Given a (compact, closed) Lie group $G$ and a (closed) subgroup $H$, what is the distance of the identity to $Hg$ (or $gH$), where $g\in G$ and $Hg$ denotes the orbit under left-multiplication? The distance is to be measured in a natural left-invariant metric. Can one calculate it? Can one give an explicit formula for the geodesic joining the identity and the minimal distance point?
This seems like a natural question to ask, however, I cannot find anything about it (but then, I'm not to familiar with the geometry of Lie groups); or is this question so easy or difficult that there is no literature about it?
EDIT: Let's do an example. Take the unitary matrices $U(n)$ as a Lie group and the matrices
$$ \begin{pmatrix}{} U(r) & 0 \\ 0 & 1 \end{pmatrix} $$
as a subgroup $H$. I have no good intuition on how to calculate the distance between a point and that subgroup. I feel the path should be something like a "lift" of a geodesics in $U(n)/H$, but I can't put it together. I've also never seen a reference where something like this is done.
These are just some ideas, not an answer at all.
This is not so easy. First of all, that geodesic is not unique. (Even if $H = \{e\}$ is the trivial group). Neglecting this problem, let $e^{tX}$ be a minimal geodesic in $G$ joining $e$ to $e^X \in gH$. Then by minimality we see that $X$ is perpendicular to $h$, the Lie algebra of $H$. So $e^X = ge^{-Y}$ for some $Y \in h$. That is, given a $g$ you want to find $X\in h^\perp$ and $Y \in h$ so that
$$g = e^X e^Y $$
and in this case $d(e, gH) = ||X||$. But how to find this $X$ and $Y$? Also write $g = e^{X' + Y'}$, where $X' \in h^\perp$ and $Y' \in h$. So
$$e^{X' + Y'} =e^X e^Y$$
This seems to be a hard problem, unless ($G$ is abelian and in this case $X' = X$, $Y'=Y$).