Very few Moore Graphs Anurag Bishnoi June 7, 0 Abstract We prove here a well known result in graph theory, originally proved by Hoffman and Singleton, that any non-trivial Moore graph of diameter is regular of degree k =, 3, 7 or 57. The existence (and uniqueness) of these graphs is known for k =, 3, 7 while it is still an open problem if there is a moore graph of degree 57 or not. Basics We ll be talking about simple undirected connected graphs. There is a natural metric d on such graphs given by the shortest distance. Definition.. Diameter d of a graph is the maximum of all shortest distances between pair of vertices. Definition.. Girth g of a non-tree graph is the length of a shortest cycle. Theorem.. (Moore s Inequality) For a graph which is not a tree g d +. Proof. We ll prove this by contradiction. Say g > d +. Let C be a cycle of length g. Then there are two vertices u, v on C which are connected by a path of length d +. Since d is the diameter, there must be a path of length at most d between u, v. But these two paths, being distinct, will form a cycle of length at most d + which is less than g. This contradicts the fact that g is the girth. Graphs that achieve equality in this bound are known as Moore Graphs. Trivial examples are odd cycles and complete graphs. A non-trivial example is the Petersen Graph with d = and g = 5. Theorem.. Moore Graphs are regular. Proof. Note that in a Moore Graph any two vertices are joined by a unique shortest path and if they are joined by a path of length at most d then that is the unique shortest path. Now, let u, v be two vertices of a Moore Graph G at distance d from each other. Then we can deduce that deg(u) = deg(v) by exhibiting a bijection between N(u) and N(v) as follows. Let x N(u). Then x is joined to v by a shortest path. Let the image of x be the neighbor of v on this path. It is straightforward to check that this is a bijection. To show that all vertices have the same degree, take a cycle of length d +. Any two adjacent vertices vertices on this cycle are distance d apart from a single vertex on the cycle and therefore have the same degree. So, all vertices on this cycle have same degree, say k. Now let x be any vertex not on this cycle. Then there exists a point y on the cycle distance d apart from x. Hence deg(x) = k.
Theorem.3. If G is a n vertex Moore Graph of degree k and diameter d, then n = + k + k(k ) +... + k(k ) d. Proof. Let u be a vertex of G. Then for all 0 i d and for each vertex x with d(u, x) = i there exists is a unique path from u to x. Hence, the number of vertices from u at distance i is k(k ) i where i, which can be proved by induction. So the total number of vertices in the graph is + k + k(k ) +... + k(k ) d since d is the maximum distance a vertex can have from u. Distance Regular Graphs Definition.. Given a graph G with diameter d we define A i, for 0 i d to be the n n matrix with rows and columns indexed by V (G) such that A i (x, y) = if d(x, y) = i and 0 otherwise. Note that A 0 is the identity matrix and A the adjacency matrix of G. We also have d i=0 A i = J. Definition.. A graph G with diameter d is called distance regular graph if there exists constants a 0,..., a d, b 0,..., b d, c,..., c d such that the following holds : Given any two vertices x, y at distance i, the number of neighbors of x at distance i /i/i + from y is c i /a i /b i. Note that a distance regular graph is k-regular with k = b 0 and hence it is a stronger condition than regularity. Also by triangle inequality we see that if d(x, y) = i and z N(x) then i d(z, y) i +. Therefore we have a i + b i + c i = k for all i, taking c 0 = b d = 0. Theorem.. Moore graphs are distance regular. Proof. It is easy to see that for k regular Moore graphs with diameter d, c = c =... = c d =, a 0 =... = a d = 0, a d = k, b 0 = k, b =... = b d = k. Let A i s be the matrices as defined before for a graph G. For distance regular graphs we can see that each A i is a polynomial in A as follows : Lets calculare A A i. Then A A i (x, z) = #{y v(g) : d(x, y) =, d(y, z) = i}. Therefore we have A A i = b i A i + a i A i + c i+ A i+, 0 i d. Hence by finite induction on i we see that Theorem.. There are polynomials p i, 0 i d s.t. A i = p i (A ). They can be found by the following recurrence relation : p 0 (z) =, p (z) = z and c i+ p i+ (z) = p (z)p i (z) b i p i (z) a i p i (z). Let q(z) = p i (z). Then q(a ) = J. Therefore, Theorem.3. If A is the adjacency matrix of a distance regular graph G with diameter d and degree k then there is an explicitly computable polynomial q of degree d such that q(a) = J. And the eigenvalues of A are either k or some root of q(z). Proof. We see that since the graph is k regular, k is an eigenvalue of A with eigenvector. Any other eigenvalue λ corresponds to an eigenvector x such that x is orthogonal to. So Jx = 0. Now, q(λ)x = q(a)x = Jx = 0 and x 0. Therefore q(λ) = 0.
Let A be the adjacency matrix of a graph G of n vertices. Then A C n n and A is self adjoint. Definition.3. The unital subring of C n n generated by A is called the adjacency algebra of G denoted by A. Clearly the adjacency algebra is a vector space over C. In fact it is a finite dimensional one since it is a subalgebra of C n n. For a distance regular graph we have the following result on this dimension. Theorem.4. Let G be a distance regular graph with diameter d and adjacency algebra A. Then dim(a) = d +. Proof. We first produce a linearly independent set of d + elements in A to show that dim(a) d + and then show that the powers of A upto d span A. Let A be the adjacency matrix of G. A 0, A,..., A d are polynomials in A by. and hence they belong to A. These matrices are 0 disjoint matrices and hence linearly independent. Now from.3 we know that A has at most d + eigenvalues. Therefore the minimal polynomial of A has degree at most d+. Hence A d+ is a polynomial in I, A, A,..., A d. Hence by induction on m, A m is a polynomial in I, A, A,..., A d for all m d +. Corollary.5. The polynomial q(z) defined in.3 has all distinct roots and all of them are eigenvalues of the adjacency matrix. So given an n vertex distance regular graph G with diameter d, it has d + distinct eigenvalues. Call them λ 0, λ,..., λ d and and let V 0, V,..., V d be the corresponding eigenspaces associated with them. Then we know that C n = V 0 V... V d. Let E is be the projection matrices corresponding to the eigenspaces (after choosing some basis). Then we have E i = E i and E i E j = 0 when i j. Theorem.6. We also have the following properties for these E is :. E 0, E,..., E d is another basis of A.. tr(e i ) = multiplicity of λ i. 3. A i = d j=0 λi je j. 4. A i = d j=0 p i(λ j )E j. The previous theorem gives us some idea of how to attack the problem of finding Moore graphs. Since we must have that the (d + ) (d + ) matrix M = (p i (λ j )) is invertible and tr(e i ) is an integer for all i. 3 Strongly Regular Graphs Definition 3.. A strongly regular graph (srg) is a distance regular graph of diameter at most. If diameter is one then it will be the complete graph so we generally ignore that case. Therefore the parameters for srg are c =, c, a 0 = 0, a, b 0 = k, b = k a. The free parameters a and c are generally denoted as λ, µ and we have an equivalent definition : 3
Definition 3.. An srg with parameters (v, k, λ, µ) is a regular graph of degree k on v vertices such that :. Any two adjacent vertices have exactly λ common neighbors.. Any two non adjacent vertices have exactly µ common neighbors. Theorem 3.. If G is an srg with parameters (v, k, λ, µ) then its complement Ḡ is also an srg. Proof. We would show explicitly the parameters that make Ḡ an srg. Firstly, it is a (v k ) regular graph. Let x, y be two adjacent vertices in Ḡ. Then x, y are non adjacent in G. A vertex z is a common neighbors to both x, y in Ḡ iff it is non adjacent to both x and y in G. But number of such vertices is a constant given by (v ) (k µ) as both x, y have k neighbors each and µ of them are common. Now let x, y to be two non adjacent vertices in Ḡ. Then by a similar argument we get that they have (v ) ((k ) λ) common neighbors. Therefore, Ḡ is also an srg with parameters (v, v k, v + µ k, v + λ k). Theorem 3.. For an srg the parameters v, k, λ, µ satisfy k(k λ ) = µ(v k ). Proof. Fix a vertex x and count the number of induced length two paths from x in two different ways. When talking of srg s we exclude complete graphs, disconnected graphs and all those connected graphs whose complements are disconnected. Which is equivalent to saying that µ must be well defined and positive. Such srg s are called primitive. Theorem 3.3. For any primitive srg v k + with equality holding iff λ = 0 and µ =. Proof. This follows from the previous theorem, since we have k(k ) k(k λ ) = µ(v k ) v k. From the above theorem we have another characterisation of Moore Graphs of diameter two. They are precisely the (k +, k, 0, ) strongly regular graphs! Computing the polynomials for a srg(v, k, λ, µ) we see that p 0 (z) =, p (z) = z and p (z) = (z λz k)/µ Therefore the eigenvalues except for k are roots of the polynomial z + (µ λ)z + µ k. So, if r, s are the its roots then we have µ = k + rs and λ = k + r + s + rs. Theorem 3.4. A primitive srg(v, k, λ, µ) has exactly three eigen values, k, [λ µ + D] and [λ µ D] with corresponding multiplicities, k+(v )(λ µ) [v D ] and k+(v )(λ µ) [v + D ] where D = (λ µ) + 4(k µ). Proof. We know that the eigenvalues are k, r, s where r, s are roots of z +(µ λ)z +µ k. So we just need to prove the multiplicities.. We show that the dimension of eigenspace corresponding to eigenvalue k is. Let A be the adjacency matrix and x = [x, x,..., x v ] t a non-zero vector such that Ax = kx. Suppose x j is the entry of x having latgest absolute value. Then we have kx j = x i where summation is over all those k i s where v i is adjacent to v j. Therefore by maximality of x j we have x j = x i for all such i. Since the graph is connected we can continue in this manner to show that all coordinates of x are equal and hence x = t. 4
. Let f, g be the multiplicities of r, s respectively. Then from Theorem.6 we see that + f + g = v and k + rf + sg = 0, since tr(e 0 ) =, tr(e ) = f, tr(e ) = g, tr(i) = v, tr(a) = 0. To solve this pair of equations, let f = (v )/ + x and g = (v )/ + y. Then we have x = y. Hence we get the result. Putting v = k +, λ = 0, µ = we get the following corollary. Corollary 3.5. If G is a Moore graph of degree k, then k k 4k 3 is an integer. Now we prove our main theorem, Theorem 3.6. A k regular graph G of diameter is a Moore graph only if k =, 3, 7 or 57. Proof. We know from previous corollary that k k 4k 3 must be an integer. One possibility is that k k = 0, this gives us k =. Now let 4k 3 = n. Then we get that k k 0 (mod n). Multiplying both sides by 6 we see that (n + 3) 8(n + 3) 0 (mod n). So n 5. Only non-trivial possibilities are n = 3, 5, 5 giving us k = 3, 7, 57. It has been proved that any non trivial Moore graph must have diameter at most two, but we don t discuss that here. And it can be proved that srg(5,, 0, ) (pentagon), srg(0, 3, 0, ) (Peterson graph) and srg(50, 7, 0, ) (Hoffman Singleton graph) are unique upto isomorphism. Therefore, the list of moore graphs is Odd cycles, Complete Graphs, Peterson Graph, Hoffman Singleton Graph and possibly an srg(350, 57, 0, ). 5