Saturday, November 26, 2005

Bridgework as Body Piercing

One of the things I'm having to get used to with my recently installed bridgework is the need to floss all the way underneath the fake tooth. And by that I mean, floss clear through underneath the bottom of the tooth, from front to back.

This is one thing I somehow never understood or foresaw before I got this "dental appliance" installed. First I take a little plastic lasso, and thread a length of dental floss through the loop of the lasso, like threading a needle. Then I stick the lasso in underneath the bridgework, and pull it on through so that the dental floss is now threaded underneath everything. And then I floss back and forth everywhere, underneath the bottom of the tooth.

The topological integrity of my body has been altered. I now have a veritable body piercing underneath my bridgework. I have pierced gums, or a pierced tooth, or whatever you want to call it— just as sure as someone who has pierced ears, or a pierced nose, or a pierced tongue, or whatever.

Body piercing. Body modification. I think I ought to qualify, you know?


Blogger The Tetrast said...

Be the Moebius strip.

Saturday, November 26, 2005 3:27:00 PM  
Blogger Paul Burgess said...

Or better yet, a Klein bottle! ;-)

Saturday, November 26, 2005 4:08:00 PM  
Blogger The Tetrast said...

Darn those Klein Bottles, expositions never clarify whether, strictly speaking, a Klein bottle is a surface twisted in 4-space or a solid twisted in 4-space.

Consider a Moebius strip from which every point has been deleted except all & only those comprising the edge. What you'd have remaining would be topologically equivalent to a circle, or a sort of floppy circle in 3-space.

It's not clear to me whether a Klein-bottle needs only the points of the surface which twists in 4-space, or needs also the points of the solid (or "hypersurface") which twists in 4-space. (It seems clear enough that a Klein bottle can be solid; consider the path of a disk carving out the Klein bottle from 4-space, just as a line-segment could carve out a Mobius strip from 3-space (imagine this Moebius-shaped spectrum thick with absorption bands (i.e. successive positions of the line segment.)

I've wondered about this once every few years for decades!

Saturday, November 26, 2005 5:31:00 PM  
Blogger Paul Burgess said...

Interesting question. Just off the top of my head (and it's been almost 25 years since questions like these used to be my bread and butter), I've always thought of the Moebius strip and the Klein bottle as examples of nonorientable two-dimensional manifolds.

If we take the rectangular region in the real plane, consisting of all points (x,y) such that

-1 ≤ x ≤ 1 and
-1 ≤ y ≤ 1,

then we can construct a Moebius strip by identifying every pair of points on the top and bottom edges

(x,1) and

where x ranges over the interval [-1,1]. This amounts to pasting the top and bottom edges of the rectangular region together with a half-twist.

We can then construct a Klein bottle if we also paste the right and left edges of the rectangular region together without a half-twist; that is, identifying every pair of points

(1,y) and

where y ranges over the interval [-1,1].

In fact, there is a third nonorientable two-dimensional manifold which we can construct, known as the projective plane. Here we identify every pair of points (x,1) and (1-x,-1) along the top and bottom edges, as with the Moebius strip; but along the right and left edges we identify every pair of points

(1,y) and

where y ranges over the interval [-1,1]. Perhaps an easier way to imagine the projective plane is to think of the unit disk in the real plane— the set of all (x,y) such that

x² + y² ≤ 1

—and identify each diametrally opposite pair of points around the circumference of the disk, that is, identify every (x,y) and (-x,-y) such that

x² + y² = 1

There's a theorem in differential geometry, known as the Whitney embedding theorem, which states that any n-dimensional manifold over the reals can be embedded in Euclidean real space of dimension at most 2n+1. Thus, any two-dimensional manifold is embeddable in (at worst) Euclidean five-space. Clearly a Klein bottle (at least as I've defined it above) requires four-space to be embedded in— there is no one-to-one mapping of a Klein bottle into Euclidean real three-space, i.e., to three-dimensional eyes the "neck" of the Klein bottle always seems to intersect its side, even though mathematically speaking it doesn't. I can't think off the top of my head of any two-dimensional manifold which can be embedded in five dimensions but not in four, but very likely there are examples.

One could also construct a projective three-space by taking the "unit ball" in three-space— that is, all (x,y,z) such that

x² + y² + z² ≤ 1

and identifying each diametrally opposite pair of points on its surface— that is, every point on the surface of the ball would be identified with the point at its antipodes. And I think one could also construct by analogy a "Klein solid," one dimension up from a Klein bottle; and perhaps other nonorientable three-dimensional manifolds. But at that point my intuition fails me; and classification theorems for n-dimensional objects sometimes get hairy very quickly as n increases.

Hope I got all that correct, I'm pulling all this off the top of my head, there may well be errors in my exposition. That's what one gets for leaving this subject matter alone for a quarter of a century! :-)

Saturday, November 26, 2005 7:50:00 PM  
Blogger The Tetrast said...

Thanks, Paul. I followed your exposition well enough to be finally led back to the idea of following the twists of the surface. A flatlander can travel along the Klein bottle's surface in such manner as to be mirror-reversed upon arriving back at his starting point. In contrast, a linelander traveling along a Moebius strip's edge would not undergo such reversal, and this corresponds to the fact that a Moebius strip's edge without the strip itself amounts to a mere circle. So a Moebius strip's edge decidedly differs, in terms of mirror-reversals, from a Klein bottles "hyper-edge" or surface. So the analogies that I use seem to be working again. In fact I don't understand how I could have gotten confused about this in the first place. I think it happened this way: When, as a child, I first read about Klein bottles, the book said that a Klein bottle has "no inside and no outside." Years later, I thought about the disk-motion carving a Klein bottle out of 4-space, analagously as a line-segment's motion could carve a Moebius strip out of 3-space, so I came to interpret a Klein bottle as indeed having an inside and an outside, and so I thought that there must be something wrong with the popular expositions that I had read. I guess the truth is that a Klein bottle can be given a non-arbitrary inside and outside, but it doesn't need such in order to be a topologically intact Klein bottle.

But if a Klein bottle can have an interior, then this (from here) would have to be the wrong way to look at it. Instead, this (which I just created) would be the way. But it seems highly doubtful that I could be right about this. I'm no topologist. Ah, well.

Sunday, November 27, 2005 12:35:00 PM  
Blogger The Tetrast said...

This comment has been removed by a blog administrator.

Sunday, November 27, 2005 1:42:00 PM  
Blogger The Tetrast said...

Love that ability to delete my own comment when it needs editing.

* * *

Here I get a bit more of the Klein bottle's usual expositional appearance, with the seeming narrowing and widening. . The traveling disk's edge traces a Klein bottle's surface out of 4-space. If the path of the disk's interior doesn't trace an interior out for the Klein, then I wonder what sort of shape instead is made by it. Anyway, that's it for me for the time being! Why does the Thanksgiving weekend have to end?

Sunday, November 27, 2005 1:44:00 PM  

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