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

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Stuart M Newberger

Posts: 479
Registered: 1/25/05
Re: Riemannian Metric Topology
Posted: Apr 2, 2012 4:57 AM
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On Apr 1, 7:49 pm, Jeff Rubin <> wrote:
> On Sunday, April 1, 2012 3:45:47 PM UTC-7, smn wrote:
> > On Mar 31, 8:12 am, Jeff Rubin <> 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

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