I'm looking for someone to check I'm doing this right. Please point out any mistakes.
I have two elements in $TSO_{3}$, $X = (g,\mathbf{v})$ and $Y = (h,\mathbf{w})$. We can write out the two tangent space elements in terms of the 'usual' Lie algebra basis $T_{1},T_{2},T_{3}$ as $\mathbf{v} = \sum_{j}v^{j}T_{j}g$ and $\mathbf{w} = \sum_{i}w^{i}T_{i}h$ since $T_{g}SO_{3}$ has basis $S_{j}=T_{i}g$ (I'm multiplying in group elements from the right in my convention). This will be relevant shortly.
Let $\exp,\log$ be the group-theoretic functions defined at the identity element. These generalise to arbitrary base points via (in my convention) :
$$\exp_{g}(\mathbf{v}) = \exp(\mathbf{v}g^{-1})g \qquad \log_{g}(h) = \log(hg^{-1})g$$
This matches the standard geodesic notions in nice Lie groups. Now I want to lift this to the tangent bundle and construct the $\textrm{Exp},\textrm{Log}$ bundle-theoretic versions.
To do this I consider the following proposed curve :
$$\gamma(\tau) = \textrm{Exp}_{Y}\big(\tau\,\textrm{Log}_{Y}(X) \big) = \textrm{Exp}_{(h,\mathbf{w})}\big( \tau \,\textrm{Log}_{(h,\mathbf{w})}(g,\mathbf{v}) \big)$$
Guessing a bit about the nature of the actions on each part :
$$\gamma(\tau) = \Big( \exp_{h}\big( \tau\,\log_{h}g \big) \, , \, \big[ (1-\tau)\,w^{i} + \tau\,v^{i} \big] T_{i} \exp_{h}\big( \tau\, \log_{h}g \big) \Big)$$
Note the tangent space basis, I've pulled out the coefficients and attached them to the appropriate $T_{\exp_{h}\big( \tau\,\log_{h}g \big)}SO_{3}$ basis, based on the definition of the basis given at the start.
We have the correct end points,
$$\gamma(0) = Y \quad,\quad \gamma(1) = X$$
The group part transforms as usual, the tangent part I'm slightly guessing at given the structure of the bundle :
- We need the tangent at $\exp_{h}\big( \tau\,\log_{h}g \big)$ to be $S_{i}(\tau) = T_{i}\exp_{h}\big( \tau\,\log_{h}g \big)$, hence that factor
- The coefficients of the basis vectors transform separately, like 'geodesics' in a nice vector space, ie linear interpolation along a straight line - hence $(1-\tau)\,w^{i} + \tau\,v^{i}$.
Unless there's some non-trivial knitting together between the group and tangent parts it seems like this is a geodesic and a valid definition for the exp map. Seeing $g=h$ or $\mathbf{v}=\mathbf{w}$ gives sensible expressions.
I haven't explicitly stated a metric for the bundle but it'll follow easily by pulling my definition for the Log map out of the above curve expression :
$$\textrm{Log}_{Y}(X) = \textrm{Log}_{(h,\mathbf{w})}\big( (g,\mathbf{v}) \big) = \Big( \log_{h}g \,,\, \big[v^{i}-w^{i}]T_{i}g \Big) = \Big( \eta^{j}T_{j}g \,,\, \big[v^{i}-w^{i}]T_{i}g \Big)$$
This is an element in $T_{Y}(TSO_{3})$ and I can easily define an inner product by viewing the coefficients as being in Euclidean $\mathbb{R}^{6}$. The path (and metric therefore) is such that it gives equal weight to disagreements in the group and tangent spaces, ie theres no warp factor which scales the tangent space metric by some function of the difference in the group elements but, aside from arbitrary preferences, nothing jumps out at me for this not being a canonical?
Is this a sensible definition of the exp (or 'Exp' in this case) map in $TSO_{3}$?