Circle immersions that can be divided into two arc embeddings

We give a complete characterization of a circle immersion that can be divided into two arc embeddings in terms of its chord diagram.


Introduction
Let S 1 be the unit circle. Let X be a set and f : S 1 → X a map. Let n be a natural number greater than one. Suppose that there are n subspaces I 1 , · · · , I n of S 1 with the following properties.
(1) Each I i is homeomorphic to a closed interval.
( 3) The restriction map f | Ii : I i → X is injective for each i. Then we say that f can be divided into n arc embeddings. We define the arc number of f , denoted by arc(f ), to be the smallest such n except the case that f itself is injective. If f itself is injective then we define arc(f ) = 1. If f cannot be divided into n arc embeddings for any natural number n then we define arc(f ) = ∞.
Note that if f can be divided into n arc embeddings then there exist n subspaces I 1 , · · · , I n of S 1 with (1), (2) and (3) above together with the following additional condition.
(4) I i ∩ I j = ∂I i ∩ ∂I j for each i and j with 1 ≤ i < j ≤ n. Namely we may assume that S 1 is covered by mutually interior disjoint n simple arcs I 1 , · · · , I n .
Let S(f ) = {x ∈ S 1 |f −1 (f (x)) is not a singleton.} and s(f ) = f (S(f )). We say that a map f : S 1 → X has finite multiplicity if S(f ) is a finite subset of S 1 . From now on we restrict our attention to maps that have finite multiplicity. The purpose of this paper is to give a characterization of a map f : S 1 → X with arc(f ) = 2. By |Y | we denote the cardinality of a set Y . Let m(f ) be the maximum of |f −1 (y)| where y varies over all points of X. It is clear that arc(f ) ≥ m(f ). Thus we further restrict our attention to a map f : S 1 → X whose multiple points are only finitely many double points. Namely f has finite multiplicity and m(f ) ≤ 2. Then we have |S(f )| = 2m for some non-negative integer m. Then the crossing number of f , denoted by c(f ), is defined by m.
Let m be a natural number. An m-chord diagram on S 1 is a pair C = (P, ϕ) where P is a subset of S 1 that contains exactly 2m points and ϕ is a fixed point free involution on P . A chord c of C is an unordered pair of points (x, ϕ(x)) = (ϕ(x), x) where x is a point in P . Let ∼ C be the equivalence relation on S 1 generated by x ∼ C ϕ(x) for every x ∈ P . Let S 1 / ∼ C be the quotient space and f C : S 1 → S 1 / ∼ C the quotient map. We call f C the associated map of C. Then the arc number of C, denoted by arc(C), is defined to be the arc number of f C . Two m-chord diagrams C 1 = (P 1 , ϕ 1 ) and C 2 = (P 2 , ϕ 2 ) are equivalent if there is an orientation preserving self-homeomorphism h of S 1 such that h(P 1 ) = P 2 and h • ϕ 1 = ϕ 2 • h. From now on we consider m-chord diagrams up to this equivalence relation. In the following we sometimes express an m-chord diagram C = (P, ϕ) by m line-segments in the plane R 2 where S 1 ⊂ R 2 , P is the set of the end points of these line-segments and x and ϕ(x) are joined by a line segment for each x ∈ P . Thus a line segment express a chord and from now on we do not distinguish them. See for example Figure 1.1.
Let f : S 1 → X be a map whose multiple points are only finitely many double points.
We call C(f ) the associated chord diagram of f . Then it is clear that arc(f ) = arc(C(f )).
A chord diagram D = (Q, ψ) is called a sub-chord diagram of a chord diagram C = (P, ϕ) if Q is a subset of P and ψ is the restriction of ϕ on Q. Then it is clear that arc(D) ≤ arc(C). We call D = (Q, ψ) a proper sub-chord diagram of C = (P, ϕ) if D is a sub-chord diagram of C and Q is a proper subset of P .
Let n be a natural number. Let C 2n+1 be a (2n + 1)-chord diagram as illustrated in Figure 1.1. To give a more precise definition we introduce the followings. Let k be a natural number greater than two. Let R k be a regular k-gon inscribed in S 1 and v k;1 , · · · , v k;k the vertices of R k that are arranged in this order on S 1 along the counterclockwise orientation of S 1 . Namely v k;i and v k;i+1 are adjacent in R k for each i where the indices are considered modulo k. Let j is a natural number less than k 2 . Let c(k; i, j) be the chord joining v k;i and v k;i+j for each i ∈ {1, · · · , k}. Then C 2n+1 is the chord diagram represented by chords c(4n + 2; 2i − 1, 2n − 1) with i ∈ {1, · · · , 2n + 1}. We will show that arc(C 2n+1 ) = 3 but arc(D) = 2 for any proper sub-chord diagram D of C 2n+1 . Then we have the following theorem. The motive for this paper was the result in [2] that every knot has a diagram which can be divided into two simple arcs. This result is re-discovered by [3] and [4]. See also [1]. Then it is natural to ask what plane closed curve can be divided into two simple arcs. Theorem 1.1 gives an answer to this question. However we still have a question whether or not do we actually need all of C 3 , C 5 , · · · . The following proposition answers this question that we actually need all of them. Proposition 1.2. For each natural number n there exist a smooth immersion f n : S 1 → R 2 with arc(f n ) = 3 that has only finitely many transversal double points such that the associated chord diagram C(f n ) of f n has a sub-chord diagram which is equivalent to C 2n+1 but has no sub-chord diagram which is equivalent to C 2m+1 for any m < n.

Proof of Theorem 1.1
First we check that arc(C 2n+1 ) = 3. Let C = (P, ϕ) be an m-chord diagram with arc(C) = 2. A pair of points p, q ∈ S 1 \ P are called a cutting pair of C if x and ϕ(x) belong to the different components of S 1 \ {p, q} for each x ∈ P . Then we have that p and q are "antipodal". Namely we have that each component of S 1 \ {p, q} contains exactly m points of P . Thus we can check whether or not a given m-chord diagram has arc number 2 by examining m pairs of antipodal points of it. Then by the symmetry of C 2n+1 we immediately have that arc(C 2n+1 ) > 2. Then it is easily seen that arc(C 2n+1 ) = 3 and arc(D) = 2 for any proper sub-chord diagram D of C 2n+1 . Then the 'only if part' of the proof of Theorem 1.1 immediately follows. The 'if part' immediately follows from the following proposition.
Proposition 2.1. Let C be a chord diagram on S 1 that satisfies the following condition ( * ).
( * ) arc(C) ≥ 3 and arc(D) = 2 for any proper sub-chord diagram D of C. Then there is a natural number n such that C is equivalent to C 2n+1 .
Note that deleting a chord will decrease the arc number at most by one. Therefore, if C is a chord diagram on S 1 that satisfies the condition ( * ), then arc(C) = 3. For the proof of Proposition 2.1 we prepare the following lemmas. Let C = (P, ϕ) be a chord diagram and c = (x, ϕ(x)) a chord of C. Let α and β be the components of S 1 \ {x, ϕ(x)}. We may suppose without loss of generality that |α ∩ P | ≤ |β ∩ P |. Then the length of c in C, denoted by l(c) = l(c, C), is defined to be |α ∩ P | + 1. By C \ c, we denote the chord diagram (P \ {x, ϕ(x)}, ϕ| P \{x,ϕ(x)} ). Let p, q, x and y be mutually distinct four points on S 1 . We say that the pair of points p and q separates the pair of points x and y if each component of S 1 \ {p, q} contains exactly one of x and y. Note that the pair of points p and q separates the pair of points x and y if and only if the chord joining p and q intersects the chord joining x and y. Lemma 2.2. Let C = (P, ϕ) be a chord diagram on S 1 that satisfies the condition ( * ). Let c = (x, ϕ(x)) be a chord of C. Let p and q be a cutting pair of C \ c. Then p and q do not separate x and ϕ(x).
Proof. If p and q separate x and ϕ(x), then p and q is a cutting pair of C itself. Then it follows arc(C) = 2. This is a contradiction. Proof. Since |P | = 2m we have 1 ≤ l(c, C) ≤ m for any chord c. First we examine the case l(c, C) = m. In this case we have that each component of S 1 \ {x, ϕ(x)} contains exactly m − 1 elements of P . Let p and q be a cutting pair of C \ c. Then by Lemma 2.2 we have that p and q do not separate x and ϕ(x). Note that each component of S 1 \ {p, q} also contains exactly m − 1 elements of P . Then it follows that p and q are next to x and ϕ(x) or ϕ(x) and x respectively. We may suppose without loss of generality that p and q are next to x and ϕ(x) respectively. Let p ′ be a point on S 1 that is next to x and such that p ′ and p separate x and ϕ(x). Then we have that p ′ and q is a cutting pair of C. This is a contradiction. Next we examine the case l(c, C) = m − 1. In this case we have that one component of S 1 \ {x, ϕ(x)} contains exactly m − 2 elements of P and the other component contains exactly m elements of P . Let p and q be a cutting pair of C \ c. Then by Lemma 2.2 we have that p and q do not separate x and ϕ(x). Note that each component of S 1 \ {p, q} contains exactly m − 1 elements of P . Then it follows that one of p and q, say p is next to x or ϕ(x), say x. Let p ′ be a point on S 1 that is next to x and such that p ′ and p separate x and ϕ(x). Then we have that p ′ and q is a cutting pair of C. This is a contradiction. Thus we have l(c, C) ≤ m − 2. Proof. Suppose that there is a chord c = (x, ϕ(x)) of C with l(c, C) ≤ m − 3. Let D = (Q, ϕ| Q ) be the maximal sub-chord diagram of C such that x, ϕ(x) ∈ Q and l(c, D) = 1. Let n be the number of chords of D. Let A (resp. B) be the point in Q such that each components of S 1 \{x, A} (resp. S 1 \{ϕ(x), B}) contains n−1 points of Q. Note that n ≥ m − (l(c, C) − 1) ≥ m − (m − 3 − 1) = 4. Therefore we have that D has at least 4 chords. Then there is a chord d = (y, ϕ(y)) of D such that {y, ϕ(y)} and {x, ϕ(x), A, B} are mutually disjoint. Suppose that y and ϕ(y) do not separate x and A. In this case there must be a chord e = (z, ϕ(z)) of D such that {z, ϕ(z)} and {x, ϕ(x), y, ϕ(y)} are mutually disjoint and z and ϕ(z) do not separate y and ϕ(y). Then we have that the chords c, d and e form a sub-chord diagram of C that is equivalent to C 3 . This is a contradiction. Suppose that y and ϕ(y) separate x and A. Let p and q be a cutting pair of C \ d. Then we have by Lemma 2.2 that p and q do not separate y and ϕ(y). Note that p and q is also a cutting pair of D \ d and they separate x and ϕ(x). Then we have that the component of S 1 \ {p, q} that contains both A and B has more points of Q \ {y, ϕ(y)} than the other. This is a contradiction.
Thus we have shown the following lemma. Lemma 2.5. Let C = (P, ϕ) be an m-chord diagram on S 1 that satisfies the condition ( * ). Then C satisfies the following condition (⋆).
Proposition 2.6. Let C = (P, ϕ) be an m-chord diagram on S 1 that satisfies the condition (⋆). If m is even then m is divisible by 4 and arc(C) = 2. If m is odd then C is equivalent to C m .
(1) If m is a multiple of 4 then (4, m−2) = 2 and therefore G 2m,m−2 is isomorphic to a disjoint union of two m-cycles.
(2) If m is congruent to 2 modulo 4 then (4, m − 2) = 4 and therefore G 2m,m−2 is isomorphic to a disjoint union of four m 2 -cycles. (3) If m is odd then then (4, m − 2) = 1 and therefore G 2m,m−2 is isomorphic to a 2m-cycle. Note that in each case C must be a complete matching of the graph G 2m,m−2 . In (1) we have up to symmetry that C is as illustrated in Figure 2.1. Then we have that arc(C) = 2. In (2) we have that G 2m,m−2 has no complete matchings because an m 2 -cycle is an odd-cycle. In (3) we have that C is equivalent to C m . This completes the proof. Proof of Proposition 2.1. By Lemma 2.5 and Proposition 2.6 we have the result.

Examples of plane curves
Let C = (P, ϕ) be a chord diagram and c = (x, ϕ(x)) and d = (y, ϕ(y)) two chords of C. We say that c and d are parallel if the pair of points x and ϕ(x) do not separate the pair of points y and ϕ(y). We say that two distinct points x and y in P are next to each other if there is a component of S 1 \ {x, y} that is disjoint from P . We say that c and d are close to each other if x and y are next to each other and ϕ(x) and ϕ(y) are next to each other, or x and ϕ(y) are next to each other and ϕ(x) and y are next to each other.
Proof of Proposition 1.2. The cases n = 1, 2 are shown in Figure 3.1. We consider the case n ≥ 3. Let G 2n+1 = S 1 / ∼ C2n+1 be the 4-regular graph obtained from S 1 by identifying the end points of each chord of C 2n+1 . It is easy to observe that G 2n+1 is isomorphic to a graph obtained from a (2n+1)-cycle Γ 2n+1 on vertices v 1 , · · · , v 2n+1 lying in this order by adding edges joining v i and v i+3 for each i such that along the counterclockwise orientation of S 1 the vertices of G 2n+1 appears Figure 3.2. Then we deform them on R 2 as illustrated in Figure 3.3. Note that they are classified into three types by 2n + 1 modulo 6. Namely G 2n+1+6 is obtained from G 2n+1 by cutting open G 2n+1 along the dotted line and inserting two pieces of a pattern as illustrated in Figure  3.3. We modify this G 2n+1 and have the image f n (S 1 ) as illustrated in Figure 3.4. Note that each vertex of G 2n+1 is replaced by two transversal double points. We call them a twin pair. The chords corresponding to them are also called a twin pair. By choosing any one of them for each twin pair we have a sub-chord diagram of C(f n ) that is equivalent to C 2n+1 by the construction. Observe that each f n (S 1 ) is made of (2n + 1)-times repetitions of "one step forward and three steps back" along the (2n + 1)-cycle Γ 2n+1 and it totally goes around Γ 2n+1 twice. Here "one step forward" corresponds to an edge of G 2n+1 joining v i and v i+1 and "three steps back" corresponds to an edge of G 2n+1 joining v i+1 and v i+1−3 . It has no local double points and each double point comes from a part and another part that is one lap behind. Therefore we have that arc(f n ) = 3. Now we will check that no sub-chord diagram of C(f n ) is equivalent to C 2m+1 for any m < n. Note that two chords in a twin pair are close to each other in C(f n ).
Let D(f n ) be a sub-chord diagram of C(f n ) obtained from C(f n ) by deleting one of two chords for each twin pair in C(f n ). Since no two chords in C 2m+1 are close to each other it is sufficient to check that no sub-chord diagram of D(f n ) is equivalent to C 2m+1 for any m < n. Suppose that E is a sub-chord diagram of D(f n ) that is equivalent to C 2m+1 for some m < n. Since no proper sub-chord diagram of C 2n+1 is equivalent to C 2m+1 we have that there is a chord c of E that does not belong to any twin pair of C(f n ). Namely c corresponds to a transversal double point of f n that comes from a double point of G 2n+1 ⊂ R 2 in Figure 3.3. Observe that for each chord d = (x, ϕ(x)) of C 2m+1 there exist exactly two chords g = (y, ϕ(y) and h = (z, ϕ(z)) of C 2m+1 that are parallel to d such that all of y, ϕ(y), z and ϕ(z) are contained in the same component of S 1 \ {x, ϕ(x)}. Therefore c must have such two chords in D(f n ). By the "one step forward and three steps back" structure of f n (S 1 ) mentioned above the double points corresponding to such chords must lie in a small neighbourhood of the double point corresponding to c. Then we can check that there are no such two chords for c in D(f n ) except the case 2n + 1 is congruent to 5 modulo 6 and c is one of the three chords of C(f n ) that come from the three double points on the same edge of G 2n+1 ⊂ R 2 in Figure 3.3. For this exceptional case we further observe that the chords g and h above intersect unless m = 1 and the end point of g (resp. h) that is next to x or ϕ(x) in C 2m+1 is not next to any end point of h (resp. g) in C 2m+1 . However in this exceptional case we can check that there are no such two chords for c in D(f n ). This is a contradiction.  G 7 G 9 G 11 G 13 G 15 G 17 Remark 3.1. It is easy to see that the graph G 2n+1 = S 1 / ∼ C2n+1 is a non-planar graph for n ≥ 2. Therefore we have that for n ≥ 2 there is no smooth immersion f : S 1 → R 2 that has only finitely many transversal double points whose associated chord diagram C(f ) itself is equivalent to C 2n+1 . G 7 G 9 G 11 G 13 G 15 G 17