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 -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading