By: Alan Bromborsky (brombo14) - 2009-04-25 11:38 |

If you want to do geometry consider conformal geometry using geometric algebra where vectors in 3D are mapped into null vectors in 5D and rotations, translations, dilations, inversions, and some compound transformations in 3D are represented as rotations in 5D. Additionally basic geometric objects such as lines, planes, circles, and spheres in 3D are blades in 5D and it is easy to calculate the intersections of any of these objects. Good references for this approach are: Chapter 10, "Geometric Algebra for Physicist" by Doran & Lasenby "Geometric Algebra for Computer Science" by Dorst, Fontijne, and Mann and the following link http://xxx.lanl.gov/abs/cs.CG/0203026 |

Yes, I have read the references that you mention and I have put my findings on this page: https://www.euclideanspace.com/maths/geometry/space/nonEuclid/conformal/index.htm Although, I don't claim to fully understand it, I certainly don't feel that I have an intuitive understanding of it. I am trying to work out how to represent the various quantities as parameters of the 5 dimensions, for example: point(x,y,z) = -n0+ (x²+y²+z²) n∞ + x n1 + y n2 + z n3 What I would like to do is derive a table of such 5D parameters for the following 3D objects: Elements (in 3D space): point[x,y,z] vector[x,y,z] two points line plane Transforms (in 3D space): pure translation pure rotation reflection in plane reflection in sphere scaling Martin |

By: Alan Bromborsky (brombo14) - 2009-04-25 19:58 |

Let the mapping from a 3D vector x to a 5D null vector X be: X = f(x) = ((x*x)*n+2*x-nbar)/2 where n is your n_0 and nbar is n_infinity. The normalization is X.n = -1. Then let x,y,z,w be point in 3D and there 5D null images are X,Y,Z,W. The we have the following 5D rotors for the corresponding 3D transformation: if u is rotation axis vector (3D) and alpha rotation angel the rotor is cos(alpha/2)+u*I_3D*sin(alpha/2) if a is translation vector (3D) the translation rotor is 1+n*a/2 if exp(-alpha) is the dilation factor and N = (nbar^n)/2 the dilation rotor is cosh(alpha/2)+N*sinh(alpha/2) Geometric objects (line,circle,plane,sphere) in 3D are represented by blades in 5D line through x and y: f(x)^f(y)^n circle through x, y, and z: f(x)^f(y)^f(z) plane through x, y, and z: f(x)^f(y)^f(z)^n sphere through x, y, z, and w: f(x)^f(y)^f(z)^f(w) If B is the 5D blade representing the object the equation of the object in 3D is given by: B^f(p) = 0 where p is a point on the object Simple geometric algebra formulas result for the intersection of geometric objects. Output of sympy test program for conformal geometry: Example: Conformal representations of circles, lines, spheres, and planes a = e0, b = e1, c = -e0, and d = e2 A = F(a) = 1/2*(a*a*n+2*a-nbar), etc. Circle through a, b, and c Circle: A^B^C^X = 0 = (-x2)e0^e1^e2^n +(x2)e0^e1^e2^nbar +(-1/2 + x0**2/2 + x1**2/2 + x2**2/2)e0^e1^n^nbar Line through a and b Line : A^B^n^X = 0 = (-x2)e0^e1^e2^n +(-1/2 + x0/2 + x1/2)e0^e1^n^nbar +(x2/2)e0^e2^n^nbar +(-x2/2)e1^e2^n^nbar Sphere through a, b, c, and d Sphere: A^B^C^D^X = 0 = (1/2 - x0**2/2 - x1**2/2 - x2**2/2)e0^e1^e2^n^nbar Plane through a, b, and d Plane : A^B^n^D^X = 0 = (1/2 - x0/2 - x1/2 - x2/2)e0^e1^e2^n^nbar Hyperbolic Circle: (A^B^e)^X = 0 = (-x2)e0^e1^e2^n +(-x2)e0^e1^e2^nbar +(-1/2 + x0 + x1 - x0**2/2 - x1**2/2 - x2**2/2)e0^e1^n^nbar +(x2)e0^e2^n^nbar +(-x2)e1^e2^n^nbar Extracting direction of line from L = P1^P2^n (L.n).nbar= (2)p1 +(-2)p2 Extracting plane of circle from C = P1^P2^P3 ((C^n).n).nbar= (2)p1^p2 +(-2)p1^p3 +(2)p2^p3 (p2-p1)^(p3-p1)= p1^p2 -p1^p3 +p2^p3 |

By: Alan Bromborsky (brombo14) - 2009-04-26 11:37 |

Additonal comment on conformal geometry - The method applies equally to N-dimensions as well as 3-dimensions. It also applies to non-euclidian geometry by letting the point at infinity be represented by e (hyperbolic geometry) or ebar (spherical geometry) instead of n. Again this can be a non-euclidian geometry in N-dimension. Of particular interest would be the non-euclidian geometries of space time. |