I had a look at paragraph 1.1 from "Basic Topology" by M.A. Armstrong.
There a very nice proof outline presented there to the Euler's theorem.


$ V-E+F=2 $

In that proof outline (towards the end) there's this note:

"The proof breaks down here for a polyhedron such as that shown in Fig. 1.3, because the dual graph $\Gamma$ will contain loops."

But I don't quite get this note since... I just think the proof breaks much earlier. Because for the polyhedron in Fig 1.3, there's simply no tree $T$ as constructed in the proof. We have 2 trees instead of a single tree $T$. Is this observation correct?

I mean... in fact the polyhedra from Figs 1.2 and 1.3 do not satisfy at all the condition (a) of the theorem, that the graph formed by the vertices and edges of the polyhedron needs to be connected. Their respective graphs are composed of 2 separate connected components and hence are not connected graphs.

Not to talk that they don't satisfy also condition (b) which the text explicitly mentions.

So the whole construction of the proof breaks much earlier, I think.
I just don't see how you can start doing the proof's construction if condition (a) is not met.


Basically I realized that my confusion comes from not being sure about the following. Let's look at this prism which has a smaller prism cut through the middle. In that proof they start by building a graph G of all vertices and edges of the polyhedron. Then they build a spanning tree T of G. Then they build the dual graph $\Gamma$ which is defined as: vertices - the faces of the polyhedron, edges - two faces are connected if they are neighbors and the neighboring side is not an edge in T.

My question is: should I consider that edges AG, BH, CI and the analogical 3 edges from the top face exist? Seems this influences significantly the Euler's characteristic of the polyhedron?! Hm... In other words: I understand this unfilled prism does not satisfy condition (b) of the theorem, but does it satisfy (a) or not?

If I consider they exist, then I can understand the note in the book. We build G, G is a connected graph, and then we build a spanning tree T of G (which is in green on my drawing).

But if these 6 edges do not exist then well... the whole graph G has no spanning tree because it's not connected.

So in this context of topology, do we consider these vertices as connected (A with G, B with H, C with I) ?


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    $\begingroup$ You're not going to get any answers if you don't tell people what these figures and conditions are. $\endgroup$ – Trevor Gunn Mar 29 at 14:19
  • $\begingroup$ @TrevorGunn Yeah, I guess so. $\endgroup$ – peter.petrov Mar 29 at 14:43
  • $\begingroup$ @peter.petrov can you copy here these figures? Or describe what we would see? $\endgroup$ – Berci Mar 29 at 14:50
  • $\begingroup$ The orange edges (in a way) can be though of as edges of the dual graph. The green edges are the edges in the spanning tree T of G. Two faces are neighbors/connected in the dual graph if they share a side which is not an edge of T. Sorry, I am somewhat bad at making pictures and visual things on a computer so I just made the drawing by hand. $\endgroup$ – peter.petrov Mar 29 at 18:03

In calculating the Euler characteristic, you need all "faces" to be discs. Meaning there are no holes in the middle of them. Or another way of looking at it: if you take two points on the boundary of your face and draw an edge between them, that should cut the face into two smaller faces. So the number of edges goes up by 1 and the number of faces goes up by 1 and the quantity

$$ V - E + F $$

does not change.

On the other hand, if you have an annular "face" like this:

Annulus with an edge between the two boundary components

then you can see that whether or not that short edge is there does not change the number of faces. So in order to make the formula work, we can't allow annular faces.

That is what the second condition is getting at. For instance consider the following torus (image by Tom Ruen at Wikipedia):

If you cut along either loop you get an annulus (you could equivalently call it a cylinder). But if you cut along both loops you get a disc. So that is why we need both loops. And now you can verify that $V - E + F = 1 - 2 + 1 = 0$. Toruses have Euler characteristic $0$. In general, for a surface $X$ of genus $g$ (orientable), $\chi(X) = 2 - 2g$. So spheres have $\chi = 2$, toruses $\chi = 0$ and so on.

So the second condition rules out higher genus polyhedra. The first condition rules out things like a disjoint union of two spheres. Imagine I have two tetrahedra side by side. Then $V - E + F = 4$ since individually $V - E + F = 2$ and I'm doubling each number. A disjoint union doesn't satisfy the first condition (because it's disjoint).

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  • $\begingroup$ Thanks a lot. So in simple terms... I do have those 6 edges which I asked about? I should consider they exist (in order to make all faces disks)? Then everything matches up, yes, and the note in the proof does make sense. Side note: is there any free software for making such nice drawings as yours? Also, you meant spheres have $\chi = 2$, I think. My "unfilled prism" is homeomorphic to a torus, I think, right? $\endgroup$ – peter.petrov Mar 29 at 16:31
  • $\begingroup$ @Peter Yes, you have those edges. I don't think you need all 6 but you do need at least one on the top and one on the bottom. But adding all 6 makes it more symmetric. The first image I made with geogebra. The second I took from Wikipedia. Yes, $\chi = 2$, thanks. Yes, the prism is homeomorphic to a torus. $\endgroup$ – Trevor Gunn Mar 29 at 16:32
  • $\begingroup$ Alright, thanks a lot once again! All the best! $\endgroup$ – peter.petrov Mar 29 at 16:33

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