10/7/2023 0 Comments Sas postulatewith a torus, if the hypothetical missing thirds segment "passes through the donut hole" for $A$ and $C$, but not for $A'$ and $C'$, then I could imagine this failing.īut would that actually be a valid counterexample? And if so, is lack of simple connectedness the only possible obstruction?Ī "global" topological obstruction (such as lack of simple connectedness) being the only possible obstruction seems conceivable to the extent that Riemannian manifolds are "locally" or "infinitesimally" Euclidean. (To be honest those are the only Riemannian manifolds I am interested in anyway.)īut even for "non-pathological" Riemannian manifolds, it still seems like the most possible obstruction to generalizing a result that holds for those with constant curvature is topological. ![]() due to the Riemannian manifold being mildly "pathological", hence my restriction to connected and geodesically complete (= complete metric space by Hopf-Rinow theorem) smooth Riemannian manifolds. There would seem to potentially be topological obstructions in general, i.e. But those are a very special class, so it's not clear to me that it would generalize. So that covers all of the ( $2$-dimensional, but the proofs probably don't depend on the dimension) Riemannian manifolds of constant curvature. And apparently it is also true in spherical geometry using the correct notion of "triangle" or "side", i.e. The SAS postulate is true in absolute geometry (i.e. no tori.)īecause this is probably a basic and well-known result, a pointer to a reference would suffice.īackground/motivation/what I've considered so far: Question 2: Same as the above but assume $M$ is additionally simply connected. Question: Assume $M$ is a connected and geodesically complete smooth Riemannian manifold. ![]() In particular the distances between their endpoints will be the length of the geodesic. By triangle I have in mind something where all sides must be length minimizing geodesics, i.e.
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