euler's theorem

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

euler's theorem

Postby elpenmaster » Sat Feb 28, 2004 5:52 am

say--what is Euler's theorem when applied to four dimensions?
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Postby Aale de Winkel » Sun Feb 29, 2004 8:41 am

according to http://mathworld.wolfram.com/EulersTheorem.html there is many an Euler theorem (not sutprinsingly given his importance to math), please state more precisely what you mean here, and what therorem you wan to allpy to the fourth dimension.
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Postby pat » Mon Mar 01, 2004 1:18 am

I'm guessing the poster meant Euler's Polyhedral Formula. And, as the link shows for a k+1-dimensional object, if N<sub>i</sub> represents the number of i-dimensional sides, then <sub>i=0</sub>∑<sup>k</sup> (-1)<sup>i</sup>N<sub>i</sub> = 1 - (-1)<sup>k</sup>.

The familiar version for three-dimensional polyhedron is V - E + F = 2. The number of vertexes minus the number of edges plus the number of faces is two. It's easy to see that the two-dimensional case is true: V - E = 0. The number of edges equals the number of vertexes in a polygon.

Of course, until I read the mathworld bit linked above, I was convinced that this was simply a simple graph-theoretical result. One can represent a convex, three-dimensional polyhedron by mapping it into a planar graph. One starts by taking all vertexes at one face and putting them on a unit circle in the plane and connecting them around the circle. Then, one continues on to put all of the vertexes which are one-edge away from that face onto a concentric circle (usually larger) connecting these vertexes to each other and to the already existing vertexes as appropriate to the polyhedron. Continue on until you've done all edges and all vertexes.

In the end, you will have a planar graph where each vertex is represented by a vertex, each edge by an edge of the graph, and each face by a region in the plane bounded by edges with no internal edges or vertexes (though one face will be represented by the infinite region outside of the graph). I thought that V - E + F = 2 was a direct graph-theoretical result then on the set of all graphs with no "loose ends" (vertexes of valence one). Hmmm...
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Postby BClaw » Tue Mar 02, 2004 4:55 am

pat wrote:The familiar version for three-dimensional polyhedron is V - E + F = 2. The number of vertexes minus the number of edges plus the number of faces is two. It's easy to see that the two-dimensional case is true: V - E = 0. The number of edges equals the number of vertexes in a polygon.


And the 4D version is V - E + F - C = 0 (C being 'cells' or 'realms').

Check it out:
5-Cell "Simplex" 5V - 10E + 10F - 5C = 0
8-Cell "Measure" or "Tesseract" 16V - 32E + 24F - 8C = 0
16-Cell "Cross" 8V - 24E + 32F - 16C = 0
24-Cell 24V - 96E + 96F - 24C = 0
120-Cell 600V - 1200E + 720F - 120C = 0
600-Cell 120V - 720E + 1200F - 600C = 0

Cool, eh? :D
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