this purpose, let k0 ∈Nbe the number defined in (8.3),
and X be an arbitrary n-tile. Since f is an expanding Thurston map, we have # post(f) ≥ 3 (see Lemma 6.1), and so ∂X contains at least three distinct n- verticesv1, v2, v3. Using these vertices, we can find three arcsα1, α2, α3⊂∂X with
pairwise disjoint interior such that ∂X = α1∪α2∪α3 and such that αi has the
endpoints vi and vi+1 fori= 1,2,3, where v4 =v1. In general, αi will not be an
n-edge, but since it lies on∂X, and its endpoints aren-vertices, it is the union of all the n-edges that it contains.
We now define
Ai = [ x∈αi
Un+k0(x)
fori= 1,2,3, whereUn+k0(x) is given as in (8.8). Then the setA
i is the union of
all (n+k0)-tiles that meet an (n+k0)-tile that has non-empty intersection with
αi. In particular,Ai is a closed set that containsαi.
We claim that the sets A1, A2, A3 do not form a cover of X. To reach a
contradiction, suppose thatX ⊂A1∪A2∪A3. We can regardX as a topological
simplex with the sidesαi,i= 1,2,3. Then the closed setsA1, A2, A3form a cover
of X such that each set Ai contains the side αi of the simplex for i = 1,2,3. A
well-known result due to Sperner [AH35, p. 378] then implies thatA1∩A2∩A36=∅.
Pick a pointx∈A1∩A2∩A3. Then by definition ofAi, there exist (n+k0)-tiles
Xi andYi with Xi∩αi6=∅, x∈Yi, andXi∩Yi 6=∅, wherei= 1,2,3. Then the
set K= 3 [ i=1 (Xi∪Yi)
consists of at most six (n+k0)-tiles, is connected, and meets each of the arcs
α1, α2, α3. Hence K′ =fn(K) is a connected set that consists of at most sixk0-
tiles, and meets each of the arcs βi = fn(αi), i = 1,2,3. Note that each arc βi
is the union of all 0-edges that it contains. Hence for i = 1,2,3 there exists a 0-edge ei ⊂βi with ei∩K′ 6=∅. Since the arcs β1, β2, β3 have pairwise disjoint
interior, it follows that the 0-edgese1, e2, e3 are all distinct. So K′ is a connected
set that meets three distinct 0-edges. Hence it joins opposite sides of C. So K′ should contain at leastDk0 ≥Dek0 ≥10 tiles of level k0. This is a contradiction,
becauseK′ is a union of at most sixk
0-tiles.
This proves the claim that the setsA1, A2, A3do not coverX, and we conclude
that we can find a point
We claim thatUn+k0(p)⊂X. Otherwise, there is a point y ∈Un+k0(p)\X,
and (n+k0)-tilesU andV withp∈U,y∈V, andU∩V 6=∅. Then the connected
setU∪V must meet∂X, and hence one of the arcsαi; but thenp∈Aiby definition
ofAi. This is a contradiction showing the desired inclusionUn+k0(p)⊂X.
From Lemma 8.10 (i) it now follows thatB̺(p,(1/C)Λ−n)⊂X, whereC≥1
is a constant independent ofnandX.
8.3. The canonical orbifold metric as a visual metric
If f:S2 → S2 is an expanding Thurston map f, one can ask whether other
standard metrics onS2 are visual metrics. To have some natural metrics available, we restrict ourselves here to rational expanding Thurston maps f defined on Cb. Recall that a rational Thurston map is expanding if and only if it does not have periodic critical points (see Proposition 2.3).
Lemma 8.12. Letf:Cb→Cb be a rational expanding Thurston map. Then the chordal metricσ onCb is not a visual metric for f.
Proof. We argue by contradiction and assume that σ is a visual metric for f with expansion factor Λ > 1. We pick a critical point c ∈ Cb of f and set d:= deg(f, c)≥2.
Consider tiles for (f,C), whereC ⊂Cbis a fixed Jordan curve with post(f)⊂ Cas usual. For eachn∈NletXnbe ann-tile that containsc. Then diam
σ(Xn)≍Λ−n
by Proposition 8.4 (ii). Since in suitable local conformal coordinates the map f nearc behaves likez7→zd near 0, forYn:=f(Xn+1) we have
diamσ(Yn) = diamσ(f(Xn+1))≍diamσ(Xn+1)d
(8.9)
≍Λ−(n+1)d≍Λ−nd
for n ∈ N0, where C(≍) is independent of n. On the other hand, Yn is an n-
tile and so diamσ(Yn)≍Λ−n by Proposition 8.4 (ii), whereC(≍) is independent
of n. Since d ≥ 2 this is irreconcilable with (8.9) for large n and so we reach a
contradiction.
We now turn to the canonical orbifold metricω =ωf of a given rational ex-
panding Thurston map f (see Section 2.5) and its relation to visual metrics. We will reformulate and prove the two implications in Proposition 8.5 separately. First we prove the “only if” statement.
Lemma 8.13. Let f:Cb →Cb be a rational expanding Thurston map such that its canonical orbifold metric ω =ωf is a visual metric for f. Then f is a Latt`es map.
Proof. The metricωis the canonical orbifold metric of the associated orbifold
Of = (Cb, αf). Note thatαf(u)<∞foru∈Cb by Proposition 2.9 (ii), becausef is
a rational expanding Thurston map and so it does not have periodic critical points. Supposeω is a visual metric forf with expansion factor Λ>1. Letp∈Cb and q∈f−1(p) be arbitrary, and setd:= deg
f(q).
We consider tiles for (f,C), whereC ⊂Cbis a fixed Jordan curve with post(f)⊂ C. For eachn∈N0 we pick ann-tileXn withq∈Xn. ThenYn:=f(Xn+1) is an