SPHERICAL SYMMETRIES -------------------- We have seen how polyhedra-with-cone-points-and-corner-points (PWCPACP) for which V - E + F = 0 give rise to plane symmetry types. PWCPACP with V - E + F < 0 or with V - E + F > 0 also have connections to symmetry types. The symmetry types associated with a PWCPACP with V - E + F < 0 encompass various patterns on a non-Euclidean hyperbolic plane. Escher's four Circle Limit prints display patterns with various symmetry types of this sort. Escher must distort the ideal non-Euclidean geometry for the sake of printing his pictures on the Eucidean surface of a piece of paper. Thus his motifs appear to shrink as your eye moves towards the circumference of the circles. A future lecture will address the fundamental concepts of non-Euclidean geometry. The symmetry types associated with most PWCPACP having V - E + F > 0 encompass the various patterns on a sphere. We shall tell this story now. At first we shall take a descriptive approach, emphasizing the analogies with plane symmetries and allowing certain phenomena to emerge. Only latter will we try to pin down the reasons behind things. MAKING A LIST ------------- We start by making a list of PWCPACP which have V - E + F > 0 . Patterns and symmetry play no role in this section. After we have our list, we will seek the interpretation of each entry. START WITH JUST CONE POINTS --------------------------- As we did when we made our list of PWCPACP having V - E + F > 0, we find it best TO START with those PWCPACP with ONLY CONE POINTS as special features. Recall that V - E + F = 2 for a sphere and the presence of a cone point of order n lowers V - E + F by (n-1)/n. As before, we prefer to say this a different way: declaring an ordinary vertex a cone point of order n ``costs'' (n-1)/n units and we have only 2 units to spend. The cone points of order 2 constitute the cheapest available modification (at least until we extend our considerations to corner points). Thus we can make no more than 3 vertices cone points lest V - E + F drop to 0 or below. First we shall list the possibilities with three cone points, then with two, then one, then none. Three cone points ----------------- To use three cone points, at least one must have order 2, otherwise our modification will cost AT LEAST AS MUCH as a 333 and that already brings V - E + F down to 0. If we have two cone points of order 2, the third can have any order (including 1 !). Indeed: A 22n has V - E + F = 2 - (1/2 + 1/2 + (n-1)/n ) = 1/n . If we have a cone point of order 2 and a cone point of order 3, the third point can have any order up to 5 (we know that a 236 (same as 632) has V - E + F = 0 ). We already have 232 (same as 223) on our list, so we get the 3 new possibilities: A 233 has V - E + F = 2 - (1/2 + 2/3 + 2/3 ) = 1/6 ; A 234 has V - E + F = 2 - (1/2 + 2/3 + 3/4 ) = 1/12 ; A 235 has V - E + F = 2 - (1/2 + 2/3 + 4/5 ) = 1/30 . If we have both a cone point of order 2 and a cone point of order 4, the third point can have any order up to 4 (we know that a 244 (same as 442) has V - E + F = 0 ). But we already have 224 and 234 (same as 223) on our list. Likewise, if we have both a cone point of order 2 and a cone point of order 5 (or 6), the third point can only have order up to 3 and we've already listed those possibilities. Finally, if we have both a cone point of order 2 and a cone point of order 7 or more, the third point can only have order up to 2 and that puts us back in the case 22n already listed. Two cone points --------------- Since no cone point costs 1 unit or more, we can have two cone points of any order. An mn has V - E + F = 2 - ((m-1)/m + (n-1)/n) = 2 - (1 - 1/m + 1 - 1/n) = 1/m + 1/n. (Later we shall see that these relate to spherical symmetry only when m=n.) One cone point -------------- Since no cone point costs 1 unit or more, we can have one cone points of any order. An n has V - E + F = 2 - (n-1)/n = 2 - (1 - 1/n) = 1 + 1/n. Since we could have m=1 just above and 1n means the same as n , in a sense we get nothing new here. (Later we shall see that none of these relate to spherical symmetry.) Zero cone points ---------------- An unmodified sphere has V - E + F = 2. Our notation for this would have no symbols. To avoid this awkwardness, we'll denote this possibility by a 1. Clearly this case relates to a decoration on a sphere with no (non-trivial) symmetry. Since we could have n=1 just above, we have already detected this possibility. NOW JUST CORNER POINTS ---------------------- Next we consider those cases which possess only corner points connected by reflective edges. Clearly we may place a * in front of any notation we've already found. The * turn cone points to corner points and thereby reduces their cost by half. Since the cost in the examples above came to less than 2 units, now it will come to less than 1 unit, leaving more than enough for us to pay 1 unit for the * itself. Thus we get A *22n has V - E + F = 2 - 1 - ( 1/2 + 1/2 + (n-1)/n )/2 = 1/2n . A *233 has V - E + F = 2 - 1 - ( 1/2 + 2/3 + 2/3 )/2 = 1/12 ; A *234 has V - E + F = 2 - 1 - ( 1/2 + 2/3 + 3/4 )/2 = 1/24 ; A *235 has V - E + F = 2 - 1 - ( 1/2 + 2/3 + 4/5 )/2 = 1/60 . A *mn has V - E + F = 2 - 1 - ( (m-1)/m + (n-1)/n )/2 = 1 - (1 - 1/m + 1 - 1/n)/2 = (1/m + 1/n)/2. Note that in every has adding the * has the net effect of reducing V - E + F by a factor of 1/2. No other possibilities exist. Otherwise we could work backwards (i.e., remove the *) and get new possibilities for our previous list. NOW CORNER POINT AND CONE POINTS -------------------------------- Since a pair of corner points of order n cost the same a one cone point of order n , whereever we have such a pair after a * we can replace them with one cone point before the start. Thus *22n gives rise to 2*n (still with V - E + F = 1/2n). *233 gives rise to 3*2 (still with V - E + F = 1/12). *nn gives rise to n* (with V - E + F = 1/n). No other possibilities exist. Otherwise we could work backwards (i.e., remove a cone point and add two corner points of the same order) and get new possibilities for our previous list. SEWING MOBIUS STRIPS -------------------- Since sewing in a Mobius strip costs the same as creating a *, whereever we have a * with no corner points, we can replace it with a x. Thus * gives rise to x (still with V - E + F =1). n* gives rise to nx (still with V - E + F = 1/n). No other possibilities exist. Otherwise we could work backwards (i.e., replace a x with a *) and get new possibilities for our previous list. CHECKING IT TWICE ----------------- We have produced a formidable list. For convenience, lets collect all the results together. 22n with V - E + F = 1/n 233 with V - E + F = 1/6 234 with V - E + F = 1/12 235 with V - E + F = 1/30 mn with V - E + F = 1/m + 1/n *22n with V - E + F = 1/2n *233 with V - E + F = 1/12 *234 with V - E + F = 1/24 *235 with V - E + F = 1/60 *mn with V - E + F = (1/m + 1/n)/2 2*n with V - E + F = 1/2n 3*2 with V - E + F = 1/12 n* with V - E + F = 1/n x with V - E + F = 1 nx with V - E + F = 1/n Certain entries on our list represent a single PWCPACP while other represent infinite families of such. Let's first have a look at just the entries which represent infinite families: 22n with V - E + F = 1/n mn with V - E + F = 1/m + 1/n *22n with V - E + F = 1/2n *mn with V - E + F = (1/m + 1/n)/2 2*n with V - E + F = 1/2n n* with V - E + F = 1/n nx with V - E + F = 1/n Of these, 22n with V - E + F = 1/n nn with V - E + F = 2/n (don't forget 11, same as 1) *22n with V - E + F = 1/2n *nn with V - E + F = 1/n (don't forget *11, same as *) 2*n with V - E + F = 1/2n n* with V - E + F = 1/n nx with V - E + F = 1/n correspond very closely with frieze symmetries. We explained the reason why at the end of the last lecture: we can coil up a decorated infinite strip to form a decorated cylinder where the motif repeats n times (n may equal any number). Shrinking the edges of the cylinder to points, we get a symmetrically decorated sphere. This leaves mn (n not equal m) with V - E + F = 1/m + 1/n (don't forget 1n, same as n) *mn (n not equal m) with V - E + F = (1/m + 1/n)/2 (don't forget *1n, same as *n) Despite the numerics, these relate to no symmetry type on the sphere (or anything else). Despite the simplicity of the reasoning, we shall postpone for now a discussion about why. This leaves us to consider 7 cases: 233 with V - E + F = 1/6 234 with V - E + F = 1/12 235 with V - E + F = 1/30 *233 with V - E + F = 1/12 *234 with V - E + F = 1/24 *235 with V - E + F = 1/60 3*2 with V - E + F = 1/12 These 7 cases relate to the 5 regular (or ``Platonic'') polyhedra, the tetrahedron (4 triangular faces meeting 3 per vertex) octahedron (8 triangular faces meeting 4 per vertex) icosahedron (20 triangular faces meeting 5 per vertex) cube (6 square faces meeting 3 per vertex) dodecahedron (12 pentagonal faces meeting 3 per vertex) Pictures, or even models, of these might help you with the following discussion, but that would exceed my abilities as a typewriter artist, so I will have to get them to you some other way. First, let's explain the relation between these polyhedra and decorations of a sphere. A given polyhedra from this list possesses a center. Imagine the polyhedron, transparent except for its edges, surrounded by a sphere which shares its center. Imagine a small bright light at the center of the sphere. The light casts shadows on the sphere. We shall describe symmetries of the sphere in terms of the configuration of these shadows. Henceforth we shall not distinguish between the polyhedron itself and the configuration of shadows that it casts. A 233 corresponds to a pattern on the sphere with gyration points of order 3 at the vertices of a tetrahedron, of order 2 at the centers of the edges and of order 3 at the centers of the (triangular) faces. A template consists of a third of a face (not unlike the template for a 3*3). A 234 corresponds to a pattern on the sphere with gyration points of order 4 at the vertices of an octahedron, of order 2 at the centers of the edges and of order 3 at the centers of the (triangular) faces. Again, a template consists of a third of a face. A 235 corresponds to a pattern on the sphere with gyration points of order 5 at the vertices of an icasohedron, of order 2 at the centers of the edges and of order 3 at the centers of the (triangular) faces. Again, a template consists of a third of a face. The 234 and 235 symmetries admit alternative descriptions in terms of the cube and dodecahedron: A 234 corresponds to a pattern on the sphere with gyration points of order 3 at the vertices of a cube, of order 2 at the centers of the edges and of order 4 at the centers of the (square) faces. A template consists of a quarter of a face. A 235 corresponds to a pattern on the sphere with gyration points of order 3 at the vertices of a dodecahedron, of order 2 at the centers of the edges and of order 5 at the centers of the (pentagonal) faces. A template consists of a fifth of a face. Patterns with symmetry *233, *234 and *235 have a similar description except with corner points instead of gyration points. For example, each edge of a tetrahedron forms an arc running along an equator on the surface of the sphere; *233 has reflections through all these equators. *234 has reflections through the equators that extend the edges of a cube. (If you want to use an octahedron here, the you need reflections across BOTH the equators that extend the edges of the octahedron AND the equators which bisect those edges at right angles.) *235 has reflections through the equators that extend the edges of an icosahedron (or equally well, a dodecahedron). The template for a *233 (resp. *234, resp. *235) could have the shape of a sixth of one of the triangular faces of a tetrahedron (resp. octahedron, resp. icosahedron), not unlike the template for a *632. One can describe patterns with 3*2 symmetry in terms of either a cube or a tetrahedron. In terms of a cube, we need just the 3 reflections that exchange one opposite pair of faces (front-back, right-left, top-bottom) and gyrations of order 3 at each vertex. A square quarter of a face can serve as a template. In terms of an octahedron, we need just the 3 reflections through the equators that extend the edges (see the description of *234 above) and order 3 gyrations at the center of each triangular face. A third of a face could serve as a template. AN INTRIGUING NUMERICAL COINCIDENCE ----------------------------------- In every case above you can verify, if you care to, that the number of copies of the template required to decorate the sphere equals 2/(V - E + F) . If we knew the truth of this and we knew that the cases mn (n not equal m) with V - E + F = 1/m + 1/n (don't forget 1n, same as n) *mn (n not equal m) with V - E + F = (1/m + 1/n)/2 (don't forget *1n, same as *n) didn't correspond to any symmetry type, then we'd know the completeness of our classification of spherical symmetry types. After all, the number of templates required to fill the sphere must exceed 0, so then 2/(V - E + F) > 0 and V - E + F > 0. But we've listed ALL the ways that we could get V - E + F > 0, so we'd be finished. (to be continued)