Given the vector space V(n+1,q), we define PG(n,q) as follows:

The objects of PG(n,q) consist of:

*points*, which are the rank 1 subspaces of V(n+1,q).*lines*, which are the rank 2 subspaces of V(n+1,q).*planes*, which are the rank 3 subspaces of V(n+1,q).- ...
*i-spaces*, which are the rank i+1 subspaces of V(n+1,q).- ...
*hyperplanes*, which are the rank n subspaces of V(n+1,q).

We can now rephrase statements about vector spaces in terms of the geometric objects of the projective geometry. For instance, the statement in the previous section about two distinct rank 1 subspaces becomes two distinct points determine a unique line. The statements about rank 2 subspaces become, in PG(2,q) every two distinct lines meet at a unique point, while in higher dimensional projective spaces two distinct lines which meet lie in a unique plane and if they do not meet (are *skew*) lie in a unique 3-space (*solid*).

**Example**
Let our field be GF(4) whose elements are 0, 1, a and a^{2}. Recall that this is a field of characteristic 2, so 1 + 1 = 0 and that a^{2} = a + 1 = 1/a. With respect to the standard basis, the vectors of V(3,4) (4^{3}= 64 in total) are represented by 3-tuples over GF(4), such as: (0,1,0), (a,0,0), (a,a^{2},1), and (a^{2}, 1, a). Now, a rank 1 subspace of this vector space containing a non-zero vector consists of the zero vector and three non-zero vectors. For instance, <(a,a^{2},1)> = {(0,0,0), (a,a^{2},1), (a^{2},1,a), (1,a,a^{2})}. A rank 2 subspace consists of all linear combinations of two vectors which are not in the same rank 1 subspace (i.e., are linearly independent). Thus, there will be 16 vectors in a rank 2 subspace, the zero vector and 15 non-zero vectors. If a vector is in this subspace, then all of its scalar multiples are as well, so the 15 non-zero vectors are divided up into 5 sets of size 3, and there are 5 rank 1 subspaces contained in a rank 2 subspace. For instance, the rank 2 subspace containing (0,1,0) and (a,0,0) consists of the vectors A(0,1,0) + B(a,0,0) = (Ba,A,0) as A and B run through GF(4). We get the following vectors:

- (0,0,0)
- (a,0,0), (a
^{2},0,0), (1,0,0) - (0,1,0), (a,1,0), (a
^{2},1,0), (1,1,0) - (0,a,0), (a,a,0), (a
^{2},a,0),(1,a, 0) - (0,a
^{2},0), (a,a^{2},0), (a^{2},a^{2},0), (1,a^{2},0)

**Coordinates**

It is easily seen from this construction that the points of the geometry have a natural relation to coordinates, namely we can take as coordinates of a point any non-zero vector written as an (n+1)-tuple in the rank 1 subspace corresponding to that point. If we do this, then unlike the situation in Euclidean geometry, one point has several possible coordinates, but they are all related by being scalar multiples of each other. Such coordinates are called *projective* (or *homogeneous*) coordinates. It is often convenient to select a standard representative from the set of equivalent coordinates for a point. Two common conventions are to select the coordinate whose last non-zero entry is a 1, and to select the coordinate whose first non-zero entry is a 1. It is always to be remembered that no matter what convention is used, the coordinates can at any time be replaced by any non-zero scalar multiple.

In terms of these point coordinates, it is easy to describe the set of points on a hyperplane by means of a linear equation. A hyperplane is a rank n subspace of V(n+1,q), and so, its orthogonal complement is a rank 1 subspace. Thus the (standard) dot product of any vector in the hyperplane with any vector in this rank 1 subspace must be 0. So, if (a_{0}, a_{1}, ..., a_{n}) is a fixed non-zero vector, and (x_{0}, x_{1}, ..., x_{n}) represents a variable non-zero vector of V(n+1,q), then the solutions of:

**Example revisited**
In the example above we calculated the vectors in a rank 2 subspace. Since a rank 2 subspace is a hyperplane, we can describe these vectors in an alternate way. Consider the vector (0,0,1). The vectors in the orthogonal complement of the subspace containing this vector must satisfy the equation (0,0,1).(x_{0},x_{1},x_{2}) = 0, i.e. the linear equation x_{2} = 0. Notice that this is the set of vectors given in the example.

**Restriction to dimensions 2 and 3**

Throughout this seminar we will be concerned only with projective geometries constructed in this way. Furthermore, we only have need to be concerned with geometries of dimensions 2 or 3, i.e., projective planes and projective 3-space (and since no confusion will occur, we'll drop the 3 and just refer to projective space). Note that in projective space the hyperplanes are planes, so planes can be described by linear algebraic equations.

There are certain types of sets of points in a projective geometry which will play a special role in this seminar. They are related in that they are all special types of *quadratic sets*, which we shall now define.

Given a set of points, S, in a projective geometry, a *tangent line* to S is a line of the geometry which either intersects S in exactly one point or is completely contained in S. The *tangent space at a point P* of the set S is the set of all points Q in the geometry such that the line joining P and Q is a tangent line of S. Finally, a set S of points in PG(n,q) is called a *quadratic set* if two conditions are satisfied:

- Any line that intersects S in more than two points is completely contained in S.
- The tangent space at any point of S is either a hyperplane or all of PG(n,q).

**Examples:**

Consider a projective plane and take any two distinct lines of that plane. Recall that a hyperplane in a projective plane is a line and that any two distinct lines of a projective plane meet. This pair of lines is easily seen to be a quadratic set. It is degenerate since the tangent space at the point of intersection of these lines is the entire plane. Similarly, in projective space, the union of two distinct planes is also a degenerate quadratic set. The empty set, singleton points, lines and planes are all examples of degenerate quadratic sets. Examples of non-degenerate quadratic sets are bit more interesting.

**Ovals**

In a projective plane, PG(2,q) a set of q+1 points with no three on the same line is called an *oval*. An oval clearly satisfies condition 1 of a quadratic set. Consider a point P on the oval **O**. As there are q+1 lines that pass through P, and q of these are determined by the other points of **O**, there is exactly one line through P which is a tangent line. The tangent space at P is therefore this line (which is a hyperplane in a projective plane), so the oval satisfies condition 2 and is therefore a non-degenerate quadratic set. It can be shown that the ovals in a projective plane are the only non-degenerate quadratic sets in projective planes.

To get a specific example of an oval we will look some sets of points determined by algebraic equations involving the coordinates. A *plane quadric* in PG(2,q) is the set of points defined in terms of their coordinates by {(x,y,z)| Ax^{2} + Bxy + Cy^{2} + Dxz + Eyz + Fz^{2} = 0}, where the coefficients of the quadratic equation come from GF(q). Because the equation is homogeneous, any scalar multiple of a solution is also a solution, so this really does define points of PG(2,q). It can be shown that any plane quadric is a quadratic set. The plane quadrics which correspond to non-degenerate quadratic sets all contain exactly q+1 points and are called *conics*. Conics are therefore ovals. Beniamino Segre (1957) has shown that if the characteristic of the field is odd, then all ovals of PG(2,q) are conics, but in even characteristic there are ovals which are not conics. The even characteristic case is different in another way as well. For any oval in a projective plane over a field of even characteristic, it can be shown that all of the tangent lines to the oval meet at a single point, called the *knot* of the oval (in odd characteristic, through a point not on the oval (conic) there pass either 0 or 2 tangent lines). The knot of a conic is called the *nucleus* of the conic.[Most authors use the words knot and nucleus interchangeably, but I prefer to maintain this distinction.] If the knot of an oval is added to the oval, the larger set is called a *hyperoval* and it still has the property that no three of its points lie on the same line. If the oval is a conic, the corresponding hyperoval is called a *hyperconic* (other terms in the literature for this are regular hyperoval, and complete conic). Note that a hyperoval is not a quadratic set, since the tangent spaces at each of its points are empty.

If we look for examples of quadratic sets in PG(3,q) we can again obtain some from algebraic considerations. We can define quadrics as sets of points that have coordinates which satisfy a homogeneous quadratic equation (in 4 variables now) and again show that they are quadratic sets. In this case however there are two types of quadrics which correspond to non-degenerate quadratic sets, the elliptic quadrics (which contain no line) and the hyperbolic quadrics (which contain lines). It can be shown that these are the only non-degenerate quadratic sets in the odd characteristic case. (Even characteristic again permits other examples.) However, we will not be concerned with these quadratic sets, rather we will be interested in a degenerate quadratic set of PG(3,q).

Let V be a point in PG(3,q) and **H** a plane which does not contain V. Let **O** be an oval in the plane **H**. By joining V to each point of **O** and taking the union of all the points on these lines we obtain an *oval cone*. The point V is called the *vertex* of the cone, the oval **O** is called the *carrier* of the cone and **H** the *carrier plane* (or base plane). The lines joining V to the points of the carrier are called the *generators* of the cone. If the oval is a conic, the cone is called a *quadratic cone*. Oval cones are quadratic sets, but they are degenerate since the tangent space at the vertex is the full space.