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Chromatic numbers of Cayley graphs of abelian groups: Cases of small dimension and rank

Jonathan Cervantes University of California, Riverside, Dept. of Mathematics, Skye Hall, 900 University Ave., Riverside, CA 92521, jcerv092@ucr.edu Mike Krebs California State University, Los Angeles, Dept. of Mathematics, 5151 State University Drive, Los Angeles, CA 91711, mkrebs@calstatela.edu
Abstract

A connected Cayley graph on an abelian group with a finite generating set SS can be represented by its Heuberger matrix, i.e., an integer matrix whose columns generate the group of relations between members of SS. In a companion article, the authors lay the foundation for the use of Heuberger matrices to study chromatic numbers of abelian Cayley graphs. We call the number of rows in the Heuberger matrix the dimension, and the number of columns the rank. In this paper, we give precise numerical conditions that completely determine the chromatic number in all cases with dimension 11; with rank 11; and with dimension 3\leq 3 and rank 2\leq 2. For such a graph without loops, we show that it is 44-colorable if and only if it does not contain a 55-clique, and it is 33-colorable if and only if it contains neither a diamond lanyard nor a C13(1,5)C_{13}(1,5), both of which we define herein. In a separate companion article, we show that we recover Zhu’s theorem on the chromatic number of 66-valent integer distance graphs as a special case of our theorem for dimension 33 and rank 22.

Keywords— graph, chromatic number, abelian group, Cayley graph, circulant graph

1 Introduction

Given an m×rm\times r integer matrix MM, let HH be the set of all linear combinations of the columns of MM with integer coefficients. Let \mathbb{Z} denote the group of integers under addition, and let m\mathbb{Z}^{m} denote the mm-fold direct product of \mathbb{Z} with itself. Let ejme_{j}\in\mathbb{Z}^{m} denote the mm-tuple (regarded as a column vector) whose iith component is 11 if i=ji=j and is 0 otherwise. Let S={H±e1,,H±em}S=\{H\pm e_{1},\dots,H\pm e_{m}\}. We may then form the Cayley graph whose underlying group is m/H\mathbb{Z}^{m}/H with respect to the generating set SS. We denote by MXSACGM^{\text{SACG}}_{X} the graph formed in this manner. We call MXSACGM^{\text{SACG}}_{X} a standardized abelian Cayley graph, and we call MM an associated Heuberger matrix. As discussed in [1], the study of chromatic numbers of Cayley graphs on abelian groups can be reduced to the study of standardized abelian Cayley graphs and their Heuberger matrices. Many, many particular cases of chromatic numbers of Cayley graphs on abelian groups have been studied; see the introduction to [1] for a long list of examples.

Our main results (Theorems 2.9 and 2.14) give precise and easily checked numerical conditions that completely determine the chromatic number when the associated Heuberger matrix is 2×22\times 2 or 3×23\times 2. These results can be summarized as follows: Suppose such a graph does not have loops. Then it is 44-colorable if and only if it does not contain a 55-clique, and it is 33-colorable if and only if it contains neither a diamond lanyard (see Def. 2.2) nor a Cay(13,{±1,±5})\text{Cay}(\mathbb{Z}_{13},\{\pm 1,\pm 5\}).

Whether such subgraphs occur, we show, can be ascertained quickly from the entries of the Heuberger matrix. After excluding some trivial exceptional cases, one first puts the matrix into a certain standardized form (lower triangular form with positive diagonal entries for 2×22\times 2 matrices, “modified Hermite normal form” for 3×23\times 2 matrices) without changing the associated graph. Theorems 2.9 and 2.14 then provide formulas for the chromatic number of graphs with matrices in this form.

We briefly sketch the method of proof. For 2×22\times 2 matrices, the main result (Thm. 2.9) follows quickly by combining Heuberger’s theorem on chromatic numbers of circulant graphs with the methods of [1]. The principal result for 3×23\times 2 matrices (Thm. 2.14) requires more work. The main idea is to apply the graph homomorphisms of [1] to obtain upper bounds on the chromatic number, utilizing the previous results from the 2×22\times 2 case.

In [8], Zhu finds the chromatic number for an arbitrary integer distance graph of the form Cay(,{±a,±b,±c})\text{Cay}(\mathbb{Z},\{\pm a,\pm b,\pm c\}), where a,b,a,b, and cc are distinct positive integers. Such graphs, as shown in [1], have associated 3×23\times 2 matrices. In another companion paper [2], we demonstrate how Thm. 2.14 yields Zhu’s theorem as a corollary of our main theorem about 3×23\times 2 Heuberger matrices.

One obvious future direction for this work will be to investigate what happens when the matrices are larger. For example, the case of a graph XX with an associated m×2m\times 2 Heuberger matrix when m4m\geq 4 seems well within reach using methods developed in this paper, and we plan to tackle that next.

As we show in the proofs of our main theorems, when an abelian Cayley graph XX has an associated 2×22\times 2 or 3×23\times 2 Heuberger matrix, an optimal coloring for XX can always be realized as a pullback, via a graph homomorphism, of a coloring of a circulant graph. We pose the question (akin to that asked in [5]): Is that statement true for all connected, finite-degree abelian Cayley graphs?

Moreover, we propose the following conjecture: Let XX be a standardized abeliam Cayley graph with an associated Heuberger matrix MXM_{X}. If XX does not have loops, and the determinant of every 2×22\times 2 minor of MXM_{X} is divisible by 33, then XX is 33-colorable.

This article depends heavily on [1], which we will refer to frequently. The reader should assume that all notation, terminology, and theorems used but not explained here are explained there.

2 Main theorems

In [1], we laid the groundwork for our main techniques, and we employed them to prove the Tomato Cage Theorem, which completely determines the chromatic number in the case where HH has rank 11. In this section, we turn our attention to our main results, which concern the case where HH has rank 22. Let XX be a standardized abelian Cayley graph, and let mm be the number of rows in an associated Heuberger matrix MXM_{X}.

In Subsection 2.1, we quickly dispense with the case where m=1m=1.

For m=2m=2, we first apply isomorphisms to XX as in [1] to put the matrix in a standard form without changing the associated graph. We then clear aside the somewhat aberrant situations where XX is bipartite, XX has loops, or the matrix has a zero row. Excluding these possibilities, we show that if the matrix entries are not relatively prime, then χ(X)=3\chi(X)=3; otherwise, XX is isomorphic to a circulant graph, and its chromatic number is given by a theorem of Heuberger’s. Subsection 2.3 contains the precise statements and proofs.

For m=3m=3, we again begin by putting the matrix into a standard form we call “modified Hermite normal form,” as detailed in Subsection 2.4. Again the “aberrant” situations can be handled quickly. We then determine the chromatic number in the remaining cases. To do so, we subtract rows to produce a homomorphism to a graph with an associated 2×22\times 2 matrix, for which we already have a complete theorem. When this fails to produce a 33-coloring, we modify the mapping. We show that whenever we are unable in this manner to properly 33-color XX, it must be that in fact XX is not properly 33-colorable, and our procedure instead yields a 44-coloring of XX. These unusual cases, we show, fall into one of six families, and we state precise numerical conditions for when they occur. Subsection 2.5 contains the precise statements and proofs.

As a by-product of our proofs, we show that for the class of graphs we consider, if they don’t have loops, then K5K_{5} (the complete graph on 55 vertices) is the only obstacle to 44-colorability, and “diamond lanyards” and “C13(1,5)C_{13}(1,5)” (both of which we define in Subsection 2.2) are the only obstacles to 33-colorability.

In Subsection 2.6, we furnish a synopsis of the procedures involved — an algorithm to determine the chromatic number whenever the Heuberger matrix is of size m×1,1×r,2×rm\times 1,1\times r,2\times r, or 3×23\times 2.

2.1 The case m=1m=1

The case m=1m=1 can be dealt with immediately.

Lemma 2.1.

Suppose XX is a standardized abelian Cayley graph defined by (y1yr)XSACG(y_{1}\;\cdots\;y_{r})^{\text{SACG}}_{X} for some integers y1,yry_{1}\,\dots,y_{r}, not all 0. Let e=gcd(y1,yr)e=\gcd(y_{1}\,\dots,y_{r}). Then XX has loops (and therefore is not properly colorable) if and only if e=1e=1; otherwise, we have that χ(X)=2\chi(X)=2 if ee is even, and χ(X)=3\chi(X)=3 if ee is odd.

Proof.

Applying column operations as in [1, Lemma LABEL:general-lemma-isomorphisms], we can in effect perform the Euclidean algorithm so as to acquire an isomorphic standardized abelian Cayley graph (e 0 0)XSACG=(e)XSACG(e\;0\;\cdots\;0)^{\text{SACG}}_{X^{\prime}}=(e)^{\text{SACG}}_{X^{\prime}}. The result follows from [1, Example LABEL:general-example-cycle]. ∎

If y1==yr=0y_{1}=\cdots=y_{r}=0, then χ(X)=2\chi(X)=2, by [1, Lemma LABEL:general-lemma-bipartite].

2.2 Diamond lanyards and C13(1,5)C_{13}(1,5)

Recall that a diamond is a graph with 44 vertices, exactly one pair of which consists of nonadjacent vertices. In other words, a diamond is a K4K_{4} (complete graph on 44 vertices) with one edge deleted. The two vertices not adjacent to one another are the endpoints of the diamond.

Definition 2.2.

An unclasped diamond lanyard of length 11 is a diamond. Recursively, we define an unclasped diamond lanyard UU of length +1\ell+1 to be the union of an unclasped diamond lanyard YY of length \ell and a diamond DD, such that YY and DD have exactly one endpoint in common. The endpoints of UU are the endpoint of YY which is not an endpoint of DD, and the endpoint of DD which is not an endpoint of YY. A (clasped) diamond lanyard of length \ell is obtained by adding to an unclasped diamond lanyard UU of length \ell an edge between the endpoints of UU. We call that edge a clasp.

Def. 2.2 does not preclude the possibiity of the diamonds in the lanyard having edges in common. For example, Let X=Cay(5,{±1,±2})X=\text{Cay}(\mathbb{Z}_{5},\{\pm 1,\pm 2\}). Then XX is a K5K_{5} graph with vertex set {0,1,2,3,4}\{0,1,2,3,4\}. We write an edge in XX as a string of two vertices. So XX contains as a subgraph a diamond with edges 01,02,12,13,2301,02,12,13,23 and endpoints 0 and 33. It also contains as a subgraph a diamond with edges 41,42,12,13,2341,42,12,13,23 and endpoints 44 and 33. The edge sets of these two diamonds are non-disjoint. Taking the union of these two diamonds produces an unclasped diamond lanyard of length two with endpoints 0 and 44. Observing that 0404 is also an edge in XX, indeed XX contains a clasped diamond lanyard as a subgraph.

We sometimes refer to a clapsed diamond lanyard simply as a diamond lanyard. A diamond lanyard of length 11 is a K4K_{4}, and a diamond lanyard of length 22 where the two diamonds have disjoint edges is called a Mosers’ spindle [7]. Figure 1 illustrates a diamond lanyard of length 44.

Refer to caption
Figure 1: A diamond lanyard of length 44

(We would have preferred to have dubbed these “diamond chain necklaces,” but the term “necklace graph” already has a standard meaning.)

We observe that diamond lanyards are not 33-colorable. For suppose we have a proper 33-coloring. Note that in any proper 33-coloring of a diamond, its endpoints must have the same color. Hence the endpoints of the diamond lanyard must have the same color, but they are adjacent, which is a contradiction. Thus, we have the following lemma.

Lemma 2.3.

Suppose XX is a graph containing a diamond lanyard as a subgraph. Then χ(X)4\chi(X)\geq 4.

We remark that a diamond lanyard of length 2\geq 2, where the defining diamonds have mutually disjoint edge sets, is a unit-distance graph; hence the chromatic number of the plane is at least 44. (Indeed, as discussed in the introduction of [1], it is now known to be at least 55.)

We now turn our attention to the other object that can stand in the way of 33-colorability for our graphs. Let C13(1,5)C_{13}(1,5) be the circulant graph Cay(13,{±1,±5})\text{Cay}(\mathbb{Z}_{13},\{\pm 1,\pm 5\}). (For more about this notation, see Def. 2.5.) Heuberger proves the following lemma in [4] using a “vertex-chasing” argument. Here we prove it by showing that the independence number is 44. We include this proof so that it might suggest generalizations.

Lemma 2.4.

The chromatic number of C13(1,5)C_{13}(1,5) is 44.

Proof.

It is straightforward to find a proper 44-coloring of C13(1,5)C_{13}(1,5). We now let AA be an independent set of vertices in C13(1,5)C_{13}(1,5). We will show that |A|4|A|\leq 4. This will prove the lemma.

Suppose that |A|5|A|\geq 5; we will show that this leads to a contradiction. First observe that we must have |A|6|A|\leq 6. For if |A|7|A|\geq 7, then AA would contain the adjacent vertices xx and x+1x+1 for some x13x\in\mathbb{Z}_{13}. Because |A|6<132|A|\leq 6<\frac{13}{2}, there must be x13x\in\mathbb{Z}_{13} such that xAx\notin A and x+1Ax+1\notin A. From |A|5|A|\geq 5 we find that

{x+2,x+3,x+4,x+5,x+6,x+7}A or {x+8,x+9,x+10,x+11,x+12}A\{x+2,x+3,x+4,x+5,x+6,x+7\}\cap A\text{ or }\{x+8,x+9,x+10,x+11,x+12\}\cap A

must contain at least 33 elements. Because zz and z+1z+1 are adjacent for all z13z\in\mathbb{Z}_{13}, these three elements must be x+y2,x+y,x+y-2,x+y, and x+y+2x+y+2 for some y{4,5,10}y\in\{4,5,10\}. Using the fact that a±1Aa\pm 1\notin A and a±5Aa\pm 5\notin A whenever aAa\in A, we see that

x+y+1,x+y+3,x+y+5,x+y+6,x+y+7,x+y+8,x+y+10,x+y+12A.x+y+1,x+y+3,x+y+5,x+y+6,x+y+7,x+y+8,x+y+10,x+y+12\notin A.

Because |A|5|A|\geq 5, we know that AA must contain at least two elements other than x+y2,x+y,x+y-2,x+y, and x+y+2x+y+2, but the only remaining elements in 13\mathbb{Z}_{13} are x+y+4x+y+4 and x+y+9x+y+9. However, x+y+4x+y+4 and x+y+9x+y+9 are adjacent and so cannot both be in AA.∎

2.3 The case m=2m=2

In this subsection, we completely determine χ(X)\chi(X) when XX has a 2×22\times 2 matrix as an associated Heuberger matrix. (Note that if the number of columns exceeds the number of rows, we can perform column operations as in [1, Lemma LABEL:general-lemma-isomorphisms] to get a zero column, and then delete it. So the results of this section, together with the Tomato Cage Theorem, completely take care of all dimension 22 cases.)

Suppose XX is a standardized abelian Cayley graph with a 2×22\times 2 Heuberger matrix. By performing row and column operations as in Lemma [1, Lemma LABEL:general-lemma-isomorphisms], one can show that XX is isomorphic to a standardized abelian Cayley graph XX^{\prime} with an associated matrix MXM_{X^{\prime}} that is lower-triangular, and such that the diagonal entries of MXM_{X^{\prime}} are non-negative. Hence we may restrict our attention to the case where MXM_{X} is a 2×22\times 2 lower-triangular matrix with nonnegative diagonal entries.

Next we delve into a class of graphs that play a crucial role in the main theorem for this subsection.

Definition 2.5.

Let a,ba,b, and nn be integers with n0n\neq 0 and gcd(a,b,n)=1\gcd(a,b,n)=1 and n\nmidan\nmid a and n\nmidbn\nmid b. We say that Cay(n,{±a,±b})\text{Cay}(\mathbb{Z}_{n},\{\pm a,\pm b\}) is a Heuberger circulant, denoted Cn(a,b)C_{n}(a,b). \square

The condition gcd(a,b,n)=1\gcd(a,b,n)=1 is equivalent to the connectedness of Cn(a,b)C_{n}(a,b), and the condition that n\nmidan\nmid a and n\nmidbn\nmid b is equivalent to the absence of loops.

In [4], Heuberger completely determines the chromatic number of all Heuberger circulants, as follows.

Theorem 2.6 ([4, Theorem 3]).

Let Cn(a,b)C_{n}(a,b) be a Heuberger circulant. Then

χ(Cn(a,b))={2if a and b are both odd, but n is even5if n=±5 and a±2b(mod 5)4if n=±13, and a±5b(mod 13)4if (i)n±5, and (ii) 3\nmidn, and (iii)a±2b(modn) or b±2a(modn)3otherwise.\chi(C_{n}(a,b))=\begin{cases}2&\text{if }a\text{ and }b\text{ are both odd, but }n\text{ is even}\\ 5&\text{if }n=\pm 5\text{ and }a\equiv\pm 2b\;(\textup{mod}\;5)\\ 4&\text{if }n=\pm 13,\text{ and }\;a\equiv\pm 5b\;(\textup{mod}\;13)\\ 4&\text{if }(i)\;n\neq\pm 5,\text{ and }(ii)\;3\nmid n,\text{ and }(iii)\;a\equiv\pm 2b\;(\textup{mod}\;n)\text{ or }b\equiv\pm 2a\;(\textup{mod}\;n)\\ 3&\text{otherwise.}\end{cases}

We note that [4] excludes the case a±b(modn)a\equiv\pm b\;(\textup{mod}\;n), in which event Cn(a,b)C_{n}(a,b) is an nn-cycle. However, the theorem as stated here includes this possibility as well.

Observe that the third case in Theorem 2.6 is precisely Lemma 2.4. Moreover, for the fourth case, Lemma 2.3 shows that χ(Cn(a,b))4\chi(C_{n}(a,b))\geq 4, for the following reason. First suppose that a2b(modn)a\equiv 2b\;(\textup{mod}\;n). Observe that for all xnx\in\mathbb{Z}_{n}, there is a diamond in Cn(a,b)C_{n}(a,b) with vertex set {x,x+a,x+2a,x+3a}\{x,x+a,x+2a,x+3a\}, where the endpoints are xx and x+3ax+3a. Because a2b(modn)a\equiv 2b\;(\textup{mod}\;n) and gcd(a,b,n)=1\gcd(a,b,n)=1, we must have that gcd(a,n)=1\gcd(a,n)=1. Moreover, because 3\nmidn3\nmid n, we have that gcd(3a,n)=1\gcd(3a,n)=1. Let mm be a positive integer such that 3am1(modn)3am\equiv 1\;(\textup{mod}\;n). Then Cn(a,b)C_{n}(a,b) contains a diamond lanyard with vertex set {ja| 0jm}\{ja\;|\;0\leq j\leq m\}, so by Lemma 2.3, we have that χ(Cn(a,b))4\chi(C_{n}(a,b))\geq 4. A similar argument works when a2b(modn)a\equiv-2b\;(\textup{mod}\;n) or when b±2a(modn)b\equiv\pm 2a\;(\textup{mod}\;n). This is essentially the reasoning in [4] for the lower bounds in these cases. Hence a Heuberger circulant is 44-colorable unless it is K5K_{5}; and it is 33-colorable unless it contains as a subgraph either a diamond lanyard or it equals C13(1,5)C_{13}(1,5). (Recall that K5K_{5} contains a diamond lanyard, so we needn’t include it in the list of obstructions to 33-colorability.)

Next, we show that when the entries of the first column of MXM_{X} are relatively prime, then with some mild additional conditions imposed, XX is isomorphic to a Heuberger circulant.

Lemma 2.7.

Let XX be a standardized abelian Cayley graph with associated Heuberger matrix

MX=(y11y12y21y22).M_{X}=\begin{pmatrix}y_{11}&y_{12}\\ y_{21}&y_{22}\end{pmatrix}.

Suppose that XX does not have loops, that det(MX)0\det(M_{X})\neq 0, and that gcd(y11,y21)=1\gcd(y_{11},y_{21})=1. Then XX is isomorphic to the Heuberger circulant Cn(a,b)C_{n}(a,b), where a=y21a=-y_{21}, b=y11b=y_{11}, and n=det(MX)n=\det(M_{X}).

Proof.

Define a homomorphism φ:2n\varphi\colon\mathbb{Z}^{2}\to\mathbb{Z}_{n} by e1a,e2be_{1}\mapsto a,e_{2}\mapsto b. Let y1y_{1} and y2y_{2} be the first and second columns, respectively, of MXM_{X}. We must show that kerφ\ker\varphi equals the \mathbb{Z}-span of y1y_{1} and y2y_{2}. It is straightforward to show that φ(y1)=φ(y2)=0\varphi(y_{1})=\varphi(y_{2})=0, giving us one inclusion. We now show the reverse inclusion. Suppose φ((x1,x2)t)=0\varphi((x_{1},x_{2})^{t})=0. Then ax1+bx20(modn)ax_{1}+bx_{2}\equiv 0\;(\textup{mod}\;n). Because aa and bb are relatively prime, there exist integers q,rq,r such that aq+br=1aq+br=1. Using elementary number theory as in [6], we find that x1=nq+kbx_{1}=\ell nq+kb and x2=nrkax_{2}=\ell nr-ka for some integers ,k\ell,k. Hence after some computations we find that

(x1x2)=ky1+n(qr)=ky1+(qy11y21qy21y12ry11y22ry21y12)=(k+qy22ry12)y1+y2.\begin{pmatrix}x_{1}\\ x_{2}\end{pmatrix}=ky_{1}+\ell n\begin{pmatrix}q\\ r\end{pmatrix}=ky_{1}+\begin{pmatrix}\ell qy_{11}y_{21}-\ell qy_{21}y_{12}\\ \ell ry_{11}y_{22}-\ell ry_{21}y_{12}\end{pmatrix}=(k+\ell qy_{22}-\ell ry_{12})y_{1}+\ell y_{2}.\qed

Note that after switching the columns of the matrix, Lemma 2.7 becomes a special case of [1, Example LABEL:general-example-arbitrary-circulant-graph]. We have included it here separately so as to give a more elementary proof for 2×22\times 2 matrices.

We remark that not every graph with an associated 2×22\times 2 Heuberger matrix is isomorphic to a circulant. The graph (4024)XSACG\begin{pmatrix}4&0\\ 2&4\end{pmatrix}_{X}^{\text{SACG}}, for instance, provides a counterexample. One computes that XX must have order 1616 and so, if circulant, would be of the form Cay(16,S)\text{Cay}(\mathbb{Z}_{16},S) for some SS. Because XX is connected and bipartite, and has degree 44, we see that up to isomorphism, we can assume S={±1,±3}S=\{\pm 1,\pm 3\} or S={±1,±7}S=\{\pm 1,\pm 7\}. Direct arguments (for example, counting the number of paths of length 22 between various vertices) show that neither choice of SS produces a circulant graph isomorphic to XX.

The following lemma characterizes 2×22\times 2 matrices whose associated graphs have loops. This occurs either when the determinant is nonzero and divides a row, or else when one row is zero and the other has relatively prime entries.

Lemma 2.8.

Let XX be a standardized abelian Cayley graph with an associated matrix

MX=(y11y12y21y22)M_{X}=\begin{pmatrix}y_{11}&y_{12}\\ y_{21}&y_{22}\end{pmatrix}

Let n=det(MX)n=\det(M_{X}). If n0n\neq 0, then XX has loops if and only if either (i) ny11n\mid y_{11} and ny12n\mid y_{12} or (ii) ny21n\mid y_{21} and ny22n\mid y_{22}. If n=0n=0, then XX has loops if and only if either (a) y11=y12=0y_{11}=y_{12}=0 and gcd(y21,y22)=1\gcd(y_{21},y_{22})=1, or (b) y21=y22=0y_{21}=y_{22}=0 and gcd(y11,y12)=1\gcd(y_{11},y_{12})=1.

Proof.

First suppose n0n\neq 0. We know that XX has loops if and only if e1e_{1} or e2e_{2} is in the \mathbb{Z}-span HH of the columns of MXM_{X}. We have e1He_{1}\in H if and only if MX1e12M_{X}^{-1}e_{1}\in\mathbb{Z}^{2}. But

MX1e1=1n(y22y21y12y11)e1=(y22/ny21/n)t.M_{X}^{-1}e_{1}=\frac{1}{n}\begin{pmatrix}y_{22}&-y_{21}\\ -y_{12}&y_{11}\end{pmatrix}e_{1}=(y_{22}/n\;\;\;\;-y_{21}/n)^{t}.

Similarly for e2e_{2}.

Now suppose n=0n=0. If (a) holds, then e2He_{2}\in H. If (b) holds, then e1He_{1}\in H. Conversely, suppose that XX has loops, so that e1He_{1}\in H or e2He_{2}\in H. First suppose e1He_{1}\in H. Then there exist r,sr,s\in\mathbb{Z} such that

r(y11y21)+s(y12y22)=e1,r\begin{pmatrix}y_{11}\\ y_{21}\end{pmatrix}+s\begin{pmatrix}y_{12}\\ y_{22}\end{pmatrix}=e_{1}, (1)

so ry21+sy22=0ry_{21}+sy_{22}=0. Because n=0n=0, the left column and right columns of MXM_{X} are linearly dependent over \mathbb{Q}. Then there exist α,β\alpha,\beta\in\mathbb{Z}, not both 0, such that

α(y11y21)=β(y12y22).\alpha\begin{pmatrix}y_{11}\\ y_{21}\end{pmatrix}=\beta\begin{pmatrix}y_{12}\\ y_{22}\end{pmatrix}. (2)

We can assume gcd(α,β)=1\gcd(\alpha,\beta)=1, for if not, replace them with α/gcd(α,β)\alpha/\gcd(\alpha,\beta) and β/gcd(α,β)\beta/\gcd(\alpha,\beta), respectively.

Then rβy21+sβy22=0r\beta y_{21}+s\beta y_{22}=0, so rβy21+sαy21=0r\beta y_{21}+s\alpha y_{21}=0, so rβ+sα=0r\beta+s\alpha=0 or y21=0y_{21}=0.

Case 1: Suppose rβ+sα=0r\beta+s\alpha=0. From (1) we also have ry11+sy12=1ry_{11}+sy_{12}=1, so gcd(r,s)=1\gcd(r,s)=1. We have rβ=sαr\beta=-s\alpha, so αr\alpha\mid r and βs\beta\mid s. Also rαr\mid\alpha and sβs\mid\beta. So either α=r\alpha=r and β=s\beta=-s, or α=r\alpha=-r and β=s\beta=s. Either way, (1) and (2) now contradict one another.

Case 2: Suppose y21=0y_{21}=0. Then 0=αy21=βy220=\alpha y_{21}=\beta y_{22}. If y22=0y_{22}=0, then using (1), we find that (b) holds. If not, then β=0\beta=0, so α0\alpha\neq 0, so y11=y21=0y_{11}=y_{21}=0. Equation (1) then implies that (b) holds.

A similar argument shows that if e2He_{2}\in H, then (a) must hold.∎

We now have all the pieces in place needed to find χ(X)\chi(X) whenever XX has an associated 2×22\times 2 matrix.

Theorem 2.9.

Let XX be a standardized abelian Cayley graph defined by

(y110y21y22)XSACG.\begin{pmatrix}y_{11}&0\\ y_{21}&y_{22}\end{pmatrix}^{\emph{SACG}}_{X}.

Suppose that y110y_{11}\geq 0 and y220y_{22}\geq 0. Let d=gcd(y11,y21)d=\gcd(y_{11},y_{21}) and e=gcd(y11,y21,y22)e=\gcd(y_{11},y_{21},y_{22}). Then:

  1. 1.

    If either (i) y22=1y_{22}=1 or (ii) y11=1y_{11}=1 and y22|y21y_{22}|y_{21} or (iii) y11=0y_{11}=0 and gcd(y21,y22)=1\gcd(y_{21},y_{22})=1, then XX has loops and is not properly colorable.

  2. 2.

    If both y11+y21y_{11}+y_{21} and y22y_{22} are even, then χ(X)=2\chi(X)=2.

  3. 3.

    If (i) neither of the conditions in the previous statements holds, and (ii) y11=0y_{11}=0 or y22=0y_{22}=0 or e>1e>1 or y22|y21y_{22}|y_{21}, then χ(X)=3\chi(X)=3.

  4. 4.

    If none of the conditions in the previous statements hold, take qq\in\mathbb{Z} such that gcd(y11,y21+qy22)=1\text{gcd}(y_{11},y_{21}+qy_{22})=1. (Such a qq necessarily exists, as we can let qq be the product of all primes pp such that p|y11p|y_{11} but p\nmiddp\nmid d. Here we adopt the convention that if there are no such primes, let q=1q=1.) Then χ(X)=χ(Cn(a,b))\chi(X)=\chi(C_{n}(a,b)), where a=y21qy22a=-y_{21}-qy_{22}, b=y11b=y_{11}, and n=y11y22n=y_{11}y_{22}.

Proof.

Lemma 2.8 implies both Statement (1) and its converse. Statement (2) follows from [1, Lemma LABEL:general-lemma-bipartite].

We now prove Statement (3). Assume that conditions (i) and (ii) in that statement both hold. Condition (i) implies that XX does not have loops, and that XX is not bipartite. If y11=0y_{11}=0, then we can delete the top row as per [1, Lemma LABEL:general-lemma-delete-row-all-zeroes] without affecting the chromatic number. Lemma 2.1 now gives us χ(X)=3\chi(X)=3. If y22=0y_{22}=0, then we can delete the second column as per [1, Lemma LABEL:general-lemma-isomorphisms(LABEL:general-item-delete-column-Z-span)] without changing XX. We then find that χ(X)=3\chi(X)=3, by [1, Theorem LABEL:general-theorem-tree-guard]. If e>1e>1, then χ(X)=3\chi(X)=3, by [1, Lemma LABEL:general-lemma-column-sums-not-coprime]. If y22|y21y_{22}|y_{21}, then after performing a column sum as in [1, Lemma LABEL:general-lemma-isomorphisms(LABEL:general-item-add-multiple_of_column)] to eliminate y21y_{21}, by [1, Lemma LABEL:general-lemma-block-structure], we have that χ(X)=3\chi(X)=3.

Finally, we prove Statement (4). By [1, Lemma LABEL:general-lemma-isomorphisms(LABEL:general-item-add-multiple_of_column)], we have that

(y110y21y22)XSACG=(y110y21+qy22y22)XSACG.\begin{pmatrix}y_{11}&0\\ y_{21}&y_{22}\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}y_{11}&0\\ y_{21}+qy_{22}&y_{22}\end{pmatrix}^{\text{SACG}}_{X}.

The result follows from Lemma 2.7.∎

It follows from Thm. 2.9 and our previous observations about Heuberger circulants that for a standardized abelian Cayley graph XX with an associated 2×22\times 2 Heuberger matrix, if XX does not have loops, then XX is 44-colorable unless it contains K5K_{5} as a subgraph; and it is 33-colorable unless it contains as a subgraph either a diamond lanyard or C13(1,5)C_{13}(1,5). We will see in §2.5 that this statement holds for the 3×23\times 2 case as well.

When MXM_{X} is a 2×22\times 2 matrix, we perform row and column operations to create a 2×22\times 2 lower triangular matrix MXM_{X^{\prime}} for which XX^{\prime} is isomorphic to XX. Observing the effect of these operations on the determinant, we find that |detMX|=|detMX||\det\;M_{X}|=|\det\;M_{X^{\prime}}|. Note that in Theorem 2.9, whenever XX has no loops and χ(X)>3\chi(X)>3, we have that |detMX||\det\;M_{X}| is not divisible by 33. Thus we have the following corollary, which provides an easily checked sufficient condition for 33-colorability.

Corollary 2.10.

Let XX be a standardized abelian Cayley graph with an associated 2×22\times 2 matrix MXM_{X}. If XX has no loops and 3detMX3\mid\det\;M_{X}, then χ(X)3\chi(X)\leq 3.

We note that Cor. 2.10 fails in general for larger matrices. For example, we have by Thm. 2.9, Thm. 2.6, [1, Lemma LABEL:general-lemma-block-structure], and Example [1, LABEL:general-example-cycle] that

χ((100250003)XSACG)=5.\chi\left(\begin{pmatrix}1&0&0\\ -2&5&0\\ 0&0&3\end{pmatrix}^{\text{SACG}}_{X}\right)=5.

However, observe that not every 2×22\times 2 minor of the matrix above has determinant divisible by 33. As mentioned in the introduction, we conjecture that if this stronger condition holds, and XX does not have loops, then XX is 33-colorable.

2.4 Modified Hermite normal form

In Subsection 2.3, we saw that it was useful to deal only with 2×22\times 2 Heuberger matrices in a certain convenient format. The same goes for 3×23\times 2 Heuberger matrices. The crux of this idea is drawn from [4], where Hermite normal form is used. For 3×23\times 2 matrices, we refine the requirements slightly for our purposes, so as to further reduce the number of exceptional cases. The purpose of the present subsection is to define this “modified Hermite normal form” for 3×23\times 2 matrices and to show that with very few exceptions every standardized abelian Cayley graph with an associated 3×23\times 2 Heuberger matrix is isomorphic to one with a matrix in this form. We do not attempt here to generalize these definitions to matrices of arbitrary size, as we do not know yet what restrictions will prove to be most useful when the rank or dimension is larger.

Definition 2.11.

Let

M=(y11y12y21y22y31y32)M=\begin{pmatrix}y_{11}&y_{12}\\ y_{21}&y_{22}\\ y_{31}&y_{32}\end{pmatrix}

be a 3×23\times 2 matrix with integer entries such that no row of MM has all zero entries. We say MM is in modified Hermite normal form if the following conditions hold:

  1. 1.

    y11>0y_{11}>0, and

  2. 2.

    y12=0y_{12}=0, and

  3. 3.

    y11y22y11y32(mod 3)y_{11}y_{22}\equiv y_{11}y_{32}\;(\textup{mod}\;3), and

  4. 4.

    y22y32y_{22}\leq y_{32}, and

  5. 5.

    |y22||y32||y_{22}|\leq|y_{32}|, and

  6. 6.

    Either (i) y22=0y_{22}=0 and 12|y32|y310-\frac{1}{2}|y_{32}|\leq y_{31}\leq 0, or else (ii) 12|y22|y210-\frac{1}{2}|y_{22}|\leq y_{21}\leq 0.

There are some departures here from the usual Hermite normal form. For example, |y21||y_{21}| cannot be more than half of y22y_{22}, a more stringent requirement than being less than y22y_{22}, as in ordinary Hermite normal form. As we shall see, we can impose this narrower condition because we have both row and column operations at our disposal, not just column operations. Moreover, this form may be the transpose of what some readers are accustomed to, but we have adopted this convention so as to be consistent with [4]. The third condition has no analogue with Hermite normal form but will turn out to be rather useful in the succeeding subsection.

We next show that every standardized abelian Cayley graph with a 3×23\times 2 Heuberger matrix of rank 2 without zero rows is isomorphic to one with a Heuberger matrix in modified Hermite normal form.

Lemma 2.12.

Let XX be a standardized abelian Cayley graph with a 3×23\times 2 Heuberger matrix MXM_{X}. Suppose that MXM_{X} has no zero rows, and that the columns of MXM_{X} are linearly independent over \mathbb{Q}. Then XX is isomorphic to a standardized abelian Cayley graph XX^{\prime} with a 3×23\times 2 Heuberger matrix MXM_{X^{\prime}} in modified Hermite normal form.

Proof.

The proof is constructive. We give an explicit algorithm for row and column operations to perform on MXM_{X}, as per [1, Lemma LABEL:general-lemma-isomorphisms], so as to result in the desired matrix MXM_{X^{\prime}}.

  1. Step Zero:

    Let α,β\alpha,\beta, and γ\gamma be the determinants of the 2×22\times 2 minors of MXM_{X}. Two of α,β,γ\alpha,\beta,\gamma must be congruent to each other or negatives of each other modulo 33. Using this fact, we can permute rows and/or multiply rows by 1-1 as needed so that the determinant of the submatrix formed by the top two rows is congruent modulo 33 to the submatrix formed by the first and third rows. This property will be preserved by all subsequent steps and will eventually lead to satisfaction of the third condition in Def. 2.11. Let MX1M_{X_{1}} be the resulting matrix.

  2. Step One:

    If the first entry of any column of MX1M_{X_{1}} is negative, multiply that column (or those columns) by 1-1. Let MX2M_{X_{2}} be the resulting matrix. The top row of MX2M_{X_{2}} has no negative entries.

  3. Step Two:

    If the first entry of the first column of MX2M_{X_{2}} is 0, then permute the two columns; otherwise, do nothing. Let MX3M_{X_{3}} be the resulting matrix. The first entry of the first column of MX3M_{X_{3}} is strictly positive. (Here is where we use the assumption that MXM_{X} has no zero rows.) If the first entry of the second column is 0, then let MX4=MX3M_{X_{4}}=M_{X_{3}} and skip to Step Four.

  4. Step Three:

    Both entries of the top row of MX3M_{X_{3}} are strictly positive. Let ee be the greatest common divisor of these two entries. Repeatedly applying [1, Lemma LABEL:general-lemma-isomorphisms(LABEL:general-item-add-multiple_of_column)], we perform column operations that effectuate the Euclidean algorithm on the entries in the top row of MX3M_{X_{3}}, so that the top row of the resulting matrix has two entries, one of which is ee, the other 0. If the first entry of the first column is 0, then permute the two columns. Let MX4M_{X_{4}} be the resulting matrix. The top row of MX4M_{X_{4}} has a strictly positive first entry, and 0 for its second entry. The matrix MX4M_{X_{4}} now meets the first three conditions in the definition of modified Hermite normal form, and these will be preserved by all subsequent steps.

  5. Step Four:

    Let zz and ww be the (3,1)(3,1) and (3,2)(3,2) entries of MX4M_{X_{4}}, respectively. If zwz\leq w and |z||w||z|\leq|w|, then do nothing. If z>wz>w and |z||w||z|\leq|w|, then multiply the second column by 1-1. If zwz\leq w and |z|>|w||z|>|w|, then switch the bottom two rows and multiply the second column by 1-1. If z>wz>w and |z|>|w||z|>|w|, then switch the bottom two rows. Whatever action was taken, let MX5M_{X_{5}} be the resulting matrix. The first five conditions in Def. 2.11 are now (and will continue to be) met.

  6. Step Five:

    Let aa and bb be the (2,1)(2,1) and (2,2)(2,2) entries of MX5M_{X_{5}}, respectively. If b=0b=0, then apply to the third rather than second row the procedure described in the rest of Step Five as well as in Step Six. (Here is where we use that the columns of MXM_{X} are linearly independent over \mathbb{Q}; this guarantees that if b=0b=0, then the (3,2)(3,2) entry of MX5M_{X_{5}} is not zero.) If b0b\neq 0, by the division theorem, there exist integers qq and rr such that r=aq|b|r=a-q|b|, where |b|<r0-|b|<r\leq 0. Applying [1, Lemma LABEL:general-lemma-isomorphisms(LABEL:general-item-add-multiple_of_column)], perform a column operations to replace the first column with the first column plus ±q\pm q times the second column, so that the second entry in the first column becomes rr. Let MX6M_{X_{6}} be the resulting matrix.

  7. Step Six:

    Let cc be the (2,1)(2,1) entry of MX6M_{X_{6}}. We still have that bb is the (2,2)(2,2) entry of MX6M_{X_{6}}. Suppose that |b|2>c-\frac{|b|}{2}>c. We then add b/|b|b/|b| times the second column to the first column; then multiply the first column by 1-1; and then multiply the first row by 1-1. Let MXM_{X^{\prime}} be the resulting matrix.

The matrix MXM_{X^{\prime}} will then satisfy all conditions in the definition of modified Hermite normal form.∎

Suppose we have a 3×23\times 2 Heuberger matrix MXM_{X}. If MXM_{X} has a zero row, then by [1, Lemma LABEL:general-lemma-delete-row-all-zeroes], we can delete it without affecting the chromatic number χ\chi, whereupon Thm. 2.9 can be used to find χ\chi. If the columns of MXM_{X} are linearly dependent over \mathbb{Q}, then as per [1, Lemma LABEL:general-lemma-isomorphisms] appropriate column operations that do not change XX will produce a zero column, which can be deleted without changing XX. [1, Thm. LABEL:general-theorem-tree-guard] can then be used to find χ(X)\chi(X). Otherwise, in light of Lemma 2.12, we lose no generality by assuming that MXM_{X} is in modified Hermite normal form.

We next show that when MXM_{X} is in modified Hermite normal form, we can determine immediately whether XX has loops. Recall that eje_{j} is the jjth standard basis vector, with a 11 as its jjth entry and 0 for every other entry.

Lemma 2.13.

Let XX be a standardized abelian Cayley graph with a Heuberger matrix MXM_{X}. Suppose that MXM_{X} is a 3×23\times 2 matrix in modified Hermite normal form. Then XX has loops if and only if either the first column of MXM_{X} is e1e_{1}, or the second column of MXM_{X} is e3e_{3}.

Proof.

Let HH be the \mathbb{Z}-span of the columns of MXM_{X}. The graph XX has loops if and only if ±ejH\pm e_{j}\in H for some jj. From the definition of modified Hermite normal form, we see that this can occur if and only if either the first column of MXM_{X} is e1e_{1} or the second column of MXM_{X} is e3e_{3}. ∎

2.5 Chromatic numbers of graphs with 3×23\times 2 matrices in modified Hermite normal form

In this section we prove the following theorem, which completely determines the chromatic number of an arbitrary standardized abelian Cayley graph with a 3×23\times 2 Heuberger matrix in modified Hermite normal form.

Theorem 2.14.

Let XX be a standardized abelian Cayley graph with a Heuberger matrix

MX=(y110y21y22y31y32)M_{X}=\begin{pmatrix}y_{11}&0\\ y_{21}&y_{22}\\ y_{31}&y_{32}\\ \end{pmatrix}

in modified Hermite normal form.

  1. 1.

    If the first column of MXM_{X} is e1e_{1} or the second column of MXM_{X} is e3e_{3}, then XX has loops and cannot be properly colored.

  2. 2.

    If y11+y21+y31y_{11}+y_{21}+y_{31} and y22+y32y_{22}+y_{32} are both even, then χ(X)=2\chi(X)=2.

  3. 3.

    If

    MX=(1001±3k1+3k) or MX=(1001±3k1+3k) or MX=(101213k2+3k) or MX=(10121+3k2+3k)M_{X}=\begin{pmatrix}1&0\\ 0&1\\ \pm 3k&1+3k\end{pmatrix}\text{ or }M_{X}=\begin{pmatrix}1&0\\ 0&-1\\ \pm 3k&-1+3k\end{pmatrix}\text{ or }M_{X}=\begin{pmatrix}1&0\\ -1&2\\ -1-3k&2+3k\end{pmatrix}\text{ or }M_{X}=\begin{pmatrix}1&0\\ -1&-2\\ -1+3k&-2+3k\end{pmatrix}
     for some positive integer k, or MX=(10013b2) for some integer b, or\text{ for some positive integer }k,\text{ or }M_{X}=\begin{pmatrix}1&0\\ 0&-1\\ 3b&2\end{pmatrix}\text{ for some integer }b,\text{ or}
    MX=(101a1a+3(k1)) for some integer a with 3\nmida and some positive integer k,M_{X}=\begin{pmatrix}1&0\\ -1&a\\ -1&a+3(k-1)\end{pmatrix}\text{ for some integer }a\text{ with }3\nmid a\text{ and some positive integer }k,

    then χ(X)=4\chi(X)=4.

  4. 4.

    If none of the above conditions hold, then χ(X)=3\chi(X)=3.

The main idea behind the proof of Theorem 2.14 is to add or subtract two rows as per [1, Lemma LABEL:general-lemma-homomorphisms] to obtain a homomorphism from XX to a graph with a 2×22\times 2 Heuberger matrix. Theorem 2.9 and its corollary provide conditions under which this latter graph (and hence XX) is 33-colorable. If those conditions are not met, then we get information about MXM_{X}. In this case we can repeat the procedure using some other homomorphism to further narrow down the possibilities for those MXM_{X} for which χ(X)>3\chi(X)>3. In particular, we are led in this way to consider three special types of matrices MXM_{X}: those with y22=0y_{22}=0 (we call these “L-shaped” matrices); those with y11=y22=1y_{11}=y_{22}=1 and y21=0y_{21}=0 (we call these “II on top” matrices); and those with y11=y21=y31=1y_{11}=y_{21}=y_{31}=1 (we call these “first column all ones” matrices). Every exceptional case traces back ultimately to one of these three. Hence, to lay the groundwork for the proof of Theorem 2.14, we first prove three technical lemmas which compute the chromatic numbers in these three circumstances. The proofs of all three use the same main idea: Add or subtract rows to map to a graph with a 2×22\times 2 matrix. If such maps fail to produce a 33-coloring, then show that XX contains a diamond lanyard.

We begin with “first column all ones” matrices.

Lemma 2.15.

Suppose we have

(101y221y32)XSACG.\begin{pmatrix}1&&0\\ 1&&y_{22}\\ 1&&y_{32}\end{pmatrix}^{\text{SACG}}_{X}.

Then XX has loops if and only if {y22,y32}\{y_{22},y_{32}\} is {0,1}\{0,-1\} or {0,1}\{0,1\} or {1}\{-1\} or {1}\{1\}. Otherwise,

χ(X)={3 if y32y22(mod 3)4 if y32y22(mod 3).\chi(X)=\begin{cases}3&\text{ if }y_{32}\equiv-y_{22}\;(\textup{mod}\;3)\\ 4&\text{ if }y_{32}\not\equiv-y_{22}\;(\textup{mod}\;3).\end{cases}
Proof.

The statement about loops is straightforward. Now suppose that XX does not have loops. Let MXM_{X} be the matrix in the lemma statement.

The first column’s sum is odd, so XX cannot be bipartite, by [1, Lemma LABEL:general-lemma-bipartite].

If y32y22(mod 3)y_{32}\equiv-y_{22}\;(\textup{mod}\;3), then we get χ(X)=3\chi(X)=3 by applying [1, Lemma LABEL:general-lemma-column-sums-not-coprime].

Now assume that y32y22(mod 3)y_{32}\not\equiv-y_{22}\;(\textup{mod}\;3), in other words that y32=y22±1+3y_{32}=-y_{22}\pm 1+3\ell for some integer \ell. First we show that χ(X)>3\chi(X)>3, by showing that XX contains a diamond lanyard. Let HH be the subgroup of 3\mathbb{Z}^{3} generated by the columns of MXM_{X}. Recall that vertices of XX are of the form (a,b,c)t+H(a,b,c)^{t}+H, which we denote by (a,b,c)t¯\overline{(a,b,c)^{t}}. The vertices (0,0,0)t¯,(0,1,0)t¯,(0,0,1)t¯,\overline{(0,0,0)^{t}},\overline{(0,1,0)^{t}},\overline{(0,0,-1)^{t}}, and (0,1,1)t¯\overline{(0,1,-1)^{t}} form a diamond in XX. Shifting this diamond x1x-1 times by (0,1,1)t¯\overline{(0,1,-1)^{t}} and concatenating, we produce an unclasped diamond lanyard L1L_{1} of length xx with endpoints (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,x,x)t¯\overline{(0,x,-x)^{t}}. In a similar vein, we have an unclasped diamond lanyard of length 22 formed by one diamond with vertices (0,0,0)t¯,(0,1,0)t¯,(0,1,1)t¯,\overline{(0,0,0)^{t}},\overline{(0,1,0)^{t}},\overline{(0,1,1)^{t}}, and (0,2,1)t¯\overline{(0,2,1)^{t}} and another diamond with vertices (0,2,1)t¯,(0,2,0)t¯,(0,3,1)t¯,\overline{(0,2,1)^{t}},\overline{(0,2,0)^{t}},\overline{(0,3,1)^{t}}, and (0,3,0)t¯\overline{(0,3,0)^{t}}. Its endpoints are (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,3,0)t¯\overline{(0,3,0)^{t}}. Assume for now that 1\ell\geq 1. Shifting 1\ell-1 times by (0,3,0)t¯\overline{(0,3,0)^{t}} and concatenating, we produce an unclasped diamond lanyard L2L_{2} of length 22\ell with endpoints (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,3,0)t¯\overline{(0,3\ell,0)^{t}}. Conjoining L1L_{1} and L2L_{2} gives us an unclasped diamond lanyard of length x+2x+2\ell with endpoints (0,x,x)t¯\overline{(0,x,-x)^{t}} and (0,3,0)t¯\overline{(0,3\ell,0)^{t}}. Taking x=±1+y22+3x=\pm 1+y_{22}+3\ell produces an edge between (0,x,x)t¯\overline{(0,x,-x)^{t}} and (0,3,0)t¯\overline{(0,3\ell,0)^{t}}, and thus we have a clasped diamond lanyard in XX. By Lemma 2.3, we have that χ(X)4\chi(X)\geq 4. A similar procedure gives the same result when 0\ell\leq 0.

Now we show that χ(X)4\chi(X)\leq 4. By [1, Lemma LABEL:general-lemma-homomorphisms] we have a homomorphism

(101y221y32)XSACG\ocirc(102y22+y32)YSACG.\begin{pmatrix}1&&0\\ 1&&y_{22}\\ 1&&y_{32}\end{pmatrix}_{X}^{\text{SACG}}\xrightarrow{\ocirc}\begin{pmatrix}1&&0\\ 2&&y_{22}+y_{32}\end{pmatrix}_{Y}^{\text{SACG}}.

The results of §2.3 imply that we fail to get a 44-coloring if and only if y22+y32{5,2,1,1,2,5}y_{22}+y_{32}\in\{-5,-2,-1,1,2,5\}.

First we consider the case where y22+y32=5y_{22}+y_{32}=-5. Then

(101y221y225)XSACG\ocirc(2y2251y22)SACG(1023y22+5)YSACG.\begin{pmatrix}1&&0\\ 1&&y_{22}\\ 1&&-y_{22}-5\end{pmatrix}_{X}^{\text{SACG}}\xrightarrow{\ocirc}\begin{pmatrix}2&&-y_{22}-5\\ 1&&y_{22}\end{pmatrix}^{\text{SACG}}\cong\begin{pmatrix}1&&0\\ 2&&3y_{22}+5\end{pmatrix}_{Y^{\prime}}^{\text{SACG}}.

Here we use [1, Lemmas LABEL:general-lemma-homomorphisms and LABEL:general-lemma-isomorphisms], and from now on we shall do this sort of thing without referring to these lemmas each time. By Theorem 2.9, we have that YY^{\prime} is 44-colorable unless y22=1y_{22}=-1. But in that case we have:

(101114)XSACG\ocirc(2114)SACGC7(1,4),\begin{pmatrix}1&&0\\ 1&&-1\\ 1&&-4\end{pmatrix}_{X}^{\text{SACG}}\xrightarrow{\ocirc}\begin{pmatrix}2&&-1\\ 1&&-4\end{pmatrix}^{\text{SACG}}\cong C_{7}(1,4),

which is 44-colorable. Here we use Theorems 2.9 and 2.6 as well as Lemma 2.7. In the sequel, usually we will simply compute chromatic numbers of graphs with 2×22\times 2 matrices using the results of §2.3 without referring to the specific theorems and lemmas used.

The other cases, where y22+y32y_{22}+y_{32} equals 2-2 or 1-1 or 11 or 22 or 55, can each be dealt with in a similar way. We omit the proofs here, but complete details can be found in our authors’ notes, which are housed on the second author’s website [3].∎

Next we tackle “L-shaped” matrices. By performing row and column operations not unlike those in §2.4, it suffices to consider only matrices with some additional restrictions imposed.

Lemma 2.16.

Suppose we have

(y110y210y31y32)XSACG,\begin{pmatrix}y_{11}&&0\\ y_{21}&&0\\ y_{31}&&y_{32}\end{pmatrix}^{\text{SACG}}_{X},

where y11,y21,y32>0y_{11},y_{21},y_{32}>0 and y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0.

Then:

  1. 1.

    We have that XX has loops if and only if y32=1y_{32}=1.

  2. 2.

    We have that χ(X)=2\chi(X)=2 if and only if y11+y21+y31y_{11}+y_{21}+y_{31} and y32y_{32} are both even.

  3. 3.

    We have that χ(X)=4\chi(X)=4 if and only if y11=y21=y31=1y_{11}=y_{21}=-y_{31}=1 and 3\nmidy323\nmid y_{32} and y32>1y_{32}>1.

  4. 4.

    Otherwise, χ(X)=3\chi(X)=3.

Proof.

The first two statements are straightforward to prove. Now suppose that XX does not have loops (i.e., that y322y_{32}\geq 2) and is not bipartite. Let MXM_{X} be the matrix in the lemma statement. We have that 33 divides either y11y_{11}, y21y_{21}, y11+y21y_{11}+y_{21}, or y11y21y_{11}-y_{21}, and we divvy into cases accordingly.

First suppose that 3y113\mid y_{11}. We have

(y110y210y31y32)XSACG\ocirc(y110y21+y31y32)YSACG.\begin{pmatrix}y_{11}&&0\\ y_{21}&&0\\ y_{31}&&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}y_{11}&&0\\ y_{21}+y_{31}&&y_{32}\end{pmatrix}^{\text{SACG}}_{Y}.

We see that 3detMY3\mid\det M_{Y}, where MYM_{Y} is the Heuberger matrix for YY shown above. By Cor. 2.10, it follows that YY is 33-colorable unless it has loops. But this cannot happen, because 3y113\mid y_{11} and y11>0y_{11}>0 and y322y_{32}\geq 2.

The case where 3y213\mid y_{21} is handled similarly, as is the case where 3y11+y213\mid y_{11}+y_{21}; in this latter case, begin with a homomorphism that “collapses” the top two rows by adding them.

Finally, suppose that 3y11y213\mid y_{11}-y_{21}. We may assume without loss of generality that y11y21y_{11}\geq y_{21}, for if not, then swap the top two rows. We have

(y110y210y31y32)XSACG\ocirc(y11y210y31y32)YSACG.\begin{pmatrix}y_{11}&&0\\ y_{21}&&0\\ y_{31}&&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}y_{11}-y_{21}&&0\\ y_{31}&&y_{32}\end{pmatrix}^{\text{SACG}}_{Y}.

By Cor. 2.10, it follows that YY is 33-colorable unless it has loops. Because y322y_{32}\geq 2 and 3y11y213\mid y_{11}-y_{21}, this occurs if and only if (i) y11y21=1y_{11}-y_{21}=1 and y32y31y_{32}\mid y_{31}, or (ii) y11y21=0y_{11}-y_{21}=0 and gcd(y31,y32)=1\text{gcd}(y_{31},y_{32})=1.

Suppose (i) occurs. By our assumption that y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0, we must have that y31=0y_{31}=0. But then by [1, Lemma LABEL:general-lemma-block-structure, Theorem LABEL:general-theorem-tree-guard, and Example LABEL:general-example-cycle], we have that XX is 33-colorable.

Now suppose that (ii) occurs. Let α,β\alpha,\beta\in\mathbb{Z} such that αy31+βy32=1\alpha y_{31}+\beta y_{32}=1. Multiply MXM_{X} on the right by the unimodular matrix

(αy32βy31)\begin{pmatrix}\alpha&-y_{32}\\ \beta&y_{31}\end{pmatrix}

to get

(y110y110y31y32)XSACG=(αy11y11y32αy11y11y3210)XSACG\ocirc(102αy112y11y32)YSACG\begin{pmatrix}y_{11}&0\\ y_{11}&0\\ y_{31}&y_{32}\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}\alpha y_{11}&-y_{11}y_{32}\\ \alpha y_{11}&-y_{11}y_{32}\\ 1&0\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ -2\alpha y_{11}&2y_{11}y_{32}\end{pmatrix}^{\text{SACG}}_{Y^{\prime}}

Provided YY^{\prime} does not have loops, YY^{\prime} is isomophic to the Heuberger circulant Cn(a,b)C_{n^{\prime}}(a^{\prime},b^{\prime}) with n=2y11y32n^{\prime}=2y_{11}y_{32}, a=2αy11a^{\prime}=2\alpha y_{11}, b=1b^{\prime}=1. So YY^{\prime} is 33-colorable unless it has loops or one of the exceptional cases in Theorem 2.6 occurs. We now deal with these possibilities one at a time.

Suppose YY^{\prime} has loops. Because y11>0y_{11}>0 and y322y_{32}\geq 2, this occurs if and only if 2y11y322αy112y_{11}y_{32}\mid 2\alpha y_{11}, which happens if and only if y32αy_{32}\mid\alpha. But then using column operations, we get that XX has loops, contrary to our assumptions. Observe that n=±5n^{\prime}=\pm 5 and n=±13n^{\prime}=\pm 13 cannot happen, because nn^{\prime} is even. In the remaining cases, we show y11=1y_{11}=1, then we proceed from there. Suppose a±2b(modn)a^{\prime}\equiv\pm 2b^{\prime}\;(\textup{mod}\;n^{\prime}). So 2αy11±2(mod 2y11y32)2\alpha y_{11}\equiv\pm 2\;(\textup{mod}\;2y_{11}y_{32}). So αy11±1(mody11y32)\alpha y_{11}\equiv\pm 1\;(\textup{mod}\;y_{11}y_{32}). So y11±1y_{11}\mid\pm 1, which implies y11=1y_{11}=1. Suppose 2a±b(modn)2a^{\prime}\equiv\pm b^{\prime}\;(\textup{mod}\;n). So 4αy11±1(mod 2y11y32)4\alpha y_{11}\equiv\pm 1\;(\textup{mod}\;2y_{11}y_{32}). So y11±1y_{11}\mid\pm 1, which implies y11=1y_{11}=1.

Thus we have:

(1010y31y32)XSACG\ocirc(10y31+1y32)Y′′SACG\begin{pmatrix}1&0\\ 1&0\\ y_{31}&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ y_{31}+1&y_{32}\end{pmatrix}^{\text{SACG}}_{Y^{\prime\prime}}

Then Y′′Y^{\prime\prime} has loops if and only if y32y31+1y_{32}\mid y_{31}+1. But then y31=ky321y_{31}=ky_{32}-1 for some integer kk. From our assumption that y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0, we must have that k=0k=0 and y31=1y_{31}=-1. The conclusion of the lemma now follows in this case from Lemma 2.15.

So now we can assume that y31<1y_{31}<-1. Then by Lemma 2.7, we have that Y′′Y^{\prime\prime} is isomorphic to the Heuberger ciculant Cn′′(a′′,b′′)C_{n^{\prime\prime}}(a^{\prime\prime},b^{\prime\prime}) with n′′=y32n^{\prime\prime}=y_{32}, a′′=y311a^{\prime\prime}=-y_{31}-1, b′′=1b^{\prime\prime}=1. Suppose Y′′Y^{\prime\prime} is not 33-colorable. So one of the exceptional cases in Theorem 2.6 occurs. We now deal with these possibilities one at a time.

»  Suppose that n′′=5n^{\prime\prime}=5 and a′′±2b′′(mod 5)a^{\prime\prime}\equiv\pm 2b^{\prime\prime}\;(\textup{mod}\;5). Then y32=5y_{32}=5 and a′′2a^{\prime\prime}\equiv 2 or 33 mod 55. Hence y31=2y_{31}=-2, using that y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0 as well as that y31<1y_{31}<-1. But then

(101025)XSACG\ocirc(1015)SACG\begin{pmatrix}1&0\\ 1&0\\ -2&5\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ -1&5\end{pmatrix}^{\text{SACG}}

gives us a 33-coloring of XX.

»  Suppose n′′=13n^{\prime\prime}=13 and one of a′′a^{\prime\prime} or b′′b^{\prime\prime} is congruent to ±5\pm 5 times the other modulo 1313: Then y32=13y_{32}=13 and a′′5a^{\prime\prime}\equiv 5 or 88 modulo 1313. From y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0 and y31<1y_{31}<-1 and a′′=y311a^{\prime\prime}=-y_{31}-1, we get y31=6y_{31}=-6. But then

(1010613)XSACG\ocirc(20613)SACG\ocirc(10313)SACG\begin{pmatrix}1&0\\ 1&0\\ -6&13\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}2&0\\ -6&13\end{pmatrix}^{\text{SACG}}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ -3&13\end{pmatrix}^{\text{SACG}}

gives us a map to a 33-colorable Heuberger circulant.

»  Suppose a′′2b′′(modn′′)a^{\prime\prime}\equiv 2b^{\prime\prime}\;(\textup{mod}\;n^{\prime\prime}). Then y32y31+3y_{32}\mid y_{31}+3. Note we cannot have y31=2y_{31}=-2, since then y321y_{32}\mid 1, but y32>1y_{32}>1.

If y31=3y_{31}=-3, then we have

(10103y32)XSACG\ocirc(203y32)ZSACG\begin{pmatrix}1&0\\ 1&0\\ -3&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}2&0\\ -3&y_{32}\end{pmatrix}^{\text{SACG}}_{Z}

The graph ZZ cannot have loops. Let a=3,b=2,n=2y32a=3,b=2,n=2y_{32}. Then ZZ is isomorphic to the Heuberger circulant Cn(a,b)C_{n}(a,b). We cannot have n=5n=5 or n=13n=13, because nn is even. If a2b(modn)a\equiv 2b\;(\textup{mod}\;n), then 2y3212y_{32}\mid-1, which cannot happen. If a2b(modn)a\equiv-2b\;(\textup{mod}\;n), then 2y3272y_{32}\mid 7, which cannot happen. If 2ab(modn)2a\equiv b\;(\textup{mod}\;n), then 2y3242y_{32}\mid 4, which implies that y32=2y_{32}=2, but this violates y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0. If 2ab(modn)2a\equiv-b\;(\textup{mod}\;n), then 2y3282y_{32}\mid 8, which implies that y32=2y_{32}=2 or y32=4y_{32}=4, both of which violate y322y310-\frac{y_{32}}{2}\leq y_{31}\leq 0.

Now assume y31<3y_{31}<-3, which implies that y31+3<0y_{31}+3<0. So from y32y31+3y_{32}\mid y_{31}+3, we get that y323y313+y322y_{32}\leq-3-y_{31}\leq-3+\frac{y_{32}}{2}. But y32>0y_{32}>0, so this cannot happen.

»  Suppose that a′′2b′′(modn′′)a^{\prime\prime}\equiv-2b^{\prime\prime}\;(\textup{mod}\;n^{\prime\prime}). Then y32y311y_{32}\mid y_{31}-1. So y321y311+y322y_{32}\leq 1-y_{31}\leq 1+\frac{y_{32}}{2}. But y32>1y_{32}>1.

»  Suppose that 2a′′b′′(modn′′)2a^{\prime\prime}\equiv b^{\prime\prime}\;(\textup{mod}\;n^{\prime\prime}). Then y322y31+3y_{32}\mid 2y_{31}+3. Because y312y_{31}\leq-2, we have 2y31+3<02y_{31}+3<0. Hence y322y313y323y_{32}\leq-2y_{31}-3\leq y_{32}-3, which is a contradiction.

»  Suppose that 2a′′b′′(modn′′)2a^{\prime\prime}\equiv-b^{\prime\prime}\;(\textup{mod}\;n^{\prime\prime}). Then y322y31+1y_{32}\mid 2y_{31}+1. So y322y311y321y_{32}\leq-2y_{31}-1\leq y_{32}-1, which is a contradiction.∎

In our final preparatory step, we contemplate “II on top” matrices. As before, we obtain upper bounds by mapping to graphs with 2×22\times 2 matrices, and we obtain a lower bound in some cases by finding diamond lanyards as subgraphs.

Lemma 2.17.

Suppose XX is a standardized abelian Cayley graph with an associated Heuberger matrix

MX=(1001y31y32),M_{X}=\begin{pmatrix}1&0\\ 0&1\\ y_{31}&y_{32}\end{pmatrix},

where y31,y32>0y_{31},y_{32}>0 and y31y32y_{31}\leq y_{32}. Then

χ(X)={2if y31 and y32 are both odd4if y31=2 and 3y324if 1y31(mod 3) and y32=1+y313otherwise.\chi(X)=\begin{cases}2&\text{if }y_{31}\text{ and }y_{32}\text{ are both odd}\\ 4&\text{if }y_{31}=2\text{ and }3\mid y_{32}\\ 4&\text{if }1\not\equiv y_{31}\;(\textup{mod}\;3)\text{ and }y_{32}=1+y_{31}\\ 3&\text{otherwise}.\par\end{cases}

Before embarking on the proof, we note that by [1, Example LABEL:general-example-arbitrary-distance-graph], we have that XX is isomorphic to the distance graph Cay(,{±1,±y31,±y32})\text{Cay}(\mathbb{Z},\{\pm 1,\pm y_{31},\pm y_{32}\}). Hence Lemma 2.17 is a special case of Zhu’s theorem, as discussed in [2]. We offer here an alternative proof using Heuberger matrices.

Proof.

[1, Lemma LABEL:general-lemma-bipartite] implies that χ(X)=2\chi(X)=2 if and only if y31y_{31} and y32y_{32} are both odd.

To show that XX is 44-colorable, consider

(1001y31y32)XSACG\ocirc(10y311+y32)YSACG\begin{pmatrix}1&0\\ 0&1\\ y_{31}&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ y_{31}&1+y_{32}\end{pmatrix}^{\text{SACG}}_{Y}

By Lemma 2.8 and 2.7, we see that YY does not contain loops and is isomorphic to the Heuberger circulant C1+y32(1,y31)C_{1+y_{32}}(1,y_{31}). So YY is 44-colorable unless y32=4y_{32}=4 and y31{2,3}y_{31}\in\{2,3\}. But in these cases respectively take

(100124)XSACG\ocirc(1023)SACG and (100134)XSACG\ocirc(1134)SACG.\begin{pmatrix}1&0\\ 0&1\\ 2&4\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ 2&3\end{pmatrix}^{\text{SACG}}\text{ and }\begin{pmatrix}1&0\\ 0&1\\ 3&4\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&-1\\ 3&4\end{pmatrix}^{\text{SACG}}.

Indeed, YY is 33-colorable unless one of the six exceptional cases in Theorem 2.6 occurs. Those cases each place restrictions on y31y_{31} and y32y_{32}, whereupon we can modify the mapping appropriately to try to get a 33-coloring. One can show that this procedure will produce a 33-coloring unless either y31=2y_{31}=2 and 3y323\mid y_{32}, or else 1y31(mod 3)1\not\equiv y_{31}\;(\textup{mod}\;3) and y32=1+y31y_{32}=1+y_{31}. The logic is quite similar to that in the proofs of Lemmas 2.15 and 2.16, so we omit it here. Complete details can be found in our authors’ notes, which are housed on the second author’s website [3].

Finally, we show that if either y31=2y_{31}=2 and 3y323\mid y_{32}, or else 1y31(mod 3)1\not\equiv y_{31}\;(\textup{mod}\;3) and y32=1+y31y_{32}=1+y_{31}, then XX contains a diamond lanyard. By Lemma 2.3, this will show that χ(X)4\chi(X)\geq 4 in these cases. We note that this is essentially what Zhu does in [8] to find a lower bound on the fractional chromatic number of distance graphs such as these. Let HH be the subgroup of 3\mathbb{Z}^{3} generated by the columns of MXM_{X}. We denote by (a,b,c)t¯\overline{(a,b,c)^{t}} the vertex (a,b,c)t+H(a,b,c)^{t}+H of XX.

Suppose y31=2y_{31}=2 and 3y323\mid y_{32}. We have that y32=3k±1y_{32}=3k\pm 1 for some positive integer kk. There is a diamond in XX with vertices (0,0,0)t¯,(0,0,1)t¯,(0,0,2)t¯\overline{(0,0,0)^{t}},\overline{(0,0,1)^{t}},\overline{(0,0,2)^{t}}, and (0,0,3)t¯\overline{(0,0,3)^{t}}. Shifting this k1k-1 times by (0,0,3)t¯\overline{(0,0,3)^{t}} and concatenating, we obtain a diamond lanyard of length kk with endpoints (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,0,3k)t¯\overline{(0,0,3k)^{t}}.

Now suppose that 1y31(mod 3)1\not\equiv y_{31}\;(\textup{mod}\;3) and y32=1+y31y_{32}=1+y_{31}. So either y31y_{31} or y32y_{32} equals 3k3k for some positive integer kk. We have a diamond in XX with vertices (0,0,0)t¯,(0,0,1)t¯,(0,0,y32)t¯\overline{(0,0,0)^{t}},\overline{(0,0,1)^{t}},\overline{(0,0,y_{32})^{t}}, and (0,0,y32+1)t¯\overline{(0,0,y_{32}+1)^{t}}. Shifting this by (0,0,y32+1)t¯\overline{(0,0,y_{32}+1)^{t}}, we obtain an unclasped diamond lanyard of length two with endpoints (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,0,2y32+2)t¯\overline{(0,0,2y_{32}+2)^{t}}. Append to this a diamond with vertices (0,0,2y32+2)t¯,(0,0,y32+3)t¯,(0,0,y32+2)t¯\overline{(0,0,2y_{32}+2)^{t}},\overline{(0,0,y_{32}+3)^{t}},\overline{(0,0,y_{32}+2)^{t}}, and (0,0,3)t¯\overline{(0,0,3)^{t}}. We thus obtain an unclasped diamond lanyard of length three with endpoints (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,0,3)t¯\overline{(0,0,3)^{t}}. Shifting this k1k-1 times by (0,0,3)t¯\overline{(0,0,3)^{t}} and concatenating, we obtain a diamond lanyard of length 3k3k with endpoints (0,0,0)t¯\overline{(0,0,0)^{t}} and (0,0,3k)t¯\overline{(0,0,3k)^{t}}. ∎

Finally, we turn our attention to proving Theorem 2.14. The essense of the proof is to show that if we do not have a homomorphism from XX to a 33-colorable graph with a 2×22\times 2 matrix, then MXM_{X} must be in a form where (perhaps after some manipulations) one of the preceding three lemmas applies.

Proof of Theorem 2.14.

The first statement follows from Lemma 2.13, and the second follows from [1, Lemma LABEL:general-lemma-bipartite].

Now suppose that MXM_{X} is one of the six types of matrices listed in the third statement. Lemma 2.17 shows that if MXM_{X} is of the form (10013k1+3k)\begin{pmatrix}1&0\\ 0&1\\ 3k&1+3k\end{pmatrix}, then χ(X)=4\chi(X)=4. For the other five, we can perform row and column operations as per [1, Lemma LABEL:general-lemma-isomorphisms] to obtain a matrix for an isomorphic graph so that either Lemma 2.17 or 2.15 proves the third statement of the corollary. For example, suppose MXM_{X} is of the form (101213k2+3k)\begin{pmatrix}1&0\\ -1&2\\ -1-3k&2+3k\end{pmatrix} for an integer kk. Add the second column to the first and then multiply the third row by 1-1 to produce the matrix (1012123k),\begin{pmatrix}1&0\\ 1&2\\ 1&-2-3k\end{pmatrix}, whereupon Lemma 2.15 applies. We leave the computations in the other four cases to the reader.

Finally, assume that none of the first three statements apply. We will show that XX has a 33-coloring.

Take the following mapping:

(y110y21y22y31y32)XSACG\ocirc(y110y21y31y22y32)YSACG\begin{pmatrix}y_{11}&0\\ y_{21}&y_{22}\\ y_{31}&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}y_{11}&0\\ y_{21}-y_{31}&y_{22}-y_{32}\end{pmatrix}^{\text{SACG}}_{Y}

Let MYM_{Y} be the 2×22\times 2 matrix given above for YY. From Def. 2.11 we have that 3detMY3\mid\det M_{Y}. So by Cor. 2.10, we have that YY (and hence XX) is 33-colorable unless YY has loops. (We remark that we imposed the third condition in Def. 2.11 specifically so that we can use Cor. 2.10 right here.) By Lemma 2.8 and Def. 2.11, we have that either (i) y22y32=1y_{22}-y_{32}=-1, or (ii) y11=1y_{11}=1 and y22y32y21y31y_{22}-y_{32}\mid y_{21}-y_{31}.

Suppose (i) holds. Because 3detMY3\mid\det M_{Y}, we must have that 3y113\mid y_{11}. Now consider

(y110y21y22y31y32)XSACG\ocirc(y110y21+y31y22+y32)SACG\begin{pmatrix}y_{11}&0\\ y_{21}&y_{22}\\ y_{31}&y_{32}\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}y_{11}&0\\ y_{21}+y_{31}&y_{22}+y_{32}\end{pmatrix}^{\text{SACG}}

By Cor. 2.10, this produces a 33-coloring unless the target graph has loops. But by Lemma 2.8, this occurs if and only if y22+y32=±1y_{22}+y_{32}=\pm 1, which gives us that y22=0y_{22}=0 and y32=1y_{32}=1. But then the first statement in the theorem holds, contrary to assumption.

Thus (ii) holds. Because 3detMY3\mid\det M_{Y}, we must have y32=y22+3ky_{32}=y_{22}+3k for some integer k0k\geq 0. (Here we use that y32y22y_{32}\geq y_{22}.) Also (y22y32)=y21y31\ell(y_{22}-y_{32})=y_{21}-y_{31} for some integer \ell, which gives us that y31=y21+3ky_{31}=y_{21}+3k\ell. So we have

MX=(10y21y22y21+3ky22+3k).M_{X}=\begin{pmatrix}1&0\\ y_{21}&y_{22}\\ y_{21}+3k\ell&y_{22}+3k\end{pmatrix}. (3)

Take the mapping

(10y21y22y21+3ky22+3k)XSACG\ocirc(102y21+3k2y22+3k)YSACG.\begin{pmatrix}1&0\\ y_{21}&y_{22}\\ y_{21}+3k\ell&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}1&0\\ 2y_{21}+3k\ell&2y_{22}+3k\end{pmatrix}^{\text{SACG}}_{Y^{\prime}}.

Either YY^{\prime} has loops, or else by Lemma 2.7 we have that YY^{\prime} is isomorphic to the Heuberger circulant Cn(a,b)C_{n^{\prime}}(a^{\prime},b^{\prime}) with n=2y22+3kn^{\prime}=2y_{22}+3k and a=2y213ka^{\prime}=-2y_{21}-3k\ell and b=1b^{\prime}=1.

»  First suppose that YY^{\prime} has loops. By Lemma 2.8, this occurs if and only if 2y22+3k2y21+3k2y_{22}+3k\mid 2y_{21}+3k\ell. Then 2y21+3k=(2y22+3k)q2y_{21}+3k\ell=(2y_{22}+3k)q for some qq\in\mathbb{Z}. So y21=qy22+32k(q)y_{21}=qy_{22}+\frac{3}{2}k(q-\ell). Letting t=qt=q-\ell, by various columm operations we have

(10y21y22y21+3ky22+3k)XSACG\displaystyle\begin{pmatrix}1&&0\\ y_{21}&&y_{22}\\ y_{21}+3k\ell&&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X} =(10qy22+32k(q)y22qy22+32k(q+)y22+3k)XSACG=(1032k(q)y2232k(q+)y22+3k)XSACG\displaystyle=\begin{pmatrix}1&&0\\ qy_{22}+\frac{3}{2}k(q-\ell)&&y_{22}\\ qy_{22}+\frac{3}{2}k(q+\ell)&&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&&0\\ \frac{3}{2}k(q-\ell)&&y_{22}\\ \frac{3}{2}k(-q+\ell)&&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}
=(1032kty2232kty22+3k)XSACG\ocirc(32kty22132kty22+3k)Y′′SACG\displaystyle=\begin{pmatrix}1&&0\\ \frac{3}{2}kt&&y_{22}\\ -\frac{3}{2}kt&&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}\frac{3}{2}kt&&y_{22}\\ -1-\frac{3}{2}kt&&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{Y^{\prime\prime}}

Either Y′′Y^{\prime\prime} has loops, or else by Lemma 2.7 we have that Y′′Y^{\prime\prime} is isomorphic to the Heuberger circulant Cn′′(a,b)C_{n^{\prime\prime}}(a^{\prime},b^{\prime}) with n′′=(3kt+1)y22+92k2tn^{\prime\prime}=(3kt+1)y_{22}+\frac{9}{2}k^{2}t and a′′=32kt+1a^{\prime\prime}=\frac{3}{2}kt+1 and b′′=32ktb^{\prime\prime}=\frac{3}{2}kt.

»  »  First suppose Y′′Y^{\prime\prime} has loops. Let MY′′M_{Y^{\prime\prime}} be the given matrix for Y′′Y^{\prime\prime}. By Lemma 2.8 we have that either the top or bottom row of MY′′M_{Y^{\prime\prime}} is zero, or else n′′n^{\prime\prime} divides every entry in a row of MY′′M_{Y^{\prime\prime}}. If the first row is zero, then kt=0kt=0 and XX had loops to begin with, contrary to assumption. If the second row is zero, then we have 132kt=0-1-\frac{3}{2}kt=0 and y22+3k=0y_{22}+3k=0, so

(1032kty2232kty22+3k)XSACG=(1013k10)XSACG(1013k10)SACG,\begin{pmatrix}1&&0\\ \frac{3}{2}kt&&y_{22}\\ -\frac{3}{2}kt&&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&&0\\ -1&&-3k\\ 1&&0\end{pmatrix}^{\text{SACG}}_{X}\cong\begin{pmatrix}1&&0\\ 1&&3k\\ 1&&0\end{pmatrix}^{\text{SACG}},

which is 33-colorable by Lemma 2.15.

Now suppose that n′′n^{\prime\prime} divides every entry in either the first or second row of MY′′M_{Y^{\prime\prime}}. We will work out here the details of the former case; for the latter, which is similar, see the authors’ notes at [3]. We have that n′′32ktn^{\prime\prime}\mid\frac{3}{2}kt and n′′y22n^{\prime\prime}\mid y_{22}. Observe that because XX does not have loops, hence t0t\neq 0 and k0k\neq 0 (and therefore k>0k>0). We split into cases according to whether tt is positive or negative.

First suppose t>0t>0. From n′′32ktn^{\prime\prime}\mid\frac{3}{2}kt we get

32kt(3kt+1)y22+92k2t32kt\displaystyle-\frac{3}{2}kt\leq(3kt+1)y_{22}+\frac{9}{2}k^{2}t\leq\frac{3}{2}kt
32kt92k2t3kt+1y2232kt92k2t3kt+1because t>0\displaystyle\frac{-\frac{3}{2}kt-\frac{9}{2}k^{2}t}{3kt+1}\leq y_{22}\leq\frac{\frac{3}{2}kt-\frac{9}{2}k^{2}t}{3kt+1}\qquad\text{because }t>0
1232k<y221232k+12+32k3kt+1<32k\displaystyle-\frac{1}{2}-\frac{3}{2}k<y_{22}\leq-\frac{1}{2}-\frac{3}{2}k+\frac{-\frac{1}{2}+\frac{3}{2}k}{3kt+1}<-\frac{3}{2}k

This cannot happen, because both kk and y22y_{22} are integers.

Now suppose t<0t<0. From n′′32ktn^{\prime\prime}\mid\frac{3}{2}kt after some calculations we get that

1232k>y223232k.\frac{1}{2}-\frac{3}{2}k>y_{22}\geq-\frac{3}{2}-\frac{3}{2}k.

Using the fact that y22y_{22} and kk are integers, this tells us that y22=32k+ϵy_{22}=-\frac{3}{2}k+\epsilon where ϵ{0,12,1,32}\epsilon\in\{0,-\frac{1}{2},-1,-\frac{3}{2}\}. We will work out here only the cases where ϵ=0\epsilon=0 and ϵ=1\epsilon=-1; the other two cases are similar, and details can be found at [3].

»  »  »  Suppose ϵ=0\epsilon=0. Then we have

(1032kty2232kty22+3k)XSACG=(1032kt32k32kt32k)XSACG=(10032k032k)XSACG.\begin{pmatrix}1&0\\ \frac{3}{2}kt&y_{22}\\ -\frac{3}{2}kt&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&0\\ \frac{3}{2}kt&-\frac{3}{2}k\\ -\frac{3}{2}kt&\frac{3}{2}k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&0\\ 0&-\frac{3}{2}k\\ 0&\frac{3}{2}k\end{pmatrix}^{\text{SACG}}_{X}.

But this would mean that XX has loops, contrary to assumption.

»  »  »  Suppose ϵ=1\epsilon=-1. So we must have y22=132ky_{22}=-1-\frac{3}{2}k, which gives us that n′′=3kt32k1n^{\prime\prime}=-3kt-\frac{3}{2}k-1. Because y22y_{22}\in\mathbb{Z} and k>0k>0, it follows that kk must be even, so y224y_{22}\leq-4. From n′′y22n^{\prime\prime}\mid y_{22}, we get that

132kn′′1+32k,-1-\frac{3}{2}k\leq n^{\prime\prime}\leq 1+\frac{3}{2}k,

which implies that 0t530\geq t\geq-\frac{5}{3} and hence t=1t=-1. Recall that

y21=qy22+32k(q)=qy22+32kt=qy2232k=(q+1)y22+1.y_{21}=qy_{22}+\frac{3}{2}k(q-\ell)=qy_{22}+\frac{3}{2}kt=qy_{22}-\frac{3}{2}k=(q+1)y_{22}+1.

By Def. 2.11, we have that y222=|y22|2y210\frac{y_{22}}{2}=-\frac{|y_{22}|}{2}\leq y_{21}\leq 0. Thus we have y22(2q+2)y22+20y_{22}\leq(2q+2)y_{22}+2\leq 0. Solving for qq and using that y224y_{22}\leq-4, we find that 14q>1-\frac{1}{4}\geq q>-1, which is a contradiction, since qq is an integer.

»  »  Now suppose that Y′′Y^{\prime\prime} does not have loops, so that Y′′Y^{\prime\prime} is isomorphic to the Heuberger circulant Cn′′(a′′,b′′)C_{n^{\prime\prime}}(a^{\prime\prime},b^{\prime\prime}). We have that Y′′Y^{\prime\prime} (and hence XX) is 33-colorable unless one of the six exceptional cases in Thm. 2.6 occurs. Here we go through only the case where b′′2a′′(modn′′)b^{\prime\prime}\equiv-2a^{\prime\prime}\;(\textup{mod}\;n^{\prime\prime}) and 3\nmidn′′3\nmid n^{\prime\prime}, and even then, only a few of the illustrative sub-cases. The others can be found at [3]—as we go along, we will no longer mention that every time.

The condition b′′2a′′(modn′′)b^{\prime\prime}\equiv-2a^{\prime\prime}\;(\textup{mod}\;n^{\prime\prime}) implies that n′′92kt+2n^{\prime\prime}\mid\frac{9}{2}kt+2. We split into cases according to whether tt is positive or negative, and we discuss here only the case t<0t<0. We have

92kt+2(3kt+1)y22+92k2t92kt2.\frac{9}{2}kt+2\leq(3kt+1)y_{22}+\frac{9}{2}k^{2}t\leq-\frac{9}{2}kt-2.

After some algebra this leads to

32k+32>y2232k2.-\frac{3}{2}k+\frac{3}{2}>y_{22}\geq-\frac{3}{2}k-2.

Because 3\nmidn′′=(3kt+1)y22+92k2t3\nmid n^{\prime\prime}=(3kt+1)y_{22}+\frac{9}{2}k^{2}t, we have that 3\nmidy223\nmid y_{22}. Hence, because kk and y22y_{22} are integers, we have that y22=32k+ϵy_{22}=-\frac{3}{2}k+\epsilon where ϵ{1,12,12,1,2}\epsilon\in\{1,\frac{1}{2},-\frac{1}{2},-1,-2\}. We cover here only the cases ϵ=12\epsilon=\frac{1}{2} and ϵ=12\epsilon=-\frac{1}{2}, as they illustrate some of the various possibilities that can arise.

» » » Suppose ϵ=12\epsilon=\frac{1}{2}. It follows that n′′=32k(1t)+12n^{\prime\prime}=-\frac{3}{2}k(1-t)+\frac{1}{2}, from which we find that kk is odd and tt is even. From n′′92kt+2n^{\prime\prime}\mid\frac{9}{2}kt+2 we get that n′′92k+12n^{\prime\prime}\mid\frac{9}{2}k+\frac{1}{2} gives us that n′′=±1n^{\prime\prime}=\pm 1. Thus

92k+1232k(1t)+12.\frac{9}{2}k+\frac{1}{2}\leq-\frac{3}{2}k(1-t)+\frac{1}{2}.

Solving for tt, we find that

423kt,-4-\frac{2}{3k}\leq t,

which gives us that t=2t=-2 or t=4t=-4. If t=2t=-2, then n′′=92k+124n^{\prime\prime}=-\frac{9}{2}k+\frac{1}{2}\leq-4, which together with n′′92k+12n^{\prime\prime}\mid\frac{9}{2}k+\frac{1}{2} gives us that n′′1n^{\prime\prime}\mid 1, a contradiction. So t=4t=-4. Then n′′=152k+127n^{\prime\prime}=-\frac{15}{2}k+\frac{1}{2}\leq-7. Thus n′′n^{\prime\prime} divides 5(92k+12)+3n′′=45(\frac{9}{2}k+\frac{1}{2})+3n^{\prime\prime}=4, a contradiction.

» » » Suppose ϵ=12\epsilon=-\frac{1}{2}. It follows that n′′=32k(1+t)12n^{\prime\prime}=-\frac{3}{2}k(1+t)-\frac{1}{2}. Thus kk is odd and tt is even. Using the same sort of logic as in the ϵ=12\epsilon=\frac{1}{2} case, we get that n′′92k+12n^{\prime\prime}\mid-\frac{9}{2}k+\frac{1}{2}, and we again find that t=2t=-2 or t=4t=-4. If t=2t=-2, we have n′′=32k12>0n^{\prime\prime}=\frac{3}{2}k-\frac{1}{2}>0. So n′′n^{\prime\prime} divides 92k+12+3n′′=1-\frac{9}{2}k+\frac{1}{2}+3n^{\prime\prime}=-1, giving us n′′=1n^{\prime\prime}=1, in turn giving us k=1k=1. But then y22=32k+ϵ=2y_{22}=-\frac{3}{2}k+\epsilon=-2 and y32=y22+3k=1y_{32}=y_{22}+3k=1, violating the condition |y22||y32||y_{22}|\leq|y_{32}| from Def. 2.11.

So t=4t=-4. This gives us that n′′=92k124n^{\prime\prime}=\frac{9}{2}k-\frac{1}{2}\geq 4. By the choice of ϵ\epsilon, we have y22=32k122y_{22}=-\frac{3}{2}k-\frac{1}{2}\leq-2. By Def. 2.11 and our previous formula for y21y_{21}, we have that

34k14q(32k12)+32k(4)0.-\frac{3}{4}k-\frac{1}{4}\leq q\left(-\frac{3}{2}k-\frac{1}{2}\right)+\frac{3}{2}k(-4)\leq 0. (4)

Solving for qq, we find that

57+83k+12q8.-5\geq-7+\frac{8}{3k+1}\geq 2q\geq-8. (5)

Because qq is an integer, this gives us that q=3q=-3 or q=4q=-4. If q=3q=-3, then from (5) and from the fact that kk is an odd positive integer, we get that k=1k=1. But then y22=32k+ϵ=2y_{22}=-\frac{3}{2}k+\epsilon=-2 and y32=y22+3k=1y_{32}=y_{22}+3k=1, violating the condition |y22||y32||y_{22}|\leq|y_{32}| from Def. 2.11. So q=4q=-4. But this contradicts (4).

»  Now suppose that YY^{\prime} does not have loops, so YY^{\prime} is isomorphic to the Heuberger circulant Cn(a,b)C_{n^{\prime}}(a^{\prime},b^{\prime}) with n=2y22+3kn^{\prime}=2y_{22}+3k and a=2y213ka^{\prime}=-2y_{21}-3k\ell and b=1b^{\prime}=1. We have that YY^{\prime} (and hence XX) is 33-colorable unless one of the six exceptional cases in Thm. 2.6 occurs. We work out here with some granularity only one of these cases, namely where 3\nmidn3\nmid n^{\prime} and a2b(modn)a^{\prime}\equiv 2b^{\prime}\;(\textup{mod}\;n^{\prime}). The other five cases are handled similarly. Complete details can be found in our authors’ notes, which are housed on the second author’s website [3].

From 3\nmidn3\nmid n^{\prime} we get that 3\nmidy223\nmid y_{22}. From a2b(modn)a^{\prime}\equiv 2b^{\prime}\;(\textup{mod}\;n^{\prime}) we get that y21=qy221+32k(q)y_{21}=qy_{22}-1+\frac{3}{2}k(q-\ell) for some integer qq. Let t=qt=q-\ell. Note kk or tt must be even. We have:

(10y21y22y21+3ky22+3k)XSACG=(10qy221+32k(q)y22qy221+32k(q)+3ky22+3k)XSACG\begin{pmatrix}1&0\\ y_{21}&y_{22}\\ y_{21}+3k\ell&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&0\\ qy_{22}-1+\frac{3}{2}k(q-\ell)&y_{22}\\ qy_{22}-1+\frac{3}{2}k(q-\ell)+3k\ell&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}
=(10qy221+32k(q)y22qy221+32k(q+)y22+3k)XSACG=(101+32k(q)y221+32k(q+)y22+3k)XSACG=(101+32kty22132kty22+3k)XSACG=\begin{pmatrix}1&0\\ qy_{22}-1+\frac{3}{2}k(q-\ell)&y_{22}\\ qy_{22}-1+\frac{3}{2}k(q+\ell)&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&0\\ -1+\frac{3}{2}k(q-\ell)&y_{22}\\ -1+\frac{3}{2}k(-q+\ell)&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}=\begin{pmatrix}1&0\\ -1+\frac{3}{2}kt&y_{22}\\ -1-\frac{3}{2}kt&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}

Suppose t=0t=0. Then y21=qy221y_{21}=qy_{22}-1. From Def. 2.11 we have that |y22|2y210-\frac{|y_{22}|}{2}\leq y_{21}\leq 0. Suppose y22>0y_{22}>0. If y22=1y_{22}=1, then y21=0y_{21}=0, and in this case the theorem now follows from Lemma 2.17. So we may assume that y222y_{22}\geq 2. Then from qy2210qy_{22}-1\leq 0 we get that q0q\leq 0. From y222qy221-\frac{y_{22}}{2}\leq qy_{22}-1 we then get that q=0q=0. From t=qt=q-\ell we then get =0\ell=0. So y21=y31=1y_{21}=y_{31}=-1. After multiplying the bottom row of MXM_{X} by 1-1, the theorem now holds by Lemma 2.15. The same is true for similar reasons if y22<0y_{22}<0. Hence we may assume that t0t\neq 0. A similar argument shows that we may assume that k0k\neq 0.

We divide now into cases according to whether t>0t>0 or t<0t<0. We write here only about the case t<0t<0, the case t>0t>0 being similar.

Consider the mapping

(101+32kty22132kty22+3k)XSACG\ocirc(32kty22132kty22+3k)Y′′SACG\begin{pmatrix}1&0\\ -1+\frac{3}{2}kt&y_{22}\\ -1-\frac{3}{2}kt&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{X}\xrightarrow{\ocirc}\begin{pmatrix}\frac{3}{2}kt&y_{22}\\ -1-\frac{3}{2}kt&y_{22}+3k\end{pmatrix}^{\text{SACG}}_{Y^{\prime\prime}}

Either Y′′Y^{\prime\prime} has loops, or else by Lemma 2.7 we have that Y′′Cn′′(a′′,b′′)Y^{\prime\prime}\cong C_{n^{\prime\prime}}(a^{\prime\prime},b^{\prime\prime}) where a′′=1+32kta^{\prime\prime}=1+\frac{3}{2}kt and b′′=32ktb^{\prime\prime}=\frac{3}{2}kt and n′′=(3kt+1)y22+92k2tn^{\prime\prime}=(3kt+1)y_{22}+\frac{9}{2}k^{2}t. Let MY′′M_{Y^{\prime\prime}} be the Heuberger matrix for Y′′Y^{\prime\prime} given above. Because kt0kt\neq 0 and ktkt\in\mathbb{Z}, it follows that neither row of MY′′M_{Y^{\prime\prime}} is a zero row. Lemma 2.8 then tells us that Y′′Y^{\prime\prime} has loops if and only if n′′n^{\prime\prime} divides every entry in either the top or bottom row of MY′′M_{Y^{\prime\prime}}. Otherwise, we have that Y′′Y^{\prime\prime} (and hence XX) is 33-colorable unless one of the six exceptional cases in Theorem 2.6 holds. This gives us a total of eight cases to consider. Of those, we write here only about the possibility that n′′n^{\prime\prime} divides both 132kt-1-\frac{3}{2}kt and y22+3ky_{22}+3k. The other seven cases can be managed using the same sort of techniques we’ve employed throughout this subsection; a full exposition can be found in [3].

To recap, we now assume that n′′132ktn^{\prime\prime}\mid-1-\frac{3}{2}kt and n′′y22+3kn^{\prime\prime}\mid y_{22}+3k and t<0t<0 and k>0k>0. So

1+32kt(3kt+1)y22+92k2t132kt.1+\frac{3}{2}kt\leq(3kt+1)y_{22}+\frac{9}{2}k^{2}t\leq-1-\frac{3}{2}kt.

Solving for y22y_{22} we find that

1232k>y22132k.\frac{1}{2}-\frac{3}{2}k>y_{22}\geq-1-\frac{3}{2}k.

Because k,y22k,y_{22}\in\mathbb{Z}, we have that y22=32k+ϵy_{22}=-\frac{3}{2}k+\epsilon where ϵ{1,12,0}\epsilon\in\{-1,-\frac{1}{2},0\}. We write here only about the case where ϵ=1\epsilon=-1. The cases where ϵ=12\epsilon=-\frac{1}{2} or ϵ=0\epsilon=0 use the same sorts of techniques we’ve seen previously. So assume y22=32k1y_{22}=-\frac{3}{2}k-1. Then n′′=3k(12+t)1n^{\prime\prime}=-3k(\frac{1}{2}+t)-1 and y22+3k=32k1>0y_{22}+3k=\frac{3}{2}k-1>0. So from n′′y22+3kn^{\prime\prime}\mid y_{22}+3k we have that

3k(12+t)132k1.-3k\left(\frac{1}{2}+t\right)-1\leq\frac{3}{2}k-1.

Solving for tt, we find that t1t\leq-1. Because tt is a negative integer, this implies that t=1t=-1. Recall that y21=qy221+32kt=(q+1)y22y_{21}=qy_{22}-1+\frac{3}{2}kt=(q+1)y_{22}. By Def. 2.11 we have that y222y21=(q+1)y220\frac{y_{22}}{2}\leq y_{21}=(q+1)y_{22}\leq 0. Here we use that y22<0y_{22}<0. Dividing by y22y_{22} we get that 12q+10\frac{1}{2}\geq q+1\geq 0, so q=1q=-1, because qq is an integer. Then y21=0y_{21}=0. From t=qt=q-\ell we get =0\ell=0, so y21=0y_{21}=0. But then XX had loops, contrary to assumption.∎

The operations performed to put a 3×23\times 2 matrix MM into modified Hermite normal form do not affect the gcd of the determinants of the 2×22\times 2 minors of MM. Hence we have the following corollary to Theorem 2.14, in analogy to Corollary 2.10.

Corollary 2.18.

Suppose XX is a standardized abelian Cayley graph with an associated 3×23\times 2 Heuberger matrix MXM_{X}. If XX does not have loops, and if 33 divides the determinant of every 2×22\times 2 minor of MXM_{X}, then XX is 33-colorable.

Proof.

If MXM_{X} has a zero row, we may delete it without affecting the chromatic number, so in this case, the result follows from Cor. 2.10. If the columns of MXM_{X} are linearly dependent over \mathbb{Q}, then after appropriate column operations we obtain a zero column, whereupon the result follows from [1, Thm. LABEL:general-theorem-tree-guard]. Otherwise, as noted just before the statement of this corollary, we may assume that MXM_{X} is in modified Hermite normal form. But for each of the six types of matrices MXM_{X} in the third statement in Theorem 2.14 for which χ(X)>3\chi(X)>3, at least one 2×22\times 2 minor has a determinant not divisible by 33. The result follows.∎

It would be interesting to know whether Cor. 2.18 holds for matrices of arbitrary size. We conjecture that it does.

Indeed, we remark that Thm. 2.14 can be recast entirely so that one can determine the chromatic number directly from an arbitrary Heuberger matrix MM, not necessarily in modified Hermite normal form. Namely, let α,β,γ\alpha,\beta,\gamma be the absolute values of the determinants of the 2×22\times 2 minors of MM, such that αβγ\alpha\leq\beta\leq\gamma. The six exceptional cases in Thm. 2.14 occur precisely when (i) one row of MM has relatively prime entries, and (ii) α>0\alpha>0, and (iii) either {α,β,γ}={1,2,3k}\{\alpha,\beta,\gamma\}=\{1,2,3k\} for some positive integer kk, or else γ=α+β\gamma=\alpha+\beta and αβ(mod 3)\alpha\not\equiv\beta\;(\textup{mod}\;3). The other cases (bipartite, loops, zero row, etc.) can easily be characterized directly from MM.

We previously noted that for a standardized abelian Cayley graph XX with an associated 2×22\times 2 Heuberger matrix MXM_{X}:

(*) If XX does not have loops, then XX fails to be 44-colorable if and only if it contains K5K_{5} as a subgraph, and it fails to be 33-colorable if and only if it contains either C13(1,5)C_{13}(1,5) or a diamond lanyard as a subgraph.

Consequently, (*) holds for a 3×23\times 2 matrix with a zero row, as the corresponding graph equals a box product of a doubly infinite path graph and a graph with a 2×22\times 2 matrix. Each of the six exceptional cases in the third statement in Theorem 2.14 contains a diamond lanyard as a subgraph, as we saw in the proofs of Lemmas 2.15, 2.16, and 2.17. Thus (*) holds also for every standardized abelian Cayley graph XX with an associated 3×23\times 2 Heuberger matrix MXM_{X}.

2.6 An algorithm to find the chromatic number for 1×r,m×1,2×r1\times r,m\times 1,2\times r, or 3×23\times 2 matrices

In this subsection, we provide a “quick-reference guide” to the results of this section. Specifically, we spell out a procedure to determine the chromatic number of a standardized abelian Cayley graph with a Heuberger matrix MXM_{X} of size 1×r,m×1,2×r1\times r,m\times 1,2\times r, or 3×23\times 2, where mm and rr are positive integers. This procedure can easily be converted into code or pseudocode. Indeed, an implementation of this algorithm in Mathematica can be found at [3].

\bullet If MXM_{X} is of size m×1m\times 1, apply [1, Theorem LABEL:general-theorem-tree-guard].

\bullet If MXM_{X} is of size 1×r1\times r, apply Lemma 2.1.

\bullet If MXM_{X} is of size 2×22\times 2, apply column operations as per [1, Lemma LABEL:general-lemma-isomorphisms] to produce a lower-triangular matrix. Then apply Theorem 2.9; if the last statement in that theorem holds, use Theorem 2.6 to complete the final step in the computation.

\bullet If MXM_{X} is of size 2×r2\times r for r>2r>2, perform column operations as per [1, Lemma LABEL:general-lemma-isomorphisms] to produce a zero column. Delete that column, and iterate this procedure until you have a 2×22\times 2 matrix. Then use the procedure from the previous bullet point.

\bullet If MXM_{X} is of size 3×23\times 2, do the following. If MXM_{X} has a zero row, delete that row, then use the procedure for 2×22\times 2 matrices to find the chromatic number of the graph with the resulting matrix. If the columns of MXM_{X} are linearly dependent over \mathbb{Q}, perform column operations as per [1, Lemma LABEL:general-lemma-isomorphisms] to produce a zero column. Delete that column, and then apply [1, Theorem LABEL:general-theorem-tree-guard]. Otherwise, use row and column operations as per Lemma 2.12 to find an isomorphic graph XX^{\prime} with a Heuberger matrix MXM_{X^{\prime}} in modified Hermite normal form. Then apply Theorem 2.14.

Acknowledgments

The authors wish to thank Tim Harris for his careful read of an early version of this manuscript and for his many helpful suggestions.

References

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