Collineation

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A collineation, roughly, is a map from one projective space to the other, preserving the geometric structure.

[edit] Definition

Let V be a vector space over a field K (of dimension at least three) and W a vector space over a field L. Consider the projective spaces PG(V) and PG(W). Call D(V) and D(W) the set of subspaces of V and W respectively. A collineation from PG(V) to PG(W) is a map $\alpha:D(V)\rightarrow D(W)$ , such that :

$α$ is a bijection.
$A\subseteq B \Longleftrightarrow A^{\alpha}\subseteq B^{\alpha}\ \forall A,B\in D(V)$

When V has dimension one, a collineation from PG(V) to PG(W) is a map $\alpha:D(V)\rightarrow D(W)$ , such that :

${0}$ is mapped onto the trivial subspace of W.
V is mapped onto W.
There is a nonsingular semilinear map $β$ from V to W such that : $\forall v\in V : (<v>)^{\alpha}=<v^{\beta}>$

The reason for the seemingly completely different definition when V has geometric dimension one will become clearer further on in this article.

When $V = W$ the collineations are also called automorphisms.

[edit] Fundamental theorem of projective geometry

Suppose $φ$ is a semilinear nonsingular map from V to W, with the dimension of V at least three.

Define $\alpha:D(V)\rightarrow D(W)$ in this way :

$Z^{\alpha}=\{\phi(z)|z\in Z\}\ \forall Z\in D(V)$

As $φ$ is semilinear, one easily checks that this map is properly defined, and further more, as $φ$ is not singular, it is bijective. It is obvious now that $α$ is a collineation. We say $α$ is induced by $φ$ .

The fundamental theorem of projective geometry states the converse :

Suppose V is a vector space over a field K with dimension at least three, W is a vector space over a field L, and $α$ is a collineation from PG(V) to PG(W). This implies K and L are isomorphic fields, V and W have the same dimension, and there is a semilinear map $φ$ such that $φ$ induces $α$ .

The fundamental theorem explains the different definition for projective lines. Otherwise, every bijection between the points would be a collineation, and then there would be no nice algebraic relationship.