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The Escher Problem

The Escher Problem. Frieze Groups. Frieze = embroidery from Friez, horizotal ornamented band (architecture). We are interested in symmetry groups of such bends. There are 7 frieze groups. We start with a rectangular stamp. Transformations. Translation Halfturn Vertical Reflection

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The Escher Problem

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  1. The Escher Problem

  2. Frieze Groups • Frieze = embroidery from Friez, horizotal ornamented band (architecture). • We are interested in symmetry groups of such bends. There are 7 frieze groups. • We start with a rectangular stamp.

  3. Transformations • Translation • Halfturn • Vertical Reflection • Glide-reflection

  4. Seven Frieze Types • Groups (notation): • 11 (translations only) • 12 (translations and halfturns) • m1 (translations and vertical reflections) • 1g (translations and glidereflections) • mg (translations, halfturns, vertical reflections and glide reflections) • 1m (translations and horizontal reflections) • mm (translations, halfturns,vertical reflections, glide reflections, horizontal reflections)

  5. 11

  6. 12

  7. m1

  8. 1g

  9. mg

  10. 1m

  11. mm

  12. The Groups • Group Elements • Identity I • Translation T • Halfturn R • Glidereflection G • Vertical mirror V • Horizontal mirror H. • Some relations: R2 = V2 = H2 = I, RV = VR = H,..

  13. Subgroups • (1) T P C1 • (2) G P,R(P) C1 • (3) T,R B C1£ D1 • (4) T,V A D1 • (5) T,H S D1 • (6) G,V A,R(A) D1 • (7) all(T,G,R,V,H) H D1£ D1

  14. Exercise • Explain how the Frieze groups can be described by the four letters (aspects of the pattern): • b, p, q, d. b d q p

  15. Discrete Isometries • Each metric space M determines the group of distance preserving maps, isometries Iso(M). • A subgroup of Iso(M) is discrete, if any isometry in it either fixes an element of M or moves is far enough. • Discrete subgroups of I(R2) fall into three classes: • 17 crystallographic groups • 7 frieze groups • finite groups (grups of rozettes).

  16. Theorem of Leonardo da Vinci • The only finite groups of isometries in the plane are the group of rosettes (cyclic groups Cn and dihedral groups Dn).

  17. The Escher problem • There is a square stamp with asymmetric motif. • By 90 degree rotations we obtain 4 different aspects. • By combining 4 aspects in a square 2 x 2 block, a translational unit is obtained that is used for plane tilnig. Such a tiling is called a pattern. • Question: What is the number of different patterns ? • Answer: 23.

  18. Example

  19. Recall Burnside Lemma. • Let G be a group, acting on space S. • For g 2 G let fix(g) denote the number of points form S fixed by g. • Let N denote the number of orbits of G on S. • Then:

  20. Application • Determine the group (G) and the sapce (S). • Pattern can be translated and rotated. • Basic observatrion: • Instead of pattern consired the block (signature). • Group operations: • H – horizontal translation • V – vertical translation • R – 90 degrees rotation.

  21. (Abstract) group G • h2 = v2 = r4 = 1. • hv = vh • hr = rv hv hvr2 hvr3 hvr h hr2 hr3 hr 1 r2 r3 r v vr2 vr3 vr

  22. Space S • Space S consists of 4 4 4 4 = 256 signatures. • Count fix(g) for g 2 G. • For instance: • fix(1) = 256. • fix(r) = fix(r3) = 4. • fix(h) = fix(v) = 16. • By Burnside Lemma we obtain N = 23.

  23. Homework • Consider the Escher problem with the motiff on the left. • H1. Determine the abstract group and its Cayley graph. • H2. What is the number of different patterns? • H3. What is the number of different patterns if we reflections are allowed? • H4. What is the number of different patterns in the original Escher problem if reflections are allowed?

  24. 1-dimensional Escher problem • Rectangular Asymmetric Motiff • Only Two Aspects. • 1 x n block (signature) • Determine the number of patterns: • Two more variations: • II two motiffs(mirror images) • III reflections are allowed.

  25. Space S • Space S consists of 2 2 .... 2 = 2n signatures. • Count fix(g) for g 2 G. • For instance: • fix(1) = 256. • fix(r) = fix(r3) = 4. • fix(h) = fix(v) = 16. • By Burnside Lemma we obtain N = 23.

  26. Solution for the basic case • where g(n) = 0 for odd n and for even n:

  27. Program in Mathematica • f[n_] := (Apply[Plus,Map[EulerPhi[#] 2^(n/#)&,Divisors[n]]] + If[OddQ[n],0,(n/2) 2^(n/2)])/(2n) • f[n_,m_] := (Apply[Plus,Map[EulerPhi[#] (2 m)^(n/#)&,Divisors[n]]] + If[OddQ[n],0,(n/2) (2 m)^(n/2)])/(2n) • g[n_] := (Apply[Plus,Map[If[OddQ[#],1,2] EulerPhi[#] 4^(n/#)&,Divisors[n]]] + If[OddQ[n],0,(n) 4^(n/2)])/(4n)

  28. Results for a tape • n III III • 1 12 1 • 2 26 4 • 3 212 6 • 4 439 23 • 5 4104 52 • 6 9366 194 • 7 10 1172 586 • 8 22 4179 2131 • 9 30 14572 7286 • 10 62 52740 26524

  29. Exercise • Determine the Cayley graph of each of the Frieze groups. • Determine the crystallographic groups that may arise from the classical Escher problem

  30. 17 CRYSTALLOGRAPHIC GROUPS 6-števna os? zrcaljenje? 4-števna os? 3-števna os? p6mm p6 zrcaljenje? zrcaljenje? Zrcala v 4 smereh? 2-števna os? p4 3-osi na zrcalih? p3 p4mm p4gm zrcaljenje? zrcaljenje? p3m1 p31m Rombska mreža? glide? glide? Drugo zrcalo? pg p1 pm Rombska mreža? cm p2 c2gg p2mg c2mm p2mm

  31. p1 • p1 = <a,b|ab=ba>

  32. p2 • p2 = <a,b,c| b2=c2=(ab)2=(ac)2=1>

  33. pm • pm = <a,b,c| b2=c2=1, ab=ba, ac=ca>

  34. pg • pg = <a,b|ab=ba-1>

  35. cm • cm = <a,b,c| b2=c2=1, ab=ca>

  36. p2mm • p2mm = <a,b,c,d| a2=b2= c2=d2= 1, (ab)2=(ad)2=(bc)2=(cd)2=1>

  37. p2mg • p2mg = <a,b,c| b2 = c2 = 1, (ab)2=(ac)2=1>

  38. p2gg • p2gg = <a,b| (ab)2=1>

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