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Article #3990

Re: What's a manifold?

#3990
From: Timothy Y. Chow
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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References: <1992Nov5.004804.24757@galois.mit.edu> <1992Nov5.035214.25991@galois.mit.edu> <1992Nov5.060400.14203@CSD-NewsHost.Stanford.EDU>

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From: tycchow@riesz.mit.edu (Timothy Y. Chow)
Subject: Re: What's a manifold?
Message-ID: <1992Nov5.174751.2086@galois.mit.edu>
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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
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