Collineations of an affine geometry

  • Thread starter samkolb
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Main Question or Discussion Point

An affine geometry is a nonempty set of points A, together with a set of lines, where each line connects one or more of the points in A.

A collineation of A is a bijection f: A --> A that carries lines to lines. That is, if P,Q are points in A lying on the same line, then f(P), f(Q) are points in A lying on the same line.

My question:

Does this definition imply that each pair of points not lying on the same line is carried to a pair of points not on the same line? That is, if P, Q do not lie on the same line, then do f(P) and f(Q) also not lie on the same line?
 

Answers and Replies

  • #2
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No, this does not follow. The condition says nothing about points which are not collinear. But, and this is somehow missing in the definition of a collineation, what about ##f^{-1}##. You would expect from a collineation as a bijection, that the inverse function is also a collineation. This should have been stated.

So if we assume, that ##f^{-1}## is a collineation, too, then the situation is a different one:
If ##f(P)## and ##f(Q)## are collinear, then ##f^{-1}(f(P))=P## and ##f^{-1}(f(Q))=Q## are collinear because ##f^{-1}## is a collineation. This is equivalent to: If ##P## and ##Q## are not collinear, then ##f(P)## and ##f(Q)## can't be collinear.

Hence we need that ##f^{-1}## is a collineation. This should either be part of the definition, or a theorem proven by other means, will say other axioms.
 

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