From: Robin Chapman
Newsgroups: uk.education.maths,sci.math
Subject: Re: GAUSS'S THEOREM EGREGIUM
Date: Thu, 25 Jun 1998 08:16:12 GMT
In article <6mrtki$nvf$1@apple.news.easynet.net>,
"Vanda Foster" wrote:
>
> Can anyone explain to me the significance of Gauss's Theorem Egregium:
>
> - "The Value of Gaussian curvature does not depend on the parametrization."
I wouldn't paraphrase this in this way. Instead I would say that Gaussian
curvature is an intrinsic propery of a surface in R^n.
> I understand what it basically means but this implies that you can have
> different parametrizations and that I don't really understand. For instance
> if we are considering a 2-dimensional manifold, M, it can be immersed into
> R^3, thereby giving M a differentiable structure. This is a
> 2-parametrization. Does the theorem mean that if we then immerse M into R^4
> the Gaussian curvature is the same as it was before? That seems
> counter-intuitive, because a two-dimensional surface embedded in R^2 should
> be flat, so the curvature should be zero, which is not necessarily true for
> "other" parametrizations. So I think my interpretation of the word "other"
> is wrong. Can anyone help?
One can immerse a 2-manifold M in many ways into R^3 or R^4 and the curvature
will differ. To prevent this one needs a metric on the manifold, and consider
only manifolds which preserve this metric. For instance on R^2 there is a
metric dx^2 + dy^2. This is a flat metric with zero curvature.
One can embed a portion of the plane in R^3 in many ways and preserve the
curvature: e.g. as a plane, or a cylinder or a cirular cone, but not
as part of a sphere. On the other hand if the metric is
(1 - x^2 - y^2)(dx^2 + dy^2) on the unit disc, then one cannot embed this
in a metric-preserving way onto a flat surface in R^3. This is what the
Theorema Egregium says: the curvature of the surface depends only on thi
metric form.
> Maybe Gaussian curvature has some special invariance property that I haven't
> yet discovered, but I still can't help feeling that M should be flat when
> embedded in R^2.
>
> Also, if M is embedded in R^2 does it's lack of differentiable structure
> (which apparently it has, according to a rather badly written book I have on
> the subject) mean that it can be treated in a "global" fashion, without
> having to worry about annoying local things like curvature? Can this space
> be treated as a metric space, or a Hilbert space if a norm is introduced?
> These latter two would appear to be spaces that are treated globally, so
> that vectors on a Hilbert space, for instance, are not manipulated with
> reference to an actual point in the space where the vector "lies" (like in
> quantum field theory, in which a vector on Hilbert space is considered, but
> without reference to any specific point in the space). Is this an accurate
> deduction? Is there such a thing as a vector-field on a Hilbert space. I
> think this is a meaningless concept, because a vector-field implies a "local
> structure" of the space which, along with a differentiable structure,
> neither of which these spaces have.
>
Robin Chapman + "They did not have proper
Department of Mathematics - palms at home in Exeter."
University of Exeter, EX4 4QE, UK +
rjc@maths.exeter.ac.uk - Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda
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