Strong and weak (1, 2) homotopies on knot projections and new invariants

Fig. 6 proves the claim.

∎

It is easy to see the canonical map $\tau(P)\mapsto|\tau(P)|$ by counting the number of circles in $\tau(P)$. In Fig. 20, $|\tau(3_{1})|$ $=$ $|\tau(6_{3})|$ $=$ $3$ but $\tau(3_{1})$ $\neq$ $\tau(6_{3})$. ∎

1. (1)

   By the definition, $|\tau(O)|$ is $1$. It is well known that an arbitrary knot projection and the simple circle $O$ can be related by a finite sequence of the first, second, and third flat Reidemeister moves, where the first and second flat Reidemeister moves are defined by Fig. 1 and the third flat Reidemeister move is defined by Fig. 9.

   

   We have already shown that $|\tau(P)|$ is invariant under RI and strong RI​I. Thus, it is sufficient to show that $|\tau(P)|$ $\equiv$ $0$ (mod $2$) under weak RI​I and the third flat Reidemeister move.

   First, let us consider weak RI​I by looking at Fig. 10. For a knot projection $P$ with an orientation (the upper-left of Fig. 10), we have $\tau(P)$ appearing as in the left figure of the second row. We consider the possibilities of the connections of arcs of finitely many simple circles which we focus on. We can draw them as locally two types of $\tau(P)$, as in the upper right of the first and second rows in Fig. 10. These two figures imply that $|\tau(P)|-|\tau(P^{\prime})|$ is either $\pm 2$ or $0$ if $P$ and $P^{\prime}$ is related by a single weak RI​I.

   

   Next, we consider the third flat Reidemeister move. The third flat Reidemeister move can be split into the two kinds of moves by the way of connection of three branches as in Fig. 9 (see Fig. 11).

   

   The upper (resp. lower) local move of Fig. 11 is called the strong (resp. weak) RI​I​I.

   These two local moves can be detected easily by giving an orientation to a knot projection. In fact, one appears in the left of the third line and the other appears in the fourth line of Fig. 10 if any orientations of knot projections are given.

   We apply non-Seifert resolutions to the two types of the third flat Reidemeister move (the left column in the third and fourth lines of Fig. 10). Therefore, it is sufficient to consider to the two cases of connections of three arcs that are parts of simple circles (the bottom line of Fig. 10). The two figures in the bottom line of Fig. 10 shows that $|\tau(P)|-|\tau(P^{\prime})|$ is either $0$ or $\pm 2$ if a knot projection $P$ is related to the other knot projection $P^{\prime}$ by a single third flat Reidemeister move. This completes the proof of (1).

2. (2)

   For two knot projections $P_{1}$ and $P_{2}$, we give them any orientations, and denote by $P_{1}^{o}$ and $P_{2}^{o}$ these oriented knot projections. The knot projection $P_{2}$ with the opposite orientation is denoted by $P_{2}^{\bar{o}}$. Now we consider a connected sum $P_{1}\sharp P_{2}$ of $P_{1}$ and $P_{2}$. We can preserve the orientation of $P_{1}^{o}$ under this connected sum by choosing either $P_{2}^{o}$ or $P_{2}^{\bar{o}}$. Note that the $\tau(P_{2}^{o})$ $=$ $\tau(P_{2}^{\bar{o}})$. Apply non-Seifert resolution at all double points $P_{1}\sharp P_{2}$. The way of smoothing them is the same as those for $P_{1}$ and $P_{2}$ since we consider either $P_{1}^{o}\sharp P_{2}^{o}$ or $P_{1}^{o}\sharp P_{2}^{\bar{o}}$.

3. (3)

   If there exists at least one part as one of the list, we have at least two circles in $\tau(P)$. By (1), we have $|\tau(P)|\geq 3$.

∎

The assumption is that there exists a knot projection $P$ such that $P$ is equivalent to the simple circle $O$ up to weak (1, 2) homotopy (i.e. $P$ $\sim_{w}$ $O$) and $P$ is also equivalent to $O$ up to strong (1, 2) homotopy (i.e. $P$ $\sim_{s}$ $O$). The condition that $P$ $\sim_{s}$ $O$ implies that $|\tau(P)|$ $=$ $1$. On the other hand, $P$ $\sim_{w}$ $O$ implies that there exists a finite sequence ($\ast$) of $1a$ and $w2a$ from $O$ to $P$ by Lemma 1.

Now, we assume that the sequence ($\ast$) contains at least one $w2a$ ($\star$). Therefore, there exists a knot projection $P^{\prime}$ with at least one incoherent $2$-gon such that $P^{\prime}\stackrel{{\scriptstyle 1}}{{\sim}}P$. Then, $|\tau(P^{\prime})|$ $=$ $|\tau(P)|$ $=$ $1$. By Theorem 4 (3), $|\tau(P^{\prime})|$ $\neq$ $1$ is a contradiction. Thus, the assumption ($\star$) is false and then the sequence ($\ast$) contains only $1a$. Then $P$ $\stackrel{{\scriptstyle 1}}{{\sim}}$ $O$.

Conversely, assume that $P\stackrel{{\scriptstyle 1}}{{\sim}}O$. Then, in particular, $P$ satisfies both $P\sim_{w}O$ and $P\sim_{s}O$. This completes the proof of Theorem 2. ∎

It is easy to see this proof by the definition of the circle arrangement. ∎

We pick up the innermost circle in a nested collection of circles and choose one of them arbitrarily, denoting it $C$. If the circle $C$ is not encircled as in the right figure of Fig. 13, then there should exist at least one circle as in the left figure of Fig. 13. Thus, the proof is completed. ∎

If a knot projection $P$ is a simple curve, it is easy to see that $4_{1}$ projection (cf. Fig. 20) is obtained by $1a$ and $s2a$. In the following, we assume that a knot projection is a non-trivial knot projection. We consider a prime decomposition as Fig. 14. If a knot projection $P$ has a non-trivial sub-curve such that the sub-curve has exactly two end points and if we can find a simple arc closing the two endpoints of the curve that produces a prime knot projection and where the closing does not create double points, then the non-trivial sub-curve is called a prime part of $P$. Let us encircle the prime part at one of the most inner regions by one circle, called a red circle, where we arbitrarily select an innermost prime part. If we replace the prime part with a simple arc, we precede to the next stage and repeat; we encircle the prime part of one of the innermost regions by a red circle that must not intersect the other red circles. We continue to draw red circles not intersecting the other red circles. Finally, we draw a circle whose part is contained by the outermost region and select a next-innermost prime part and repeat the above process. For each step, if the outermost red circle is already given by the former process, we omit drawing it (e.g., Fig. 14). Since the number of double points in a knot projection is finite, every encircling must eventually stop. As a result, we can choose a decomposition where each red circle $r$ contains a prime part where we replace prime parts $t_{1},t_{2},\dots,t_{m}$ with simple arcs $a_{1},a_{2},\dots,a_{m}$ if the other red circles $r_{1},r_{2},\dots,r_{m}$ are contained in $r$.

Let us look at Fig. 15. Every red circle can be drawn as in Fig. 15. Call the two arcs of Fig. 15 marked arcs. Every arc of the sub-diagram within the red circle of Fig. 15 is called a prime part’s arc. Choose two non-marked arcs $\alpha_{1}$ and $\alpha_{2}$ in the same region where one is picked from a prime part’s arc $\alpha_{1}$ and the other is picked from a distinct neighboring prime part’s arc $\alpha_{2}$. Apply the operation shown in Fig. 16 to arcs $\alpha_{1}$ and $\alpha_{2}$. This operation is one $s2a$ or a pair of $s2a$ and $1a$; if two arcs $\alpha_{1}$ and $\alpha_{2}$ cannot make a coherent $2$-gon directly, we make the allowable situation by applying one $1a$ (see Fig. 16).

As a result, the relation of the two neighboring prime parts is not a connected sum and so we should delete the red circle between the two prime parts. We apply this operation to every red circle, and so we can eventually resolve all the connected summations to obtain a prime knot projection. Thus, the proof is completed. ∎

If $|\tau(P)|\leq 2$, $|\tau(P)|=1$ by Theorem (1). We have $|\tau(4_{1})|=1$ (the symbol $4_{1}$ presents a knot projection, see Fig. 20). Thus, we can assume that $|\tau(P)|\geq 3$ in the following. Let $Q_{1}$ (resp. $Q_{2}$) be the left (resp. right) figure of Fig. 13. By Lemma 4, we can find at least one copy of either $Q_{1}$ or $Q_{2}$ within any arbitrary circle arrangement. For such an arbitrary circle arrangement $s=s_{1}$, obtain $s_{2}$ by deleting this part, denoted by $R_{1}$ $=$ $Q_{1}$ or $Q_{2}$. In $s_{2}$, we can find at least one copy of $Q_{1}$ or $Q_{2}$ which we denoted by $R_{2}$. Repeating the deletions, we eventually have exactly one circle $s_{m+1}$ and a sequence $R_{1}R_{2}R_{3}\dots R_{m-1}R_{m}$ for some positive integer $m$.

Next, we will construct any circle arrangement from one circle $O$. For one circle $O$, we put $R_{m}$ in the two regions given by $O$ on $S^{2}$. Using Lemma 3, if $R_{m}$ $=$ $Q_{1}$ (resp. $Q_{2}$), we consider the connected sum of the trefoil (resp. $6_{3}$) projection and $O$ as in Fig. 12. Next, for $R_{m-1}$, we consider an appropriate connected sum by specifying $R_{m-1}$ and $s_{m-1}$. By applying connected sums step by step as above, we finally obtain a knot projection $P$ such that $\tau(P)$ $=$ $s_{1}$ $=$ $s$ and $P$ consists of finitely many trefoil and $6_{3}$ projections.

Next we consider the goal of obtaining a prime knot projection such that $\tau(P)$ $=$ $s$. To obtain such a knot projection, every time we consider a connected sum in the above step $R_{k}\to R_{k-1}$, we apply an operation $s2a$ or a pair $(1a)(s2a)$ as Fig. 16 (e.g., Fig. 17).

∎