I'm trying to imagine the geometry "operations" here:
Angular deficit
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?

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    $\begingroup$ Quick remark: being completely flat everywhere is exactly equivalent to having no angular deficit for any loop. $\endgroup$ – David Z Apr 26 '11 at 23:39
  • $\begingroup$ @David But Riemann=0 is a local flatness test, and angular deficit is a global test. Like in the linked question, a path that is everywhere flat can tell something is up with how a vector rotates. So I don't immediately see why I can't have both. I guess that is a separate question though. $\endgroup$ – Ginsberg Apr 27 '11 at 0:10
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    $\begingroup$ @David: that's true for differentiable manifolds, but I think this question is asking about a $C^{1}$ discontinuous space. You could still have deficit angles in such a space if you take it as the limit of some smooth curved manifold with the curvature nonzero over a compact support of its domain, and then you let the volume of that compact support go to zero while keeping $\int d^{n}x \,R$ constant $\endgroup$ – Jerry Schirmer Apr 27 '11 at 1:14
  • $\begingroup$ @Jerry: OK, thanks for the correction. I've seen the equivalence of flatness and zero angle deficit stated in several (relatively nontechnical) sources, but I didn't realize there would be exceptions. Guess I should have known though, there are always exceptions ;-) $\endgroup$ – David Z Apr 27 '11 at 1:24
  • $\begingroup$ @David: You don't even need Jerry's construction. Your statement is only true if every loop is homotopic to zero. So non trivial topologies will give you easy candidates. $\endgroup$ – Willie Wong Apr 27 '11 at 1:41

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.

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    $\begingroup$ I don't quite follow. What prevents me from cutting out a slice from a pizza and then putting it back in place? Ok, for a pizza you can't reverse the process - but for a smooth manifold (such as a disk) you can. I just did it in my head five time over ;-) I don't know of any restriction where a smooth manifold can be cut up but not sown back together to its original shape (i.e. without creating or removing any singularities) as long as it is sown back exactly along the lines where the cuts were made. $\endgroup$ – user346 Apr 27 '11 at 8:50
  • $\begingroup$ @Deepak: My interpretation of the query is not that the you remove a piece and put it back, I interpret it as a generalization of "making a cone": you remove the "pizza slice", eat it, and try to force the remainder together to look like a whole pizza. $\endgroup$ – Willie Wong Apr 27 '11 at 16:13
  • $\begingroup$ @Deepak: ah, I see, you were referring to the last sentence I wrote in my answer. Sorry I wasn't being clear: I meant "sew it back up" in the sense of surgical removal of something, you sew up whatever is left. I changed the wording now and hope it is clearer. $\endgroup$ – Willie Wong Apr 27 '11 at 16:16

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?)

  • $\begingroup$ Well, at least in principle this can get me to an answer. Can someone with access to Mathematica plug this in and tell me what the requirement on f(r) is to obtain no curvature? $\endgroup$ – Ginsberg Apr 28 '11 at 20:02

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