
Re: Riemannian Metric Topology
Posted:
Apr 3, 2012 12:31 AM


On Monday, April 2, 2012 1:57:45 AM UTC7, smn wrote: > On Apr 1, 7:49 pm, Jeff Rubin <JeffBRu...@gmail.com> wrote: > > On Sunday, April 1, 2012 3:45:47 PM UTC7, smn wrote: > > > 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 , Unless you assume the manifold topology is Hausdorf you will > > > not be able to show that the pseudo metric is a metric with the same > > > topology since a metric space is Hausdorf.It still may be the case > > > that the pseudo metric topology is the same as as the manifold > > > topology anyway.Maybe an almost same proof works for this ,I havent > > > tried . Good Luck smn . > > > > Yes, I did stipulate early in the post that the original manifold topology > > was assumed to be Hausdorff. Hide quoted text  > > > >  Show quoted text  > > Hello again. Yes I think you need to assume that the Hausdorf manifold > topology is regular .As you might know it would sufice to assume that > the topology had a countable base (hence metrizable) ;or that the > manifold was finite dimensional ,thus locally compacct,thus regular. > A Hilbert space is a Reimannian manifold and not necessarily 2nd > countable ,what page does AM say this in Foundations of mechanics .His > manifolds are all 2nd countble ,in fact finite dimensional I think. > Infinite dimensional manifolds are often subsets of a Banach space > hence metrizable. > Anyway if assuming regular gets you through all this by all means > assume it neatly adding it as an additional hypothesis .Lang > sometimes forgets to say thingsthe book you are reading grew from an > older one and the older one didn't have the Riemannian part but he > didn't rewrite the older part. > Its good stuff though. Regards,smn
Hi smn.
AM Foundations of Mechanics 2nd Edition Copyright 1978 Sixth Printing October 1987 page 128:
"Recall that we include second countable in our definition of a manifold. It is interesting that a manifold which admits a Riemannian metric (or a connection) must be second countable (see Abraham [1963])."
The only entry in the references for Abraham 1973 is:
Abraham, R. 1963.a Transversality in manifolds of mappings. Bull. Am. Math. Soc. 69 (4):470474
So your point is well taken: a Hilbert space as a manifold has a trivial Riemannian metric, yet needn't be separable and therefore needn't be second countable. So what could AM have been assuming?
Now, are you saying that a Hausdorff space which has a countable base is metrizable? There are counterexamples in "Steen and Seebach, Counterexamplex in Topology", for example #60 Relative Prime Integer Topology and #61 Prime Integer Topology. Or are you saying that being a manifold or having a Riemannian metric adds some other condition? What would that condition be? AM's definition of manifold does not include being finite dimensional.
I'm hesitant to assume regularity since not only Lang but AMR also doesn't assume it. I feel like I'm missing something obvious.

