Planar Ternary Rings

Arthur Busch

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 ={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 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] [QI]) and y = f([PX] [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 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)] [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 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 . 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 . 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 *: x .

Traditional addition and multiplication are obvious examples of binary operations.

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

Define F: x x : 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] [m, k]. Since P is a projective plane, the intersection must be a unique point. And since [x] [ ] = ( ) while [m, k] [ ] = (m), this point cannot be on [ ]. Additionally, recall that the points of [x] all have first coordinate x. So [x] [m, k] = (x, a) for some unique a . 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 .
  2. F(1,x,0) = F(x,1,0) = x for all x .
  3. For all x, y, z, w x<>z, there exists a unique a such that F(a,x,y) = F(a,z,w).
  4. For all x, y, z , there exists a unique a st F(x,y,a) = z.
  5. For all x, y, z, w x<>z, there exists a unique (a,b) x such that F(x,a,b) = y, F(z,a,b) = w.



  1. [x] [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] [x, z] = (0, z) and F(0,x,z) = z.
  2. [1] [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] [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 . 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 ( ,F), where 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 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 x x and F satisfies 1-5, then we can describe a plane of order | | based on the structure of F.


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 .

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

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

  • (m)I[ ] for all m

    ( )I[ ]

    ( )I[k] for all k

  • 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 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 and \{0} respectively. In fact, they turn out to form a structure called a Loop, or possibly non-associative, possibly non-abelian Group.

    First, ( ,+) 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, ( \{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 , 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 . Clearly, the projective plane derived from the Euclidean Affine Plane is a linear PTR, and so is any finite Desargian plane.



    A word about Collineations

    Two non-isomorphic projective planes will obviously produce non-isomorphic planar ternary rings. The converse, however is false. Our coordinatization of a plane is based on the arbitrary choice of our four points, no three of which are collinear. A different choice of these points can produce a second planar ternary ring that is not isomorphic to the original.

    The relationship between our two sets of four points is based on the collineation group of the plane. If and only if a collineation exists that maps X to X’, Y to Y’, Q to Q’ and I to I’, then the resulting planar ternary rings are isomorphic.

    There are two particularly interesting consequences of this. First of all, in PG(2,pe) there exists a collineation between any two ordered sets of four points no three of which are collinear. Therefore, any two planar ternary rings of PG(2,pe) are isomorphic.

    Secondly, if it were possible to enumerate all the non-isomorphic planar ternary rings of a given order, then we have also determined a set of projective planes. This set contains all the non-isomorphic planes of the given order, along with possibly isomorphic copies of these planes. So the search for non-isomorphic projective planes is roughly equivalent to the search of planar ternary rings. As such, we would expect this problem to be rather hard.