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Archimedean Tilings -19- Archimedean Tilings Archimedes enumerated the convex polyhedra with regular faces and only one type of vertex (because the symmetry group is transitive on the vertices). Many people have independently enumerated the tilings of the Euclidean plane that satisfy the same condition, and a few have tried to enumerate the corresponding tilings of the hyper- bolic plane and achieved some partial results. Often there is some subgroup H of the full symmetry group G of such a tiling that remains transitive on the vertices; in this case we say that the tiling is Archimedean relative to H, to contrast it with the absolute case when H = G. For example, the Archimedean snub square tessellation (top right) has full symmetry group 4∗2, with respect to which it is absolute. However, coloring the tiles as in the second marginal figure, we see that the group 442 still acts transitively on its vertices, and so it is Archimedean relative to this subgroup. The complete classification of all Archimedean tilings, both rel- ative and absolute, appears for the first time in this book. As usual the guiding idea is that of orbifold, since once we have specified the orbifold we can recover the tiling by “unwrapping.” The orbifold of an Archimedean tiling is hard to visualize, in view of all those faces. Fortunately, we don’t have to visualize the faces, because we can remove them from the tiling with no loss of information if we’re careful. What we do is enlarge the vertices to blobs and the edges joining them to strips by slightly shrinking the faces. (One can think of these blobs and strips as the grout between the tiles.) Then we simply remove these shrunken faces! The result is like a doily made of paper (opposite page) The thirty-five relative Archimedean tilings of the Euclidean plane by squares. 251 © 2016 by Taylor & Francis Group, LLC 252 19. Archimedean Tilings circles and strips. This doily still contains all the information, since we can recover the original tessellation simply by sewing polygonal patches onto all the boundary curves. The orbifold of the doily is really quite a simple thing, since the Archimedean property of the tessellation implies that it consists of just one blob, possibly folded and probably connected to itself by various strips, again possibly folded. Below we see the doily for the snub square tessellation, with its full symmetry group 4∗2; the blob in it has five arms which are numbered 0, 1, 2, 2 ,and1 , because in the orbifold arms 1 and 2 have been folded onto 1 and 2, respectively. Also, arm 0 has been folded onto itself to become a half arm 0 . Finally, the outer ends of arms 1 and 2 have been identified, turning those arms into a band (12) that joins the blob to itself. 1 2 (12) 0 {4} 2 1 { } 0 3 What has become of the two types of faces? The square face has become the brown boundary curve, which now consists of just one edge in the orbifold; since this goes around the order-4 gyration point, we know it must have had four sides in the original, as in- dicated by our {4}. The triangle has become the purple boundary, 1 which has 1 2 sides (the full side at arms 1 and 2 and the half side at 0). But, since it ends on the red mirror line, its original polygon Downloaded by [University of Bergen Library] at 04:54 26 October 2016 was a {3}. We can convey the topological information here by the permuta- tion symbol 0 (12), (which conveys the topology), together with the face-code 34343, starting with the face between arms 0 and 1. We combine both parts into a single symbol, here 0 (12)∗(3, 4, 3, 4, 3), with the separating symbol (here ∗) indicating the local symmetry at the vertex. © 2016 by Taylor & Francis Group, LLC The Permutation Symbol 253 The Permutation Symbol In general, neighboring blobs in an archimedean tiling can be con- nected to one another by arms in just a few ways. Suppose that arm i of one blob is connected to arm i of another vertex. Then there is a symmetry fixing the central point of this edge. If there is only a rotational symmetry, of order 2, then in the orbifold we see the arm grasp an order 2 cone point: we call this a rotary arm (i). If there is a reflection symmetry interchanging the two blobs, in the orbifold we see the arm run from the vertex to a boundary of the orbifold; this is a folded band [i]. Finally, if there is a symmetry ∗2 at the center of this edge, in the orbifold we have a half arm i ,justaswesawinthesnubsquaretessellation.The type of brackets used here hints at the shape at the end of the arm. i i i i i i rotary arm (i) folded band [i] half arm i Or, we might have arm i of a blob connected to arm j = i of another blob. If there is a symmetry interchanging the two sides of this edge in the tiling, we have a half band ij , running along a boundary of the orbifold. If not, and the blobs are of the same orientation, we see an untwisted band (ij) in the orbifold, just as in the snub square tessellation. Finally, if the blobs are of opposite orientations, the orbifold will contain a twisted band [ij]. Downloaded by [University of Bergen Library] at 04:54 26 October 2016 i i i j j j untwisted band (ij) twisted band [ij] half band ij © 2016 by Taylor & Francis Group, LLC 254 19. Archimedean Tilings The parts of the orbifold that correspond to the faces we removed must each be disks; these may contain at most one cone point or kaleidoscopic point according to whether they lie in the interior of the orbifold or on its boundary, as shown on the left in the figure below. · ∗ b a n n {6a}{6b} Our separating symbol is a · if the blob lies in the interior of the orbifold, when it might become a cone point of order n,anda ∗ if it lies on a boundary of the orbifold, when it might become a Euler Characteristic kaleidoscopic point ∗n. The key observation is that an arbitrary permutation symbol If a filled in doily has f precisely describes the topology of the orbifold, up to the orders of faces and g bands (of all kinds—untwisted, twisted, cone and kaleidoscopic points! This is simply because there is just half, and folded), the Euler one way to zip disks to the boundary of the doily to get a closed − characteristic is 1+f g. surface. Half arms and rotary arms contribute nothing. This The orbifold for the symbol [0](12)[34]· is drawn below: we see number, together with the that the blob may have a cone point of order n ≥ 1 at its center. number of boundaries and the orientability of the orb- There are two boundary curves of the doily, one of which must meet ifold, determines the orb- the boundary of the orbifold. Consequently, the orbifold has exactly ifold’s topology. two faces, one with (perhaps) a cone point of order a and the other with (perhaps) a kaleidoscopic point of order b. We shall see that the values of a, b,andn can be deduced from the face code. Downloaded by [University of Bergen Library] at 04:54 26 October 2016 1 0 4 a n 2 3 b [0](12)[34]·(8b, a, 8b, 8b, 8b)n © 2016 by Taylor & Francis Group, LLC The Permutation Symbol 255 This orbifold is nonorientable with one boundary component; its Euler characteristic is 0, so the orbifold must be a M¨obius band and the tiling must have symmetry a∗b×. Moreover, counting the edges of the two faces, we see that the face code can only be (8a, 8a, 8a, b, 8a)n. Conversely, given the face code, we can deduce a, b,andn, and so know the symmetry. Consider the more complicated example 08 (1)[26](34)(57)∗. What is the corresponding orbifold? Beginning at arm 0 and tracing around, we find that the doily has two boundaries, colored purple and yellow, and so the orbifold will be filled in with two faces. It’s difficult to draw these, though! 2 1 2 0 3 9 ∗2 4 8 7 5 6 2 08(1)[26](34)(57)∗(143, 9, 148, 9, 143)2 What can we deduce about the local features of the orbifold from the symbol? There is an order-2 cone point at the end of rotary arm Downloaded by [University of Bergen Library] at 04:54 26 October 2016 (1). The boundary of the half band might have a kaleidoscopic point n at the blob, and two faces may each have cone points of orders, say, a for yellow and b for purple. The yellow face has seven edges and so will be a (7a)-gon in the tiling. The purple face has just one edge andsowillbeab-gon in the tiling. Any tiling with the permutation symbol 08 (1)[26](34)(57)∗ must therefore have a face code of the form ((7a)3,b,(7a)8,b,(7a)3)n; this immediately implies that for the face code we were given, (143, 9, 148, 9, 143)2,wehaven =2,a =2, and b =9. © 2016 by Taylor & Francis Group, LLC 256 19. Archimedean Tilings What is the orbifold’s topology? The orbifold is nonorientable because it includes the twisted band [26].
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