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On the contrary, real intrinsic singularities exist on surfaces. They are true geometric singularities:
(55)
(56)
(57)
And so on...
By the way, a fold is a particular region of a surface where linear curvature is concentrated. In figure (57), on the left, we have negative linear curvature; on the right, positive linear curvature.
In each sub-figure, we have used two portions of a sphere. The overall object has the same topology as the sphere, which means that its total angular curvature is $4\pi$.
Suppose the object on the left was constructed from two portions of a sphere, each containing an angular curvature of $3\pi$:
$$
3\pi + 3\pi = 6\pi
$$
This is too much. Therefore, the (negative) linear curvature must compensate for this in order to obtain the final required value $4\pi$:
In conclusion, our fold contains a negative curvature of:
$$
-2\pi
$$
This curvature is uniformly distributed along the circular curve, along the fold.
Returning to figures (57), we have represented triangles constructed from geodesic lines. However, you can cross a fold without any problem with a (narrow) piece of tape. You know how to calculate, and predict, the sum of the three angles of the triangle. You just need to compare the area of the triangle to the area of the sphere. The excess of curvature is:
$$
\text{(58)}
$$
But you have to take into account the (negative or positive) curvature contained in the portion of the fold, that is, in the arc $mn$. This curvature is:
$$
\text{(59)}
$$
Suppose a sort of lens, on the right of figure (57), is built from two portions of a sphere, each containing an angular curvature of $\pi$. Therefore, if we ignore the fold, this set of two portions of a sphere contains an angular curvature of $2\pi$. However, this lens has a spheroidal topology. The angular curvature contribution must therefore be $2\pi$. Therefore:
$$
2\pi + 2\pi = 4\pi \quad \text{(total curvature of the sphere)}
$$
You can also predict the sum of the angles of this strange triangle, formed by three geodesic lines. The arc $mn$ contains the following linear angular curvature:
$$
\text{(60)}
$$
When measuring the amount of angular curvature contained in the fold, inside the triangle, one can evaluate the deviation from the Euclidean sum, which is $\pi$.
You thus see that you can relatively easily handle these curvature problems on surfaces.
A surface may have conical points or fold lines. These are intrinsic singularities, not these artificial singularities, due to a particular choice of coordinates. Note that we can smooth the fold; we then obtain a shape resembling a peanut:
$$
\text{(61)}
$$
This is similar to smoothing the point-like summit of a cone (concentrated angular curvature), transforming the object into a blunted cone (angular curvature spread over a portion of a sphere).
Suppose the two portions of sphere, above, in figure (61), each correspond to $2/3$ of a sphere, i.e., a curvature:
$$
\text{(62)}
$$
The gray portion of the "peanut" contains negative curvature. Precisely:
$$
\text{(63)}
$$