SECRETARIA DE PLANEACIÓN GUACA PROYECTOS OBJETIVOS
DIMENSIÓN ECONÓMICA
5.1 SECTOR AGROPECUARIO.
5.1.2 LINEA ESTRATEGICA “GUACA DESPENSA AGROPECUARIA 2015”.
We can find an inner product onR2such thatv
1andv2form an orthonormal
basis ofR2 with respect to this inner product. Ifk · kis the norm induced by this
product, then
kav1+bv2k= p
a2+b2
fora, b∈R. If L has a matrix representation given byM1 orM2, it is clear that kL(v)k ≥ |λ1| · kvk and so
(6.6) kLn(v)k ≥ |λ1|n· kvk
for all v ∈ R2 and n
∈ N. If L is represented by the matrix M3, then a similar
estimate is harder to obtain, but one can show that
kLn(v)k ≥ |λ1| n+1 p 2λ2 1+n2 kvk.
To see this, one bounds the operator norm of the matrixM3−nby its Hilbert-Schmidt
norm. We leave the details to the reader.
Since all norms onR2 are comparable, it follows that |Ln(v)| ≥ρn|v|
for all sufficiently largenindependent ofv∈R2withρ=|λ
1|1/2>1.
Recall that ifGis a planar crystallographic group, then we say that a continuous map Θ : R2
→ S2 is induced by G if Θ(u) = Θ(v) for u, v
∈ R2 if and only
if there exists g ∈ G such that v = g(u) (see Section A.7). In this case, G is necessarily of non-torus type (see Theorem 3.7) and Θ is essentially the quotient map Θ :R2
→R2/G∼=S2 (see the discussion after Proposition 3.9).
Lemma 6.14. Let Gbe a planar crystallographic group and Θ :R2
→S2 be a continuous map induced by G. Suppose Kn ⊂ R2 is a connected set for n ∈ N. Then
lim
n→∞diam(Kn) = 0if and only if nlim→∞diam(Θ(Kn)) = 0.
Here diam(Kn) is the Euclidean diameter ofKn, and diam(Θ(Kn)) the diam-
eter of Θ(Kn) with respect to some base metricdonS2.
Proof. “⇒” For this implication it is enough to show that Θ is uniformly continuous onR2. This follows from the fact that Θ is induced byGand thatGacts
isometrically and cocompactly onR2; indeed, we can find a compact fundamental
domainF ⊂R2 for the action ofGonR2. Now supposex, y ∈R2andδ:=|x−y|
is small. Then there exists g ∈Gsuch that g(x)∈F. Ifδ is small enough, then g(x), g(y) ∈ U, where U is a compact neighborhood of F. Since Θ is uniformly continuous onU, and |g(x)−g(y)|=|x−y|=δ, it follows that
d(Θ(x),Θ(y)) =d(Θ(g(x)),Θ(g(y))
is small only depending on δ. The uniform continuity of Θ follows.
“⇐” We argue by contradiction and assume that the statement is false. Then there exist connected sets Kn ⊂ R2 such diam(Θ(Kn)) → 0 as n → ∞, but
diam(Kn)≥ǫ0forn∈N, where ǫ0>0.
We pick a point xn ∈ Kn for n ∈ N. If we replace each set Kn with its
image K′
n =gn(Kn) for suitable gn ∈ G (note that diam(Kn′) = diam(Kn) and
Θ(K′
n) = Θ(Kn)), and pass to a subsequence if necessary, then we may assume
Let p:= Θ(x). Then the set Θ−1(p) is equal to the orbit Gx of x under G.
Since the action of Gon R2 is properly discontinuous, the set Θ−1(p) = Gx has
no limit point in R2. Since Galso acts cocompactly on R2, this implies that the
distance of distinct points in Θ−1(p) is bounded away from 0; so there exists a
constantm >0 such that |u−v| ≥mwheneveru, v∈Θ−1(p) andu6=v.
Pick a constantc with 0 < c <min{ǫ0/2, m}. The set Kn is connected, and
has diameter diam(Kn) ≥ ǫ0 > 2c. Hence Kn cannot be contained in the disk {z ∈ R2 :
|z−xn| < c}, and so it meets the circle {z ∈ R2 : |z−xn| = c}. It
follows that there exists a pointyn∈Knwith|xn−yn|=c. By passing to another
subsequence if necessary, we may assume that the sequence {yn} converges, say
yn →y ∈R2 asn→ ∞. Then |x−y|=c < m. Note that Θ(xn),Θ(yn)∈Θ(Kn)
forn∈N, and diam(Θ(Kn))→0 asn→ ∞. So
p= Θ(x) = lim
n→∞Θ(xn) = limn→∞Θ(yn) = Θ(y),
and x, y ∈ Θ−1(p). Since |x−y| = c > 0, we have x 6=y. So xand y are two
distinct points in Θ−1(p) with |x−y|=c < m. This contradicts the choice ofm,
and the claim follows.
Proof of Proposition 6.12. In the proof metric notions onR2will refer to
the Euclidean metric. Letf:S2
→S2 be the given Latt`es-type map, and A, Θ, G be as in Defini-
tion 3.3. Then Θ : R2
→ S2 is a branched covering map induced by the crystal-
lographic group G. We know that here G is not isomorphic to Z2, because the
quotient space R2/G ∼= S2 is not a torus. So by Proposition 3.9 there exists a
holomorphic branched covering map Θ :e R2∼=C
→Cb induced byG, and a unique homeomorphismϕ:S2
→Cb such thatΘ =e ϕ◦Θ (note that the roles of the maps
e
Θ and Θ are reversed in Proposition 3.9).
We now conjugatef byϕto obtain a Thurston mapfe:=ϕ◦f◦ϕ−1defined on b
C. Thenfeis a Latt`es-type map with the tripleA,Θ,e Gas in Definition 3.3. Note that the affine mapA has not changed here and thatf is expanding if and only if
e
f is expanding. In other words, in order to prove the statement, we can make the additional assumptions that the Latt`es-type map f is defined on Cb and the map Θ :C→Cb is holomorphic.
Then Θ :C→Cb is the universal orbifold covering map of the parabolic orbifold
Of = (Cb, αf) off (see Proposition 3.6, Corollary 3.17, and Theorem 3.10).
Let ω be the canonical orbifold metric of Of. Since Of is parabolic, ω is
essentially the push-forward of the Euclidean metric onR2 by Θ (see Section A.10
and Section 2.5). The metricωis a length metric that induces the standard topology onCb (so it can be used as a base metric onCb as in Section 6.1 or Lemma 6.14) and the map Θ :R2→Cb is a path isometry in the sense that
length(α) = lengthω(Θ◦α)
for each pathαinR2.
If L = LA is the linear part of A, then the mapLn is the linear part of An
for eachn∈Z. SoLn and An only differ by a translation. Since translations are
isometries, it follows that
6.3. LATT`ES-TYPE MAPS AND EXPANSION 155
for alln∈Zwheneverαis a path inR2. Ifγ:= Θ◦α, then
fn
◦γ=fn
◦Θ◦α= Θ◦An ◦α forn∈N. Since Θ is a path isometry, we conclude that
lengthω(fn◦γ) = length(An◦α) = length(Ln◦α)
forn∈N.
Now suppose that Lis expanding. Then Lemma 6.13 implies that there exist N ∈Nand a constantρ >1 such that
length(LN ◦α)≥ρlength(α)
for all pathsαinR2. Ifγis an arbitrary path inCb, then it has a lift by the branched
covering map Θ (see Lemma A.18); so it can be written in the form γ = Θ◦α, whereαis a path inR2. Hence
lengthω(fN ◦γ) = length(LN ◦α)≥ρlength(α)
=ρlengthω(Θ◦α) =ρlengthω(γ).
Lemma 6.7 implies thatfN is an expanding Thurston map. Hencef is expanding
(see Lemma 6.5).
To prove the converse, we assume that f is expanding, but L is not. Then one of the eigenvalues of L has absolute value ≤ 1. Since the product of these eigenvalues is equal to det(L) = deg(f)≥2 (see Lemma 3.16), it follows from the considerations in the beginning of the proof of Lemma 6.13 that both eigenvalues ofLare real. So Lhas a real eigenvalueλwith|λ| ≤1. Note that λ6= 0, because Lis invertible.
Then there exists u ∈ R2 with |u| = 1 such that L(u) = λu. Let αbe the
parametrized line segment joining 0 and u, and define αn := A−n◦α for n∈ N.
Then
diam(αn) = diam(A−n◦α) = length(A−n◦α)
(6.8)
= length(L−n◦α) = 1
|λ|nlength(α) ≥length(α) = 1
for alln∈N.
Suppose for n ∈ N the path γn is a lift of some path γ in Cb by fn. Since
f is expanding, we then have diamω(γn) →0 as n → ∞. Indeed, if γ is a path
whose diameter is less than the Lebesgue numberδ >0 of an open cover U as in Proposition 6.4 (iii), then there exists U ∈ U such thatγ ⊂U. Then γn lies in a
connected component off−n(U) and so
diamω(γn)≤mesh(f−n(U))→0
as n → ∞. The statement diamω(γn)→ 0 asn → ∞ remains true for arbitrary
pathsγ, because we can breakγup into finitely many paths of diameter< δ. (We will later see that with respect to avisual metricforf(see Chapter 8) the diameters of lifts of any path by fn actually shrink to 0 exponentially fast asn
→ ∞ (see Lemma 8.9)).
We apply this toγ:= Θ◦α, andγn:= Θ◦αn forn∈N. The pathγn is a lift
ofγ byfn, because
We now obtain a contradiction from Lemma 6.14, because the sets αn are
connected and
diamω(γn) = diamω(Θ(αn))→0
asn→ ∞, but diam(αn)≥1 for alln∈Nby (6.8).
It follows thatLis expanding iff is. Together with the first part of the proof, we conclude that f is expanding as a Thurston map if and only if L = LA is
expanding as a linear map.
We finish this chapter by giving an example of a Thurston map that is eventu- ally onto, but not expanding. The example is due to K. Pilgrim.
Example 6.15. Let Gbe the crystallographic group consisting of all maps g of the form
u∈R2
7→g(u) =±u+γ,
whereγ∈Z2. SoGis of type (2222) (see Theorem 3.7). Let Θ :R2
→R2/G∼=S2
be the quotient map. We consider the matrix
A= 4 2 2 2 .
and the map A: R2 → R2, u ∈ R2 7→ Au given by left-multiplication of u ∈R2
(written as a column vector) by the matrixA. For simplicity, here (and also below) we not not distinguish in our notation between a matrix and the linear map it induces onR2by left-multiplication.
The map Ahas the form (3.23). The linear part L=LA ofA agrees with A.
SoL is also represented by the matrix A. Since det(LA) = det(A)≥2, we know
that the mapAinduces a Latt`es-type mapf:S2
→S2on the quotientS2=R2/G
according to Proposition 3.21. We claim that f is not expanding, but eventually onto.
Recall that the latter property means that for any non-empty open setU ⊂S2
there is an iteratefn such that fn(U) =S2 (see also Lemma 6.6).
The map L = A has the eigenvalues λ1 = 3−√5 and λ2 = 3 +√5. Since |λ1|<1, the mapf is not expanding by Proposition 6.12.
Now consider the linear maps B and C given by left-multiplication ofu∈R2
with the matrices
B= 2 1 1 1 andC= 2 0 0 2 ,
respectively. ThenA=B◦C=C◦B. The mapsB and C again have the form (3.23) and so descend to mapsg:S2
→S2andh:S2
→S2respectively. Note that
det(B) = 1, and so g is a homeomorphism with an inverse induced by B−1 (see
Proposition 3.21).
These maps satisfy f = g◦h = h◦g. Indeed, note that f ◦Θ = Θ◦A, g◦Θ = Θ◦B, andh◦Θ = Θ◦C. Thus
f◦Θ = Θ◦A= Θ◦B◦C=g◦Θ◦C=g◦h◦Θ.
Since Θ is surjective, it follows that f = g◦h. A similar argument shows that f =h◦g. Sincef =g◦h=h◦g, we havefn=gn◦hn for alln∈N.
Now let U ⊂ S2 be an arbitrary non-empty open set. Then V := Θ−1(U)
is a non-empty open set in R2. Since Cn(V) = 2nV this set Cn(V) will contain
6.3. LATT`ES-TYPE MAPS AND EXPANSION 157
such that Cn(V) contains a translate γ+R with γ ∈ Z2 of the rectangle R =
[0,1]×[0,1/2], which is a fundamental domain (see Section A.7) for the action of G. For suchnwe have
S2= Θ(R) = Θ(γ+R) = Θ(Cn(V)) =hn(Θ(V)) =hn(U), and so, sinceg is a homeomorphism,
fn(U) = (gn
◦hn)(U) =gn(S2) =S2.
CHAPTER 7