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Euler's formula?

Joppy

Well-known member
MHB Math Helper
Mar 17, 2016
256
Hi, I have been working through a book on algebraic topology (an introduction) and am (embarrassingly) stuck on the following exercise.

Given a polyhedron (doesn't have to be regular), let $F_n$ be the number of $n$-gon faces, and let $V_n$ be the number of vertexes at which exactly $n$ edges meet. Verify the following (there are several similar relations, but I'll just give one):

$(2V_3 + 2V_4 + 2V_5 + ...) - (F_3 + 2F_4 + 3F_5 + ...) = 4$.

I don't understand what we mean when we are given a polyhedron and then we sum an infinite number of vertices and faces? Would someone mind providing insight into what these sums are representing? :)
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi Joppy ,

In those sums, all but finitely many terms are zero. So the sums are really finite sums.
 

Joppy

Well-known member
MHB Math Helper
Mar 17, 2016
256
Hi Joppy ,

In those sums, all but finitely many terms are zero. So the sums are really finite sums.
Thanks Euge.

Do you have any ideas on how to verify the expression?

Because $F_n$ and $V_n$ represent an arbitrary number of $n$-gon faces and vertices (at which $n$ edges meet) I'm quite confused as to how we even work with these expressions.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
The way the question is phrased, the author must be using a more general definition of polyhedron. What is the definition of polyhedron in your text?

Euler's polyhedral formula states that for a polyhedron with $V$ vertices, $F$ faces, and $E$ edges, $V - E + F = 2$.
 

Joppy

Well-known member
MHB Math Helper
Mar 17, 2016
256
Thanks!

The way the question is phrased, the author must be using a more general definition of polyhedron. What is the definition of polyhedron in your text?
A polyhedron is a complex that is topologically equivalent to a sphere. A complex is built from cells (topologically equivalent to a disk) by 'gluing' the edges of cells together.

Euler's polyhedral formula states that for a polyhedron with $V$ vertices, $F$ faces, and $E$ edges, $V - E + F = 2$.
I was expecting to apply this.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,648
Leeds, UK
The total number of vertices in the polyhedron is $V_3 + V_4 + V_5 + \ldots$. The total number of faces is $F_3 + F_4 + F_5 + \ldots$. To apply Euler's formula, you then need to know how many edges there are.

Notice that an $n$-gon face has $n$ edges, and that each edge in the polyhedron is shared between two faces.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
A polyhedron is a complex that is topologically equivalent to a sphere. A complex is built from cells (topologically equivalent to a disk) by 'gluing' the edges of cells together.

OK. Then show that $2E = 3F_3 + 4F_4 + 5F_5 + \cdots$. Using $2V - 2E + 2F = 4$, obtain the result.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The simplest example is a tetrahedron. It has 4 faces, 4 vertices, and 6 edges. 4- 6- 4= 2.

There is also "Euler's formula" on a surface such as a plane. With F, E, and V the faces, edges, and vertices of a polygon, F- E+ V= 1. For example, a triangle has 1 face, 3 edges, and 3 vertices: 1- 3+ 3= 1. Given Euler's formula for a surface, you can then prove Euler's formula for three dimensions, as Joppy suggested. In the surface of a sphere we would need to treat the "outside" as a face so that "F" is increased by 1 and instead of F- E+ V= 1, we have F- E+ V= 2.