Re: [math-fun] linked spheres
Fred wrote: << On 10/20/09, Dan Asimov <dasimov@earthlink.net> wrote: << Linking is a fascinating phenomenon in topology, but the only chance of proving anything about it will require the use of a definition of linking.
Which is not supplied; although at a later stage (ex 5.8) a definition is requested! It turns out that for standard k-spheres, a simple and intuitive criterion is available.
Yes. In fact, for disjoint round spheres S^p and S^q in R^n where p+q+1 = n, they link exactly when there exists a line L in R^n that intersects each sphere twice in alternating order. (In other words, L intersects the spheres in two linked 0-spheres.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On 10/20/09, Dan Asimov <dasimov@earthlink.net> wrote:
... Yes. In fact, for disjoint round spheres S^p and S^q in R^n where p+q+1 = n, they link exactly when there exists a line L in R^n that intersects each sphere twice in alternating order.
(In other words, L intersects the spheres in two linked 0-spheres.)
--Dan
Completely different from the criterion I had in mind --- that they are not linked just when there exists a prime disjoint from and separating them! And Dan's criterion does yield an algorithm --- the line in question simply joins their centres. WFL
A "prime" is a subspace of codimension 1 --- at any rate, when flat. I've never hit upon nomenclature for these apparently common concepts that seems entirely satisfactory. "Hyperplane", though more popular, is etymological nonsense. "0-flat" for point, "1-flat" for line, etc is often useful when the dimension of space is fixed; but becomes clumsy when instead the subspace codimension is specified. "Coline" seems to work well for codimension 2; but "copoint" for 1? Nah! And when the space under discussion is a space of spheres rather than points, clear unambiguous nomenclature becomes twice as difficult ... WFL
Dan has very politely pointed out that I've been talking through my hat again. To start with, there are unlinked cicles in 3-space which are not separated by a plane (prime), which demolishes my proposed "criterion". Secondly, the line in Dan's criterion is the meet of the planes in which the respective circles lie --- what I had in mind as a decision "algorithm". Thirdly, in general a k-sphere and l-sphere lie in a (k+1)-flat and (l+1)-flat, which again meet in a line in (k+l+1)-space: there's no need for any induction. Fred Lunnon On 10/20/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/20/09, Dan Asimov <dasimov@earthlink.net> wrote:
...
Yes. In fact, for disjoint round spheres S^p and S^q in R^n where p+q+1 = n, they link exactly when there exists a line L in R^n that intersects each sphere twice in alternating order.
(In other words, L intersects the spheres in two linked 0-spheres.)
--Dan
Completely different from the criterion I had in mind --- that they are not linked just when there exists a prime disjoint from and separating them!
And Dan's criterion does yield an algorithm --- the line in question simply joins their centres. WFL
Two nits ... (a) There's an edge case in the intersecting-plane test: The two circles could lie in the same plane. In this case, they might intersect (or not), or be tangent, but they can't be linked. This isn't a theory problem, but it might be a programming problem: Computing the intersection of two nearly-parallel planes is a numerical challenge. We could have two circles that have no chance of linkage, because they are moderately far apart, but still cause a "can't tell" in the linked-p subroutine. (b) Fred's separability test seems to work if we allow spheres as separators (instead of just allowing planes). Rich --------- Quoting Fred lunnon <fred.lunnon@gmail.com>:
Dan has very politely pointed out that I've been talking through my hat again.
To start with, there are unlinked cicles in 3-space which are not separated by a plane (prime), which demolishes my proposed "criterion".
Secondly, the line in Dan's criterion is the meet of the planes in which the respective circles lie --- what I had in mind as a decision "algorithm".
Thirdly, in general a k-sphere and l-sphere lie in a (k+1)-flat and (l+1)-flat, which again meet in a line in (k+l+1)-space: there's no need for any induction.
Fred Lunnon
On 10/20/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/20/09, Dan Asimov <dasimov@earthlink.net> wrote:
...
Yes. In fact, for disjoint round spheres S^p and S^q in R^n where p+q+1 = n, they link exactly when there exists a line L in R^n that intersects each sphere twice in alternating order.
(In other words, L intersects the spheres in two linked 0-spheres.)
--Dan
Completely different from the criterion I had in mind --- that they are not linked just when there exists a prime disjoint from and separating them!
And Dan's criterion does yield an algorithm --- the line in question simply joins their centres. WFL
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On 10/21/09, rcs@xmission.com <rcs@xmission.com> wrote:
Two nits ...
(a) There's an edge case in the intersecting-plane test: The two circles could lie in the same plane. In this case, they might intersect (or not), or be tangent, but they can't be linked. This isn't a theory problem, but it might be a programming problem: Computing the intersection of two nearly-parallel planes is a numerical challenge. We could have two circles that have no chance of linkage, because they are moderately far apart, but still cause a "can't tell" in the linked-p subroutine.
Sure --- I've implicitly been considering things "in general". The planes might be coincident, or distinct but parallel. There are also limiting cases where "circles" and "spheres" include lines and planes resp, or degenerate to points or (inversive) infinity. The circles might have one or two common points, be tangent, coincide ...
(b) Fred's separability test seems to work if we allow spheres as separators (instead of just allowing planes).
Sketch proof of (b) --- which was also suggested privately by Dan --- If neither circle meets the plane of the other, any other plane containing the line L where their planes meet will separate them. Otherwise, consider the coaxal system of circles through the pair of points (in general) where one circle A meets L. Choose a circle C of this system which does not meet the other circle B --- this is possible provided A and B do not both meet L in pairs of points which overlap. Construct the unique sphere containing A and C, then offset its radius down or up, depending upon whether B lies within or outside C. The new sphere S separates A and B. Conversely, if A and B do both meet L in pairs of points which overlap, any sphere separating A and B would have to meet the plane of B in circle C, with C disjoint from B, and C separating B from both points of A --- which I'll take to be evidently impossible --- though if put up against a wall, I'd have to admit to a certain vagueness about exactly what my axioms might be here ... Fred Lunnon
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