Euler Number of Polyhedra Joined Along An Edge Or Face?

[RuthlessOptimism]

Hello,

 

I have been looking around lately for literature about a specific topic but have not found anything directly relevant to what I am looking for, so I was wondering if anyone had any "keywords" regarding this topic for me to try out.

 

Basically if we have a polyhedron, and we were to join it to another polyhedron along an edge or face, does this change its euler characteristic? I would like to find some kind of proof that the euler characteristic stays the same no matter how many joined polyhedra you have, even with the joining possibly occuring sometimes between edges, other times between faces, maybe also between completely different polyhedra. Does anyone know if this already exists somewhere and what field it would be a part of?

 

Its fairly simple to try an example of a "cluster" of Tetrahedrons joined along a single edge.

 

For N Tetrahedrons joined along a single edge we have:

 

X = V - E + F

 

with 2 = 2(N + 2) - 6(N + 1) + 4(N + 1), with the obvious limit that N has to be less than or equal to six. Also its important to note that the center axis that the Tetrahedrons in this example are joined along is counted as a different edge in between each of the Tetrahedrons (is it correct to do that?).

 

For a chain of Tetrahedrons joined by their faces we have:

 

2 = (N + 4) - 3(N + 2) + 2(N + 2)

 

You could easily prove these as being true for integers via induction, but what I am interested in is the actual geometries.

[RuthlessOptimism]

For anyone who might have been interested in this question I answered it myself, sort of. This source has a more general definition of the euler characteristic of a surface as well as the euler characteristic of "polyhedral sets".

 

"A New Look at Euler's Theorem for Polyhedra", Branko Grunbaum, G.C. Sheppard, The American Mathematical Monthly, Vol. 101, No. 2, Feb 1994

 

The reason I write "sort of", is because it is unknown to me how this particular definition, which is slightly different from the one I am familiar with affects the Gauss Bonnet Theorem. For example, using their approach you would instead calculate the euler characteristic of a cube to be 1 (as opposed to 2), but the joining of arbitrary number of cubes along edges (or faces) to also be 1. But what I still wonder is whether or not the arbitrary number of cubes joined along edges or faces still has 4 pi total curvature?

 

Does anyone know?

[RuthlessOptimism]

Ok I have once again sort of answered my own question. This reference;

 

Curvatures of Smooth and Discrete Surfaces, John M. Sullivan, Discrete Differential Geometry, Vol. 38 175-188, 2008 .

 

Describes how one can preserve the Gauss Bonnet theorem for discrete surfaces.

 

[latex] \int\int K dA = \sum_{p \in D} K_p [/latex]

 

the integral of the gauss curvature is equal to the summation of the curvature defined at every vertex of a surface composed of flat triangular faces.

The curvature at each vertex is defined as:

 

[latex] K_p = 2 \pi – \sum_{i} \theta_i [/latex]

 

Ok I can't figure out what is wrong with the syntax here but Kp is supposed to be two pi minus the sum of all of the interior angles of the triangular faces that meet at the vertex i if they were all pushed down into a plane.

 

Where [latex] \theta_i [/latex], are the interior angles that meet at each vertex i, if you were to flatten this vertex into a plane, ending up with a bunch of triangles all meeting at a common point. This gives a value of four pi for my example of 3 tetrahedrons joined along an edge at a central axis. But I am not sure if this special case is actually applicable to what the author has defined. In my example I take the center vertex where we really have one vertex of each individual tetrahedron meeting on the central axis as having 9 triangles each with interior angles of 60 degrees.

 

Thus we end up with:

 

[6*(360 - 3*60)] + [2*(360 - 9*60)] = (6*180) - (2*180) = 4*180 = 4 pi

 

But obviously 9 triangles cant meet at a central point without overlap. As I've said, I am not sure if that breaks what the author has defined, but it did give me the answer I was hoping for lol.

Edited November 13, 2015 by RuthlessOptimism