Simple topological question, slightly OT

Simple topological question, slightly OT

Post by David Turne » Sat, 28 Jun 2003 00:17:58



Hello

The following question(s) occurred to me, and I wonder if anyone can answer
it (them):

1. I read that it is fairly widely believed, but not proved, that [the
surfaces of] all convex polyhedra can be unfolded to form a planar surface.
I think this is called an isometric embedding in R^2?

2. At any rate: It is well established that a sphere cannot be similarly
embedded in R^2.

3. A sphere is supposed to be the limiting case of a convex regular
polyhedron as the number of faces gets larger.

So: why is it that every c.r.p. is embeddable, but the limit is not?

I realise that this is not a contradiction per se, but it certainly seems
interesting to me.

Regards
David Turner

 
 
 

Simple topological question, slightly OT

Post by David Turne » Sat, 28 Jun 2003 00:30:10


Postscript:

The answer that seems sensible to me is, "because locality vanishes as the
size of the facet tends to zero".  But I'd like to hear what everybody else
thinks.

A slightly less satisfying way of putting it is that it's pointless to call
it an isometry if it's everywhere discontinuous.

Regards
David Turner

 
 
 

Simple topological question, slightly OT

Post by David Turne » Sat, 28 Jun 2003 00:52:49


Hi


On Thu, 26 Jun 2003 17:17:58 +0200, "David Turner"


> First, it's not really a topology issue, but rather an issue of local
> (Riemannian) geometry.  The key is in the word "isometric"--you are
> looking for a mapping that preserves distance, as measured on the
> surface.   Indeed, the impossibility of isometry between the sphere
> and the plane applies even locally.  You cannot map even a tiny patch
> of the sphere isometrically to the plane.

Unless said patch is degenerate?

Imagine wrapping an infinitely thin string around a ball to form a sphere.
You could then unwind the string onto a plane.  Another way of putting it is
that there exists a bijection between R and S^2.  Does there?  ;-)

Regards
David Turner

 
 
 

Simple topological question, slightly OT

Post by Hans-Bernhard Broeke » Sat, 28 Jun 2003 01:19:53





[...]
>> and the plane applies even locally.  You cannot map even a tiny patch
>> of the sphere isometrically to the plane.
> Unless said patch is degenerate?

Patches of the sphere cannot be degenerate --- only coordinate systems
or mappings constructed over their domain can.

Quote:> Imagine wrapping an infinitely thin string around a ball to form a sphere.
> You could then unwind the string onto a plane.  Another way of putting it is
> that there exists a bijection between R and S^2.  Does there?  ;-)

That such a mapping must exist is clear to anyone who ever heard about
Mr. Cantor and his theories about \aleph and \aleph_0.

But of course, such a mapping from the sphere via R to the plane would
find it every bit as impossible to be isometric as a direct mapping
is.  Taking a detour doesn't invalidate geometry.

--

Even if all the snow were burnt, ashes would remain.

 
 
 

Simple topological question, slightly OT

Post by David Turne » Sat, 28 Jun 2003 01:30:53


Hi


Quote:

> Patches of the sphere cannot be degenerate --- only coordinate systems
> or mappings constructed over their domain can.

Of course a patch can be degenerate, in the sense that it can have zero
area.  It "degenerates" into a point.

Quote:

> > Imagine wrapping an infinitely thin string around a ball to form a
sphere.
> > You could then unwind the string onto a plane.  Another way of putting
it is
> > that there exists a bijection between R and S^2.  Does there?  ;-)

> But of course, such a mapping from the sphere via R to the plane would
> find it every bit as impossible to be isometric as a direct mapping
> is.  Taking a detour doesn't invalidate geometry.

Yes, of course, I realise that.  There is a strong parallel between this
detour and, say, lim as x->0 of x/x.  In both cases the property ("value" or
"can be cut to have a locally isometric embedding in R^2") fails to have
meaning in the limit, but is everywhere else uniform.

One might argue that a sphere _can_ be cut to have such an embedding, if one
is prepared to weaken the definition of "isometric" to include limiting
cases.  But of course this is all pointless.  I just found it interesting.

Regards
David Turner

 
 
 

Simple topological question, slightly OT

Post by Just d' FAQ » Sat, 28 Jun 2003 09:42:21


On Thu, 26 Jun 2003 17:17:58 +0200, "David Turner"


>1. I read that it is fairly widely believed, but not proved, that [the
>surfaces of] all convex polyhedra can be unfolded to form a planar surface.
>I think this is called an isometric embedding in R^2?

>2. At any rate: It is well established that a sphere cannot be similarly
>embedded in R^2.

>3. A sphere is supposed to be the limiting case of a convex regular
>polyhedron as the number of faces gets larger.

>So: why is it that every c.r.p. is embeddable, but the limit is not?

You are mixing apples and oranges, as the expression goes. Shall we
look at a few facts?

If we are discussing topology, there is no difference between a sphere
and the surface of a convex polyhedron. In order to flatten either one
you must first disrupt its topology with at least a puncture, if not
some cuts. But in topology we do not speak of isometric embedding.

We must be talking about some kind of geometry, but what kind? If it's
differential geometry, there's no problem applying it to a sphere; but
how do we treat a polyhedron as a differentiable surface? For example,
we know how to compute "curvature" on a sphere, and it's one of the
things that tells us a sphere is *intrinsically* not flat (whereas a
cylinder *is* intrinsically flat). But how should we compute curvature
on a polyhedron?

Actually, there is a technology that says all the curvature of the
polyhedron is concentrated at the vertices, though this is not a kind
of theory one often encounters. In such a formalism you could take a
limit as the number of vertices increases, and find that curvature
distributes across the surface in the limit. But in such a formalism
you could not flatten a corner, because that would change curvature.

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