On the Chromatic Number of \(\mathbb {R}^4\)
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Abstract
The lower bound for the chromatic number of \(\mathbb {R}^4\) is improved from \(7\) to \(9\). Three graphs with unit distance embeddings in \(\mathbb {R}^4\) are described. The first is a \(7\)chromatic graph of order \(14\) whose chromatic number can be verified by inspection. The second is an \(8\)chromatic graph of order \(26\). In this case the chromatic number can be verified quickly by a simple computer program. The third graph is a \(9\)chromatic graph of order \(65\) for which computer verification takes about one minute.
Keywords
Unit distance graph Chromatic number1 Introduction
The chromatic number of the \(n\)dimensional Euclidean space, denoted \(\chi (\mathbb {R}^n)\), is the minimum number of colors with which the entire space \(\mathbb {R}^n\) can be painted so that every two points at distance one have different colors. The problem has a long and celebrated history and is one of the most famous unsolved problems in discrete geometry.
A unit distance graph in \(\mathbb {R}^n\) is a graph whose vertices are points in \(\mathbb {R}^n\) and whose edges are the pairs of vertices that are at distance one from each other. Assuming the axiom of choice, a result of de Bruijn and Erdős [5] implies that the chromatic number of \(\mathbb {R}^n\) is equal to the maximum chromatic number of any finite unit distance graph in \(\mathbb {R}^n\).
The structure of \(\mathbb {R}^n\) does not change under scaling. Therefore, instead of forbidding two points which are at distance one from having the same color, we can equivalently use any \(r>0\) as the forbidden distance. This simple observation will be useful later, since the unit distance graphs we will construct will have simpler vertex coordinates if regarded as \(r\)distance graphs for an appropriate choice of \(r\).
2 Background
It is almost immediate that \(\chi (\mathbb {R})=2\), since we can \(2\)color the real line by assigning, for each integer \(k\), the color red to the interval \([2k1,2k)\) and blue to the interval \([2k,2k+1)\). Under this coloring, no pair of points at distance one have the same color.
The situation becomes more complex in higher dimensions. Although the origins of the problem are unclear, Edward Nelson is usually credited with having been the first who raised the question for the plane in 1950. A few years earlier, Hadwiger [7] published a related result, showing that any cover of the plane by five congruent closed sets contains a unit distance in one of the sets. Soifer [14] offers an extensive history of the problem.
In this note, we improve the lower bound for \(\chi (\mathbb {R}^4)\). As mentioned earlier, Cantwell [2] proved that \(\chi (\mathbb {R}^4)\ge 7\). His proof is quite elaborate and hinges on the existence of a finite order unit distance graph of chromatic number \(7\) that can be embedded in \(\mathbb {R}^4\). Cantwell did not aim to optimize the size of his unit distance graph, which has more than \(2000\) vertices.
In 2008, Cibulka [3] used some of Cantwell’s ideas to find a much smaller unit distance graph of chromatic number \(7\) that can be embedded in \(\mathbb {R}^4\): his graph has \(32\) vertices. Cibulka did not provide an explicit embedding.
In 2006, Ivanov [8], unaware of Cantwell’s result, again proved the bound \(\chi (\mathbb {R}^4)\ge 7\) using a completely new construction; Ivanov’s graph has \(33\) vertices.

a \(4\sqrt{2}\)distance graph \(G\) with \(14\) vertices and chromatic number \(\chi (G)=7\),

a \(4\sqrt{2}\)distance graph \(H\) with \(26\) vertices and chromatic number \(\chi (H)=8\),

a \(4\sqrt{2}\)distance graph \(K\) with \(65\) vertices and chromatic number \(\chi (K)=9\).
3 \(\chi (\mathbb {R}^4) \ge 7\)
At its core, our method uses the ideas of Cantwell and Cibulka.
Theorem 3.1
Proof
Suppose that \(G\) is \(6\)colorable. Since \(\alpha (G)=3\), it follows that no four vertices can receive the same color. There are \(14\) vertices and \(6\) colors, so by the pigeonhole principle there are two colors, each of which has to be used for an (independent) set of three vertices. However, this is impossible since every pair of maximum independent sets have nonempty intersection. This proves that \(\chi (G)\ge 7\).
Vertex  \(v_1\)  \(v_2\)  \(v_3\)  \(v_4\)  \(v_5\)  \(v_6\)  \(v_7\)  \(v_8\)  \(v_9\)  \(v_{10}\)  \(v_{11}\)  \(v_{12}\)  \(v_{13}\)  \(v_{14}\) 

Color  1  2  3  3  4  2  5  1  6  7  4  7  4  5 
\(\square \)
4 \(\chi (\mathbb {R}^4) \ge 8\)
The above construction suggests that there exist \(4\sqrt{2}\) distance graphs whose vertices are points in \(\mathbb {R}^4\) with coordinates in the quadratic extension field \(\mathbb {Q}(\sqrt{5})\) and having high chromatic number. This intuition turns out to be correct as we will next present a relatively small unit distance graph which improves the lower bound for \(\chi (\mathbb {R}^4)\) from \(7\) to \(8\).
Theorem 4.1
There exists a graph \(H\) with chromatic number \(8\) which can be represented as a unit distance graph in \(\mathbb {R}^4\).
The theorem follows from the construction of a graph of order \(26\), given below. We begin by first building a graph with \(35\) vertices and an automorphism group of order \(24\). Then our graph \(H\) is obtained by deleting nine of the \(35\) vertices.
Points used in the construction of the graph in Theorem 4.1
Type 1  \( 1 + 1\cdot \sqrt{5} \),  \( 1 + 1\cdot \sqrt{5} \),  \( 1 + 1\cdot \sqrt{5} \),  \( 1 + 1\cdot \sqrt{5} \)  1 
Type 2  \( 4 + 0\cdot \sqrt{5} \),  \(0 + 0\cdot \sqrt{5} \),  \( 0+0\cdot \sqrt{5} \),  \( 0+0\cdot \sqrt{5} \)  4 
Type 3  \(4 + 0\cdot \sqrt{5} \),  \(0+0\cdot \sqrt{5} \),  \( 0+0\cdot \sqrt{5} \),  \( 0+0\cdot \sqrt{5} \)  4 
Type 4  \( 0+2\cdot \sqrt{5} \),  \(2 + 0\cdot \sqrt{5} \),  \(2 + 0\cdot \sqrt{5} \),  \(2 + 0\cdot \sqrt{5} \)  4 
Type 5  \( 1  1\cdot \sqrt{5} \),  \(1 + 1\cdot \sqrt{5} \),  \(1 + 1\cdot \sqrt{5} \),  \(1 + 1\cdot \sqrt{5} \)  4 
Type 6  \( 0+2\cdot \sqrt{5} \),  \( 2+ 0\cdot \sqrt{5} \),  \( 2 + 0\cdot \sqrt{5} \),  \( 2 + 0\cdot \sqrt{5} \)  12 
Type 7  \( 1 + 1\cdot \sqrt{5} \),  \( 1 + 1\cdot \sqrt{5} \),  \(1  1\cdot \sqrt{5} \),  \(1  1\cdot \sqrt{5} \)  6 
Total  35 
In order to decrease the time it takes to complete an exhaustive search for a \(7\)coloring, one can employ the usual degree saturation procedure. In this case, an ordering of the vertices is generated by picking vertices adjacent to the maximum number of previously selected vertices, and in case of ties, by picking the vertex of largest degree. This idea was first used in the context of chromatic numbers by Breláz [1]. Taking advantage of this, a \(2013\) era computer can easily complete the exhaustive search for a \(7\)coloring of the \(26\)vertex graph in a fraction of a second.
The adjacency matrix of this graph, along with a computer program for verifying its chromatic number, can be found at the web site [15].
5 \(\chi (\mathbb {R}^4) \ge 9\)
Points used in the construction of the graph in Theorem 5.1
\( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\)  1 
\( 4+ 0 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\)  4 
\( 4+ 0 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\)  4 
\( 0+ 2 \cdot \sqrt{5}\),  \( 2+ 0 \cdot \sqrt{5}\),  \( 2+ 0 \cdot \sqrt{5}\),  \( 2+ 0 \cdot \sqrt{5}\)  4 
\( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\)  4 
\( 1 1 \cdot \sqrt{5}\),  \( 1 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\)  6 
\( 1+ 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 3+ 1 \cdot \sqrt{5}\),  \( 3+ 1 \cdot \sqrt{5}\)  6 
\( 2+ 0 \cdot \sqrt{5}\),  \( 2+ 0 \cdot \sqrt{5}\),  \( 2+ 0 \cdot \sqrt{5}\),  \( 0+ 2 \cdot \sqrt{5}\)  12 
\( 0+ 0 \cdot \sqrt{5}\),  \( 2+ 2 \cdot \sqrt{5}\),  \( 2+ 2 \cdot \sqrt{5}\),  \( 0+ 0 \cdot \sqrt{5}\)  12 
\( 1+ 1 \cdot \sqrt{5}\),  \( 3 1 \cdot \sqrt{5}\),  \( 1+ 1 \cdot \sqrt{5}\),  \( 3+ 1 \cdot \sqrt{5}\)  12 
65 
Using the programs provided at [15], one can verify, in under one minute, that there are no proper \(8\)colorings of \(K\), thereby establishing that the chromatic number is at least \(9\). A proper coloring of \(K\) using \(9\) colors is available at [15]. Hence we have the following theorem.
Theorem 5.1
The graph \(K\) has chromatic number \(9\) and can be represented as a unit distance graph in \(\mathbb {R}^4\).
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