Wedge Product on $C\ell^+(0,3,0)$ Relationship to Quaternion Cross Product 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, $u \wedge v = \frac{1}{2} (uv - vu)$.


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*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?

 A: 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.
A: (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.
