Quasisymmetric Rigidity Of Carpets In A C~*-Cylinder

The rigidity of carpets is a hot problem on the study of quasisymmetric mappings.For example,M.Bonk,B.Kleiner,and S.Merenkov proved that a quasisymmetric mapping between two round carpets of measure zero in the plane is the restriction of Mobius transformation.M.Bonk and S.Merenkov proved that a quasisymmetric mapping of a standard Sierpinski carpet onto itself is an isometry,etc.Also,they studied the quasisymmtric rigidity of C*-square carpets of measure zero in a C*-cylinder.Let C*=C\{0} be the complex plane omitted the origin.Let ds=|dz|/|z| be its length element and da=dxdy/|z|2 be its area element.Then C*is a metric measure space.Let 0<r<R<∞.We call the set A={z:r≤|z|≤R}a C*-cylinder.Let r<a<b<R,0 ≤α<β<2π,β-α<2π.We call the set{ρeiθ:a<ρ<b,α<θ<β}a C*-rectangle in A.In the case β-α=log(b/a),we also call this C*-rectangle a C*-square.We say that S is a C*-square carpet in C*-cyclinder A,if S=A\∪Qi,i∈N,where Qi’s are C*-squares in A and their closures are pairwise disjoint.M.Bonk and S.Merenkov proved the following theorem:Let S be a C*-square carpet of measure zero in A and f be a quasisymmetric mapping of S onto itself satisfying f({z:|z|=r})={z:|z|=r} and f({z:|z|=R})={z:|z|=R}.Then f is the restriction of a rotation.We continue to study this question in this thesis.The following theorem will be proved.Let S=(A\R0)\(?)Qi,i∈N,be a carpet of measure zero in A,where R0 is a C*-rectangle in A,Qi’s are C*-squares in A,and their closures are pairwise disjoint.If f is a quasisymmetric homeomorphism of S onto itself with f({z:|z|=r})={z:|z|=r},f({z:|z|=R})={z:|z|=R},and f((?)R0)=(?)R0,The f is the identity.