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# Structure of Affine Geometries

Chloe M. Landry

This course on projective geometry introduced projective spaces axiomatically and then described the nature of affine spaces in relation to the projective space. This presentation is a way to build affine spaces using the following set of axioms: Let S =(P,G,I) be a geometry of rank 2 having the following properties [6]:
(1) Any two distinct points are on precisely one line.
(2) S has a parallelism.
(3) (Triangle Axiom) Let A,B,C be 3 noncollinear points, and let A',B'be points such that . If g is a line through A' parallel to AC, and h is the line through B' parallel to BC then g and h intersect in a point C'.
(4) There is a line with at least two points; there are at least two lines.

The following conjectures should be proved using these axoims. By doing this a structure for the construction of projective spaces from affine spaces will be established. The first property is the Trapezoid Axiom [4].

Conjecture 1   Let A,B,C be noncollinear points. Then for any point B' on AB, the line through B' parallel to BC intersects AC.

The proof of this axiom is a direct use of Axiom (3). Here are some properties of projective spaces as defined in Section 1.3 of the text.

Definition 1   A subset U of a point set P is called linear if for any two points P and Q that are contained in U each point of the line PQ is contained in U as well

In affine spaces, however, if one has two lines in the same parallel class, the subset of both lines is also linear.

Definition 2   A subspace of P is a linear set U which consists of the lines of P which are completely contained in U and I is the induced incidence.

A subspace of an affine plane is also called a flat.

Definition 3   A span of a subset X of P is the smallest linear set containing X.

The span of a subspace and a point of an affine plane can be described in three ways:

Conjecture 2   Let U be a subspace, and let P be a point outside of U. Then one can describe the span as follows: Let Q be an arbitrarypoint of U. Then consists of the points X of U that are parallel to PQ.

Conjecture 3   Let U be a subspace, and let P be a point outside of U. Then for each point Q of U we have: the span consists precisely of the points in the planes where g runs through all lines of U through Q.

Conjecture 4   Let U be a subspace, and let P be a point outside of U. Then the span consists precisely of the points on the lines PX with and the points on the lines through P that are parallel to some line of U.

These conjectures can be shown to correspond to the definition of the span of a projective space.

The following conjecture asks for corresponding definitions for certain notions in projective spaces:

Conjecture 5   Define independence of points and the notions of 'basis' and 'dimension'.

The following definitions are from Section 1.3 of the text.

Definition 4   A set B of points in the affine space A is called independent if for any subset and any point we have that When we say that B is dependent.

Definition 5   An independent set B of points that spans A ( ) is called a basis of A.

Definition 6   Let A be a finitely generated affine space (that is, there is a finite set of points that spans A). If d+1 denotes the number of elements in a basis, then we call d the dimension of A.

The next conjecture involves Playfair's parallel axiom. It states that if g is a line and P is a point outside g then there is precisely one line through P that has no point in common with g.

Conjecture 6   We say that two subspaces U and U' are parallel if each line of U' is parallel to some line of U. Show that the parallelism defined in this way is an equivalence relation satisfying Playfair's parallel axiom.

This can be shown to be an equivalence relation by showing it is reflexive, symmetric, and transitive. It can also be easily shown to satisfy the parallel axiom. The next three conjectures are mentioned in the text but will not be discussed here.

Conjecture 7   Let U,U' be distinct t-dimensional subspaces. If U,U' are parallel, then has dimension t+1.

Conjecture 8   Let g,g' be two skew lines, then has dimension 3. Generalize to skew t-dimensional subspaces.

Let A = (P,L,). If A satisfies:
(1) Two distinct points are on exactly one line.
(2) If l is a line and P is a point not on l, then there is exactly one line through P that does not intersect l.
(3) There are four points, no three of which are collinear.
then A is an affine plane [2]. Since the relationship between affine planes and projective planes has been established, we can now prove the following theorem [2]:

Theorem 1 (a)   If one line and all the points on it are removed from a projective plane, the remaining incidence structure is an affine plane.
(b) Given an affine plane, there is a projective plane that determines it.

(Proof) (a) If the line is removed, then is the unique line required by axiom 2. The other conditions are clear.
(b) Let A = (P,L,) be an affine plane. Call two lines l,m of A parallel if either l = m or l and m have no common points. This defines an equivalence relation on L. Add one new point to P for each equivalence class of parallel lines to form the set P'. Add one new line to L to form the set L'. Define ' as follows: If P P and l L, then P ' l iff P l; if P is new and l L, then P ' l iff P corresponds to the equivalence class containing l; iff P is new. It is easy to check that (P',L', ') is a projective plane and that if and all its points are removed, then A remains.

An interesting topic viewed from the perspective of affine planes instead of projective planes is the affine analogues of the Theorem of Desargues and the Theorem of Pappus [5].

Theorem 2   First Affine Desargues Theorem (Minor Desargues Theorem)
If abc and a'b'c' are two triangles such that aa', bb', and cc' are parallel and and , then .

Theorem 3   Second Affine Desargues Theorem (Major Desargues Theorem)
If abc and a'b'c' are centrally perspective from any given point p and if and , then .

Theorem 4   Affine Pappus Theorem
If a,b,c and a',b',c' are triples of collinear points in an affine plane such that and , then .

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Bill Cherowitzo 2001-12-10