From:  (Geoffrey Mess)
Newsgroups:     sci.math
Subject:        Re: Borromean Rings don't exist!
Date:           5 Mar 1995 03:31:18 GMT
Organization:   University of California, Los Angeles

In article <3j8ait$326@jupiter.SJSU.EDU> (Roger  
Alperin) writes:
> What is the proof that the Borromean Rings configuration of circles can't
> exist in R^3? There is some proof using hyperbolic geometry I heard, but
> can't recall,  that embeds the rings at infinity in H^3 and then using some
> group generated by reflections get a contradiction. Are there other proofs?
> It can be done with squares though.
Or with three ellipses, all with arbitrarily small eccentricity.

3 round circles in R^3 (or its conformal compactification S^3) with pairwise
linking number 0  bound  3 mutually disjoint hemispheres in the 4 ball
H^4 union S^3.
Each hemisphere is the compactification of a hyperbolic plane.
Consider a round 4 ball with radius R(t) increasing linearly with time and with center not lying on one of the three hyperbolic planes. Let T be the last of the times when it touches one (say N) of the three planes L. M, N. Then there is a hyperplane Q tangent to the ball B(T) at the point of tangency with  the plane 
N. There's an isotopy, within the class of totally geodesic planes which 
are disjoint from L and M,
of  the plane N within the hyperplane Q which ends at a plane P' which is is separated from L and M  by a hyperplane disjoint 
from L, M and P'. Therefore a link of 3 round circles which have linking 
numbers zero is split. (In fact by a further isotopy one can see that the link is the trivial link.) So the Borromean rings is not a link of round circles.

I heard this at MSRI in 1983. I think the argument is Freedman's. I don't 
think reflection groups have anything to do with it, though there may be a
variant argument.

Geoffrey Mess
Department of Mathematics, UCLA.

From:  (Daniel A. Asimov)
Newsgroups:     sci.math.research
Subject:        Arrangements of Geometric k-Spheres in R^(2k+1)
Date:           Tue, 10 Jan 1995 21:45:22 GMT
Organization:   NAS - NASA Ames Research Center, Moffett Field, CA

(This is a generalization of a problem I posed around two years ago.)

Throughout this article, let n = 2k+1, where k is some fixed integer > 0;  

Define a k-hoop in R^n to be any geometric k-sphere of radius 1 in R^n. 

Denote the space of all k-hoops in R^n by H(k).  The space H(k) is endowed with
a natural topology (as the product of R^n and the Grassmannian G(k,n) of 
k-planes in R^n).

Define an r-hoople to be a set of r disjoint k-hoops.

Denote the space of all such r-hooples by H(k;r).  This space H(k;r) also has
a natural topology on it (from the deleted rth Cartesian power of H(k), which 
is then factored out by the symmetric group S(r)).

		*		*		*

QUESTION:  How many arcwise-connected components does H(k:r) have, and what is 
a representative r-hoople for each one?

		*		*		*

For a simple case, what is the answer for k = 2, r = 3 ???  (I don't know.)

The original question considered the case k = 1, r = 3, and was answered 
empirically rather than with a proof.  The apparent answer here is that there 
are exactly 5 components, represented by the following cases:

(The word "hoop" is used here to mean a 1-hoop.  In R^3, of course.)

a) three mutually unlinked hoops,
b) two linked hoops and the third far away, 
c) a chain of three hoops, 
d) a circular chain of mutually linked hoops, and 
e) left to the reader as a puzzle.

All circular chains of three hoops seem to lie in the same component of H(1;3)
(and there is a theorem that the Borromean rings cannot be represented by
round circles).

The case k = 1, r = 4 has been studied by Stein Kulseth of Norway, who came up
with the empirical result of 33 distinct components (last I heard).

Dan Asimov
Mail Stop T27A-1
NASA Ames Research Center
Moffett Field, CA 94035-1000
(415) 604-4799 w
(415) 604-3957 fax