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

a telcs
SMART_READER_LITE
LIVE PREVIEW

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.:TMOP-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


slide-1
SLIDE 1

Supporting research schools at the University of Pannonia 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.

  • A. Telcs

AFRT December 2012 Hong Kong

slide-2
SLIDE 2

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Di¤usive limits on the Penrose tiling

András Telcs Hong Kong December 2012.

slide-3
SLIDE 3

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).

slide-4
SLIDE 4

Introduction Preliminaries Prelimineries The result Open problems Bonnus The beginings

slide-5
SLIDE 5

Introduction Preliminaries Prelimineries The result Open problems Bonnus The Penrose tiling

The Penrose tiling

Penrose 1978

slide-6
SLIDE 6

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)

slide-7
SLIDE 7

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

P (x; y) = 1 d (x) if x y P (Xn = yjXn1 = x) = P (x; y) de…nes a d (x)-reverzibe Markov chain

slide-8
SLIDE 8

Introduction Preliminaries Prelimineries The result Open problems Bonnus Random walk

Conventions

with d (x; y) ; graph distance called Penrose graph with jx yj embedded into R2 called Penrose net Z2 with d (x; y) the graph on the integer lattice Z2 with jx yj embedded into R2 called integer net

slide-9
SLIDE 9

Introduction Preliminaries Prelimineries The result Open problems Bonnus Questions

Questions

Random walks on graphs

x;y = 1; d (x; y) , pn (x; y) ? 1 CneC d2(x;y)

n

pn (x; y) + pn+1 (x; y) C n e d2(x;y)

Cn

slide-10
SLIDE 10

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; jx yj 1 paXbatc = ) WD (t) as a ! 1 as a ! 1: D pos. def.

slide-11
SLIDE 11

Introduction Preliminaries Prelimineries The result Open problems Bonnus Questions

Why is this a question? put Zd grid on Rd then the invariance principle holds for the ssnn RW Put any periodic (shift invariant) grid on Rd , still we have the IP. What if we put a non-periodic net on it? What if .... (comes later).

slide-12
SLIDE 12

Introduction Preliminaries Prelimineries The result Open problems Bonnus Pentagrid

The construction of the Penrose tiling (deBruijn)

Pentagrid

slide-13
SLIDE 13

Introduction Preliminaries Prelimineries The result Open problems Bonnus Pentagrid

The construction of the Penrose tiling (deBruijn)

Pentagrid

slide-14
SLIDE 14

Introduction Preliminaries Prelimineries The result Open problems Bonnus Pentagrid

slide-15
SLIDE 15

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 d0 ( (x) ; (y)) ad (x; y) + b; 1 c d0 ( (x)) d (x) cd0 ( (x)) ; for all y0 2 0 : d0 ( () ; y0) < M: Let us use 1.

slide-16
SLIDE 16

Introduction Preliminaries Prelimineries The result Open problems Bonnus Stability

Stability against rough isometry

Stability of (GE;2) c 1 n=2 exp

  • C d2 (x; y)

t

  • e

pn (x; y) C 1 n=2 exp

  • c d2 (x; y)

t

  • e

pn = pn + pn1, Theorem

(Delmotte 1997.) Yes. On Zd we have,(GEd;2)

slide-17
SLIDE 17

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 Z2, then

c 1 n exp

  • C d2 (x; y)

t

  • e

pn (x; y) C 1 n exp

  • c d2 (x; y)

t

  • n the Penrose graph.
slide-18
SLIDE 18

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry

Rough isometry

Theorem Solomon The Penrose graph is bi-lipschitz to Z2.

slide-19
SLIDE 19

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry

Corollary A Penrose graph is rough isometric to Z2. Direct proof. :

slide-20
SLIDE 20

Introduction Preliminaries Prelimineries The result Open problems Bonnus Rough isometry

Random walk on the Penrose graph

Theorem c 1 n exp

  • C d2 (x; y)

t

  • e

pn (x; y) C 1 n exp

  • c d2 (x; y)

t

slide-21
SLIDE 21

Introduction Preliminaries Prelimineries The result Open problems Bonnus Invariance principle

Preparation

c 1 n exp

  • C d2 (x; y)

t

  • e

pn (x; y) C 1 n exp

  • c d2 (x; y)

t

slide-22
SLIDE 22

Introduction Preliminaries Prelimineries The result Open problems Bonnus Invariance principle

On the Penrose net

the random walk Xn = Pn

i=1 xi :xi vectors between centers of

tiles. !n the environment seen from Xn Zn = (!n; Xn) ; Xn =

n

X

i=1

V (Zi1; Zi) V (Zi1; Zi) = Xi Xi1 = xi:

slide-23
SLIDE 23

Introduction Preliminaries Prelimineries The result Open problems Bonnus Invariance principle

Penrose graph

Theorem (deMasi, Ferrari, Goldstein, Wick) Zn reversible ergodic Markov chain with stationary probability measure, Xn as above, anti-symmetric function of Z then 1 p A XbAtc = ) WD (t) : D ???.

slide-24
SLIDE 24

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. Zn egrodic stac. (M. Kunz,A. Robinson)
  • 2. D pos. def. so the limiting process is non-degenerate.
slide-25
SLIDE 25

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 1.

The …ve directions ek =

  • cos

2 + (k 1) 2 5

  • ; sin

2 + (k 1) 2 5 2 + (k 1) 2 5

  • k = 1::5 and the grids

Gk = fx 2 R2 : xek? = z + k; z 2 Zg

slide-26
SLIDE 26

Introduction Preliminaries Prelimineries The result Open problems Bonnus

A cross-points Gi \ Gj are center of a tiles, the position is defned by the phases l; m mod 1 so that

5

X

i=1

i = 0 and z:

slide-27
SLIDE 27

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 1.

Xn is at a crosspoint of the pentagrid Gi \ Gj : i 6= j 2 f1; 2; 3; 4; 5g that is the reference point for !, i.e.: i = j = 0. For the pentagrid we know that

5

X

i=1

i = 0: ( 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 .

slide-28
SLIDE 28

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 1.

f!ng is dense with null Lebesgue measure on the compact . Let ! 2 i;j; ji jj = 1; !0 2 i;j; ji jj = 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.

slide-29
SLIDE 29

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 2.

D non-degenerate. 0 6= e 2 R2 eDe > 0:

slide-30
SLIDE 30

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 2.

D non-degenerate. 0 6= e 2 R2 eDe > 0:

slide-31
SLIDE 31

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 2.

D non-degenerate. 0 6= e 2 R2 eDe > 0: Let A = B

  • x0; C2

pn

  • nB
  • x0; C1

pn

  • ; C cone at e with angle

: =2 > > 0. The intersection H = A \ C. E (eXnX

n EejX0 = x0)

= E

  • (eXn)2 jX0 = x0
slide-32
SLIDE 32

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Proof 2.

Let A = B

  • x0; C2

pn

  • nB
  • x0; C1

pn

  • ; C cone about x0;with

=2 > > 0. H = A \ C. E (eXnX

n ejX0 = x0)

= E

  • (eXn)2 jX0 = x0
  • X

x2H

(ex)2 Pn (x0; x)

  • c
  • cos () pn

2 c0 exp

  • C (aC2

pn+b)

2

n

  • n

c > 0 PN ' PG!

slide-33
SLIDE 33

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Other tilings?

What conditions needed for the IP?

slide-34
SLIDE 34

Introduction Preliminaries Prelimineries The result Open problems Bonnus

Quenched invariance principle

Is the invariance principle true for almos all starting point?

slide-35
SLIDE 35

Introduction Preliminaries Prelimineries The result Open problems Bonnus Projection

The construction of the Penrose tiling (deBruijn)

Pentagrid

slide-36
SLIDE 36

Introduction Preliminaries Prelimineries The result Open problems Bonnus Projection

The construction of the Penrose tiling (deBruijn)

Pentagrid

slide-37
SLIDE 37

Introduction Preliminaries Prelimineries The result Open problems Bonnus Projection

The construction of the Penrose tiling (deBruijn)

Pentagrid

Z5 R5

Let = exp (2i=5) ; the plain P consists of points x; z 2 R5 :

  • rthogonal to 1 = (1; 1; 1; 1; 1) and

a =

  • 1; 2; 4; 6; 8

K is the set of unit cubes in R5 identi…ed with their centre: C = fK 2 K : K \ P 6= ;g and project vertices and edges to P.

slide-38
SLIDE 38

Introduction Preliminaries Prelimineries The result Open problems Bonnus Projection

Thanks for the attention!

slide-39
SLIDE 39

Introduction Preliminaries Prelimineries The result Open problems Bonnus Local modi…cation

Local modi…cation

Let 0 a local modi…cation of the Penrose graph, Clearly and 0 rough isometric,(GE2;2) holds: c n exp

  • C d2 (x; y)

n

  • p0

n (x; y) C

n exp

  • c d2 (x; y)

n

slide-40
SLIDE 40

Introduction Preliminaries Prelimineries The result Open problems Bonnus Local modi…cation

Local modi…cation

Does the invariance hold? (Doma Szász’s question)

slide-41
SLIDE 41

Introduction Preliminaries Prelimineries The result Open problems Bonnus Further questions

Further questions

Is it true that D = 2 2

  • ?

For local modi…cation is D invariant? Is the invariance principle stabil agains rough isometry? How the percolation behaves for the Penrose graph? And the RW on it??