Mike wrote:
> On Apr 27, 4:02 am, "W. Dale Hall"
> <wdunderscorehallatpacbelldotnet@[EMAIL PROTECTED]
> wrote:
>> Mike wrote:
>>> Are there metric space or otherwise manifolds that are not
>>> locally Euclidean? Thanks.
>> If you mean "locally homeomorphic to Euclidean space", meaning that
>> every point has a neighborhood homeomorphic to an open subset of
>> R^n for some n, take the space formed by taking three rays meeting
>> at the origin in R^2.
>>
>> No neighborhood of the origin in this space is homeomorphic to an
>> open subset of R^n for any n.
>>
>> More pathological examples can be constructed (such as spaces for
>> which no point has a neighborhood homeomorphic to an open subset of
>> R^n), of course.
>>
>> Dale
>
> I guess what I mean is that the basis vectors at a point in some
> coordinate system are not orthogonal to each other as they are in
> local euclidean coordinate systems? Or is it that ANY coordinate
> system can be parameterized locally with euclidean geometry? Thanks.
Let's suppose we're talking about a space for which every point
has a neighborhood homeomorphic to an open subset of R^n (where
we fix n, for convenience). This is then an n-dimensional manifold,
or n-manifold. I'll note that this is an extremely special condition:
there are metric spaces that aren't manifolds. However, it seems that
this may be the most general situation where your question makes sense.
One then can define charts, systems of these "Euclidean" neighborhoods
(I'll use U to represent such a neighborhood in the space, which I'll
call M), with each U equipped via a homeomorphism (I'll call this a
coordinate mapping) to an open subset of R^n. Since the balls in R^n
form a basis of its standard topology, we can assume all of these
Euclidean neighborhoods are mapped to balls in R^n.
Next, one obtains (with a minor amount of work) mappings
A_ij ---> U_i . U_j ---> B_ij
where A_ij represents the image in R^n, via the coordinate mapping from
U_i to R^n, of the intersection U_i . U_j, and B_ij the image in R^n,
of the same intersection, via the coordinate mapping from U_j.
Since the coordinate mappings are all homeomorphisms, the composition
A_ij ---> B_ij
is a homeomorphism (from one open subset of R^n to another) . If all
these so-called "transition mappings" are differentiable, then one
obtains a differential structure on M. It is worthwhile to note that
not all manifolds admit a differential structure. All manifolds of
dimension <= 3 do, and I know of a 10-manifold with no differential
structure, but beyond that I'm too far beyond my day-to-day interaction
with the topic to speak authoritatively.
A differential structure enables one to define the tangent bundle, for
which the above transition mappings can be used to produce a GL(n)
structure, meaning that you can obtain the tangent bundle from the
trivial (i.e., product) R^n-bundle over the sets U_i, by attaching the
****tion over U_i to the ****tion over U_j using the differential of
the above mapping A_ij to B_ij . In general, local sections of the
tangent bundle are as close as you can get to the notion of "basis
vectors". If your manifold is compact and has nonzero Euler
characteristic, there is not even a single "basis vector" that
can be defined globally. Is this unusual? Just think of the 2-
dimensional sphere: Euler characteristic = 2 ==> no globally
nonzero vector field. That is, you can't define a single continuous
nonzero tangent vector on the whole space.
At long last, the issue of "orthogonal coordinates".
Points of M have what we call "local coordinates", defined via
the local "coordinate mappings" as I described above. As such,
there is no such thing as "orthogonal coordinates", until you
specify some more information. The way this is done is via the
imposition of structure on the tangent bundle, and as it turns
out, this is equivalent to finding a set of transition mappings
for the tangent bundle that lie in the *orthogonal group* O(n),
as contrasted with the full general linear group GL(n).
Fortunately, this is always possible (we assume manifolds are
paracompact), since (by the Gram-Schmidt process) the Lie groups
GL(n) and O(n) are homotopy-equivalent. The procedure turns out
to be equivalent to specifying a smooth metric on M, where the
term "metric" is only tangentially (heh, heh) related to the
topological notion of metric. Rather than a distance measure
on M, it is an inner product defined on each tangent space, in
such a way that it varies smoothly with the corresponding base
point in M.
The bottom line is that (1) you usually can't define a single
basis over the whole space, let alone an orthogonal one, and (2)
there is always a reduction of the structure group of the tangent
bundle to the orthogonal group, so yes, one *can* get an orthonormal
basis locally. In order to be able to extend a *single* basis over
the whole space, you need the manifold to have a trivial tangent
bundle (i.e., TM must be M x R^n). This condition is called
"parallelizability", of we say M is parallelizable.
All orientable 3-manifolds are parallelizable. This follows from
the fact that the second homotopy group of GL(3) is trivial.
All Lie groups are parallelizable. This follows from the
group property, and from the fact that translation is a smooth
self map of any Lie group.
I'm not sure if I've even addressed your question.
Dale
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