Characterization of bounded geometry - Reference-request - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:52:29Z http://mathoverflow.net/feeds/question/113781 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113781/characterization-of-bounded-geometry-reference-request Characterization of bounded geometry - Reference-request ngrosse 2012-11-18T23:14:20Z 2012-11-19T14:12:52Z <p>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.</p> <p>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).</p> <p>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.</p> <p>In the paper of Shubin it is written that there are the following equivalent characterizations: (Let $(M,g)$ always have positive injectivity radius.)</p> <p>(i) $(M,g)$ is of bounded geometry.</p> <p>(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 $x′$).</p> <p>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 $\alpha$).</p> <p>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.</p> <p>Thank you in advance.</p> http://mathoverflow.net/questions/113781/characterization-of-bounded-geometry-reference-request/113833#113833 Answer by Sergei Ivanov for Characterization of bounded geometry - Reference-request Sergei Ivanov 2012-11-19T14:12:52Z 2012-11-19T14:12:52Z <p>I assume that by "all derivatives" you mean derivatives of every order.</p> <p>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.)</p> <p>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.</p> <p>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.</p>