LAST STEPS TO CLASSIFY PLANE SYMMETRY TYPES ------------------------------------------- Let's begin with an outline of our classification strategy: >From a PLANE SYMMETRIC PATTERN we extract a TEMPLATE and then ``tape together'' certain edges to form a decorated POLYHEDRON-WITH-DISTINGUISHED-CONE-POINTS-AND-CORNER-POINTS for which the original plane pattern could serve as a map. Not every polyhedron-with-distinguished-cone-points-and-corner-points can arise from a plane symmetric pattern because THE POLYHEDRON-WITH-... MUST HAVE HAVE ALL _LOCAL ANGLE DEFECTS_ EQUAL 0. That means that THE POLYHEDRON-WITH-... has _TOTAL ANGLE DEFECT_ EQUAL 0. Now the refined Gauss-Bonnet theorem says that _TOTAL ANGLE DEFECT_ = 360(V - E + F) at least when we compute V and E in appropriate ways. Thus for our polyhedron-with-... we have V - E + F = 0. But we (take on faith the fact that) we may construct the surface underlying any polyhedron-with-... by starting with a sphere and adding handles, holes, Mobius strips, declaring boundary cycles reflective, and declaring vertices to be cone points or corner points of various orders. Moreover, EULER'S THEOREM TELLS US THE VALUE FOR V - E + F FOR ANY POLYHEDRON-WITH-... ONCE WE KNOW THE CONSTRUCTION OF ITS UNDERLYING SURFACE. In detail, for a polyhedron with underlying surface a sphere, V - E + F = 2 and: adding a handle brings V - E + F down by 2; adding a hole brings V - E + F down by 1; adding a Mobius strip brings V - E + F down by 1; adding an order n cone point brings V - E + F down by (n-1)/n; adding an order n corner point brings V - E + F down by (n-1)/2n. In addition Declaring a boundary cycle reflective doesn't change V - E + F. >From this we may deduce a list of 17 (topological) surfaces with V - E + F = 0. If two plane symmetric patterns lead to the same topological surface, then they have the same symmetry type, so we can have no more than 17 patterns. Conversely, we have constructed 17 different symmetry types, so all 17 topological surfaces with V - E + F = 0 really do yield distinctive plane symmetric patterns. FINDING THE LIST OF 17 TOPOLOGICAL SURFACES WITH V - E + F = 0 -------------------------------------------------------------- Adding features can only bring down V - E + F. Since we start with a sphere, we start with V - E + F = 2. We begin by classifying surfaces with no holes and so, necessarily, no corner points. Thus, to start with, we only consider adding handles, Mobius strips and cone points. Of these, the modification which brings down V - E + F the least is adding a cone point of order 2; doing so brings V - E + F down by 1/2. Thus we may add a maximum of four features. Indeed to ADD EXACTLY FOUR FEATURES we must add four cone points of order 2, which gives us a 2222. HOW CAN WE ADD EXACTLY THREE FEATURES? We must add at least one cone point of order 2 or 3, otherwise each addition brings V - E + F down by at least 3/4 and 3(3/4)=9/4 > 2. Now if we don't use a cone point of order 2, then each addition brings V - E + F down by at least 2/3 (the cost of a cone point of order 3) and 3(2/3)=2. This makes three cone points of order 3 the only possibility, giving a 333. So now we must consider possibilities of the form 2rs . Here, once we decide on a feature r, that will determine how much feature s must bring down V - E + F. r=2 (cone point of order 2) forces s to bring down V - E + F by 1, which leads to a 22x. r=3 (cone point of order 3) forces s to bring down V - E + F by 5/6 (since 1/2 + 2/3 + 5/6 = 2) which leads to a 236, which we usually write instead as 632. r=4 (cone point of order 4) forces s to bring down V - E + F by 3/4 (since 1/2 + 3/4 + 3/4 = 2) which leads to a 244, which we usually write instead as 442. r=5 (cone point of order 5) forces s to bring down V - E + F by 7/10 (since 1/2 + 4/5 + 7/10 = 2). Since no feature brings down V - E + F by 7/10, r=5 yields no surface. r=6 makes s=3, so we've already discovered that possibility. r>6 would force s to bring down V - E + F by less than 2/3 but more than 1/2 and no possibilities of that sort occur. HOW CAN WE ADD EXACTLY TWO FEATURES? One feature must bring down V - E + F by at least 1 but less than 2 , and only feature x (Mobius strip) does the trick. Then for the other feature we must must use another x (since 1+1=2) and we get xx. Finally HOW CAN WE ADD EXACTLY ONE FEATURE? A o (handle) brings V - E + F down by 2. That gives o. Now one fears that allowing holes and corner points might make things much more complex, but fortunately not! A * already brings V - E + F down by 1 (equals half of 2) and a corner point of order n brings down V - E + F by (n-1)/2n (equals half of (n-1)/2, the cost of a cone point of order n.) Thus THE POSSIBILITIES WITH ONE * AND ONLY CORNER POINTS (no cone points) consist of *2222; *333; *442; *632; by exactly the logic we used above, but with all the numbers cut in half. How about ONE * AND A MIX OF CONE POINTS AND (possibly no) CORNER POINTS? Given such a possibility, we could always convert any cone point of order n into two corner points of order n without affecting V - E + F. By eliminating all cone points in this manner eventually we'd have to get one of the 4 surfaces above. Working backwards, we might wind up with *2222 starting from 2*22 or 22*. We might wind up with *333 starting from 3*3 and wind up with *442 starting from 4*2. To finish, we must consider possibilities with a * and no corner points or cone points. Since a * costs the same as a x, we can change any x to * in a surface we've already found, such as xx or 22x. This leads just to *x; **. Thus we've got the classification we seek. ---------------------------------------------------------------------------- The next set of notes will explain how EULER'S THEOREM TELLS US THE VALUE FOR V - E + F FOR ANY POLYHEDRON-WITH-... ONCE WE KNOW THE CONSTRUCTION OF ITS UNDERLYING SURFACE. Is anyone reading this stuff??? Send me and e-mail (David.Feldman@unh.edu) if you get this far.