Conformal geometry with Clifford algebra

To cite the clifford package in publications please use Hankin (2022). This short document shows how conformal geometry may be implemented using the clifford R package; it follows Hildenbrand (2013) and Perwass (2009). Here we work in $Cl (p, q)$ , which Perwass denotes $G_{p, q}$ . The set of grade- $k$ objects is denoted $G_{p, q}^{k}$ . First we define the IPNS and OPNS [inner product null space and outer product null space] of $A \in G_{p, q}^{k}$ .

$\begin{array}{rcl} IPNS (A) & = & {x \in G_{p, q}^{1} : x \cdot A = 0} \\ OPNS (A) & = & {x \in G_{p, q}^{1} : x \land A = 0} \end{array}$

Thus if $A = {⟨ A ⟩}_{k} = ⋀_{i = 1}^{k} a_{i}$ , then $OPNS (A) = span (a_{1}, \dots, a_{k}) \subseteq G_{p, q}$ . Further, $OPNS (A)$ is linear in the sense that $x, y \in OPNS (A)$ implies $α x + β y \in OPNS (A)$ for any real numbers $α, β$ . We will use this system to express a geometrical object $G \subset R^{3}$ (such as a sphere or a line) as a point $P$ in $Cl (4, 1)$ ; the idea is that the IPNS or OPNS of $P$ is $G$ . This allows one to express geometrical ideas in pure Clifford formalism, without needing to take a dangerous and unsightly basis.

To work in R, we set up some basic features of conformal geometry. Specifically, we consider the three Euclidean basis vectors $e_{1}$ , $e_{2}$ , $e_{3}$ together with two additional basis vectors $e_{+}$ and $e_{-}$ obeying $e_{+}^{2} = 1$ , $e_{-}^{2} = 0$ , $e_{+} \cdot e_{-} = 0$ . If we define

$e_{0} = \frac{1}{2} (e_{-} - e_{+}), e_{\infty} = e_{-} + e_{+}$

then we may say that $e_{0}$ represents “the origin” and $e_{\infty}$ represents “the point at infinity”. We see that $e_{0}^{2} = e_{\infty}^{2} = 0$ and $e_{\infty} \cdot e_{0} = - 1$ ; the geometric product is given by $e_{\infty} e_{0} = - E - 1$ and $e_{0} e_{\infty} = E - 1$ where $E = e_{\infty} \land e_{0}$ . It is straightforward to implement these objects using the clifford package:

dimension <- 3
options("maxdim" = dimension+2)  # paranoid safety measure
signature(dimension + 1,1)
eplus <- basis(dimension+1)
eminus <- basis(dimension + 2)

e0 <-  (eminus - eplus)/2
einf <- eminus + eplus
E <- e0 ^ einf

e0

## Element of a Clifford algebra, equal to
## - 0.5e_4 + 0.5e_5

einf

## Element of a Clifford algebra, equal to
## + 1e_4 + 1e_5

## Element of a Clifford algebra, equal to
## - 1e_45

Points

With these definitions, we can consider Euclidean vectors $a, b \in R^{3}$ and conformal embeddings $A, B$ given by

$A = C (a) = a + a^{2} e_{\infty} / 2 + e_{0} B = C (b) = b + b^{2} e_{\infty} / 2 + e_{0}$

This is straightforward in package idiom:

point <- function(x){ as.1vector(x) + sum(x^2)*einf/2 + e0 }

Thus point() takes an R vector of length 3 and returns its conformal embedding. For example, we may translate points $(1, 2, 5)$ and $(2, 2, 2)$ to their conformal equivalent:

a <- c(1,2,5)
b <- c(2,2,2)
point(a)

## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 5e_3 + 14.5e_4 + 15.5e_5

point(b)

## Element of a Clifford algebra, equal to
## + 2e_1 + 2e_2 + 2e_3 + 5.5e_4 + 6.5e_5

It can be shown that $A \cdot B = - {‖ a - b ‖}^{2} / 2$ [where the dot product is the Clifford inner product, %.%]. Package idiom to verify this would be

c(conformal=drop(point(a) %.% point(b)), Euclidean = -sum((a-b)^2)/2)

## conformal Euclidean 
##        -5        -5

showing that the two results match.

Sphere, IPNS

We can define a sphere with center $a$ and radius $ρ$ as

$S = C (a) - ρ^{2} e_{\infty} / 2$

and then the sphere is just the inner product null space of $S$ , that is ${x : S \cdot x = 0}$ . This is straightforward to implement computationally. Suppose we have a sphere of radius 5, center $(1, 2, 3)$ :

sphere <- function(x,r){ point(x) - r^2*einf/2}
S <- sphere(1:3,5)  # center (1,2,3) radius 5:
S

## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3 - 6e_4 - 5e_5

Then object S is a conformal representation of such a sphere. The radius $ρ$ can be calculated from $S^{2} = ρ^{2}$ . Package idiom:

drop(S^2)   # 5^2 = 25

## [1] 25

Finding the center of the sphere is slightly more involved; Hildenbrand calls this the sandwich product given by $P = S e_{\infty} S$ :

S*einf*S

## Element of a Clifford algebra, equal to
## - 2e_1 - 4e_2 - 6e_3 - 13e_4 - 15e_5

Hildenbrand shows that the scaling factor is $- 2$ so this gives us

-S*einf*S/2

## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3 + 6.5e_4 + 7.5e_5

which we recognise as the point $(1, 2, 3)$ :

point(1:3)

## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3 + 6.5e_4 + 7.5e_5

Sphere, OPNS

We can also consider the OPNS for a sphere, defined in terms of four points that lie on it. For example, if our four points are $(0, 0, 0)$ , $(1, 0, 0)$ , $(0, 1, 0)$ , $(0, 0, 1)$ we would have a rather involved algebraic calculation resulting in a sphere of radius $\frac{\sqrt{3}}{2}$ and center $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ .

The conformal representation for a sphere in OPNS would be

$S^{*} = P_{1} \land P_{2} \land P_{3} \land P_{4}$

and points that lie on the sphere are the set

${x : x \land S^{*} = 0} .$

Observe again that this parameterization does not require one to take a basis of $R^{3}$ , as all the results are presented in terms of vector quantities: at no point does one need to consider components or elements of any vector. The R idiom for defining the sphere touching ${P_{1}, P_{2}, P_{3}, P_{4}}$ is straightforward:

origin <- point(c(0,0,0))
px <- point(c(1,0,0))
py <- point(c(0,1,0))
pz <- point(c(0,0,1))

(S <- origin ^ px ^ py ^ pz)

## Element of a Clifford algebra, equal to
## + 0.5e_1234 - 0.5e_1235 - 0.5e_1245 + 0.5e_1345 - 0.5e_2345

And this is the representation for a sphere (translating this into a center and radius is not yet implemented). Slightly slicker R idiom might be:

spherestar <- function(...){Reduce(`^`,list(...))}
spherestar(origin, px, py, pz)

## Element of a Clifford algebra, equal to
## + 0.5e_1234 - 0.5e_1235 - 0.5e_1245 + 0.5e_1345 - 0.5e_2345

As a verification, we may check that point $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2} + \frac{\sqrt{3}}{2})$ is on the sphere as well:

p <- point(c(1,1,1+sqrt(3))/2)
Mod(p ^ S)

## [1] 1.110223e-16

Above we see that $p \land S$ is zero to within numerical error. Again observe that this is established without taking a basis of $R^{3}$ .

Planes

A plane is defined as $\hat{n} + d e_{\infty}$ , where $\hat{n}$ is the unit normal and $d$ the distance to the origin.

plane <- function(n,d){ as.1vector(n/sqrt(sum(n^2))) + d*einf}

For example, we consider the plane $Π$ with normal $n = (1, 2, 5)$ and distance 7. We may use plane() above but first need to calculate $\hat{n} = n / | n |$ :

n <- c(1,2,5)
nhat <- n/sqrt(sum(n^2))
d <- 7
Pi <- plane(nhat,d)
Pi

## Element of a Clifford algebra, equal to
## + 0.1825742e_1 + 0.3651484e_2 + 0.9128709e_3 + 7e_4 + 7e_5

Just to check, we may verify that some points known to lie in $Π$ are in fact in its IPNS. We observe that the point $7 \hat{n} \in Π$ , and also that vectors $u = (2, - 1, 0)$ and $v = (5, 0, - 1)$ are orthogonal to the normal of $Π$ . So $7 \hat{n} + α u + β v \in Π$ for any real numbers $α, β$ . In R:

u <- c(2,-1,0)
v <- c(5,0,-1)
P1 <- point(d*nhat)                
P2 <- point(d*nhat + 1.3*u + 3.44*v)
P3 <- point(d*nhat - 6.1*u + 1.02*v)

Above we use some made-up values of $α, β$ to generate P1, P2, P3 which are known to lie in plane $Π$ . Then to verify that these points lie in the IPNS of $Π$ we need to take the inner product with the conformal representation of $Π$ :

c(drop(Pi %.% P1),drop(Pi %.% P2),drop(Pi %.% P3))

## [1]  3.191891e-16 -1.110223e-16 -3.330669e-16

Above we see zero (to numerical precision), showing that we do indeed have $7 \hat{n} + α u + β v \in IPNS (Π)$ for at least these values of $α$ and $β$ . Alternatively, a plane can be thought of as a sphere that touches the point at infinity; thus the OPNS of plane is given by

$π^{⋆} = P_{1} \land P_{2} \land P_{3} \land e_{\infty}$

where $P_{1}, P_{2}, P_{3}$ are any points that lie in the plane. To illustrate this we will use the three points used in the IPNS above:

Pi2 <- P1 ^ P2 ^ P3 ^ einf

Then for verification we can create another point known to be in the plane and check that this is in the OPNS. Below we will use p4 which is $d \hat{n} + 7.6 u - 9.23 v$ :

p4 <- point(d*nhat + 7.6*u - 9.23*v)
Mod(Pi2 ^ p4)

## [1] 3.005929e-13

Above we see a very small result showing that p4 is indeed in the OPNS of plane Pi2.

Circle

A circle is defined as the intersection of two spheres:

circle <- function(S1,S2){  # IPNS
    S1 ^ S2
}

For example

circle(sphere(1:3,5),sphere(c(1.1,2.1,3.4),6))

## Element of a Clifford algebra, equal to
## - 0.1e_12 + 0.1e_13 + 0.5e_23 - 3.31e_14 - 7.22e_24 - 9.33e_34 - 3.41e_15 -
## 7.32e_25 - 9.73e_35 + 3.91e_45

A circle may be represented in the OPNS by specifying three points that lie on it:

$Z^{*} = P_{1} \land P_{2} \land P_{3}$

circlestar <- function(...){  # OPNS; A^B^C
    jj <- list(...)
    stopifnot(length(jj) == dimension)
    Reduce(`^`,lapply(jj,point))
}

(CIRC <- circlestar(c(1,2,3),c(5,6,3),c(8,8,-2)))

## Element of a Clifford algebra, equal to
## + 8e_123 - 38e_124 - 110e_134 - 54e_234 - 42e_125 - 130e_135 - 74e_235 +
## 40e_145 + 68e_245 + 140e_345

verify:

poc  # point on circle, found numerically [chunk omitted]

## Element of a Clifford algebra, equal to
## - 0.7152127e_1 - 0.2498563e_2 + 0.3267822e_3 - 0.159628e_4 + 0.840372e_5

poc ^ CIRC

## Element of a Clifford algebra, equal to
## - 0.003402925e_1234 - 0.003402866e_1235 - 0.001701745e_1245 - 0.00850813e_1345
## - 0.008507714e_2345

Above we see that poc is at least close to the circle from the small magnitude of the terms in the wedge product.

Lines and point pairs

A line is the intersection of two planes; in R:

line <- function(P1,P2){ P1 ^ P2 }

and a “point pair” is the intersection of three spheres; in R:

pointpair <- function(S1,S2,S3){ S1 ^ S2 ^ S3 }

It is not at all obvious that three spheres intersect in a pair of points; and still less obvious that the process is associative. However, we may verify associativity explicitly:

S1 <- sphere(c(3,2,4),3)
S2 <- sphere(c(3,1,4),4)
S3 <- sphere(c(1,3,3),3)
(S1^S2)^S3

## Element of a Clifford algebra, equal to
## - 5e_123 + 31e_124 + 25e_134 - 40.5e_234 + 29e_125 + 25e_135 - 39.5e_235 -
## 10e_145 + 10e_245 - 5e_345

(S1^S2)^S3 == S1^(S2^S3)

## [1] TRUE

References

Hankin, R. K. S. 2022. “Clifford Algebra in R.” arXiv. https://doi.org/10.48550/ARXIV.2209.13659.

Hildenbrand, D. 2013. Foundations of Geometric Algebra Computing. Springer.

Perwass, C. 2009. Geometric Algebra with Applications in Engineering. Springer.

Robin K. S. Hankin