## Lecture Notes 17: Projective Points and Homogeneous Equations in D3

Let D be a division ring (skewfield).

Def: D3 is a left vector space over D. That is the vectors of D3 are the triples (a,b,c) with a,b and c in D. Addition of vectors is componentwise addition. Scalar multiplication by elements of D is defined from the left only by k(a,b,c) = (ka,kb,kc).

We define an incidence structure (P,L,I) by:
P = the point set = set of all 1-dimensional subspaces of D3.
L = the line set = set of all 2-dimensional subspaces of D3.
I = the incidence relation = containment.

Note: A 1-dimensional subspace of D3 consists of all the scalar multiples of a non-zero vector. If v is a non-zero vector, we denote the 1-dimensional subspace containing v by <v>. The zero vector does not generate a 1-dimensional subspace.

Theorem: The incidence structure (P,L,I) defined above is a projective plane.

We shall show later that this plane is Desarguesian.

### Homogeneous Linear Equations

Def: A homogeneous linear equation in three variables over D is an equation of the form,
xa + yb + zc = 0
where x,y and z are the variables and a,b and c are in D.(Note that the scalars are on the right)

From linear algebra, we know that the solution set of such a homogeneous equation is a 2-dimensional vector subspace (1 equation, 3 unknowns), that is, it is a line of our projective plane. We can also see this directly. Consider the homogeneous linear equation given by xa + yb + zc = 0. If (x1,y1,z1) and (x2,y2,z2) are both solutions to this equation, then (x1+x2)a + (y1+y2)b + (z1+z2)c = x1a + y1b + z1c + x2a +y2b + z2c = 0 + 0 = 0. Thus, (x1+x2, y1+y2,z1+z2) is also a solution. If (x1,y1,z1) is a solution then (kx1)a + (ky1)b + (kz1)c = k(x1+y1+z1) = k(0) = 0, for any k in D. Thus, the solution set is closed under addition and scalar multiplication from the left, i.e., it is a subspace of the left vector space D3. If the dimension of this solution set is 3, then every vector is a solution, and this can only happen if a = b = c = 0. So, if not all the coefficients of the homogeneous equation are 0, the solution set will be a subspace of dimension less than 3. We will now show that the dimension of the solution set is 2 by showing that it contains at least two independent vectors. If two of the coefficients are 0, we may assume without loss of generality that a = b = 0. Then, for any c, (1,0,0) and (0,1,0) will be independent solutions. Now, we have that at most one of the coefficients is 0 and so we can assume WLOG that neither a nor b is 0. The points (0,-c(b-1),1) and (-c(a-1),0,1) are then independent solutions if c is not 0. If c = 0 then the points (0,0,1) and (a-1, -(b-1),1) are independent solutions.

Observe that if a point <p,q,r> satisfies a linear homogeneous equation xa + yb + zc = 0, then it also satisfies the equation xam + ybm + zcm = 0 for any m in D, since pam + qbm + rcm = (pa + qb + rc)m = 0m = 0. Thus, the equation xam + ybm + zcm = 0 has the same solution set as the equation xa + yb + zc = 0.

We can represent the homogeneous linear equation xa + yb + zc = 0 by the vector [a,b,c]. These vectors form a vector space, but this time it is a right vector space. Lines of our projective plane (P,L,I) are thus represented by 1-dimensional subspaces of this right vector space. This observation leads to a representation of a duality (a collineation which interchanges points and lines) of our projective plane. This duality is given by (a,b,c) <=> [a,b,c] where scalar multiplication is reversed from left to right or vice-versa.

### Canonical Forms

It is convenient to select a specific non-zero vector to represent a 1-dimensional subspace. For projective points we select vectors of the form (p,q,1), (1,m,0) or (0,1,0).[Note that this is different from the selection in the text] For lines we make the selections [a,1,b], [1,0,c] or [0,0,1].

Example: Let D = Z3 (the integers modulo 3). We will represent the elements of D by {0,1,-1} [notice that one often uses the symbols {0,1,2}, but note that 2 = -1 in this field]. There are 27 vectors in D3 (3 choices for each coordinate position), but one of these is the all zero vector. Since there are only two non-zero scalars, each 1-dimensional subspace will contain just two non-zero vectors. Thus, there will be (27-1)/2 = 26/2 = 13 1-dimensional subspaces (projective points). These are (without writing the zero vector which is a member of every subspace) the columns of:
 001 011 0-11 101 111 1-11 -101 -111 -1-11 100 110 1-10 010 00-1 0-1-1 01-1 -10-1 -1-1-1 -11-1 10-1 1-1-1 11-1 -100 -1-10 -110 0-10
The canonical choices are given in the first row of the table above.

The line [0,0,1], whose points are the solutions to the equation z = 0, contains the 4 points <1,0,0>, <1,1,0>, <1,-1,0> and <0,1,0>. The line [1,0,-1], corresponding to the equation x + z(-1) = 0 [Note that this is x = z since D is a field] consists of the points <1,0,1>, <1,1,1>, <1,-1,1> with z = 1 and <0,1,0> with z = 0. Finally, the line [-1,1,1], with equation x(-1) + y + z = 0 (or y = x - z, again because D is a field) contains the points (with z = 1) <0,-1,1>, <1,0,1>, <-1,1,1> and with z = 0, <1,1,0>.