\documentclass{article}

\title{All Convex Polyhedra are Homeotopic to the Sphere}
\author{Beau Dabbs}
\date{10-12-07}
\begin{document}
\maketitle


\begin{center}
\section*{Definitions and Assumptions}
\end{center}
Define the following characteristics to be:
$\newline$
V(K) = the number of vertices within a graph K$\newline$
E(K) = the number of edges within a graph K$\newline$
F(K) = the number of faces a graph K separates the plane or surface into$\newline$
Definition:  A graph G is a \textit{tree} if G is connected and has no simple cycles, i.e. a cycle with no repeated vertices besides the starting and ending vertex.$\newline$
$\newline$
Assumptions:$\newline$
Jordan Curve Theorem: Any simple closed curve separates the space it's contained within into two distinct parts
\section{Lemma: Euler's Theorem in 2 Dimensions}

Definition:  The Euler characteristic for any graph K in 2 dimensions is defined by:
$\newline$
$$
\chi(K)=V(K)-E(K)
$$
\textbf{Claim 1}: $\forall$ connected graphs K $\chi$(K)$\leq$1
$\newline$
\textbf{Claim 2}: $\chi$(K)=1 $\Leftrightarrow$ K is a \textit{tree}
$\newline$
\textbf{Proof of Claim 1}: Define $G_{k}$ to be a connected graph with k edges. For $G_{0}$ there are two possible graphs:$\newline$
The graph consisting of a single vertex satisfying:
$$
\chi(G_{0})=1-0=1
$$
And the Grapch consisting of no vertices satisfying:
$$
\chi(G_{0})=0-0=0
$$
Thus for k=0, $\chi$($G_{k}$)$\leq$1.  Now we will induct on the number of edges in order to prove Claim 1.
$\newline$
Assume that $\chi$($G_{n}$)$\leq$1 with V($G_{n}$)=$l$ and E($G_{n}$)=n.  In order to create any graph $G_{n+1}$ there are only two possible methods for adding an additional edge:$\newline$
(1) Add an edge between $v_{i}$ and $v_{j}$ with $v_{i},v_{j}$ $\in$ $G_{n}$.  Thus:
$$
V(G_{n+1})=V(G_{n})=l
$$
$$
E(G_{n+1})=E(G_{n})+1=n+1
$$
$$
\chi(G_{n+1})=V(G_{n+1})-E(G_{n+1})=l-(n+1)=(l-n)-1=\chi(G_{n})-1<\chi(G_{n})
$$
(2) Add an edge between $v_{i}$ and $v_{j}$ with $v_i$ $\in$ $G_n$ and $v_j$ $\notin$ $G_n$.  Thus:
$$
V(G_{n+1})=V(G_{n})+1=l+1
$$
$$
E(G_{n+1})=E(G_{n})+1=n+1
$$
$$
\chi(G_{n+1})=V(G_{n+1})-E(G_{n+1})=(l+1)-(n+1)=(l-n)+1-1=\chi(G_{n})
$$
Hence $\forall$k,  $\chi$($G_k$)$\leq$1//
$\newline$
\textbf{Lemma}: Adding an edge to a connected graph G using method (1) creates a simple cycle.
$\newline$
Proof:  G is connected $\Rightarrow$ $\exists$ a path from $v_i$ to $v_j$ s.t. no vertex is repeated.  If a new edge is added connecting $v_i$ to $v_j$, this edge will add allow the path to extend from $v_i$ to $v_i$ without crossing any other vertices twice.//
$\newline$
\textbf{Proof of claim 2}:  We shall assume that $G_0$ with 0 vertices is not a tree, though it does make intuitive sense that the nonexistent vertices could have infinite cycles between nothingness.  Thus our base case shall be $G_0$ with V($G_0$)=1, and hence as shown above $\chi$($G_0$)=1.
$\newline$
Assume $G_k$ is a tree.$\Rightarrow$ $G_k$ is generated from $G_0$ solely by method(2) by the Lemma.  Thus:
$$
\chi(G_k)=\chi(G_{k-1})=...=\chi(G_1)=\chi(G_0)=1
$$
Since method (1) preserves the Euler Characteristic.
$\newline$
 
$\newline$
Now assume $\chi$($G_k$)=1.  We shall also assume that $G_k$ is generated by at least one edge of type (1).
$$
\Rightarrow \chi(G_k)=\chi(G_{k-1})=...\chi(G_{j+1}<\chi(G_j)=...=\chi(G')=\chi(G)=1
$$
$$
\Rightarrow \chi(G_k)<1
$$
A contradiction of our initial assumption, thus all edges of $G_k$ must be added by method (2)$\Rightarrow$ $G_k$ is a tree.//












\section{Proof of Euler's Theorem in 3 Dimensions}










\textbf{Definition}:  For any surface or solid K in 3 Dimensions the euler characteristic $\chi$ of K is denoted:
$$
\chi(K)=V(K)-E(K)+F(K)
$$
\textbf{Definition}:  Let a net on a convex surface be defined as a graph of connected vertices and edges separating the surface into faces.
$\newline$

\textbf{Claim}:  $\forall$ nets P on a convex surface $\chi$(N)=2$\newline$
\textbf{Proof}:  Let $P_k$ be a net with k edges, and $P_0$ be a net on a surface consisting of a single vertex be the smallest possible net on a surface.  Thus:
$$
\chi(P_0)=1-0+1=2
$$
Thus $\chi$($P_k$)=2 for k=0.  We will now procede by induction on the edges.$\newline$
Assume $P_n$ is a net with V($P_n$)=$l$, E($P_n$)=$n$, and F($P_n$)=$m$ and that $\chi$($P_n$)=2.$\newline$
As in the previous section, in order to create $P_{n+1}$ from $P_n$ there are two methods for adding an additional edge, with the same consequences for the vertices and edges as above, with the addition that for method (1) creates a simple closed curve, and thus separates 1 of the existing faces into 2 distinct faces, though method (2) creates no additional faces.  Thus the following is obtained by each method:
$\newline$
Method (1):
$\newline$
$$
V(P_{n+1})=V(P_{n})=l
$$
$$
E(P_{n+1})=E(P_{n})+1=n+1
$$
$$
F(P_{n+1})=F(P_{n})+1=m+1
$$
$$
\chi(P_{n+1})=V(P_{n+1})-E(P_{n+1})+F(P_{n+1})=l-(n+1)+(n+1)=(l-n+m)-1+1=\chi(P_{n})
$$
Method (2):
$\newline$
$$
V(P_{n+1})=V(P_{n})+1=l+1
$$
$$
E(G_{n+1})=E(G_{n})+1=n+1
$$
$$
F(P_{n+1})=F(P_{n})=m
$$
$$
\chi(P_{n+1})=V(P_{n+1})-E(P_{n+1})+F(P_{n+1})=(l+1)-(n+1)+m=(l-n+m)+1-1=\chi(P_{n})
$$
Using our inductive assumption we find that $\chi(P_{n+1})$=$\chi(P_{n})$=2.$\newline$
Thus $\chi$(P)=2 for all nets on convex surfaces, and hence $\chi$(P)=2 for all $\newline$polyhedra, a subset of those nets.//








\section{Euler Characteristic an Homeotopy to the Sphere}
For this section we will be showing the equivalency of 3 statements:
$\newline$
(1)$\chi$(K)=2
$\newline$
(2)Any embedded loop in K separates K into two distinct parts
$\newline$
(3)K is homeotopic to the sphere
$\newline$
Proof:
\textbf{(3)$\Rightarrow$(2)}:  First by the Jordan Curve Theorem we can say already that (3)$\Rightarrow$(2) since the sphere has one surface, any closed curve would have to separate it.
$\newline$
Now we will show that (2)$\Rightarrow$(1)$\Rightarrow$(3) to complete the proof.  In doing this we will use the lemma stating that the euler characteristic of any tree is equal to 1.
$\newline$
To do this we will consider a closed combinatorial surface K, essentially a triangulation of the surface, or a net composed solely of triangles.
$\newline$
\textbf{(2)$\Rightarrow$(1)}:  First create a maximal tree T, in K.  Since the tree is maximal $\Rightarrow$ 
$$
V(T)=V(K)
$$
$\newline$
Then create a dual graph $\Gamma$ with each face of K being a vertext of $\Gamma$, and adding an edge between every pair of vertices $v_i, v_j\in\Gamma$ s.t. the faces $f_i, f_j\in K$ corresponding to $v_i, v_j$ satisfy:
$$
f_{i}\cap f_{j}\notin T
$$
Thus $\Gamma$ will have the property 
$$
F(K)=V(\Gamma)
$$
and the two graphs together will satisfy:
$$
E(K)=E(T)+E(\Gamma)
$$
Thus combining these results we obtain:
$$
\chi(K)=V(K)-E(K)+F(K)=V(T)-[E(T)+E(\Gamma)]+V(\Gamma)=
$$
$$
[V(T)-E(T)]+[V(\Gamma)-E(\Gamma)]=\chi(T)+\chi(\Gamma)
$$
By the Lemma we know that $\chi$(T)=1 since T is a tree, so to show that $\chi$(K)=2, we simply need to show that $\chi$($\Gamma$)=1, or that $\Gamma$ is a tree.
$\newline$
Suppose $\Gamma$ is a tree$\newline$
$\Rightarrow$ $\Gamma$ has an embedded loop$\newline$
$\Rightarrow$ $\Gamma$ separates K, and hence that T isn't connected, but T is connected, a contradiction.
Thus $\Gamma$ must be a tree, and $\chi$(K)=2
\textbf{(1)$\Rightarrow$(3)}:  Now K itself is composed of a neighborhood around T and a neighborhood around $\Gamma$, which we have just shown are both trees, and the neighborhood around a tree is homeomorphic to a disk.  If we take the neighborhood around T and $\Gamma$ and combine them together we obtain an area homeomorphic to a sphere.

\section{All Convex Polyhedra are Homeomorphic to the sphere}
Proof:
In the previous two sections we found that:
$\newline$
1)$\chi$(K)=2 $\forall$ convex polyhedra
$\newline$
2)$\chi$(K)=2$\Leftrightarrow$ All embedded loops in K separate$\Leftrightarrow$ K is homeotopic to the sphere.
$\newline$
Thus we can now say that all convex polyhedra are homeomorphic to the sphere//


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