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K5 and K3,3 Violate Planar Density in Different Ways by Vicfred

Last updated: Sun 15 September 2024

Euler bounds prove both classical obstruction graphs are nonplanar. This is a compact note, but the quantifiers and hypotheses stay on the page.

Start locally

A planar embedding divides the sphere into vertices, edges, and faces. Euler's relation \(|V|-|E|+|F|=2\) constrains density, while the dual \(G^\ast\) records adjacency of faces.

$$ |E(K_5)|=10,\qquad|E(K_{3,3})|=9 $$

There are two layers here: the object \(\mathsf D\) and the law \(\mathsf C\). Writing them separately makes the direction of \(\Longrightarrow\) visible and keeps an accidental converse from slipping in.

$$ K_5,\ K_{3,3}\ \text{are nonplanar} $$

Compute before generalising

The computation below is not a second theorem. It is a checksum for the definitions and a place to inspect the difficult LaTeX at full size.

$$ \begin{cases}10>3\cdot5-6=9,&K_5,\\9>2\cdot6-4=8,&K_{3,3}\text{ is bipartite and triangle-free}.\end{cases} $$

The global view

The abstraction earns its keep by explaining why the same computation reappears. The notation compresses repeated reasoning without erasing the hypothesis that licenses it.

$$ \begin{aligned} \mathsf{D}\;&:\quad |E(K_5)|=10,\qquad|E(K_{3,3})|=9,\\[5pt] \mathsf{C}\;&:\quad K_5,\ K_{3,3}\ \text{are nonplanar}. \end{aligned} $$

Edge conditions

Planarity is a property of a graph, while a plane graph includes a chosen embedding. The dual depends on that embedding.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] K_5,\ K_{3,3}\ \text{are nonplanar} \end{gathered}} $$

The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.

This article was posted on Sat 02 April 2016. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.