Generalized Quadrangles

Definition: A finite generalized quadrangle is a 1-(v,k,λ) design whose blocks are called lines such that
  1. any two points are on at most one line (and hence any two lines meet in at most one point), and
  2. if P is a point not on a line l, then there is a unique point on l collinear with P.
A generalized quadrangle (GQ) is said to have order (s,t) where k = s + 1 and λ = t + 1.

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 K6 and the lines of the GQ are the 1-factors of K6. 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 s2(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 t2 and t s2.

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)s2. Simplifying this inequality gives (s2 -t)(t+1)(s-1) 0. Since s > 1 this implies t s2. Applying the same argument to the point graph of the dual quadrangle gives s t2.

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.

Line Systems

Definition: A line system (local definition) is a set S of lines through the origin in some Euclidean space d having the property that any two lines in S make an angle of 90° or 60° with each other.

Definition: A star is a system of 3 lines in the plane 2, mutually meeting at 60°. A line system is star-closed if, whenever two lines of S make an angle of 60° the third line of the star they determine also belongs to S.

(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 2½ 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, {ei} denotes a set of mutually orthogonal unit vectors (which we will take as the standard unit vectors).

An: This system consists of the lines determined by all the vectors ei - ej where 0 i < j n in n+1. Thus, in 3, A2 consists of the lines determined by the vectors e0 - e1 = (1,-1,0), e0 - e2 = (1,0,-1), and e1 - e2 = (0,1,-1). Note that the lines all lie in a plane (the plane orthogonal to the vector (1,1,1)). This is true in general, that is, the lines of An will always lie in the n which is orthogonal to the vector j = e0 + e1 + ... + en.

Dn: This system consists of the lines determined by all the vectors ±ei ± ej where 1 i < j n in n. Thus, in 3, D3 consists of the 6 lines determined by the 12 vectors (1,-1,0),(1,0,-1),(0,1,-1), (1,1,0), (1,0,1), (0,1,1), (-1,-1,0), (-1,0,-1), (0,-1,-1), (-1,1,0), (-1,0,1) and (0,-1,1). Note that An is always contained in Dn+1. To see that Dn is indecomposable, suppose that it isn't. Then the lines of Dn can be partitioned into two sets, where a line in one set is perpendicular to all the lines of the other set. Now, the lines determined by the vectors e1 + ei for i > 1 have pairwise inner products equal to 1, so all of these lines must be in the same partition set. But every other line is represented by a vector which has a nonzero inner product with at least one of these vectors, and so can not be in the other partition set. Thus, the other set is empty and Dn is indecomposable. It is easy to see that Dn is star-closed. Any two vectors which are not orthogonal have exactly one common non-zero coordinate. If this coordinate is the same in both vectors (that is, the inner product is 1) then replace one of the vectors by its negative. The sum of the two vectors will now have exactly two non-zero coordinates and so will be in Dn.

E8: There are several descriptions of this system of 120 lines (240 vectors) which is contained in 8, but we shall only give one. The vectors of E8 are the 112 vectors of D8 together with the 128 vectors ½(ε1e1 + ε2e2 + ε3e3 + ε4e4 + ε5e5 + ε6e6 + ε7e7 + ε8e8) where the εi = ±1 and an even number of terms are positive.

E7 and E6: These are subsystems of E8. E7 is the subsystem consisting of all lines of E8 that are perpendicular to a fixed line, while E6 is the subsystem of lines perpendicular to a fixed star.

(3.7) Theorem: Any indecomposable star-closed system of lines at 90° and 60° is isomorphic to An (n 1), Dn (n 4), E6, E7 or E8.

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

<x,a> + <x,b> + <x,c> = 0.
Since each of these terms is in {-1,0,1} we either have that all three are 0 or the three terms are -1, 0, 1 in some order. Now, fix a star represented by a, b and c. Partition the other lines of S into the sets Aa, Ab, Ac and B, where Aa consists of all the lines perpendicular only to a, Ab those perpendicular only to b, Ac those perpendicular only to c and B the set of lines perpendicular to all three. In the following we will adopt the convention that a vector w chosen to represent a line in Aa will be chosen so that <w,b> = 1 and <w,c> = -1.

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

Claim 1: The vectors representing any two lines of Aa 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 Aa. Now u + v + w = b - c, so <t,u> + <t,v> + <t,w> = <t,b> - <t,c> = 2 for all t representing lines in Aa. 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 Aa, 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 An, Dn, E8, E7, E6, 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.