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The Escher Problem. Frieze Groups Frieze = embroidery from Friez, horizotal ornamented band (architecture). We are interested in symmetry groups of such.

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Presentation on theme: "The Escher Problem. Frieze Groups Frieze = embroidery from Friez, horizotal ornamented band (architecture). We are interested in symmetry groups of such."— Presentation transcript:

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: R 2 = V 2 = H 2 = I, RV = VR = H,..

13 Subgroups (1) T PC 1 (2) G P,R(P)C 1 (3) T,R BC 1 £ D 1 (4) T,V AD 1 (5) T,H SD 1 (6) G,V A,R(A)D 1 (7) all(T,G,R,V,H) HD 1 £ D 1

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

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( R 2 ) 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 C n and dihedral groups D n ).

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 h 2 = v 2 = r 4 = 1. hv = vh hr = rv v vr vr 2 vr 3 1 r r2r2 r3r3 h hr hr 2 hr 3 hv hvr hvr 2 hvr 3

22 Space S Space S consists of = 256 signatures. Count fix(g) for g 2 G. For instance: fix(1) = 256. fix(r) = fix(r 3 ) = 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 n signatures. Count fix(g) for g 2 G. For instance: fix(1) = 256. fix(r) = fix(r 3 ) = 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 I II III

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? p6mmp6 Zrcala v 4 smereh? zrcaljenje? p4 3-števna os? p4mmp4gm zrcaljenje? 2-števna os? zrcaljenje? glide?Rombska mreža? glide? Drugo zrcalo? Rombska mreža? zrcaljenje? p1pg pm cm p2 c2gg p2mg 3-osi na zrcalih? p31m p3m1 p3 c2mmp2mm

31 p1 p1 =

32 p2 p2 =

33 pm pm =

34 pg pg =

35 cm cm =

36 p2mm p2mm =

37 p2mg p2mg =

38 p2gg p2gg =


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