aa r X i v : . [ m a t h . C V ] F e b ON THE MODULI SPACEOF THE STANDARD CANTOR SET
HIROSHIGE SHIGA
Abstract.
We consider a generalized Cantor set E ( ω ) for an infinitesequence ω = ( q n ) ∞ n =1 of positive numbers with 0 < q n <
1, and examinethe quasiconformal equivalence to the standard middle one-third Cantorset E ( ω ). We may give a necessary and sufficient condition for E ( ω )to be quasiconformally equivalent to E ( ω ) in terms of ω . Introdution
Let ω = ( q n ) ∞ n =1 = ( q , q , . . . ) be an infinite sequence of positive numberswith 0 < q n < n ∈ N ). We may create a Cantor set, which is denotedby E ( ω ), from the sequence ω . The standard middle one-third Cantor set E ( ω ) is given by ω = ( ) ∞ n =1 = ( , , . . . ).The construction of E ( ω ) is the same as that of E ( ω ). Namely, we startwith an interval I := [0 ,
1] and remove an open interval J with the length q so that I \ J consists of two closed intervals I and I of the same length.We put E = ∪ i =1 I i = I \ J . We remove an open interval of length | I i | q from each I i so that the remainder E consists of four closed intervals of thesame length, where | J | is the length of an interval J . Inductively, we definethe set of the ( k + 1)-th step, E k +1 , from E k = ∪ k i =1 I ik by removing an openinterval of length | I ik | q k +1 from each closed interval I ik of E k so that E k +1 consists of 2 k +1 closed intervals of the same length. The generalized Cantorset E ( ω ) for ω is defined by E ( ω ) = ∩ ∞ k =1 E k . We say that two generalized Cantor sets E ( ω ) , E ( ω ) are quasiconfor-mally equivalent if there exists a quasiconformal mapping ϕ : b C → b C suchthat ϕ ( E ( ω )) = E ( ω ). For each ω = ( q n ) ∞ n =1 , we define the moduli space M ( ω ) of ω as the set of ω ′ = ( q ′ n ) ∞ n =1 so that E ( ω ′ ) is quasiconformallyequivalent to E ( ω ). In this paper, we completely determine the modulispace M ( ω ) of ω . Date : February 17, 2021.2010
Mathematics Subject Classification.
Primary 30C62; Secondary 30F45.
Key words and phrases.
Cantor set, Quasiconformal mapping.The author was partially supported by the Ministry of Education, Science, Sports andCulture, Japan; Grant-in-Aid for Scientific Research (B), 16H03933, 2016–2020.
In the previous paper [4], we consider conditions for generalized Cantorsets to be quasiconformally equivalent, and we obtain the following theorem.
Theorem 1.1 ([4]) . Let ω = ( q n ) ∞ n =1 and e ω = (˜ q n ) ∞ n =1 be sequences with δ -lower bound. We put (1.1) d ( ω, e ω ) = sup n ∈ N max (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) log 1 − ˜ q n − q n (cid:12)(cid:12)(cid:12)(cid:12) , | ˜ q n − q n | (cid:27) . (1) If d ( ω, e ω ) < ∞ , then there exists an exp( C ( δ ) d ( ω, e ω )) -quasiconformalmapping ϕ on b C such that ϕ ( E ( ω )) = E ( e ω ) , where C ( δ ) > is aconstant depending only on δ ; (2) if lim n →∞ log − ˜ q n − q n = 0 , then E ( e ω ) is asymptotically conformal to E ( ω ) , that is, there exists a quasiconformal mapping ϕ on b C with ϕ ( E ( ω )) = E ( e ω ) such that for any ε > , ϕ | U ε is (1+ ε ) -quasiconformalon a neighborhood U ε of E ( ω ) . Here, we say that a sequence ω = ( q n ) ∞ n =1 has δ -lower bound if thereexists δ ∈ (0 ,
1) such that q n > δ for any n ∈ N . Therefore, it follows fromTheorem 1.1 that for ω = ( q n ) ∞ n =1 , E ( ω ) and E ( ω ) are quasiconformallyequivalent if 0 < inf n q n ≤ sup n q n < n q n = 1, then E ( ω ) and E ( ω ) are not quasiconformally equivalent ([4]). Thus, it is a problem whathappens if inf n q n = 0.At the first glance, the author considered that E ( ω ) and E ( ω ) might notbe quasiconformally equivalent if inf n q n = 0. However, we find that it is nottrue. Indeed, we obtain a necessary and sufficient condition for ω so that E ( ω ) is quasiconformally equivalent to E ( ω ). We established the followingtheorems (see § Theorem I.
Let E ( ω ) be a generalized Cantor set for ω = ( q n ) ∞ n =1 . Then, D ( ω ) := b C \ E ( ω ) is a uniform domain if and only if N ( ω, δ ) < ∞ for some δ ∈ (0 , . The above theorem enables us to determine M ( ω ). Theorem II.
An infinite sequence ω = ( q n ) ∞ n =1 with < q n < n =1 , , . . . ) belongs to the moduli space M ( ω ) of ω = ( ) ∞ n =1 if and only if (1) sup n q n < , and (2) N ( ω, δ ) < ∞ for some δ ∈ (0 , . Preliminaries
First of all, we define the number N ( ω, δ ) in Theorems I and II.For ω = ( q n ) ∞ n =1 and δ ∈ (0 , ω ( δ ; i ) ( i ∈ N ) by ω ( δ ; i ) = inf { k ∈ N | q i + k ≥ δ } and N ( ω, δ ) by N ( ω, δ ) = sup i ∈ N ω ( δ ; i ) . For example, N ( ω , δ ) = 1 for any δ ∈ (0 , ]. It is easy to see that if q n → q > n → ∞ , then N ( ω, δ ) < ∞ for any δ ∈ (0 , q ). On the otherhand, if q n → n → ∞ , then N ( ω, δ ) = ∞ for any δ >
0. Also, for ω = ( , , , , . . . , , (2 n ) − , , (2 n + 2) − , . . . ), we have N ( ω, δ ) = 1 forany δ ∈ (0 , ] but inf n q n = 0.Next, we define uniform domains (cf. [2], [6]).A domain D ⊂ C is uniform if there exists a constant c ≥ a, b ∈ D can be joined by a curve γ in D so that for each z ∈ γ , (U1): | γ | ≤ c | a − b | ; (U2): min {| γ | , | γ |} ≤ c dist( z, ∂D ),where γ , γ are connected components of γ \ { z } , | α | is the length of a curve α , and dist( · , · ) stands for the Euclidian distance.Typical uniform domains are the upper half plane H and the unit disk.In fact, for a, b ∈ H , the hyperbolic geodesic γ connecting a and b satisfies(U1) and (U2) for c = π . Namely, inequalities(2.1) | γ | ≤ π | a − b | and(2.2) min {| γ | , | γ |} ≤ π z, ∂ H )hold for any z ∈ γ .Finally in this section, we define the uniform perfectness. A closed set in C is uniformly perfect if there exists a constant c ∈ (0 ,
1) such that(2.3) E ∩ { z ∈ C | cr < | z − a | < r = ∅} for any a ∈ E and r ∈ (0 , diam E ), where diam( A ) is the Euclidean diameterof a set A in C .It is easy to see that the standard Cantor set E ( ω ) is uniformly perfect.It is also known (cf. [4]) that the limit sets of some Kleinian groups areuniformly perfect. 3. Proof of Theorem I (Part 1)
In this section, we shall prove the sufficiency of the condition. Namely, weshow that if N ( ω, δ ) < ∞ for some δ >
0, then D ( ω ) is a uniform domain.Suppose that N := N ( ω, δ ) < ∞ for some δ ∈ (0 , a, b be anarbitrary pair of points of D ( ω ). We find a curve γ connecting a and b which satisfies the conditions (U1) and (U2) of the definition of uniformdomains. Case 1: a, b ∈ H (or in the lower half plane L ).Let γ be the hyperbolic geodesic γ in H connecting a and b . From (2.1),we have | γ | ≤ π | a − b | . HIROSHIGE SHIGA
Moreover, since dist( z, ∂ H ) ≤ dist( z, ∂D ( ω )) for z ∈ γ , we have from (2.2)min { γ , γ } ≤ π z, ∂ H ) ≤ π z, ∂D ( ω )) . Hence, (U1) and (U2) are satisfied for c = π . Case 2: a ∈ D ( ω ) \ R and b ∈ D ( ω ) ∩ R (or b ∈ D ( ω ) \ R and a ∈ D ( ω ) ∩ R ).We may assume that a ∈ H . Let γ be a hyperbolic geodesic in H whoseendpoints are a and b ∈ ∂ H . It is obvious that | γ | ≤ π | a − b | . Hence, the condition (U1) holds for c = π .To consider the condition (U2), we take z ∈ γ and put γ \ { z } = γ ∪ γ .We label γ as γ ∋ a . For a sufficiently small ε >
0, we take z ε ∈ γ so that | γ ε | = ε , where γ ε is the connected component of γ \ { z ε } containing b . Wemay assume that z ∈ γ − γ ε .By applying (2.2) to the geodesic γ − γ ε , we obtainmin {| γ | , | γ − γ ε |} ≤ π z, ∂ H ) ≤ π z, ∂D ( ω )) . Thus, we have(3.1) min {| γ | , | γ |} = lim ε → min {| γ | , | γ − γ ε |} ≤ π z, ∂D ( ω )) . Hence, the condition (U2) still holds for c = π . Case 3: a, b ∈ D ( ω ) ∩ R .Let γ be the hyperbolic geodesic joining a and b . Then, the same argumentas in Case 2 works and we may show that the conditions (U1) and (U2) holdfor c = π . Case 4: a ∈ H and b ∈ L (or a ∈ L and b ∈ H ).We will show that there exists a constant c ≥ a ∈ H and b ∈ L there exists a curve γ connecting a and b in D ( ω ) so that (U1)and (U2) hold for c and γ . We show it by contradiction.Suppose that for any n ∈ N there exist a n ∈ H , b n ∈ L such that anycurve γ connecting a n and b n in D ( ω ) does not satisfy (U1) or (U2) for c = n . Taking subsequences of { a n } and { b n } , we may assume that { a n } converges to some a ∞ and { b n } converges to b ∞ . Case 4-(i). a ∞ , b ∞ ∈ D ( ω ) and a ∞ = b ∞ .Take a curve γ connecting a ∞ and b ∞ in D ( ω ). Then, the curve γ satisfies(U1) and (U2) for some c ≥
1. Since a n is in a small neighborhood of a ∞ in D ( ω ) and so is b n for a sufficiently large n , it is not hard to see that for asufficiently large n , we may find a curve γ n connecting a n and b n such that γ n satisfies (U1) and (U2) for some c ′ ≥
1. Therefore, it is a contradiction.
Case 4-(ii). a ∞ , b ∞ ∈ D ( ω ) and a ∞ = b ∞ .Take a small disk D in D ( ω ) centered at a ∞ . Then, for a sufficientlylarge n , both a n and b n are contained in D . We verify that the hyperbolicgeodesic γ with respect to the hyperbolic metric of D connecting a n and b n satisfies (U1) and (U2) for some c ≥
1. Thus, this case also does nothappen.
Case 4-(iii). a ∞ or b ∞ ∈ E ( ω ) and a ∞ = b ∞ .We may assume that b ∞ ∈ E ( ω ). From the construction of E ( ω ), thereexists an infinite sequence of closed intervals { I n } ∞ n =1 such that(1) for each n ∈ N , I n is a closed interval appearing at the n -th step ofthe construction of E ( ω );(2) b ∞ ∈ · · · ⊂ I n +1 ⊂ I n ⊂ · · · ⊂ I ;(3) { b ∞ } = T ∞ n =1 I n .Take a sufficiently large n so that ε n := | b n − b ∞ | < − | a ∞ − b ∞ | and | a n − a ∞ | < − | a ∞ − b ∞ | .Here, we exhibit some fundamental estimates for the proof.From now on, we use C ( N ) as a positive constant depending only on N = N ( ω, δ ) (hence on ω , δ ) and C as an absolute positive constant whilethe both of them may change in each place.Then, we have | a n − b n | ≥ | a ∞ − b ∞ | − | a n − a ∞ | − | b n − b ∞ |≥ | a ∞ − b ∞ | ≥ ε n . Thus, we obtain(3.2) 0 < ε n ≤ C | a n − b n | . Let K be the minimal number of k ∈ N with | I k | < ε n . From the definitionof N ( ω, δ ), there exists n ∈ { , , . . . , N (= N ( ω, δ )) } such that q K + n ≥ δ but 0 < q K + i < δ for i = 1 , , . . . , n −
1. From the construction of E ( ω ), | I K + n − | = (cid:18) (cid:19) n − n − Y i =1 (1 − q K + i ) | I K | Hence, we have(3.3) (cid:18) − δ (cid:19) N − ε n ≤ (cid:18) − δ (cid:19) n − ε n ≤ | I K + n − | ≤ ε n . Let J n be the removed open interval from I n − in the construction of E ( ω ).Since q K + n ≥ δ , we see from (3.3)(3.4) | J K + n | = q K + n | I K + n − | ≥ C ( N ) ε n . Since J K + n ⊂ I K + n − , we also see(3.5) | J K + n | ≤ ε n . Namely, we verify that | I K + n − | and | J K + n | are comparable with ε n .Now, we proceed to prove the theorem. HIROSHIGE SHIGA
Let c n be the middle point of J K + n . We put b ′ n := c n − | J K + n |√− ∈ L .It follows from (3.3) and (3.5) that | b n b ′ n | ≤ | b n − b ∞ | + | b ∞ − c n | + | c n − b ′ n |≤ ε n + | I K + n | + 2 | J K + n | ≤ ε n , where ab is the line segment between a and b . For the hyperbolic geodesic b γ in L connecting b n and b ′ n , we have(3.6) | b γ | ≤ πε n , | b γ | ≤ C ( N ) | J K + n | from (2.1) and (3.4). We also have | b γ | ≤ C | a n − b n | from (3.2).Let e γ be the hyperbolic geodesic in H connecting a n and a ′ n := c n + | J K + n |√− ∈ H . Then, γ := e γ ∪ a ′ n b ′ n ∪ b γ is a curve in D ( ω ) from a n to b n .It follows from (2.1) that | e γ | ≤ π | a n − a ′ n | . As we have seen that | b n b ′ n | ≤ ε n , we have | a n − a ′ n | ≤ | a n − b n | + | b n b ′ n | + | a ′ n b ′ n | ≤ | a n − b n | + Cε n . From (3.2) | a n − a ′ n | ≤ C | a n − b n | and we get | e γ | ≤ C | a n − b n | . Combining those inequalities, we obtain | γ | = | e γ | + | a ′ n b ′ n | + | b γ | ≤ C | a n − b n | , Namely, γ satisfies (U1) for some C .Next, we consider the condition (U2) for γ .Let z be an arbitrary point of γ and γ \ { z } = γ ∪ γ . We label γ as γ ∋ a n .Suppose that z ∈ e γ . Then, γ ⊂ e γ , γ = γ ′ ∪ a ′ n b ′ n ∪ b γ , where γ ′ := γ ′ \ γ is the hyperbolic geodesic in H connecting z and a ′ n .If | γ | ≤ | γ ′ | , then | γ | ≤ | γ | and it follows from (2.2) thatmin {| γ | , | γ |} = | γ | ≤ π z, ∂ H ) ≤ π z, ∂D ( ω )) . In this case, the condition (U2) holds for c = π .If | γ | ≥ | γ ′ | , then we also have from (2.2) | γ ′ | ≤ π z, ∂ H ) ≤ π z, ∂D ( ω )) . Moreover, if | γ ′ | ≥ | a ′ n c n | , then we see that | γ | = | γ ′ | + | a ′ n c n | + | b γ | ≤ | γ ′ | + | b γ | . We also see | b γ | ≤ C ( N ) | J K + n | = C ( N ) | a ′ n c n | from (3.6). Therefore, weverify thatmin {| γ | , | γ |} ≤ | γ | ≤ C ( N ) | γ ′ | ≤ C ( N ) π z, ∂D ( ω ))and (U2) also holds.If | γ ′ | < | a ′ n c n | (= | J K + n | ), then z is in the disk centered at a ′ n withradius | J K + n | . Hence,dist( z, ∂D ( ω )) ≥ Im z ≥ | J K + n | . We obtain | γ | = | γ ′ | + | a ′ n b ′ n | + | b γ | ≤ | J K + n | + | b γ | ≤ C ( N ) | J K + n | from (3.6), and | γ | ≤ C ( N )dist( z, ∂D ( ω )) . The condition (U2) also holds.Next, we suppose that z ∈ a ′ n b ′ n . Then, it is easily seen thatdist( z, ∂D ( ω )) ≥ | J K + n | , since c n is the middle point of J K + n . As we have seen, | a ′ n b ′ n | + | b γ | ≤ C ( N ) | J K + n | . Since γ ⊂ a ′ n b ′ n ∪ b γ , we obtain | γ | ≤ C ( N )dist( z, ∂D ( ω )) , as desired.Finally, we suppose that z ∈ b γ and we may put b γ \ { z } = b γ ∪ γ .If | γ | ≤ | b γ | , then we have | γ | ≤ π z, ∂ H ) ≤ π z, ∂D ( ω ))from (2.2).If | γ | ≥ | b γ | , then we see that | Im z | ≥ | Im b ′ n | = | J K + n | because b γ isunder the line segment b n b ′ n in L . Sincedist( z, ∂D ( ω )) ≥ | Im z | and | γ | ≤ | b γ | ≤ C ( N ) | J K + n | , we obtain the condition (U2):min {| γ | , | γ |} ≤ | γ | ≤ C ( N )dist( z, ∂D ( ω )) . Case 4-(iv): a ∞ = b ∞ ∈ E ( ω ).We may assume that ε n := | a n − b n | is sufficiently small, and put { x } := a n b n ∩ R .If x ∈ E ( ω ), then there exists an infinite sequence { I n } ∞ n =1 of closedintervals such that HIROSHIGE SHIGA (1) for each n ∈ N , I n is a closed interval appearing at the n -th step ofthe construction of E ( ω );(2) x ∈ · · · ⊂ I n +1 ⊂ I n ⊂ · · · ⊂ I ;(3) { x } = T ∞ n =1 I n .Let K be the minimal number of k with | I k | ≤ ε n . From the definitionof N ( ω, δ ), there exists n ∈ { , , . . . , N ( ω, δ ) } such that q K + n ≥ δ but0 < q K + i < δ for i = 1 , , . . . , n − J n be the removed open interval from I n − in the construction of E ( ω ).We define a ′ n := x n + | J K + n |√− b ′ n = x n − | J K + n |√−
1, where x n is the middle point of J K + n . Then, we put γ = e γ ∪ a ′ n b ′ n ∪ b γ , where e γ isthe hyperbolic geodesic in H connecting a n and a ′ n , and b γ is the hyperbolicgeodesic in L connecting b n and b ′ n . By using the same estimates as (3.3)–(3.5), we see that | a n − a ′ n | ≤ | a n − x | + | x − x n | + | x n − a ′ n | ≤ Cε n . Similarly, we have | b n − b ′ n | ≤ Cε n . It follows from (2.1) that | e γ | ≤ Cε n , | b γ | ≤ Cε n . Thus, we have | γ | = | e γ | + | a ′ n b ′ n | + | b γ | ≤ C | a n − b n | , that is, the condition (U1) holds.As for (U2), repeating the same argument as in Case 4-(iii), we can provethat the condition (U2) hold for some c ≥ x ∈ D ( ω ).If dist( x , ∂D ( ω )) ≥ ε n , then it is easy to see that γ := a n b n satisfies theconditions (U1) and (U2) for c = 1.If dist( x , ∂D ( ω )) < ε n , we take x ∈ E ( ω ) so that | x − x | = dist( x , ∂D ( ω )) < ε n . Since x ∈ E ( ω ), we may take an infinite sequence { I n } ∞ n =1 for x as before.Repeating the same argument as above for { I n } ∞ n =1 to get K, n ∈ N for ε n ,we obtain an open interval J K + n and x n , the middle point of J K + n . Weput a ′ n := x n + | J K + n |√− b ′ n := x n − | J K + n |√−
1. Applying the sameargument as above, we verify that γ := e γ ∪ a ′ n b ′ n ∪ b γ satisfies the conditions(U1) and (U2) for some c = C ( N ), where e γ is the hyperbolic geodesic in H connecting a n and a ′ n , and b γ is the hyperbolic geodesic in L connecting b n and b ′ n .Overall, we get a contradiction in any case in the arguments of Case 4.Therefore, we conclude that D ( ω ) is a uniform domain. Proof of Theorem I (Part 2)
In this section, we show that if N ( ω, δ ) = ∞ for any δ >
0, then D ( ω ) isnot a uniform domain.Suppose that there exists c ≥ a, b ∈ D ( ω ) and for some γ .Take N ∈ N so that c < N . Since x − | log(1 − N − x ) | − x − → ∞ ( x → ∞ ) , there exist L > K ∈ N such that(4.1) L − N − < K < L − | log(1 − N − L ) | log N − . For those
K, L , we take M ∈ N so that q M + i < N − L for i = 0 , , , . . . , K .The condition N ( ω, N − L ) = ∞ guarantees the existence of M . We mayassume that K, L, M, N are sufficiently large, more than 10, for example.Now, we consider a closed interval I M − in the ( M − E ( ω ). We remove an closed interval J M with length q M | I M − | from I M − and we have two closed intervals I iM ( i = 1 ,
2) with the samelength. Hence,(4.2) | I M | = | I M | = 12 (1 − q M ) | T M − | ≥
12 (1 − N − L ) | I M − | . and(4.3) | J M | = q M | I M − | < N − L | I M − | . Let x ∈ R be the middle point of J M . We define a ∈ H and b ∈ L by a = x + N − L +2 | I M − |√− , b = x − N − L +2 | I M − |√− . From our assumption, there exists a curve γ in D ( ω ) connecting a and b such that it satisfies (U1) and (U2) for c ( < N ). We look for z ∈ γ ∩ R andconsider γ \ { z } = γ ∪ γ with γ ∋ a .First of all, if z I M − , then | γ | > | I M − | ≥ N (2 N − L +2 ) | I M − | > c | a − b | . The condition (U1) does not hold for c . Hence, z must be in I M − = I M ∪ J M ∪ I M .If z ∈ J M , then c dist( z, ∂D ( ω )) ≤ N | J M | ≤ N − L +1 | I M − | . On the other hand, since | γ | , | γ | ≥ | a − x | = N − L +2 | I M − | ,min( | γ | , | γ | ) > c dist( z, ∂D ( ω )) . The condition (U2) does not hold. Hence, z J M and z ∈ I M ∪ I M .We may assume that z ∈ I M . Since z ∈ D ( ω ), z is contained in some J M + m ⊂ I M ( m ≥ J n ⊂ I is a closed interval obtained in the n -thstep of the construction of E ( ω ). We have(4.4) | I kM + i | = (cid:18) (cid:19) i +1 i Y j =0 (1 − q M + j ) | I M − | ( k = 1 , , where I kM + i ( k = 1 ,
2) are closed intervals adjoining J M + i . and(4.5) | J M + i | = q M + i | I kM + i − | . If m ≤ K , then q M + i < N − L for i = 0 , , . . . , m . From (4.4) and (4.5), wehave (cid:18) − N − L (cid:19) m +1 | I M − | ≤ | I kM + m | ≤ (cid:18) (cid:19) m +1 | I M − | and | J M + m | ≤ q M + m | I M + m − | ≤ N − L (cid:18) (cid:19) m +1 | I M − | . Since z ∈ J M + m , | γ | ≥ | a − z | , | γ | ≥ | b − z | . Hence, we have(4.6) | γ | , | γ | ≥ | I kM + m | ≥ (cid:18) − N − L (cid:19) m +1 | I M − | . On the other hand,(4.7) dist( z, ∂D ( ω )) ≤ | J M + m | ≤ (cid:18) (cid:19) m +1 N − L | I M − | . However, from (4.1), we have m ≤ K ≤ L − | log(1 − N − L ) | log N − . This implies(4.8) N − L +1 ≤ (1 − N − L ) K +1 ≤ (1 − N − L ) m +1 . From (4.6)–(4.8), we have c dist( z, ∂D ( ω )) < N dist( z, ∂D ( ω )) ≤ | γ | , | γ | . Therefore, γ does not satisfy the condition (U2) if γ ∩ J M + m = ∅ for m ∈{ , , . . . , K } .Finally, we suppose that z ∈ J M + m for some m > K . Since J M + m ⊂ I kM + K ( k = 1 or 2), we have | J M + m | ≤ | I M + K | ≤ (cid:18) (cid:19) K +1 | I M − | . Hence,(4.9) dist( z, ∂D ( ω )) ≤ | J M + m | ≤ (cid:18) (cid:19) K +2 | I M − | . On the other hand,(4.10) | γ | , | γ | ≥ | a − b | = 2 N − L +2 | I M − | . However, from (4.1), we have L − N − < K. This implies(4.11) N (cid:18) (cid:19) K +2 < N − L +2 . From (4.9)–(4.11), we have c dist( z, ∂D ( ω )) < N dist( z, ∂D ( ω )) ≤ | γ | , | γ | . Hence, we verify that if γ ∩ J M + m = ∅ for m > K , then γ does not satisfythe condition (U2) for c < N .Thus, there is no curve γ in D ( ω ) for a, b satisfying the conditions (U1)and (U2), and we conclude that D ( ω ) is not a uniform domain.5. Proof of Theorem II
Theorem II is an immediate consequence of the following proposition (cf.[1] Corollary 2.1, [3] Theorem D).
Proposition 5.1.
Let D be a domain in C whose complement is a Cantorset. The domain D is quasiconformally equivalent to D ( ω ) if and only if C \ D is uniformly perfect and D is a uniform domain. It is easy to see that E ( ω ) is uniformly perfect if sup n q n <
1. It followsfrom Theorem I that D ( ω ) is uniform if N ( ω, δ ) < ∞ for some δ > D ( ω ) is quasiconformallyequivalent to D ( ω ).Conversely, if sup n q n = 1, then it is already shown that D ( ω ) is not qua-siconformally equivalent to D ( ω ). From Theorem I, D ( ω ) is not uniform if N ( ω, δ ) = ∞ for any δ >
0. Hence, D ( ω ) is not quasiconformally equivalentto D ( ω ) from Proposition 5.1.Thus, we complete the proof of the theorem. References [1] P. Bonfert-Taylor and E. C. Taylor, Quasiconformally homogeneous planar domains,Conf. Geom. & Dyn. 12 (2008), 188–198.[2] F. W. Gehring and K. Hag, The Ubiquitous Quasidisk, Mathematical Surveys andMonographs 184, Amer. Math. Soc., 2012.[3] P. MacManus, R. N¨akki and B. Palka, Quasiconformally bi-homogeneous compactain the complex plane, Proc. London Math. Soc. (3) 78 (1999), 215–240.[4] H. Shiga, On the quasiconformal equivalence of Dynamical Cantor sets, to appear inJ. d’Analyse Math.[5] T. Sugawa, Uniform perfectness of the limit sets of Kleinian groups, Trans. Amer.Math. Soc. 353 (2001), 3603–3615.[6] J. V¨ais¨al¨a, Uniforma domains, Tˆohoku Math. 40 (1988), 101–118.
Department of Mathematics, Kyoto Sangyo University, Motoyama, Kamig-amo, Kita-ku Kyoto, Japan
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