From: lrudolph@panix.com (Lee Rudolph)
Newsgroups: sci.math
Subject: Re: Moebius strip
Date: 28 Jun 1998 10:31:17 -0400
Allen Adler writes:
>We work with a Moebius strip and a pair of scissors in three dimensional
>Euclidean space.
>
>If you cut a Moebius strip M lengthwise, you get a single strip S.
Which is an annulus. (If we imagine the "cutting" operation to
have actually dissolved away a tiny tubular neighborhood of the
circle C which is the "core" of your strip M, then S is what we
call a "collar" of the boundary of M--that boundary being one circle,
S is the product of a circle with a closed interval, i.e., an annulus.)
>That is easy to prove from the identification diagram of a Moebius
>strip.
>
>If you do the same to the strip S, you get two strips that are
>linked. That one gets two strips is again easy to see from the
>identification diagram. But why are they linked?
Let's be more precise. When you say "a Moebius strip ... in
three dimensional space", you haven't specified the embedding
of M in R^3. (There is an issue of language here: some people,
and for all I know Moebius himself, use[d] "Moebius strip" to
refer to a specific [ambient isotopy type of] embedding in R^3
of the 2-manifold-with-boundary which is produced "from the
identification diagram" you allude to. I prefer what I think
is now the general usage among topologists, which reserves
"Moebius strip" for the space in the abstract.) If we stay
(say) within the piecewise smooth world, such an embedding
(in an oriented R^3) is specified by two pieces of data:
a knot(type) K, the core of the embedded strip; and an odd
integer m, the number of "half-twists" in the embedded strip.
Similarly, an embedding of an annulus is specified by a knot
K and an integer n, the number of (whole) twists.
In the case I assume you're thinking of, for your embedding
of M we see that K = O is an unknotted (plane) circle, and
m is either 1 or -1 (depending on your taste); while for S,
K is again O (but not the same O: rather, a "cable of type
(2,m)" on the original O), and n is m. (If your original
M had been of type (O,3), then S would have been of type
(O{2,3},somethingorother), where the cable O{2,3} is a
trefoil knot, and I'm too lazy to calculate somethingorother.
And so on.)
Now, when you cut S, you are again creating a collar of *its*
boundary: S being an annulus, this boundary is two circles,
so the collar is a pair of annuli. Their status as "linked"
or "unlinked" will be the same as the status of the boundary
components of S to which they correspond. So your question
is, why are the two components of the boundary of an annulus
of type (O,\pm 1) linked? Well, that's just homology (specifically,
I guess, Alexander duality?): each component represents a
generator of the (infinite cyclic) 1-dimensional integer homology
of the complement of the other (in R^3). You can ad lib a proof
in any of a dozen ways, I'm sure!
More generally (and no more subtly), the two boundary components
of an annulus of type (O,n) are *unlinked* only when n = 0.
Much more generally (and considerably more subtly in the case
n = 0), the two boundary components of an annulus of type (K,n),
with K not O, are *always* linked whatever n may be. In particular,
whatever embedding of M in R^3 you start with, the two strips
(annuli) you get after two cuts are always linked (even homologically)
because 0 is not odd.
>I know they are
>linked because I have actually done the cutting with a scissor,
>but I don't know how to give a mathematical proof. Can someone
>explain this to me?
Lee Rudolph