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- From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
- Subject: Re: What's a manifold?
- Message-ID: <1992Nov10.013653.9305@infodev.cam.ac.uk>
- Sender: news@infodev.cam.ac.uk (USENET news)
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- Organization: U of Cambridge, England
- References: <SMITH.92Nov5132141@gramian.harvard.edu> <1992Nov6.024142.6758@galois.mit.edu> <1992Nov6.190913.18507@nas.nasa.gov> <1992Nov10.000228.14551@samba.oit.unc.edu>
- Date: Tue, 10 Nov 1992 01:36:53 GMT
- Lines: 32
-
- 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]
-