Veblen-Wedderburn Systems

In this section, we will look at the definition of a specialized planar ternary ring. We shall then build a specific example of a planar ternary ring which is a Veblen-Wedderburn system, and show that a projective plane coordinatized by this system is actually a non-Desarguesian plane of order 9.

Example

The field:

Let us begin with the Galois field of 9 elements, which looks like where . Addition is defined as

and we have multiplication as

This uses the standard foil method.

A square in this field is one of the following: . These squares behave much as even versus odd numbers in the real field. For example, and .

The act of cubing elements creates an automorphism of the group. This will be useful presently.

The ternary operation:

Define T as follows:

is a planar ternary ring:

This can be shown by proving that T1 - T5 hold:

T1

We have even if b is not square. Also, since 0 is a square.

T2

We have since 1 is square. Also, even if a is not square.

T3

We want to show that has a unique solution when . This breaks into cases depending on whether m and m' are square or not:

If m and m' are square: we can solve for a unique using the field operations.
If m and m' are not square: we can solve for a unique .
If m is square and m' is not (or alternately, if m' is square and m is not) we can use field operations to obtain the following equation: . Using the fact that the field is characteristic 3, this equation has a unique solution when is not square. Since m is square and is not, we have a unique solution.

T4

We want to show that and have a unique solution when . We can combine these equations and solve for x using field operations depending on whether x is square or not. This in turn depends on a,b,a', and b' in the following way:

If and are both square or both not square, then x is a square, and the unique solution to the above equations is .
If only one of or are square, then x is not square, and the unique solution to the above equations is .

T5

If m is square, . If m is not square, , and these are unique solutions.

is a Veblen-Wedderburn System: (Problem D.7)

This can be shown by proving that VW1 - VW4 hold.

VW1

Note that for ,

thus addition is the regular field operation, and forms an abelian group by field properties.

VW2

We need to check both conditions of the definition of a loop.

and both by T2.
Suppose . Given a and b, is unique. Given a and c, we have . If a and c are both square or not square, then x is square, and is unique. If only one of a or c is not square, then x is not square, and is unique. Given b and c, we have which has a unique solution by T3.

VW3

We have and or , both equal 0. This is done using field properties.

VW4

We want to prove right distributivity. Note that

Also,

The last equality is due to mulitiplication and addition being modulo 3.

We have now shown that our field along with the described ternary operation is not only a planar ternary ring, but also a Veblen-Wedderburn system. We can pause for a moment here to show an additional property of .

is a group:

We have the properties of a loop, thus the only property to prove is that of associativity, ie, .

Note that

If b is square:

If c is square:

If c is not square:

If b is not square:

If c is square:

If c is not square:

And

If c is square:

If b is square:

If b is not square:

If c is not square:

If b is square:

If b is not square:

Note that the last line is true since any element from our field raised to the ninth power is itself.

The projective plane:

We can define the lines and points of our projective plane in the following way.

Lines

First, let us define a line at infinity, . Other lines will be of the form for , and

Points

First there will be a point at infinity, which will be denoted as . This point will be on the line , and on lines of the form . There will be usual points of the form which will be on lines for which they satisfy the equation . There will also be points on of the form for . A point will also be on any line with slope m. Thus, we have

with points , [0], [1], [2],...

with points (x,y), and [m]

with points (a,y), and .

Problem D.8

Proof:

We will begin by choosing three lines in our projective plane which meet at one point. Let our lines be:

All three meet at the point .

 
Figure 6: Triangles perspective from a point

The points denoted above are given by the coordinates:

    ¯  

   		   

   		    

Note that ABC and A'B'C' form triangles. We can find the lines making up the triangles, AB, AC, BC, A'B', A'C', and B'C' by looking at the two points on each line. For example, if we wish to find the line through A' and B', form the two equations

Solve for b in one, and substitute in the other to obtain . We can now solve for the slope, m,

Solving in turn for b, . The line through A' and B' is thus .

This can be done for each of the above lines, yielding

    ¯  

   		   

   		   

   		   

   		   

   		   

We can now find the intersection points of the lines AB and A'B', BC and B'C', and AC and A'C'. For example, the intersection point of the lines BC and B'C' can be found by solving the two equations and simultaneously in the following manner.

Thus, the point of intersection is .

This can be done for each of the above intersection points, which gives us:

    = ¯   = ¯  

   = 		    = 		   

   = 		    = 		   

If we can show that these points are not collinear, then our three triangles are perspective from a point, but not perspective from a line, thus the projective geometry is non-Desarguesian. This can be easily accomplished by finding the line through and , and showing that cannot lie on it. The line through and turns out to be . We can check and note:

Thus, our triangles are not perspective from a line.