What started as a simple matter of string tying has ended in a quagmire of topology and polynomials, far beyond my education; so I’ll throw it out here to anyone interested:
Doubtless you’ve seen, even if you didn’t know the name, the Borromean Rings, where you have three rings no two of which are linked to each other but the assembly as a whole is linked. My dilemma is as follows:
Preserving the central part where the three rings intertwine, let’s suppose that each ring is cut at its outermost part away from the epicenter. Is there any way to cross and join the loose ends together such that the result will be a single “unknot”- topologically equal to an unknotted loop?
The thing is, I thought I accomplished this once by random chance, but could never reproduce it- unless I was wrong about starting with true Borromean rings. If someone can either do it or confidently say it’s impossible, I’d be grateful.
Picture 1: Starting point.
Picture 2: Open the loops.
Picture 3: Uncross the three crossings.
Picture 4: Expand the central triangle.
Picture 5: Uncross the remaining crossings.
It is not clear what you are proposing. Call the rings A, B, C. Open each and let the ends be A1, A2, B1, B2, C1, and C2. If you insist on finally rejoining A1 to A2, etc., then the answer is certainly not. There were three separate pieces at the beginning and there will be three at the end no matter what you do in between. The mathematical statement is that the number of connected components is a topological invariant and your cut and paste is a homeomorphism (continuous along with its inverse).
On the other hand, if you are allowed to join A1 to B2, B1 to C2, C1 to A2, do so at random. The result will likely be knotted but no problem. Simply cut it somewhere, unknot it and repair the cut. Or simply unknot it before you make the last join.
The link worked for me a couple hours after it was posted. Doesn’t work now.
The key point of markn+'s decomposition was that once, per OP, you cut each loop out at the periphery what you now have topologically speaking is three line segments. Which you are free to slide back and forth or to to cross over or under each other at will.
So simply recross them over so as to undo the existing crossovers. That leaves three disjoint uncrossed segments laying on your table.
Then slide the ends around on your table until you have the three segments arranged roughly as 3 segments of a single circle. Then connect the ends together (A2 to B1, B2 to C1, C2 to A1 in Hari’s notation) to form a single closed circle. QED.
So as markn+ said, the problem is either trivial or the OP is imagining some limitations on how the three segments can be legally rearranged. Limitations that he left out of his OP.
Yeah, sorry about that. It turns out that my hosting provider changed my DNS to point to a different server slightly after the time that I uploaded the photo to the (old) server. I’ve copied it to the new server so it should work again now.
