Some notes on geometric algebra in 4 dimensions

Working in $\mathbb{R}^4$ here with the geometric algebra corresponding to the usual norm, i.e. $e_i e_j = - e_j e_i$ for $i \neq j$ and $e_i e_i = 1$. The inverse of a vector $v$ is just $v / \|v\|^2$. A geometric algebra is an algebra with zero divisors, for if $v$ is a vector of norm 1 so $v v = 1$ then $\frac{1}{2}(1 + v)$ is idempotent and is also a zero divisor (by multiplying by $(1 - v)$. An element of grade $k$ is often called a $k$ vector. Now the product of two vectors $a,b$ is $g(a, b) + a\wedge b$. The product of $a$ and $b - a g(a,b)/g(a,a)$ is therefore $g(a,b) - g(a,a)g(a,b)/g(a,a) + (a \wedge (b - ag(a,b)/g(a,a)) ) = a \wedge b$ since $a \wedge a = 0$. So the grade 2 part of a 2 blade is also always a 2 blade. Put differently, $a\wedge b$ is always a 2 blade. Now given a 2 blade $a\wedge b$ then as above we can assume without loss of generality that $g(a,b) = 0$ and we can also assume that $a$ and $b$ are unit vectors. Then for any scalar $\alpha$ the two blade $a (b + \alpha a) = g(a,b+\alpha a) + a\wedge (b + \alpha a) = \alpha + a \wedge b$. So the grade 0 part is irrelevant to whether something is a 2 blade. If $A = u v$ for othogonormal vectors $u,v$ then $AA = uvuv = -uuvv = -1$. Let $A = r_{12} e_1e_2 + r_{13} e_1e_3 + r_{14} e_1e_4 + r_{23}e_2e_3 + r_{24}e_2e_4 + r_{34} e_3 e_4$. Then $AA = -1$ means $(r_{12} e_1e_2 + r_{13} e_1e_3 + r_{14} e_1e_4 + r_{23}e_2e_3 + r_{24}e_2e_4 + r_{34} e_3 e_4)(r_{12} e_1e_2 + r_{13} e_1e_3 + r_{14} e_1e_4 + r_{23}e_2e_3 + r_{24}e_2e_4 + r_{34} e_3 e_4) = -1$ Now $r_{ij} e_ie_j r_{ij} e_ie_j = r^2_{ij} e_ie_je_ie_j = -r^2_{ij} e_ie_ie_je_j = -r^2_{ij}$. For the mixed terms things are uglier. The $e_1e_2e_3e_4$ term is $r_{12}r_{34}e_1e_2e_3e_4 + r_{13}r_{24}e_1e_3e_2e_4 + r_{14}r_{23}e_1e_4 e_2e_3 + r_{23}r_{14}e_2e_3e_1e_4 + r_{24}r_{13}e_2e_4e_1e_3 + r_{34}r_{12}e_3e_4e_1e_2$ This is equal to $(r_{12}r_{34} - r_{13}r_{24} + r_{14}r_{23} + r_{23}r_{14} - r_{24}r_{13} + r_{34}r_{12})$. Thus $r_{12}r_{34} - r_{13}r_{24} + r_{14}r_{23} = 0$ for a blade. The $e_1e_2$ part of $AA$ is $r_{13}r_{23}e_1e_3e_2e_3 + r_{14}r_{24}e_1e_4e_2e_4 + r_{23}r_{13} e_2e_3e_1e_3 + r_{24}r_{14}e_2e_4e_1e_4 = (-r_{13}r_{32} - r_{14}r_{24} + r_{13}r_{23} + r_{24}r_{14})e_1e_2 = 0$. Let $B = r_{12} e_1e_2 + r_{13} e_1e_3 + r_{14} e_1e_4 + r_{23}e_2e_3 + r_{24}e_2e_4 + r_{34} e_3 e_4$ in $\mathbb{R}^4$ and assume that $r_{12}r_{34} - r_{13}r_{24} + r_{14}r_{23} = 0$. How big is the space of vectors $u$ with $u \wedge B = 0$? If $u = x_1 e_1 + x_2 e_2 + x_3 e_3 + x_4 e_4$ then $u\wedge B = 0$ becomes $x_1 r_{23} e_1e_2e_3 + x_1 r_{24} e_1e_2e_4 + x_1 r_{34} e_1e_3e_4 + x_2 r_{13} e_2 e_1e_3 + x_2r_{14} e_2 e_1 e_4 + x_2 r_{34}e_2e_3e_4 + x_3 r_{12} e_3 e_1e_2 + x_3 r_{14} e_3e_1e_4 + x_3 r_{24} e_3 e_2e_4 + x_4 r_{12} e_4 e_1 e_2 + x_4 r_{13} e_4 e_1 e_3 + x_4 r_{23} e_4 e_2e_3 = 0$ This then becomes the system of 4 homogeneous equations: $x_1 r_{23} - x_2 r_{13} + x_3r_{12} = 0$, $x_1 r_{24} - x_2 r_{14} + x_4r_{12} = 0$, $x_1 r_{34} - x_3 r_{14} + x_4r_{13} = 0$, $x_2 r_{34} - x_3 r_{24} + x_4r_{23} = 0$. Taking $r_{14}$ times the first equation minus $r_{13}$ times the second we obtain $x_1(r_{14}r_{23}-r_{13}r_{24}) + x_3r_{14}r_{12} - x_4r_{13}r_{12} = 0$ If $r_{12}r_{34} - r_{13}r_{24} + r_{14}r_{23} = 0$ then we can rewrite this as $-x_1 r_{12}r_{34} + x_3 r_{14}r_{12} - x_4 r_{13}r_{12} = 0$ which is $-r_{12}$ times the third equation. Similarly if we take the linear combination of the first two equations which eliminates the $x_1$ term we should obtain the fourth equation. Thus the system has rank at most two. If $B$ is a wedge of two orthonormal vectors then we can obtain the span of those vectors as the space of vectors $u$ satisfying $u\wedge B = 0$.