The net created from the Penrose tiling is biLipschitz to the integer lattice

THE NET CREA TED FR OM THE PENROSE TILING IS BILIPSCHIT Z TO THE INTEGER LA TTICE Y AAR SOLOMON AD VISOR: PROF. BARA K WEISS 1. Intro duct ion Definition 1.1. A sep ar ate d net is a discrete set Y ⊆ R 2 with t wo parameters c 1 , c 2 > 0 s uc h that for ev ery x, y ∈ Y d ( x, y ) ≥ c 1 , and for ev ery x ∈ R 2 there is a y ∈ Y with d ( x, y ) ≤ c 2 . Remark 1.2. F or inf ormation ab out tilings se e [G S8 7] . Let τ 1 b e the P enrose tiling on R 2 [G77] (using kites and darts) with edges o f lengths 1 and φ , where φ is the golden ratio 1+ √ 5 2 . Le t U be a square of the f orm [ j 1 , j 1 + l ] × [ j 2 , j 2 + l ] where l = 2 i , i, j 1 , j 2 ∈ N . Notice that when w e ha v e a tiling of the plane (so that the diameter of its tiles is b ounded) w e can pro duce a separated net from it. W e do it by placing o ne p oin t in each tile, and remem b er to maintain the minimal distance pro p erty from (1.1). Ob viously , by placing the p oin ts differen tly in the tiles w e will get diff erent separated nets, but a ll these nets will b e in the same biLipsc hitz equiv alence class. Th us w e can say that ev ery tiling cr e ates a separated net. Let Y b e the separated net whic h is created by the tiling τ 1 . F or a num b er ρ > 0 w e define as in [BK02] (1) e ρ ( U ) = max  ρ |U | ♯ ( U ∩ Y ) , ♯ ( U ∩ Y ) ρ |U |  , E ρ (2 i ) = sup  e ρ ( U ) : U as ab ov e  Our ob jectiv e is to sho w tha t there is a ρ > 0 such that the pro d- uct Q ∞ i =1 E ρ (2 i ) con v erges. Then, by [BK02], we conclude that Y is biLipsc hitz to Z 2 . 2. The convergence of the product Assume that w e ha ve a P enrose tiling of the plane σ 1 . Let σ n b e the tiling that is obtained from applying the deflation op eratio n (see [G77], [B81]) to σ 1 n − 1 times. Let A ⊆ R 2 b e a set whic h is co v ered b y exactly K 1 half kites and D 1 half darts from the tiling σ 1 (w e consider half with resp ect to the symmetry axis). In other w ords, if B is a half 1 THE NET CREA TED FROM THE PENROSE TILING I S BILIPSCHITZ T O THE INTEGER LA TTICE 2 kite or a half dart in σ 1 then A ∩ B = ∅ or B ⊆ A . D enote b y K n and D n the n umber o f half kites and half dar t s in σ n resp ectiv ely that ar e wholly con tained in A . W e will show that the ratio K n D n con v erges to φ = 1+ √ 5 2 when n a pproac hes infinit y . Prop osition 2.1. F or every n ≥ 3     K n D n − φ     ≤ 1 2 n − 1 Pr o of. By using the deflation o p eration on a half kite and a half da r t one can see that every half kite is divided in to tw o half kites and one half dart , and ev ery half dart is divided in to one half kite and one half dart [G77]. F rom this r ule w e obtain the follo wing recursiv e formulas: K n +1 = 2 K n + D n , D n +1 = K n + D n Define x n = K n D n , then x n +1 = K n +1 D n +1 = 2 K n + D n K n + D n = 2 K n + D n D n K n + D n D n = 2 x n + 1 x n + 1 No w lo ok at the function f : [0 , ∞ ) → [0 , ∞ ), tha t is defined by f ( x ) = 2 x +1 x +1 (as in [P79]). Notice that for ev ery x , 1 ≤ f ( x ) ≤ 2 . Then fo r ev ery x, y ∈ Rng ( f ):   f ( x ) − f ( y )   =   2 x + 1 x + 1 − 2 y + 1 y + 1   =   x − y ( x + 1 )( y + 1)   ≤ 1 4   x − y   Hence f is a con traction mapping with a Lipsch itz constan t 1 4 ( < 1). Since [1 ,2] is a complete space, a ccording to the contraction mapping theorem, f has a unique fixed p oint p , and for ev ery x ∈ [1 , 2] | f n ( x ) − p | ≤ 1 4 n | x − p | ≤ 1 4 n . f ( p ) = p ⇒ p = 2 p + 1 p + 1 ⇒ p 2 − p − 1 = 0 ⇒ p = φ = 1 + √ 5 2 Since f w a s define d as f ( x n ) = x n +1 , meaning f ( K n D n ) = K n +1 D n +1 , we conclude that     K n D n − φ     =     f n − 1 ( K 1 D 1 ) − p     K 2 D 2 ∈ [1 , 2)= Rng ( f ) =     f n − 2 ( K 2 D 2 ) − p     ≤ 1 4 n − 2 n ≥ 3 < 1 2 n − 1  No w let τ 1 b e the P enrose tiling on R 2 with edges o f lengths 1 and φ . Let Y b e the separated net which is created ( a s described in (1)) b y the tiling τ 1 . Let U b e a square with a n edge of length l = 2 i as THE NET CREA TED FROM THE PENROSE TILING I S BILIPSCHITZ T O THE INTEGER LA TTICE 3 defined ab o v e. Denote b y K and D the n umbers of p oin ts in U ∩ Y that a re obtained from kites and darts resp ectiv ely (eve r y p oint in Y is obta ined fro m a dart or a kite in τ 1 ). Prop osition 2.2. F or i ≥ 5 0     K D − φ     ≤ φ − i 3 Pr o of. Let m ∈ N suc h that φ m ≤ l < φ m +1 . Let τ 2 b e the tiling o f R 2 whic h is obtained aft er  m 2  inflations (see [G7 7], [B81]) of τ 1 . No w lo ok at the tiling τ 2 and define k 2 and d 2 to b e the n umber of half kites and half darts resp ectiv ely , whic h are wholly contained in U , and let V b e the a rea whic h is cov ered b y these k 2 + d 2 shap es. In a similar w ay w e define K 2 and D 2 to b e the num b er of half kites and half darts resp ectiv ely , that in tersect with U , and let W b e the area whic h is co v ered b y t hese K 2 + D 2 shap es. First, lets lo ok at V . Notice that if w e p erform  m 2  iterations of deflation on τ 2 , we’ll g o bac k t o τ 1 . Denote b y k 1 and d 1 the n um b er of half kites and half darts respectiv ely , from τ 1 , whic h are wholly con tained in V . Then those k 1 + d 1 shap es a r e exactly t he shap es that a re o btained from the k 2 + d 2 shap es of τ 2 after  m 2  deflations. By (2 .1) w e conclude that (2)     k 1 d 1 − φ     ≤ 1 2 ⌊ m 2 ⌋ No w lets lo ok at the area of W − V . Notice that w e can b ound K D as follo ws: (3) k 1 2 d 1 2 + ♯ { darts that fit in W − V } ≤ K D ≤ k 1 2 + ♯ { kites that fit in W − V } d 1 2 Since τ 2 is the tiling that is obtained by  m 2  inflations on τ 1 , the length of a long edge of a tile in τ 2 is φ ⌊ m 2 ⌋ +1 . Simple geometry show s that t his is also t he diameter of a tile in τ 2 . Denote this n umber b y a . By the definition of V a nd W w e conclude that the distance b et w een U and V , and b etw een U and W , can’t b e more than a . That is V con tains a square ˜ V with an edge of length l − 2 a , and W is contained in a square ˜ W with an edge of length l + 2 a . F rom these estimations w e get b ounds for the areas of V and W − V : (4) |V | ≥ | ˜ V | ≥ ( l − 2 a ) 2 = l 2 − 4 al + 4 a 2 ≥ l 2 − 4 al THE NET CREA TED FROM THE PENROSE TILING I S BILIPSCHITZ T O THE INTEGER LA TTICE 4 |W − V | ≤ | ˜ W | − | ˜ V | = ( l + 2 a ) − ( l − 2 a ) = 8 al Here w e can see that the area of W − V is insignifican t in comparison to the area of V , when i is large. W e’ll use this la ter. Denote b y ψ the area of the tile of the dart in the tiling τ 1 . Then the kite’s area is ψ φ . By (4) ♯ { darts that fit in W −V } ≤ 8 al ψ , ♯ { kites that fit in W −V } ≤ 8 al ψ φ ( 3 ) ⇒ (5) k 1 d 1 + 32 al ≤ K D ≤ k 1 + 16 al d 1 In o r der t o estimate the difference b etw een the r atio K D and φ , w e’ll first estimate d 1 : V is cov ered b y k 1 half kites and d 1 half darts, a nd by (2), the ratio k 1 d 1 is v ery close to φ . W e also kno w that the ratio b et w een the areas of the shap es is exactly φ . Therefore: (6) |V | = k 1 ψ φ + d 1 ψ ( 2 ) ,i ≥ 50 ≤ 4 d 1 ( 4 ) ⇒ l 2 5 i ≥ 50 ≤ |V | 4 ≤ d 1 Th us: K D ( 3 ) ≥ k 1 d 1 + 32 al =  k 1 d 1 + 32 al − k 1 d 1  +  k 1 d 1 − φ  + φ ( 2 ) , ( 6 ) ,i ≥ 50 ≥ φ − φ − i 3 and in a similar w ay: K D ( 3 ) ≤ k 1 + 16 al d 1 =  k 1 + 16 al d 1 − k 1 d 1  +  k 1 d 1 − φ  + φ ( 2 ) , ( 6 ) ,i ≥ 50 ≤ φ + φ − i 3 Therefore     K D − φ     ≤ φ − i 3  In order to choose ρ correctly , w e’ll assume that t he ratio K D is exactly φ , and that the K + D shap es cov er the whole square U , and nothing more. In this case it is easy to calculate that (7) ρ = φ 2 (1 + φ 2 ) ψ Prop osition 2.3. With ρ as in (7), and for i ≥ 50 : E ρ (2 i ) − 1 ≤ 10 · φ − i 3 THE NET CREA TED FROM THE PENROSE TILING I S BILIPSCHITZ T O THE INTEGER LA TTICE 5 Pr o of. According to (2.2),     K D − φ     ≤ φ − i 3 ⇒ D ( φ − φ − i 3 ) ≤ K ≤ D ( φ + φ − i 3 ) Let U b e a square with an edge o f length l = 2 i . Since ♯ ( U ∩ Y ) = K + D (b y definition), we can get b o unds for ♯ ( U ∩ Y ): (8) D (1 + φ − φ − i 3 ) ≤ ♯ ( U ∩ Y ) ≤ D (1 + φ + φ − i 3 ) No w lets estimate |U | : U is mostly cov ered by D darts and K kites, the only problem is the b oundary of U . Since the length of the long edge of a tile is φ , as we did in the pro o f of (2.2), w e can lo ok at a frame with an area of ( l + 2 φ ) 2 − ( l − 2 φ ) 2 = 8 φl around the b oundary of U , and b ound U b y thro wing and not thro wing this frame. W e conclude that: (9) D ψ (1 + φ 2 − φ 1 − i 3 ) − 8 φl ≤ |U | ≤ D ψ (1 + φ 2 + φ 1 − i 3 ) + 8 φl Hence ♯ ( U ∩ Y ) ρ |U | ( 7 ) , ( 8 ) , ( 9 ) ≤ (1 + φ 2 )( φ 2 + φ − i 3 ) φ 2 (1 + φ 2 − φ 1 − i 3 ) − 8 φlρ D ψ ⇒ ♯ ( U ∩ Y ) ρ |U | − 1 ( 6 ) ,D ≥ d 1 ,i ≥ 50 < 10 φ i 3 On the other hand: ρ |U | ♯ ( U ∩ Y ) ( 7 ) , ( 8 ) , ( 9 ) ≤ φ 2 (1 + φ 2 + φ 1 − i 3 ) + 8 φlρ D ψ (1 + φ 2 )( φ 2 − φ − i 3 ) ⇒ ρ |U | ♯ ( U ∩ Y ) − 1 ( 6 ) ,D ≥ d 1 ,i ≥ 50 < 10 φ i 3 Therefore e ρ ( U ) − 1 ( 1 ) = max  ρ |U | ♯ ( U ∩ Y ) , ♯ ( U ∩ Y ) ρ |U |  − 1 < 10 φ i 3 W e get this inequalit y fo r ev ery U with an edge of length l = 2 i , hence E ρ (2 i ) − 1 ≤ 10 φ i 3 as required.  Corollary 2.4. With ρ = φ 2 (1+ φ 2 ) ψ , Q ∞ i =1 E ρ (2 i ) < ∞ THE NET CREA TED FROM THE PENROSE TILING I S BILIPSCHITZ T O THE INTEGER LA TTICE 6 Pr o of. It is enough to sho w tha t ln  Q ∞ i =1 E ρ (2 i )  < ∞ : ln  ∞ Y i =1 E ρ (2 i )  ≤ ∞ X i =1 ln  E ρ (2 i )  ln x ≤ x − 1 ≤ ∞ X i =1  E ρ (2 i ) − 1  ( 2.3 ) ≤ 50 X i =1  E ρ (2 i ) − 1  + ∞ X i =51 10 φ i 3 = 50 X i =1  E ρ (2 i ) − 1  + 10 φ 51 3 (1 − φ − i 3 ) < ∞  3. Subs titution Tilings Definition 3.1. A substitution tiling is a tiling of R n , with finitely man y prototiles T 1 , . . . , T n , that can b e obta ined by applying a certain dissection rule on its prototiles. In other w ords, every T i comes with a division rule tha t sho ws ho w to divide it into prototiles of the same tiling, with a smaller scale (with these rules one can create the tiling). Question 3.2. A r e al l s ubstitution tilings of R 2 cr e ate se p ar ate d nets that ar e biLipschitz to Z 2 ? These da ys w e are w or king o n an answ er for Question 3 .2 . The same argumen t that w e used should w ork as w ell for substitution tilings, but this w ork is still in progress. Reference s [B81] N. G. de Bruijn, A lgebr aic the ory of Penr ose’s non-p erio dic tilings of the plane I and II , Non Nederl. Ak ad. W etensch A 84 (19 81) p.39- 66. [BK02] D. Bur ago and B. K leiner, R e ctifying sep ar ate d net s , Ge o m. func. a na l. V ol.1 2 (20 02) 80 -92. [G77] M. Gardner, Extr aor dinary nonp erio dic t iling that enriches t he the ory of tiles , Scientific Amer ic an, (Ja n 1977), p.10 9 -121 . [GS87] Branko Grubaum and G. C. Shephard, Tilings and p atterns , W. H. F ree- man and Company , New Y ork, 19 87. [P79] Roger Penrose, Pentaplexity - A class of non-p erio dic t ilings of t he plane , Math. In tellig e ncer vol 2 . (1), p.32-3 7 (1979)