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It is well known a Riemannian manifold with constant sectional curvature is a quotient of the Euclidean space, hyperbolic space or sphere. In particular we know how their metric looks like locally.

My question is, is there a quantitative version of the above result? By this, I mean for example, given $\varepsilon>0$, there exists $\delta$ such that if $(M,g)$ satisfies $\left|Rm\right|_g<\delta$ `, does there exist a local diffeomorphism $\phi$ to $\mathbb R^n$ such that $|\phi_*g-g_0|_{g_0}< \varepsilon$, where $g_0$ is the standard metric on $\mathbb R^n$?

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3 Answers 3

up vote 5 down vote accepted

This question is two-sided, and I'm not sure what you mean by "local diffeomorphism" so I'll treat both aspects. There is a local and a global version :

Local version :

Q1 : Given a point $x$ in a Riemannian manifold $(M,g)$, can we find a constant curvature metric on a neighborhood of $x$ which is close to $g$ ?

First remark : since we want a local statement, zero curvature is as good as constant curvature here.

The answer to Q1 is "yes", without any restriction on the curvature. This can be seen using normal coordinates centered at $x$ : in these coordinates, the metric at $x$ is euclidean and its distortion from being euclidean as one moves away from $x$ can be controlled (using that the curvature is bounded in a neighborhood of $x$, and that the injectivity radius at $x$ is positive).

Edit : The above statement is too complicated. As Anton's say in his comment to Agol's answer, use the exponential map ! The pull back of $g$ by the exponential map defines a riemannian metric on some neighborhood of the origin in $T_xM$ which is equal to $g_x$ at $x$, by continuity, this pullback stay close to the euclidean metric $g_x$ on $T_xM$ and this does the job.

We can refine the question then :

Q1' : What can be said about the size of the neighborhood we obtained ?

In this case we need to impose geometric restrictions on $(M,g)$. For instance, Cheeger and Anderson proved the following :

For $n\in\mathbb{N}$, $k\in\mathbb{R}$, $i>0$ and $\varepsilon>0$, one can find $\delta>0$ such that in a $n$-manifold of Ricci curvature greater than $k$ and injectivity radius greater than $i$, any ball of radius $\delta$ admits a flat metric which is $\varepsilon$-close to $g$ in $C^0$-norm.

See "$C^\alpha$-compactness for manifolds with Ricci curvature and injectivity radius bounded below."

The proof uses more elaborate machinery than just normal coordinates : harmonic coordinates are used. If you stick to normal coordinates you can obtain a similar result but with stronger geometric assumptions.

Global version :

Q2 : Under which condition does a manifold with almost $k$ curvature admit a $C^0$-close metric of constant curvature $k$ ?

If you consider large spheres, they have almost zero cuvrvature but don't admit any flat metric, so you need to put some restrictions on the side of the manifolds.

An example of theorem you can get is the following :

For any $n\in\mathbb{N}$, $k\in\mathbb{R}$, $V>0$, $D>0$ and $\varepsilon>0$, there is a $\delta>0$ such that any $n$-manifold $(M,g)$ of diameter less than $D$, volume more than $V$, and sectional curvature between $k-\delta$ and $k+\delta$ admits a metric af constant sectional curvature $k$ which is $\varepsilon$-close to $g$ in $C^0$-norm.

The proof relies on Cheeger-Gromov compactness theorem for sequences of Riemannian manifolds. A (really) sketchy goes like that : we argue by contradiction, you take a sequence $\delta_i$ going to $0$, and you assume you can find a sequence of manifolds $(M_i,g_i)$ satisfying the hypothesis of the theorem with $\delta=\delta_i$ and not satisfying the conclusion of the theorem. Then up to a subsequence, the sequence has a limit which is a manifold of constant curvature $k$, by the very definition of Cheeger-Gromov convergence, this imply the for some $i$ large enough, $M_i$ admit a constant curvature $k$ metric $\varepsilon$-close to $g_i$, a contradiction.

The lower bound on the volume is necessary (at least in the $k=0$ case) the so called "infranilmanifolds" admit metrics of curvature as close as wanted to $0$ with diameter bounded above but no flat metric.

For the $k=1$ case, the bounds on the diameter is unnecessary because of Myers theorem.

For the $k=-1$ case, I don't know if the hypothesis can be weakened.

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For $k=-1$ you do not need the lower volume bound, but you need the upper diameter bound. See Pinching constants for hyperbolic manifolds. by Gromov and Thurston math.psu.edu/petrunin/teach-old/minicourse-china-2008/… –  Anton Petrunin Sep 9 '12 at 16:15
    
Thnks ! It's funny that the situation is opposite to that of positive curvaure. Another question came to my mind about the positive case : if we assume that $M$ is simply connected, then the lower bound on the volume isn't necessary (Klingenberg's Lemma). With Synge, it also shows that the volume is not needed in even dimensions. What about odd dimensions ? –  Thomas Richard Sep 9 '12 at 16:27
    
Thanks for the very detailed reply! The question in the "global version" comes close to what's in my mind. Can you suggest a reference for me to look up your stated (or similar) result? (I am indeed not very familiar with Cheeger-Gromov compactness, but I can take it for granted and take a look at the proof of these types of results. ) –  Kwong Sep 10 '12 at 13:40
    
In fact, almost everything is in the proof of the Cheeger-Gromov compactness theorem. The only additional observation is to show that the limit space has constant curvature, which is not that obvious because the limit space is only a $C^{1,\alpha}$ riemannian manifold. The trick is to see that the comparison results with spaceforms of constant curvature $k$ hold in both directions, and to use this to build a local isometry with a constant curvature space. For the proof of Cheeger-Gromov compactness, see the papers by Greene and Wu, and by S. Peters. I don't remember the titles, I'll check. –  Thomas Richard Sep 11 '12 at 17:52

There's an unpublished preprint of Tian which gives a criterion on when a manifold is close to being Einstein may actually be deformed to being Einstein (see Theorem 6.1 of his paper - sorry, this copy has only odd pages!). In 3 dimensions, this means that if one is close to being hyperbolic in his sense, then there is a deformation of the metric to a hyperbolic metric. '

There's also the $1/4$ pinching theorem of Brendle-Schoen.

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Ian, the answer to Kwong's question is "exponential map" and you are answering different (and more advanced) question. –  Anton Petrunin Sep 9 '12 at 13:19
    
@Anton : I didn't see your comment when I was writing my answer. –  Thomas Richard Sep 9 '12 at 13:27

For the global question, and hyperbolic metrics, in dimension > 3 this is a result of Gromov, stated in his 1978 JDG paper, and in dimension 3 it is an unpublished result of Daryl Cooper, from the late nineties, and Gromov, independently, so while Tian might have a more general result, he is far from the first.

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