Geodesics and paths for non-unit quaternions

Question

Given two unit quaternions $\mathbf{q}_1$ and $\mathbf{q}_2$ $\in \mathbb{H}_{1}$, it is well known that the unique geodesic parametrized by the length, $t \mapsto \gamma(t)$, joining the two quaternions is

$\gamma(t) = \mathbf{q}_{1}\left(\mathbf{q}_{1}^{-1}\mathbf{q}_{2}\right)^{t}$

It corresponds to the geodesic in $\mathbb{S}^3$. The corresponding geodesic distance can be written using the logarithm of quaternions

$d(\mathbf{q}_1 , \mathbf{q}_2 )=||\log\left( \mathbf{q}_{1}^{-1} \mathbf{q}_{2} \right) ||$

Let now us consider that we are dealing with general quaternions in $\mathbb{H}$, of type $\mathbf{q}_i = |\mathbf{q}_i |U\mathbf{q}_i $, where $\mathbf{q}_i$ is the norm and $U\mathbf{q}_i$ the corresponding versor. Ìf we consider now the same expression than above for a parametrized path between $|\mathbf{q}_1 |U\mathbf{q}_1$ and $|\mathbf{q}_2 |U\mathbf{q}_2$, i.e.,

$\gamma(t) = |\mathbf{q}_1 |^{1-t} |\mathbf{q}_2 |^t U\mathbf{q}_{1}\left(U\mathbf{q}_{1}^{*}U\mathbf{q}_{2}\right)^{t}$

Thus, a path defined by the product of a geodesic in $\mathbb{R}^+$ (weighted geometric mean of the norms) and a geodesic in $\mathbb{S}^3$. The length of the path involves also a decoupling between both manifolds, i.e.,

${\cal l}(\mathbf{q}_1,\mathbf{q}_1 )^2 =$ $|\log(|\mathbf{q}_2|) - \log(|\mathbf{q}_1|)|^2 + ||\log\left( U\mathbf{q}_{1}^{*} U\mathbf{q}_{2} \right) ||^2 $

Curved paths $\gamma(t)$ are of course longer than the straight lines in $\mathbb{R}^4$ and therefore this is not the minimal geodesic of $\mathbb{H}$

Questions:

• Is the path $\gamma(t)$ a geodesic of the embedding of $\mathbb{H}$ in $\mathbb{R}^+ \times \mathbb{S}^3$?

• What happens in particular if $|\mathbf{q}_i| \leq 1$ (i.e., quaternions lying inside the sphere $\mathbb{S}^3$)

• Is the corresponding line element given by $ds^2 = (d\log(|\mathbf{q}_{i}|))^2 + (dU\mathbf{q}_{i})^2$ ?

• More generally, what are the strucutre of the manifold $\mathbb{R}^+ \times \mathbb{S}^3$ with this kind of metric?... compact, completeness, bounds of curvature, etc.

Answer 1

The metric you have written down is the standard bi-invariant metric on the Lie group of nonzero quaternions. I.e., it is the metric such that the left-invariant 1-form $\omega = \mathbf{q}^{-1}\ d\mathbf{q}$ with values in $T_{\mathbf{1}}\mathbb{H}\simeq\mathbb{H}$ is an isometry at every point. I.e., for every $\mathbf{p}$ a nonzero quaternion, $\omega_{\mathbf{p}}:T_{\mathbf{p}}\mathbb{H}\to \mathbb{H}$ induces an isometry between $T_{\mathbf{p}}\mathbb{H}$ under this metric with $\mathbb{H}$ given the standard quaternion norm.

This metric (which is complete and homogeneous) is a product metric and the Riemannian manifold is isometric to $\mathbb{R}\times\mathbb{S}^3$. Thus, it is not compact, it is complete, the sectional curvature everywhere is bounded by $1$, etc.