Let $C_N$ denote the labelled configuration of $N^{th}$ roots of unity with $p_J = e^{\frac{2\pi iJ}{N}}$ for $J = 1\ldots N$.

As a corollary of something else I was playing around with, I recently proved the following:

Theorem: Every tame knot (or link) $K$ has a (not necessarily minimal) stick presentation which such that the sticks can be projected onto the set of chords of some $C_N$ with the following crossing condition: Whenever the projection has chords $p_{J_1}p_{J_3}$ and $p_{J_2}p_{J_4}$ such that $1\leq J_1 < J_2 < J_3 < J_4\leq N$ then $p_{J_1}p_{J_3}$ crosses in front of $p_{J_2}p_{J_4}$. (In other words for any two intersecting chords in the projection, the chord which has the lowest numbered endpoint passes in front of the other chord).

Germane to this question is the fact that the theorem leads to a knot invariant which, for lack of a better name, I will call the circular stick number of $K$ and which is defined to be the minimum $N$ needed to obtain a projection of $K$ with the above properties.

My proof of the above theorem was very non-constructive, so I wanted to see some concrete realizations of such projections. And after some work, I was able to find that for the left trefoil knot, the circular stick number is 7. After much more playing around on $C_7$ and not finding a projection of the right trefoil, I moved on to $C_8$ where I was able to find a projection of the right trefoil.

Main Question: Does this invariant actually distinguish chiralities of the trefoil, or can someone find a projection of the right trefoil with the above properties on $C_7$? If it doesn't distinguish chiralities in this case, is it possible it distinguishes chiralities of some other pair of knots, or does anyone see a slick way to see that it cannot distinguish chiralities?

Secondary Question: (Mainly for knot theorists, or anyone with a deeper knowledge of knots than I) Has this invariant been studied before, and if so, what is the terminology used for it?

I am fairly sure that the general answer to the main question is strongly related to the characterization (and in particular the disjointness) of the sets of forbidden minors of the following two sets (which are almost certainly not known in general as the Robertson-Seymour theorem is non-constructive):

$\bullet$ Graphs which are $L$-lessly embeddable for a given chiral knot $L$

$\bullet$ Graphs which are $L^+$-lessly and $L^-$-lessly embeddable for a given chiral knot $L$ with chiralities $L^+$ and $L^-$

Note that graphs in the second set may admit embeddings in $\mathbb{R}^3$ containing either $L^+$ or $L^-$, but have at least one embedding which does not contain both $L^+$ and $L^-$. As of yet, I have not been able to flesh out a proof using this approach however.

For those who don't want to search for the solutions I found, for the left trefoil the points of $C_7$ are connected in the following order:

$p_1 \rightarrow p_3 \rightarrow p_5 \rightarrow p_7 \rightarrow p_2 \rightarrow p_4 \rightarrow p_6 \rightarrow p_1$

For the right trefoil the points of $C_8$ are connected in the following order:

$p_1 \rightarrow p_3 \rightarrow p_7 \rightarrow p_5 \rightarrow p_2 \rightarrow p_8 \rightarrow p_4 \rightarrow p_6 \rightarrow p_1$

EDIT: Here are pictures of these two projections to help clarify the situation:


With regards to Dylan's question about how this projection arose, I was considering the parametric curve $S(t) = (t,t^2,t^3) \subset \mathbb{R}^3$. In a comment or answer on a past MO post which for the love of me I cannot find now, someone had shown in a very simple way that no two chords of $S$ intersect one another anywhere in $\mathbb{R}^3$ (unless they share an endpoint). Thus, for each $n$, $K_n$ can be embedded in $\mathbb{R}^3$ as the set of chords connecting the points $(1...n)$. Now for a fixed tame knot or link $K$, the Robertson-Seymour theorem implies there is a finite set of forbidden minors for graphs which are $K$-lessly embeddable. Hence, every embedding of $K_n$ for $n$ sufficiently large contains $K$, so one may ask (like I did) "what $n$ is sufficient for a given $K$, and what order must I connect the points $(1...n)$ in order to realize $K$?" The crossing condition comes from looking at the chords projected onto the $yz$-plane as viewed from the $-x$ direction (which I think is the direction I did my crossing calculations from). Finally, I moved the $yz$-projections of the integral $t$-valued points around so that they lay on the unit circle to make the pictures easier/clearer.

UPDATE: I was just skimming through a paper on some results on knots which were unrelated to this one. That paper referenced the Ramsey number $r(L)$ of a link $L$, and upon following its references I found this paper which proves the existence of $r(L)$ using essentially the representation I layed out above:


So it would appear the answer to Question 2 is this is known (in at least one paper) as a plat representation (much to my surprise, Negami derives the existence of such representations using essentially the argument I gave above). So that would seem to settle this thread entirely...

  • $\begingroup$ Interesting. I think the figure I drew for mathoverflow.net/questions/95477 is accidentally an illustration? If I am not misinterpreting, then perhaps the relationship to torus knots might be helpful... $\endgroup$ Apr 30, 2012 at 15:45
  • $\begingroup$ Your illustration is close to what I am describing. Hopefully later this evening I will have time to create and upload some pictures of the two trefoils I described above. $\endgroup$
    – ARupinski
    Apr 30, 2012 at 23:27
  • $\begingroup$ Here's a quick proof of the non-intersecting chord result. Denote the moment curve by $\mu(t) = (t, t^2, t^3)$, and consider any four points $\mu(t_1), \mu(t_2), \mu(t_3), \mu(t_4)$ with $t_1 < t_2 < t_3 < t_4$. The volume of the convex hull of these points is $1/6$ the determinant of the matrix whose $i$th row is $[1, t_i, t_i^2, t_i^3]$. This determinant is an antisymmetric polynomial with total degree $6$ in the $t_i$'s, and thus must be $\prod_{i<j} (t_j-t_i) \ne 0$. The same result also holds for one turn of the standard helix $(t, \cos t, \sin t)$ where $0 \le t < 2\pi$. $\endgroup$
    – JeffE
    May 3, 2012 at 11:29
  • $\begingroup$ Isn't the non-trivial knot with $N=7$ a right-handed trefoil, not a left-handed trefoil? Or am I getting some convention backwards? $\endgroup$ May 3, 2012 at 15:36
  • $\begingroup$ Yeah, must have messed up that when I looked it up originally. But all my other knots' chiralities were computed by comparing to the reduced form I did of that one, so the chiralities are at least consistent (so if that one is wrong, all of the $N=8$ ones I calculated below will change chirality too). $\endgroup$
    – ARupinski
    May 4, 2012 at 1:45

1 Answer 1


I'm very curious where this came up. In any case, the answer to the first question is yes, it does distinguish these trefoils; you found the minimal representatives.

Let $a_0,\dots,a_{N-1}$ be the roots of unity that are visited along the knot, in (cyclic) order. Suppose we have a minimal representative for some non-trivial knot. Then we cannot have $|a_k - a_{k+1}| = 1$ for any $k$, as otherwise we could replace this pair $a_k, a_{k+1}$ by a single root of unity (for $N-1$), adjusting the other roots of unity as appropriate. A little more subtly, we cannot have $|a_{k-1} - a_{k+1}| = 1$ either, as then we could again delete $a_k$ from the sequence to get a smaller representation. With these simple constraints, the smallest possible sequence for a non-trivial knot is the one you found for one of the trefoils with $N=7$. There are several possibilities for $N=8$, including the one you found for the other trefoil. I've included a very short Haskell program below that computes this. The possibilities for $N=8$ are $$ (2,7,5,3,1,6,4,0)\quad (2,5,7,3,1,6,4,0)\quad (3,6,1,4,7,2,5,0)\quad (2,6,4,1,7,3,5,0) $$ $$ (3,1,6,4,2,7,5,0)\quad (2,4,6,1,3,7,5,0)\quad (3,5,1,7,4,2,6,0)\quad (4,2,7,5,1,3,6,0) $$ $$ (3,1,5,7,2,4,6,0)\quad (5,3,1,6,4,2,7,0)\quad (2,4,6,1,3,5,7,0) $$

For the second question, I have never heard of this representation before.

Here is the code, for anyone interested.

 -- A (partial) circular stick representation is a list of integers,
 -- the order of the roots of unity to visit in order
 type CircStick = [Int]

 -- The next element ak after a partial representation a1, ..., a{k-1} 
 -- must satisfy
 --   (a) ak has not already been seen
 --   (b) |ak - a{k-1}| > 1
 --   (c) |ak - a{k-2}| > 1
 -- There are a few more "easy" constraint, eg the first and last entries
 -- cannot differ by one.  We do not impose those constraint here.
 nexts :: Int -> CircStick -> [Int]
 nexts n [] = [0]
 nexts n [a1] = filter (\a -> abs (a-a1) > 1) [0..n-1]
 nexts n (a1:a2:as) =
   filter (\a -> not (elem a as)) $
       filter (\a -> abs (a-a1) > 1) $
   filter (\a -> abs (a-a2) > 1) $

 completions :: Int -> CircStick -> [CircStick]
 completions n as | length as >= n = [as]
 completions n as =
   concat [completions n (a:as) | a <- nexts n as]

 -- Impose final constraints:
 --    (a) Last entry cannot be 1
 --    (b) Take entry that is lexicographically less than its reverse
 --    (c) first and next-to-last entries cannot differ by one
 circSticks :: Int -> [CircStick]
 circSticks n = 
   filter (\as -> abs ((as!!0) - (as!!(n-2))) > 1) $
       filter (\as -> as < tail (reverse as)) $
   filter (\as -> head as /= 1) $
   (completions n [])

Edit: For those interested, here are the 108 possibilities for $N=9$. I hope there's some way of checking what these are more efficiently than just going through them by hand.


  • $\begingroup$ 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 $|a_k-a_{k-2}|$. $\endgroup$
    – ARupinski
    May 1, 2012 at 0:16
  • 1
    $\begingroup$ 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 $K = 9$) $\endgroup$
    – ARupinski
    May 2, 2012 at 0:23
  • 1
    $\begingroup$ $\mod 8$ 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 $|a_k - a_{k-2}| = 1$. $\endgroup$
    – ARupinski
    May 3, 2012 at 1:06
  • 1
    $\begingroup$ 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. $\endgroup$
    – ARupinski
    May 9, 2012 at 0:04
  • 1
    $\begingroup$ 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. $\endgroup$
    – ARupinski
    May 9, 2012 at 0:08

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