THE TOPOLOGICAL CLASSIFICATION OF POLYHEDRA (continued) ------------------------------------------------------- At then end of Step 4 we obtain a planar model whose vertices ALL represent the same vertex on the original polyhedron. (Thus, a circumnavigation of the planar model would correspond to a path on the original polyhedron consisting of a series of loops leaving and returning to a single point, with no other point visited twice.) Step 6 (Collecting Twisted Pairs) Suppose we have a nonconsecutive twisted pair of edges. We can picture this as follows: a .______\______. (As always, the dots on the sides represent . / . an undetermined number of edges about which . . we just don't happen to care at the moment.) . . . . . . . . . . . . . . . . . . . . . a . .______/______. \ Now slice from the foot of one edge a to the foot of the other. Give the new edge formed by the cut an unused name (we'll use f here) and pick an orientation for it, represented in the picture by an arrow. Comments: a) Slicing from the head of one edge a to the head of the other would work just as well, or course, since changing the orientation of BOTH edges would not alter the way we taped together the planar model to recover the original polyhedron. b) For the same reason, either orientation of the cut works fine. a .______\______. . \ / . . \ . . \ . . \ . . \ . . \ f . . _| . . \ . . \ . . \ . . \ . . \ . . a \ . .______/______. \ Now separate the planar model into two pieces along the cut, a . .______\______. . \ \ / . . \ \ . . \ \ . . \ \ . . \ \ . . \ f \ . . _| _| . . \ f \ . . \ \ . . \ \ . . \ \ . . \ \ . . a \ \ . .______/______. . \ tape together the original twisted pair (you must turn one piece over), . . \ . \ . \ . \ . \ . \ f . _| . \ . \ . \ . \ . \ . a \ .______/______. . \ / . / . / . / . / . / . |_ . / f . / . / . / . / . / . and eliminate the original edge which now run internally: . . \ . \ . \ . \ We have now replaced a nonconsecutive twisted . \ pair of edges (labelled a ) with a consecutive . \ f pair (labelled f ) without changing the total . _| number of edges. . \ . \ . \ Doing this repeated, we can make ALL the twisted . \ pairs consecutive. . \ . \ . . . / . / . / . / . / . / . |_ . / f . / . / . / . / . / . Step 7 (Collecting pairs of opposing pairs) Before we can describe the next cut-and-paste operation, we must make a logical point. In the following diagram a .______\______. . / . . . . . . . . . . . . . . . . . . . . . . . . a . .______\______. / the dots running up the left cannot consist ONLY of twisted pairs. Why? If they did, then MORE THAN ONE vertex from the original polyhedron must occur on the boundary. Why? Look at this typical example and count vertices (of course, no harm comes from assuming that all the edges on the left point down): a .______\______. | / . | . b V . | . | . . . | . | . b V . | . | . . . | . | . c V . | . | . . . | . | . c V . | . | a . .______\______. / We computer as follows: foot of a = foot of b = head of b = foot of c = head of c = head of a so all the vertices on the left represent the same vertex in the original polyhedron (we expected that), but NO OTHER VERTEX does, CONTRARY to how we arranged things after Step 4. So the picture can't look like this. That means that some other opposing pair dovetails with the two edges marked a , as in the following picture: a ._____\____. . / . . . . . | | | | b V V b | | | | . . . . . a . ._____\____. / Now we cut-and-paste twice. The pictures will tell the story. Here's the first cut-and-paste: a a ._____\____. ._____\____. .________\_______. . / . . / . | / | . . . . | c | .________\_______. .________\_______. b V V b | / | / | | | c | c | | b V V b . . | | .________\_______. . . | | | / | . a . . . | c | ._____\____. . . b V V b / . a . | | ._____\____. | | a / . . ._____\____. . . . / . . a . . . ._____\____. .________\_______. / / c .________\_______. .________\_______. .________\_______. | / | | / | | / | | c | | c | | c | b V V b b V V b b V V b | | | | | | | | | | | | . . . . . . . . . . . . . a . . . . . ._____\____. . . . . . / . . . . . . . . . . . .________\_______. .________\_______. .________\_______. / / / c c c For the sake of the typewriter art, prior to the second cut-and-paste distort the diagram in a topological harmless way: .________\_______. .__________\_______. | / | | / | | c | | c | b V V b | | | | | | | | becomes | | . . V b b V . . | | . . | | . . | | . . | | . . | | .________\_______. . . / . . c ...._____\___... / c and now .__________\_______. .__________\_______. . | / /| | / / /| | c / | | c / / | | / | | / / | | / | | / / | | / | | / / | V b / b V V b / / b V | d |_ | | d |_ d |_ | | / | | / / | | / | | / / | | / | | / / | | / | | / / | . / . . / / . . / . . / / . .../_____\___... .... ._____\___... / / c c . .____________\______. .____________\______. /| | / / /| / / / | | c / / | c / / | | / / | / / | | / / | / / | | / / | / / b V V b / / V b / d |_ | | d |_ d |_ | d |_ / | | / / | / / | | / / | / / | | / / | / / | | / / | / / . . / / . / / . . / / . . / ._____\___... .... ._____\___... .... / / c c and finally: .__________\_______. (Observe that if you tape the edges / / / marked c together and then do / c / the same for the edges marked d , / / you'll get a torus (=bagel surface) / / with a hole in its surface.) / / / / |_ d |_ d / / / / / / / / / / / / ._____\___........ / c Let us call a configuration of edges such as the c's and d's here a _tight pair of opposing pairs. -------------------------------------------- Now we've almost come to the end of the story. So far we've learned that any closed (=no boundary) polyhedron has a planar model which consists, (when you circumnavigate its boundary) ONLY OF CONSECUTIVE TWISTED PAIRS and TIGHT PAIRS OF OPPOSING PAIRS. This already means that WE CAN BUILD *ANY* POLYHEDRON BY STARTING WITH A SPHERE AND ATTACHING HANDLES AND MOBIUS STRIPS. If, perhaps, you don't find this claim obvious, even after the work we've just done, maybe the following observations will help. We usually think of a Mobius strip in terms of the following sort of planar model: ............ | | | | | | | | | | a V ^ a | | | | | | | | ............ Now a very simple cut-and paste gives us a different model: ............ ............ . ............ ............ | /| | / /| \ | | / | / | | / / | \ | | / | / | | / / | \ | | / | / | | / / | \_ | | / | |_ | | |_ |_ | | | | |_ a V / b ^ a a V / b / b ^ a b \ a V a V / b | / | | / / | \ | | / | / | | / / | \ | | / | / | | / / | \ | | / |/ | |/ / | \| |/ ............ . ............ . . ....................... ....................... \ | / \ / \ | / \ / \ | / \ / \_ | / \_ / | | |_ | |_ b \ a V / b b \ / b \ | / \ / \ | / \ / \ | / \ / \|/ \ / . . Now we can see that the operation of "sewing in a Mobius strip" amounts to inserting a consecutive twisted pair. ____________________________ | | | | | | | | | | This box represents some (small) region of | | our polyhedron where we wish to sew in a | | Mobius strip. | | | | | | | | | | | | | | ---------------------------- ____________________________ | | | | | | | . | Now we've made a slit. When doing this | | | in terms of a planar model, we should | | | arrange for one endpoint of the slit | | | to appear as a vertex in the planar model | V a | (either by moving the slit or changing | | | the model). In terms of the planar model, | | | we've inserted a consecutive opposing pair | . | | | | | | | ---------------------------- ____________________________ | | | | | . | | / \ | Now we've opened up the slit. In terms | / \ | of the planar model, the consecutive | | | | opposing pair has become a BIT OF BOUNDARY | | | | we DON'T plan to tape back up. Of course | a V V a | this changes the topology! We've made | | | | a hole in the polyhedron; we've inserted | \ / | some boundary (not part of any pair) in | \ / | the planar model. | . | | | | | ---------------------------- Now we sew the boundary of the Mobius strip to the boundary of hole we've open up. In terms of the planar model, we just attach a triangle of the form attach it here ....................... \ / \ / Don't try to picture this \ / in three dimensions. \_ / | |_ b \ / b \ / \ / \ / \ / . to the BIT OF BOUNDARY we'd just opened up. As the net effect on the planar model, we've inserted a consecutive twisted pair. If, instead of the triangle, we attached something of the form: b a .__\__.__/__. | / \ | | | | | | | | | a V V b | | | | | | | | ............. then we'd have added a handle (=bagel surface with a hole; see the observation above.) What's left to do? Turns out that when a planar model has a CONSECUTIVE TWISTED PAIR followed by a TIGHT PAIR OF OPPOSING PAIRS, a cut-and-paste (that doesn't affect the topology) and replace this with three consecutive CONSECUTIVE TWISTED PAIRS. Instead of showing all the steps, we'll explain why this shouldn't surprise us. The diagram below shows a portion a polyhedron which contains a handle (on the left) and a Mobius strip (on the right). I've avoided the difficulty a picturing how the Mobius strip sits in the rest of the polyhedron because all the action will take place in the portion shown. ____ _______________ |\ \ | | | \ \ | ___________ | | \ \ | | | | | |\ \ | | | | | | \ \ |----| |-----------| |------------- | \ \ (In my drawings I make | | | | | | \ /| the handle opaque but | / \ / \ | \/ | the floor transparent.) | | / | | left foot right foot | / | | | / / | | / / |---------------------------------- |/ / | | / | | / | | / ___|/ The handle has two feet. Drag one of them (say the right) towards the Mobius strip and take it for a round tour. Going around a Mobius strip takes you from one side to the other. So when you come around you've got the right foot attached to the back, as it. ____ _______________ |\ \ | | | \ \ | ___________ | | \ \ | | | | | |\ \ | | | | | | \ \ |----| |--------------------------- | \ \ | | | | | | \ /| | / \ / \ | \/ | | | / | | | / | | | / / | | / / |---------------------------------- |/ / | | / | | / | | / ___|/ The left portion of the diagram, a punctured Klein bottle actually, amounts to two sewed in Mobius strips. _______________ Perhaps you can just visualize the | | two Mobius strips directly; One | ___________ | runs along the front of the tube, | | | | the other along the back. | | | | |----| |----------------------| If not, you can use a planar model | | | | | | First sever the tube and contract | / \ / \ | the two ends thereby created so as | to leave two holes. Then make a | | slice to connect the holes. You | | get something like: | | | | |-----------------------------| |--------------------------| | | Make sure you understand the | b | picture at the left. The OUTER | .______\______. | rectangle represent the same | | / | | thing as the OUTER rectangle in | | | | the picture above. The edge b | | | | represents the slice which connect | | | | the whole, and the edge a | a V ^ a | represents cut that severs the | | | | tube. | | | | | | | | The next step is easier to see if | | b | | you first turn the picture inside | .______\______. | out to get something like | / | | | |--------------------------| b .______\_______. Now the small inner rectangle |\ / | corresponds to the large outer | \ | rectangles in the previous | \ ----- | pictures (and we've moved it | \ | | | off to one side to get it out | \ | | | of the way). Now slice down the | \ | | | diagonal and reattach | \ ----- | a V \ ^ a | \ | | \ | b | \ | .______\_______. | \ | |\ / | | b \ | | \ | .______\______\. | \ ----- | / | \ | | | | \ | | | | \ | | | | \ ----- | a V \ ^ a | c \ | | _| | | \ | | \ | | b \ | .______\______\. / and you get b b .______\_______.______\_______. \ / | / / \ | / \ ----- | / \ | | | _/ \ | | | /| \ | | | / c \ ----- | / \ ^ a / c \ | / _| | / which, when you erase the \ | / internal line, square it up, \ | / and turn it inside out again, \ | / looks like \./ |--------------------------| | | | c | | .______\______. | | | / | | | | | | | | | | | | | | | b V V c | | | | | | | | | | | | | | | b | | | .______\______. | | / | | | |--------------------------| as claimed. Now we know that WE CAN BUILD *ANY* POLYHEDRON BY STARTING WITH A SPHERE AND *EITHER* ATTACHING HANDLES *OR* ATTACHING MOBIUS STRIPS.