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Let $\{v_1,\cdots,v_k\}$ be the vertices of a regular $(k-1)$-simplex $\Delta(k,\ell)$, with a given metric such that the pairwise distance between the vertices is $\ell$.

Given a Riemannian manifold $M^n$. Is there any criterion such that $\Delta(k,\ell)$ can be isometrically embedded in $M$(i.e. preserve the distance, here we only consider the embedding of the vertexes and 1-skeleton).

For example $\Delta(3, \ell)$ can be isometrically embedded into $S^2$ for $\ell$ not too large. $\Delta(4,\ell)$ can only be embedded into $S^2$ with $\ell=\arccos(-1/3)$. What can we say about $\mathbb{CP}^n$ or other symmetric spaces?

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  • $\begingroup$ I believe if $k \leq n+1$ and $l$ is small enough (depending on the injectivity radius) it should be possible. You can think of build a $k$ simplex by taking successive intersections of $l$ spheres. If $l$ is smaller than $inj/4$, say, the topology of the intersections of balls should behave how you expect in $\mathbb{R}^n$. $\endgroup$ – Tim Carson Sep 7 '17 at 16:21
  • $\begingroup$ Are you asking about an embedding of a simplex or its vertex set? $\endgroup$ – Jarek Kędra Sep 7 '17 at 19:17
  • $\begingroup$ @JarekKędra Just the vertex set. Since the distance is preserved, so all the 1- skeleton is embedded. $\endgroup$ – J. GE Sep 7 '17 at 22:25
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    $\begingroup$ To be clear, the 1 skeleton is not embedded isometrically. In $\mathbb R^2$ the distance from the vertex of an equilateral triangle to the midpoint of the opposite segment is smaller than the sidelength $s$ of the triangle. On a sphere the distance is $s$. If you put the intrinsic metric on the embedded 1 skeleton you do get an isometry. $\endgroup$ – Tim Carson Sep 7 '17 at 22:45
  • $\begingroup$ @TimCarson Yes, you are right. What I actually care about is the complex consists only the vertices and edges, nothing else. I don't know of correct name of this complex other than it is part of the regular simplex. $\endgroup$ – J. GE Sep 8 '17 at 0:01

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