NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/sci/math/18785

Text File | 1993-01-27 | 3.6 KB | 74 lines

Newsgroups: sci.math
Path: sparky!uunet!enterpoop.mit.edu!galois!riesz!jbaez
From: [e-mail] (John C. Baez)
Subject: Re: Smooth manifolds and function extensions
Message-ID: <[e-mail]>
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Organization: MIT Department of Mathematics, Cambridge, MA
Date: Wed, 27 Jan 93 02:26:48 GMT
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In article [e-mail] (Charles Yeomans) writes:
>In article <[e-mail]>, [e-mail] (M.R.Watkins) writes:
>>
>> Given a manifold M with a smooth (C-infinity) structure, it is possible to
>> define E(M), the linear space of all smooth functions on M.  Now is it
>> possible to reconstruct the smooth structure (that is the atlas on M) from
>> the knowledge of E(M) alone?

>I believe the answer is no, in general.  Consider , in C^2, the two sets
>X = unit ball z^2 + w^2 < 1 and Y = X - (0,0).  Let H(X), H(Y) be the
>spaces of holomorphic functions on X, Y.  Hartogs' theorem implies that
>any function holomorphic on Y extends (uniquely) to a holomorphic
>function on X.  Thus H(X) is isomorphic to H(Y).  But X is very diferent
>from Y.

>But for a manifold with a C-infinity structure, what you ask might be
>possible. However, I think you need more structure on X, say that of a ring or
>algebra.

The holomorphic case is utterly different from the C-infinity case,
because no theorem like Hartog's theorem holds in the C-infinity case.
One can formalize this by saying that smooth functions form a fine sheaf
and holomorphic functions don't.

You definitely need to know more than the structure of the smooth
functions as a *vector* space to reconstruct the manifold!!!  All vector
spaces of the same dimension are isomorphic!  A topological vector space
would be better, but really you want to be able to use the algebra
structure.  Using the tricks Yeomans outlines -- which are formalized in
the Gelfand-Naimark theorem -- one should be able to see that you can
reconstruct M up to homeomorphism from its ring of smooth functions.
(Sketch: give this ring E(M) a norm by taking the sup of |rho(f)| over
all homomorphisms f: E(M) -> C (throw out those elements for which this
is infinite.  Give it a *-algebra structure by pointwise complex
conjugation - which can be defined purely algebraically using the 1-1
correspondence between points and the above homomorphisms (or
equivalently, maximal ideals.  Complete it to form a C*-algebra B(M);
stting inside this is the C*-algebra of cont fns on M that vanish at
infinity.  Figure out how to characterize this subalgebra purely
algebraically if you can.  Then hit it with the Gelfand-Naimark theorem
and you would be done.)  Then one has to show one can actually
reconstruct M up to diffeomorphism.  Since tangent spaces, etc. can all
be defined purely algebraically in terms of E(M), I think this should be
doable too.

Note I'm using complex smooth functions, while Yeomans uses real ones.
Watkins didn't say which he meant, but it's easy to get the complex ones
from the real ones by tensoring with C, or to pick out the real ones
from the complex ones as those fixed by complex conjugation, which as I
note above can be defined purely algebraically.

>With this, you ought to be able to do the following:  Suppose E(M) has the
>structure of a commutative R-algebra.  Let I be a maximal ideal in E(M). What
>you'd like is for I to have the form {f in E(M) | f(p) = 0}, for some p
>in M. This would be the hard part.  Given such a fact, you could reconstruct
>M using the set of all maximal ideals.
>
>I seem to recall that there are such theorems in Banach algebra.
>
>Charles Yeomans