- #1

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That is, for every finite metric space (X,d), does there exist some manifold such that there are |X| many points on it which is isometric (with the length of geodesics as metric) to (X,d)?

- Thread starter Dragonfall
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- #1

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- 4

That is, for every finite metric space (X,d), does there exist some manifold such that there are |X| many points on it which is isometric (with the length of geodesics as metric) to (X,d)?

- #2

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Can you do a set with 4 points?

- #3

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What do you mean?

- #4

Hurkyl

Staff Emeritus

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(P.S. I assume you mean "connected and Riemannian" manifolds?)

- #5

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o

\

o ---- o

/

o

where the metric is the number of steps between nodes, can't be done. But I can't prove it conclusively.

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