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Suppose $M$ is a smooth complete Riemannian manifold and $x$ is a point in $M$. For any positive radius $r$ we consider the open ball $B(x,r)$ centered at $x$ with radius $r$.

If we ignore the Riemannian structure on $B(x,r)$ and consider it only as a smooth manifold. Are there only a countable number of diffeomorphism classes it can belong to?

An afirmative answer includes as a particular case the fact that there are only countably many compact manifolds up to diffeomorphism. This follows from Cheeger's finiteness theorem because on any such manifold has bounded curvature, volume, and injectivity radius.

It is well known that there are uncountably many diffeomorphism classes of smooth manifolds (even homeomorphic to $\mathbb{R}^4$; Are there uncountably many surfaces?).

My motivation for this is that I'm trying to study a particular concept of a random complete Riemannian manifold with a distinguished point. The starting point is to topologize the space of such manifolds (and I need something stronger then Lipschitz) and things would be simpler if the answer to this question was yes (and I had a short proof or reference :) ). Also, the question seems interesting in its own right.

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5  
It is not true that there are countably many surfaces, even up to homeomorphism. –  Alexandre Eremenko Nov 28 '12 at 20:22
    
@Alexandre: Why not? –  Pablo Lessa Nov 28 '12 at 21:44
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Agol explains in mathoverflow.net/questions/25009/… why there are uncountably many surfaces. –  Igor Belegradek Nov 28 '12 at 22:46
    
@Igor: Thanks! –  Pablo Lessa Nov 28 '12 at 23:13

1 Answer 1

The answer to this question should be negative for $M=S^2$. I will sketch a tentative proof of this claim that would use a couple of simple lemmas.

Lemma 1. The set of closed totally disconnected subsets $X_{dis}$ in the interval $[0,1]$ modulo diffeos of $[0,1]$ is uncountable.

Lemma 2. The set of surfaces $S^2\setminus X_{dis}$ where $X_{dis}$ as above is uncountable (we consider here $X_{dis}$ lying in a segment $I\cong [0,1]$ in $S^2$, $I\subset S^2$.)

Lemma 3. For any $X_{dis}\subset [0,1]$ there is a smooth non-negative function $f$ on $[0,1]$ whose set of zeros is $X$.

All these lemmas are very straightforward (Lemma 2 requires a bit of thinking but it should be correct). I want to use them to construct for any $X_{dis}\subset S^2$ a metric on $S^2$ such that for a certain point $O\in S^2$ the ball $B(O,1)$ is $S^2\setminus X_{dis}$. In order to do this start with a metric on $S^2$ such that $B(O,1)$ is $S^2\setminus [0,1]$, where $[0,1]$ is an interval smoothly embedded in $S^2$. Note that the equidistants emanating from $O$ come to $[0,1]$ from two sides. So just "speed up" them from one side so that they hit all the points from $[0,1]\setminus X_{dis}$ before the "other side" of equidistant hits them; at the same time do so that all points of $X_{dis}$ are hit at the same time from both sides. (to do this explicitly I would use something like Lemma 3).

I am afraid that this sketch is not so well phrased, but I hope that the idea is more-less clear.

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Thanks for the answer! Lemmatas 1 and 2 are addressed in Agol's answer to mathoverflow.net/questions/25009/… (thanks to Igor Belegradek for pointing me to this answer). Lemma 3 can be proved using smooth bump functions and in fact any compact set in $\mathbb{R}^n$ is the set of zeros of a smooth function. However, the hard part seems to be what follows since the "speeding up" of geodesics cannot be done without affecting other geodesics. I'm still not convinced there is a smooth metric on the sphere with the properties you suggest. Still, nice ideas. –  Pablo Lessa Nov 29 '12 at 0:23
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@Pablo, do the bumps closer to the segment. –  Anton Petrunin Nov 29 '12 at 2:50

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