Recall that $(M_{k}, g_k,O_k)$ of complete pointed Riemannian manifolds converges smoothly in the sense of Cheeger-Gromov to $(M_{\infty}, g_{\infty}, O_{\infty})$ if there exists an exhaustion of open sets $U_k$ of $M_{\infty}$ containing $ O_{\infty}$, and a sequence of diffeomorphism $\phi_k$ from $U_k$ to $V_k=\phi_k(U_k) \subset M_k$ with $\phi_k (O_{\infty})=O_k$, such that the pull back $(U_k, \phi_k^{\ast} (g_k))$ converges to $(M_{\infty}, g_{\infty})$ uniformly on compact sets in $M_{\infty}$.

These days I am interested in how wild the diffeomorphism $\phi_k$ can be. Consider a sequence of rotationaly invariant metrics on $\mathbb{R}^n$ with uniformly bounded geometry and injective radius at the origin has a uniform lower bound (which can be ensured by for example, choosing a uniform bounded positive cone angle at infinity for the sequence metrics), fix the sequence of points to be the origin, by the compactness theorem we expect a limit metric which is still rotationaly invariant. However, I am not sure if the limit is necessarily rotationaly invariant. If so does it mean that in this special case, we can choose the sequence of diffeomorphism to be a sequence of rotations around the origin?

I understand that a way to think about this symmetric case is to go over the general construction on diffeomorphism and the limit metric in the proof of Compactness Theorem. It is a little involved and I am still on the way to understand it. Any suggestions or help will be appreciated.

Added after posted:

To make the point of "understand the diffeomorphism $\phi_k$" clear, let me continue on the rotationally symmetric example, Let us put one more restrition on the sequence $(M_k, g_k, O_k)=(R^n, g_k,O)$ that the scalar curvature $R(O)$ attain the maximum for each $g_k$, then after the taking limit we can get a metric $(M_\infty, g_\infty,O_\infty)$ which also have $R(O_\infty)$ as a maximum on $M_\infty$, But if we take $O_k$ to be a sequence points $P_k$ suitably close to the origin $O$, the limit will be a new metric $(N_\infty, h_\infty,P_\infty)$. It is reasonably to believe the new metric could be the same as $(M_\infty, g_\infty,O_\infty)$. However $R(P_\infty)$ might not be the maximum of scalar curvature on $N_\infty$ in general, So it is natural to find a maximum of scalar curvature on $N_\infty$ first, name it $Q_\infty \in N_\infty$, then check how far the pull back sequence $\Phi_k(Q_\infty) \in M_k$ is from $O$. It sounds to me that it is important to understand the behavior of $\Phi_k$. However I feel that it is a little wild even for this example.

  • $\begingroup$ Not sure what you mean by "wild". As for rotational symmetry, $g_k$ has a subsequence that converges to a rotational symmetric metric because of precompactness in the equivariant GH-topology (described in papers of Fukaya). By uniqueness of the limit $g_\infty$ is rotationally symmetric. However, I do not see how it would help you learn about the diffeomorphisms $\phi_k$ because as you say they are defined not on the whole manifold $M_\infty$ but on its compact domains. $\endgroup$ – Igor Belegradek Dec 28 '12 at 14:27
  • $\begingroup$ The diffeomorphisms $\phi_k$ defined on compact domains won't be equivariant though, they will be almost equivariant, which is why it is unclear how this restricts $\phi_k$. $\endgroup$ – Igor Belegradek Dec 28 '12 at 14:50
  • $\begingroup$ Thanks Igor for suggesting equivariant version of convergence. Sorry for the vague question. Just add one more example to clarify. $\endgroup$ – Bo_Y Dec 28 '12 at 22:05
  • $\begingroup$ I do not understand the added paragraph. I think it makes no sense in several places. For example the maximum of scalar curvature need not be preserved under GH convergence. Also changing the base point can change the limit metric, so your "reasonably to believe" seems unfounded. As far as I can see your question is not precisely stated, and hence is not appropriate for MO. You need to think what you wish to ask, and then repost. $\endgroup$ – Igor Belegradek Dec 29 '12 at 3:10

I also don't understand the question at all. The diffeomorphisms are constructed to "normalize" the Riemannian metrics, so that the metrics differ as little as possible. So by construction they are as far fron being wild as possible. Cheeger-Gromov compactness is the statement that such "tame" diffeomorphisms exist and, since they behave so nicely, subsequences of both the diffeomorphisms and metrics converge on any compact domain.

If you know something more about the metrics, then you can often use the additional information to construct particularly nice diffeomorphisms.

If all of the metrics are rotationally symmetric on $\mathbb{R}^n$, then there is no need to construct diffeomorphisms at all. Cheeger-Gromov compactness, for example assuming bounded sectional curvature, in this situation is easily verified using the metrics written with respect to polar co-ordinates and the Jacobi equation.

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  • $\begingroup$ I'm not sure about you last statement. If you take a metric $g$ on $\mathbb{R}^2$ which looks like paraboloid, and define $g_i$ to be the pullback of $g$ by the diffeomorhism $x\mapsto \frac{x}{i}$, the diffeomorhisms are needed I think ? $\endgroup$ – Thomas Richard Dec 29 '12 at 4:47
  • $\begingroup$ Ok, now I understand : in the case you mentionned the diffeomorhisms are just give by the exponential maps at the origin and isometric identification of the tangent spaces at the origin. $\endgroup$ – Thomas Richard Dec 29 '12 at 9:29

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