# 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.

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Is your edit intended to rule out the following answer to your question? Consider R^2 with the flat metric, and let X be the universal cover of the punctured plane R^2 \ 0. Note that any two preimages in X of the same point in the plane are close together in X (a universally bounded distance, going to zero as your point goes to the origin). Thus the completion of X adds only one point, but is not even a manifold (it is not locally compact at the singular point you add). If so, can you explain why this is unsatisfying? –  Tom Church Oct 21 '09 at 7:06
Yes, my assumptions are intended to rule out possibilities like that. The goal was a geometric analogue of Uhlenbeck's removable singularity theorem for Yang-Mills connections. –  Deane Yang Oct 22 '09 at 20:03
Can you prove that the boundary of a sufficently small neighborhood of the point is topolgically a disk? –  Dmitri Nov 20 '09 at 23:33
I think you mean a sphere and not a disk? No, I don't know how to do that first. The approach we took was to start with a single geodesic ray emanating from the singularity and show that it could be extended to a differentiable family of geodesic rays covering a conic open set. Then we showed that the set of all such geodesic rays is both open and closed. The curvature bound and standard Jacobi field estimates then give the space of geodesic rays a differentiable structure and identify it with the unit sphere. –  Deane Yang Nov 21 '09 at 1:06

## 5 Answers

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.

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Thanks! I'll take a look at your paper and contact you directly if I have any comments or questions. Offhand, I would be surprised if Berger spheres can give a counterexample. It seems to me that the bounded sectional curvature condition on the space forces the sectional curvature of the geodesic sphere to approach $1$ everywhere. –  Deane Yang Jan 3 '10 at 13:23
Yes, it seems you are right --- the picture around "zero" and arround "infinity" is different. But the method we use should work. For any dimensions $d\not=4$, it should give that tangent cone at zero is Euclidean d-space --- the rest should be easy. The example I described is the only place where you can expect something funny (I mean the method might not work or require something extra). –  Anton Petrunin Jan 3 '10 at 18:41
I just took a quick look at your paper. My impression is that the question above can be addressed properly only by a very careful understanding of collapsed Riemannian manifolds. Is there some way to see why this is necessary without going through the guts of the proof? –  Deane Yang Jan 4 '10 at 19:21
Deane, I will better write an e-mail :) –  Anton Petrunin Jan 5 '10 at 6:12

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.)

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Rafe, Doesn't the "localy simply connected" assumption (see EDIT) preclude your example? And although using an Einstein metric as the "geodesic sphere" makes the Ricci curvature bounded, doesn't the sectional curvature still blow up? –  Deane Yang Oct 21 '09 at 12:11

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.

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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.

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No, it is not reasonable to expect the metric to extend smoothly assuming only bounded sectional curvature. Morally speaking, curvature is the second derivative of the metric. So the most you can hope for is that the metric is C^{1,1}. But nobody knows whether even that is always possible. The best known results, using harmonic co-ordinates and Schauder estimates, is that if a metric has bounded sectional curvature, there exists coordinates such that the metric is C^{1,a}, for any a < 1. That's all I'm asking for above. –  Deane Yang Oct 22 '09 at 1:07
I am not familiar with the details of the consistency conditions you refer to, but I am confident that they would hold if you also assume bounds on higher covariant derivatives of the curvature tensor. –  Deane Yang Oct 22 '09 at 1:09

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 ...?

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Yes, this is the approach we took. But without a "no small loops" assumption, I couldn't see how to make sure that a triangle (defined as two geodesics emanating from the singularity and a third joining the two geodesics at some distance away from the singularity) is really a triangle (i.e., a surface foliated by a differentiable family of geodesic segments with a common endpoint at the singularity). –  Deane Yang Nov 21 '09 at 1:01
I mean singularity is inside the triangle. No small loop just for n=2, where my suggestion would not work. For n>2 (simply connected "annulus") - we may compare - map the whole tetraedr to the similar one in complete manifold with bounded curvature. Then take the foliation from the image (pull back) - the distortion is under control -> we have "nice" foliation of the completion of the target manifold. Does it work? (valeri) –  valeri Nov 21 '09 at 1:12
Sure. So how do you construct this map from the given punctured manifold with the point singularity to the complete manifold? Once this map is constructed, the rest is easy. But where does this map come from? My first response to you was perhaps an overly oblique way of asking you this question. –  Deane Yang Nov 21 '09 at 4:14
What if we measure distances between vertices of the tetraedr in incomplete manifold M^n and then find points in complete - say R^n having the same mutual distances, and then continue this correspondence by linearity? Bounded curvature then implies that this is "nice" map (C{1,\alpha} or {2} - not sure) ? –  valeri Nov 21 '09 at 8:55