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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 1, 2012 6:45 PM
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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 .

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