Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an interesting thought when I draw polygon and 3D polyhedron. My question is: Can I know the number of Face, Edge, and Vertex from a given 'space angle constraint'? For example, a vertex a cube consist of 3 faces which each of the angle near the vertex sum up to $270^\circ$. Can I tell if there is a 'thing'(simple connected closed surface) on space which each of its vertex is $270^\circ$(summing up the angle near it), then that 'thing' has 6 faces?

Here I tell you why I think about this silly question. Everyone should know that sum of exterior angle of polygon is $360^\circ$. Of course I know the proof, but I try to think it in a different way: Every point have $360^\circ$ 'to be occupied' and every two line can span a space which included angle $<180^\circ$. When you draw a from a point, you actually 'occupied some angle'. Every time you bent your line, you span another direction( here the bending $<180^\circ$). Once the line reaches the original point, all $360^\circ$ are 'occupied'. Then If I tell you this 'thing'(simple closed curve) consist of $n$ angles which EACH of it is $108^\circ$, then you can immediately tell me $n=5$.

I generalize it into 3D spaces. If there is a point in space and you span some surface from it, actually you occupied some 'space angle' of the point. Here I do not know how to define the sharpness of a vertex and my thought is just adding the angle of faces around that point just like '$270^\circ$' in a vertex of a cube. The smaller the 'space angle'(my definition) is, the sharper it is. Just like drawing polygon, planes can be spaned from a vertex which 'space angle'$<360^\circ$. Every time the plane reach a vertex, it occupied some 'space angle'. May be when the plane reach the original vertex, all 'space angle' are occupied and the drawing is completed. In this case, can I tell about the amount of faces I draw?

The whole idea is just like Gauss-Bonnet Thm. For a simpe connected closed orientable regular surface, $$\oint_SKdA=2\pi\chi(S)$$ where K is the Gaussian curvature and $\chi(S)$ is the Euler chracteristic of S. If $S$ is homeomorphic to a sphere(in this case a polyhedron), Then $\oint_SKdA=4\pi$. Just like in a 2D plane, the simple closed plane curve have the property $\oint_C\kappa ds=2\pi$ where $\kappa$ is the curvature of the curve $C$.Is it useful to answer my question? Maybe in polygon, the most angle you can 'occupied'$=\oint_C\kappa ds=2\pi$ and in polyhedron, maybe the 'space angle' you can occupied are related to Gauss Bonnet Thm.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The Gauss-Bonnet theorem is certainly your friend here. If the angle sum around a vertex is $\alpha$ then there is a point mass of Gaussian curvature of magnitude $2 \pi - \alpha$ at the vertex. Thus, if every vertex of a polyhedron with $n$ vertices has the same $\alpha$, then Gauss-Bonnet tells us that $n (2 \pi - \alpha) = 4 \pi$, and thus $$ n = \frac{4 \pi}{2 \pi - \alpha}.$$

Thus we know the number of vertices. However, this does not tell us the number of faces unless we have more information about the polyhedron - for example, the cuboctahedron and truncated tetrahedron both have 12 vertices each with angle sum $5\pi/3$; but they have different face and edge counts.

CuboctahedronTruncated Tetrahedron

share|improve this answer
    
For a polygon, once you find out the amount of vertexes, then you know the amount edges. Why this cannot generalize to polyhedron? What information should be added to determined the faces and edges? and why a point has Gaussian curvature of magnitude $2\pi-\alpha$?(sorry that I have only learnt about local Gauss Bonnet Thm which is $\sum \theta_i+\oint_C\kappa_g ds+\iint_{\Sigma_0}KdA=2\pi$) –  Unem Chan Sep 5 '13 at 13:59
    
@UnemChan: The two-dimensional case is very special - if you know the Euler characteristic then the vertex and edge counts always differ by a known constant. For polyhedrons we can e.g. add in the diagonal of a square, which increases both the edge and face counts by one but leaves the vertices the same. I think the easiest way to derive the curvature at a vertex is finding the total curvature of the Gauss map over a region containing one vertex - see en.wikipedia.org/wiki/Gauss_map#Total_curvature –  Anthony Carapetis Sep 5 '13 at 14:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.