7. ANÁLISIS SECTORIAL
7.11. SECTOR SERVICIOS PÚBLICOS Y SANEAMIENTO BASICO
be preperiodic under iteration off. This implies that in (3.14) we may in addition assume thatx andy are periodic points for A, and that pis a periodic point for f. Moreover, by replacing the mapsA andf with suitable iterates, we are further reduced to the case wherexandyare fixed points ofA, andpis a fixed point off. If we introduce a suitable holomorphic coordinate w near psuch that w= 0 corresponds top, thenf has a local power series representation of the form
f(w) =λwd+. . . ,
whereλ6= 0 andd∈N. Sincef has no periodic critical points by Lemma 3.12 (i), we actually haved= 1, and so
f(w) =λw+. . . .
As we discussed before the proof, the map A lifts to a map A: C → C of the form A(z) = αz+β for z ∈ C, where α, β ∈ C, α 6= 0, such that we have a commutative diagram as in (3.13). This implies that by introducing a suitable holomorphic coordinate u near x such that u= 0 corresponds to x, the maps A and Θ can be given the formsA(u) =αuand
w= Θ(u) =buk+. . .
nearu= 0, whereb6= 0 andk= deg(Θ, x). Similarly, by using a suitable holomor- phic coordinatevneary such thatv= 0 corresponds toy, we can writeA(v) =αv and
w= Θ(v) =cvn+. . .
near v = 0, where c6= 0 and n= deg(Θ, y). Since f◦Θ = Θ◦Anear u= 0, we obtain
f(Θ(u)) =λbuk+· · ·= Θ(A(u)) =bαkuk+. . . .
In particular, λ =αk. Similarly, by considering the relation f(Θ(v)) = Θ(A(v))
nearv= 0, we obtainλ=αn. We conclude thatαk=λ=αn. Now 2
≤deg(f) = deg(A) =|α|2 by (3.11), and so |α| >1; but then αk =αn impliesk =n. This
contradicts our assumption thatk= deg(Θ, x)6= deg(Θ, y) =n.
This finishes the proof of Theorem 3.1.
3.3. Classifying Latt`es maps
Theorem 3.1 allows us to explicitly construct each Latt`es map as a quotient of a holomorphic automorphismA:C→Cby a crystallographic groupG. For such a mapAto pass to the quotientC/Git has to beG-equivariant. In this section we study this condition on A and some related questions in more detail. Our results essentially provide a classification of all Latt`es maps.
LetGbe a crystallographic group not isomorphic toZ2, and let Θ :C
→Cbbe a holomorphic map induced byGas provided by Proposition 3.9. LetA:C→Cbe a map of the formA(z) =αz+β, whereα, β∈Cwith|α|>1. Then by Lemma A.24 there is a (unique) mapf:Cb →Cb that satisfiesf◦Θ = Θ◦Aif and only ifA is G-equivariant. In this case,f is a Latt`es map by condition (ii) in Theorem 3.1.
Recall from Proposition 3.9 that for a given crystallographic groupG, the map Θ is unique up to postcomposition with a M¨obius transformation. So suppose
e
Θ = ϕ◦Θ is another (holomorphic) map induced by G, where ϕ: Cb → Cb is a M¨obius transformation. Then if f ◦Θ = Θ ◦A it is immediate to check that
e
map induced by a mapAthat is equivariant for a given crystallographic groupG is unique up to conjugation by a M¨obius transformation.
For a given Latt`es map f: Cb → Cb, the map A: C → C and the group G provided by Theorem 3.1 (ii) are not unique. Indeed, we can conjugate G and A by an arbitrary map h ∈ Aut(C). Then Ge = {h−1
◦g◦h : g ∈ G} is also a crystallographic group. IfAe=h−1
◦A◦handΘ = Θe ◦h, then we obtain the same map f in Theorem 3.1 (ii), if we replace G,Θ, A with G,e Θ,e A, respectively. Thee situation is illustrated in the following commutative diagram:
(3.15) C Ae // h e Θ C h e Θ ~ ~ C A // Θ C Θ b C f //Cb.
By using such a conjugation, in Theorem 3.1 (ii) we can always assume that the groupGis one of the groupsGe listed in Theorem 3.7.
The requirement that the map A(z) = αz+β is G-equivariant can then bee explicitly analyzed and puts strong restrictions onαandβ as the following propo- sition shows.
Proposition3.14. Let G=Ge be a crystallographic group as in Theorem 3.7 not isomorphic to Z2. LetΓ =Z⊕Zτ be the underlying lattice, whereτ ∈Cwith
Im(τ)>0 in the case (2222), τ =i in the case (244), and τ =ω =eπi/3 in the cases (333)and (236). ThenA: C→Cgiven by A(z) =αz+β (where α, β∈C,
α6= 0) is G-equivariant if and only if
α, ατ,2β α,(1 +i)β α,(1 +ω)β α, β
are elements of Γwhen Gis of type (2222), (244), (333), (236).
As we will see in the proof, the condition onαis equivalent to the requirement thatαΓ⊂Γ. According to this proposition, we can always chooseα∈Z\ {0}and β = 0 for any lattice Γ. This and Theorem 3.1 (ii) imply that Latt`es maps exist for all signatures (2,2,2,2), (2,4,4), (3,3,3), and (2,3,6). We will discuss more explicit examples for each of these signatures later in Section 3.6.
Proof. Recall from (3.2) that a mapz7→A(z) =αz+β as in the statement isG-equivariant if and only ifA◦g◦A−1∈Gfor allg∈G.
The mapsg∈Ghave the formg(z) =λz+γ, whereγ∈Γ andλis a root of unity depending on the type ofG. An elementary computation shows that
(A◦g◦A−1)(z) =λz+αγ+ (1 −λ)β. Using this first forγ= 0∈Γ, we see that (3.2) can only be valid if
(3.16) (1−λ)β ∈Γ,
for the appropriate roots of unityλ; in addition, it is necessary that
3.3. CLASSIFYING LATT`ES MAPS 67
Conversely, if (3.16) and (3.17) are true, A satisfies (3.2). Thus A is G- equivariant if and only ifαandβ satisfy (3.16) and (3.17).
Note that (3.17) is equivalent to the condition that αΓ ⊂ Γ. Since 1 and τ generate the lattice Γ, this in turn is the same as the requirement that
(3.18) α∈Γ andατ ∈Γ.
If Gis of type (244), (333), or (236), then we can omit the second condition here. Indeed, in these casesτΓ = Γ, and so α∈Γ if and only ifατ ∈Γ.
This discussion shows that A is G-equivariant if and only if α satisfies the conditions as in the statement of the proposition, andβ satisfies (3.16). To analyze this latter condition further, we consider several cases depending on the type ofG. IfGis of type (2222), then we haveλ=±1. Thus (3.16) is true if and only if 2β∈Γ.
If G is of type (244), then we have Γ = Z⊕Zi and λ = 1,i,−1,−i. Thus
(3.16) implies that (1 +i)β∈Γ.
Conversely, suppose this last condition is true. Note that −iΓ = Γ and (1− i)Γ⊂Γ. So we conclude that−i(1+i)β= (1−i)β ∈Γ and (1−i)(1+i)β = 2β∈Γ.
Therefore, (3.16) holds forλ= 1,i,−1,−i.
IfGis of type (333), then we have Γ =Z⊕Zω andλ= 1, ω2, ω4(recall that
ω=eπi/3). So (3.16) implies that (1
−ω4)β = (1 +ω)β ∈Γ.
Conversely, if this last condition is true, then usingωΓ = Γ we obtainω5(1 +
ω)β= (1−ω2)β
∈Γ; so (3.16) holds forλ=ωj wherej= 0,2,4.
Finally, ifGis of type (236), then Γ =Z⊕Zω andλ=ωj wherej= 0, . . . ,5.
Thus (3.16) implies (1−ω5)β = (1 +ω2)β
∈Γ. Note that 1 +ω2=ω, and so we
obtainωβ∈Γ. SinceωΓ = Γ, this shows thatβ∈Γ.
Conversely, assume thatβ ∈Γ. UsingωΓ = Γ again, we conclude thatωjβ ∈Γ forj= 0, . . . ,5. This impliesβ−ωjβ= (1
−ωj)β
∈Γ. Thus (3.16) is satisfied for allλ=ωj withj = 0, . . . ,5.
The claim follows.
Suppose a Latt`es mapf:Cb→Cb is given as in Theorem 3.1 (ii). Thenf◦Θ = Θ◦A, where Θ :C→Cb is induced by a crystallographic groupG(not isomorphic to Z2) and A(z) =αz +β (withα, β
∈C and |α| >1) is G-equivariant. As we discussed, here we can always assume thatG=Ge is as in Theorem 3.7.
One can make another reduction. Namely, we can replace A with any map
e
A=g◦A, whereg∈G. Indeed, we then have Θ◦Ae= Θ◦A=f◦Θ, because Θ is induced byGand so Θ◦g= Θ. In particular,Aeis alsoG-equivariant and induces the same Latt`es mapf.
One can use this remark to substantially restrict the values of the coefficientβ of the mapA.
Proposition 3.15. Let f:Cb → Cb be a Latt`es map as above obtained from a map A: C → C given by A(z) = αz+β, a crystallographic group G = Ge as in Theorem 3.7, and a holomorphic mapΘ : C→Cb induced by G. Then by postcom- posing Awith a suitable translation g∈Gtr⊂Gwe can always assume thatβ has
one of the following forms: β∈ {0,1 2, 1 2τ, 1 2(1 +τ)} β∈ {0,1 2(1 +i)} β∈ {0,1 3 + 1 3ω, 2 3+ 2 3ω} β= 0 whenGis of type (2222), (244), (333), (236).
Proof. As before, we denote by Γ the underlying lattice ofG. Let∼be the equivalence relation onCdefined byz∼wif and only ifw−z∈Γ forz, w∈C. This is the equivalence relation induced by the action of the subgroup of all translations Gtr ⊂G. As we have seen, ifg(z) =z+γ with γ ∈Γ, then g ∈Gtr and we can
replace theG-equivariant mapA(z) =αz+β with
e
A(z) = (g◦A)(z) =αz+ (β+γ).
This means that we can changeβ to any elementβ′ withβ′∼β without affecting
the Latt`es mapf. We now analyze this in combination with the condition onβ in Proposition 3.14 for the different types ofG.
IfGis of type (2222), then 2β ∈Γ by Proposition 3.14 or equivalently,β∈ 1 2Γ.
As we can replaceβ with any β′ satisfyingβ′∼β, we may assume that (3.19) β= 12(k+ℓτ), wherek, ℓ∈ {0,1}.
Soβ has the desired form.
IfGis of type (244), then by Proposition 3.14 the relevant condition is (1+i)β∈
Γ. This implies that
β ∈(1 +i)−1Γ = 1
2(1−i)Γ⊂ 1 2Γ.
So again we may assume thatβ is as in (3.19) withτ =i. For such β we have
(1 +i)β =1
2((k−ℓ) + (k+ℓ)i)∈Γ
precisely ifk=ℓ∈ {0,1}. The statement follows in this case.
IfGis of type (236), then by Proposition 3.14 the condition onβ isβ ∈Γ, or equivalentlyβ∼0. This means that we can always takeβ = 0 in this case.
Finally, ifGis of type (333), then by Proposition 3.14 the relevant condition is (1+ω)β∈Γ, whereω=eπi/3
and Γ =Z⊕Zω. SinceωΓ = Γ and (1+ω)(1+ω5) = 3,
this implies
β ∈(1 +ω)−1Γ = 13(1 +ω5)Γ⊂ 1 3Γ.
Hence we may assume thatβ has the form
β =1
3(k+ℓω), where k, ℓ∈ {0,1,2}.
If we useω2=ω
−1, we see that for suchβ we have (1 +ω)β =1
3((k−ℓ) + (k+ 2ℓ)ω)∈Γ