Quasisymmetric Embeddings of Slit Sierpi nski Carpets into R 2 - PowerPoint PPT Presentation

Quasisymmetric Embeddings of Slit Sierpi´ nski Carpets into R 2 Wenbo Li Graduate Center, CUNY CAFT, July 21, 2018 Wenbo Li 1 / 17

Sierpi´ nski Carpet S 3 - the standard Sierpi´ nski carpet in [ 0 , 1 ] 2 . Figure: Finite generation of standard Sierpi´ nski carpet ≈ S 3 if Whyburn’s Theorem: X = S 2 ∖ ⋃ ∞ homeo i D i D i ∩ D j = ∅ , ∀ i ≠ j . 1 diam ( D i ) → 0 , as i → ∞ . 2 ( ⋃ i D i ) = S 2 . 3 A metric space ( X , d ) is a (metric) carpet if X ≈ S 3 . homeo Wenbo Li 2 / 17

Quasiconformality and Quasisymmetry f ∶ X � → Y - homeomorphism f is (metrically) quasiconformal if ∃ K ≥ 1 s.t. sup { d Y ( f ( x ) , f ( y )) ∶ d x ( x , y ) ≤ r } inf { d Y ( f ( x ) , f ( y )) ∶ d x ( x , y ) ≥ r } H f ( x ) = limsup r → 0 ≤ K for all x ∈ X . f is quasisymmetric if ∃ homeomorphism η ∶ [ 0 , ∞) → [ 0 , ∞) s.t. d Y ( f ( x ) , f ( y )) d Y ( f ( x ) , f ( z )) ≤ η ( d X ( x , y ) d X ( x , z )) for all x , y , z ∈ X with x ≠ z . Wenbo Li 3 / 17

Geometry of Carpets: Questions ≈ S 3 ? BL Question 1 Is every carpet X NO. Easy. S 5 . Using Hausdorff dimension. ≈ S 3 ? QS Question 2 Is every carpet X NO. Hard. S 5 . (Bonk Merenkov) QS ↪ R 2 ? Question 3 Is every carpet X NO. A self-similar slit carpet. (Merenkov, Wildrick) QS ↪ R 2 ? Question 4 Is every group boundary carpet X Not known. Kapovich-Kleiner conjecture (simplified version): Every group boundary carpet X can be quasisymmetrically embedded into R 2 . General Problem: Characterize carpets which can be quasisymmetrically embedded into R 2 . Our Result: We give a complete characterization for a special kind of carpets. Wenbo Li 4 / 17

Slit Domains Dyadic slit domain of n th-generation corresponding to r = { r i } ∞ i = 0 : S n ( r ) = [ 0 , 1 ] 2 /⎛ ⎞ 2 i ⋃ n ⋃ ⎝ ⎠ , s ij i = 0 j = 1 s ij ⊂ ∆ ij where ∆ ij is a dyadic square of generation i The center of s ij coincides with the center of ∆ ij l ( s ij 1 ) = l ( s ij 2 ) = r i ⋅ 1 2 i for j 1 , j 2 ∈ { 1 , . . . , 2 i } . Figure: Slit domains corresponding to ( 1 2 , 1 2 , 1 2 , 1 2 ) and ( 1 10 , 2 5 , 1 8 , 1 2 ) Wenbo Li 5 / 17

Slit Sierpi´ nski Carpets Let r = { r i } ∞ i = 0 be a sequence satisfying: r i ∈ ( 0 , 1 ) , ∀ i ∈ N 1 lim sup i →∞ r i < 1 2 S n ( r ) = S n ( r ) equipped with the path metric. The limit S ( r ) = S n ( r ) lim n →∞ G-H limit is called a dyadic slit Sierpi´ nski carpet corresponding to r . S( r ) = [ 0 , 1 ] 2 /( ⋃ ∞ j = 1 s ij ) . i = 0 ⋃ 2 i S ( r ) is a metric carpet. Wenbo Li 6 / 17

Transboundary Modulus Let Γ be a family of curves in metric measure space ( X , d ,µ ) . The transboundary 2-modulus of Γ with respect to { K i } is defined as tr-mod X , { K i } ( Γ ) = inf { ∫ X / ⋃ i K i ρ 2 d µ + ∑ i } ρ 2 i ∈ I where infimum is taken over all Borel function ρ ∶ X / ⋃ i K i → ( 0 , ∞] , and weights ρ i ≥ 0 such that ρ ds + ρ i ≥ 1 , ∀ γ ∈ Γ . ∑ ∫ γ ∩ X / ⋃ i K i γ ∩ K i ≠ φ Transboundary n -modulus is a conformal invariant and quasiconformal quasi-invariant on R n . Wenbo Li 7 / 17

Transboundary Modulus Estimation Theorem 1 (Upper bound, Hakobyan-Li, 2017) Let Γ = Γ ( L , R , I ) and K n be the collection of all slits in S n ( r ) . Then tr-mod I , K n ( Γ ) ≤ n ( 1 − ǫ r 2 i ) + 3 ǫ ∏ i = 0 for ∀ ǫ small enough. Wenbo Li 8 / 17

Transboundary Modulus Estimation Lemma 2 Let Γ io = Γ ( s 0 ,∂ I , I / s 0 ) . If { r i } ∉ ℓ 2 , then n → ∞ tr-mod I / s 0 , K n /{ s 0 } ( Γ io ) = 0 . lim Wenbo Li 9 / 17

Proof of Theorem 1 B ϵ 0 O ϵ 0 B ϵ 0 } ϵ l(s 0 ) Figure: Slit collar n ( x ) = χ B ǫ n ( x ) = χ I ∖ O ǫ n ( x ) A ( ρ,ρ i ) ≤ ∏ n i = 0 ( 1 − ǫ r 2 i ) + 3 ǫ. ρ ǫ n ∪ R ǫ n = { ǫ l ( v j ) , v j are chosen slits ρ j 0 , otherwise Wenbo Li 10 / 17

Proof of Theorem 1 Figure: admissible Replacing each curve with a new one. the new curve is in I ∖ O ǫ n . the new curve is “shorter” than the old one. Wenbo Li 11 / 17

Main Theorem Main Theorem (Hakobyan-Li, 2017) S ( r ) can be quasisymmetrically embedded into R 2 if and only if r = { r i } ∞ i = 0 ∈ ℓ 2 . Wenbo Li 12 / 17

r ∉ ℓ 2 � ⇒ ∄ ϕ ∶ S ( r ) ↪ R 2 quasisymmetrically Suppose not, i.e. ∃ ϕ ∶ S ↪ R 2 , where ϕ is η -QS. 1 ∃ f n ∶ S n → R 2 is quasiconformal. 2 Figure: An illustration of f 2 = ψ 2 ○ ϕ 2 ○ i 2 . Wenbo Li 13 / 17

r ∉ ℓ 2 � ⇒ ∄ ϕ ∶ S ( r ) ↪ R 2 quasisymmetrically If { r i } ∉ ℓ 2 then tr-mod ( Γ io ) → 0 as n → ∞ .(Lemma 2) 3 tr-mod ( f n ( Γ io )) = > ǫ > 0, since quasisymmetry on inner and 2 π 4 log Rn rn outer boundary. 0 < ǫ < tr-mod ( f n ( Γ io )) ≤ C ⋅ tr-mod ( Γ io ) → 0. Contradiction. 5 Figure: An illustration of the proof. Wenbo Li 14 / 17

Application of Main Theorem Corollary 3 There exists a doubling, linearly locally contractible metric space X which is homeomorphic to R 2 or S 2 such that X is not quasisymmetrically embedded into R 2 or S 2 , respectively. Every weak tangent of X is quasisymmetric equivalent to R 2 with a uniformly bounded distortion function. Wenbo Li 15 / 17

Further Questions and Extensions What is a general slit carpet? A metric carpet which is the closure of a slit domain equipped with path metric and uniformly relatively separated. What is the equivalent criterion of planar quasisymmetric embeddability for general slit carpets? One guess is that the length of slits in each ball should be ℓ 2 -uniformly relatively bounded. Are the statements true for higher dimensions? A similar statement for necessity is valid for higher dimensions. The sufficiency is not known. A similar statement for corollary 3 in higher dimensions is also valid and is working in progress. Wenbo Li 16 / 17

Thank you. Wenbo Li 17 / 17