Understanding V-E+F=2-2p: An Explanation from What is Mathematics by Courant

[Serious Max]

Could someone explain what's happening here with subtracting $2p$? Was thinking it could result from adding and/or subtracting edges of faces on the left-hand side but doesn't seem to work.

This is from Courant's "What is Mathematics"

 

[scottdave]

The term p refers to the type of surface it is. I believe it is the "number of holes" the surface has.
I found something, but really is talking about polyhedrons (special type of surface where p = 0). https://www.wolframalpha.com/input/...rawformassumption={"MC",+""}+->+{"MathWorld"}

 

[Serious Max]

Yes, p is the number of holes, for sphere p=0, for torus p=1, for the shape they give p=2.

But the problem I have is I don't understand what they do in the highlighted step to arrive at the formula.

They go something like:
1) For a regular sphere ##V-E+F=2##
2) Now we have a sphere with ##2p## holes
3) Therefore ##V-E+F=2-2p##

It is not at all obvious to me what's happened in between 2 and 3.

 

[Serious Max]

Ah, I think I got it. In proving ##V-E+F=2##, we first proved that if we removed one face from a polyhedron we could prove that ##V-E+F=1##, and then adding the removed face we'd get 2: ##V-E+F=1+1##. Then this result was shown to apply to a sphere, which is just an inflated polyhedron in this context.

So a hole in a sphere corresponds to one face, in this case we transformed our initial shape into a sphere with ##2p## holes, i.e. with ##2p## faces removed. We know for a sphere with no faces removed ##V-E+F=2##, so for a sphere with ##2p## faces removed the right-hand side should be ##2p## less, since ##F## is ##2p## less.

Should make sense for those who've read the book or correct me if I'm wrong.

 

[mathwonk]

that's exactly right. cutting open a surface with p holes along each hole, adds p edges, but gives you a sphere from which p faces have been subtracted and 2p edges have been added. hence subtracting p from the result X for our surface, gives us 3p less than the result for a sphere, so X-p = 2-3p, so X = 2-2p.

 

[lavinia]

Ah, I think I got it. In proving ##V-E+F=2##, we first proved that if we removed one face from a polyhedron we could prove that ##V-E+F=1##, and then adding the removed face we'd get 2: ##V-E+F=1+1##. Then this result was shown to apply to a sphere, which is just an inflated polyhedron in this context.

So a hole in a sphere corresponds to one face, in this case we transformed our initial shape into a sphere with ##2p## holes, i.e. with ##2p## faces removed. We know for a sphere with no faces removed ##V-E+F=2##, so for a sphere with ##2p## faces removed the right-hand side should be ##2p## less, since ##F## is ##2p## less.

Should make sense for those who've read the book or correct me if I'm wrong.

What you are saying is true but it does not complete the proof. You need to show that the steps that leave you with a sphere minus 2p holes leave ##F-E+V## unchanged.

Step 1: Deform the surface into a sphere with p handles. Why does this deformation leave ##F-E+V## unchanged?
Step 2: Cut the handles along the curves A,B... Since each curve can be triangulated with 3 edges and three vertices, each cut adds 3 edges and 3 vertices so ##F-E+V## is unchanged.

Now you have a sphere with p cylinders sticking out of it.

Step 3: Here is where the book leaves something out. It says,

"Next, we deform the surface by flattening out the projecting handles, until the resulting surface is simply a sphere from which 2p regions have been removed."

The book assumes without proof that this flattening out doesn't change the count. It implies that it is just another deformation as in step 1 but in fact, it is not. The deformation in step 1 does not flatten anything. It just changes the shape of the faces and edges. None of them are lost so the count is preserved. In this last deformation the entire tube of the cylinder is lost. It is crushed out of existence. So some of the edges and faces in the triangulation of the cylinder are removed. You need to account for these.

An equivalent way of doing this is to lop off all of the cylinders to get a sphere with 2p holes plus p cylinders. This collection of pieces has the same ##V-E+F## as the original surface. Now you need to show that each cylinder has a ##V-E+F## equal to zero.

 

[Serious Max]

Yes, I assume that's because it's a very introductory book. But thank you for pointing it out, had to think for a bit about this one.

Step 1: Deform the surface into a sphere with p handles. Why does this deformation leave F−E+VF−E+VF-E+V unchanged?

Seems like you answered this yourself? Or is there something else?

The deformation in step 1 does not flatten anything. It just changes the shape of the faces and edges. None of them are lost so the count is preserved.

Step 3:
A cylinder has ##V=2n##, ##E=2n+n=3n##, ##F=n##, hence ##V-E+F=0##. Basically each loose edge of the cylinder has as many vertices as it has edges, which cancel out, and the middle portion of the cylinder has as many edges as it has faces, so they cancel out.

 

[lavinia]

Yes, I assume that's because it's a very introductory book. But thank you for pointing it out, had to think for a bit about this one.

Seems like you answered this yourself? Or is there something else?
Step 3:
A cylinder has ##V=2n##, ##E=2n+n=3n##, ##F=n##, hence ##V-E+F=0##. Basically each loose edge of the cylinder has as many vertices as it has edges, which cancel out, and the middle portion of the cylinder has as many edges as it has faces, so they cancel out.

- I did answer it myself. Sorry.I wanted to point out the difference between flattening out and the first deformation. These are completely different.

- It is always true that a flattening out as in this example preserves the count. One does not need to have a surface. It also is true for higher dimensional surfaces - aka n-dimensional manifolds - and more generally, for any space that can be triangulated. Even more generally, the space need not be a polyhedron although one then needs to redefine what is meant by edges, and faces and their higher dimensional analogues.The general statement that I know is formulated using Algebraic Topology. I am willing to go through that with you.

- Another way to do this is to start with a torus and then build up surfaces with more handles by successively adding tori. For instance to make a surface with two handles, start with two tori and remove the face of a triangle from each. Then glue the two together along the edges of the two empty triangles. The formula can then be demonstrated by induction.

 

[Serious Max]

The general statement that I know is formulated using Algebraic Topology. I am willing to go through that with you.

I guess that wouldn't hurt.

 

[mathwonk]

A nice approach I like is to equate the euler characteriatic with the alternating sum of the number of max, min, and saddle points, or pits + peaks-passes.

If you turn a surface with p holes on end you see there is one max, one min, and 2p saddle points. Now just convince yourself that deforming the surface does not change that number. e.g. raising up a new (local) max also introduces exactly one saddle point so adds one and subtracts one from the result.

To equate it with the triangulation method, just take a triangulated surface and push up and down on it so that every vertex becomes a local max, every face has a local min at its center, and every edge has a saddle point at its center.

 

[lavinia]

Mathwonk's point in post #10 and also the point in your book, is that ##V-E-F## is always the same no matter how the surface is triangulated. In Mathwonk's method one always gets the sum of the local maxes and mins minus the number of saddles. In your book's way of doing it one always gets the sphere minus the same number of holes.

This means that the Euler characteristic is an invariant of the surface itself and is not dependent on how it is triangulated. The triangulation just gives one method of calculating it.

The book also brushes over this point because it assumes that the curves A,B... are covered by edges in the triangulation. But it is possible to triangulate with these curves lying partially inside faces.

In the method of gluing together successive tori, one still needs to know that no matter how the torus is triangulated, the count is always zero. Then independence of the triangulation follows.

There are many proofs that the Euler characteristic is an invariant. This is true of any polyhedron.