🚀 go-pugleaf

In article <1992Nov10.000228.14551@samba.oit.unc.edu>, Bruce.Scott@launchpad.unc.edu (Bruce Scott) writes:
> asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:
> 
> >As has been mentioned, it *is* a theorem that all manifolds do occur
> >as subspaces of some Euclidean space: in fact, any n-manifold embeds in R^(2n).
> >
> >This is a pleasant theorem, but it would be inappropriate to take this as a 
> >definition.
> 
> What about the one-dimensional helix embedded in R^3. Is the helix not a
> 1-manifold? Note that if you compact it along its axis you get a circle
> (also a 1-manifold--is this S^1?) embedded in R^2. But a circle is a
> degenerate case of a helix ("stretching" along the axis having vanished)
> and is not in general a helix.
> 
> Someone educate me on this.

The helix, as a manifold, is just the same as, say, an infinite straight
line; and as such it does embed in R^2; indeed, in R [:-)].

The thing is that manifolds are only defined up to an appropriate kind of
equivalence; namely diffeomorphism; we say that A and B are diffeomorphic
if there is a map from A to B which is smooth and which has a smooth
inverse (exactly what "smooth" means depends on exactly what kind of manifold
you're dealing with; say, infinitely differentiable.)

A diffeomorphism between the helix parametrised as (cos t, sin t, t) and the
real line is given by (cos t, sin t, t) -> t.

-- 
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@cus.cam.ac.uk  Cambridge University, England.    [Research student]