To show that a quotient geometry is a projective space we need the concept of isomorphism.

Def: Let G = (,I) and G' = (', I') be rank 2 geometries. G and G' are ** isomorphic** if there exists a map :' such that

- () = ' and () = ' with restricted to these domains being a bijection, and
- P and B,
P I B (P) I' (B).

An ** automorphism** is an isomorphism from G onto itself. If the blocks of G are called lines, then an automorphism is usually called a

*Example*: The points of a 3-dimensional projective space **P** are given by the 15 non-zero binary vectors of length 4 written as row vectors. The planes of this space are the same vectors written as column vectors. A point and a plane are incident if their dot product is 0 (mod 2). A line is the set of points incident with two distinct planes. Thus, the plane (0,0,0,1)^{T} = {0010,0100,0110,1000,1010,1100,1110} and the plane 1000^{T} = {0001,0010,0011,0100,0101,0110,0111}, and so, the set of points {0010,0100,0110} form a line. Incidence of points and lines, and lines and planes is given by set inclusion. Let Q be the point 0001, then the set of planes containing Q is {0010^{T},0100^{T},0110^{T},1000^{T},1010^{T}, 1100^{T},1110^{T}} and the set of lines containing Q is {a ={0001,1000,1001}, b={0001,0100,0101}, c={0001,0010,0011}, d={0001,1100,1101}, e={0001,1010,1011}, f={0001,0110,0111}, g={0001,1110,1111}}. We form the quotient geometry **P**/Q:

POINTS | LINES Through POINT |
---|---|

a | 0010^{T}, 0100^{T}, 0110^{T} |

b | 1000^{T}, 0010^{T}, 1010^{T} |

c | 1000^{T}, 0100^{T}, 1100^{T} |

d | 0010^{T}, 1100^{T}, 1110^{T} |

e | 0100^{T}, 1010^{T}, 1110^{T} |

f | 1000^{T}, 0110^{T}, 1110^{T} |

g | 1100^{T}, 0110^{T}, 1010^{T} |

*Example*: Continuing with the above example. Define a map on the points of **P** by (x,y,z,w) = (w,y,z,x). Notice that maps the points of 0001^{T} to the points of 1000^{T} bijectively. One then checks to see that the points of a line of 0001^{T} are mapped to the points of a line of 1000^{T}. For instance, the points of the line {0010,1100,1110} which is the intersection of 0001^{T} and 1100^{T}, are mapped to {0010,0101,0111} which lie on the line which is the intersection of 1000^{T} and 0101^{T}. This is therefore an isomorphism between the planes 0001^{T} and 1000^{T}.

**Proposition 1.4.2**: Let **P **be a d-dimensional projective space, Q a point of **P**. Then there exists a hyperplane of **P** that does not pass through Q.

*Pf*: By Lemma 1.3.8, one can extend Q to a basis of **P**, {Q,P_{1},P_{2},...P_{d}}. The subspace H = <P_{1},P_{2},...,P_{d}> is spanned by d independent points. Hence H has dimension d-1 and is therefore a hyperplane. Since the original basis is an independent set, Q H.

**Theorem 1.4.1**: Let **P** be a d-dimensional projective space, Q a point of **P**. Then the quotient geometry **P**/Q is a projective space of dimension d-1.