INCREASING CHROMATIC NUMBER AND GIRTH YO JOONG CHOE Abstract. This expository paper deals with some interesting concepts related to the problem of increasing the chromatic number and the girth of a graph. Specifically, the proof of Kneser's conjecture and also related concepts such as Gale's distribution of points on the sphere and the Borsuk graph will be studied in order to show various ways of constructing a graph with infinite chromatic number and girth. Contents 1. Introduction 1 2. Preliminaries 3 3. General position and moment curve kissing lemma 3 4. Gale's distribution of points on the r-sphere 5 5. The Borsuk graph 7 6. The Kneser graph 10 Acknowledgments 13 References 13 1. Introduction In graph theory, many intriguing problems revolve around graph coloring. One common type of question here is about how to reduce the chromatic number of a certain graph. For example, the famous Four (used to be Five) Color Theorem states that you can always color a map with no more than four different colors. Another common type, on the opposite side, is about how to increase the chromatic number. There is an interesting question of this type that we will deal with in this paper: the problem of finding a graph with infinite chromatic number and girth. Definition 1.1 (chromatic number). A k-coloring of a graph G is a labeling of the vertices of G with k colors such that adjacent vertices have different colors. When this is the case, we say that G is k-colorable. The chromatic number χ of a graph G is the minimum value of k such that G is k-colorable. Definition 1.2 (girth). A walk of length k in a graph G is a sequence of vertices v0; v1; : : : ; vk of G such that vi−1 and vi are adjacent for 1 ≤ i ≤ k. A walk is closed if v0 = vk.A cycle is a closed walk without repeated vertices except for v0 = vk. The girth γ of a graph G is the length of its shortest cycle. A graph without a cycle has infinite girth. 1 2 YO JOONG CHOE It is not at all a difficult problem to find a graph satisfying one of the two conditions. For some large n, any complete graph with n vertices has chromatic number of n. Any n-gon has girth of n. But it is not as easy to find a graph satisfying both conditions. A triangle has girth of 3 and requires at least 3 colors. From here, one can increase the girth by simply adding two vertices to make a pentagon. (In fact, any (2k + 1)-gon has chromatic number of 3.) Now, here is a more challenging question: find a graph without a triangle - that is, with girth greater than 3 - and with chromatic number of 4. (Hint: the graph requires at least 11 vertices and has 5-fold rotational symmetry. This graph is referred to as ’Gr¨otzsch graph'.) These mini experiments already hint the existence of a graph with large girth and chromatic number. The proof was given by Paul Erd¨osusing the classic probabilistic method. Theorem 1.3 (Erd¨os,1959). For any positive integers γ and χ, there exists a graph G with girth γ and chromatic number χ. We will not show the proof of this theorem, however, since this paper will focus on introducing significant examples of this result. Mainly, we will study a specific kind of graph with high girth and chromatic number, called the Kneser graph, which was introduced by Martin Kneser. The chromatic number of the Kneser graph, which was conjectured in 1955 and proved much later in 1978, is a great example of the Erd¨ostheorem. Here is a formal definition of the graph: Definition 1.4 (the Kneser graph). For n ≥ 2m + 1, the Kneser graph K(n; m) is the graph with the vertex set as the set of m-subsets of f1; : : : ; ng. Two vertices are adjacent if their corresponding m-subsets are disjoint. Figure 1.1. The Kneser graph K(5,2) and the Petersen graph The most common example is K(5; 2), which is isomorphic (i.e. each vertex in one graph can be mapped to a unique vertex in another graph) to the Petersen graph (Figure 1.1). Ask a seven-year-old kid, and after struggling for a few minutes, the child will tell you that you need at least 3 colors to color this graph properly, and that the shortest cycle is 5. That part is easy, but the proof of the generalized version requires some difficult steps. We will state the theorem here just to clarify that proving this theorem is the goal of this paper. INCREASING CHROMATIC NUMBER AND GIRTH 3 Theorem 1.5 (Kneser's conjecture, 1955) (Lov´asz,1978). The chromatic number of the Kneser graph K(2m + r; m) is r + 2. 2. Preliminaries One important fact about this paper is that many parts of the proofs heavily rely on concepts in linear algebra and related geometry. These may seem totally unrelated at first sight, but they all give important insights to the proof of Kneser's conjecture. Before we move on, there are some preliminary concepts to note about through- out this paper. For statements related to graph theory, all graphs are simple graphs, meaning that we will not be concerned with loops or multiple edges. (This is be- cause such concepts are not important in dealing with chromatic numbers.) Also, all variables (e.g. n, m, r, k, i), unless specified, are nonnegative integers. For statements related to linear algebra, all vectors, unless specified, are column vec- tors. Lastly, this paper contains diagrams containing various different colors in order to illustrate graph coloring more effectively, so please print in color in case you are printing out this document on paper. 3. General position and moment curve kissing lemma We will first introduce the moment curve kissing lemma. This is necessary in proving the next theorem. Definition 3.1 (general position). Let S ⊆ W be a set of points, where W is the linear space of dimension n. S is in general position if any n of the points in S are linearly independent. For example, the vectors (1; 1); (1; 2); (1; 3) are in general position, as any 2 of them are linearly independent. Informally speaking, the concept of general position is the opposite to the Murphy's Law: elements coincide only when necessary. Now, there is a particularly useful instance of a set in general position. Definition 3.2 (the moment curve). The moment curve in Rn is the set 2 n−1 Mn = fmn(α) = (1; α; α ; : : : ; α ) j α 2 Rg: To be clear, the points of the moment curve will be row vectors in this paper. Proposition 3.3. The points of the moment curve are in general position. Proof. Let α1; α2; : : : ; αn be n distinct elements in R. Consider the matrix (mn(α1); mn(α2); : : : ; mn(αn)). This matrix is a Vandermonde matrix, so the determinant Q is 1≤i<j≤n(αj − αi). Because all αi are distinct, the determinant is nonzero. Therefore, the row vectors mn(ai) are linearly independent. With this property established, we can now move on to the actual lemma that uses the concept of the moment curve. 4 YO JOONG CHOE Definition 3.4 (hyperplane and normal vector). In an n-dimensional space, a hyperplane is an (n − 1)-dimensional flat subset. In other words, a hyperplane P in Rn is n T P = fx 2 R j c x = βg where c 2 Rn, c 6= 0, and β 2 R. Here, c is the normal vector of the hyperplane P . A linear hyperplane is a hyperplane crossing the origin. This is precisely when β = 0. d Lemma 3.5 (moment curve kissing lemma). Let d ≥ 1. Let 0 ≤ k ≤ 2 and α1; α2; : : : ; αk 2 R. Then, there exists a linear hyperplane P such that the mo- d+1 ment curve Md+1 ⊂ R lies entirely on one side of P . Here, P \ Md+1 = fmd+1(α1); md+1(α2); : : : ; md+1(αk)g. Proof. A linear hyperplane P is defined precisely by a linear equation cx = 0 where d+1 c = (γ0; γ1; : : : ; γd) 2 R and c 6= 0. (c is a row vector.) We want to find c such that T (1) c · md+1(z) > 0 where z 2 R and z2 = fα1; α2; : : : ; αkg, and T (2) c · md+1(ai) = 0 where i = 1; 2; : : : ; k. Let f be a polynomial of degree k with roots α1; α2; : : : ; αk. Thus, f(z) = Qk i=1(z − αi). 2 We know that the degree of f does not exceed d + 1, so let γi (the components of c) be the coefficients of f 2. That is, d 2 γ0 + γ1z + ::: + γdz = (f(z)) : T 2 In other words, we have c · md+1(z) = (f(z)) for any z 2 R. Thus, if z2 = T 2 T fα1; α2; : : : ; αkg, then c · md+1(z) = (f(z)) > 0, and otherwise, c · md+1(αi) = 2 (f(αi)) = 0. Here is a nice restatement of the lemma in terms of linear algebra that will be used in the next section. Corollary 3.6 (moment curve kissing lemma restated). Let 1 ≤ d ≤ n−1 and 0 ≤ d n×(d+1) k ≤ 2 . Then, there exists a matrix A 2 R such that the n rows are in general position.