# christhomson/lecture-notes

MATH 239: added March 18, 2013 lecture.

 @@ -3691,7 +3691,99 @@ \end{tikzpicture} \end{center} - There are also two more graphs ($k = 3, l = 5$ and $k = 5, l = 3$), which we won't show here for simplicity. + There are also two more graphs ($k = 3, l = 5$ and $k = 5, l = 3$), which we won't show here for simplicity. + + % Eric Katz gave this lecture. + \subsubsection{Review of Planar Graphs} \lecture{March 18, 2013} + \begin{itemize} + \item $\deg(f)$ is the length of the boundary walk about the face $f$. + \item Handshake Lemma: $$\sum_{v \in V} \deg(v) = 2q$$ + \item Dual Handshake Lemma: $$\sum_{f \in \text{faces}} \deg(f) = 2q$$ + \item Euler's Formula: $$p - q + s = 1 + c$$ (where $p$ is the number of vertices, $q$ is the number of edges, $s$ is the number of faces (earlier, we denoted this as $f$), and $c$ is the number of components.) + \end{itemize} + + \subsubsection{Planar Solids} + There are solids that correspond to the five platonic graphs. What do they have in common? Every vertex has the same degree \emph{and} every face has the same degree. + \\ \\ + We'll now finally define planar graphs formally. + \begin{defn} + A \textbf{planar graph} is a graph such that there are integers $k, l \ge 3$ such that every vertex has degree $k$ and every face has degree $l$. + \end{defn} + + Does the data of $(k, l)$ determine the graph? Yes! When drawing these graphs, they are unique because there are no ambiguous decisions to be made. + \\ \\ + Which values of $(k, l)$ are possible? If we try drawing the graphs for $(4,4)$ and $(5,4)$, we'll see that as we're constructing them, they'd each need to be infinite in size to satisfy the $k$ and $l$ constraints. Since these graphs cannot be finite, we say $(4,4)$ and $(5,4)$ are two particular instances of $(k, l)$ for which the constraints are not satisfied. + \\ \\ + \underline{Idea}: use the facts about planar graphs to constrain $(k, l)$. + + \begin{itemize} + \item By the handshake lemma, $\sum \deg(v) = kp = 2q$. + \item By the dual handshake lemma, $\sum \deg(f) = ls = 2q$. This implies that $p = \frac{2q}{k} = s = \frac{2q}{l}$. This is known as the duality of graphs. + \item By Euler's formula, $p - q + s = 2$. + \end{itemize} + + This gives us: + \begin{align*} + \frac{2q}{k} - q + \frac{2q}{l} = 2 &> 0 \\ + \frac{2}{k} - 1 + \frac{2}{l} = \frac{2}{q} &> 0 \\ + \frac{2}{k} + \frac{2}{l} &> 1 \\ + 2l + 2k &> kl \\ + kl - 2k - 2l &< 0 \\ + kl - 2k - 2l + 4 - 4 &< 0 \\ + (k-2)(l-2) &< 4 + \end{align*} + + So, we know that $k \ge 3$ and $l \ge 3$. This gives us: + \begin{align*} + &(k-2)(l-2) < 4 \\ + \implies& k - 2, l - 2 \ge 1 \text{ and } k-2, l-2 < 4 \\ + \implies& k, l \le 5 + \end{align*} + + All of this gives us: + \begin{itemize} + \item If $k = 3$, then $l$ can be 3, 4, or 5. + \item If $k = 4$, then $l$ can only be 3. + \item If $k = 5$, then $l$ can only be 3. + \end{itemize} + + \begin{theorem} + If $G$ is a platonic graph, then $(k, l)$ must be one of (3, 3), (3, 4), (3, 5), (4, 3), or (5, 3). + \end{theorem} + + Does each of these possibilities of $(k, l)$ occur for a platonic graph? That is, we know these are all \emph{possibilities}, but are they all \emph{actually} possible? For any choice of $(k, l)$, how many platonic graphs are there? + \\ \\ + \underline{Claim}: each of the five possibilities for $(k, l)$ can be achieved by a platonic graph. + \begin{proof} + Just use the five platonic solids that we know. % This is literally the entire proof as written on the board. + \begin{itemize} + \item (3, 3): tetrahedron. + \item (3, 4): cube. + \item (3, 5): dodecahedron. + \item (4, 3): octahedron. + \item (5, 3): icosahedron. + \end{itemize} + \end{proof} + + \underline{Claim}: the only platonic graphs are the five coming from platonic solids. + \begin{proof} + Just try to draw six platonic graphs. As you keep drawing, the graph will be uniquely determined. + \end{proof} + + How many faces, edges, and vertices does each have? + \begin{align*} + q \left( \frac{2}{k} + \frac{2}{l} + 1 \right) &= 2 \\ + p &= \frac{2q}{k} \\ + s &= \frac{2q}{l} \\ \\ + \implies q &= \frac{2}{\frac{2}{k} + \frac{2}{l} - 1} \\ + &= \frac{2kl}{2l + 2k - kl} + \\ \\ + \implies s &= \frac{2q}{l} \\ + &= \frac{4k}{2l + 2k - kl} + \end{align*} + + We can verify this. If $(k, l) = (3, 5)$, then $s = \frac{12}{10 + 6 - 15} = 12$, as expected. + \newpage \section*{Clicker Questions}