Search results
Results from the WOW.Com Content Network
Given two projective frames of a projective space P, there is exactly one homography of P that maps the first frame onto the second one. If the dimension of a projective space P is at least two, every collineation of P is the composition of an automorphic collineation and a homography. In particular, over the reals, every collineation of a ...
In projective geometry, a homography is an isomorphism of projective spaces, ... In synthetic geometry, a projective space S can be defined axiomatically as a set P ...
Most significantly, representation of P 1 (R) in a projective space over a division ring K is accomplished with a (K, R)-bimodule U that is a left K-vector space and a right R-module. The points of P 1 ( R ) are subspaces of P 1 ( K , U × U ) isomorphic to their complements.
A homography (or projective transformation) of PG(2, K) is a collineation of this type of projective plane which is a linear transformation of the underlying vector space. Using homogeneous coordinates they can be represented by invertible 3 × 3 matrices over K which act on the points of PG(2, K ) by y = M x T , where x and y are points in K 3 ...
If the geometric dimension of a pappian projective space is at least 2, then every collineation is the product of a homography (a projective linear transformation) and an automorphic collineation. More precisely, the collineation group is the projective semilinear group , which is the semidirect product of homographies by automorphic collineations.
Given another projective space P(V) of the same dimension n, and a frame F of it, there is exactly one homography h mapping F onto the canonical frame of P(K n+1). The projective coordinates of a point a on the frame F are the homogeneous coordinates of h(a) on the canonical frame of P n (K).
The projective plane cannot be embedded (that is without intersection) in three-dimensional Euclidean space. The proof that the projective plane does not embed in three-dimensional Euclidean space goes like this: Assuming that it does embed, it would bound a compact region in three-dimensional Euclidean space by the generalized Jordan curve ...
Frequently cross ratio is introduced as a function of four values. Here three define a homography and the fourth is the argument of the homography. The distance of this fourth point from 0 is the logarithm of the evaluated homography. In a projective space containing P(R), suppose a conic K is given, with p and q on K.