Recall that a graph is *planar* if it can be drawn on the plane without edge crossings. In the last post, I mentioned Kuratowski’s characterization of planar graphs:

**Kuratowski’s Theorem.** A graph is planar if and only if it has neither nor as a minor:

(Here refers to the complete graph on 5 vertices, and is the complete bipartite graph on two sets of three vertices.)

I don’t know a really elegant proof of this theorem, but I do have some sense of why these two graphs are the right forbidden minors.

** and the Jordan Curve Theorem**

The basic topological property of the plane is the Jordan Curve Theorem — any closed loop separates the plane into two components. This has some interesting consequences for graph drawing.

The above graph consists of a single loop with two red edges attached. If we wish to avoid crossings, the red edges must be drawn on opposite sides of the loop. By the Jordan Curve Theorem, it follows that the red edges are in different components of the complement of the loop. In particular, the blue edge in the following graph always crosses the loop:

This graph is exactly . As you can see, the statement “ is not planar” is a graph-theoretic version of the Jordan Curve Theorem.

**The ****Möbius Strip**

One surface for which the Jordan Curve Theorem fails is the Möbius strip. This is the surface obtained by gluing joining two ends of a rectangle with a half-twist:

If you cut the Möbius Strip along its center loop, you get a single twisted cylinder:

(There is a nice video of this on YouTube.) In particular, cutting the Möbius strip along the center loop doesn’t seperate it into two components. This makes it possible to embed in the Möbius strip without crossings:

**What About ?**

If represents the Jordan Curve Theorem, what is the role of the other forbidden minor, ?

It turns out that is somewhat less interesting than . To understand why, let us examine the proof that isn’t planar.

**Theorem.** is not planar.

*Proof:* This is a proof by contradiction. Suppose that you *could* draw on the plane without crossings, and consider a neighborhood of one of the vertices:

We don’t know much about how the graph is drawn, but we do know that there are four edges coming out of this vertex. Consider what happens if we “split” the vertex in two:

This result is new graph drawn on the plane, with one more vertex and one more edge than . This new graph is isomorphic to the middle graph shown below:

The contradiciton is that the new graph contains a copy of . Since can’t be drawn without crossings, neither can the splitting, and so neither can .

As you can see, is essentially just a tagalong to . Its failure to be planar doesn’t really convey any new topology.

Before ending this post, I should mention that can be drawn on the Möbius strip fairly easily. (Try this for yourself.) This leaves us with an important question: is there any graph that *can’t* be drawn on a Möbius strip?

July 6, 2007 at 3:31 pm |

A Möbius strip satisfies the six-color theorem (reference: Mathworld), so cannot be drawn on a Möbius strip.

July 6, 2007 at 6:35 pm |

That is true, cannot be drawn on a Möbius strip, and neither can any other graph with chromatic number 7.

I haven’t talked about graph coloring in any of my posts, but it is known that every surface possesses a “color theorem” analogous to the four color theorem for the plane. For example, graphs on the Möbius strip require up to 6 colors, and graphs on the torus require up to 7 colors. (Click here for the general formula.) This provides a large class of graphs that cannot be drawn on any given surface.

April 12, 2009 at 5:42 am |

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