Graphs with smallest eigenvalue -2

Let Γ be a graph whose adjacency matrix A has smallest eigenvalue -2 or greater. Then A + 2I is positive semi-definite and symmetric (the eigenvectors of A + 2I are also eigenvectors of A with eigenvalues that are larger by 2). By (2.22) it is the Gram matrix of a set {x1, ..., xn} of vectors in Euclidean space. These vectors satisfy <xi,xi> = 2, <xi,xj> = 0 or 1 for i j, and so they span a system of lines with angles 90° and 60°. The decomposition of this system into minimal pairwise orthogonal subsystems corresponds to the decomposition of the graph Γ into connected components. So if Γ is connected, the system is indecomposable, and its star-closure S is one of those given by (3.7). We say that Γ is represented in the system S.

We can therefore determine Γ by looking in the systems of (3.7) for sets of vectors with all inner products non-negative. We will first look at how this is done for An.

Definition: The line graph L(Γ) of a graph Γ is the graph whose vertices are the edges of Γ and two vertices are adjacent if the corresponding edges of Γ have a common vertex.

Example: It is easy to see that L(K3) = K3 = T(3), and it follows directly from the definition that T(m) = L(Km). Below we have an example of a generic graph Γ and its line graph:

As another example we have that L(Kn,n) = L2(n), that is the line graph of a complete bipartite graph with bipartite sets of size n is the square lattice graph of order n.(Verify this!)

Proposition: Γ is represented in An if and only if it is the line graph of a bipartite graph on n+1 vertices.

Proof: The vertex set of Γ is a set S of vectors of the form ei - ej, 0 i, j n, i j. Since all the inner products are non-negative, any basis vector ei must appear with the same sign in every member of S that contains it. Now define a graph Γ0 with vertex set {0,1,...,n} and join i to j whenever ±(ei - ej) ε S. Γ0 is bipartite, with the bipartition corresponding to the indices of basis vectors which appear with positive and negative signs. The line graph of Γ0 is clearly Γ.

Definition: Let Γ be a graph with vertex set {v1,...,vr} and let m1,m2,..., mr be non-negative integers. The generalized line graph L(Γ m1,...,mr) is the disjoint union of L(Γ) and cocktail party graphs CP(mi), i = 1,...r, together with all edges between a vertex {vi,vj} of L(Γ) and the cocktail party graphs CP(mi) and CP(mj).

Example: Let Γ be the graph in our last example, the generalized line graph L(Γ;0,1,2,3) is (note CP(0) is the empty graph):

A cocktail party graph CP(m) of order 2m can be represented in Dm+1 by the vectors e1 ± ei for i = 2,...m+1.

The generalized line graph L(Γ m1,...,mr) can be represented in Dn, where n = r + m1 + m2 + ... + mr in the following way. Number the n basis vectors by ei,j, where i = 1,...,r and j = 0,...,mi; and take

S = {ei,0 + ej,0 : {vi,vj} is an edge of Γ} {ei,0 ± ei,j : j = 1,...,mr, i = 1,...,r}.
Proposition: Γ is represented in Dn if and only if it is a generalized line graph.

Proof: The above example shows how to represent a generalized line graph in Dn, so we need only prove the converse. Let Γ be a graph represented in Dn. Let S0 be the set of basis vectors which always occur with the same sign in vertices of Γ. WLOG we can assume that this sign is positive. Then the vectors {ei + ej ε S : ei, ej ε S0} represent a line graph, say L(Γ0). Any basis vector ei not in S0 occurs together with at most one other basis vector ej which must be in S0 since if we have ei ± ej and -ei ± ek in S their inner product will be negative unless j = k, and even in that case it must appear with the same sign in both vectors. As this is just the representation given above, Γ must be a generalized line graph.

Proposition: Only finitely many graphs are represented in E8.

Proof: As there are only 120 lines in E8 this result is obvious. However, we can go further and show that a graph represented in E8 can have no more than 36 vertices. For any vector x ε 8, let Px be the 8 × 8 matrix xTx. The matrices Px are symmetric and span a subspace of the real vector space formed by the 8 × 8 symmetric matrices which has dimension 36. We will prove the result by showing that the matrices Px for x in the representation of a graph are linearly independent.
Suppose that there are real numbers ax such that Σ axPx = 0. Then 0 = tr((Σ axPx)2) = Σ axay tr(PxPy) = Σ axay tr(xTxyTy) = Σ axay (xyT) tr(xTy) = Σ axay (xyT)2 = Σ axay <x,y>2. If G is the Gram matrix of the vectors in the representation and a is the vector of the ax's, then this last sum is just a(GºG)aT. Since G = A + 2I, is positive semidefinite, GºG = A + 4I = G + 2I is positive definite (recall that A is a 01-matrix). Therefore, a = 0 and the Pxs are linearly independent.

These last three propositions can be summarized in:
(3.8) Theorem: With finitely many exceptions, any connected graph whose least eigenvalue is -2 or greater is a generalized line graph. The exceptions are represented in E8.

If we now insist that the graph be regular then

(3.9) Theorem: With finitely many exceptions, a connected regular graph with least eigenvalue -2 is either a line graph or a cocktail party graph. The exceptions are all represented in E8.

Proof: By examining the degrees of a vertex in a cocktail party subgraph and a vertex in the line graph subgraph of a generalized line graph, we see that the graph can not be regular. Thus, either there are no cocktail party subgraphs, and the graph is a line graph, or Γ0 has no edges, and being connected can have only one vertex, so the generalized line graph consists of a single cocktail party graph.

All the exceptional graphs were determined in 1978. There are 187 regular connected graphs represented in E8 which are not line graphs or cocktail party graphs.

If we further insist that the graph be strongly regular then we can prove:
(3.10) Theorem: With finitely many exceptions, a connected strongly regular graph with least eigenvalue -2 is isomorphic to L2(m), T(m) or CP(m) for some m.

Proof: By Theorem (3.9), if Γ is not one of the exceptional graphs, then it is a line graph or a cocktail party graph. To prove this result we must show that if Γ is a line graph then with only finitely many exceptions it must be a T(m) or an L2(m). Suppose then that Γ = L(Γ0). We must show that (with finitely many exceptions) Γ0 = Km (so that Γ = L(Km) = T(m)) or Γ0 = Km,m (so that Γ = L(Km,m) = L2(m)).
Since L(Γ0) is strongly regular, it is regular of degree k. Each vertex of L(Γ0) is contained in exactly two cliques which only have this vertex in common. Since L(Γ0) is connected and regular, only two clique sizes are possible. Each of the cliques corresponds to a vertex of Γ0 and the size of the clique is the degree of the vertex. If the two clique sizes are the same, Γ0 is a regular graph. If the sizes are different, then Γ0 is bipartite with the bipartition consisting of all vertices with the same degree in the same part.
Since Γ is strongly regular, the constancy of λ means that each edge of Γ is contained in exactly λ triangles. An edge of Γ corresponds to a pair of edges meeting at a vertex of Γ0. A triangle in Γ corresponds to either three edges meeting at a common vertex or a triangle in Γ0. Suppose that a triangle a,b,c of Γ corresponds to a triangle of Γ0 with edges a and b meeting at vertex A. In another triangle of Γ containing edge ab, say a,b,c', the vertex c' corresponds to an edge in Γ0 which meets edges a and b. Since edge c joins the non-A vertices of a and b, c' must contain vertex A. In Γ0, vertex A has degree λ+1 (a and b and the λ -1 edges coming from triangles containing the edge ab in Γ other than c.) Now consider the edge c' in Γ0. Since it meets edge a at A, ac' is an edge in Γ, and so, contained in λ triangles. The third points of these triangles give edges in Γ0, not all of which can pass through A, so there exists one of these edges which meets a and c' in their non-A vertices. Repeating this for all pairs of edges through A shows that the closed neighborhood of A is a clique. Γ0 is connected since Γ is, so there is a path from A to any other vertex of Γ0. Every vertex on this path is adjacent to A since their closed neighborhoods are all cliques, so A is adjacent to all vertices of Γ0. As A was arbitrary, Γ0 is a complete graph. Thus, either Γ0 is a complete graph or Γ0 contains no triangles. In the no triangle case, the degree of each vertex is λ+2, so the graph is regular in all cases.
We can now assume that Γ0 is not a complete graph and so it is a regular triangle free graph. Again, since Γ is strongly regular, the number of edges of Γ0 which meet two non-adjacent edges is a constant μ. As Γ0 is triangle fee, we must have μ 2. If μ = 2 then Γ0 must be a complete bipartite graph and the regularity forces both partition sets to have the same size. If μ = 1 then Γ = Γ0 = C5. And if μ = 0 then Γ = Γ0 = K3, which is not a strongly regular graph.

We will determine all the exceptions in the next section.