I know that rotors are the exponentials of bivectors. If B is simple, then $e^B = cos|B| + \hat{B}sin|B| $. All Bivectors in 3D are simple, but the same is not true in 4D. Is there a matching formula for the exponent of a general 4D bivector?
I wikipedia they write that if B1 and B2 are orthogonal then: $$R=e^{{{\frac {{\mathbf {B}}_{1}+{\mathbf {B}}_{2}}{2}}}}=e^{{{\frac {{\mathbf {B}}_{1}}{2}}}}e^{{{\frac {{\mathbf {B}}_{2}}{2}}}}=e^{{{\frac {{\mathbf {B}}_{2}}{2}}}}e^{{{\frac {{\mathbf {B}}_{1}}{2}}}}$$
And that it is possible to separate any Bivector to a sum of two simple orthogonal bivectors but I didn't understand how to do that. It also seems to me there should a simpler way of doing that.
Nicholas refered me to https://arxiv.org/abs/2107.03771. In section 6 it is explained how to to split any bivector into orthogonal blades. Let $B$ be a bivector in a geometric algebra with dimension 4, we aim to find two orthogonal bovectors $b_1, b_2$ such that $B = b_1 + b_2$.
It is shown there that:
$$b_1,_2=\frac {\lambda_i + \frac{1}{2} B \wedge B} {B}$$
where
$$\lambda_1,_2 = \frac{1}{2}B \cdot B \pm \frac{1}{2}\sqrt{(B \cdot B)^2 - (B \wedge B)^2}$$
I've used Wolfram Alpha to make sure that $\Delta = (B \cdot B)^2 - (B \wedge B)^2$ is never < 0 in a 4D euclidian space.
It is said in that paper that the only case the above is not defined, is when $\Delta=0$, because $B$ is not invertible in that case.
However, in this case $B^3 = 2(B \cdot B)B$ and the exponent can be calculated like of a simple bivector with $e^B = cos|B| + \hat{B}sin|B|$.
Once we have $b_1$ and $b_2$ we can calculate the exponent like this (wikipedia): $$ e^{b_1 + b_2} = e^{b_1} e^{b_2}$$