Exponential of a Multivector in Geometric Algebra: $\exp (xe_1 + ye_2 + be_1\wedge e_2)$

Question

I'm working on understanding the exponential function applied to multivectors in the context of Geometric Algebra, specifically for the multivector $xe_1 + ye_2 + be_1\wedge e_2$. I have found expressions for the exponential of a vector and the exponential of a bivector separately:

1. For the vector part $xe_1 + ye_2$, the exponential is given by: $$ \exp (xe_1+ye_1) = \cosh \sqrt{x^2+y^2} + \frac{xe_1+ye_2}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2} $$

2. For the bivector part $be_1\wedge e_2$, the exponential is:

$$ \exp {b e_1\wedge e_2} = \cos b +(e_1\wedge e_2)\sin b $$

Given these expressions, how do we compute the exponential of the combined multivector $xe_1 + ye_2 + be_1\wedge e_2$? I understand that the components are not commutative, which complicates the computation. Could someone guide me through the process or provide insights into how to approach this problem?

Answer 1

$ \newcommand\G{\mathscr G} $I'm assuming that you are interested in a positive-definite signature $e_1^2 = e_2^2 = 1$, but the following can easily be adapted to other signatures.

Without loss of generality we can assume that we have a two-dimensional geometric algebra $\G_{2,0}$ generated by $e_1, e_2$. You can notice that $(xe_1 + ye_2 + be_1e_2)^2$ is always a scalar, $$ (xe_1 + ye_2 + be_1e_2)^2 = x^2 + y^2 - b^2 $$ and from this you can classify the exponentional based on the sign of this scalar just as you would for just a bivector or just a vector. However, it is enlightening to look at the problem from the following perspective:

The algebra $\G_{2,0}$ is isomorphic to the even subalgebra of a larger geometric algebra, for example $\G_{2,0} \cong \G^+_{2,1} \cong \G^+_{1,2}$. (In general $\G_{q,p-1} \cong \G^+_{q,p} \cong \G^+_{p,q} \cong \G_{p,q-1}$.) The latter $\G_{1,2}$ is the (2+1)D spacetime algebra. Let $f_0, f_1, f_2$ be the generators of $\G_{1,2}$ with $f_0^2 = 1$ and $f_1^2 = f_2^2 = -1$. Then we can identify $$ e_1 = f_1f_0,\quad e_2 = f_2f_0,\quad e_1e_2 = -f_1f_2. $$ The span of $e_1, e_2, e_1e_2$ is then simply the set of all bivectors in $\G_{1,2}$, so the exponentials $$ e^{xe_1 + ye_2 + be_1e_2} = e^{xf_1f_0 + yf_2f_0 - bf_1f_2} $$ are just the bivector exponentionals of $\G_{1,2}$. Because this is a 3D algebra, all bivectors are simple, and so all bivectors square to a scalar.