From: wdh@linus.mitre.org (Dale Hall)
Subject: Re: What's a manifold?
Followup-To: sci.math
Summary: abunchastuff
Keywords: manifolds
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Distribution: na
Date: Thu, 5 Nov 1992 17:31:44 GMT

In article <1992Nov5.060400.14203@CSD-NewsHost.Stanford.EDU>
		pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes: 
>In article <1992Nov5.035214.25991@galois.mit.edu>
		tycchow@riesz.mit.edu (Timothy Y. Chow) writes: 
>>
>>Well, this definition has the advantage of being easily motivated and
>>concrete, but as people started studying manifolds more deeply they
>>found that it was often a nuisance to have to be tied to a particular
>>function f to define their manifolds.  There were certain geometric
>>properties of manifolds that were "intrinsic" to the surface and didn't
>>really depend on the function f in any essential way.  It seemed that
>>what was needed was a way of defining manifolds without having to pick
>>a space R^n for the manifold to live in and specify an explicit function.
>>
>>Hence the modern approach to manifolds is to define them as objects in
>>their own right, without reference to a space that they're imbedded
>>in.  This is what motivates the definition that John Baez gives.
>>Notice that he doesn't require the manifold to live in some R^n.  The
>>extra abstraction is a small price to pay for the simplification and
>>logical clarity that it yields, as you will appreciate if you study the
>>subject more deeply.  Similarly, smooth manifolds and algebraic
>>varieties (and later schemes) are nowadays defined without making them
>>live in R^n.
>
>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.)

The problem, as I see it, with thinking of manifolds as living in some
ambient Euclidean space, is that it causes confusion when the manifold
arises full-grown from the head of Zeus, as it were.  For instance,
imagine a manifold M (maybe it's in R^n), and you have some notion of
symmetry that applies to M.  Further, you know that (for instance) for
any x in M, the set of all things related to x by symmetry (for the
sake of discussion, let's call it the cohort of x [non-standard
terminology here]) is particularly nice: the cohort of x is always a
circle, say, and all of the symmetry operations treat nearby cohorts
in a continuous fashion.  That is, if operation \theta sends x to the
"circle element" x_\theta, and if y is near x, then the operation
\theta yields y_\theta which is similarly near x_\theta (I'm being
purposely informal about things that can indeed be made precise).  

If the symmetry is regular in a suitable sense, then one can construct
a manifold M/~ where ~ is the equivalence relation defined by x ~ x'
if x and x' belong to the same cohort.  (I mean, you can always define
the quotient of a space w.r.t. an equivalence relation, but in this
case the quotient inherits a manifold topology)

Now, the symmetry may not apply to all of R^n, so you're left with M/~
that is homeless in the sense of not having a Euclidean space to
reside in.  If your definition of manifold forces you to think only of
manifolds within Euclidean spaces, then you're stuck until you can
find an equivariant embedding of M in some R^m (perhaps different from
its earlier home of R^n). 

A more insidious problem arises when you think of manifolds that are
thought of as the zero-set of some smooth function f: R^n --> R^k.  If
you use the implicit function theorem to guarantee that you have a
real-live manifold (i.e., you restrict the Jacobian to have rank k
along the zero set),then you have condemned your manifolds to a very
nearly trivial existence from some perspectives: each will have its
characteristic (Stiefel-Whitney, Pontrjagin, Chern) classes <<all>>
zero, for the reason that the normal bundle of the manifold in R^n
will be trivial, since it is the pullback of the normal bundle of 0 in
R^k (a patently trivial bundle if ever there was one) and thus the
tangent bundle will be stably trivial.  No projective spaces RP^n or
CP^n, no cobordism theory because every manifold you ever see will
bound (in the next higher dimension).  Pretty dull, if you ask me.

If you're a complex kind of guy, you'll likely stick to holomorphic
functions, and will notice that you never ever see compact manifolds,
since no compact complex manifold embeds holomorphically in C^n.  Even
the Riemann sphere, that old friend from complex analysis, stubbornly 
refuses to visit.  Not to mention the global picture of Riemann
surfaces. 

My sense is that it is the notion of freeing manifolds from the bonds
of the Euclidean captors that has enabled the theory to blossom as it
has.  That other stuff?  Yeah, it's important too, but it could have
been accomplished with embedded manifolds, even if it would have been
clumsy.  

My opinion only.
						Dale.
Newsgroups: sci.physics