Algebraic properties of chromatic roots Peter J. Cameron p.j.cameron@qmul.ac.uk 7th Australian and New Zealand Mathematics Convention Christchurch, New Zealand December 2008 Co-authors root . (For example, the complete graph K m + 1 can- not be coloured with m colours.) The problem was suggested by Sir David Wal- lace, director of the Isaac Newton Institute, during On the other hand, no negative integer is a chro- the programme on “Combinatorics and Statistical matic root . Mechanics” during the first half of 2008. Apart from him, others who have contributed include Real chromatic roots Vladimir Dokchitser, F. M. Dong, Graham Farr, Bill Jackson, Kerri Morgan, James Sellers, Alan Sokal, • There are no negative chromatic Theorem 1. and Dave Wagner. roots, none in the interval ( 0, 1 ) , and none in the interval ( 1, 32 27 ] . Chromatic roots • Chromatic roots are dense in the interval [ 32 27 , ∞ ) . A proper colouring of a graph G is a function from the vertices of G to a set of q colours with The non-trivial parts of this theorem are due to the property that adjacent vertices receive differ- Bill Jackson and Carsten Thomassen. ent colours. The chromatic polynomial P G ( q ) of G is the func- Complex chromatic roots tion whose value at the positive integer q is the For some time it was thought that chromatic number of proper colourings of G with q colours. roots must have non-negative real part. This is It is a monic polynomial in q with integer coeffi- true for graphs with fewer than ten vertices. But cients, whose degree is the number of vertices of Alan Sokal showed: G . A chromatic root is a complex number α which is Theorem 2. Complex chromatic roots are dense in the a root of some chromatic polynomial. complex plane. This is connected with the Yang–Lee theory of Integer chromatic roots phase transitions. An integer m is a root of P G ( q ) = 0 if and only if the chromatic number of G (the smallest number Algebraic properties, I of colours required for a proper colouring of G ) is We first observe that any chromatic root is an alge- greater than m . braic integer . The main question is, which algebraic Hence every non-negative integer is a chromatic integers are chromatic roots? 1
Let G + K n denote the graph obtained by adding If α is irrational, then the set { α + n : n ∈ Z } n new vertices to G , joined to one another and to is the set of all quadratic integers with given dis- all existing vertices. Then criminant. So it is enough to show that, for any non-square d congruent to 0 or 1 mod 4, there is P G + K n ( q ) = q ( q − 1 ) · · · ( q − n + 1 ) P G ( q − n ) . a quadratic integer with discriminant d which is a chromatic root. We conclude that if α is a chromatic root, then so is I will sketch the ideas behind the proof of this α + n, for any natural number n . and partial results for higher-degree algebraic in- However, the set of chromatic roots is far from tegers. being a semiring; it is not closed under either ad- dition or multiplication. (Consider α + α and αα , Rings of cliques where α is non-real and close to the origin.) A ring of cliques is the graph R ( a 1 , . . . , a n ) whose vertex set is the union of n + 1 complete subgraphs Algebraic properties, II of sizes 1, a 1 , . . . , a n , where the vertices of each We were led to make two conjectures, as follows. clique are joined to those of the cliques immedi- ately preceding or following it mod n + 1. Conjecture 1 (The α + n conjecture). Let α be an algebraic integer. Then there exists a natural number n Theorem 4 (Read). The chromatic polynomial of such that α + n is a chromatic root. R ( a 1 , . . . , a n ) is a product of linear factors and the poly- nomial � � n n 1 ∏ ∏ ( q − a i ) − ( − a i ) . Conjecture 2 (The n α conjecture). Let α be a chro- q i = 1 i = 1 matic root. Then n α is a chromatic root for any natural number n. We call this the interesting factor . If the α + n conjecture is true, we can ask, for Examples given α , what is the smallest n for which α + n is a chromatic root? • If a i = 1 for all i (so that the graph is an ( n + 1 ) -cycle), the interesting factor is (( q − 1 ) n − ( − 1 ) n ) / q = ( x n − ( − 1 ) n ) / ( x + 1 ) , where x = An example √ q − 1. Its roots are 2 n th roots of unity which The golden ratio α = ( 5 − 1 ) /2 is not a chro- are not n th roots (for n odd), or n th roots (for matic root, as it lies in ( 0, 1 ) . n even). In particular, if n is prime, this factor Also, α + 1 and α + 2 are not chromatic roots is irreducible and its Galois group is cyclic of since their algebraic conjugates are negative or in order n − 1. ( 0, 1 ) . However, there are graphs (e.g. the trun- • If n = 3, the interesting factor of R ( 1, 1, 5 ) is cated icosahedron) which have chromatic roots √ q 2 − 7 q + 11, with roots ( 7 ± 5 ) /2. This is very close to α + 2, the so-called “golden root”. the eight-vertex graph promised earlier. We do not know whether α + 3 is a chromatic root or not. Quadratic integers However, α + 4 is a chromatic root (the smallest For n = 3, the interesting factor of R ( a , b , c ) is x 2 − ( a + b + c ) x + ( ab + bc + ca ) . The discriminant such graph has eight vertices), and hence so is α + of this quadratic is ( a + b + c ) 2 − 4 ( ab + bc + ca ) . n for any natural number n ≥ 4. It takes but a little ingenuity to show that this Quadratic roots discriminant takes all possible values congruent to 0 or 1 mod 4. Theorem 3. Let α be an integer in a quadratic number field. Then there is a natural number n such that α + n For n = 4, we have a four-parameter family is a quadratic root. of cubics for the interesting factors. Are these 2
enough to prove the α + n conjecture for cubic inte- Note also that, if n is prime, then the interesting gers? (We have a long list of cubics obtained from factor is n th cyclotomic polynomial in x = q − 1, this construction but don’t seem to have hit every- so that the cyclic groups of prime order all occur thing!) as Galois groups. The next table shows what happens for small A higher-dimensional family values. Let G be a graph whose vertex set is the union of two cliques, of sizes n and m . For i = 1, . . . , m , let Small rings of cliques F i be the set of neighbours in the first clique of the For given n , we test all non-decreasing n -tuples i th vertex of the second. We may assume without ( a 1 , . . . , a n ) of positive integers with gcd 1 and loss of generality that the union of all the sets F i a n ≤ l . G is the Galois group, in case the polyno- is the whole n -clique, and that their intersection is mial is irreducible. S n and A n are the symmetric empty. and alternating groups of degree n , C n the cyclic The chromatic polynomial can be computed by group of order n , V 4 the Klein group of order 4, D n inclusion-exclusion in terms of the sizes of the F i the dihedral group of order 2 n , and ≀ denotes the and their intersections. wreath product of permutation groups. If m = 2, | F 1 | = a and | F 2 | = b , we have a ring of • n = 4, l = 20: 774 reducible, 3 with G = A 3 , cliques R ( 1, a , b ) . 7215 with G = S 3 . For m = 3, we get a six-parameter family of cu- bics as the “interesting factors”. We have not been • n = 5, l = 20: 586 reducible, 6 with C 4 , 5 with able to find suitable specialisations to prove the V 4 , 360 with D 4 , 6 with A 4 , and 39250 times α + n conjecture using this family. S 4 . So every transitive permutation group of degree up to 4 occurs as a Galois group. A remark on the n α conjecture • n = 6, l = 30: 23228 reducible, one dihedral The only small piece of evidence is the follow- group of order 10, two Frobenius groups of ing. If α is a root of the interesting factor of order 20, three A 5 , 1555851 times S 5 . In this R ( a 1 , . . . , a m ) , then for any natural number n , n α is case, we are missing C 5 . a root of the interesting factor of R ( na 1 , . . . , na m ) . However, this does not generalise to arbitrary More small rings chromatic roots. Problem 3. Is there a graph-theoretic construction n l red S n − 1 Other G �→ F ( G , n ) such that, if α is a chromatic root of G, 7 15 734 113401 C 6 , S 2 ≀ S 3 ( 6 ) , then n α is a chromatic root of F ( G , n ) ? S 3 ≀ S 2 ( 52 ) , PGL ( 2, 5 )( 5 ) 8 10 1132 22630 Galois groups 9 8 152 11054 S 4 ≀ S 2 ( 3 ) A weaker form of our conjecture (modulo the 10 8 1061 18089 Inverse Galois Problem(!)) would assert: 11 6 29 4248 C 10 12 6 592 5492 Conjecture 4. Every finite permutation group of de- 13 6 33 8415 C 12 gree n is the Galois group of an extension of Q gener- 14 6 884 10609 ated by a chromatic root. 15 6 307 15045 This conjecture is amenable to computation. We 16 6 1366 18813 computed the Galois groups of many of the in- There are 16 transitive groups of degree 6. We teresting factors of rings of cliques R ( a 1 , . . . , a n ) . have only found five of them as Galois groups. Note that we can assume without loss that gcd ( a 1 , . . . , a n ) = 1. Not overwhelming support for our conjecture! 3
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