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Re: torus?



Tanks!

I can understand your answer (it's even intuitive) to my question. The
remaining I'll have to get a book to translate the hieroglyphics!

bc


David Bowman wrote:

Regarding bc's and Joel's questions:

Does one have to cut it to lay it flat?

Is the above the test?

bc

"RAUBER, JOEL" wrote:

If I remember correctly:

An ordinary torus, (the ideal donut), T2 =3D S1XS1, has a surface =
that
is two
dimensional and the surface has intrinsic curvature.

Is this correct?

The torus T^2 =3D (S^1)X(S^1) =3D (T^1)X(T^1) is constructed as a dir=
ect
product of 2 1-d tori (i.e. 2 1-d 'spheres' or circles) has *no*
intrinsic curvature as a 2-d surface. *But* if you wanted to imbed
such a beast in the Euclidean space E^3 (i.e. R^3 endowed with a
Euclidean quadratic metric form) you would find out that it is
impossible to do without significant distortion. The distortion does
nothing to the *topology* of the surface, but it *does* affect its
metric and curvature. A doughnut shape is *topologically* a torus,
but as it sits in E^3 it *does* have significant curvature (both
intrinsic *and* extrinsic). Also, as Chuck mentioned a coffee cup
surface is also *topologically* equivalent to T^2, but its actual
geometric structure and metric is even more complicatedly curved than
the doughnut. Remember, topology and metric geometry (and its
associated curvature) are *different* things. Topological properties
are *global* properties and the curvature effects of metric geometry
are *local* (but not necessarily asymptotically infinitesimal)
properties.

If you wanted to embed T^2 into a Euclidean space in such a way that
the embedding is undistorted and remains (intrinsically) curvature-
free, you would have to use a Euclidean space of *at least 5*
dimensions to imbed it in.

One way to see that an ordinary doughnut is a *distorted* T^2 torus
is to imagine constructing it from a flat *square* piece of paper.
First you tape/paste together a pair of opposite edges of the square
(thus forming an open ended cylinder) and then you bend the two
end-circles around so they are connected to each other in such a way
that the cylinder is not axially twisted with any screw twist.
Unfortunately, if you try doing this in ordinary 3-space you will
discover that it cannot be done without tearing or stretching or
otherwise violating the internal geometry of the original square
piece of paper. If you used a square rubber sheet, and did not mind
some internal distortion of the sheet, such a construction can
certainly be made. The result is a doughnut shape (or maybe, even
something more complicated such as a coffee cup shape).

If the torus was left *undistorted* and it was made of an original
square of edge length L, its circumference would be L along each of
the orthogonal directions parallel to the 2 circular seams *no matter
where* you started from. However, its circumference would be
sqrt(2)*L along a diagonal direction at 45 degrees to a direction
that has a circumference of L. Such a torus is truly homogeneous in
that all points of the surface (i.e. manifold) are equivalent and the
surface is invariant under arbitrary (internal) translations.

OTOH, if you considered a doughnut surface in ordinary 3-space you
immediately notice that the circumference around the outermost
diameter circle is longer than that of the parallel concentric circle
going around the inner diameter. Additionally, if you intersected
the doughnut with a plane containing the the doughnut's axis of
symmetry, each of the two circles of intersection would, in general,
have a circumference which is different than either of the inner
diameter and outer diameter circumferences. If the torus had been
left undistorted all 4 of these circles would be the same
circumference L.

Bc asked if cutting to lay flat on a plane is the test for flatness.
The answer is no. If we start with an intrinsically flat T^2 torus
we can lay it flat on a plane by cutting along the 2 original
orthogonal seams (or make cuts parallel to them elsewhere) and get a
nice square that lays flat on a plane. The criterion is not whether
a cut or two is needed, but whether or not the already cut opened out
surface really does lay flat on a plane. If it doesn't lay flat it
is not an intrinsically flat manifold. If it does lay flat, it is
intrinsically flat. If we cut an ordinary doughnut surface in
3-space so that it could be opened up if it was flexible (but not
stretchable) we would find that it will not lay flat without further
nonuniform stretching, shrinking or without making an infinite number
of other cuts into an infinite number of pieces. Such a doughnut
surface *is* intrinsicaly curved. Likewise, if we took an ordinary
spherical surface we could open it up by cutting it into a pair of
hemispheres. These hemispheres will not lay flat on a plane without
further stretching, shrinking, or cutting into an infinite number of
longitudinal strips. This is because the spherical surface is also
intrinsically curved. But if we took a circular cylinder and opened
it up by unzipping it along its length, it *will* lay flat on a plane
because it *is* intrinsically flat. Similarly a right circular cone
is intrinsically flat and will lay flat when opened up.

(BTW, in case anyone is interested, if you wish to make a Klein
bottle you start with a square piece of paper and you again attach a
pair of opposing edges. But this time you give the end edges a
mutual 1/2 twist before attaching them (making a Mobius cylinder).
Next, you bend the ends around and attach them to each other. This
is much easier said than done in 3 dimensions.)

David Bowman
David_Bowman@georgetowncollege.edu