Conformal geometry with Clifford algebra

Robin K. S. Hankin

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 \(\operatorname{Cl}(p,q)\), which Perwass denotes \(\mathbb{G}_{p,q}\). The set of grade-\(k\) objects is denoted \(\mathbb{G}_{p,q}^k\). First we define the IPNS and OPNS [inner product null space and outer product null space] of \(\mathbf{A}\in\mathbb{G}_{p,q}^k\).

\[ \begin{eqnarray} \operatorname{IPNS}(\mathbf{A}) &=& \left\lbrace\mathbf{x}\in\mathbb{G}^1_{p,q}\colon\mathbf{x}\cdot\mathbf{A} = 0\right\rbrace\\ \operatorname{OPNS}(\mathbf{A}) &=& \left\lbrace\mathbf{x}\in\mathbb{G}^1_{p,q}\colon\mathbf{x}\wedge\mathbf{A} = 0\right\rbrace \end{eqnarray} \]

Thus if \(\mathbf{A}=\left\langle \mathbf{A}\right\rangle_{k}=\bigwedge_{i=1}^k\mathbf{a}_i\), then \(\operatorname{OPNS}(\mathbf{A})=\operatorname{span}(\mathbf{a}_1,\ldots,\mathbf{a}_k)\subseteq\mathbb{G}_{p,q}\). Further, \(\operatorname{OPNS}(\mathbf{A})\) is linear in the sense that \(\mathbf{x},\mathbf{y}\in\operatorname{OPNS}(\mathbf{A})\) implies \(\alpha\mathbf{x}+\beta\mathbf{y}\in\operatorname{OPNS}(\mathbf{A})\) for any real numbers \(\alpha,\beta\). We will use this system to express a geometrical object \(\mathcal{G}\subset\mathbb{R}^3\) (such as a sphere or a line) as a point \(\mathbf{P}\) in \(\operatorname{Cl}(4,1)\); the idea is that the IPNS or OPNS of \(\mathbf{P}\) is \(\mathcal{G}\). 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 \(\mathbf{e}_1\), \(\mathbf{e}_2\), \(\mathbf{e}_3\) together with two additional basis vectors \(\mathbf{e}_+\) and \(\mathbf{e}_-\) obeying \(\mathbf{e}_+^2=1\), \(\mathbf{e}_-^2=0\), \(\mathbf{e}_+\cdot\mathbf{e}_-=0\). If we define

\[ \mathbf{e}_0=\frac{1}{2}\left(\mathbf{e}_--\mathbf{e}_+\right),\qquad \mathbf{e}_\infty=\mathbf{e}_- +\mathbf{e}_+ \]

then we may say that \(\mathbf{e}_0\) represents “the origin” and \(\mathbf{e}_\infty\) represents “the point at infinity”. We see that \(\mathbf{e}_0^2=\mathbf{e}_\infty^2=0\) and \(\mathbf{e}_\infty\cdot\mathbf{e}_0=-1\); the geometric product is given by \(\mathbf{e}_\infty\mathbf{e}_0=-E-1\) and \(\mathbf{e}_0\mathbf{e}_\infty=E-1\) where \(E=\mathbf{e}_\infty\wedge\mathbf{e}_0\). 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

So

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

E

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

Points

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

\[ \mathbf{A}=C(\mathbf{a})=\mathbf{a}+\mathbf{a}^2\mathbf{e}_\infty/2 +\mathbf{e}_0\qquad\mathbf{B}=C(\mathbf{b})=\mathbf{b}+\mathbf{b}^2\mathbf{e}_\infty/2 +\mathbf{e}_0 \]

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 \(\mathbf{A}\cdot\mathbf{B}=-\left\lVert\mathbf{a}-\mathbf{b}\right\rVert^2/2\) [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 \(\mathbf{a}\) and radius \(\rho\) as

\[ \mathbf{S}=C(\mathbf{a}) - \rho^2\mathbf{e}_\infty/2 \]

and then the sphere is just the inner product null space of \(\mathbf{S}\), that is \(\left\lbrace\mathbf{x}\colon\mathbf{S}\cdot\mathbf{x}=0\right\rbrace\). 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 \(\rho\) can be calculated from \(\mathbf{S}^2=\rho^2\). 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 \(P=\mathbf{S}\mathbf{e}_\infty\mathbf{S}\):

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 \(\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\).

The conformal representation for a sphere in OPNS would be

\[ S^*=P_1\wedge P_2\wedge P_3\wedge P_4 \]

and points that lie on the sphere are the set

\[\left\lbrace\mathbf{x}\colon\mathbf{x}\wedge S^*=0\right\rbrace.\]

Observe again that this parameterization does not require one to take a basis of \(\mathbb{R}^3\), 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 \(\left\lbrace P_1,P_2,P_3,P_4\right\rbrace\) 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 \(\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}+\frac{\sqrt{3}}{2}\right)\) 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 \(p\wedge S\) is zero to within numerical error. Again observe that this is established without taking a basis of \(\mathbb{R}^3\).

Planes

A plane is defined as \(\hat{\mathbf{n}} + d\mathbf{e}_\infty\), where \(\hat{\mathbf{n}}\) 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 \(\Pi\) with normal \(\mathbf{n}=\left(1,2,5\right)\) and distance 7. We may use plane() above but first need to calculate \(\hat{\mathbf{n}}=\mathbf{n}/\left|\mathbf{n}\right|\):

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 \(\Pi\) are in fact in its IPNS. We observe that the point \(7\hat{\mathbf{n}}\in\Pi\), and also that vectors \(\mathbf{u}=(2,-1,0)\) and \(\mathbf{v}=(5,0,-1)\) are orthogonal to the normal of \(\Pi\). So \(7\hat{\mathbf{n}} + \alpha\mathbf{u} + \beta\mathbf{v}\in\Pi\) for any real numbers \(\alpha,\beta\). 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 \(\alpha,\beta\) to generate P1, P2, P3 which are known to lie in plane \(\Pi\). Then to verify that these points lie in the IPNS of \(\Pi\) we need to take the inner product with the conformal representation of \(\Pi\):

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 \(7\hat{\mathbf{n}} + \alpha\mathbf{u} + \beta\mathbf{v}\in\operatorname{IPNS}(\Pi)\) for at least these values of \(\alpha\) and \(\beta\). Alternatively, a plane can be thought of as a sphere that touches the point at infinity; thus the OPNS of plane is given by

\[ \pi^\star = P_1\wedge P_2\wedge P_3\wedge\mathbf{e}_\infty \]

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 \(d\hat{\mathbf{n}} + 7.6\mathbf{u} - 9.23\mathbf{v}\):

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:

\[ Z^*=P_1\wedge P_2\wedge P_3\]

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.