From:geoff@math.ucla.edu (Geoffrey Mess)Newsgroups:sci.mathSubject:Re: Borromean Rings don't exist!Date:5 Mar 1995 03:31:18 GMTOrganization:University of California, Los Angeles

In article <3j8ait$326@jupiter.SJSU.EDU> alperin@sjsumcs.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. geoff@math.ucla.edu

From:asimov@nas.nasa.gov (Daniel A. Asimov)Newsgroups:sci.math.researchSubject:Arrangements of Geometric k-Spheres in R^(2k+1)Date:Tue, 10 Jan 1995 21:45:22 GMTOrganization: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 asimov@nas.nasa.gov (415) 604-4799 w (415) 604-3957 fax