This is community wiki; this also does not answer the question. I just want to post some quotes from Burago, Burago, Ivanov which seem possibly relevant here. If upon further reflection I can think of more to say about them, or realize they are irrelevant to the hand, I will modify or delete this CW "answer" as appropriate. Cluttering up the question further seemed inadvisable.
pp. 140-143
(discussing/referring to arbitrary Finsler manifolds)
As usual, the length structure $L$ gives rise to an intrinsic metric $d$. It is easy to show that the length of a curve induced by $d$ coincides with $L$ for all smooth curves (Exercise 5.1.1). However, whereas the existence of a shortest path for $d$ connecting any two points follows from Theorem 2.5.23, its smoothness is less obvious. Moreover, a shortest path for $d$ can fail to be smooth unless a certain additional assumption of strict convexity is imposed on $\lambda_p$.
Such a pathology (nonsmoothness and nonuniqueness of shortest paths, even in an arbitrary small neighborhood of a point) is due to the fact that balls of the norm (which are "diamonds") are not strictly convex [emphasis mine]. An interested reader can prove that for strictly convex norms all shortest paths are smooth, and a shortest path connecting two sufficiently close points is unique.
Hint: Use an analog of Lemma 5.1.13 with an appropriate two-dimensional normed space instead of Euclidean space.
...
Remark 5.1.5. Isometric Riemannian regions are obviously isometric as length spaces. The converse is also true: if Riemannian regions $\Omega$ and $\Omega'$ are isometric as length spaces, then they are isometric in the sense of Definition 5.1.4 (that is, there exists a smooth isometry that respects their Riemannian structures). Moreover, every isometry map from $\Omega$ to $\Omega'$ is smooth, and (as a consequence) can be taken as $\varphi$ in Definition 5.1.4.
We leave this fact as an (not so obvious!) exercise. The easiest proof we know is based on the results of Section 5.2 (namely, smoothness of shortest paths and properties of exponential maps).
Smoothness of isometries allows us to give a metric definition of a Riemannian manifold (for we were so far looking only at regions with Riemannian metrics):
Definition 5.1.6. A Riemannian manifold is a length space such that every point has a neighborhood isometric to a region with Riemannian metric.
Remark 5.1.7. This definition is not standard. In most textbooks Riemannian manifolds are defined as smooth manifolds equipped with Riemannian structures. The definitions are equivalent: this easily follows from Remark 5.1.5 (in particular, a length space that is locally isometric to a Riemannian region naturally carries a structure of a smooth manifold). If you are familiar with smooth manifolds, we recommend you prove this equivalence as an exercise.
What I take away from all of this is (at least) the following:
The fact that Riemannian manifolds are length spaces is crucial to Palais's construction.
The fact that the metric on Riemannian manifolds is not only intrinsic but even strictly intrinsic is also crucial to Palais's construction. (Also called a geodesic space? (cf.) BBI uses different terminology.) Anyway, this also implies (I think) that Riemannian manifolds are convex metric spaces (not to be confused with geodesic convexity).
The fact that the metric balls of Riemannian manifolds are strictly convex, while those of arbitrary Finslerian manifolds are not, may also be crucial.
The fact that all isometries (and thus presumably also all local similitudes) of Riemannian manifolds are smooth is crucial.
The fact that all geodesics (or at least of shortest paths) are smooth is crucial. Namely, we can replace as representatives of (many) germs in the standard construction of the tangent space with geodesics (of possibly varying speeds). (Although the fact that every germ has as a representative some geodesic of some speed is surprising to me, but anyway if this is true I believe that it is almost certainly related to the fact that geodesics are smooth.)
Riemannian manifolds have this nice length functional $L(\gamma, a , b ) = \int_a^b \lambda_{\gamma(t)}(\gamma'(t))dt$, where $\lambda_p(v) = \sqrt{Q_p(v,v)}$ for some positive definite (non-degenerate?) quadratic form $Q_p$ -- although this is ridiculously circular, because this already assumes the existence of a smooth manifold structure, i.e. the existence of tangent spaces at every point and thus tangent vectors $v$ to populate those tangent spaces and thus the ability to define length functionals which take tangent vectors as arguments.
So for me, the primary concerns/confusions are the following:
- These concerns help to show how Palais's construction coincides with the standard construction for smooth manifolds. They do not show how Palais's construction will be degenerate for metric spaces which are not Riemannian manifolds.
- What about Finslerian manifolds where the norms have balls which are strictly convex, but which aren't induced by an inner product. Is there a theorem which says that the balls of a normed space are strictly convex if and only if the norm is induced by an inner product? (I know that the norm is isotropic if and only if it is induced by an inner product, but I don't see how that is related to strict convexity of its unit balls.)
- I vaguely recall there being a relationship between sub-additivity and lower semi-continuity of the length functional (this being mentioned in Burago, Burago, Ivanov), as well as a relationship with convexity of the unit balls and all of this being related to (semi- or quasi- or asymmetric) norms somehow (mentioned elsewhere). The answers might be found in pp. 200-208 of Volume 2 of Spivak (the Addendum about Finsler geometry). Maybe there is a relationship with being CAT(0)?
(Wikipedia 1 2)
Furthermore, [a convex set] $C$ is strictly convex [emphasis mine] if every point on the line segment [geodesic?] connecting $x$ and $y$ other than the endpoints is inside the interior of $C$.
The definition of a convex set and a convex hull extends naturally to geometries which are not Euclidean by defining a geodesically convex set [link added by me] to be one that contains the geodesics joining any two points in the set.
