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Is the Euler characteristic $\chi =2$ for the prism with a hole?

In attempting to lead a bunch of high school students to an understanding of the Gauss-Bonnet theorem, I've run into a problem determining the Euler characteristic of a cube with a square hole through it. The problem is that the surface is like that of a torus, and so ought to have $\chi = 0$, but a simple counting of $V$, $F$, and $E$ yields $\chi = 2$.

I have a wide range of abilities in my class, so I need to motivate this theorem without the use of calculus. My hope was that by examining polyhedra of different types, they would come to "discover" a connection between the Euler characteristic, the total angle defect, and the number of holes through the surface, but now I'm stuck trying to come up with a way to rig the counting to make $\chi$ come out right.

I would love some insight on

  1. how to make the count come out to $\chi = 0$
  2. how to justify such a counting method so that it does not seem arbitrary to my students.

marked as duplicate by Rahul, J. M. is a poor mathematician, t.b., Zev Chonoles Dec 22 '11 at 2:24

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    $\begingroup$ Very likely you're making the mistake that your faces have holes -- they shouldn't! See here $\endgroup$ – t.b. Dec 21 '11 at 15:08
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    $\begingroup$ Also: You may want to have a look at Lakatos's Proofs and Refutations. $\endgroup$ – t.b. Dec 21 '11 at 15:15
  • $\begingroup$ Thanks! I guess I should have dug deeper for an older post... $\endgroup$ – Jacob Nazeck Dec 21 '11 at 16:09

You just have to count carefully.
$V=16$: 8 on the hole, 8 on the outside box. $F=16$: 4 groups of 4. $E=32$: 12 on top, 12 on bottom, 8 vertical.

$$V-E+F=16-32+16=0 \;.$$

In terms of the Gauss-Bonnet formula, there are 8 vertices with curvature $\pi/2$ (the outside corners), and 8 vertices with curvature $-\pi/2$ (the hole vertices), yielding a total curvature of $0 = 2 \pi \chi$.

For justification, you might look at Discrete and Computational Geometry, from which the figure above is taken, Chapter 6.


your space should have some kind of simplicial or cw structure. the interior of your faces should be 2-cells, but the top and bottom "faces" of what you are describing have "holes." one way to present this would be to triangulate your surfaces (this gives a solid "rule" to explain to students) or if you dont want to chop it up that much, explain that the faces have to "look like" disks, or cant have holes or whatever.

  • $\begingroup$ I'll have to look up "simplicial" and "cw structure," but this is helpful. Triangulation seems like the best bet here for a reasonable, well-defined count for every surface. Thanks. $\endgroup$ – Jacob Nazeck Dec 21 '11 at 16:11

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