.
A Homogeneous Framework
for
Computational Geometry and Mechanics
David Hestenes
Department of Physics and Astronomy
Arizona State University, Tempe, Arizona, USA
Need: Euclidean geometry supplies essential mathematical
underpinnings for physics, engineering and
Computer-Aided Geometric Design.
Question: How should we formulate Euclidean Geometry to
facilitate geometric modeling and analysis
optimize computational efficiency?
A fundamental problem in the Design of Mathematics
GEOMETRIC ALGEBRA Website:
http://modelingnts.la.asu.edu
Can access related websites from this one
.
Standard model for Euclidean 3-space:
~
E3 R3 = {x} = 3d real vector space
=
Rigid body motions and symmetries described by
Euclidean group E(3)
x - x = Rx + a
yy
Invariant: (x )2 =(x )2
Computations: Require coordinates and Matrices
[x ] =[R][x] +[a]
Drawbacks:
Extraneous information
(from arbitrary choice of coordinates)
Redundant matrix elements
(9 elements for 3 degrees of freedom)
Difficult to interpret
(parameters mixed in matrix elements)
Computationally inefficient
(9 9 vs. 3 3 multiplications)
2
Geometric algebra eliminates need for coordinates:
Canonical form for orthogonal transformations:
+1
R parity R =
Rx = RxR1
-
Reflection represented by a vector a
a
x
1
Ax = axa
Ax
Rx
Rotation = double reflection
a
b
Rx = BAx
x
1 1
= b( axa )b
1 1 1
=
(ba) x (a b ) =RxR
Ax
Rotation represented by a versor (rotor, spinor, quaternion)
R = ba
Reduces composition of rotations/reflections to
geometric product of versors:
R2R1 = R3 R2R1 = R3
!
Extensive treatment in (Hestenes 1986)
3
Symmetries of the Cube
'
b
2
c'
c
2
a' 3
Ą/2
Ą/3
Ą/4
a
b
2
22
a, b, c
3 Generators: a = b = c 2 = 1
( a b) 4 = 1
Relations:
3
( b c) = 1
2
( c a) = 1
4
Problem:
~
Vector space model for E3 R3 = {x}
=
singles out one point for special treatment
as the origin x =0
Drawbacks:
Often need to shift origin to
simplify calculations
avoid dividing by zero
prove results independent of origin
Rigid body displacements combine
rotations multiplicatively and
translations additively,
destroying the simplicity of both
Solution:
Design a homogeneous model for that
E3
treats all points equally
treats both rotations and translations multiplicatively
[ homogeneous in the sense of homogeneous coordinates]
5
Homogeneous Euclidean Geometry
Arena: Minkowski Algebra Rn+1,1 = G(Rn+1,1)
= vector space with signature (n + 1,1)
Rn+1,1
{x | x2 =0}
Null cone:
.
Hyperplane: {x | x e =1}
Homogeneous model of Euclidean space:
~
.
En {x | x2 =0, x e =1} e2 =0
=
x called a point in e = point at infinity
En
Horosphere (F. A. Wachter 1792-1817), (originally with coordinates)
.
x e
= 1
e0
e
.
x
_
e
Geometric algebra essential
to make the horosphere
coordinate-free and computationally efficient
6
Hermann Gnther Grassmann (1809-1877)
Laid the foundations for geometric algebra
Set a direction for future development
But he failed to reach his main goal,
ultimately concluding that it was impossible
Grassmann s Goal: To formulate
Euclidean geometry as an algebra of geometric objects
(points, lines, planes, circles, etc.)
integrated
Computational
Synthetic
{ }
}
{
geometry
geometry
with
married
Interpretation Computation
to
The homogeneous model for En within geometric algebra
reaches Grassmann s Goal with many surprises!!!
7
Euclidean Geometry is inherent in
algebraic properties of homogeneous points
I. Point-to-point distance from inner products:
00
.
(x y)2 = x2 2x y + y2
. = Euclidean distance
(x y)2 = 2x y
=!
Points must be null vectors! Why Grassmann failed!
=!
Pythagorean theorem, triangular inequality, etc.
II. Lines and Planes from outer products
A= a '"b '"e =line segment or line thru pts ab
and
2
.
A =
(a '" b '" e)2 =(a b)2 = 2a b = (length)2
a '" b '" c '" e = plane segment or plane
0 111
. .
1 0 a b a c
2
P = (a '" b '" c '" e)2 = = 4(area)2
. .
1 b a 0 b c
. .
1 c a c b 0
Cayley-Menger determinant (another historical curiosity)
Cayley, 1841 (Volumes) Menger, 1931 (distancegeometry)
Dress and Havel, 1993 (relation to GA)
III. Geometric relations from algebraic products
e.g.: Point x lies on the line A or plane P if and only if
x '" A =0 x'"P =0
8
=
P
En
The homogeneous model of is related
to the standard inhomogeneous model by a
Conformal split:
A 1-to-1 mapping of
En = {x}
onto
the pencil of lines Rn = {x}
e0,
through a fixed point
defined by
x - x = x '" E = x '" e0 '" e
and the inverse mapping
1
x - x = xE x2e + e0
2
where
e0 '" e = E =! E2 =1
and (absorption)
eE = e = Ee
~
This implies the isomorphism:
En Rn
=
and determines a split of the whole algebra:
Rn+1,1 = Rn R1,1
,
where Rn = G(Rn)
R1,1 {1, e, e0, E = e0 '"e}
and has the basis:
9
Conformal Splits for Simplexes
1 1
Point: a =(a a2e e0) E=E(a + a2e + e0)
2 2
1
=aE a2e+e0
2
1 1
Product: ab =(aE)(Eb) =(a + a2e+e0)(b b2e e0 )
2 2
1 1
= (a b)2 + a '" b + (a2b b2a)e +(b a)e0 1(b2 a2)E
2 2 2
.b
a '" b
a
Similar to the mess from the spacetime split in STA
.
Line (or line segment) thru a, b, e
b
a '" b '"e = a '" be +(b a)E
a
moment tangent
a b
Plcker coordinates
0
.
Plane (or plane segment)
b
a '" b '" c '" e = a '" b '" ce +(b a) '" (c a)E
tangent
moment tangent
a
a '" [(b a) '" (c a)] = a '" b '" c
c
10
A Surprising algebra of spheres
All spheres and hyperspheres in Enare uniquely
represented by
Rn+1
positive vectors in
.
Sn 1( ,p) =sphere with radius and center p
represented by vector s with
s2 s
1
2
2
= p = e =! p2 =0
2
. .
(s e)2 s e
.
Can simplify with constraint s e =1, so
.
x
1
2
2
s2 = >0, p = s e
2
.p
.
Eqn. for sphere: x s =0 , x2 =0
1
Conformal split: s = pE + (2 p2)e + e0
2
1
. 2
x p = =! 2 =(x p)2
2
.
n e =0
.
Hn 1(n) =hyperplane rep. by vector n with
n2 =1
Conformal split: n = nE e =! n2 = n2 =1
.
n
Eqn. for hyperplane: x n =0 , x2 =0
n -1
.
or: x n =
(n)
H
> 0
.
0
11
The homogeneous model of En
maps all spheres and hyperplanes Rn
into hyperplanes thru the origin in Rn+1,1:
. . >0 , x2 =0; x, s "Rn+1,1}
>
{x | x s =0, s2 0 , s e
fl
.
x s = 0 is the eqn for a hyperplane thru the origin
i.e., a (n + 1)-dim subspace of Rn+1,1
.
s e = 0 A sphere thru e = " is a hyperplane
.
Sphere determined by n + 1 points: a0, a1, a2, an
. . .
. . .'" an = 0
s a0 '" a1 '" a2 '" tangent form
Radius and center p given by duality:
. . .
normal form
s = (a0 '" a1 '" a2 '" '" an)
I
2
s s1
= 2 ,p = 2e
( )
2
. .
s e s e
.
Hyperplane = sphere thru ", say a0 =e
. . .
tangent form
ń =e '" a1 '" a2 '" '" an
. . .'" an)
n = (e '" a1 '" a2 '" normal form
I
.
Eqn for sphere: x '" s =0 !! x s =0
12
Example: Simson s construction
C1
A
..
D
.
.
.
B1
P
. .
.
A1
C
B
s = A '" B '" C = circumcircle of triangle e '" A '" B '" C
2
.
s s s (C '" B '" A) (A '" B '" C)
2 = = =
(
)
.
s e (s e)2 (e '" A '" B '" C)2
'"
A '" B '" C '" D
Identity: (H. Li)
e'"A1 '"B1 '"C1 =
22
A '" B '" C '" D =0 !! e '" A1 '" B1 '" C1 =0
Dlies on circumcircle !! A1, B1, C1 are collinear
13
Rigid Displacements and Symmetries
Versor group
Conformal group
Lorentz group
~
~
=
}
{
= }
} {
{
in R
2 n+1,1
on E n (or )
on R n+1,1 Rn
Lorentz Trans. G ( = orthogonal trans.)
1
+
G(x) = Gx G = x parity = 1
G = versor representation of G ( = spin rep. if =+1)
For homogeneous points x = x 2 =0,
. .
is a scale factor to enforce e x = e x =1
(not a Lorentz invariant)
.
Conformal split
1 1 1
[ + x2e + e0]G =
x [x + (x )2e + e0]
GE E
2
2
x = g (x) is a conformal trans. on Rn
.
Versor rep. reduces composition of conformal transformations
to versor multiplication
g3(x) =g2[g1(x)] !! G3 = G2G1
.
Versor factors G = sk . . . s2 s1 vectors s2 =0
i
1
.
Euclidean group E(n) defined by G e G = e
14
Rotations & translations generated
multiplicatively from reflections
Reflection in the n-(hyper)plane n2 =1
1
n(x) = nxn = x =1
n
Split form: = nE e n2 =1
. .
For plane thru pt c: n c =0 =! n c
=
.
Rotation from planes n and m intersecting at point c:
. .
G = mn =(mE+em c)(nE en c)
m
.
= mn e(m '" n) c
= R e(R c)
.
n
c
mn = R
R
m =
n
Spinors as directed arcs
.
.
Translation from parallel planes m, n
a
G = mn =(mE 0)(nE + e)
n
1
=1 + ae =Ta
2
n
where a =2n
.
generated by sphere vector s
Inversion
15
SE(3) = Special Euclidean group on E3
= { rigid displacements D} {spinors D}
=
2
1
D(x) =Dx D D = TR
1
Translation by c: T = Tc =1 + ec
2
Rotation axis: n = RnR
.
Chasles Theorem: Any rigid displacement can be
expressed as a screw displacement
Proof: Find a point b on the screw axis so that
D = Tc TcĄ"R = Tc Rb = RbTc
where
.
Rb = R + eb R, c =(c n)
1 R2
2 1 1
Solution: b = cĄ"(1 R ) = cĄ"
2
1 R2
1
.
Screw form: D = e2 S S = a screw
se(3) = Lie algebra of SE(3) = {S = in + em}
is a bivector algebra,
1
closed under S1 S2 = (S1S2 S2S1)
2
16
Screw theory follows automatically!
Screws:
Sk = imk + enk
.
Product: S1S2 = S1 S2 + S1 S2 + S1 '" S2
1
Transform: Sk = USk = USkU = AdU Sk
S1S2 = U(S1S2) Product preserving
1
= U(S1 .S2 + S1 S2 + S1 '" S2)U
Invariants: Ue = e, Ui = i
. .
=! U(S1 '"S2) =S1 '"S2 = ie(m1 n2 + m2 n1)
. .
S1 S2 = S1 S2 = m1 .m2 (Killing Form)
Coscrew (Ball s reciprocal screw)
1
Sk*= Sie0 = (Sie0 + ie0S) =ink +mke0
2
2
* *
Invariant: S1 .S2 = S2 .S1 = (S1 '" S2)ie0
.
= m1 n2 + m2 .n1
.
*
S S
1
.
Pitch: h = = n m1
2
S .S
1 7
Screw Mechanics (of a rigid body)
1
I Kinematics of body pt. x = Dx0D D = D(t)
.
1
.
= VD x =V x
2
.
V = i + ev =! = x + v
= instantaneous screw v = CM velocity
.
P = M V = iI + mve = i + pe
= comomentum (a coscrew)
II Dynamics
.
e0
V = W = i+ f = coforce (wrench)
Ł
+
=! K = V .W = . v .f = Power
1 1
.
K = V .P = + v .p = K.E.
2 2
More in NFCM
1
III Change of Frame x - x = Ux = UxU
Ł
=! V = UV Covariant U =0
P =UP Contravariant
P .V = P .V Invariant
18
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