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# PROBLEMS IN ALGEBRAIC COMBINATORICS

PROBLEMS IN ALGEBRAIC COMBINATORICS
C. D. Godsil 1 Combinatorics and Optimization University of Waterloo Waterloo, Ontario Canada N2L 3G1 chris@bilby.uwaterloo.ca Submitted: July 10, 1994; Accepted: January 20, 1995.

an algebraic avour. Except for 6.1, 7.1 and 12.2 they are either folklore, or are stolen from other people. AMS Classi cation Number: 05E99

Abstract: This is a list of open problems, mainly in graph theory and all with

1. Moore Graphs
We de ne a Moore Graph to be a graph with diameter d and girth 2d + 1. Somewhat surprisingly, any such graph must necessarily be regular (see 42]) and, given this, it is not hard to show that any Moore graph is distance regular. The complete graphs and odd cycles are trivial examples of Moore graphs. The Petersen and Ho man-Singleton graphs are non-trivial examples. These examples were found by Ho man and Singleton 23], where they showed that if X is a k-regular Moore graph with diameter two then k 2 f2; 3; 7; 57g. This immediately raises the following question: Support from grant OGP0093041 of the National Sciences and Engineering Council of Canada is gratefully acknowledged.
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1.1 Problem. Is there a regular graph with valency 57, diameter two and girth ve?
We summarise what is known. Bannai and Ito 4] and, independently, Damerell 12] showed that a Moore graph has diameter at most two. (For an exposition of this, see Chapter 23 in Biggs 7].) Aschbacher 2] proved that a Moore graph with valency 57 could not be distance transitive and G. Higman (see 9]) proved that it could not even be vertex transitive. By either a square-counting or an interlacing argument, one can show that the maximum number of vertices in an independent set in the Ho man-Singleton graph is 15. If S is an independent set of size 15 in this graph then each vertex not in S is adjacent to exactly three vertices in S , and so the graph induced by the vertices not in S is 4-regular. This leads to a construction of the Ho man-Singleton graph. Let G be the graph formed by the 35 triples from a set of seven points, with two triples adjacent if they are disjoint. (This is the odd graph O(4), with diameter three.) Call a set of seven triples such that any pair meet in exactly one point a heptad. It is not too hard to show that there are exactly 30 heptads, all equivalent under the action of Sym(7) and falling into two orbits of length 15 under Alt(7). Choose one of these two orbits and then extend G to a graph H by adding 15 new vertices, each adjacent to all seven vertices in a heptad in the selected orbit. Then H is the Ho man-Singleton graph. Note that we may view the vertices in S as points and the vertices not in S as lines, with a point and line incident if the corresponding vertices are adjacent. This gives us a 2-(15; 7; 1) design and in the construction above this design is the design of points and lines in PG(3; 2). Now consider a possible Moore graph with valency 57. In this case an independent set has at most 400 vertices. If S is an independent set of cardinality then the incidence structure formed by the vertices in S and the vertices not in S is a 2-(400; 8; 1) design. The points and lines of PG(3; 7) form a design with these parameters. The Ho man-Singleton graph contains many copies of the Petersen graph. It is easy to show that there are subgraphs in it isomorphic to Petersen's graph with one edge deleted, and it can be shown that any such subgraph must actually induce a copy of the Petersen graph. (See 10: Theorem 6.6], where this is used to show that the Ho man-Singleton graph is unique, i.e., it is the only graph of diameter two, girth ve and valency seven.) As far as I know, it has not been proved that Moore graph with valency 57 must contain even one copy of the Petersen graph (to say nothing of the Ho man-Singleton graph).

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2. Triangle-free Strongly Regular Graphs
A graph is strongly regular if it is not complete or empty and the number of common neighbours of two vertices is determined by whether they are equal, adjacent or neither equal nor adjacent. An (n; k; a; c) strongly regular graph is a k-regular graph on n vertices such that any pair of adjacent vertices has exactly a common neighbours while a pair of distinct non-adjacent vertices has exactly c common neighbours. We are concerned with strongly regular graphs with no triangles, i.e., with a = 0. Any Moore graph with diameter two is an example. Three have already appeared|the cycle on ve vertices, the Petersen graph and the Ho man-Singleton graph|but only four more are known. We describe them. The rst is the Clebsch graph, which we build from Petersen's graph. We may view the vertices of the Petersen graph as the unordered pairs from the set

F := f0; 1; 2; 3; 4g;
where two unordered pairs are adjacent if and only if they are disjoint. It is not hard to show that the maximum size of an independent set in Petersen's graph is four, and that any such set consists of the four pairs containing a given point from our F . Let Si be the independent set formed of the four pairs containing i. Now construct a graph C as follows. If P denotes the vertex set of the Petersen graph, vertex set of C is

1; F; P:
The vertex 1 is adjacent to each of the points of F and the vertex i in F is adjacent to all vertices in Si . Thus C is a 5-regular triangle-free graph on 16 vertices, and it is not di cult to show that it is strongly regular. The Higman-Sims graph is also very easy to construct. Let W22 be the Witt design on 23 points. This is a 3-(22; 6; 1) design with 77 blocks. Let V be the point set of W22 and let B be its block set. The vertex set of the Higman-Sims graph is the set

1 V B:
The adjacencies are as follows. The vertex 1 is adjacent to all vertices in V and each block is adjacent to the six points in V which lie in it, and to all the blocks in B which are disjoint from it. With some e ort it can be shown that this is a (100; 22; 0; 6) strongly regular graph.

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It is possible to partition the vertices of the Higman-Sims graph into two sets of size 50, with the subgraph induced by each set isomorphic the Ho man-Singleton graph. (See, e.g., Exercise 2 in Chapter 8 of 10] or Chapter VI of 41].) The graph induced by the vertices at distance two from a chosen vertex in the HigmanSims graph form a subgraph isomorphic to the complement of the block graph of W22. (The block graph of a design has the blocks of the design for vertices, with two blocks adjacent if and only if the have a point in common.) Thus the complement of the block graph of W22 is triangle-free. It is also a (77; 16; 0; 4) strongly regular graph. (This follows from standard results on quasi-symmetric designs.) The 21 blocks in W22 containing a given point form an incidence structure isomorphic to the projective plane of order four. The remaining 56 blocks form another quasi-symmetric design and the complement of its block graph is a (56; 10; 0; 2) strongly regular graph, known as the Gewirtz graph. Now we have seen seven triangle-free strongly regular graphs, which leads naturally to the question for this section.

2.1 Problem. Is there an eighth triangle-free strongly regular graph?
Biggs 5: Section 4.6]] shows that if a (n; k; 0; c) strongly regular graph exists and c 2 f2; 4; 6g then k is bounded by a function of c. This bounds n too, since =

? n = 1 + k + k(k c 1) :
The smallest open case appears to be the existence of a strongly regular graph with parameters (162; 21; 0; 3). Triangle-free strongly regular graphs are of interest in knot theory. For more information about the connection see, e.g., Jaeger 24]. Unfortunately for the knot theorists the strongly regular graphs they need must not only be triangle-free, they should also be \formally self-dual". For what this means see 24], (or 17: p. 249]); this extra condition does imply that the set of vertices at distance two from a xed vertex must also be a strongly regular graph. The Higman-Sims graph is formally self-dual.

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3. Equiangular Lines
A set of lines through the origin in Rn is equiangular if the angle between any two lines is the same. Our general problem is to determine the maximum size of a set of equiangular lines in Rm. The diagonals of the icosahedron provide a set of six equiangular lines in R3. Let L be a set of equiangular lines in Rm and let x1 ; : : : ; xm be a set of unit vectors such that xi spans the i-th line of L. Let U be the matrix with i-column equal to xi and let denote jxT xj j?1, for i 6= j . Then i

U T U = I + ?1S
where S is a symmetric matrix with all diagonal entries equal to zero, all o -diagonal entries equal to 1 or ?1, rank m and least eigenvalue ? . Since S is an integer matrix this implies that is an algebraic integer. Further, if is not rational then its multiplicity n ? m can be at most n=2. Thus must be rational if n > 2m. Since the only rational algebraic integers are the plain old-fashioned integers, we deduce that if n > 2m then is an integer. In fact must be an odd integer, as we now show. 1 To see this let A be 2 (S + J ? I ). Then A is a symmetric 01-matrix and S = 2A + I ? J . If n ? m > 2 then is an eigenvalue of S + J (with multiplicity at least n ? m ? 1. Hence ( + 1)=2 is a rational eigenvalue of A. Since A is an integer matrix and is rational, this implies that ( + 1)=2 is an integer, and must be an odd integer. Let Xi be the matrix xxT , which represents orthogonal projection on the line spanned by xi . (Note that replacing xi by ?xi does not change Xi .) Finally suppose that the square of the cosine of the angle between any two distinct lines in L is . The ? matrices Xi lie in the +1 vector space of all symmetric m m matrices, which has dimension m2 . The mapping (A; B) ! tr AB is an inner product on this space and the Gram matrix of X1; : : : ; Xn with respect to this inner product is (1 ? )I + J: Since < 1 this is the sum of a positive de nite and a positive semide nite matrix. Hence it is positive de nite and therefore invertible. Consequently the matrices X1 ; : : : ; Xn are ? +1 linearly independent and therefore n m2 . Before discussing how good this bound is, we examine?what happens in the case of equality. +1 If n = m2 then X1 ; : : : ; Xn is a basis for the space of symmetric m m matrices. Hence there are constants ci such that

I=

n X i=1

ciXi :

(3:1)

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Now

1 ? ; if i = j ; tr(Xi ? I )Xj = 0; otherwise and therefore Equation (3.1) yields that tr(Xi ? I ) = ci(1 ? ):

Thus ci = (1 ? m )=(1 ? ) and taking the trace of both sides of Equation (3.1) yields that n = m(1 ? ) : (3:2) 1?m ? +1 Substituting n = m2 here and solving for yields = (m + 2)?1, with the consequence that m + 2 must be the ? square of an odd integer when m 6. +1 Examples of sets of m2 equiangular lines in Rm are known to exist, and be unique, when m = 2; 3; 7 and m = 23. (When m = 2 we may take the diagonals of a regular hexagon and, when m = 3, the diagonals of a regular isohedron. For the remaining two cases, see pages 129{130 and pages 166{167 in 40].)

3.1 Problem. Is there a set of

?m+1

2

equiangular lines in Rm when m > 23?

In nitely many examples of sets of equiangular lines with cardinality of order m3=2 ? +1 are known 40: Theorem 10.5]; it may be that the m2 bound is not even asymptotically correct.

4. Two-graphs
There is another bound on sets of equiangular lines. Consider the matrix S above. Its least eigenvalue is ? , and this eigenvalue has multiplicity n ? m. Let 1; : : : ; m be its remaining eigenvalues. Since tr S = 0, (n ? m) = and, since tr S 2 = n(n ? 1), These two equations imply that n(n ? 1) ? (n ? m) m
X

i

i
X

n(n ? 1) ? (n ? m) 2 =
2

i

i:
2

(n ? m) m

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from which it follows that m(n ? 1) ? (n ? m) 2 0. If equality holds then the eigenvalues 2 1 ; : : : ; m must all be equal. If m < then this implies that 2 n m(2 ?? 1) : m We also obtain the following. spanning these lines with Gram matrix I + ?1S , where > 0. Then 2 m(n ? 1) n?m and, if equality holds then S has exactly two distinct eigenvalues.

4.1 Lemma. Let L be a set of n equiangular lines in Rm , let x1 ; : : : ; xn be unit vectors

A set of n equiangular lines such that S has only two eigenvalues?is the same thing +1 as a regular two-graph on n vertices. Note that an equiangular set of m2 lines in Rm will give equality in this lemma. (But this only gives two examples.) The matrix S in the lemma is symmetric with zero diagonal and entries 1 o the diagonal. If D is diagonal matrix of the same order with diagonal entries 1 then DSD and S are similar and DSD still is symmetric with zero diagonal and entries 1 o the diagonal. We may choose D so that all o -diagonal entries of the rst row and column of DSD are positive. If S has only two eigenvalues then S2 + S + I = 0 (4:1) for some and . Since tr S = 0 and tr S 2 = n(n ? 1), taking the trace of Equation (4.1) yields that = n ? 1. If, as we may assume, S has the form 0 jT S= T j T then Equation (4.1) implies that TJ = ? J; T 2 + T ? (n ? 1)I = ?J: (4:2) 1 From Equation (4.2) we deduce that 2 (T + J ? I ) is the adjacency matrix of a strongly regular graph on n ? 1 vertices. Such a graph must have k = 2c and, conversely, any strongly regular graph with k = 2c on n ? 1 vertices determines a regular two-graph on n vertices. Two surveys on regular two-graphs appear in 40]. We mention one question.

4.2 Problem. Is there a regular two-graph on 76 or 96 vertices?

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5. Hamilton Cycles
We consider the existence of Hamilton cycles in vertex transitive graphs. Ignoring K2, there are only four known vertex transitive graphs without Hamilton cycles. Two of these are the Petersen and Coxeter graphs. The Coxeter graph can be de ned as follows. An anti ag in a projective plane is an ordered pair (p; `), where p is a point and ` is a line such that p 2 `. The vertices of the Coxeter graph are the 28 anti ags from the projective = plane of order two. Two such anti ags (p; `) and (q; m) are adjacent if the set

fp; qg ` m
contains all points of the plane. For more on the Coxeter graph, see Section 12.3 in 8]. The remaining two non-Hamiltonian vertex transitive graphs are obtained from the Petersen and Coxeter graphs by `blowing up' each vertex to a triangle. (Formally we take the line graph of the subdivision graphs of the Petersen and Coxeter graphs. The subdivision graph S (G) of G is obtained by installing one vertex in the middle of each edge of G.) The problem with the blowing-up process is that the graphs produced by applying it a second time are no longer vertex transitive, although they are still cubic and have no Hamilton cycle. For proofs that the Coxeter graph has no Hamilton cycle, see 6, 47].

5.1 Problem. Are there any more connected vertex-transitive graphs without Hamilton
cycles? Still ignoring K2, the known non-Hamiltonian vertex-transitive graphs are not Cayley graphs. Thus we are lead to ask whether all connected Cayley graphs have Hamilton cycles. All known connected vertex-transitive graphs have Hamilton paths, and Lovasz has conjectured that all connected vertex-transitive graphs have Hamilton paths. Witte 49] has proved that all Cayley graphs of p-groups have Hamilton cycles. For a survey of results on Hamilton cycles in Cayley graphs see, e.g., 50]. Babai 3] gives an ingenious proof that p connected vertex-transitive graph on n vera tices must contain a cycle of length at least 3n; no better lower bound is known. Mohar 34] derives an algebraic technique for showing that certain graphs do not have Hamilton cycles. We present a simpli ed version of this for the Petersen graph. Let P denote the Petersen graph and suppose that C is a cycle of length ten in it. Then the edges not in C form a perfect matching in P and the vertices in the line graph of P corresponding to the edges of C induce a cycle of length ten. Thus P has a Hamilton cycle if and only if there is an induced copy of C10 in L(P ). For any graph X

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let i (X ) denote the i-th largest eigenvalue of the adjacency matrix of X . By interlacing 17: Theorem 5.4.1], we know that if Y is an induced subgraph of X then i (Y ) i (X ). Since 7 (C10) > 7(L(P )), the Petersen graph cannot have a Hamilton cycle. The only problem with this argument is that there seems to be no other interesting case where it works. It fails on the remaining three vertex-transitive graphs with no Hamilton cycles. It would be very interesting to nd a modi cation of this technique which could be used to show that Coxeter's graph has no Hamilton cycle.

6. The Matchings Polynomial
Let p(X; k) denote the number of k-matchings in the graph X , i.e., the number of matchings with exactly k edges. If X has n vertices then its matchings polynomial of X is de ned to be X (X; x) = (?1)k p(X; k)xn?2k :
k 0

It is known that the zeros of (X; x) are all real 17: Corollary 6.1.2] and that, if X has a Hamilton path, they are all simple. This leads us to ask: have only simple zeros?

6.1 Problem. Is there a connected vertex-transitive graph X such that (X; x) does not

This question is discussed at some length in 16]. There a graph X is de ned to be -critical if, for each vertex u of X , the multiplicity of as a zero of (X n u; x) is less than its multiplicity as a zero of (X; x). All vertex transitive graphs are -critical, by a straightforward argument. Therefore we could solve Problem 6.1 by showing that if X were a connected -critical graph then must be a simple zero of (X; x). This is known to be true if = 0 (Gallai, see 33: Section 3.1]) or if X is a tree (Neumaier 35]). For details, see 16].

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7. Characterising Line Graphs
If X is a line graph then the least eigenvalue of its adjacency matrix A(X ) is at least ?2. Cameron, Goethals, Seidel and Shult proved a converse to this, which we want to discuss. First, some de nitions. A root system is, more or less, a set of vectors in Rm which is invariant under re ection in the hyperplane orthogonal to any vector in it. Let e1; : : : ; em be the standard basis in Rm . Then the root system Am is the set of vectors

ei ? ej ; i 6= j
and the root system Dm is the set of vectors

ei ej ; i 6= j:
1 We de ne E8 to be D8 together with all vectors in R8 with entries 2 and an even number of positive entries. It is not hard to see that a graph X is the line graph of a bipartite graph if and only if A(X )+2I is the Gram matrix of a subset of Am . We de ne X to be a generalised line graph is it is the Gram matrix of a subset of Dm . Every line graph lies in Dm . Cameron et al. proved that if X is a graph with least eigenvalue at least ?2 then it is either a line graph, a generalised line graph or A(X ) + 2I is the Gram matrix of subset of E8. This extended earlier work, in particular of Alan Ho man. Now Ho man 22] has also proved that a graph with least eigenvalue greater than

p ?1 ? 2

and su ciently large minimum valency is a generalised line graph. (Here `su ciently large' is determined by Ramsey theory, which means that it is only nite in a fairly technical sense :-) .)

7.1 Problem. Is there a classi cation of the graphs X with gous to that of the graphs with least eigenvalue at least ?2?

min (X ) > ?1 ?

p

2, analo-

Let 2 (X ) denote the second-largest eigenvalue of X . Then for the complement X of X we have min (X ) ?1 ? 2 (X ): (This follows because we obtain A(X ) by adding the matrix J , with rank one, to ?I ? p p A(X ).) Hence if min (X ) > ?1 ? 2 then 2 (X ) < 2. This indicates that it should also p be interesting to classify the graphs X such that 2 (X ) < 2.

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Woo and Neumaier 51] have recently proved that, if X is a graph with least eigenvalue greater than the smallest root of the polynomial x3 +2x2 ? 2x ? 2 (approximately ?2:4812) p and the minimum valency of X is large enough, then min (X ) ?1 ? 2 and X has a well-determined structure. This result is very interesting, but it still makes use of Ramsey theory and requires a lower bound on the minimum valency of X .

8. Shannon Capacity
The strong product X Y of the graphs X and Y has vertex set V (X ) V (Y ), with (u; v) adjacent to (u0 ; v0 ) if and only if (a) u u0 and v v0 , (b) u = u0 and v v0 , or (c) u u0 and v = v0 . We denote the strong product of n copies of X by X (n). Let (X ) denote the maximum number of vertices in an independent set in X . It is not hard to show that, for any graphs X and Y we have (X Y ) (X ) (Y ) and from this it follows by Fekete's lemma (see Lemma 11.6 in 27]) that the limit always exists. We call it the Shannon capacity of X . This quantity is of some interest in coding theory, but for further information about this we refer the reader to 30] and the references given there. Let !(X ) be the size of the largest clique in the graph X . We recall that X is perfect if, for any induced subgraph Y of X , the chromatic number of Y is equal to !(Y ). All bipartite graphs are perfect, and there are many other classes of perfect graphs, almost as many as there are graph theorists who have studied them. (For more information see, e.g., 31].) Shannon showed that if X is perfect then its Shannon capacity is equal to (X ). However, using the fact that C5 is self-complementary, it is not hard to show that (C5 C5) = 5 31] and so the Shannon capacity of C5 is at least 5, while (C5 ) = 2. In 1979 Lovasz p settled a long-standing open problem by proving that the Shannon capacity of C5 is 5. His methods enabled the Shannon capacity of many other graphs to be determined, but the following question is still open.
n!1

lim (X (n))1=n

p

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8.1 Problem. What is the Shannon capacity of C7? 9. Perfect Codes
The ball of radius m about a vertex v in a graph X is the set of all vertices in X at distance at most m from v. If C is a subset of V (X ), the packing radius of C is maximum integer e such that the balls of radius e about the vertices in C are pairwise disjoint. An e-code in X is a subset with packing radius at least e and an e-code is perfect if the balls of radius e about its vertices partition V (X ). The Hamming graph H (n; q) has the set of all sequences of length n from f0; : : : ; q ? 1g as its vertices, with two sequences adjacent if they agree on all but one coordinate. If X is the Hamming graph, e-codes and perfect codes are precisely the e-codes and perfect codes of standard coding theory. If e 3 and q is a prime power then there are only two perfect e-codes (see, e.g., 8: p. 355], one in H (11; 3) and the other in H (23; 2). The Johnson graph J (v; k) has all k-subsets of a xed v-subset as its vertices, with two k-subsets adjacent if and only if they intersect in exactly k ? 1 elements. Two k-subsets are then at distance i if and only if they have exactly k ? i elements in common. The graphs J (v; k) and J (v; v ? k) are isomorphic, so we will assume that v 2k. When v = 2k and k is odd, the pair formed by a given k-subset and it complement is a perfect code with e equal to bk=2c. Delsarte 13: p. 55] raised the following question.

9.1 Problem. Is there a perfect code with more than two vertices in a Johnson graph?
The strongest result is due to Roos 37], who proved that if there is a perfect code in J (v; k) with packing radius e then + v 2e e 1 (k ? 1): Hammond 21] proved that there are no perfect codes in J (2v + 1; v) and J (2v + 2; v). Perfect codes in other classes of distance regular graphs can also be very interesting. Perfect codes in the Hamming graphs H (n; q) are part of classical coding theory|if e 3 and q is a prime power then the only perfect codes are binary and ternary Golay codes. Chihara 11] has proved that most of the known families of distance regular graphs do not contain perfect codes. The Johnson graphs are one family of exceptions here, and the closely related odd graphs are another. The odd graph O(k + 1) has the k-subsets of a (2k + 1)-set as its vertices, with two k-subsets adjacent if and only if they are disjoint. (Thus it has the same vertex set as

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J (2k +1; k).) The lines of a Fano plane form a perfect 1-code in O(4) and the blocks of the Witt design on 11 points forms a perfect 1-code in O(6). No other examples are known. It not hard to show that there is a perfect 1-code in O(m + 1) if and only if there is a Steiner system with parameters (m ? 1)?(2m + 1; m; 1):
Hence these codes will not be easy to nd. Smith 43] has proved that there are no perfect 4-codes in the odd graphs. Perhaps it can be proved that there are no perfect e-codes in the odd graphs for e su ciently large (ideally for e 2). Of course we should not forget that there may be interesting classes of codes which are not perfect. Completely regular codes (see Chapter 11 in 8]) provide one example.

10. p-Ranks
Let Hv (k; `) be the 01-matrix with rows and columns respectively indexed by the k- and `-subsets of a xed v-set, and with ij -entry equal to one if and only if the i-th k-subset is contained in the j -th `-subset. When k ` v ? `, the rows of Hv (k; `) are linearly independent over the rationals. A surprisingly large number of the applications of linear algebra to combinatorics rest on this fact. (Some of these are presented in 18].) For the earliest proof of independence known to me, see 20]. More recently Richard Wilson 48] determined the rank of Hv (k; `) over all nite elds. It is not clear what the combinatorial implications of this will be, but surely it will be useful in time. There is a so-called q-analog of this problem. Consider the incidence structure formed by the k- and `-subspaces of a v-dimensional vector space over the eld with q elements (where a k-space is incident with the `-spaces which contain it). Then it is natural to want to know the rank of this matrix. Over what eld? There are three cases. Over the rationals this has been known at least since Kantor 25]. For primes not dividing q the answer appears in 15]; the result is in fact analogous to Wilson's for the rank of Hv (k; `) in positive characteristic. The most interesting case is still open. of a v-dimensional vector space over a eld of order pr ?

10.1 Problem. What is the p-rank for the incidence matrix of k-spaces versus `-spaces

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11. Homomorphisms
Let X and Y be graphs. A mapping f from V (X ) to V (Y ) is a homomorphism if f (u) is adjacent to f (v) in Y whenever u is adjacent to v in X . Since we do not allow vertices to be adjacent to themselves, f must map edges of X to edges of Y . If Y is a complete graph with r vertices then the homomorphisms from X into Y correspond to the proper colourings of X using at most r vertices. The product X Y of the graphs X and Y has vertex set V (X ) V (Y ) and (u; v) is adjacent to (u0 ; v0 ) if and only if u is adjacent to u0 and v is adjacent to v0 . (This is the natural product in the category of graphs and homomorphisms, if it helps.) S. Hedetniemi has made the following conjecture.

11.1 Conjecture (Hedetniemi). For any two graphs X and Y
(X Y ) = minf (X ); (Y )g: When n = 3 we can verify the conjecture by showing that the product of two odd cycles contains an odd cycle. For n = 4, it was proved true by El-Zahar and Sauer in 1985 14]. The remaining cases are still open, in fact we cannot exclude the possibility that (X Y ) is less than 16 for some pair of graphs X and Y with arbitrarily large chromatic number! (See 36] for this and, for some recent work on this problem, with more references, see 38].) The Kneser graph K (v; k) has all k-subsets of a xed v-set as its vertices, with two ksubsets adjacent if they are disjoint. So K (v; 1) is the complete graph Kv and K (2v +1; v) is the odd graph O(v + 1). (In particular, K (5; 2) is Petersen's graph.) The following question was raised by Pavol Hell.

X to Y ?

11.2 Problem. For which pairs (X; Y ) of Kneser graphs is there a homomorphism from
Stahl 44: Section 2] proved that if v > 2k there is a homomorphism from K (v; k) to K (v ? 2; k ? 1). Hence there is a homomorphism from K (v; k) to Kv?2k+2, i.e., the chromatic number of K (v; k) is at most v ? 2k + 2. Lovasz 29] proved that equality holds here, from which it follows that if v ? 2k > v0 ? 2k0 then there is no homomorphism from K (v; k) to K (v0 ; k0 ). Further K (v0 ; k) is an induced subgraph of K (v; k) whenever v0 v.

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For any integer r there is an obvious homomorphism from K (v; k) into K (rv; rk), but none into K (v0 ; kr) when v0 < rv. (See the corollary to Theorem 9 in Stahl 44].) Let X be a graph and let M be the matrix whose columns are the characteristic vectors of the maximal independent subsets of V (X ). The value of the linear program min 1T x Mx 1 x 0 is the fractional chromatic number of X . We denote it by (X ). Note that (X ) is the value of the 01-integer program obtained from this LP (by requiring the entries of x to be 0 or 1), and that this is also the value obtained if we require that x be an integer vector. Perles observed that if there is homomorphism from X to Y then (X ) (Y ). It is not hard to show that if X is vertex transitive then (X ) = jV (X )j= (X ), whence (K (v; k)) = v=k. Thus we conclude, for example, that there is no homomorphism from K (8; 3) into K (11; 4). Nothing else seems to be known about existence or non-existence of homomorphisms between Kneser graphs. Dennis Stanton has raised the following problem. Let Kq (v; k) be the graph whose vertices are k-subspaces of the n-dimensional vector space over GF (q), with two subspaces adjacent if and only if their intersection is zero. What is the chromatic number of Kq (v; k)? Clearly we are only interested in the case where n 2k. When q = 1 the graph Kq (v; k) reduces to the Kneser graph K (v; k). Remark: I am indebted to Pavol Hell, who has had to explain much of the above material to me on more than one occasion. I hope I have it right by now.

12. Compact Graphs
Let G be a graph with adjacency matrix A and let ? be the set of all permutation matrices which commute with A. (Thus ? is isomorphic to Aut(G).) By S (A) we denote the set of all doubly stochastic matrices which commute with A. Then S (A) is is the set of all matrices X such that

XA = AX; X 1 = X T 1 = 1; X 0
and therefore it is a convex polytope. We call G compact if S (A) is equal to the convex hull of ? or, equivalently, if the extreme points of S (A) are all permutation matrices. The following problem is raised implicitly by Tinhofer at the end of 46].

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12.1 Problem. Is there a good characterisation of compact graphs?
Tinhofer 45, 46] has proved a number of results concerning compact graphs. In particular he has shown that trees and cycles are compact, and that the disjoint union of isomorphic compact graphs is compact. It is an easy observation that a compact regular graph must be vertex transitive. The converse to this is false|in 19] it is noted that the line graph of the complete graph Kn is not compact, at least when n 7, and that the automorphism group of a compact regular graph is a multiplicity-free permutation group with rank equal to to the number of distinct eigenvalues of G. Schreck and Tinhofer 39] prove that a regular graph G on p vertices, p prime, is compact if and only if Aut(G) is isomorphic to the dihedral group of order 2p. Using this it can be shown (see 19]) that there is a polynomial time algorithm for determining if a regular graph on a prime number of vertices is compact. The graphs with S (A) = fI g can be recognised in polynomial time 19]. The set S (A) is a semigroup, and the convex hull of ? is a sub-semigroup of it. In 19] it is shown that each equitable partition of G determines an idempotent element X of S (A).

X , where is equitable?

12.2 Problem. Is S (A) generated (as a semigroup) by the convex hull of ? and matrices

A compact graph G has the property that the cells of any equitable partition are the orbits of some group of automorphisms of G. It is not clear if the converse is true. If false then the answer to the previous problem is no.

13. Edge-di erence Polynomials
Let G be a graph with vertex set V = f1; : : : ; ng and edge set E . De ne the edge-di erence polynomial pG by Y pG(x1 ; : : : ; xn ) = (xi ? xj ): The zero set of this polynomial is a set of jE j hyperplanes through the origin in Rn. The number of regions into which Rn is divided by these hyperplanes is equal to the absolute value of the chromatic polynomial of G, evaluated at ?1. Note that pG is actually a function on oriented graphs, but changing the orientation leads at worst to a change in sign. If we expand pG as a sum of monomials then the number of terms in the result equals the number of orientations of G and each term has degree jE j.
(i;j )2E; i<j

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Now let U (n; k) denote the set of all vectors in Rn which have k entries equal, and let V (n; k) denote the set of all vectors with at most k ? 1 distinct entries. We have the following obvious result.

13.1 Lemma. The graph G has independence number less than k if and only if pG is
zero on U (n; k). Further, the chromatic number of G is at least k if and only pG is zero on V (n; k). This suggests that we should be able to obtain information about the independence and chromatic numbers of G by analysing pG but, it seems fair to say, no great progress has been made in this direction yet. Lovasz 32] proves that the ideal of polynomials which vanish on V (n; k) is generated by the polynomials pH , where H is any graph on V (G) consisting of a k-clique and n ? k isolated vertices. He also shows that (G) k if and only if we can write pG in the form

pG = pH1 +

+ pHN ;

where each Hi is a graph on V (G) containing a k-clique. We mention one problem, raised in both 26] and 32]. of terms in the above expansion increases exponentially?

13.2 Question. Is there a sequence of graphs Gi such that the minimum possible number

Analogous results holds for the independence number, see 26, 32, 28]. De Loera 28] shows that certain natural bases for the ideals of polynomials vanishing on U (n; k) and V (n; k) are Grobner bases, which means that there is an e ective algorithm for testing whether pG lies in one of these ideals.

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