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

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

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