THE TOPOLOGICAL CLASSIFICATION OF POLYHEDRA ------------------------------------------- Why do we want a topological classification of polyhedra? --------------------------------------------------------- Our main goal remains the classifying the symmetry types of plane patterns. Given a plane pattern, we can extract a template. By ``taping'' together certain edges (possibly none) of the template we get a polyhedron (possibly with certain vertices marked as cone points and corner points). The original pattern then serves as a map of the polyhedron. Now two plane patterns have the same symmetry type if we can stretch one over the other in such a way as to make the symmetries coincide, in other words, so that a reflection, rotation or translation that leaves one pattern unchanged does the same for the other. If we can stretch one pattern over the other in this way, then we can stretch the polyhedron associated with it over the polyhedron associated with the other. Topology captures the idea of regarding two polyhedra as the same if we can stretch one over the other. Compared to the geometrical viewpoint, the topological view of polyhedra represents a simplication, a course classification into broad categories which ignore many geometrical complexities. As such the classification possesses an elegant structure: every closed (= no boundary) polyhedron comes about by starting with a sphere and either adding a certain number of handles, or sewing in a certain number of Mobius strips. Knowing this allows us, for instance, to enumerate all the polyhedra _with cone points and corner points_ which have V - E + F = 0, a major step in out project. A normal form for polyhedra --------------------------- Given a polyhedron, we would like to modify its geometry (but without changing its topology) until we have brought into a standard configuration or, to use a common mathematical parlance, a ``normal form.'' By the end of high school, most students have met several times the idea of bringing a mathematical structure into a normal form. Consider the following familiar examples: a) We have equality of fractions a c - = - b d whenever a x d = c x b. For example 4/6 = 10/15. Even though we have an easy test for the equality of a pair of fractions, we often find it useful to put a given fraction in a normal form call ``lowest terms.'' We say we have a fraction a/b reduced to lowest terms if there exists no number d>1 such that d | a (read this as `` d divides a '' ) and d | b. This does the trick for positive fractions. For negative fractions we must still choose between, say (-3)/5 and 3/(-5). Once we insist on only positive denominators, every fraction has a unique expression in lowest terms. b) We have equality of real numbers (decimals) .999999999... = 1.000000000... . One sees that these are equal by substracting one from the other and getting 0: 1.000000000000... - .999999999999... -------------------- .000000000000... . These two forms can both arise as the solution of problems. For example, adding 1/3 + 2/3 decimal style gives: .3333333333333333... + .6666666666666666... -------------------- .9999999999999999... and we must recognize that this equals 1.0 . The normal form for decimals simply excludes those decimals which end in a trail of 9's. Now we describe the procedure by which we shall put a polyhedron P in a normal form. We proceed to do this in a number of steps: Step 1 (Planar model) Pick any face f_1 of the polyhedron P and draw a picture of f_1 on a plane. Pick a face f_2 (if any) that shares an edge with f_1 and draw f_2 on the plane showing f_2 sharing that one edge with f_1. Then pick a face f_3 (if any) sharing an edge with either f_1 or f_2 and draw f_3 on the plane show f_3 sharing that one edge with one of the previously drawn faces (even if f_3 actually shares additional edges with the other faces). Continue this process until you have exhausted all the faces of P. Your picture on the plane now has the form of a polygon (divided, perhaps, into many faces). Each edge of P belongs to two faces. Therefore, to reconstruct the ORIGINAL POLYHEDRON, we would need to tape together IN PAIRS the various outside edges of this polygon. Instead of actually taping them together, we shall content ourselves with marking the pairs we ought to tape, also indicating how we might need to line them up before the taping. The following picture illustrations this a ______\_______ / / \ / \ / \ / \ . . . . . . \ / \ / \ a / \______\_______/ / Here, we must match the top and bottom edge since they both have the label a . We must not twist as we match them because one edge has a clockwise marking and the other a counterclockwise marking. (Here we speak of an OPPOSING PAIR OF EDGES.) a ______\_______ / / \ / \ / \ / \ . . . . . . \ / \ / \ a / \______/_______/ \ Here we match the same edges, but this time with a twist, since they both carry clockwise markings. (It would make no differenct if they both carried counterclockwise markings. (Here we speak of a TWISTED PAIR OF EDGES.) (All the other edges also carry labels and the ...'s may represent many edges no currently under discussion.) Step 2 (Eliminate adjacent opposing pairs) From a a _____\______._____/______ / / \ \ / \ / \ / \ pass to /|\ / | \ / |a \ . | Since this amounts to taping together edges that really represent the same edge in the original model, this does not change the topology. Repeat as many times as possible. Step 3 (Consolidate edge sequences) As we have set things up, the outside edges of the polygonal planar model occur in pairs sharing a common label. It may happen that sequences of labelled edges of length greater than one occur twice. If so replace these sequence with a single edge. Repeat as many times as possible. Examples: a b c d ____\_____.___/____.____\____ ____\_____ . / \ / . . / . . . . . . . . . . . BECOMES . . . . . . . a b c . . d . ____\_____.___/____.____\____ ____\_____ / \ / / Repetition occurs with in the opposite order with opposite orientation as one goes around the boundary (clockwise, say). a b c d ____\_____.___/____.____\____ ____\_____ . / \ / . . / . . . . . . . . . . . BECOMES . . . . . . . c b a . . d . ____/_____.___\____.____/____ ____/_____ \ / \ \ Repetition occurs with in the same order with same orientation as one goes around the boundary (clockwise, say). Step 4 (Reduce to a single outside vertex) At a glance one can't easily detect the number of distinct vertices of the original polyhedron from looking at the outside vertices of the planar model which represent them. For example, consider the following planar model, shown after Step 3: a b .___\___.___\___. | / / | | | | | e V V c | | | | | | . . | | | | | | e V V a | | | | | c b | .___/___.___\___. \ / Each edge has a head and foot. Observe that, head of a = foot of b (look at the top middle vertex) foot of b = foot of c (look at the bottom middle vertex) foot of c = head of b (look at the top right vertex) head of b = head of a (look at the bottom right vertex) so these four vertices in the picture represent one vertex, call it x , in the original polyhedron. Similarly head of c = foot of a (look at the right middle vertex) foot of a = foot of e (look at the top left vertex) foot of e = head of e (look at the left middle vertex) head of e = head of c (look at the bottom left vertex) so these four vertices in the picture also represent one vertex, call it y , in the original polyhedron. Thus the eight vertices in the picture represent only two vertices. Here we label them according to which vertex they represent in the original polyhedron a x b x y.___\___.___\___. | / / | | | | | e V V c | | | | | | y. .y | | | | | | e V V a | | | | | c b | .___/___.___\___. y \ x / x Now we'd like to cut and paste the planar model, BUT WITHOUT CHANGING THE POLYHEDRON THAT IT REPRESENTS, and reduce, ultimately to 1, the number of vertices from the original polyhedron visible as outer vertices. In this example we could get rid of the y's or the x's. Since it doesn't matter, let's get rid of the y's. The method consists of cleverly cutting off certain triangles and reattaching them, and occasionally reapplying STEP 2 or STEP 3. We illustrate the process here in detail. (Because of the limitations of ASCII art, my triangles may have some crooked edges, but that shouldn't matter.) a x b x y.___\___.___\___. | / / /| | / | | / | e V / V c | | | Here I've cut a triangle on the | | | left with sides c , a and the | | | new side f . y. V f .y | | | | | | e V \ V a | \ | | \ | | c b \| .___/___.___\___. y \ x / x a x b x x Now I've sectioned off the y.___\___.___\___. . triangle (and straightened | / / | /| one of the crooked edges in | | / | the picture). Notice that I | | / | now have two more vertice e V | / V c than before, but both labelled x. | | | | Now if I reattach the triangle | | | | along EITHER edge a or c then | | | | I'll lose one x and one y . y. V f V f .y As the net result, I'll gain one | | | | x and lose one y. | | | | e V | \ V a | | \ | | | \ | | c b | \| .___/___.___\___. . y \ x / x x a x b x The reattachment completed. Note that we y.___\___.___\___. had to turn the triangle over, so we see | / / | its edges in the opposite order now if we | | take a clockwise tour. You should regard | | the next step as merely cosmetic. We'll e V | eliminate FROM OUR DRAWING the edge c | | because it now belongs to the interior of | | the planar model so we no longer care about | | it. Also, we'll square everything off y. V f to reduce the visual complexity of the picture. | | | | e V | | | | | | c b | .___/___.___\___. y| \ /x / x | / | / a V |_ | / f | / |/ . x a x b x Tour the boundary of this square to check y.___\___.___\___. that you see the same edges in the same | / / | order with the same orientations as when | | you tour the boundary of the irregular | | polyhedron just above. Also observe e V V f the sequence of vertex names. Now we | | have 5 x's and 3 y's, so we've made | | progress. (No one said we could do this | | fast!) y. .x | | | | e V ^ b | | | | | a f | .___\___.___/___. y / x \ x a x b x y.___\___.___\___. Now we've cut another triangle. When | / / | we section it off (not pictured this time) | | we'll create one new vertex x and and one | | new vertex y . To make progress we need e V V f to reattach in such a way that we lose | | two y's. This time we MUST reattch along | | edge e , or we'd just spin our wheels. | | y. .x |\ | | \ | e V \ g ^ b | _| | | \ | | a \ f | .___\__\.___/___. y / x \ x g y a x b x x.___/___.___\___.___\___. Now we've cut another triangle. When \ \ | / / | we section it off (not pictured this time) \ | | we'll create one new vertex x and and one \ | | new vertex y . To make progress we need - V e V f to reattach in such a way that we lose a |\ | | two y's. This time we MUST reattch along \ | | edge e , or we'd just spin our wheels. \| | y. .x \ | \ | \ g ^ b _| | \ | \ f | \.___/___. x \ x g y a x b x x.___/___.___\___.___\___. We erase the internal edge e which no \ \ / / | longer matters and ... \ | \ | - V f a |\ | \ | \ | y. .x \ | \ | \ g ^ b _| | \ | \ f | \.___/___. x \ x y a x b x .___\___.___\___. ... square everything up. Now we could cut | / / | off a triangle in the upper left and reattach | | it on the lower left to reduce the number of | | y's to 1. But it's faster to notice that g V V f the sequence | | | | .___/___.___\___. | | \ / x. .x g a | | occurs twice and replace it both place with | | a single edge. | ^ b a ^ | | | | g f | .___\___.__/___. y / x \ x x b x .___\___. ... square everything up. Now we could cut / / | off a triangle in the upper left and reattach / | it on the lower left to reduce the number of h|_ | y's to 1. But it's faster to notice that / V f the sequence / | / | .___/___.___\___. / | \ / x. .x g a \ | occurs twice and replace it both place with \ | a single edge. \ ^ b h _| | \ | \ f | \.__/___. x \ x Observations: a) It would never help to slice from one y vertex to another. This would create two new vertices y and we'd loss at most two after pasting, so no gain. b) If we'd followed the path that brought us down to one vertex y, we'd have to use STEP 2 to eliminate it and finish. (to be continued)