Exponential of a 3D multivector in Clifford's Geometric Algebra

Question

An arbitrary 3D multivector can be written as the superposition: $M=Z+F$, where $Z=a+bi$ with $a,b\in\mathbb R$ and $i$ been the pseudo scalalar of $(Cl_3)$. The remaining term $F=v+iw$, where $v$ and $w$ are vectors of $Cl_3$, have vector and bivector parts, respectively. Since $Z\in$ Cen $(Cl_3)$, the exponential of $M$ becomes $$\exp(M)=\exp(Z+F)=\exp(Z)\exp(F).$$ To obtain a closed form of $\exp(M)$, I would like to ask if there is a way to show that $$\exp(F)\stackrel{?}{=}\exp(v)\exp(iw)\ldots.$$ Thank you very much in advance.

Answer 1

I am note sure whether you can generally find such an exponential factorization. However, by the Baker-Campbell-Hausdorff theorem, we have $$ e^{A + B + \frac{1}{{2}} \left[{A},{B}\right] + \cdots} = e^A e^B,$$ where $ \left[{A},{B}\right] $, is the commutator, the antisymmetric sum of the objects $ A, B $. Consider that commutator for your case $$\begin{aligned}\left[{v},{iw}\right]&=v i w - i w v \\ &=i \left( { v w - w v} \right) \\ &= 2 i \left( { v \wedge w } \right).\end{aligned}$$ Note that this commutator expansion is specific to $\mathbb{R}^{3}$ since the pseudoscalar $ i $ happens to commute with all grades in that case.

This expansion shows that if $ v, w $ are colinear, one must have $$e^{v + iw} = e^v e^{iw}.$$ If $ v, w $ were not colinear, but the higher order commutators happended to be zero, one could say $$ e^{v + iw + i (v \wedge w)} = e^v e^{iw},$$ but that is not the type of decomposition that you are seeking.

Answer 2

As noted by Peeter Joot, $e^F$ cannot be split into $e^{v}e^{iw}$ since $[v,iw] \neq 0$. So the hunt is on for a decomposition of $F$ into two commuting parts. Fortunatelly, by staring at Cayley tables we can discover that $\mathbb{R}_3$ is isomorphic to $\mathbb{R}^+_{3,1}$, the even subalgebra of the spacetime algebra, and thus that your $F$ can be viewed as a bivector in spacetime. And for bivectors, we can use the invariant decomposition to decompose them into commuting terms.

So adapting the invariant decomposition formula slightly using the isomorphism, we can find an $F_1$ and $F_2$ satisfying $F = F_1 + F_2$, $\lambda_i := F_i^2 \in \mathbb{R}$, and $[F_1, F_2] = 0$:

$$F_i = \frac{\lambda_i + \tfrac{1}{2} \langle F^2 \rangle_3}{F},$$

where the $\lambda_i$ are given by

$$\lambda_i = \tfrac{1}{2} \langle F^2 \rangle_0 \pm \tfrac{1}{2} \sqrt{ \langle F^2 \rangle_0^2 - \langle F^2 \rangle_3^2 }.$$

So we have now found two commuting elements $F_1$ and $F_2$, which both square to scalars so they follow Euler's formula. Specifically, since $\lambda_1 \geq 0$ and $\lambda_2 < 0$, we find

$$ \begin{aligned} e^{F_1} &= \cosh(\sqrt{\lambda_1}) + \frac{F_1}{\sqrt{\lambda_1}} \sinh(\sqrt{\lambda_1}) \\ e^{F_2} &= \cos(\sqrt{-\lambda_2}) + \frac{F_2}{\sqrt{-\lambda_2}} \sin(\sqrt{-\lambda_2}) \end{aligned} $$

To conclude, the exponential of any multivector $M = Z + F$ is $$ e^{M} = e^{a}e^{bi}e^{F_1}e^{F_2}. $$

So the invariant decomposition allows us to bypass the Baker-Campbell-Hausdorff formula. For more details and to see how this generalises to higher dimensions I would highly recommend reading the invariant decomposition paper.