Euler characteristic in 4-polytopes

Question

My question is simple. It is well-known that in three dimensions every planar graph/convex polyhedron has the property that if its number of faces is $F$, its number of edges is $E$ and its number of vertices is $V$: $$F-E+V=2$$

This has quite a number of proofs and I feel comfortable with this result. However with what I'm stumped is the 4-dimensional analog: In a polychoron (4-polytope) with $C$ cells, $F$ faces, $E$ edges and $V$ vertices: $$C-F+E-V=0$$

I assume this may also be seen in a graph embedded in three dimensions, as such:

However, how can I even define what are faces and cells in this diagram? And if I can and I can use this to prove the theorem in 4D, for what polychora would this be valid? (since some non-convex polychora don't have a zero Euler characteristic).

I am aware that the Euler characteristic is a very topological property, but as someone not versed in topology I'd like the most combinatorial approach possible.

Answer 1

It is well-known that the Euler characteristic of a convex polytope $P$ is $1$. We need to define exactly what this is. We need to regard $P$ as a $d$-dimensional face of itself. Then $\chi(P)=a_0-a_1+a_2-\cdots+(-1)^d a_d$ where $a_k$ is the number of $k$-dimensional faces. By our convention $a_d=1$, and the theorem is that $\chi(P)=1$.

Several of the proofs of Euler's formula collected by David Eppstein as his Geometry Junkyard are valid for $d$-dimensional polytopes.

Answer 2

One way to answer is to consider the Euler characteristic of the border of a simplex.

The $2$-simplex is a triangle. If we represent the triangle by labeling its vertices 1, 2, and 3, we list all of the faces, edges, and vertices: $\{123,12,13,23,1,2,3\}.$ To take its border, we remove $123$ and see there are $3$ vertices and $3$ edges. So the Euler characteristic would be $3-3=0$.

The $3$-simplex is a tetrahedron, $$T=\{1234,123,124,134,234,12,13,14,23,24,34,1,2,3,4\}$$. Removing the cell, we see $$V-E+F=4-6+4=2$$, which is what we expected.

How do we extend this to $4$-simplices? As you can expect, we just add the vertex $5$ to all of the cells and append the result to the original set. So $$S=\{12345,1235,1245,1345,2345,125,135,145,235,245,345,15,25,35,45,1234,123,124,134,234,12,13,14,23,24,34,1,2,3,4,5\}$$

We see there are $\binom{5}{1}$ vertices, $\binom{5}{2}$ edges, $\binom{5}{3}$ triangles, $\binom{5}{4}$ 3-simplices. Taking the alternating sum of these gives zero.

In general, the Euler characteristic of the border of a simplex is always $2$ or $0$. We can see this according to the binomial theorem:

$$\chi(\partial S^{n})= \sum_{k=1}^{n-1} (-1)^{k-1} \binom{n}{k} = 1 - (-1)^n$$