Euler's Formula Applied to a Torus

Date: 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/