Problem 36: The Syntheme-Duad Geometry

GENERALIZED QUADRANGLE 1. Two distinct points are contained in at most one line. 2. Given a line l and a point p not on l, there is exactly one line k through p that intersects l (in some point q)

We have already seen some great examples of generalized quadrangles, as well as applications of them (problem 33: Circular Walks). You'll probably remember that the doily is one of the most famous examples of a generalized quadrangle, as well as being the smallest non-trivial example.

Recall that a generalized quadrangle has order (s, t) if every line contains s + 1 points and every point is contained in t + 1 lines.

We will get back to this shortly. But first, lets look at an example of a generalzed quadrangle in which each line does not contain an equal number of points. We will call a generalized quadrangle such that each point is on two lines, a grid. Certainly we can have a square grid, in which case every line is incident with the same number of points. But this is not a neccesary condition. Here is an example of a grid:

If the grid happens to be square, then it has a definite order (s, t). Specifically, because each point is incident with two lines, it will have order (s,1).

Recall from earlier problems the concept of dual geometries. To obtain a dual from a given geometry, we interchange points with lines. Let' s look at an example of a grid and its dual.

If a generalized quadrangle has order (s, t), then the dual of the quadrangle has order (t, s). Note that a dual grid is really just a complete bipartite graph. As a matter of a fact, any generalized quadrangle with order (1, n) will be a complete bipartite graph on 2n vertices.

We have seen several types of examples of generalized quadrangles, and it is obvious that every point need not be incident with an equal number of lines and every line need not be incident with an equal number of points. It is true however that if a geometry is a generalized quadrangle, one of the following statements must hold:

1. S is a grid with each point on two lines such that the lines are partitioned into two sets, L₁ and L₂. Each line of the first set meets every line of the second set, but no two lines from the same set meet. Lines of L₁ are incident with the same number of points s₁, and lines of L₂ are also incident with the same number of points s₂. (Note that s₁ and s₂ are not neccesarily equal. If they are equal, we simply have case 3).
2. S is a dual grid.
3. S has order (s, t) such that each line is incident with s + 1 points and each point is incident with t + 1 lines.

In case 3 we will call s and t the parameters of S. A great property of generalized quadrangles is the following: if we are given parameters s and t, it follows that v = ( s + 1 )(st + 1) and b = ( t + 1 ) (st + 1). [Recall that b is the total number of lines and v is the total number of points of the geometry.]

Let's see why this is true. Suppose that we have a line l such that it will have s + 1 many points that are incident with it. Then there are v - (s + 1) points not on l. Each point that is not on l is on a unique line meeting l. Thus, lets count all the points not on l by counting all the points on lines that meet l. There are (s + 1)t - many lines (because l has s +1 points and each of these points are incident with t +1 lines. However, we do not count l, so we only have t - many lines). Each of these lines has s points not on l. So there are a total of st (s + 1) points not on l. So v - (s + 1) = st (s + 1). So v = (s + 1) + st (s + 1) = (s + 1)(st + 1). Dually, we can see that b = (t + 1)(st +1).

Now that we've seen the basic properties behind all generalized quadrangles, lets look at a way to generate the generalized quadrangle on fifteen points and fifteen lines (we know that this is the doily!).

Let's define a duad as an unordered pair of elements (i, j) [meaning (i, j ) = (j, i)] such that i and j are from the set {1, 2, 3, 4, 5, 6} and . In addition, define a syntheme as a set {(i ,j), (k, l), (m, n)} of three duads, such that i, j, k, l, m and n are all distinct.

It is a simple exercise to see that there exist 15 duads as well as 15 synthemes. To pick an unordered pair from our set {1, 2, 3, 4, 5, 6} such that no number of the set is repeated, there are choices. Thus, there are 15 duads. In order to pick a symtheme we must pick 3 duads. Thus, first we have choices for the first duad. Then we have choices for the second duad, and finally, choices for the last duad. However, it is possible that we will repeat choices in this calculation, as {(1,2), (3,4), (5,6)} is the same as {(3,4), (1,2), (5,6)}. There are 3! many ways to order any given syntheme. Thus, to get the total number of synthemes we must divide through by 3!. So we have synthemes.

The entire set of duads is:

{(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6)}.

The set of synthemes is:

{(1,2), (3,4), (5,6)},
{(1,2), (3,5), (4,6)},
{(1,2), (3,6), (4,5)},
{(1,3), (2,4), (5,6)},
{(1,3), (2,5), (4,6)},
{(1,3), (2,6), (4,5)},
{(1,4), (2,3), (5,6)},
{(1,4), (2,5), (3,6)},
{(1,4), (2,6), (3,5)},
{(1,5), (2,3), (4,6)},
{(1,5), (2,4), (3,6)},
{(1,5), (2,6), (3,4)},
{(1,6), (2,3), (4,5)},
{(1,6), (2,4), (3,5)},
{(1,6), (2,5), (3,4)}

It follows that S = (P,L), where P is the set of duads and L is the set of synthemes, is a generalized quadrangle. In addition it is the only generalized quadrangle of order (2, 2) and is isomorphic to the doily with which we are already familiar. We will leave it as an exercise to find an isomorphism between the two.

There are very interesting ways to represent this syntheme-duad geometry. Let's look at one in three space.

Below is the icosahedron:

Now we will take the six axes that connect opposite vertices of the icosohedron to be the elements of our set {1, 2, 3, 4, 5, 6}. Each color below corresponds to one of the six axes:

Now to create our duads we must take any pair of axes. There is one pair of edges that connect the vertices of the pair of axes. We represent the duad by the rotation axis through the centers of the edges. The rotation axis is in red below:

If we take every possible pair of axes and do the same thing, we get the following fifteen axes of rotation through the center of the icosahedron.

As stated, we can also represent these axes of rotation by the two edges that the axis meets. You can see that if we were to show all 15 duads on the icosohedron, we would have the entire set of edges of the icosohedron.

These 15 axes (or dual edges) represent the points of the quadrangle. To represent the lines of the quadrangle, we must look at groups of 3 of these axes. Certainly it is possible to pick many different triples of these axes. However, we will only pick three axes such that no two axes meet an edge that meets a similar line which corresponds to our original 6 axes. It follows that there are 15 such triplets. Five of the triplets consist of axes that are mutually perpendicular. To illustrate the other 10, we pick an axis that passes through the center of two opposite faces. This axis does not correspond to one of the points on the line, but rotation through 120 degrees around this axis leaves the line invariant (note that the line corresonds to the three axes of rotation).

Recall from Problem 35 what the polarities corresponding to the quadrangle of order (2,2) were: there were six of them. The absolute points of a given polarity formed an ovoid, and the absolute lines formed a spread. There were 5 absolute points and 5 absolute lines.

When we look at the representation of the quadrangle on the icosohedron, the ovoids are represented by 5 edges meeting at a given point (as well as the opposite image of this on the icosohedron). Similarly, we can represent this by the 5 axes of rotation to which the dual edges correspond. Here is an example.

Notice that the "absolute points" of the ovoid correspond to the dual edges of the icosohedraon represented by (i, j) (in the figure above they are colored purple). For a given ovoid, i is a fixed integer from the set {1, 2, 3, 4, 5, 6} and j is also from the set such that i and j are distinct. (i.e., a given ovoid will contain the edges that correspond to the duads, (1,2), (1,3), (1,4), (1,5), (1,6)).

The spread on the icosohedron will correspond to a good coloring of the edges of the figure. Here is an example.

So now we have seen a representation of the generalized quadrangle of order (2,2) in two different ways that are really isomorphisms of each other (the doily and the syntheme-duad geometry shown on the icosohedron). Its very exciting that we can represent this geometry is so many ways. Let's look at another model of the genearlized quadrangle that we again arrive at through the syntheme-duad geometry.

Its obvious that the generalized quadrangle of order (2,2) is self dual. Every point is incident with 3 lines and each line is incident with 3 points. Being self-dual means that it is possible to switch the points with the lines and keep the incidence structure. Thus, as we have seen that it is possible to represent the points of a generalized quadrangle with duads, then it is also possible to represent the lines of a generalized quadrangle with duads. We can then represent the points of the generalized quadrangle with synthemes.

Let's represent the duads as the edges of a complete graph on 6 vertices. We know that this graph will have 15 edges. The synthemes will then be represented by the collection of 3 lines that represent a spread over the graph. Recall that spreads partition our point set, thus the syntheme will contain 3 lines, no two of which contain a similar point.

There are certainly different ways to represent 6 points, both on the plane and in space. Let's look at a couple. First, we could represent the 6 points with a pentagon and a point in the center. If we do this, our synthemes are illustrated in the set below:

Notice that each row really just represents a set of 5 rotations for a given graph.

Now that we have these synthemes, it is possible to create the generalized quadrangle from them. We make each of these 15 synthemes a distinct point. A line will represent a given duad, or edge. Thus, three synthemes lie on a line if they contain a similar duad. Let's see how this looks:

But remember that it is possible to create these synthemes and duads with other representations of the six points in the plane or in space - for example, the hexagon.

PROBLEM 36: (a) Show that the doily is the unique generalized quadrangle with order (2,2). (b) Find an isomorphism between the generalized quadrangle of order two and the syntheme-duad geometry. (c) Find the set of of synthemes for the generalized quadrangle represented as above (such that synthemes are points and duads are lines) using the hexagon as our arrangement of six points on the plane. Once you find all 15 synthemes, use this to model the generalized quadrangle of order (2,2). Solutions

Reading Project: Read chapter 8 from Polster.

References: Batten and Polster