## Lecture Notes 23: Dimension and Desargues Theorem

Theorem 22.4 permits us to define the concept of dimension of a subspace in a recursive manner.

Def: A point is a subspace of dimension 0. A line is a subspace of dimension 1. In general, if a subspace M has dimension m and P is a point not in M, then the dimension of <M,P> is m + 1. If U is a subspace, we denote the dimension of U by dim(U).

We see from Proposition 22.1 that a plane has dimension 2. To be consistent we also define the dimension of the empty set (which is a subspace) to be -1.

Theorem 23.1: For any two subspaces U and V of a projective space, we have:

dim(<U,V>) = dim(U) + dim(V) - dim(UV).

We will not give a proof of this important result, but we note that it is consistent with our definition of dimension. Suppose that U is a subspace with dim(U) = m, and P is a point. If P is not in U, then UP = empty set, and the formula gives dim(<U,P>) = dim(U) + dim(P) - dim(empty set) = m + 0 -(-1) = m+1. If P is in U, then <U,P> = <U> = U and UP = P, so the formula gives dim(<U,P>) = dim(U) + dim(P) - dim(P) = dim(U) + 0 - 0 = dim(U).

One quick application of this dimension formula is:

Theorem 23.2: Two distinct planes contained in the same 3-dimensional space intersect in a line.

Pf: Let U and V be distinct planes contained in the subspace W, with dim(W) = 3. Since the planes are distinct their span is larger than either of them and in fact <U,V> = W. By the dimension formula, dim(<U,V>) = dim(W) = dim(U) + dim(V) - dim(UV), so,3 = 2 + 2 - dim(UV) and we have that dim(UV) = 1. That is, UV is a line.

With this theorem we can prove the somewhat surprising result that Desargues theorem, which can not be proved to be true in a projective plane, is true in a projective space of dimension at least 3.

Theorem 23.3: Let P be a projective space in which not all the points are coplanar. Let A,B, C and A',B',C' be two triples of noncollinear points such that the lines AA', BB' and CC' all meet at a point P. Suppose that AB meets A'B' at point C", AC meets A'C' at B" and BC meets B'C' at A". Then A", B" and C" are collinear.

Pf: First note that the points A", B" and C" all exist since the lines that determine them are in a plane, namely, AB and A'B' are in the plane <P,A,B>, AC and A'C' in <P,A,C> and BC and B'C' in <P,B,C>.

Case I: Suppose that the planes <A,B,C> and <A',B',C> are distinct. Both planes are in the 3-dimensional space <P,A,B,C>, so by Theorem 23.2 they meet in a line. The points A", B" and C" are each in both planes, so they must lie on this line of intersection.
Case II: Now suppose that the planes are the same, say plane M. Choose a point Q not in M and a point R on PQ but not equal to either P or Q. Note that R can not be in M, else the line PQ would be in M which contradicts the choice of Q. Let A* be the intersection of QA' and RA, B* the intersection of QB' and RB and C* the intersection of QC' and RC. These points all exist since QR and each of AA', BB' and CC' meet at P, so QR and each of the lines determine a plane. Let N = <A*,B*,C*>. As R is not in M, M and N can not be the same plane. Thus, we may apply Case I to the triangles A,B,C and A*,B*,C* which are perspective from Q and conclude that the lines AC and A*C*, AB and A*B* and BC and B*C* meet at the collinear points D,E and F respectively. Now triangles A',B',C' and A*,B*,C* are perspective from Q and so do not lie in the same plane. Apply Case I again to obtain collinear points G,H and J where G is the intersection of A'C' and A*C*, H is the intersection of A'B' and A*B*, and J is the intersection of B'C' and B*C*. Let m be the line of intersection of M and N. Observe that E and H are on the line A*B* (and so, in N) and on AB and A'B' respectively (and so, in M), thus E = H being the intersection of m with A*B*. Hence, AB and A'B' meet at this point on m, i.e., C" is on m. Similarly, D = G = B" on m and F = J = A" on m. Thus, A", B" and C" are collinear.

The immediate consequence of this theorem is that a non-Desarguesian projective plane can not exist in a projective space of dimension greater than 2. A second consequence, although not as immediate, is that all projective spaces of dimension greater than 2 can be obtained in the same way (as in the following example).

In the following example, the term "dimension" has two different meanings. In the vector space setting, dimension means the number of elements in a basis, and we shall refer to this as the algebraic dimension. The term, when applied to a projective space will be referred to as the geometric dimension. There is a simple relationship between the two.

Example: Let D be a skewfield and V a (left) vector space of algebraic dimension n over D. The projective space PG(n-1,D) is obtained from V by:
The points of PG(n-1,D)(of geometric dimension 0) are the vector subspaces of V of algebraic dimension 1.
The lines of PG(n-1,D)(of geometric dimension 1) are the vector subspaces of V of algebraic dimension 2.
The planes of PG(n-1,D)(of geometric dimension 2) are the vector subspaces of V of algebraic dimension 3.
...
The hyperplanes of PG(n-1,D)(of geometric dimension n-2) are the vector subspaces of V of algebraic dimension n-1.