- any two points are on at most one line (and hence any two lines meet in at most one point), and
- if P is a point not on a line l, then there is a unique point on l collinear with P.

A GQ with t = 1 is a grid (two sets of parallel lines whose points of intersection are the points of the GQ) and a GQ with s = 1 is a complete bipartite graph (also called a *dual grid*). These two cases are usually considered to be *trivial* GQ's. A non-trivial example is given by: the vertices of the GQ are the edges of the complete graph K_{6} and the lines of the GQ are the 1-factors of K_{6}. This is a GQ of order (2,2).

Given a GQ we can form a graph called the * point graph* (or collinearity graph) of the GQ whose vertices are the points of the GQ and two vertices are adjacent if the points are collinear.

(3.a) **Lemma**: The point graph of a GQ of order (s,t) is strongly regular with parameters ((s+1)(st+1), s(t+1), s-1, t+1).

*Proof*: Each point P of the GQ lies on t+1 lines of size s+1, and none of these lines can have another point in common. Thus, the point graph is regular of degree s(t+1). By condition b) of the definition of a GQ, there can exist no triangles. So, if two points are collinear, no points off of the line they determine can be collinear with both, so the only points in the graph adjacent to both are the other points on the line, that is λ = s - 1. If P and Q are non-adjacent (non-collinear) then Q is collinear with exactly one point on each line through P, so μ = t + 1.

Finally, we determine the number of points in the graph. Let *l* be a line of the GQ. Each point not on *l* is collinear with a unique point of *l*. Thus, there are st points off of *l* and collinear with a fixed point of *l*. This gives st(s+1) points off of *l*, and (s+1) + st(s+1) = (s+1)(st+1) points in total.

(3.b) **Lemma**: The eigenvalues of the point graph of a GQ(s,t) are s(t+1), s-1 and -t-1 with respective multiplicities 1, st(s+1)(t+1)/(s+t), and s^{2}(st+1)/(s+t).

*Proof*: We have derived formulas for the eigenvalues and multiplicities of strongly regular graphs. (See (2.16)).

The integrality conditions on the multiplicities give restrictions on the parameters of a GQ.

(3.c) **Lemma**: The parameters of a GQ with s > 1 and t > 1 satisfy s t^{2} and t s^{2}.

*Proof*: The second of the Krein bounds (2.26b) gives -t( s(t+1) + s -1 + 2(s-1)(-t-1)) ( s(t+1) + -t -1)s^{2}. Simplifying this inequality gives (s^{2} -t)(t+1)(s-1) 0. Since s > 1 this implies t s^{2}. Applying the same argument to the point graph of the dual quadrangle gives s t^{2}.

We now consider the special case of GQ's with s = 2.

(3.d) **Lemma**: If a GQ(2,t) exists, then t ε {1,2,4}.

*Proof*: With s = 2 the multiplicity of the eigenvalue -t -1 of the point graph is 4(2t+1)/(t+2) = 8 - 12/(t+2). Thus, t+2 divides 12, which yields that t ε {1,2,4,10}. The case t = 10 is excluded by the previous lemma (Krein bound).

The parameters of the point graphs of these (putative) GQ's are (6t+3, 2t+2, 1, t+1) which we may rewrite as (6u-3, 2u, 1, u) for u = 2,3, or 5. By Proposition (2.5), these graphs exist and are unique for their parameter sets. Given a strongly regular graph with these parameters, define an incidence structure whose points are the vertices of the graph and whose blocks (lines) are the triangles of the graph. It is easily checked that this incidence structure is a GQ, so the existence and uniqueness of the graphs imply the existence and uniqueness of the GQ's.

**Definition**: A ** star** is a system of 3 lines in the plane

(3.1) **Lemma**: Any line system S is contained in a star-closed line system.

*Proof*: We represent the lines of S by vectors of length 2^{½} lying on the lines. For each of these vectors we have <x,x> = 2, and if x and y are different vectors then <x,y> = 0 (if the corresponding lines are perpendicular) or <x,y> = ±1 (if the lines meet at 60°, the sign depends on the choice of orientation of the vectors). Now suppose that <x,y> = -1, then x + y is a vector representing the third line of the star determined by the lines corresponding to x and y (and we scale this vector so that it has the same length as the others). Let w be another vector representing a line of S, then <w,x+y> = <w,x> + <w,y> ε {-2,-1,0,1,2}. If <w,x+y> = ±2, then <w,x> = <w,y> and we can assume w.l.o.g. that this value is 1. But then, <w-x-y, w-x-y> = <w,w> - 2<w,x+y> + <x+y,x+y> = 0, so w = x+y. Thus, either the line corresponding to x+y is in S, or <w,x+y> ε {-1,0,1} and this line makes an angle of 90° or 60° with all the lines of S and so can be added to S. After finitely many steps we reach a star-closed line system.

**Definition**: A line system S is * indecomposable* if its lines are not contained in the union of two non-zero perpendicular subspaces.

**Examples**: We now provide some examples of indecomposable star-closed systems. In what follows, {e_{i}} denotes a set of mutually orthogonal unit vectors (which we will take as the standard unit vectors).

**A _{n}**: This system consists of the lines determined by all the vectors e

**D _{n}**: This system consists of the lines determined by all the vectors ±e

**E _{8}**: There are several descriptions of this system of 120 lines (240 vectors) which is contained in

**E _{7}** and

(3.7) **Theorem**: Any indecomposable star-closed system of lines at 90° and 60° is isomorphic to A_{n} (n 1), D_{n} (n 4), E_{6}, E_{7} or E_{8}.

*Proof*: Let S be such a system. First note that for any star in the system, every other line of the system is perpendicular to either one or all three of the lines of the star. Choose the vectors a,b and c to represent the lines of the star so that <a,b> = <b,c> = <c,a> = -1. Then, as in the proof of 3.1 we have that c = -a -b. Thus, for any other vector x representing a line in S, we have

Now form a graph Γ whose vertex set is A_{a}, and two of these lines are adjacent if and only if they are perpendicular.

Claim 1: The vectors representing any two lines of A_{a} have a non-negative inner product.

If u and v are two such vectors with <u,v> = -1 then u + v ε S by star closure. But
then <u+v,b> = 2 a contradiction.

Claim 2: Γ satisfies the hypothesis (*) of (2.3).

Let u and v represent adjacent lines. Then b - u and c + v ε S by star closure and since <b-u, c+v> = 1, w = b-c - u - v ε S by star closure. Now <w,a> = <b,a> - <c,a> - <u,a> - <v,a> = -1 + 1 - 0 - 0 = 0, and <w,b> = <b,b> - <c,b> - <u,b> - <v,b> = 2 + 1 - 1 - 1 = 1 so w represents a line in A_{a}. Now u + v + w = b - c, so <t,u> + <t,v> + <t,w> = <t,b> - <t,c> = 2 for all t representing lines in A_{a}. Since by Claim 1, these three inner products can only be 0 or 1, exactly one of them is 0. Thus the lines represented by u, v and w give the required triangle in Γ.

Claim 3: Γ determines S uniquely.

The graph structure of Γ determines the Gram matrix of the vectors of A_{a}, which in turn determines the Gram matrix of the entire line system. This Gram matrix then determines the line system up to isometry and S is the star-closure of this set.

It is readily checked that, for the line systems A_{n}, D_{n}, E_{8}, E_{7}, E_{6}, the graph Γ is isomorphic respectively to a null graph, a windmill, or one of the point graphs of the GQ's with 3 points on a line. Thus, (2.3) and the above claims prove the theorem.