0
$\begingroup$

True or false: "Any edge-weighted undirected graph can be isometrically embedded into some Riemannian manifold".

"isometric embedding" here means that for any pair of nodes, their shortest path distance in the graph equals their geodesic distance in the Riemannian manifold.The question seems equivalent to asking whether any finite discrete metric space can be isometrically embedded into a Riemannian manifold.

$\endgroup$
2
$\begingroup$

This is already false for the graph which consists of a root and three nodes connected to it by edges. The proof follows from uniqueness of continuation of geodesics in Riemannian manifolds.

$\endgroup$
  • $\begingroup$ Thanks for the answer. Can you please elaborate on why such a 4-node star graph cannot be isometrically embedded into a Riemannian manifold? I know that there exist such 4-node star graphs with uniform edge weights that cannot be isometrically embedded into a Euclidean space. I am actually working on a proof that such embedding in general is always possible. $\endgroup$ – user47456 Mar 7 '14 at 21:46
  • $\begingroup$ Such an isometric embedding would extend to an isometric embedding of the entire graph so that the union of each pair of edges maps to a geodesic. But geodesics in Riemannian manifolds cannot branch. $\endgroup$ – Moishe Kohan Mar 7 '14 at 21:53
  • $\begingroup$ Do you mean that it's not possible for two geodesics AB and AC in a Riemannian manifold to share an initial segment and then branch out to two different destinations B and C? $\endgroup$ – user47456 Mar 7 '14 at 22:33
  • $\begingroup$ Exactly! This is a standard result in Riemannian geometry, corollary of uniqueness of solutions of ODEs with smooth coefficients. $\endgroup$ – Moishe Kohan Mar 8 '14 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.