Maths - Forum Discussion

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

By: Martin Baker (martinbakerProject Admin) - 2009-04-25 18:01
Yes, I have read the references that you mention and I have put my findings on this page: 
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): 
two points 
Transforms (in 3D space): 
pure translation 
pure rotation 
reflection in plane 
reflection in sphere 

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  
if a is translation vector (3D) the translation rotor is  
if exp(-alpha) is the dilation factor and N = (nbar^n)/2 the dilation rotor is  
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  
+(-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  
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 
+(-1/2 + x0 + x1 - x0**2/2 - x1**2/2 - x2**2/2)e0^e1^n^nbar 
Extracting direction of line from L = P1^P2^n 
(L.n).nbar= (2)p1  
Extracting plane of circle from C = P1^P2^P3 
((C^n).n).nbar= (2)p1^p2  
(p2-p1)^(p3-p1)= p1^p2 

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.


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