So the wedge product can be defined by simply setting a basis $e_i$ and requiring that $e_i \wedge e_j = - e_j \wedge e_i$ for $i \neq j$ and that $e_i \wedge e_i = 0$ along with commuting with scalars and distribution and associativity. It can be shown then that for any vector $a$ one has $a \wedge a = 0$. The wedge product defines a bit of signed area (or volume) on an oriented plane, hyperplane, etc.... just as a vetor represents a signed quantity in a specific direction. The (typical) geometric product is defined as follows for an orthonormal basis: $e_i e_j = - e_i e_j$ for $i\neq j$ and $e_i e_i = 1$. If $a$ is a vector then $a a = a\cdot a$ the usual dot product of a with itself. In general for vectors $a$ and $b$ the geometric product is $ab = a\cdot b + a \wedge b$. If $b$ has unit length then $b b = 1$, so $b$ is its own inverse. Consider the conjugation $\tau_v \colon v \mapsto e_1 v e_1$. Then for any basis vector other than $e_1$ the result is $e_i \mapsto e_1 e_i e_1 = - e_1 e_1 e_i = - e_i$. But $e_1 \mapsto e_1 e_1 e_1 = e_1$. Thus $\tau_v$ preserves the $e_1$ axis, but is an inversion on the perpendicular space. The same is true for $\tau_a \colon v \mapsto a v a$ for any unit vector $a$. The line through $a$ is fixed pointwise, and the perpendicular space to $a$ undergoes an involution where everything is negated. The combination of $\tau_a$ followed by $\tau_b$ then fixes everything perpendicular to the $a,b$ plane. However, on the $a,b$ plane, if $h$ is rotation from $a$ to $b$ then $\tau_b\circ \tau_a$ is $h$ applied twice, giving twice the rotation of a direct rotation from $a$ to $b$. Thus the map $v \mapsto (ba)v(ab)$ rotates the $a,b$ plane as if a single rotation directly from $a$ to $b$ had been applied twice, whereas it leaves the subspace perpendicular to the $a,b$ plane pointwise fixed. The scalar part of $ab$ is the cosine of $\psi$ where $\psi$ is the angle from $a$ to $b$. Put differently, the angle of rotation is $\theta$ and the scalar part of this rotation is $\cos(\theta/2)$. Because $abab = -aabb = -1$ then if we write out the taylor series expansion for $exp(t ab)$ where t is a scalar we get $cos(t) + sin(t) ab$. The inverse of $cos(t) + sin(t)ab$ is $cos(t) + sin(t) ba = cos(t) - sin(t)ab$. Thus there is a rotation subgroup cos(t) + sin(t)ab, i.e. $v \mapsto (cos(t) - sin(t) ab) v (cos(t) + sin(t) ab)$ When $t=0$ this is the identity map and when $t=\pi/2$ this is the rotation $v\mapsto (ba) v (ab)$ Now $ab = a\cdot b + a \wedge b$. If $a,b$ are unit vectors, then multiplying $[\cos(t) + \sin(t)ab][\cos(s) + \sin(s)ab]$ yields $\cos(t) \cos(s) + \sin(t)\sin(s)abab + \cos(t)\sin(s) ab + \sin(t)\cos(s) ab = (\cos(t) \cos(s) - \sin(t)\sin(s)) + [ \cos(t)\sin(s) + \sin(t)\cos(s)]ab$ but this is just $\cos(t + s) + \sin(t +s) ab$ In 4 dimensions a rotation is determined by the element $( \cos(s/2) + \sin(s/2) e_1 e_2 ) ( \cos(t/2) + \sin(t/2) e_3 e_4 )$ Expanding gives $\cos(s/2) \cos(t/2) + \cos(s/2) \sin(t/2) e_3 e_4 + \cos(t/2) \sin(s / 2) e_1 e_2 + \sin(s/2) \sin(t/2) e_1 e_2 e_3 e_4$ So there is a vector and a pseudovector and one is $\cos(s/2) \cos(t/2)$ and the other is $\sin(s/2) \sin(t/2) e_1 e_2 e_3 e_4$ There is also a bivector part which is where we get real direction (because the vector and pseudovector are directionless). So given a unit versor for $\mathbb{R}^4$ is there necessarily a corresponding rotation, and if so, can we read off the planes of rotations and their rotation angles as we can in 3 dimensions? That is, given $a + b_{12) e_1 e_2 + b_{13} e_1 e_3 + b_{14} e_1 e_4 + b_{23} e_2 e_3 + b_{24} e_2 e_4 + b_{34} e_3 e_4 + c e_1 e_2 e_3 e_4$ a unit vector where the sum of squares of the coefficients is one, does this always give a rotation, and if so, what are the planes and their rotations? Now the grade two part should be the sum of two different blades, each one the orthogonal complement of the other. A single 2 blade in $\mathbb{R}^4$ has grade two part satisfying $u_{12}u_{34} + u_{13}u_{24} + u{14}u_{23} = 0$. The orthogonal blade grade 2 part has $v_{12} = u_{34}$ and $v_{13} = u_{24}$ etc.... One would expect that every possible choice of $b_{ij}$ can be represented nearly uniquely as a weighted sum of two orthogonal 2 blades, i.e. choose 1, then the other is detemined, and one can choose the coefficient of the other which gives just the right dimension.