Lecture Notes 22: Projective Space

To this point we have been studying affine and projective planes. It is now time to investigate higher dimensional objects. We start with an axiomatic treatment of projective spaces.

Def: A projective space is an ordered pair (P,L), where P is a nonempty set whose elements are called points and L is a nonempty collection of subsets of P called lines such that:
PS1: Given any two distinct points P and Q, there is one and only one line (l(P,Q)) containing them.
PS2: (Veblen's Axiom) If A, B, C and D are distinct points such that there is a point E in l(A,B) l(C,D), then there is a point F in l(A,C) l(B,D).
PS3: Each line contains at least three points; not all points are collinear.

Note that the only difference between the axioms for a projective space and those for a projective plane is axiom PS2. This axiom, which has also been called Pasch's Axiom (but this attribution is incorrect and is losing favor), is a clever way of permitting an extension to higher dimensions. Every projective plane is a projective space, since the projective plane axiom P2 (every pair of lines intersect) implies Veblen's Axiom, but the converse does not hold. We will provide other examples later.

Def: A subspace of a projective space (or flat) is a set of points U with the property that if two points are in U then all the points of the line determined by these points are also in U.

Def: The span (or projective closure) of a set of points S of a projective space, denoted by <S>, is the intersection of all subspaces which contain S.

Note that a span of a set is a subspace, since the intersection of any number of subspaces is a subspace.

We now define a plane in a projective space, but we will have to prove that this definition gives the same object that we have previously called a projective plane.

Def: A plane in a projective space is the span of a set of 3 noncollinear points.

Proposition 22.1: The plane <P,Q,R> consists of the points which are on lines through P which intersect the line QR.

Pf: Let N be the set of points which are on lines through P which intersect the line QR. If X is a point in N, and suppose the line PX intersects QR in the point S. Since <P,Q,R> is a subspace containing Q and R, it contains all the points on the line QR, in particular the point S. Since P and S are in <P,Q,R>, all the points on PS are in <P,Q,R>, in particular X. Thus, N <P,Q,R>. We will now show that N is a subspace. Let S and T be two points in N. If P is on the line ST, then by the definition of N, all the points on PS = ST are in N. So, we may assume that ST does not contain P. Since S and T are in N, there exist points U and V on QR with U on PS and V on PT. Now, let X be any point of ST other than S or T. Now the lines US and VT meet at P, so by Veblen's axiom, the lines ST and UV (= QR) meet at a point Y. Also, the lines PU and XY (= ST) meet at the point S, so again by Veblen's axiom, PX and UY must meet. But, UY = QR, so X is on a line through P which meets QR. That means that X is in N. Thus, the entire line ST is contained in N and so, N is a subspace. Since N clearly contains P, Q and R, and is a subspace, it must contain the span of P,Q and R. So, we have <P,Q,R> N, and so N = <P,Q,R>.

Theorem 22.2: Two distinct lines in a projective space intersect if and only if they are in the same plane (coplanar).

Pf: Suppose that lines l and m intersect in a point P. By PS3, there are points Q on l and R on m, neither equal to P. P,Q and R are not collinear by PS1. The plane <P,Q,R> contains all the points of l and m since it is a subspace.
Now suppose that the lines l and m are contained in the plane <P,Q,R>. By PS3, we can find points A and B on l and C and D on m. Since these 4 distinct points lie in the plane <P,Q,R>, we know, by Proposition 22.1, that they lie on lines through P which meet the line QR. Let U, V, S and T be the points of QR at which the lines PA, PB, PC and PD meet QR respectively. Since lines AU and BV meet at P, by PS2 we have that AB ( = l) meets UV (= QR) at a point X. Now, PA meets SX (=QR) at U, so PS meets AX (= l) at some point, say W. Also, PB meets TX (=QR) at V, so PT meets BX (= l) at some point, say Y. Finally, since CW meets DY at P, we have that CD (= m) meets WY (= l) at some point.

Corollary 22.3: Any plane in a projective space is a projective plane.

Proposition 22.1 is really a special case of a more general result which we present now.

Theorem 22.4: If M is a subspace and Q a point not in M, then the span <M,Q> is the union of all the points on all the lines joining Q to some point of M.

Pf: Let U = the union of all the points on all the lines joining Q to some point of M. It is easy to see that M Q U <M,Q>. Thus, to prove the result we need to show that U is a subspace. Suppose then that B and C are points of U. By the definition of U, there are points B' and C' in M with B on QB' and C on QC'. Now, let A be any point (other than B or C) on the line BC. The points Q, A, B' and C' are all in the plane <Q,B,C>, so the lines QA and B'C' are in this plane, since a plane is a subspace. By Theorem 22.2, these lines meet at a point F, which being on the line B'C', is in M. Hence, A is in U, and so, the entire line BC is in U. That is, U is a subspace.