Timeline for Riemannian manifold of bounded geometry has a normal bundle of bounded geometry

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Mar 3, 2011 at 4:38	history	edited	Deane Yang	CC BY-SA 2.5	added 783 characters in body
Mar 2, 2011 at 20:41	comment	added	Deane Yang		Alex, that appears to be exactly the Riemannian metric on the normal bundle that I used. In other words, there is a natural immersion of the normal bundle into $R^N$, and the metric on the normal bundle is just restriction of the metric on $R^N$ to $T_N_M$.
Mar 2, 2011 at 19:12	comment	added	AlexE		@Deane Yang: The normal bundle $NM \subset TR^N$ is the orthogonal complement of $TM$ w.r.t. the standard Euclidean metric on R^N. Here the embedding $TM \subset TR^N$ comes from the isometric embedding $M \to R^N$. Now we use on $NM \subset TR^N$ the induced metric and the induced connection (which is given by the standard Euclidean connection on R^N composed with the orthogonal projection onto NM). Perhaps my words "pull-back metric/connection" were not well chosen - "induced" is better.
Mar 2, 2011 at 18:52	comment	added	Deane Yang		Anton, if the Riemannian metric is the one induced by the embedding of, say, a small tubular neighborhood of the zero section in Euclidean space, then it is almost tautological to say that the metric is flat. But maybe I'm using the wrong metric on the normal bundle?
Mar 2, 2011 at 17:31	comment	added	Anton Petrunin		Do you want to say that normal bundle has zero curvature? (I do not think so --- say normal bundle of $\mathbb C\mathrm P^2$...)
Mar 2, 2011 at 15:25	history	answered	Deane Yang	CC BY-SA 2.5