Let $M$ be a compact orientable $n$ dimensional Riemannian manifold.
Is there a triangulation of $M$ such that every $k$ dimensional face of each simplex is a totally geodesic submanifold, $\forall k \;\;1\leq k \leq n$?
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4$\begingroup$ I don't think so. Even totally geodesic subsurfaces are really rare. For example, if you take a compact quotient of NIL or SOL, I don't believe that there is such a triangulation. related : mathoverflow.net/questions/18108/… and mathoverflow.net/questions/54635/… $\endgroup$– HenrikRüpingCommented Feb 17, 2019 at 8:58
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1 Answer
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For surfaces, see
MR1151746 Colin de Verdière, Yves. Comment rendre géodésique une triangulation d'une surface? Enseign. Math. (2) 37 (1991), no. 3-4, 201–212.
answered Feb 17, 2019 at 13:53