Date: Tue, 26 Mar 1996 13:28:37 -0800 (PST)
From: ctm@math.berkeley.edu (C. T. McMullen)
To: rusin@math.niu.edu
Subject: Re: Diffeomorphism invariant subspaces of functions

> >	You might find techniques to prove this kind of statement 
> >	in Gromov's book "Partial Differential Relations".
> >	It contains an exercise asking for a proof that there
> >	are only a finite number of Diff(R) invariant subspaces
> >	of C^\infty(R).  (It turns out this exercise is wrong -- there
> >	are uncountably many; but that has to do with noncompactness
> >	of R).
> 
>So where do the many invariant subspaces come from -- is
> it the lack of closure of this invariant subspace? Or the lack of
> compact support?
> 
	It is definitely from the lack of closure.  Here is
a typical exotic subspace -- take some linear space S of sequences,
say those that are bounded and satisfy \sum |a_i-a_{i+1}| < \infty.
Now consider the set of smooth functions f on R such that under all 
proper embeddings Z -> R, sending n to x_n, the sequence
a_n = f(x_n) is in S.  This is clearly a linear subspace and by
definition it is Diff(R) invariant.  By varying the choice of S
you can easily making uncountably many examples.

	As for the case where M is compact, I would be interested to
hear about the proof if you find it.