
Re: Riemannian Metric Topology
Posted:
Apr 4, 2012 5:29 AM


On Mar 31, 8:12 am, Jeff Rubin <JeffBRu...@gmail.com> wrote: > This is a question about a step in a proof that appears in two of my textbooks: > > Lang[1999] Fundamentals of Differential Geometry, Serge Lang > AMR[1988] Manifolds, Tensor Analysis, and Applicxations, second edition, > R. Abraham, J.E. Marsden, T. Ratiu > > The setting is that we have a connected Hausdorff manifold, X, and a > Riemannian metric, g, on X. No other assumptions are made about the > manifold, so in particular I don't know that it is paracompact, normal, > regular, or if it admits partitions of unity. I don't even know if the > Hilbert space (or spaces) it is modeled on is separable or not. For > simplicity, I assume X us a manifold without boundary. > > Given a point x of X and a chart (U, \phi) at x for X, where \phi(U) > is open in a Hilbert space E, one easily gets the positive definite, > invertible, symmetric operator A(x) on E which corresponds to g(x). > Given an element z of the tangent space above x, one also easily gets the > real value (g(x)(z, z))^{1/2}. We then go on > to define a length function L_g which assigns a real number L_g(\gamma) > to each piecewise C^1 path \gamma:J=[a,b] \to X as follows: > > L_g(\gamma) = \int_J (g(\gamma(t))(\gamma'(t),\gamma'(t)))^{1/2} dt. > > We then define a function dist_g: X x X \to R by > > dist_g(x,y)=inf{L_g(\gamma) : \gamma is a piecewise C^1 path in X > from x to y, defined on the closed interval J=[a,b]} > > Without any difficulty, dist_g is a pseudo metric. However, we have not yet > shown that the topology it induces on X is the same as the original manifold > topology. The first main point of the proofs in both books (Lang p189190 > and AMR p381 Proposition 5.5.10) is to show that dist_g is not just a > pseudo metric but is in fact a metric. So we start with distinct points > x and y of X and set out to show dist_g(x,y) > 0. We have the chart (U, \phi) > at x, as above, and we can arrange U to be small enough that y is not in U, > since the manifold is assumed to be Hausdorff. Working in \phi(U) we find an > r>0 such that the closed ball D(\phi(x),r) is contained in \phi(U) and such > that certain other properties hold. Let S(\phi(x),r) be the boundary of > D(\phi(x),r). Then we define D(x,r)=\phi^{1}(D(\phi(x),r)) and > S(x,r)=\phi^{1}(S(x,r)), both subsets of U. > > Since \phi is a homeomorphism, D(x,r) and S(x,r) are closed in U (not > necessarily closed in X). To me, this is a key stumbling point, as I'll > explain. We next let \gamma:J \to X be any piecewise C^1 path in X from > x to y. Both proofs make the following assumption: since x is in D(x,r) > and since y is not in U, the path \gamma must cross S(x,r). Neither author > explicitly proves this assumption (and AMR doesn't even state it). > > When I set out to prove this, using the continuity of \gamma and the > connectedness of J, I quickly run into the need to show that D(x,r) > is closed in X, not just in U. If X were known to be regular, it would > not be a problem to take r small enough that D(x,r) was closed in X. > But as I mentioned at the beginning, I don't know that X is regular. > If I could show that the pseudometric topology for X induced by dist_g > was the same as the original manifold topology, I would also get that > X was regular. But I don't see how to do that without first completing > the first part of the proof. > > The whole question seems to be, can I make r small enough that D(x,r) stays > away from the topological (in the original manifold topology of X) boundary > of U? But this does not seem to be a local issue, since it depends on what is > closed in X which in turn, depends on what is open everywhere in X including > outside of U. > > In Abraham's "Foundation of Mechanics", I found a statement to the effect that > a manifold which admits a Riemannian metric is necessarily second countable. > However, I don't see how that could be applied here (nor do I immediately see > why it is true). > > Now, assuming that the statements that the authors are trying to prove is > actually true, then X will turn out to be a metric space and therefore > regular. So how do I get this regularity early enough in the proof to > noncircularly use it to show \gamma must cross S(x,r)? Alternatively, > how do I directly show that \gamma must cross S(x,r)? > > Interestingly, a third reference I have (Kobayashi & Nomizu, Foundations > of Differential Geometry, Volume 1, 1963 and 1991) gives a proof (Chapter IV > Proposition 3.5, p166) which doesn't seem to proceed the same exact way, > but it is completely impenetrable.
Hello ,its me again . Let B= D\S be the open disk ,its open in the manifold since its open in U and U is open in the manifold .So M\B is closed in the manifold . So if g(t) ,t e [0,1] is the curve gamma joining x=g(0) to y=g(1) ,then the set of t with g(t) in M\B is closed in [0,1] so there is a minimum ,t=a >o with g(t) in M\B so g(t) is in B for all t<a and g(t)> g(a) as as t increases to a which shows that g(a) is is in the closure in U of B and not in B ,that is g(a) is in S . I believe this gets you through your stumbling block ,without using regularity ,Regards smn

