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Article #3990Re: What's a manifold?
From: Timothy Y. Chow
Date: Thu, 05 Nov 1992 17:47
Date: Thu, 05 Nov 1992 17:47
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In article <1992Nov5.060400.14203@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes: >Ah, now this is starting to sound very interesting and helpful. What I >don't understand here is how "living in R^n" is of itself creating >complexity and obscurity. I can see that the arbitrariness of f might >get in the way. But where does the complexity and obscurity creep in >if for example we define a manifold to be a smooth retract of an open >subset of R^n? (This is essentially taking the existence of tubular >neighborhoods as definitive of manifolds, and is how Bill Lawvere likes >to think of them.) Well, for a start there are some definitions of manifolds that allow things that don't embed in R^n. But even if we exclude these cases from consideration, there is still good reason not to define manifolds as subsets of R^n. The point is that there are many different subsets of R^n that represent the "same" manifold. So we really should talk about equivalence classes of subsets of R^n, under some suitable equivalence relation. Actually it is worse than that since we might want to identify certain subsets of R^3 with certain subsets of R^4. This is why I think it's cleaner to lay down the axioms for a topological space and use that as the basis for the definition of a manifold, rather than have to bring in this huge equivalence class of subsets every time you want to talk about a manifold. In fact, I think that despite appearances your retract definition of manifold actually contains topological axioms implicitly. Don't you really want to say that a manifold is something *homeomorphic* to a smooth retract of an open subset of R^n? Once you do this you've already taken the implicit step of embracing the axiomatic approach to topology. None of this is to say that it isn't often convenient to take a subset of R^n to *represent* a manifold. Your point is well taken---we don't necessarily want to avoid R^n altogether once after we've abstracted away from it. But I still think it's simpler to *define* a manifold axiomatically rather than make the definition essentially dependent on subsets of R^n. -- Tim Chow tycchow@math.mit.edu Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs 30 tons, computers in the future may have only 1,000 vacuum tubes and weigh only 1 1/2 tons. ---Popular Mechanics, March 1949
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Newsgroups: sci.physics,sci.math Path: arc1-ca.pugleaf.net!sparky!uunet!snorkelwacker.mit.edu!galois!riesz!tycchow From: tycchow@riesz.mit.edu (Timothy Y. Chow) Subject: Re: What's a manifold? Message-ID: <1992Nov5.174751.2086@galois.mit.edu> Sender: news@galois.mit.edu Nntp-Posting-Host: riesz Organization: None. This saves me from writing a disclaimer. References: <1992Nov5.004804.24757@galois.mit.edu> <1992Nov5.035214.25991@galois.mit.edu> <1992Nov5.060400.14203@CSD-NewsHost.Stanford.EDU> Date: Thu, 5 Nov 92 17:47:51 GMT Lines: 40