Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Topic: Riemannian Metric Topology
Replies: 10   Last Post: Apr 4, 2012 3:52 PM

 Messages: [ Previous | Next ]
 Stuart M Newberger Posts: 475 Registered: 1/25/05
Re: Riemannian Metric Topology
Posted: Apr 3, 2012 2:52 PM

On Apr 2, 9:31 pm, Jeff Rubin <JeffBRu...@gmail.com> wrote:
> On Monday, April 2, 2012 1:57:45 AM UTC-7, smn wrote:
> > On Apr 1, 7:49 pm, Jeff Rubin <JeffBRu...@gmail.com> wrote:
> > > On Sunday, April 1, 2012 3:45:47 PM UTC-7, 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 p189-190
> > > > > 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 pseudo-metric 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
> > > > > non-circularly 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 things-the 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):470-474
>
> 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.- Hide quoted text -
>
> - Show quoted text -

Hello. I am sorry,I misquoted Urysons' theorem,- a REGULAR ,Hausdoff
topological space with a countable base (i.e. 2nd countable) is
metrizable . Thats not too relavent.
The quote on P128 of AM must be assuming that the manifold is modeled
on a 2nd countable (ie separable) Banach space which would let my
example out of the picture .The Banach space might even be finite
dimensional. Good Luck smn

Date Subject Author
3/31/12 Jeff Rubin
3/31/12 HOPEINCHRIST
4/1/12 Stuart M Newberger
4/1/12 Jeff Rubin
4/2/12 Stuart M Newberger
4/3/12 Jeff Rubin
4/3/12 M. Vaeth
4/4/12 M. Vaeth
4/3/12 Stuart M Newberger
4/4/12 Guest
4/4/12 Stuart M Newberger