Question: Assume $M$ is a connected, smooth, and geodesically complete Riemannian manifold. (Hence $M$ is a complete metric space by the Hopf-Rinow theorem.)
Then (assuming that $M$ satisfies the SAS postulate if and only if it satisfies the five segment axiom), does $M$ satisfy the SAS postulate if and only if $M$ is an isotropic Riemannian manifold?
If not, does being isotropic imply the SAS postulate? Or vice versa?
Notes: Cf. the definitions below.
A pointer to a reference would suffice. The only reference I know of for isotropic manifolds is chapters 8 and 12 of Wolf's Spaces of Constant Curvature, but it doesn't seem to mention the five-segment axiom or the SAS postulate.
My understanding is that the five-segment axiom and the SAS postulate are equivalent when given the remaining axioms of Tarski's axiom system for Euclidean geometry. I don't actually know whether they are equivalent for arbitrary Riemannian manifolds. The five-segment axiom has the advantage of being expressed only in terms of the metric, i.e. not requiring angle measure to be defined from the metric first. (That is possible even in general Riemannian manifolds, cf. this MO answer.)
Definition: A Riemannian manifold $M$ is called isotropic if for any point $p \in M$ and any unit tangent vectors $v,w \in T_p M$ there exists an isometry $\varphi: M \to M$ such that $\varphi(p) = p$ and $\varphi_{\ast}(v)=w$.
(Note: this definition might be slightly off or vary between sources, cf. this related question.)
Definition: A Riemannian manifold $M$ is said to satisfy the SAS postulate if for any points $p, q, r, p', q', r' \in M$ we have that $$d(p,q) = d(p',q'), \quad d(q,r)=d(q',r'), \quad \angle (p,q,r) = \angle(p',q',r') ,$$ always implies that $$d(p,r)=d(p',r'), \quad \angle(q,r,p)=\angle(q',r',p'), \quad \angle(r,p,q)=\angle(r',p',q') .$$
Definition: A metric space $M$ is said to satisfy the five-segment axiom if for any points $p,q,r,s,p',q',r',s'$ with $p \not= q$ we have that $$d(p,q) + d(q,r) = d(p,r), \quad d(p',q') + d(q',r') = d(p',r'), $$ $$ d(p,q)=d(p',q'), \quad d(q,r) = d(q',r'), \quad d(p,s)=d(p',s'), \quad d(q,s)=d(q',s'), $$ always implies that $d(r,s) = d(r',s')$.
Related questions:
Do general Riemannian manifolds satisfy the SAS (side-angle-side) postulate?
Isotropic Manifolds
What is the difference between an homogeneous and an isotropic space
How to derive the five-segment axiom of Tarski's geometry from Hilbert's axioms?
Tarski-like axiomatization of spherical or elliptic geometry
Is there a way to prove that absolute geometry must take place on a Riemannian manifold?
Riemannian Triangles Third Side
Bounding angles in Riemannian triangles with bounded sides
MO: characterization of Riemannian metrics
Motivation: The accepted answer to a previous question pointed out that the primary obstruction to general Riemannian manifolds satisfying the SAS postulate is metric in nature, not topological. It also points out how "[the SAS postulate] is saying in some sense that your space looks the same at every point and from every direction". My understanding is that in physics such a hypothesis for the universe is called "isotropy", and it seems to match the precise mathematical definition of "isotropic manifold". In particular, the accepted answer points out that the SAS postulate can be thought of as a homogeneity statement, and apparently it is a straightforward theorem that every isotropic manifold is additionally a homogeneous manifold.
Also according to the Wikipedia article, examples of manifolds that are homogeneous but not isotropic include the flat torus. So apparently there can be topological obstructions to isotropy as well. In particular, while all simply connected spaces of constant curvature are isotropic Riemannian manifolds, not all spaces of constant curvature are isotropic.
Apparently, besides the simply-connected spaces with constant curvature, there are other isotropic Riemannian manifolds, such as complex projective spaces with the Fubini-study metric. I know that (cf. the "background" section of the previous question) the former probably satisfies the SAS postulate and five-segment axioms, i.e. the intersection of Riemannian manifolds that are isotropic, and that satisfy the SAS postulate / five-segment axiom, contains the simply connected spaces of constant curvature. So I am curious about how large that intersection is (e.g. does it also include the complex projective spaces with the Fubini-Study metric), and which spaces (if any) satisfy one property but not the other.
Because betweenness and equidistance can both be expressed in terms of a metric function, Tarski's axioms for Euclidean geometry actually gives a fairly concise intrinsic characterization of Euclidean (and possibly even general Hilbert) spaces in terms of their metric. Based on Nikolaev's result, it is known that something similar is possible for general Riemannian manifolds, albeit in a much more complicated manner.
Because Riemannian geometry can be considered "infinitesimally Euclidean" geometry, I am curious which of Tarski's axioms are valid as "global" properties only for Euclidean spaces (and thus would need to be "infinitesimalized" first to be used to characterize connected, smooth, geodesically complete Riemannian manifolds). E.g. those of Tarski's axioms that are basically the definition of a semimetric space obviously apply "globally" to arbitrary Riemannian manifolds. But the "segment builder" axiom implies the space has infinite metric diameter, and thus e.g. fails "globally" for the sphere and thus would need to be "infinitesimalized". So basically I am curious about which class of Riemannian manifolds satisfies the property that the SAS postulate / five-segment axiom is still a "global" property (such as for the simply-connected spaces of constant curvature).
