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 A_{n}.

**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(K_{3}) = K_{3} = T(3), and it follows directly from the definition that T(m) = L(K_{m}). Below we have an example of a generic graph Γ and its line graph:

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

**Definition**: Let Γ be a graph with vertex set {v_{1},...,v_{r}} and let m_{1},m_{2},..., m_{r} be non-negative integers. The * generalized line graph* L(Γ m

**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 D_{m+1} by the vectors e_{1} ± e_{i} for i = 2,...m+1.

The generalized line graph L(Γ m_{1},...,m_{r}) can be represented in D_{n}, where n = r + m_{1} + m_{2} + ... + m_{r} in the following way. Number the n basis vectors by e_{i,j}, where i = 1,...,r and j = 0,...,m_{i}; and take

*Proof*: The above example shows how to represent a generalized line graph in D_{n}, so we need only prove the converse. Let Γ be a graph represented in D_{n}. Let S_{0} 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 {e_{i} + e_{j} ε S : e_{i}, e_{j} ε S_{0}} represent a line graph, say L(Γ_{0}). Any basis vector e_{i} not in S_{0} occurs together with at most one other basis vector e_{j} which must be in S_{0} since if we have e_{i} ± e_{j} and -e_{i} ± e_{k} 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 E_{8}.

*Proof*: As there are only 120 lines in E_{8} this result is obvious. However, we can go further and show that a graph represented in E_{8} can have no more than 36 vertices. For any vector x ε ^{8}, let P_{x} be the 8 × 8 matrix x^{T}x. The matrices P_{x} 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 P_{x} for x in the representation of a graph are linearly independent.

Suppose that there are real numbers a_{x} such that Σ a_{x}P_{x} = 0. Then 0 = tr((Σ a_{x}P_{x})^{2}) = Σ a_{x}a_{y} tr(P_{x}P_{y}) = Σ a_{x}a_{y} tr(x^{T}xy^{T}y) = Σ a_{x}a_{y} (xy^{T}) tr(x^{T}y) = Σ a_{x}a_{y} (xy^{T})^{2} = Σ a_{x}a_{y} <x,y>^{2}. If G is the Gram matrix of the vectors in the representation and a is the vector of the a_{x}'s, then this last sum is just a(GºG)a^{T}. 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 P_{x}s 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 E_{8}.

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 E_{8}.

*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 Γ_{0} 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 E_{8} 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 L_{2}(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 L_{2}(m). Suppose then that Γ = L(Γ_{0}). We must show that (with finitely many exceptions) Γ_{0} = K_{m} (so that Γ = L(K_{m}) = T(m)) or Γ_{0} = K_{m,m} (so that Γ = L(K_{m,m}) = L_{2}(m)).

Since L(Γ_{0}) is strongly regular, it is regular of degree k. Each vertex of L(Γ_{0}) is contained in exactly two cliques which only have this vertex in common. Since L(Γ_{0}) is connected and regular, only two clique sizes are possible. Each of the cliques corresponds to a vertex of Γ_{0} and the size of the clique is the degree of the vertex. If the two clique sizes are the same, Γ_{0} is a regular graph. If the sizes are different, then Γ_{0} 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 Γ_{0}. A triangle in Γ corresponds to either three edges meeting at a common vertex or a triangle in Γ_{0}. Suppose that a triangle a,b,c of Γ corresponds to a triangle of Γ_{0} 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 Γ_{0} which meets edges a and b. Since edge c joins the non-A vertices of a and b, c' must contain vertex A. In Γ_{0}, 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 Γ_{0}. 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 Γ_{0}, 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. Γ_{0} is connected since Γ is, so there is a path from A to any other vertex of Γ_{0}. Every vertex on this path is adjacent to A since their closed neighborhoods are all cliques,
so A is adjacent to all vertices of Γ_{0}. As A was arbitrary, Γ_{0}
is a complete graph. Thus, either Γ_{0} is a complete graph or Γ_{0}
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 Γ_{0} 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 Γ_{0} which meet two
non-adjacent edges is a constant μ. As Γ_{0} is triangle fee, we must have μ
2. If μ = 2 then Γ_{0} must be a complete bipartite graph and the regularity forces both partition sets to have the same size. If μ = 1 then Γ = Γ_{0} = C_{5}. And if μ = 0 then Γ = Γ_{0} = K_{3}, which is not a strongly regular graph.

We will determine all the exceptions in the next section.