How does the singular surfaces obtained when the border of a Euclidean set becomes a point look like?

I'm curious to understand in several manner, what is the metric geometry of the metric space homeomorphic to a sphere, obtained from a compact convex set $K\subset R^2$ with the Euclidean distance, when one makes the following quotient that the border $\partial K$ is one point, say $p$.

The space is certainly geodesic and even locally Euclidean except in a neighborhood of $p$. The space of directions around this point seems to have an infinite angle.

Can the space be represented is some nice manner as a (non-convex) surface of $R^3$? What happens if $K$ is a disk? Or $K$ is a square?

• Since the space is homeomorphic to a sphere, it can be represented as a nice surface in $\mathbb R^3$ (namely, the sphere). What kind of metric (or other structure) do you want to put on the quotient space and what do you want from its embedding in $\mathbb R^3$? – Joonas Ilmavirta Jul 11 '15 at 13:05
• Assuming you're talking about the length metric, then since the interior of $K$ is a flat Riemannian manifold, it can only isometrically embed in the obvious way, ie, $K \to \mathbb{R}^2 \to \mathbb{R}^3$. Then obviously $\partial K$ cannot be mapped to a single point. – John Harvey Jul 12 '15 at 12:48
• Yes, I'm talking about the length metric and concerning surfaces of $R^3$ the lengh metric for rectifiable curves on the surface. For instance cylinder with basis an injective curve of finite length is isometric to $R^2$. A tetraedon a locally a Euclidean geometry everywhere except in 4 points – Nicolas Juillet Jul 12 '15 at 14:02
• Sorry, my comment was patently ridiculous. Obviously not every flat manifold is contained in a plane. But is it the case that the embedding will always give you a piecewise ruled surface? I think that should be the case for a $C^2$ embedding at least, and it should not be too hard to go from there to see that the boundary cannot be mapped to a single point. – John Harvey Jul 12 '15 at 19:32
• Do you know if it is even possible to embed a neighborhood of $p$? It would have to look pretty crazy. – John Harvey Jul 13 '15 at 8:31