Some progresses on Lipschitz equivalence of self-similar sets - PowerPoint PPT Presentation

Some progresses on Lipschitz equivalence of self-similar sets Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Zhejiang University Chinenes University of Hong Kong – Dec 10-14, 2012 Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Part I. Lipschitz equivalence of dust-like self-similar sets Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Definition Let E , F be compact sets in R d . We say that E and F are Lipschitz equivalent, and denote it by E ∼ F , if there exists a bijection g : E − → F which is bi-Lipschitz, i.e. there exists a constant C > 0 such that for all x , y ∈ E , C − 1 | x − y | ≤ | g ( x ) − g ( y ) | ≤ C | x − y | . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question Under what conditions, two self-similar sets are Lipschitz equivalent? Necessary condition: same Hausdorff dimension. The condition is not sufficient even for dust-like case. (The generating IFS satisfies the strong separation condition.) Example Let E be the Cantor middle-third set. Let s = log 2 / log 3 and 3 · r s = 1. Let F be the dust-like self-similar set generated as the following figure. Then E �∼ F . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question Under what conditions, two self-similar sets are Lipschitz equivalent? Necessary condition: same Hausdorff dimension. The condition is not sufficient even for dust-like case. (The generating IFS satisfies the strong separation condition.) Example Let E be the Cantor middle-third set. Let s = log 2 / log 3 and 3 · r s = 1. Let F be the dust-like self-similar set generated as the following figure. Then E �∼ F . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question Under what conditions, two self-similar sets are Lipschitz equivalent? Necessary condition: same Hausdorff dimension. The condition is not sufficient even for dust-like case. (The generating IFS satisfies the strong separation condition.) Example Let E be the Cantor middle-third set. Let s = log 2 / log 3 and 3 · r s = 1. Let F be the dust-like self-similar set generated as the following figure. Then E �∼ F . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question Under what conditions, two self-similar sets are Lipschitz equivalent? Necessary condition: same Hausdorff dimension. The condition is not sufficient even for dust-like case. (The generating IFS satisfies the strong separation condition.) Example Let E be the Cantor middle-third set. Let s = log 2 / log 3 and 3 · r s = 1. Let F be the dust-like self-similar set generated as the following figure. Then E �∼ F . r r r 1/3 1/3 Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Let E , F be dust-like self-similar sets generated by the IFS { Φ j } n j = 1 , { Ψ j } m j = 1 on R d , respectively. ρ j (resp. τ j ) is the contraction ratio of Φ j (resp. Ψ j ). Q ( a 1 , . . . , a m ) : subfield of R generated by Q and a 1 , . . . , a m . sgp ( a 1 , . . . , a m ) : subsemigroup of ( R + , × ) generated by a 1 , . . . , a m . Theorem (Falconer-Marsh, 1992) Assume that E ∼ F . Let s = dim H E = dim H F . Then (1) Q ( ρ s 1 , . . . , ρ s m ) = Q ( τ s 1 , . . . , τ s n ) ; (2) ∃ p , q ∈ Z + , s.t. sgp ( ρ p 1 , . . . , ρ p m ) ⊂ sgp ( τ 1 , . . . , τ n ) and sgp ( τ q 1 , . . . , τ q n ) ⊂ sgp ( ρ 1 , . . . , ρ m ) . Using (2), we can show that E �∼ F in the above example. Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Let E , F be dust-like self-similar sets generated by the IFS { Φ j } n j = 1 , { Ψ j } m j = 1 on R d , respectively. ρ j (resp. τ j ) is the contraction ratio of Φ j (resp. Ψ j ). Q ( a 1 , . . . , a m ) : subfield of R generated by Q and a 1 , . . . , a m . sgp ( a 1 , . . . , a m ) : subsemigroup of ( R + , × ) generated by a 1 , . . . , a m . Theorem (Falconer-Marsh, 1992) Assume that E ∼ F . Let s = dim H E = dim H F . Then (1) Q ( ρ s 1 , . . . , ρ s m ) = Q ( τ s 1 , . . . , τ s n ) ; (2) ∃ p , q ∈ Z + , s.t. sgp ( ρ p 1 , . . . , ρ p m ) ⊂ sgp ( τ 1 , . . . , τ n ) and sgp ( τ q 1 , . . . , τ q n ) ⊂ sgp ( ρ 1 , . . . , ρ m ) . Using (2), we can show that E �∼ F in the above example. Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Let E , F be dust-like self-similar sets generated by the IFS { Φ j } n j = 1 , { Ψ j } m j = 1 on R d , respectively. ρ j (resp. τ j ) is the contraction ratio of Φ j (resp. Ψ j ). Q ( a 1 , . . . , a m ) : subfield of R generated by Q and a 1 , . . . , a m . sgp ( a 1 , . . . , a m ) : subsemigroup of ( R + , × ) generated by a 1 , . . . , a m . Theorem (Falconer-Marsh, 1992) Assume that E ∼ F . Let s = dim H E = dim H F . Then (1) Q ( ρ s 1 , . . . , ρ s m ) = Q ( τ s 1 , . . . , τ s n ) ; (2) ∃ p , q ∈ Z + , s.t. sgp ( ρ p 1 , . . . , ρ p m ) ⊂ sgp ( τ 1 , . . . , τ n ) and sgp ( τ q 1 , . . . , τ q n ) ⊂ sgp ( ρ 1 , . . . , ρ m ) . Using (2), we can show that E �∼ F in the above example. Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Let E , F be dust-like self-similar sets generated by the IFS { Φ j } n j = 1 , { Ψ j } m j = 1 on R d , respectively. ρ j (resp. τ j ) is the contraction ratio of Φ j (resp. Ψ j ). Q ( a 1 , . . . , a m ) : subfield of R generated by Q and a 1 , . . . , a m . sgp ( a 1 , . . . , a m ) : subsemigroup of ( R + , × ) generated by a 1 , . . . , a m . Theorem (Falconer-Marsh, 1992) Assume that E ∼ F . Let s = dim H E = dim H F . Then (1) Q ( ρ s 1 , . . . , ρ s m ) = Q ( τ s 1 , . . . , τ s n ) ; (2) ∃ p , q ∈ Z + , s.t. sgp ( ρ p 1 , . . . , ρ p m ) ⊂ sgp ( τ 1 , . . . , τ n ) and sgp ( τ q 1 , . . . , τ q n ) ⊂ sgp ( ρ 1 , . . . , ρ m ) . Using (2), we can show that E �∼ F in the above example. Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Let E , F be dust-like self-similar sets generated by the IFS { Φ j } n j = 1 , { Ψ j } m j = 1 on R d , respectively. ρ j (resp. τ j ) is the contraction ratio of Φ j (resp. Ψ j ). Q ( a 1 , . . . , a m ) : subfield of R generated by Q and a 1 , . . . , a m . sgp ( a 1 , . . . , a m ) : subsemigroup of ( R + , × ) generated by a 1 , . . . , a m . Theorem (Falconer-Marsh, 1992) Assume that E ∼ F . Let s = dim H E = dim H F . Then (1) Q ( ρ s 1 , . . . , ρ s m ) = Q ( τ s 1 , . . . , τ s n ) ; (2) ∃ p , q ∈ Z + , s.t. sgp ( ρ p 1 , . . . , ρ p m ) ⊂ sgp ( τ 1 , . . . , τ n ) and sgp ( τ q 1 , . . . , τ q n ) ⊂ sgp ( ρ 1 , . . . , ρ m ) . Using (2), we can show that E �∼ F in the above example. Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Let E , F be dust-like self-similar sets generated by the IFS { Φ j } n j = 1 , { Ψ j } m j = 1 on R d , respectively. ρ j (resp. τ j ) is the contraction ratio of Φ j (resp. Ψ j ). Q ( a 1 , . . . , a m ) : subfield of R generated by Q and a 1 , . . . , a m . sgp ( a 1 , . . . , a m ) : subsemigroup of ( R + , × ) generated by a 1 , . . . , a m . Theorem (Falconer-Marsh, 1992) Assume that E ∼ F . Let s = dim H E = dim H F . Then (1) Q ( ρ s 1 , . . . , ρ s m ) = Q ( τ s 1 , . . . , τ s n ) ; (2) ∃ p , q ∈ Z + , s.t. sgp ( ρ p 1 , . . . , ρ p m ) ⊂ sgp ( τ 1 , . . . , τ n ) and sgp ( τ q 1 , . . . , τ q n ) ⊂ sgp ( ρ 1 , . . . , ρ m ) . Using (2), we can show that E �∼ F in the above example. Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question What’s the necessary and sufficient condition? How about for two branches case? τ ρ ρ τ 2 1 1 2 WLOG, we may assume that ρ 1 ≤ ρ 2 , τ 1 ≤ τ 2 and ρ 1 ≤ τ 1 . Conjecture. Lipschitz equivalent iff ( ρ 1 , ρ 2 ) = ( τ 1 , τ 2 ) . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question What’s the necessary and sufficient condition? How about for two branches case? τ ρ ρ τ 2 1 1 2 WLOG, we may assume that ρ 1 ≤ ρ 2 , τ 1 ≤ τ 2 and ρ 1 ≤ τ 1 . Conjecture. Lipschitz equivalent iff ( ρ 1 , ρ 2 ) = ( τ 1 , τ 2 ) . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets

Question What’s the necessary and sufficient condition? How about for two branches case? τ ρ ρ τ 2 1 1 2 WLOG, we may assume that ρ 1 ≤ ρ 2 , τ 1 ≤ τ 2 and ρ 1 ≤ τ 1 . Conjecture. Lipschitz equivalent iff ( ρ 1 , ρ 2 ) = ( τ 1 , τ 2 ) . Huo-Jun Ruan (With Hui Rao, Yang Wang and Li-Feng Xi) Some progresses on Lipschitz equivalence of self-similar sets