Euler's Formula Applied to a TorusDate: 06/05/2001 at 23:24:03 From: Jim Vinci Subject: Geometry Can you explain why Euler's characteristic is zero for a torus? If, for example, I drew an arc with two vertices on top of the torus and connected another arc to it to form a circle, wouldn't V=2, E=2, and F=1, so that V-E+F=1? What I am I missing? Isn't this an admissible graph? Date: 06/07/2001 at 08:37:53 From: Doctor Peterson Subject: Re: Geometry Hi, Jim. I'm not sure I picture exactly how your vertices are connected, but most likely your mistake is that one of the "faces" includes the "hole" of the torus, and therefore is not a valid face. A face must be topologically equivalent to a disk; you should be able to flatten it out into a plane. If this doesn't clear it up, please write back and tell me more precisely where your "other arc" goes; also give me the defintion you are using of an "admissible graph," so I can use the same terms you are familiar with - there are several ways to describe this. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/08/2001 at 08:21:36 From: Jim Vinci Subject: Re: Geometry Dr. Peterson, Thanks for the reply. Actually, the term "admissible graph" comes from Steven Krantz's book, _Techniques in Problem Solving_. He defines it simply as a connected configuration of arcs and his example focuses on a sphere. He also defines a face as any two-dimensional region, without holes, that is bordered by edges and vertices. One of his problems is to determine the Euler characteristic for a torus and to show that it will work for any admissible graph on the torus. So let's say you drew two arcs on top of the torus so they formed a circle around the torus (it would look like three concentric circles if you viewed the torus from the top with the hole looking like one of the circles). What is F for this configuration? Also, if you drew one arc from the outer "edge" of the torus to the hole and back up and around (this would sever the torus if a cut was applied along the arc), what is F for this configuration? Is F always zero for a torus? If so, why? I guess my real problem is that I don't understand how Euler's formula applies to a torus. I understand how it applies to figures with pointy edges like a cube, pyramid, etc. Maybe a good comparison to a torus is a nut (square with a hole in the middle). In this case, would F=4 because two of the six sides have a hole and therefore do not count as valid faces? Jim Date: 06/08/2001 at 12:16:27 From: Doctor Peterson Subject: Re: Geometry I've found that just about every place I look for a definition of the Euler characteristic and related concepts is either too vague (as yours is), or too deeply embedded in topology (and dependent on definitions given elsewhere, or perhaps never clearly stated) to make a good reference to answer questions like this. The general idea is simply that either we are making an actual polyhedron that is topologically equivalent to, say, a torus, or we are making a graph on the surface that is "polyhedral" in a topological sense. But exactly what this means is seldom stated. Here is one answer I found to a similar question, which is better than most in stating the requirements fairly carefully: How many edges (lines) are in a cylinder? - Final Answers, Geometry and Topology - Gerard P. Michon http://home.att.net/~numericana/answer/geometry.htm#edges In discussing V-E+F=2 for a cylinder with no vertices, two "edges," and three "faces," this says: Nothing is wrong if things are precisely stated. Edges and faces are allowed to be curved, but the Descartes-Euler formula has 3 restrictions, namely: 1. It only applies to a (polyhedral) surface which is topologically "like" a sphere (imagine making the polyhedron out of flexible plastic and blowing air into it, and you'll see what I mean). Your cylinder does qualify (a torus would not). 2. It only applies if all faces are "like" an open disk. The top and bottom faces of your cylinder do qualify, but the lateral face does not. 3. It only applies if all edges are "like" an open line segment. Neither of your circular edges qualifies. This is good enough to answer your specific question. Your edges are valid; but your single "face" is not equivalent to a disk; if you cut along the edges and spread it out flat, it becomes an annulus. That's the problem. There are several ways in which a "face" may fail the test. One kind of "hole" is that in your example, where the "face" is like an annulus or cylinder; its set of edges is not connected. In your example, the inner and outer edges of the annulus are glued together when you put it on the torus, but they are distinct when you view the "face" by itself. It is easy to miss this! You must picture taking the "face" off the surface, so you can see what it really is. Another way is for the "face" to have only one edge, but have a hole in the same sense that the torus has a whole; this is what happens if you simply draw a circle (with one vertex to make it a valid edge) on the side of the torus, so that the inner face is a valid disk, but the outer "face" is all the rest of the torus, including the "hole," which you can also picture as a "handle." This can't be flattened out at all. In either case, the "face" is not simply connected; you can draw a circle in it that can't be shrunk to a point. The problem in the definition you are using seems to be that he defines an admissible graph without reference to the faces, which really are the determining feature in this context; and perhaps also he has not clearly defined what he means by a "hole" in a face. If you remember that this can mean either a hole with an edge, or a "handle," it might be clearer. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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