TITLE: Does this orbifold embed into $\mathbb{R}^3$?
QUESTION [5 upvotes]: Let $X$ be the space obtained by gluing together two congruent equilateral triangles along corresponding edges.
Note that $X$ has the structure of a Riemannian manifold except at the three cone points.  In particular, $X$ is a Riemannian orbifold.
Is there an isometric embedding of $X$ into $\mathbb{R}^3$?

REPLY [2 votes]: The common boundary of the two triangles is a curve in the plane whose interior is isometric to each of the 2-simplices making up your orbifold $X$.  By the solution of Is an isometric embedding of a disk determined by the boundary?, the imbedding of the boundary extends in a unique way to an isometric imbedding of its interior.  Hence the images of both 2-simplices must be identical.  In particular, any isometric map $X\to\mathbb{R}^3$ cannot be an imbedding.
{"set_name": "stack_exchange", "score": 5, "question_id": 631800}