You're right --- it must have been the Mathworld page, which I think has been revamped rather unsuccessfully. Ralph Martin at Cardiff has repeated my computations using Mathmatica, and to my chagrin he finds 4 nodes as classically predicted. I haven't so far managed to establish whether the discrepancy results from Maple's incompetence or mine. What concerns me more than that is the implications for invariants --- if there are two pairs of conical nodes, there should be two angles (one or both imaginary / hyperbolic) associated with them, and invariant under the Moebius group; but I cannot see how to associate any (real) geometric feature with the second angle. Similarly, there should be two associated (tangent) lengths, invariant under the Laguerre group; once again --- bearing in mind that the generating spheres are now oriented --- I can find only one candidate. And since there are but 9-6 = 3 degrees of freedom for shape and size, there has to be a constraint relating these 4 quantities --- what could this be? The whole business has me tied in knots at the moment, I must admit. Fred Lunnon On 7/22/11, Allan Wechsler <acwacw@gmail.com> wrote:
I can't contribute to the mathematics; at least, I can't do it easily, without a lot of studying. But I did take ten minutes to page through *all* previous revisions of the Wikipedia article, and I can see no sign that it ever sported any graphic other than the one it now has. Perhaps you are conflating your memory of the Wikipedia article with another source.
On Fri, Jul 22, 2011 at 12:55 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
I have read in a number of places --- though now I come to check up, I'm having difficulty tracking down a definitive source --- that the quartic surface known as the "cyclide of Dupin" has 4 nodes: conical singularities at which the gradient vector vanishes.
Furthermore, it is averred that one pair of these may may be real or may be complex; the other pair is always complex. Like the trusting innocent I am, I have for years cheerfully accepted this scenario, without ever properly understanding why it might --- or might not --- obtain.
The first pair can easily be seen in action in spindle or horn (real pair) versus ring (conjugate complex pair) cyclides --- at least, they could easily be seen before some brainstorm overtook the two pages at http://en.wikipedia.org/wiki/Dupin_cyclide http://mathworld.wolfram.com/Cyclide.html which I was sure used to show informative plots of the various forms!
Pinning down the second pair is more problematic. In the special "parabolic" case --- an infinite cubic "full twist" surface --- there are indeed 4 complex nodes: for example differentiating y(x^2 + y^2 + z^2 - 1) - 2 x z = 0 wrt x,y,z and solving, we find solutions at [x, y, z] = [i, 1, i], [-i, 1, -i], [-i, -1, i], [i, -1, -i], where as usual i^2 = -1.
However, in the general quartic case, Maple is quite insistent that there is only a single pair of solutions; and in consideration of the contradictions in which my earlier credulity was beginning to embroil me, I am beginning to think that Maple might just be right about this.
In the case of the torus in standard postion with central and tube radii p and q, we (at least, Maple and me) find (x^2 + y^2 + z^2 - p^2 - q^2)^2 - 4 p^2(q^2 - z^2) = 0 has just the two nodes [x, y, z] = [0, 0, +sqrt(q^2-p^2)] , [0, 0, -sqrt(q^2-p^2)] . Inverting this with centre x = e say, we get a general shape with again just two nodes [x, y, z] = [e(e^2-1+q^2-p^2), 0, (+/-)sqrt(q^2-p^2)] / (e^2+q^2-p^2) unless e^2 = p^2 - q^2 (parabolic case).
Can anyone cast light on my confusion? Fred Lunnon
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun