## The chromatic number of the plane, part 2: lower bounds

In a previous post I explained the Hadwiger-Nelson problem—to determine the chromatic number of the plane—and I claimed that we now know the answer is either 5, 6, or 7. In the following few posts I want to explain how we know this! The proofs are not too hard to understand (with the possible exception of Aubrey de Grey’s new proof, but I’ll get to that later) and they are really cool.

In this post I’ll talk about lower bounds. That is, how do we know that the chromatic number of the plane (CNP) is at least 5?

## CNP > 1

Let’s start with a smaller number. Can we argue that the CNP is at least 2? Sure, this is easy: if the entire plane is colored with only one color, then obviously any two points separated by a distance of 1 unit have the same color. So one color is not enough, and hence CNP $\geq 2$.

## CNP > 2

Is two colors enough? If you play around with it a bit, most likely you will quickly get an intuition that this won’t work either. Here’s an intuitive argument why we need at least three colors. Suppose some region of the plane is colored, say, blue.

(It doesn’t have to be a circle, but let’s just think about a circular-ish region for simplicity’s sake.) The diameter of the region has to be smaller than $1$, because otherwise there would be two points within the region that are $1$ unit apart. But that means the entire area outside the region has to be, say, red.

But if the diameter of the blue region is less than 1 and the area outside the region is all red, then intuitively, we can find two red points which are a distance of 1 apart, as illustrated in the above picture.

This isn’t really a formal argument at all (what about really weirdly-shaped regions)? But we can give a different argument which is completely formal yet still quite simple. Consider the graph consisting of three vertices connected in a triangle.

As mentioned in my previous post, the chromatic number of this graph is $3$—each vertex needs its own color since each vertex is adjacent to both other vertices. But this actually tells us something about the plane, since this graph can be drawn as an equilateral triangle with all its edges having length $1$. If we pick any three points in the plane arranged in a unit equilateral triangle, they must all have different colors; hence at least three colors are needed to color the plane.

## CNP > 3

Generalizing from the previous proof, we can see a general plan of attack to prove that three colors are not enough to color the plane. We want to find some graph $G$ such that

• the chromatic number of $G$ is $4$, and
• $G$ is a unit-distance graph, that is, it can be drawn in such a way that every edge has length $1$.

If we can find such a graph, it immediately shows that we need at least four colors for the plane, since we can pick a set of points in the plane in exactly the same configuration as the vertices of the graph; since the vertices of the graph need four colors, so do the chosen points of the plane.

Well, such a graph $G$ exists! In fact, there are many such graphs, but the simplest one is called the Moser spindle. (It’s named after William and Leo Moser, who published it in 1961; I have no idea why it is called a “spindle”.) Here it is:

As you can see, it’s made of four equilateral triangles, with two pairs glued together along an edge, a shared vertex at the top, and an extra edge of length exactly $1$ connecting the bottom vertices. You can check for yourself that this is a unit-distance graph. But what is its chromatic number? Well, suppose we try coloring it with only three colors. Suppose the top vertex is red. That vertex is connected to two equilateral triangles; of course the other two vertices of each triangle must have the other two colors (say, green and blue). Then the two bottom vertices must again be colored red, since each is part of an equilateral triangle with one green and one blue vertex. But this isn’t allowed: the two bottom vertices are connected by an edge, so they can’t be the same color.

So, three colors aren’t enough to color the Moser spindle, and hence three colors aren’t enough to color the plane either!

## CNP > 4?

So the above proofs are not that hard to grasp, and we have known that CNP > 3 since 1961 or so. So how did Aubrey de Grey prove that CNP > 4, and why did it take almost 60 years to go from 3 to 4? I’ll explain that in my next post!

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 3 Responses to The chromatic number of the plane, part 2: lower bounds

1. Eric Burgess says:

A tighter argument for n>2: let the origin be colored blue; then the unit circle is not blue. But we can find two points on the unit circle at unit distance from one another, therefore there are at least two not-blue colors.

• Eric Burgess says:

Never mind, I didn’t read far enough! The three-point graph proof is even simpler.