I think this information should be well known, and in a textbook somewhere but I have been unable to find it so I apologise if this is standard material. I am a physicist and trying to do some calculations which require integrating some function of the geodesic distance over (subsets of) the projective unitary group. For this of course I need to understand what the geodesic distance is.
For $U \in\mathrm{U(n)}$, let $[U] = \{e^{i\phi} U | \phi\in\mathbb{R}\}$ be the corresponding equivalence class (i.e. $[U] \in\mathrm{PU(n)}$). By standard considerations (translating one of the points to the identity and diagonalizing) it suffices to answer the following question. Let $I$ be the $n\times n$ identity matrix, $\phi_j$ be a bunch of real numbers and \begin{align} U = \begin{pmatrix}1 \\ & e^{i\phi_1} \\ & & \ddots \\ & & & e^{\phi_{n-1}}\end{pmatrix}, \end{align} what is the geodesic distance $d([I], [U])$ in the projective unitary group? My guess would be something like \begin{align} d([I], [U]) = \inf_{V\in [U]} \delta(I, V) = \inf_{a\in\mathbb{R}} \delta(e^{ia} I, U) , \end{align} where $\delta$ is the geodesic distance in $\mathrm{U}(n)$, but I don't have any justification for this other than intuition. Maybe it is simpler to give a description in terms of $\mathrm{SU}(n)$ rather than $\mathrm{U}(n)$ because then we're only quotienting by a finite subgroup.
Any help would be gratefully appreciated!