Timeline for Does this knot invariant distinguish trefoil chiralities?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 11, 2012 at 12:27 | comment | added | Dylan Thurston | Thank you for both pointing out the mistake in both my example and giving the general argument; I agree now. | |
| May 9, 2012 at 0:08 | comment | added | ARupinski | Your figure 8 example is actually a right trefoil (in my original notation, which seems to be the opposite of the standard convention); contracting the 0-8 edge reduces it to the fourth knot in the second row of your $N=8$ list. | |
| May 9, 2012 at 0:04 | comment | added | ARupinski | Suppose $0$ and $N-1$ are connected by a chord and let C be the other chord incident to $N-1$. It is easy to see that C must cross behind any chord it intersects (in the projection). Contracting the chord between $0$ and $N-1$, C becomes a chord incident to $0$ (and thus should cross in front of, not behind, any other chord it intersects). But this is not a problem; since C crossed behind all other chords, it can be slid around the figure to the front to become a valid chord incident to $0$. Thus the chord between $0$ and $N-1$ contracts. | |
| May 4, 2012 at 2:50 | comment | added | Dylan Thurston | The edge does not contract. Please think things through more. | |
| May 4, 2012 at 1:49 | comment | added | ARupinski | Right, but even with the index shift, 0 and N-1 are still adjacent points on the circle, correct? So shouldn't the edge between them contract to the 0 point on $N-1$ vertices? Or am I misinterpreting how your Haskell program is looking at the data? | |
| May 3, 2012 at 15:33 | comment | added | Dylan Thurston | To reduce, you really need $|a_k - a_{k-1}| = 1$, not just that it's equivalent $\textrm{mod} N$. For instance, on the list with $N=9$, I find that $(6,3,1,5,7,2,4,8,0)$ represents the figure 8 knot, which according to your calculations was not representable with $N=8$. One point that is possibly confusing is that I switched the indexing, so that the points run from $0$ to $N-1$ rather than $1$ to $N$. | |
| May 3, 2012 at 1:06 | comment | added | ARupinski | $\mod 8$ they differ by 1, and in fact that is enough to reduce 5 of the cases on your list to the $N = 7$ case via one of the two conditions $|a_k - a_{k-1}| = 1$ or $|a_k - a_{k-2}| = 1$. | |
| May 2, 2012 at 11:39 | comment | added | Dylan Thurston | I don't think adjacent 0-7 pairs reduce, do they? In any case, I added the $N=9$ possibilities. | |
| May 2, 2012 at 11:38 | history | edited | Dylan Thurston | CC BY-SA 3.0 |
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| May 2, 2012 at 0:23 | comment | added | ARupinski | With regards to Mariano's question, I just checked the 11 you listed... along the rows they are: Trivial Right Left Right; Trivial Trivial Right Right; Trivial Left Left. The last two reduce to the $N = 7$ case because of the adjacent 0-7 pairs. So nothing new here. I would suspect that with $N = 9$ or $10$ one could obtain the figure 8 knot (with a little bit of playing I found at least one of the cinquefoil knots on $K = 9$) | |
| May 1, 2012 at 16:23 | comment | added | Dylan Thurston | I haven't checked them all, but several seem to be unknots. Maybe there are other conditions on these variables besides the ones I found above. | |
| May 1, 2012 at 4:46 | comment | added | Mariano Suárez-Álvarez | What are the other knots with $N=8$? | |
| May 1, 2012 at 0:16 | comment | added | ARupinski | Nice slick proof... I had used the fact that $|a_k-a_{k-1}|\neq 1$ to narrow down the search, but I hadn't noticed the condition on $|a_k-a_{k-2}|$. | |
| Apr 30, 2012 at 23:23 | vote | accept | ARupinski | ||
| Apr 30, 2012 at 18:13 | history | answered | Dylan Thurston | CC BY-SA 3.0 |