FROM GEOMETRY TO TOPOLOGY ------------------------- Suppose I place an object in front of you and you ask me ``what is it?'' I could truthfully answer, perhaps : ``it's the Cortland apple I picked last Saturday afternoon at my grandmother's orchard'' or ``it's a Cortland apple''; or simply ``it's an apple''; or ``it's a piece of fruit''; or just ``it's food.'' How could all these answers could be correct? The question ``what is it?'' constitute a request to locate a certain object within a particular category. If the category is _Varieties of Fruit_, the ``it's an apple'' gives an appropriate reply. If the category is _Varieties of Apple_, then I'm required to be just a bit more specific, so I might say ``it's a Cortland''. Now in ordinary conversation we often understand the categories implicitly. After all, if someone doesn't like our answers, they can request more information or ignore superfluous detail. Mathematicians, on the other hand, find it useful to declare their categories explicitly. Here's an example. Suppose, instead of an apple, the object I place in front of you is an (idealized) cube. To the question ``what is it?'' a metric geometer might answer ``a cube 3 inches on a side'' (metric geometry concerns distances); a conformal geometer might answer ``a cube'' (conformal geometry concerns angles); a combinatorial geometer might answer ``8 points connected by 12 edges forming 6 quadrilaterals'' (combinatorial geometry concerns configurations) a topologist might answer ``a sphere.'' A conformal geometer sees no distances, and so no difference between a cube 3 inches on a side and another 5 inches on a side. A combinatorial geometer sees no distance or angles, just the connections between points, edges, faces, etc. Thus a combinatorial geometer makes no distinction between a cube and a truncated pyramid, say. A topologist sees only basic shape. Imagine the cube made out of rubber. Inflating the cube would gradually efface all the edges and corners, rounding everything out. This process would represent no change at all to a topologist. Here is the immediate significance of all this to us. We are about to meet a quantity, called _Total Angle Defect_ associated to polyhedra. As the name suggests, computing _Total Angle Defect_ involves measuring angles, so the notion would seem to belong to conformal geometry. This makes it a remarkable fact that _Total Angle Defect_ depends not an the conformal geometry of a polyhedron, nor even on its combinatorial geometry, but *only* on its topology. To define _Total Angle Defect_ we first need the notion of _Local Angle Defect_. _Local Angle Defect_ measures the flatness of a polyhedron at a particular point and one computes it with the formula 360 degree - (sum of the angles that meet at that corner). If the _Local Angle Defect_ equals 0, one could cut out the corner and flatten it completely. (Warning: later we will refine this definition to take account of cone points and corner points.) The _Total Angle Defect_ of a polyhedron equals the sum of all _Local Angle Defects_ at all points. EXAMPLES: A. Consider a regular tetrahedron (4 equilateral triangles forming a pyramid with a triangular base). The _Local Angle Defect_ at any corner equals 360- (60 + 60 + 60) = 180 degrees since three 60 degree angles meet there. Since a tetrahedron possesses 4 corners, its _Total Angle Defect_ equals 4 x 180 = 720. (Note that the _Local Angle Defect_ at any point within one of the triangular faces equals 0 (360 - 360) and the _Local Angle Defect_ at any point within an edge equals 0 (360 - (180 + 180)). So even though the definition of _Total Angle Defect_ has us summing the _Local Angle Defects_ at *all* points, only the corners matter.) B. Consider a regular octahedron (8 equilateral triangles forming two pyramids joined along their square bases.) The _Local Angle Defect_ at any corner equals 360- (60 + 60 + 60 + 60) = 120 degrees since four 60 degree angles meet there. Since a octahedron possesses 6 corners, its _Total Angle Defect_ equals 6 x 120 = 720. C. Consider a standard cube. The _Local Angle Defect_ at any corner equals 360- (90 + 90 + 90) = 90 degrees since three 90 degree angles meet there. Since a cube possesses 8 corners, its _Total Angle Defect_ equals 8 x 90 = 720. So far each time we get the answer 720. This is no coincidence. 720 turns out to be the answer whenever the polyhedron has the topology of a sphere. D. Consider a cube with a rectangular tunnel from the top face the the bottom. The eight corners of the origal cube still have _Local Angle Defect_ equal 90 degrees. The eight corners of the tunnel have _Local Angle Defect_ equal 360- (90 + 90 + 270) = -90 degrees (You'll find the 270 degree angle on the top or bottom of the cube and the 90 degree angles on the walls of the tunnel.) Thus, in this, case the _Total Angle Defect_ equals 0. As we shall see, you'd also have _Total Angle Defect_ equal 0 for any other surface with a donut-type topology. Now we come to our main result: SUPPOSE YOU HAVE A POLYHEDRON P WITH V VERTICES, E EDGES AND F FACES. THEN _Total Angle Defect_(P) = 360*(V-E+F). Proof: Suppose that P has a non-triangular face U. The addition one of the diagonals of U as an edge increases the total number of edges by 1 , but it also increase the total number of faces by 1 , so it has no overall effect on the quantity V - E + F, nor on the _Local Angle Defect_ at any point. This allows us to assume that P has only triangular faces, for otherwise we could add diagonals as we please until we'd cut up all the faces into triangles, and not change either side of the equation in the process. We interrupt the argument for a SLOGAN: In mathematics you learn something interesting whenever you can compute one quantity in two really different ways. In this case, let's consider two ways to count the number of ``sizes of edges'' in P. Since each edge has two sides, P has 2E sides of edges. On the other hand, each triangle face contains 3 sides of edges, so P has 3F sides of edges. This gives us the equation 2 E = 3 F . Now _Total Angle Defect_(P) equals the sum of all the _Local Angle Defects_ at each vertex, each of which is 360 minus the sum of the angles which meet at the vertex. Thus _Total Angle Defect_(P) = 360 V - (sum of all angles in P). But observe that sum of all angles in P = 180 F because each angle belongs to exactly one triangle and the sum of the angles in a triangle equals 180 degrees. (This is another example of the SLOGAN, because we can add up all the angles either a vertex at a time or a face at a time.) Putting these last two fact together gives _Total Angle Defect_(P) = 360 V - 180 F. On the other hand, 2 E = 3 F implies 360 E = 540 F (multiply both sides by 180), so 360 E - 540 F = 0. Hence _Total Angle Defect_(P) = 360 V - 180 F - (360 E - 540 F) = 360*(V-E+F) as we sought to prove. This is a version of the celebrated Gauss-Bonnet theorem. Even stating the full version requires calculus, but fortunately this simplified version will satisfy our needs. How will we use this fact? Here is the basic outline of our whole project: 1. Plane are flat, so 2. The surfaces you get from symmetric plane patterns are also flat (in some sense), so 3. All their _Local Angle Defects_ equal 0 (if you compute the right way), so 4. Their _Total Angle Defects_ equal 0, but 5. _Total Angle Defect_ depends only on topology and 6. Only 17 surfaces can have _Total Angle Defect_ equal 0. What's left to do: The Gauss-Bonnet theorem shows us that _Total Angle Defect_ depends only on the combinatorial structure. Euler's magic trick (which we push it a little harder) will show that V - E + F depends only on topology, as we will see Wednesday. So that will give us step 5. Knowing that all surfaces can be manufacture from a sphere by adding holes, handles and Mobius strips will allow us to deduce 6. We'll also have to modify our definition of _Local Angle Defect_ to account for reflective edges, gyration/cone points and corner points. Then we'll have to modify the Gauss-Bonnet formula. We'll be spending some time on this, but here's a quick illustration of the idea. A 22* pattern gives a surface with the shape of a rectangular pillowcase open on one side. From an ordinary perspective such a pillowcase ISN'T flat, because the two cone points have _Local Angle Defects_ of 180 degrees. But these points come from gyration points of order 2, so we should have expected them to be only 180 degrees around. Gyration points of order 2 become cone points of order 2, so for these we should compute the _Local Angle Defect_ as 180 - sum of the angles We should do the same for ordinary points on reflective edges. Then the pillowcase would have all its _Local Angle Defects_ equal 0, just as a map of a plane pattern should.