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The even Clifford sub-algebra $C\ell^+(0,3,0)$ is isomorphic to the quaternion algebra. The mapping between terms is $e_0 \mapsto 1$, $e_{23} \mapsto i$, $e_{31} \mapsto j$, $e_{12} \mapsto k$. In quaternion literature for spacecraft control, the quaternion cross product between $a = a_0 + a_1 i + a_2 j + a_3 k = (a_0,\vec{a})$ and $b = b_0 + b_1 i + b_2 j + b_3 k = (b_0,\vec{b})$ is given by $a \times b = (0,a_0 \vec{b} + b_0\vec{a} + \vec{a} \times \vec{b}) = \frac{1}{2}(ab - b^*a^*)$.

This last expression, $a \times b = \frac{1}{2}(ab - b^*a^*)$ is different from the well-known expression for Clifford algebra elements $u$ and $v$, of grade-0 or grade-1, $u \wedge v = \frac{1}{2} (uv - vu)$.

  • Question 1. What is the correct way, if any, to reconcile the two?
  • Question 2. Is $e_{23}$, for example, considered a bivector? Or would a term like $e_{23} \wedge e_{31}$ be the bivector of the Clifford algebra?
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  • $\begingroup$ I believe $C\ell^+(0,3)$ is isomorphic to $C\ell^+(3,0)$. What is the third number for? vectors that square to $0$, instead of $1$ or $-1$? $\endgroup$ – mr_e_man Aug 3 '18 at 5:40
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In the geometric algebra of $R^3$ the bivectors are ismorfic to the complex numbers since any basis bivector $e_{ij} = e_i \wedge e_j = e_i * e_j$ multiplied by itself is $e_{ij}*e_{ij} = -1$.

$(e_i*e_j) * (e_i*e_j) = -e_j*(e_i *e_i)*e_j = -e_j*e_j = -1$

The product involved is the geometric product. Geometric product is defined for two vectors $a$ and $b$ as:

$a b = a \cdot b + a \wedge b$

Conversely, it is easy to show that:

$a \wedge b = 0.5( a b - b a)$

It is also easy to show that the cross product of two vectors is:

$a \times b = (a \wedge b) * e_{321}$

The dual of any bivector $B$ is a vector $c$ which is normal to the plane defined by $B$ itself and is defined as $c = B * e_{321}$.

Since

$e_1 = e_{23} * e_{321}$

$e_2 = e_{31} * e_{321}$

$e_3 = e_{12} * e_{321}$

An even grade multivector of the form $\alpha + B$, where $\alpha$ is a scalar (0-vector) and $B$ is a bivector or $2-vector$, is called a rotor. Rotors are isomorphic to quaternions as you pointed out. A unit rotor represent a rotation in 3D space.

The conversion between basis bivectors and conplex numbers $i$, $j$, $k$ is as you described.

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  • $\begingroup$ Thank you! Can you take a wedge product of two rotors? As $(a_0 + \vec{a}) \wedge (b_0+\vec{b})$? $\endgroup$ – Balavs Jul 21 '18 at 16:48
  • $\begingroup$ The wedge product of two rotors is $(a_0 + A) \wedge (b_0 + B) = a_0 b_0 + a_0 B + b_0 A + A \wedge B$ but $A \wedge B = 0$ so you get $(a_0 + A) \wedge (b_0 + B) = a_0 b_0 + a_0 B + b_0 A$ $\endgroup$ – Mauricio Cele Lopez Belon Jul 22 '18 at 0:35
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(This builds on Mauricio's answer & comments.)

The wedge product of bivectors (the grade 4 part of their geometric product) does not correspond to the cross product. Instead, we use the "commutator" $A\times B=(AB-BA)/2$, which is the bivector (grade 2) part of their product. For example,

$$(2e_1e_2-e_2e_3)\times(e_2e_3) = \langle(2e_1e_2-e_2e_3)(e_2e_3)\rangle_2 = \langle2e_1e_2e_2e_3-e_2e_3e_2e_3\rangle_2$$ $$= \langle2e_1e_2e_2e_3+e_3e_2e_2e_3\rangle_2 = \langle2e_1e_3+1\rangle_2 = 2e_1e_3$$

For rotors, which are scalars plus bivectors, the commutator results in

$$(a_0+A)\times(b_0+B)$$ $$= \frac{a_0b_0-b_0a_0}{2}+\frac{a_0B-Ba_0}{2}+\frac{Ab_0-b_0A}{2}+\frac{AB-BA}{2}$$ $$= 0+0+0+A\times B$$

which is a pure bivector; the scalar part disappears. But this still doesn't agree with your quaternion formula, which depends on $a_0$ and $b_0$.

The quaternion or Complex conjugate $A^*$ corresponds to the "reverse" $\tilde A$, which reverses the multiplication order of the vectors in $A$. For example,

$$(e_1e_2)^\sim = e_2e_1 = -e_1e_2$$

and any bivector's reverse is its negation, while vectors and scalars are unchanged. In general, a grade $k$ blade's reverse is $\tilde A=(-1)^{k(k-1)/2}A$.

So your formula is translated to

$$\frac12\Big((a_0+A)(b_0+B)-(b_0+B)^\sim(a_0+A)^\sim\Big)$$ $$= \frac12\Big((a_0+A)(b_0+B)-(b_0-B)(a_0-A)\Big)$$ $$= \frac12\Big((a_0b_0+a_0B+Ab_0+AB)-(b_0a_0-b_0A-Ba_0+BA)\Big)$$ $$= \frac12\Big(0+2a_0B+2b_0A+AB-BA\Big)$$ $$= a_0B+b_0A+A\times B$$

An easier way to remember this formula is to simply multiply them (with quaternion or geometric product) and drop the scalar part.

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  • $\begingroup$ In the commutator example, I assumed that the vectors square to ${^+}1$. With the opposite signature, the result would be $-2e_1e_3$. $\endgroup$ – mr_e_man Aug 3 '18 at 5:50
  • $\begingroup$ This is very useful and really clears up many of my confusions. Would the dot product behave similarly, except we just keep the grade-0 element of the multiplication? $\endgroup$ – Balavs Aug 4 '18 at 22:29
  • $\begingroup$ What kind of dot product? If you mean $a_0b_0-a_1b_1-a_2b_2-a_3b_3$, then yes. $\endgroup$ – mr_e_man Aug 7 '18 at 7:18

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