Point singularity of a Riemannian manifold with bounded curvature Suppose you have an incomplete Riemannian manifold with bounded sectional curvature such that its completion as a metric space is the manifold plus one additional point. Does the Riemannian manifold structure extend across the point singularity?
(Penny Smith and I wrote a paper on this many years ago, but we had to assume that no arbitrarily short closed geodesics existed in a neighborhood of the singularity. I was never able to figure out how to get rid of this assumption and still would like someone better at Riemannian geometry than me to explain how. Or show me a counterexample.)
EDIT: For simplicity, assume that the dimension of the manifold is greater than 2 and that in any neighborhood of the singularity, there exists a smaller punctured neighborhood of the singularity that is simply connected. In dimension 2, you have to replace this assumption by an appropriate holonomy condition. 
EDIT 2: Let's make the assumption above simpler and clearer. Assume dimension greater than 2 and that for any r > 0, there exists 0 < r' < r, such that the punctured geodesic ball B(p,r'){p} is simply connected, where p is the singular point. The precludes the possibility of an orbifold singularity.
ADDITIONAL COMMENT: My approach to this was to construct a differentiable family of geodesic rays emanating from the singularity. Once I have this, then it is straightforward using Jacobi fields to show that this family must be naturally isomorphic to the standard unit sphere. Then using what Jost and Karcher call "almost linear coordinates", it is easy to construct a C^1 co-ordinate chart on a neighborhood of the singularity. (Read the paper. Nothing in it is hard.)
But I was unable to build this family of geodesics without the "no small geodesic loop" assumption. To me this is an overly strong assumption that is essentially equivalent to assuming in advance that that differentiable family of geodesics exists. So I find our result to be totally unsatisfying. I don't see why this assumption should be necessary, and I still believe there should be an easy way to show this. Or there should be a counterexample.
I have to say, however, that I am pretty sure that I did consult one or two pretty distinguished Riemannian geometers and they were not able to provide any useful insight into this.
 A: Once we considered a similar problem but around infinity,
try to look in our paper "Asymptotical flatness and cone structure at infinity".
Let us denote by $r$ the distance to the singular point.
If dimensions $\not= 4$ then the same method shows that at singular point we have Euclidean tangent cone even if curvature is "much less" than $r^{-2}$ (say if $K=O(\tfrac{1}{r^{2-\varepsilon}})$ for some $\varepsilon>0$, but one can make it bit weaker).
In dimension 4 there might be some funny examples: Your singular point has tangent space $\mathbb R^3$,
the $r$-spheres around this point are  Berger spheres, so its curvature is very much like curvature of $r\cdot(S^2\times \mathbb R)$, the size of Hopf fibers goes to $0$ very fast. 
However if you know that dimension of the tangent space is $4$ then it has to be Euclidean.
All this can happen if curvature grows slowly.
If it is bounded then one can extend Bishop--Gromov type inequality for balls around singular point.
It implies that the dimension of the tangent space is $4$. That will finish the proof.
A: Take a cone over a finite quotient S^{2n-1}/\Gamma. The curvature is 0, but the manifold
structure does not even extend. (More generally, you can take the cone over any compact 
Einstein manifold of dimension n-1 with Einstein constant n-2.)  
A: Here is what seems to be a counterexample. Let (M,g) be a simply-connected closed Riemannian manifold. Then M times (0,infinity) with the warped product metric dr^2 + r^2 g has bounded curvature and the completion at r=0 is a point. If the metric is smooth, then M is diffeomorphic to a sphere, so any other M gives a counterexample. 
EDIT: Sorry, this does not work as curvature blows up at zero unless g has constant curvature 1.
A: If by "extends across the point singularity" you mean extends smoothly, then I think you may just start with the Euclidean space thought of as a warped product over (0,infinity) with sphere as a fiber and replace the warping function r by any smooth function f(r) that is near r in C^2-topology. Then the curvature will not change much, while for the metric to be smooth at the origin f must satisfy consistency conditions on the higher derivatives of f at r=0 that surely will be violated for nearly every f. The consistency conditions are like those that can be found eg in Peterson's book, page 13 (in first edition).
It might be possible to build a multiple warped product example in which the link at the singular point is not a sphere but I am not sufficiently motivated to attempt the computation. Handy formulas for curvature tensor of multiple warped products can be found in my paper arXiv:0711.2324 in appendix C. 
A: What if you try a family of triangles, parallel to some two-direction s.t. their union contains singularity? (Like tetraedr for n=3)? Then their geometry (angles, sides, etc) are controlled from "outside" the singularity, so they all have uniformly bounded cirvature - including that one which contains singularity. Let the size then goes to zero. Does it mean that the tangent plane is defined at singularity and is R^n and so on ...? 
