🚀 go-pugleaf

In article <1992Nov10.185338.29742@samba.oit.unc.edu> Bruce.Scott@launchpad.unc.edu (Bruce Scott) writes:
>asimov@nas.nasa.gov (Dan Asimov) writes:
>>Yet a third notion is a manifold which happens to be a metric space which
>>arose through embedding the manifold in Euclidean space, by measuring the
>>distance across the Euclidean space.  Interestingly enough, this is not
>>very often considered in math research.
>
>>But as *such*, with the extrinsic metric (the one obtained from the
>>distance across R^3), your helix will, of course, *not* embed
>>"isometrically" in R^2, which I believe is your point.
>
	[...]
>The thing I really misunderstood seems to have been the theorem. In my
>sense of embedding, is there a known minimum N for which *any* 1-D
>manifold will extrinsically embed in R^N?
>
>Dr Bruce Scott                             The deadliest bullshit is

To answer this question, we need to be very careful about *what type
of manifold* we are talking about.  Since you use the word extrinsic,
I am going to guess that you mean any 1-manifold that has been smoothly
embedded in some Euclidean space R^n, and which has then been supplied
with the metric which comes from oridnary distance across R^n.

THE ANSWER:
In that case, then there is no maximum N such that every such 
1-manifold- metric-space embeds "extrinsically."

Proof:  The finite metric space consisting of any n+1 points in general
	position in R^n (i.e., they lie on no (n-1)-dimensional plane) cannot be
	isometrically embedded in R^N for any N < n.  For example, the origin
	(0,...,0) and the n points (1,0,...,0),(0,1,...,0),...,(0,...,0,1) in R^n.

	Now simply find a smooth embedding of the line R^1 into R^n which passes
	through all n+1 of these points.  (This is not hard, but I leave the 
	details as an exercise for those interested.) If that could be "extrinsically" 
	embedded in any lower dimension, that would violate the impossibility of 
	embedding the finite metric space isometrically.


Dan Asimov
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