A. Telcs Supporting research schools at AFRT December 2012 Hong - PowerPoint PPT Presentation

A. Telcs Supporting research schools at AFRT December 2012 Hong the University of Pannonia Kong Reg.No.:TÁMOP-4.2.2/B-10/1-2010-0025 Project period: 2011.08.01-2013.07.31. Name and address of Contractor: University of Pannonia Egyetem utca 10, Veszprém H-8200 Name and Address of Subcontractor: Agricultural Research Institute of the Hungarian Academy of Sciences Brunszvik utca 2, Martonvásár H-2462 Total funding granted by the European Union and the Hungarian Government: 669 779 225.- HUF.

Introduction Preliminaries Prelimineries The result Open problems Bonnus Di¤usive limits on the Penrose tiling András Telcs Hong Kong December 2012.

Introduction Preliminaries Prelimineries The result Open problems Bonnus The beginings Figure: Islamic tiling from the 15th century Peter J. Lu and Paul J. Steinhardt "Decagonal and Quasi-crystalline Tilings in Medieval Islamic Architecture," Science 315, 1106 (2007).

Introduction Preliminaries Prelimineries The result Open problems Bonnus The beginings

Introduction Preliminaries Prelimineries The result Open problems Bonnus The Penrose tiling The Penrose tiling Penrose 1978

Introduction Preliminaries Prelimineries The result Open problems Bonnus Random walk on the Penrose graph Random walk on the Penrose graph The graph and the net (the dual of the Penrose tiling)

Introduction Preliminaries Prelimineries The result Open problems Bonnus Random walk The graph and the net (the dual of the Penrose tiling) � countably in…nite graph, with edges x � y ; and trivial edge weight 1, degree of nodes, d ( x ) = P y I ( x � y ) = 1 4 1 P ( x ; y ) = d ( x ) if x � y P ( X n = y j X n � 1 = x ) = P ( x ; y ) de…nes a d ( x ) -reverzibe Markov chain

Introduction Preliminaries Prelimineries The result Open problems Bonnus Random walk Conventions � with d ( x ; y ) ; graph distance called Penrose graph � with j x � y j embedded into R 2 called Penrose net Z 2 with d ( x ; y ) the graph on the integer lattice Z 2 with j x � y j embedded into R 2 called integer net

Introduction Preliminaries Prelimineries The result Open problems Bonnus Questions Questions Random walks on graphs � � x ; y = 1 ; d ( x ; y ) , p n ( x ; y ) ? 1 � p n ( x ; y ) + p n + 1 ( x ; y ) � C Cne � C d 2 ( x ; y ) n e � d 2 ( x ; y ) n Cn

Introduction Preliminaries Prelimineries The result Open problems Bonnus Questions Questions Domokos Szász problem: Do we have invariance principle on the Penrose net? � � x ; y = 1 ; j x � y j 1 p aX b at c = ) W D ( t ) as a ! 1 as a ! 1 : D pos. def.

Introduction Preliminaries Prelimineries The result Open problems Bonnus Questions Why is this a question? put Z d grid on R d then the invariance principle holds for the ssnn RW Put any periodic (shift invariant) grid on R d , still we have the IP. What if we put a non-periodic net on it? What if .... (comes later).

Introduction Preliminaries Prelimineries The result Open problems Bonnus Pentagrid The construction of the Penrose tiling (deBruijn) Pentagrid

Introduction Preliminaries Prelimineries The result Open problems Bonnus Pentagrid The construction of the Penrose tiling (deBruijn) Pentagrid

Introduction Preliminaries Prelimineries The result Open problems Bonnus Pentagrid

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry Rough isometry Given � ; � 0 ; rough isometric if , there are � : � ! � 0 ; a ; b ; c ; M > 0 : 1 ad ( x ; y ) � b � d 0 (� ( x ) ; � ( y )) � ad ( x ; y ) + b ; 1 c d 0 (� ( x )) � d ( x ) � cd 0 (� ( x )) ; for all y 0 2 � 0 : d 0 (� (�) ; y 0 ) < M : Let us use � � 1 .

Introduction Preliminaries Prelimineries The result Open problems Bonnus Stability Stability against rough isometry Stability of ( GE �; 2 ) � � � � � C d 2 ( x ; y ) � c d 2 ( x ; y ) c 1 1 � e n �= 2 exp p n ( x ; y ) � C n �= 2 exp t t e p n = p n + p n � 1 , Theorem ( Delmotte 1997.) Yes. On Z d we have, ( GE d ; 2 )

Introduction Preliminaries Prelimineries The result Open problems Bonnus Random walk on the Penrose graph Random walk on the Penrose graph � Penrose graph If � roughly isometric to Z 2 , then � � � � � C d 2 ( x ; y ) � c d 2 ( x ; y ) c 1 p n ( x ; y ) � C 1 � e n exp n exp t t on the Penrose graph.

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry Rough isometry Theorem Solomon The Penrose graph is bi-lipschitz to Z 2 .

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry Corollary A Penrose graph is rough isometric to Z 2 . Direct proof. � :

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry Random walk on the Penrose graph Theorem � � � � � C d 2 ( x ; y ) � c d 2 ( x ; y ) c 1 p n ( x ; y ) � C 1 � e n exp n exp t t

Introduction Preliminaries Prelimineries The result Open problems Bonnus Invariance principle Preparation � � � � � C d 2 ( x ; y ) � c d 2 ( x ; y ) c 1 p n ( x ; y ) � C 1 n exp � e n exp t t

Introduction Preliminaries Prelimineries The result Open problems Bonnus Invariance principle On the Penrose net the random walk X n = P n i = 1 x i : x i vectors between centers of tiles. ! n the environment seen from X n Z n = ( ! n ; X n ) ; X n X n = V ( Z i � 1 ; Z i ) i = 1 V ( Z i � 1 ; Z i ) = X i � X i � 1 = x i :

Introduction Preliminaries Prelimineries The result Open problems Bonnus Invariance principle Penrose graph Theorem (deMasi, Ferrari, Goldstein, Wick) Z n reversible ergodic Markov chain with � stationary probability measure, X n as above, anti-symmetric function of Z then 1 p X b At c = ) W D ( t ) : A D ???.

Introduction Preliminaries Prelimineries The result Open problems Bonnus The Penrose graph Theorem For the random walk on the Penrose net the invariance principle holds. Proof. 1. Z n egrodic � stac. (M. Kunz,A. Robinson) 2. D pos. def. so the limiting process is non-degenerate.

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 1. The …ve directions e k = � � � � � � � � � �� 2 + ( k � 1 ) 2 � 2 + ( k � 1 ) 2 � 2 + ( k � 1 ) 2 � cos ; sin 5 5 5 k = 1 :: 5 and the grids G k = f x 2 R 2 : xe k ? = z + � k ; z 2 Z g

Introduction Preliminaries Prelimineries The result Open problems Bonnus A cross-points G i \ G j are center of a tiles, the position is defned by the phases � l ; � m mod 1 so that X 5 � i = 0 i = 1 and z :

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 1. X n is at a crosspoint of the pentagrid G i \ G j : i 6 = j 2 f 1 ; 2 ; 3 ; 4 ; 5 g that is the reference point for ! , i.e.: � i = � j = 0. For the pentagrid we know that X 5 � i = 0 : i = 1 ( � i mod 1 ) Still we have two "free" �: So ! $ ( i ; j ; � k ; � l ) identi…es our position. We get 10 tori � i ; j : Let � = [ � i ; j : The dinamics: ! n ! ! n + 1 on � .

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 1. f ! n g � � is dense with null Lebesgue measure on the compact � . Let ! 2 � i ; j ; j i � j j = 1 ; ! 0 2 � i ; j ; j i � j j = 2 ; d � ( ! ) = � d � ( ! ) d � ( ! 0 ) d � ( ! 0 ) = � the golden ratio. � is …nite, can be normalised to probability measure. From the density theorem of topological groups it follows that if A � � invarian and � ( A ) > 0 then � ( A ) = 1 ,i.e. ergodic.

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 2. D non-degenerate. 0 6 = e 2 R 2 e � De > 0 :

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 2. D non-degenerate. 0 6 = e 2 R 2 e � De > 0 :

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 2. D non-degenerate. 0 6 = e 2 R 2 e � De > 0 : � � � � p n p n Let A = B x 0 ; C 2 n B x 0 ; C 1 ; C cone at e with angle � : �= 2 > � > 0. The intersection H = A \ C . E ( e � X n X � n Ee j X 0 = x 0 ) � � ( e � X n ) 2 j X 0 = x 0 = E

Introduction Preliminaries Prelimineries The result Open problems Bonnus Proof 2. � � � � p n p n Let A = B x 0 ; C 2 n B x 0 ; C 1 ; C cone about x 0 ; with �= 2 > � > 0. H = A \ C . E ( e � X n X � n e j X 0 = x 0 ) � � X ( e � X n ) 2 j X 0 = x 0 ( ex ) 2 P n ( x 0 ; x ) = E � x 2 H � � p n + b ) 2 � C ( aC 2 � 2 c 0 exp � cos ( � ) p n n � c � c > 0 n P N ' P G !

Introduction Preliminaries Prelimineries The result Open problems Bonnus Other tilings? What conditions needed for the IP?

Introduction Preliminaries Prelimineries The result Open problems Bonnus Quenched invariance principle Is the invariance principle true for almos all starting point?

Introduction Preliminaries Prelimineries The result Open problems Bonnus Projection The construction of the Penrose tiling (deBruijn) Pentagrid