Planar Ternary Rings

December 5, 2001

Coordinatizing an Arbitrary Projective Plane

Given a Projective Plane, P , of order q, we develop a technique to label the points and lines of P using q distinct symbols. There are q2 + q + 1 of each, and when the plane is Desargian, we can do so using homogeneous coordinates (ordered triples of linearly independent elements from GF(q)). But non-Desargian planes can’t be represented as easily.

Take any four points of P , no three collinear. Call these points X, Y, Q and I.

First, we name our points. To do so we need two things:

1. A set of q symbols containing 0 and 1. WLOG we’ll use Â ={0,1,…,q-1}
2. A bijection f from the points of the line connecting Q and I (except the point where this line meets [XY]) and the elements of Â such that f(Q) = 0 and f(I) = 1.

We can name the q2 points of our plane not on [XY] in the following way:

Given an arbitrary point P not on [XY], assign an ordered pair of elements (x, y) such that x = f([PY] Ç [QI]) and y = f([PX] Ç [QI]). (Note that [PX] cannot intersect [QI] on [XY] since that would force P to be on [XY])

 Figure 1. Labeling the points of P not on [XY]

By this procedure the points on [QI] are clearly labeled (x, x) for some x Î Â and Q is labeled (0,0) and I is labeled (1,1). Furthermore, the points on [QX] are labeled (x, 0) and the points of [QY] are labeled (0,x).

 Figure 2. The points of [QI], [XQ] and [YQ]

As a consequence of the way we’ve named these points, the lines through the point X all have the same second coordinate, while the lines through Y all have the same first coordinate.

 Figure 3. The lines through X and Y

This property of the lines through Y will be used to name the points on [XY]. Consider the line [YI]. As noted, all the points on this line except Y have the same first coordinate. Since I is one of these points, the second coordinate of all these points is 1. Next, connect each of these points to Q. These lines all must intersect [XY] in distinct points not = Y. Label the point [Q, (1,a)] Ç [XY] as (a).

 Figure 4. The points on [XY]

The only point not yet labeled is Y. We can just call this point Y, or we can call it (¥ ).

In a similar way, we will label the q2 + q + 1 lines of P . We’ll give q2 of the lines an ordered pair as a name, q points a single element of Â and one line ([XY]) will be named [¥ ]. These last two categories will be the q+1 lines incident with Y.

Those lines not through Y intersect [XY] at a point labeled (m) and also intersect [QY] at a point labeled (0,k) for some m and k in Â . Label such a line [m, k].

Label [XY] as [¥ ], and the remaining q lines through Y intersect [QX] at some point (x, 0) for some x Î Â . These lines will be called [x].

 Figure 5. The lines of P

The Algebraic Viewpoint

Now that we’ve given the points and lines of our plane labels, we can try to examine the structure of P algebraically. Binary operations are central to most algebraic structures, but for our purposes we will work with a ternary operation.

A Binary Operation is a function *: Â x Â ® Â .

Similarly, a Ternary Operation is a function *: Â x Â x Â ® Â .

Define F: Â x Â x Â ® Â : F(x, m, k) = y if and only f (x, y)I[m, k]

In order to show that F is a ternary operation, we must show that it is well defined.

Consider [x] Ç [m, k]. Since P is a projective plane, the intersection must be a unique point. And since [x] Ç [¥ ] = (¥ ) while [m, k] Ç [¥ ] = (m), this point cannot be on [¥ ]. Additionally, recall that the points of [x] all have first coordinate x. So [x] Ç [m, k] = (x, a) for some unique a Î Â . Therefore if F(x, m, k) = a and F(x, m, k) = c, then a = c.

Much of the structure of P carries through to F.

1. F(x,0,z) = F(0,y,z) = z for all x,y,z Î Â .
2. F(1,x,0) = F(x,1,0) = x for all x Î Â .
3. For all x, y, z, w Î Â x<>z, there exists a unique a Î Â such that F(a,x,y) = F(a,z,w).
4. For all x, y, z Î Â , there exists a unique a Î Â st F(x,y,a) = z.
5. For all x, y, z, w Î Â x<>z, there exists a unique (a,b) Î Â x Â such that F(x,a,b) = y, F(z,a,b) = w.

Proof:

1. [x] Ç [0,z] = (x, z) since the points of [x] all have x as their first coordinate and the points of [0,z] all have second coordinate z. So (x, z)I[0, z] and F(x,0,z) = z. Similarly, [0] Ç [x, z] = (0, z) and F(0,x,z) = z.
2. [1] Ç [x,0] = (1,x) by the same reasoning, so F(1,x,0) = x. And [1,0] = [QI], so all these points have their first and second coordinates equal. So [x] Ç [1,0] = (x, x) and F(x,1,0) = x.
3. Since x<>z, [x, y] and [z, w] intersect [¥ ] in distinct points. That means that their unique point of intersection is affine (i.e. of the form (s, t)). Then F(s,x,y) = t = F(s,z,w).
4. Consider the unique line in P connecting (y) and (x, z). This line intersects [0] at a unique affine point whose first coordinate is 0. By definition, this point (0, a) is connected uniquely with the point (y) by [y, a]. Thus (0, a), (y) and (x, z) are collinear. Then F(x,y,a) = z.
5. Let g be the line determined by (x, y) and (z, w). If (¥ ) is on g, then g = [c] (or g = [¥ ] which is immediately ruled out since no affine points are on [¥ ]) for some c Î Â . But then all the points of g have the same first coordinate, contradicting the assumption that x<>z. So g = [a, b] and F(x,a,b) = y as well as F(z,a,b) = w.

Definition: A Planar Ternary Ring is a pair (Â ,F), where Â is the coordinatizing set of a projective plane and F is the associated ternary operation F.

The close relationship between Algebra and Geometry is well known. So it’s probably not surprising that any set Â and ternary operation F that satisfies properties 1-5 above is a Planar Ternary Ring. In other words, if F is a ternary operation defined Â x Â x Â and F satisfies 1-5, then we can describe a plane of order |Â | based on the structure of F.

Proof:

The first q2 incidences are defined by (x, y)I[m, k] iff F(x,m,k) = y.

Additionally, we define [x] by (x, y)I[x] for all x,y Î Â .

And points (m) by (m)I[m,k] for all m,k Î Â .

Finally, we define the line [¥ ] and the point (¥ ) by:

• (m)I[¥ ] for all m Î Â

(¥ )I[¥ ]

(¥ )I[k] for all k Î Â

• We prove the two critical axioms of a projective plane are satisfied.

1. Any two points determine a unique line.
2. Given two distinct arbitrary points of the form (x, y) and (z, w), there are two cases:

-z = x. Then (x, y)I[x] and (z, w)I[x] and for any m and k, F(x,m,k) = F(z,m,k). But this requires that y=w, contradicting the assumption that the two points were distinct.

-z <> x. By property 5, there exists a unique (a,b) such that F(x,a,b) = y and F(z,a,b) = w. This implies that the line [a,b] is the unique line incident to (x, y) and (z, w).

Next consider two points (x, y) and (m). Property 4 guarantees a unique element a such that F(x,m,a) = y, which in turn guarantees that [m,a] is the unique line incident to both (m) and (x,y).

The remaining cases are almost trivial. Consider the point (¥ ) and a point (x, y). Both are incident to [x] and no other line. And (¥ ), (m) and (n) are all incident to [¥ ].

3. Any two lines determine a unique point.
• If the two lines are of the form [m, k] and [n, j], the same two cases result:

-m = n. Then (m)I[m, k] and (m)I[n, j]. If (a, b)I[m, k], then F(a,m,k) = b <> F(a,m,j) by the definition of incidence and the well-formedness of F.

-m <> n. Property 3 implies that there exists a unique a Î Â such that F(a,m,k) = F(a,n,j). Then (a,F(a,m,k)) is the unique point incident to both lines.

Two lines [m,k] and [n] are clearly both incident to (n,F(n,m,k)) and this point only.

Finally, two lines [n] and [m] are both incident to the unique point (¥ ), as are the lines [m] and [¥ ].

• #

Ternary Operations seem very unnatural. But the ternary operation of a Planar Ternary Ring has some similarities to the more familiar binary operations that we are used to.

Namely, if we define

a + b = F(a,1,b)

a * b = F(a,b,0)

It turns out that these binary operations very nearly form groups on Â and Â \{0} respectively. In fact, they turn out to form a structure called a Loop, or possibly non-associative, possibly non-abelian Group.

First, (Â ,+) is a loop with identity 0.

1. 0 + a = a + 0 = 0 by the first and second properties of a PTR.
2. Given x and z, property 4 of a PTR produces a unique a such that F(x,1,a) = z (or equiv. x + a = z).
3. Given 0, x, 1 and z, property 3 of a PTR produces a unique b such that F(b,0,z) = F(b,1,x). And by property 1 F(b,0,z) = z, so F(b,1,x) = z (and b + x = z).

Second, (Â \{0},*) is a loop with identity 1.

1. a * 1 = 1 * a = a by the second property of a PTR.
2. Given x and z, the unique line determined by (x, z) and (0, 0) intersects [¥ ] at some unique point (a). Then (x, z)I[a,0], F(x,a,0) = z and x * a = z.
3. Given x and z, the lines [0,z] and [x,0] intersect in a unique point whose second coordinate is z. This point (b, z)I[x,0] so F(b,x,0) = z and b * x = z.

Additionally, a * 0 = 0 * a = 0 for all a Î Â , as we would expect.

A Linear Planar Ternary Ring is a a Planar Ternary Ring where F(x,m,k) = (x*m) + k for all x,m,k Î Â . Clearly, the projective plane derived from the Euclidean Affine Plane is a linear PTR, and so is any finite Desargian plane.