Analogon of the polyhedral formula in three dimensions

Question

I was recently wondering if there is an analogous formula to the polyhedral formula on the (two-dimensional) sphere: $$V - E + F = 2,$$ where $V$ is the number of vertices, $E$ the number of edges and $F$ the number of faces of polygons on a sphere.

This is rather nice and intuitive. But now I was wondering if there is such a formula for three-dimensional closed manifolds involving basic concepts. I've read about the generalization involving Betti-numbers, but Betti-numbers do not seem to have any basic analogy (at least to me, I don't know much geometry).

My question thus is: Is there any generalization of such a formula for polyhedra in a three-dimensional manifold?

Answer 1

The polyhedral formula $V-E+F=2$ on the two-dimensional sphere $S^2$ is a formula for a particular 2-manifold, namely, the two-dimensional sphere. What it says is that however you subdivide $S^2$ into triangles so that any two triangles meet along an edge, a vertex, or not at all, the formula $V-E+F=2$ is true.

Although you say in your comment that you are not asking for an analogue of the Euler characteristic, you may be missing an important fact, namely: the left hand side of the formula for $V-E+F=2$ for $S^2$ IS the Euler characteristic of $S^2$. What this equation says, in words, is

The Euler characteristic of $S^2$ equals $2$.

So, by asking for "an analogous formula to the polyhedral formula on the two-dimensional sphere", you are asking for an analogous formula for the Euler characteristic of $S^2$, whether you know it or not.

Before moving on to 3-dimensional manifolds which is the topic of your question, let me first point out the further aspects of the Euler characteristic for 2-dimensional manifolds.

There is a different polyhedral formula for every particular compact 2-manifold. What that formula tells you is what the Euler characteristic of that 2-manifold is equal to.

For example, the formula for the 2-dimensional torus $T^2$ is $V-E+F=0$, and this formula says, in words,

The Euler characteristic of $T^2$ equals $0$.

Also, for the genus 2 surface the formula is $V-E+F=-2$, and this formula says

The Euler characteristic of the genus 2 surface equals $-2$.

So, on to 3-dimensional manifolds. Every compact manifold 3-dimensional manifold $M$ has a similar polyhedral formula, and what that formula tells you is what the Euler characteristic of $M$ is equal to. What you do is to write $M$ as a union of 3-dimensional simplices or tetrahedra so that any two tetrahedra meet along a common 2-dimensional triangular face, or a common 1-dimensional edge, or a common 0-dimensional vertex, or not at all. The left hand side of the formula is always the same: $V-E+F-T$.

As it turns out, for any compact 3-dimensional manifold $M$ the result is the same, namely $V-E+F-T=0$. That is, the right hand side of the formula is always $0$, no matter what $M$ is. What this formula says, in words, is:

The Euler characteristic of every compact 3-dimensional manifold is equal to zero.

For a particular example, let $S^3$ be the unit sphere in $\mathbb R^4$, namely the set of all $p \in \mathbb R^4$ such that $|p|=1$. Since $S^3$ is a compact 3-dimensional manifold, we get

The Euler characteristic of $S^3$ is equal to zero.