From: Douglas Zare Subject: Re: Hypersphere Cross-Section Date: Wed, 22 Sep 1999 21:05:54 -0400 Newsgroups: sci.math Keywords: Reeb foliation of S^3 Kevin Foltinek wrote: > In article <8e77fxsprcg8@forum.swarthmore.edu> qqquet@hotbot.com > (Leroy Quet) writes: > > > I heard somewhere that a 3-dimensional cross-section of the 4-sphere > > is a torus. > > The equation of the 4-sphere is r^2=x^2+y^2+z^2+t^2. > > This equation describes the 3-sphere S^3 (embedded in R^4). > > I'm embarassed to admit that I've forgotten the name of this > construction and the details, but isn't there something like a > "foliation" of the 3-sphere by 2-tori? (I think it's actually a pair > of foliations of disjoint pieces of the 3-sphere, each of the two > pieces being a solid torus.) No (is DA about?), but you mean the Reeb foliation of S^3. Incidentally, there should still be a nice painting of this on the 9th floor of Evans Hall of UC Berkeley, IIRC. In the Reeb foliation, there is one torus leaf, and the other leaves are topological disks which fill two solid tori. The disks can be taken to be perpendicular to the core of each solid torus, but bending so that they spin infinitely about the solid torus in one direction, so that they are close to flat near the torus. In other words, one can spin the graphs of sec(x)+c between -pi/2 and pi/2 about the y-axis, then quotient out by a translational symmetry. Any time one has a circle transverse to a 2-dimensional foliation, one can "spoil" the foliation by replacing a neighborhood of the circle by a neighborhood of one solid torus of the Reeb foliation. Now people primarily study foliations which do not have such solid tori. I've been told that Reeb was originally given the thesis problem of showing that there was no 2-dimensional foliation of S^3. Douglas Zare ============================================================================== From: Chris Hillman Subject: Re: Hypersphere Cross-Section Date: Tue, 28 Sep 1999 21:00:24 -0700 Newsgroups: sci.math Keywords: foliation of S^3 by Hopf tori On 22 Sep 1999, Kevin Foltinek wrote: > In article <8e77fxsprcg8@forum.swarthmore.edu> qqquet@hotbot.com > (Leroy Quet) writes: > > > I heard somewhere that a 3-dimensional cross-section of the 4-sphere > > is a torus. > > The equation of the 4-sphere is r^2=x^2+y^2+z^2+t^2. > > This equation describes the 3-sphere S^3 (embedded in R^4). > > I'm embarassed to admit that I've forgotten the name of this > construction and the details, but isn't there something like a > "foliation" of the 3-sphere by 2-tori? (I think it's actually a pair > of foliations of disjoint pieces of the 3-sphere, each of the two > pieces being a solid torus.) Someone else suggested the Reeb foliation, but I suspect that you have in mind something much simpler, the foliation of S^3 by Hopf tori. It is not hard to see that you can parametrize S^3 as a surface in E^4 by (cos w cos u, cos w sin u, sin w cos u, sin w sin u) To verify this, take the norm and notice that it has constant value! Next, differentiate to obtain the tangent vectors d/dw = (-sin w cos u, -sin w sin u, cos w cos u, cos w sin u) d/du = (-cos w sin u, ... ) d/dv = (....) Now take the pairwise inner products to obtain the metric matrix, which gives the line element of S^3 in Hopf coordinates (very useful!) ds^2 = dw^2 + cos^2 w du^2 + sin^2 w dv^2 Now observe that surfaces of constant w are flat tori T^2; for instance w = w0 can be cut apart to give a rectangle with horizontal edge length cos w0 and vertical edge length sin w0. This Hopf tori are pairwise linked! The degenerate cases w = 0, w = pi are a pair of linked great circles: every other Hopf torus passes "between" these circles. The "equatorial" Hopf torus is the one which can be made from a square. The others are all "rectangular". If you're having a hard time picturing this, take a tetrahedron and consider a pair of "opposite edges" (notice they are "parallel" but run in mutually orthogonal directions, so to speak). Slice the tetrahedron into rectangles "parallel" to this pair of edges. The central rectangle should be a square. Now imagine gluing together the pair of faces of the tetrahedron which meet in one of the pair of edges. (Imagine mentally rotating one face around the edge to meet the other--- that's how you want to glue them.) Then do the same thing with the remaining pair of faces. Result: a topological sphere S^3, but with the curvature concentrated in two linked circles. Still, this is a pretty fair model of the Hopf tori; indeed your rectangles with be metrically flat tori as they should be. However, the way "adjacent" tori relate to one another is not quite right--- the curvature should be spread out uniformly instead of being concentrated in the two linked circles formed from the original pair of edges. In the very important geometric Hopf map S^1 = SO(2) >--> S^3 = SU(2) -->> S^2 the Hopf tori are the inverse images of latitude circles on S^2. The inverse images of points are the Hopf circles, also called Clifford parallels, which form a system of great circles which fill up S^3 in such a way that each pair is linked. In the Hopf torus with horizontal edge cos w0 and vertical edge sin w0 draw the obvious "diagonal". This is one of the Hopf circles which lies in that Hopf torus. The others are translates of this one. In several books you can find nice pictures of the Hopf tori and Hopf circles; see for instance the article by Roger Penrose in Mathematics Today, ed. by Lynn Arthur Steen, Basic Books, 1978 or thereabouts. The geometric Hopf map can be defined using quaternions and the fact that if you identify E^4 with the algebra of real quaternions, then S^3 is the group of quaternions of unit norm, the analogue of the unit circle for the complex numbers. I've discussed this beautiful object many times in the past (try looking up posts by me to sci.physics.relativity on the appearance of the night sky), so I won't repeat myself any further here. The Reeb foliation, by the way, is something much complex which one can get by modifying the Hopf fibration (which is a really "nice" foliation!), in a way which I don't feel up to describing in words right now, other than to suggest that you imagine taking an inner tube, cutting it crosswise, and imagine taking an inner tube of a slightly smaller radius, likewise cut crosswise, and gluing the two tubes so that you now have a tube which continues "inside itself". Repeat infinitely often. This gets you started on the way to constructing the Reeb foliation. I think there might be a picture of it in the book by George Francis, A Topological Picture Book, in addition to the famous murals in Evan's Hall. Chris Hillman Home Page: http://www.math.washington.edu/~hillman/personal.html