# Stitching together piece of flat space

If I start with an infinite flat sheet of graph paper, and in polar coordinates cut out a piece according to: $r>0, \ \ -f(r) < \theta < f(r)$

Now I want to stitch the remaining graph paper together, by associating each point $(r,f(r))$ to $(r,-f(r))$ on the seam.

How do I calculate what the curvature is of the stitched up paper along the seam?

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Are you sure it is possible to stitch the paper together? – Fabian Apr 30 '11 at 7:39
I don't see why not. The top and bottom seam is a mirror image across $\theta=0$, so I could easily stitch these in real life. So I don't see any problem doing it theoretically. Is there some requirement I should know about for stitching seams together? Is it more than assigning points on the seam smoothly to each other? – John Apr 30 '11 at 8:01
Unless I am misunderstanding, one has the following problem: If $f(r)$ is not constant (i.e. if the top and bottom seams are not straight) then in real life you would have to essentially arrange the paper so that near the seams the top and bottom of the paper were parallel (so that you match the two seams). The sewn piece of paper will then have a cusp all the way along the seam (so the curvature will be infinite along the seam). If you try to do it any other way, you will tear the paper while trying to sew up the seams. (But if the seam is a straight line, then you can match the two ... – Matt E Apr 30 '11 at 8:19
So the curvature is either zero or infinite (a delta function like limit)? And the only way to get zero is with a straight line? This doesn't seem intuitive to me. Can you show the math in an answer so I can learn the details? That would be great! – John Apr 30 '11 at 8:32
@John In a sufficiently small neighbourhood of any point not on your seam, the (Gaussian) curvature is zero, since it is isometric to the plane. So all the curvature has to be concentrated in the seam; a set of measure zero. The integral of the curvature in a small disk that intersects the seam cannot be nonzero if the curvature on the seam is represented by finite values. – yasmar Apr 30 '11 at 10:16

Assume that if we did the cut with a straight line, then we'd get a wedge described by the lines $\pm \tilde{\theta}$. Now, we can imagine replacing those straight lines by the squiggly one described by your function. To keep from making a mess, we'd need constraints on the function, corresponding to your squiggly line on the cone not wrapping around and intersecting itself. Say $f'(0) = 0$, and $|f(r)| < \pi r$, but we might have to think about that more. Now we have a wedge cut out by curves described by $\pm \tilde{\theta} + f(r)$. In other words, we're removing the piece according to $$-\tilde{\theta} + f(r) \leq \theta \leq \tilde{\theta} + f(r),$$ which is not the same as you have in your question. If we get it right, and we have a seam with no curvature, then it will be represented by a curve on the cone we get, so it makes sense to go backwards like this.
If you make a cone (i.e., if $f(r)$ is constant in your question), then the discrete curvature at the cone point is the angle defect, i.e. $2\pi -2\tilde{\theta}$, where I've reverted to the notation in my answer. This lets you recover the Gauss-Bonnet theorem for polyhedra, for example. – yasmar Apr 30 '11 at 13:30
If you make a cusp along a seam, as you describe, then things will get more complicated. If the cusp ends, i.e. $f(r) = C$ for $r \geq r_0$, for some constants $C$ and $r_0$, then presumably you can determine the curvature by integrating something around a loop containing the cusp, as Willie's comment to your question above suggests. We could maybe argue that it'll be something simple like $2\pi r_0 - \ell(\gamma)$, where $\gamma$ is the curve defined the circular arc a distance $r_0$ from the cone point. But this probably isn't quite right, because the geodesic distance to the cone point ... – yasmar Apr 30 '11 at 13:41
won't be $r_0$ when the cusp gets in the way. I am not sure why @Willie suggested that we have a $C^1$ manifold here. Maybe he will elaborate on that. – yasmar Apr 30 '11 at 13:43