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I already asked this question at stackexchange three days ago. Since I got no answer, I want to try mathoverflow now. I hope that you can help.

I'm looking for a proof of an equivalence that can e.g. be found in a paper by Shubin 'Spectral theory of elliptic operators on non-compact manifolds' (Appendix A.1.1 below Def. 1.1).

It's about manifolds of bounded geometry, where bounded geometry means here: positive injectivity radius and the curvature tensor and all its covariant derivatives are bounded.

In the paper of Shubin it is written that there are the following equivalent characterizations: (Let $(M,g)$ always have positive injectivity radius.)

(i) $(M,g)$ is of bounded geometry.

(ii) Let $r>0$ be smaller than the injectivity radius. For any $x,x′\in M$ let $y$ (resp. $y′$) be geodesic normal coordinates around $x$ (resp. $x′$) with an r-ball as domain. If the balls of radius $r$ around $x$ and $x′$ have a nonempty intersection, then the transition function $y^{−1}\circ y$ and all its derivatives are uniformly bounded (where uniformly means here independent of $x$ and $x′$).

I know that (i) is equivalent to the following: (iii) Let $g^\alpha_{ij}$ be the representation of the metric coefficienst on a ball of radius $r$ around $x_\alpha$ with respect to geodesic normal coordinates. Then $g^\alpha_{ij}$ and all its derivatives are uniformly bounded (where uniformly means independent on $\alpha$).

I can see that (iii) implies (ii) (by looking at the geodesic flow and using Gronwall's inequality). But I have no idea how to get the converse. Is there any reference for the proof? I'm grateful for any idea for the proof as well.

Thank you in advance.

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  • $\begingroup$ Have a look at: MR2343536 (2008i:58001) Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp. I do not have my copy at hand, so I cannot check whether a proof is there. Since the whole book uses bounded geometry, it might be there. $\endgroup$ Nov 19, 2012 at 10:04
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    $\begingroup$ I have to check, don't have time now, if the statement you want is actually in there, but a lot of these type of theorems are proven in the PhD thesis of Jaap Eldering arxiv.org/pdf/1204.1310v2.pdf . See for example Lemma 2.6 $\endgroup$
    – Thomas Rot
    Nov 19, 2012 at 12:33
  • $\begingroup$ Indeed to confirm what @Thomas Rot said: the converse statement is not in my PhD thesis. I don't know of another place where you might find a citable reference. $\endgroup$ Nov 19, 2012 at 22:38

1 Answer 1

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I assume that by "all derivatives" you mean derivatives of every order.

Suppose that all transitions between normal coordinates have uniformly bounded derivatives within some radius $r$. For every point $p$ we have a unit radial vector field $V=V(p)$ whose derivatives are uniformly bounded at distances between, say $r/10$ and $r$ from $p$. (This holds in normal coordinates centered at $p$ and hence in normal coordinates centered at any nearby point.)

Now fix a point $q\in M$ and arrange points $p_k$, $k=1,\dots,\frac{n(n+1)}2$, where $n=\dim M$, so that the distances from $q$ to $p_k$ equal $r/2$ and the corresponding vector fields $V_k=V(p_k)$ at $q$ are generic in the sense that the symmetric tensors $V_k\otimes V_k$ are linearly independent. Then the same holds at every point in a neighborhood of $q$. Actually the points $p_k$ should have some prescribed coordinates in normal coordinate system centered at $q$, then all subsequent estimates are uniform.

We have a system of equations $g(V_k,V_k)=1$, $k=1,\dots,\frac{n(n+1)}2$. At every point near $q$ this is a nondegenerate linear system on the components of the metric tensor $g$. Hence it uniquely determines $g$ via some explicit formulae in terms of $V_k$. Since the derivatives of $V_k$ are uniformly bounded, so are the derivatives of the metric tensor.

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  • $\begingroup$ Thanks a lot. Do you know any ("citeable") reference for this? (The book of Eichhorn (mentioned above by Peter Michor) does not work with transition functions and in the PhD-Thesis of Elderling I only found the implication "(i) implies (ii)") $\endgroup$
    – ngrosse
    Nov 19, 2012 at 19:15
  • $\begingroup$ No, I have no reference. It may be hard to find one as this implication does not look very useful. $\endgroup$ Nov 19, 2012 at 20:27

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