If bivector exponentials are rotations, what are translation? It is well known that a rotation is given by the bivector exponentials, such as:
$$
e^{\frac{ \mathbf{B}}{2}} \mathbf{x} e^{-\frac{ \mathbf{B}}{2}}
$$
where $\mathbf{B}$ is a bivector and $\mathbf{x}$ is a vector.
I am trying to find the geometric effect of:
$$
e^{\frac{\mathbf{v}}{2}} \mathbf{x} e^{-\frac{\mathbf{v}}{2}}
$$
where $\mathbf{v}$ is a vector. Is it a translation? If not, what is it?
 A: Define $\mathbf{j}=\dfrac{\mathbf{v}}{\left|\mathbf{v}\right|}$ and $\theta=\left|\mathbf{v}\right|$. We may want to write $\mathbf{x}=\mathbf{x}_{\parallel}+\mathbf{x}_{\perp}$ where $\mathbf{x}_{\parallel}=\left(\mathbf{x}\cdot\mathbf{j}\right)\mathbf{j}$. Note that $\mathbf{jxj}=\mathbf{jxj}^{-1}$ is the reflection of $\mathbf{x}$ in $\mathbf{j}$: $\mathbf{x}_{\parallel}-\mathbf{x}_{\perp}$, or $2\mathbf{x}_{\parallel}-\mathbf{x}$. In analogy with rotations where the convention is to have the minus sign in the first factor, consider:
\begin{align}&\phantom{=}e^{-\mathbf{v}/2}\mathbf{x}e^{\mathbf{v}/2}\\&=e^{-\mathbf{j}\theta/2}\mathbf{x}e^{\mathbf{j}\theta/2}\\
&=\left(\cosh\dfrac{\theta}{2}-\mathbf{j}\sinh\dfrac{\theta}{2}\right)\mathbf{x}\left(\cosh\dfrac{\theta}{2}+\mathbf{j}\sinh\dfrac{\theta}{2}\right)\\
&=\mathbf{x}\cosh^{2}\dfrac{\theta}{2}-\mathbf{jxj}\sinh^{2}\dfrac{\theta}{2}+\left(\mathbf{xj}-\mathbf{jx}\right)\sinh\dfrac{\theta}{2}\cosh\dfrac{\theta}{2}\\
&=\mathbf{x}\cosh^{2}\dfrac{\theta}{2}-\mathbf{jxj}\sinh^{2}\dfrac{\theta}{2}+\left(\mathbf{xj}-\mathbf{jx}\right)\sinh\dfrac{\theta}{2}\cosh\dfrac{\theta}{2}\\
&=\mathbf{x}\cosh^{2}\dfrac{\theta}{2}-\mathbf{jxj}\sinh^{2}\dfrac{\theta}{2}+\left(\mathbf{x}\wedge\mathbf{j}\right)2\sinh\dfrac{\theta}{2}\cosh\dfrac{\theta}{2}\\
&=\mathbf{x}\cosh^{2}\dfrac{\theta}{2}-\left(\mathbf{x}_{\parallel}-\mathbf{x}\right)\sinh^{2}\dfrac{\theta}{2}+\left(\mathbf{x}\wedge\mathbf{j}\right)2\sinh\dfrac{\theta}{2}\cosh\dfrac{\theta}{2}\\
&=\mathbf{x}\left(\cosh^{2}\dfrac{\theta}{2}+\sinh^{2}\dfrac{\theta}{2}\right)-2\mathbf{x}_{\parallel}\sinh^{2}\dfrac{\theta}{2}+\left(\mathbf{x}\wedge\mathbf{j}\right)2\sinh\dfrac{\theta}{2}\cosh\dfrac{\theta}{2}\\
&=\mathbf{x}\cosh\theta-2\mathbf{x}_{\parallel}\sinh^{2}\dfrac{\theta}{2}+\left(\mathbf{x}\wedge\mathbf{j}\right)\sinh\theta\\
&=\mathbf{x}\cosh\theta-\mathbf{x}_{\parallel}\left(\cosh\theta-1\right)+\left(\mathbf{x}\wedge\mathbf{j}\right)\sinh\theta\\
&=\mathbf{x}\cosh\theta+\mathbf{x}_{\parallel}\left(1-\cosh\theta\right)+\left(\mathbf{x}\wedge\mathbf{j}\right)\sinh\theta\\
&=\left(\mathbf{x}-\mathbf{x}_{\parallel}\right)\cosh\theta+\mathbf{x}_{\parallel}+\left(\mathbf{x}\wedge\mathbf{j}\right)\sinh\theta\\
&=\boxed{\mathbf{x}_{\perp}\cosh\theta+\left(\mathbf{x}_{\perp}\wedge\mathbf{j}\right)\sinh\theta+\mathbf{x}_{\parallel}}\end{align}
So it has aspects reminiscent of a Lorentz boost, but there is a bivector there and the parallel component is unaffected.

We can also compare this formula to the rotation formulas. To rotate $\mathbf x$ in a plane $\mathbf I$ by an angle $\theta$, we use $e^{-\mathbf{I}\theta/2}\mathbf{x}e^{\mathbf{I}\theta/2}=\mathbf{x}_{\parallel}e^{\mathbf{I}\theta}+\mathbf{x}_{\perp}$ (here $\mathbf{x}_{\parallel}$ means the part that's in the plane $\mathbf I$). And the above calculation shows that we have an analogous $e^{-\mathbf{j}\theta/2}\mathbf{x}e^{\mathbf{j}\theta/2}=\mathbf{x}_{\perp}e^{\mathbf{j}\theta}+\mathbf{x}_{\parallel}$. If $\mathbf j=\mathbf J^*$, we might also write $\mathbf x_{\perp}\wedge\mathbf J^*$ as $\left(\mathbf x_\perp\cdot\mathbf J\right)^*$ (using the left contraction).
But no matter what we do, since the result is a sum of a bivector part and a vector part, it can't really have as tidy a visualization as a vector or a bivector, etc.
