Sewing together flat spacetime pieces = flat spacetime?

Question

I'm trying to imagine the geometry "operations" here:
Angular deficit
and
Curvature of Conical spacetime

If we sew flat spacetime pieces together, what is the requirement for the sewing to not create curvature at that seam?

Clearly there has to be some condition, because in the examples above, sewing together the wedge "sides" doesn't produce cuvature, but somehow at the corner of the removed wedge, curvature appears.

So, more concretely, if we remove this chunk of flat spacetime: $$(x>0) \mathrm{\ and \ } (0<y<x)$$ and now sew it together, somehow this creates flat spacetime everywhere except at $x=y=0$. Is there some intuitive physical meaning to this?

If the issue is the sharp angle there, then what if we remove instead a nice smooth parabola: $$(x>0) \mathrm{\ and \ } (-\sqrt{x}<y<\sqrt{x})$$ or maybe a wedge that is completely smooth to zero angle $$[x>0] \mathrm{\ and \ } [0<y<\exp(-1/x)]$$

Are these completely flat everywhere (no string defect) AND have angular deficit?

Answer 1

The mathematics actually tells you that your second and third constructions are physically impossible!

I shall assume here the "lack of string defects" is equivalent to your space-time being globally a $C^2$-Riemannian manifold. As David mentioned in the comments, for any null-homotopic loop in such a manifold (null-homotopic just means that the loop can be continuously shrunken to a point) there can be no angular defect. (In two dimensions this is a Corollary of the Gauss-Bonnet theorem.)

Now, the objects that you want to construct by removing a wedge and sewing up will obviously be simply connected, so every loop is null homotopic. So if you were actually able to do the sewing without creating a lack of smoothness (a string defect), the above will imply that there cannot be an angular defect.

However, by construction, if this object were to exist, it is easy to argue, as in the cone case, that there must be an angular defect.

So by eliminating all the impossibilities, the only option left is that whatever the shape you cut out of the plane, you can not "sew up the new edges" in a way without introducing a singularity.

Answer 2

what is the requirement for the sewing to not create curvature at that seam?

In Ted's answer to my question, he wrote:
"In general, you only have curvature if you have to "crumple" or "stretch" the paper. When you form a cone, you don't have to do that, anywhere except at the vertex."

I don't understand why the vertex counts as crumpling or stretching, but the comment makes intuitive sense. So one condition is that you need to sew the pieces together without stretching. If sewing is mapping a point on one edge to a point on the other edge, then once one pair of points are equated there will be curvature if the other points aren't equated with equal path length along the sewing seam.

I'm not sure how to go further in general, so I'll simplify the problem some. Let's cut the following out of flat space (in polar coordinates): $0<r<\infty, -f(r)<\theta<f(r)$

Sewing it up we can change coordinates
$k \phi=\theta, k=\frac{2\pi - 2f(r)}{2\pi} = 1- f(r)/\pi$
giving the line element in these coordinates as
$$d\theta = \phi \frac{\partial k}{\partial r} dr + k\ d\phi = \phi F(r) dr + k\ d\phi$$ $$ds^2 = dr^2 + r^2 (\phi^2 F^2\ dr^2 + 2\phi F k\ dr\ d\phi + k^2\ d\phi^2)$$ and the coordinates go from $0 \le r < \infty$ and $0 \le \phi \le 2\pi$

Oh god this is going to be messy. But you can in principle get the Riemann curvature and a condition on f(r) for the curvature to be zero. (Does anyone know how to make Mathematica or something calculate the answer here?)