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

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Jeff Rubin

Posts: 17
Registered: 11/25/07
Riemannian Metric Topology
Posted: Mar 31, 2012 11:12 AM
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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.

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