Does this orbifold embed into $\mathbb{R}^3$?

Question

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 $\mathbb{R}^3$?

Answer 1

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.

Answer 2

If you want the image to be convex, then no. Otherwise, my guess is that Kuiper's theorem gives you an embedding...