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A quantitative Birman-Menasco finiteness theorem and its application to crossing number

Tetsuya Ito Department of Mathematics, Kyoto University, Kyoto 606-8502, JAPAN [email protected]
Abstract.

Birman-Menasco proved that there are finitely many knots having a given genus and braid index. We give a quantitative version of Birman-Menasco finiteness theorem, an estimate of the crossing number of knots in terms of genus and braid index. This has various applications of crossing numbers, such as, the crossing number of connected sum or satellites.

1. Introduction

In [7] Birman-Menasco proved (in)finiteness theorem on closed braid representatives of knots and links; a knot or link KK has only finitely many closed braid representatives, up to conjugacy and exchange move. Here an exchange move is an operation that converts an nn-braid of the form ασn1βσn11\alpha\sigma_{n-1}\beta\sigma_{n-1}^{-1} (α,βBn2\alpha,\beta\in B_{n-2}) to a braid ασn11βσn1\alpha\sigma_{n-1}^{-1}\beta\sigma_{n-1} and its converse (see Figure 1). As a byproduct they proved another remarkable finiteness result;

Theorem 1 (Birman-Menasco finiteness theorem [7]).

For given g,n>0g,n>0, there are only finitely many knots/links with genus gg and braid index nn.

Refer to caption
Figure 1. Exchange move.

In this paper, we give a quantitative version of Theorem 1, an estimate of the crossing number by braid index and genus. For a link LL in S3S^{3}, let b(L),c(L)b(L),c(L) and χ(L)\chi(L) be the braid index, the minimum crossing number, and the maximum euler characteristic of Seifert surface, respectively.

Theorem 2 (Quantitative Birman-Menasco finiteness theorem).

Let

f(n)={1(n=2)53(n=3)(2n5)(n>3).f(n)=\begin{cases}1&(n=2)\\ \frac{5}{3}&(n=3)\\ (2n-5)&(n>3).\end{cases}

For a link LL in S3S^{3},

χ(L)+b(L)c(L)f(b(L))(χ(L)+b(L))-\chi(L)+b(L)\leq c(L)\leq f(b(L))(-\chi(L)+b(L))

holds.

The case b(L)=2b(L)=2 is easy and the case b(L)=3b(L)=3 is proven in [13] by using a special property of closed 3-braid that a closed 3-braid always bounds a minimum genus Bennequin surface, a minimum genus Seifert surface whose braid foliation has only aa-singular points [1, 5].

In the following, we restrict our attention to knots, and use the knot genus instead of the maximum euler characteristic.

Theorem 2 implies that 2g(K)1+b(K)2g(K)-1+b(K) is a linear approximation of the crossing number. This has several applications to the crossing numbers since the genus is one of the most familiar and well-studied invariant whereas the minimum crossing number is one of the most difficult invariant whose basic properties are still unknown.

It is a famous conjecture that c(K#K)=c(K)+c(K)c(K\#K^{\prime})=c(K)+c(K^{\prime}) [14, Problem 1.65]. Theorem 2 immediately gives an estimate of the crossing number of composite knots.

Corollary 1.

For a knot K,KK,K^{\prime} in S3S^{3} let N=max{b(K),b(K)}N=\max\{b(K),b(K^{\prime})\}.

c(K#K)1f(b(K))c(K)+1f(b(K))c(K)1f(N)(c(K)+c(K))c(K\#K^{\prime})\geq\frac{1}{f(b(K))}c(K)+\frac{1}{f(b(K^{\prime}))}c(K^{\prime})\geq\frac{1}{f(N)}(c(K)+c(K^{\prime}))
Proof.

Recall that g(K)g(K) is additive under connected sum, and that b(K)1b(K)-1 is also additive under connected sum [6]. Thus

c(K#K)\displaystyle c(K\#K^{\prime}) b(K#K)+2g(K#K)1\displaystyle\geq b(K\#K^{\prime})+2g(K\#K^{\prime})-1
=(2g(K)1+b(K))+(2g(K)1+b(K))\displaystyle=(2g(K)-1+b(K))+(2g(K^{\prime})-1+b(K^{\prime}))
c(K)f(b(K))+c(K)f(b(K))\displaystyle\geq\frac{c(K)}{f(b(K))}+\frac{c(K^{\prime})}{f(b(K^{\prime}))}
1f(N)(c(K)+c(K))\displaystyle\geq\frac{1}{f(N)}(c(K)+c(K^{\prime}))

In [15] Lackenby showed that c(K#K)1152(c(K)+c(K))c(K\#K^{\prime})\geq\frac{1}{152}(c(K)+c(K^{\prime})). Although the estimate given in Corollary 1 gets worse as b(K)b(K) increases, when b(K)b(K) is small it gives a better estimate.

Our proof gives an estimate of regularity introduced in [18] which plays an important role to relate two different problem; a genericity of hyperbolic knots/links and the additivity of crossing numbers.

Definition 3.

For λ\lambda\in\mathbb{R}, a knot KK is λ\lambda-regular if c(K)λc(K)c(K^{\prime})\geq\lambda c(K) if KK^{\prime} contains KK as its prime factor.

Corollary 2.

For a knot KK in S3S^{3}, KK is 1f(b(K))\frac{1}{f(b(K))}-regular.

Similarly, we have an estimate of the crossing number of satellite knots and an estimate of the asymptotic crossing number. The asymptotic crossing number ac(K)ac(K) of KK is defined by111The definition of the asymptotic crossing number given here is a version appeared in Kirby’s problem list [14]. In [8] Freedman-He called this the quadratic asymptotic crossing number and use different definition for the assymptotic crossing number, which we denote by acFH(K)ac_{FH}(K). Since Freedman-He proved ac(K)acFH(K)2g(K)1ac(K)\geq ac_{FH}(K)\geq 2g(K)-1, the inequality in Corollary 3 is valid for acFH(K)ac_{FH}(K).

ac(K)=infw1{c(Kw)w2}ac(K)=\inf_{w\geq 1}\left\{\frac{c(K_{w})}{w^{2}}\right\}

where the infimum is taken over all satellites KwK_{w} of KK with winding number w0w\geq 0. It is conjectured ac(K)=c(K)ac(K)=c(K) [14, Problem 1.68].

Corollary 3.

Let KK be a satellite knot with companion knot K0K_{0} and pattern CC, whose winding number (homological degree) is ww. Then

c(K)w2f(b(K0))c(K0)w2b(K0)c(K)\geq\frac{w^{2}}{f(b(K_{0}))}c(K_{0})-w^{2}b(K_{0})

In particular,

ac(K)1f(b(K))c(K)b(K).ac(K)\geq\frac{1}{f(b(K))}c(K)-b(K).
Proof.

By [8] it is shown that c(K)w2(2g(K0)1)c(K)\geq w^{2}(2g(K_{0})-1). Therefore

c(K)\displaystyle c(K) w2(2g(K0)1)w2(1f(b(K0))c(K0)b(K0))\displaystyle\geq w^{2}(2g(K_{0})-1)\geq w^{2}\left(\frac{1}{f(b(K_{0}))}c(K_{0})-b(K_{0})\right)

Although Corollary 3 is meaningful only if b(K)b(K) is small compared with c(K)c(K), it gives a linear lower bound of ac(K)ac(K) in terms of c(K)c(K).

A famous satellite crossing number conjecture asserts c(K)c(K0)c(K)\geq c(K_{0}) [14, Problem 1.67]. Corollary 3 says that the satellite crossing number conjecture is true when ww and c(K0)c(K_{0}) are large enough compared with b(K0)b(K_{0}).

A weaker and more tractable version of the satellite crossing number conjecture, c(Kp,q)c(K)c(K_{p,q})\geq c(K) for the (p,q)(p,q)-cable Kp,qK_{p,q} of KK which we call the cabling crossing number conjecture, is still open. For a braided pp-cable, a satellite knot whose pattern CC is a closed pp-braid in the solid torus, we have another estimate that gives more supporting evidences for the cabling crossing number conjecture and the satellite crossing number conjecture which works even if b(K)b(K) is large;

Corollary 4.

If KpK_{p} is a braided pp-cable of KK, then

c(Kp)pf(b(K))c(K)+(p1)c(K_{p})\geq\frac{p}{f(b(K))}c(K)+(p-1)

Thus, if pf(b(K))p\geq f(b(K)) then c(Kp)c(K)c(K_{p})\geq c(K).

Proof.

Since g(Kp)=pg(K)+g(C)pg(K)g(K_{p})=pg(K)+g(C)\geq pg(K) and b(Kp)=pb(K)b(K_{p})=pb(K) [19] we get

c(Kp)\displaystyle c(K_{p}) 2g(Kp)1+b(Kp)p(2g(K)1+b(K))+(p1)\displaystyle\geq 2g(K_{p})-1+b(K_{p})\geq p(2g(K)-1+b(K))+(p-1)
pf(b(K))c(K)+p1\displaystyle\geq\frac{p}{f(b(K))}c(K)+p-1

In particular, since f(3)=53<2f(3)=\frac{5}{3}<2, the cabling crossing number conjecture is true for closed 3-braids:

Corollary 5.

If b(K)3b(K)\leq 3 then c(Kp,q)c(K)c(K_{p,q})\geq c(K) for all (p,q)(p,q)-cables Kp,qK_{p,q} of KK.

It is interesting to ask whether Corollary 4 can be extended to general satellite knot. We used the braided pp-cable assumption is to guarantee b(Kp)=pb(K)b(K_{p})=pb(K). It looks to be reasonable to expect b(Kw)wb(K)b(K_{w})\geq wb(K) for satellite KwK_{w} of KK with winding number ww so the same conclusion as Corollary 4 holds for general satellite knot KwK_{w} with winding number ww.

In [16] Lackenby showed c(K)11013c(K0)c(K)\geq\frac{1}{10^{13}}c(K_{0}) if KK is a satellite of K0K_{0}. Since this 11013\frac{1}{10^{13}} estimate uses an estimate for (2,t)(2,t)-cable of K2,tK_{2,t} of K0K_{0} and Corollary 4 gives an improvement of Lackenby’s estimate 1119024(c(K0)|t|)c(K2,t)\frac{1}{119024}(c(K_{0})-|t|)\leq c(K_{2,t}) [16, Theorem 5.1], one can improve Lackenby’s 11013\frac{1}{10^{13}} bound when b(K0)b(K_{0}) is not so large. Indeed, combining Lackenby’s another result and our result we give a dramatic improvement of the estimate of the crossing number of satellite knots when b(K0)b(K_{0}) is not large.

Corollary 6.

If KK is a satellite of K0K_{0},

c(K)176f(b(K0))c(K0)c(K)\geq\frac{1}{76f(b(K_{0}))}c(K_{0})
Proof.

By [16, Theorem 1.5], there is a knot LL which is either K0K_{0} or a cable of K0K_{0} such that c(K)1152c(L)c(K)\geq\frac{1}{152}c(L). By Corollary 4, if LL is a cable of K0K_{0} then c(L)2f(b(K0))c(K0)c(L)\geq\frac{2}{f(b(K_{0}))}c(K_{0}) hence

c(K)1152c(L)1152(2f(b(K0))c(K0))=176f(b(K0))c(K0)c(K)\geq\frac{1}{152}c(L)\geq\frac{1}{152}\left(\frac{2}{f(b(K_{0}))}c(K_{0})\right)=\frac{1}{76f(b(K_{0}))}c(K_{0})

Finally we give another application. The braid index problem is a problem to algorithmically compute the braid index of a link LL from a given diagram DD of LL. Strictly speaking, to make it as a decision problem, by the braid index problem we mean the problem to determine, for a given diagram DD of LL and a natural number nn, determine whether b(L)nb(L)\leq n or not. Since KK is the unknot if and only if b(K)=1b(K)=1, this problem includes the unknotting problem, a problem to recognize the unknot.

Theorem 4.

The braid index problem is solvable.

Proof.

Our proof of Theorem 2 actually shows that a link LL has a closed b(L)b(L)-braid diagram that has at most f(b(L))(χ(L)+b(L))f(b(L))(-\chi(L)+b(L)) crossings. There are finitely many closed nn-braid diagram with less than or equal to cc crossings. Since one can effectively enumerate such closed braid diagrams, by checking all the possibilities (this is possible since one can algorithmically check a given diagram DD represents LL or not [9, 10, 11]) one can determine b(L)nb(L)\leq n or not algorithmically. ∎

It is desirable to give more direct algorithm of the braid index problem that avoids to use knot recognition, in a spirit of the Birman-Hirsch unknotting algorithm [3]. Also, it is an interesting problem to determine the computational complexity of the braid index problem.

Acknowledgement

The author has been partially supported by JSPS KAKENHI Grant Number 19K03490,16H02145. He would like to thank Joan Birman for stimulating comments for earlier version of the paper, and would like to thank to Sorsen Rebecca and Keiko Kawamuro pointing out an error in the earlier version of the paper.

2. Proof of Theorem 2

2.1. Quick review of braid foliation

In this section we briefly review Birman-Menasco’s braid foliation techniques. For details of braid foliation theory, we refer to [17] or [2].

Let AA be the unknot in S3S^{3} and fix a fibration π:S3AS1\pi:S^{3}\setminus A\rightarrow S^{1} whose fiber St=π1(t)S_{t}=\pi^{-1}(t) is a disk bounded by AA. An oriented link LL is a closed nn-braid with axis AA if LL positively transverse to every fiber StS_{t} at nn points.

Let FF be an oriented incompressible Seifert surface of a closed nn-braid LL. We put FF so that {FSt}\{F\cap S_{t}\} induces a singular foliation \mathcal{F} of FF satisfying the following properties;

  • (i)

    AA and FF transversely intersect. Moreover, each vAFv\in A\cap F is an elliptic singular point of \mathcal{F}; there is a disk neighborhood UvU_{v} of vv in FF such that the foliation \mathcal{F} on UvU_{v} is the radial foliation with node vv.

  • (ii)

    All but finitely many fibers StS_{t} transversely intersect with FF. Each exceptional fiber StS_{t} is tangent to FF at exactly one point hh in the interior of FF, as a saddle tangency that appears as a hyperbolic singular point of \mathcal{F}.

  • (iii)

    Each leaf of \mathcal{F}, a connected point of StFS_{t}\cap F, transverse to L=FL=\partial F.

We call the foliation \mathcal{F} satisfying these properties braid foliation.

An elliptic singular point vv is positive (resp. negative) if the sign of intersection of AFA\cap F at vv is positive (resp. negative). A hyperbolic singular point hFSth\in F\cap S_{t} is positive (resp. negative) if the positive normal direction nF\vec{n}_{F} of FF at hh agrees (resp. desagrees) with that of StS_{t} at hh (see Figure 2)222Although in the following argument the sign of hyperbolic point does not appear explicitly, it plays a crucial role in the proof of Proposition 1 below..

Refer to caption
Figure 2. Singular points and its signs of braid foliation

If a connected component cc of StFS_{t}\cap F is a simple closed curve, it bounds a disk in StS_{t}. Since FF is incompressible, by standard innermost circle argument one can remove such leaves by isotopy of FF. Thus in the following, we always assume that the leaf of \mathcal{F} is either

  • a-arc: an arc connecting a positive elliptic point and a point of L=FL=\partial F, or,

  • b-arc: an arc connecting two elliptic points of opposite signs.

Moreover, with no loss of generality, we may assume that each b-arc bb in StS_{t} is essential as a properly embedded arc in the punctured disk St(StL)S_{t}\setminus(S_{t}\cap L). In other words, both components of StbS_{t}\setminus b are pierced by the braid L=FL=\partial F.

According to the types of nearby leaves, hyperbolic points of \mathcal{F} are classified into three types, aaaa, abab and bbbb and each hyperbolic point has a canonical foliated neighborhood as shown in Figure 3 which we call aaaa-tile, abab-tile, and bbbb-tile.

Refer to caption
Figure 3. Tiles

A decomposition of FF into tiles induces a cellular decomposition of FF; The 22-cells are tiles, the 11-cells are aa-arcs and bb-arcs that appear as the boundary of tiles, and the 0-cells are elliptic points and the end points of aa-arc 1-cells.

2.2. Euler characteristic equality

We say that an elliptic point is of type (α,β)(\alpha,\beta) if the valence is α+β\alpha+\beta and it is a boundary of α\alpha aa-arc 1-cells and β\beta bb-arc 11-cells. We denote by V(α,β)V(\alpha,\beta) be the number of elliptic points of type (α,β)(\alpha,\beta).

By a standard euler characteristic counts we have the following equality which we call the euler characteristic equality, or, the fundamental equality of braid foliation;

(2.1) 2V(1,0)+2V(0,2)+V(0,3)+V(1,1)4χ(F)\displaystyle 2V(1,0)+2V(0,2)+V(0,3)+V(1,1)-4\chi(F)
=V(2,1)+2V(3,0)+v4α=0v(v+α4)V(α,vα)\displaystyle\qquad\qquad=V(2,1)+2V(3,0)+\sum_{v\geq 4}\sum_{\alpha=0}^{v}(v+\alpha-4)V(\alpha,v-\alpha)

The following result plays the fundamental role in the braid foliation theory and explain why exchange move is a fundamental operation in braid foliation; using exchange move one can simplify the braid foliation (or braid itself, by destabilization) so that it contains no low-valency vertices. In a light of the euler characteristic equality (2.1), this gives a strong constraint for V(α,β)V(\alpha,\beta).

Proposition 1.

[7, Lemma 4] If n=b(L)n=b(L), by applying exchange move we may assume that V(1,0)=V(0,2)=V(0,3)=V(1,1)=0V(1,0)=V(0,2)=V(0,3)=V(1,1)=0.

2.3. Braid foliation and crossing number

One can read the braid from the braid foliation [3, 4, 12]; one aaaa-tile gives rise to a braid of the form

σi,j:=(σiσi+1σj2)σj1±1(σiσi+1σj2)1(1i<jn)\sigma_{i,j}:=(\sigma_{i}\sigma_{i+1}\cdots\sigma_{j-2})\sigma_{j-1}^{\pm 1}(\sigma_{i}\sigma_{i+1}\cdots\sigma_{j-2})^{-1}\quad(1\leq i<j\leq n)

and one abab-tile gives rise to a braid of the form,

(σiσi+1σj1)±1,or, (σj1σj2σi)±1(1i<jn).(\sigma_{i}\sigma_{i+1}\cdots\sigma_{j-1})^{\pm 1},\mbox{or, }(\sigma_{j-1}\sigma_{j-2}\cdots\sigma_{i})^{\pm 1}\quad(1\leq i<j\leq n).

A bb-tile does not affect the boundary the braid L=FL=\partial F.

One useful way to see this is to use a movie presentation, a sequence of slice StFS_{t}\cap F that describes how the arcs in StS_{t} changes as tt increase as we illustrate in Figure 4.

Refer to caption
Figure 4. How aa- and ab-singularity affect the braid L=FL=\partial F

Let RaaR_{aa} and RabR_{ab} be the number of aaaa and abab tiles in the braid foliation, respectively. Since the above discussion shows that one aa-tile provides a braid with at most 2(n2)+1=2n32(n-2)+1=2n-3 crossings and one ab-tile provides a braid with at most (n1)(n-1) crossings, we have an estimate of the crossing number via the braid foliation

c(L)(2n3)Raa+(n1)Rab.c(L)\leq(2n-3)R_{aa}+(n-1)R_{ab}.

With a little more argument we slightly improve the estimate as follows;

Proposition 2.

Let LL be a closed nn-braid and FF be an incompressible Seifert surface admitting a braid foliation.

  1. (i)

    If Rab=0R_{ab}=0 then

    c(L){Raa(n=2)53Raa(n=3)(2n5)Raa(n>4)c(L)\leq\begin{cases}R_{aa}&(n=2)\\ \frac{5}{3}R_{aa}&(n=3)\\ (2n-5)R_{aa}&(n>4)\\ \end{cases}
  2. (ii)

    If Rab0R_{ab}\neq 0 then

    c(L)(2n5)Raa+(n3)Rabc(L)\leq(2n-5)R_{aa}+(n-3)R_{ab}
Proof.

(i): Assume that Rab=0R_{ab}=0. Then case n=2n=2 is obvious so we assume that n>2n>2. Let R1,R2,,RnR_{1},R_{2},\ldots,R_{n} be the number of aa-tiles that gives rise to a braid σi,i+1\sigma_{i,i+1} and let R=R1+R2++RnR^{\prime}=R_{1}+R_{2}+\cdots+R_{n}. Here indices are understood by modulo nn so we regard σn,n+1\sigma_{n,n+1} as σ1,n\sigma_{1,n}. Since

(σ1σ2σn1)σi,j(σ1σ2σn1)1=σi+1,j+1,(\sigma_{1}\sigma_{2}\cdots\sigma_{n-1})\sigma_{i,j}(\sigma_{1}\sigma_{2}\cdots\sigma_{n-1})^{-1}=\sigma_{i+1,j+1},

by taking conjugates we may assume that RnR1,,Rn1R_{n}\leq R_{1},\ldots,R_{n-1}. Thus Rn1nRR_{n}\leq\frac{1}{n}R^{\prime}. Each ab-tile counted in R1,,Rn1R_{1},\ldots,R_{n-1} provides a braid with one crossing. Each of rest of the aa-tiles provides a braid with at most 2(n1)3=2n52(n-1)-3=2n-5 crossings so

c(L)\displaystyle c(L) (2n3)Rn+n1nR+(2n5)(RaaR)\displaystyle\leq(2n-3)\frac{R^{\prime}}{n}+\frac{n-1}{n}R^{\prime}+(2n-5)(R_{aa}-R^{\prime})
=3n4nR+(2n5)(RaaR)\displaystyle=\frac{3n-4}{n}R^{\prime}+(2n-5)(R_{aa}-R^{\prime})
{3n4nRaa(n=3)(2n5)Raa(n4)\displaystyle\leq\begin{cases}\frac{3n-4}{n}R_{aa}&(n=3)\\ (2n-5)R_{aa}&(n\geq 4)\\ \end{cases}

(ii) When Rab0R_{ab}\neq 0, the braid foliation of FF contains at least one negative elliptic point. We put a negative elliptic point vv so that it is leftmost among all the ellitpic poionts. At each tt, a b-arc from vv always separates StS_{t} into two components so that both component contains at least one puncture point LStL\cap S_{t}. Since vv is leftmost, the leftmost puncture point, the 1st strand of the braid, and the rightmost puncture point, the nn-th strand of the braid, are always separated by the b-arc from vv. Hence no aa-tile can provide a braid σ1,n\sigma_{1,n}, so one aaaa-tile can provide a braid having at most 2n52n-5 crossings (see Figure 5 (i)).

By the same reason no abab-tile can provide a braid of the form

(σ1σn1)±1,(σn1σ1)±1.(\sigma_{1}\cdots\sigma_{n-1})^{\pm 1},(\sigma_{n-1}\cdots\sigma_{1})^{\pm 1}.

Moreover, an abab-tile cannot produce a braid of the form

(σ1σn2)±1,(σ2σn1)±1,(σn2σ1)±1,(σn1σ2)±1(\sigma_{1}\cdots\sigma_{n-2})^{\pm 1},(\sigma_{2}\cdots\sigma_{n-1})^{\pm 1},(\sigma_{n-2}\cdots\sigma_{1})^{\pm 1},(\sigma_{n-1}\cdots\sigma_{2})^{\pm 1}

either, because if such a braid is produced by an abab-tile, the b-arc from vv becomes boundary-parallel (see Figure 5 (ii)). Thus one ab-tile actually can provide a braid having at most (n3)(n-3) crossings. Therefore we conclude

c(L)(2n5)Raa+(n3)Rab.c(L)\leq(2n-5)R_{aa}+(n-3)R_{ab}.
Refer to caption
Figure 5. (i): The b-arc from the leftmost negative elliptic point vv prevents to form an aaaa-singular point producing σ1,n\sigma_{1,n}. (ii): An abab-singular point producing (σn11σ21)(\sigma_{n-1}^{-1}\cdots\sigma_{2}^{-1}) makes the b-arc from vv boundary-parallel.

2.4. Completion of proof

Once all the necessary backgrounds or results from braid foliation theory are prepared, the proof of Theorem 2 is essentially an elementary counting argument.

Proof of Theorem 2.

Let DD be a minimum crossing diagram of LL. By Seifert’s algorithm, we get a Seifert surface ΣD\Sigma_{D} of LL with χ(ΣD)=s(D)c(L)\chi(\Sigma_{D})=s(D)-c(L), where s(D)s(D) denotes the number of Seifert circles. Since b(L)s(D)b(L)\leq s(D) we get

χ(L)χ(ΣD)=s(D)c(L)b(L)c(L).\chi(L)\geq\chi(\Sigma_{D})=s(D)-c(L)\geq b(L)-c(L).

We prove the upper bound of c(L)c(L). By Proposition 2, it is sufficient to give an upper bound of RaaR_{aa} and RabR_{ab}. In the following argument, we give an estimate of 2Raa+Rab2R_{aa}+R_{ab}.

Let FF be a Seifert surface of LL admitting a braid foliation such that χ(F)=χ(L)\chi(F)=\chi(L) and that F\partial F is a closed b(L)b(L)-braid. By Proposition 1 we may assume that V(1,0)=V(0,2)=V(0,3)=V(1,1)=0V(1,0)=V(0,2)=V(0,3)=V(1,1)=0. In particular, V(α,vα)=0V(\alpha,v-\alpha)=0 whenever α+v4<0\alpha+v-4<0.

Let V±(α,β)V^{\pm}(\alpha,\beta) be the number of positive and negative elliptic points of type (α,β)(\alpha,\beta). Note that V+(α,β)=V(α,β)V^{+}(\alpha,\beta)=V(\alpha,\beta) whenever α>0\alpha>0 since the elliptic point is a boundary of aa-arc only if it is positive. Since F\partial F is a closed b(L)b(L)-braid, the algebraic crossing number of the axis AA and FF is b(L)b(L) so

(2.2) b(L)=v0α=0vV+(α,vα)V(α,vα)b(L)=\sum_{v\geq 0}\sum_{\alpha=0}^{v}V^{+}(\alpha,v-\alpha)-V^{-}(\alpha,v-\alpha)

Each tile contains exactly two positive ellitpic points so

2Raa+2Rab+2Rbb=v0α=0vvV+(α,vα).2R_{aa}+2R_{ab}+2R_{bb}=\sum_{v\geq 0}\sum_{\alpha=0}^{v}vV^{+}(\alpha,v-\alpha).

Similarly, each ab-tile contains one negative elliptic point, and each bb-tile contains two negative elliptic points so

Rab+2Rbb=v0α=0vvV(α,vα).R_{ab}+2R_{bb}=\sum_{v\geq 0}\sum_{\alpha=0}^{v}vV^{-}(\alpha,v-\alpha).

Thus

(2.3) 2Raa+Rab=v0α=0vvV+(α,vα)vV(α,vα).2R_{aa}+R_{ab}=\sum_{v\geq 0}\sum_{\alpha=0}^{v}vV^{+}(\alpha,v-\alpha)-vV^{-}(\alpha,v-\alpha).

By (2.3) and (2.2) we get

2Raa+Rab4b(L)\displaystyle 2R_{aa}+R_{ab}-4b(L) =v0α=0v(v4)V+(α,vα)(v4)V(α,vα)\displaystyle=\sum_{v\geq 0}\sum_{\alpha=0}^{v}(v-4)V^{+}(\alpha,v-\alpha)-(v-4)V^{-}(\alpha,v-\alpha)
=v0α=1v(v4)V(α,vα)+v>0(v4)V+(0,v)v>0(v4)V(0,v).\displaystyle=\sum_{v\geq 0}\sum_{\alpha=1}^{v}(v-4)V(\alpha,v-\alpha)+\sum_{v>0}(v-4)V^{+}(0,v)-\sum_{v>0}(v-4)V^{-}(0,v).

Since we have seen that V(0,v)=0V(0,v)=0 for v<4v<4,

(2.4) 2Raa+Rab4b(L)v0α=0v(v4)V(α,vα)2R_{aa}+R_{ab}-4b(L)\leq\sum_{v\geq 0}\sum_{\alpha=0}^{v}(v-4)V(\alpha,v-\alpha)

On the other hand, let EaE_{a} be the number of 1-cells which are aa-arcs. By counting EaE_{a} as a boundary of 2-cells, Ea=2Raa+RabE_{a}=2R_{aa}+R_{ab}. Similarly by counting EaE_{a} as edges connecting 0-cells we get

(2.5) 2Raa+Rab=Ea=v>0α=0vαV(α,vα)2R_{aa}+R_{ab}=E_{a}=\sum_{v>0}\sum_{\alpha=0}^{v}\alpha V(\alpha,v-\alpha)

Thus by (2.4), (2.5), and the euler characteristic equality (2.1)

2(2Raa+Rab)4b(L)v>0α=0v(v+α4)V(α,vα)=4χ(L)2(2R_{aa}+R_{ab})-4b(L)\leq\sum_{v>0}\sum_{\alpha=0}^{v}(v+\alpha-4)V(\alpha,v-\alpha)=-4\chi(L)

so we eventually arrive at the inequality

2Raa+Rab2χ(L)+2b(L)2R_{aa}+R_{ab}\leq-2\chi(L)+2b(L)

By Proposition 2 we conclude

c(L){(χ(F)+b(L))b(L)=253(χ(F)+b(L))b(L)=3,(2b(L)5)(χ(F)+b(L))b(L)>4.c(L)\leq\begin{cases}(-\chi(F)+b(L))&b(L)=2\\ \frac{5}{3}(-\chi(F)+b(L))&b(L)=3,\\ (2b(L)-5)(-\chi(F)+b(L))&b(L)>4.\end{cases}

References

  • [1] D. Bennequin, Entrelacements et équations de Pfaff, Astérisque, 107-108, (1983) 87-161.
  • [2] J. Birman and E. Finkelstein, Studying surfaces via closed braids, J. Knot Theory Ramifications, 7, No.3 (1998), 267-334.
  • [3] J. Birman and M. Hirsch, A new algorithm for recognizing the unknot, Geom. Topol. 2 (1998), 175–220.
  • [4] J. Birman, W. Menasco, Studying links via closed braids. I. A finiteness theorem. Pacific J. Math. 154 (1992), no. 1, 17-36.
  • [5] J. Birman and W. Menasco, Studying links via closed braids. II. On a theorem of Bennequin. Topology Appl. 40 (1991), no. 1, 71–82.
  • [6] J. Birman, W. Menasco, Studying links via closed braids. IV. Composite links and split links. Invent. Math. 102 (1990), no. 1, 115-139.
  • [7] J. Birman, W. Menasco, Studying links via closed braids. VI. A nonfiniteness theorem. Pacific J. Math. 156 (1992), no. 2, 265-285.
  • [8] M. Freedman, an Z.-X. He, Divergence-free fields: energy and asymptotic crossing number. Ann. of Math. (2) 134 (1991), no. 1, 189-229.
  • [9] W. Haken, Theorie der Normalflächen, Acta Math. 105 (1961), 245–375.
  • [10] W. Haken, Über das Homöomorphieproblem der 3-Mannigfaltigkeiten. I. Math. Z. 80 (1962), 89–120.
  • [11] G. Hemion, On the classification of homeomorphisms of 2-manifolds and the classification of 3-manifolds, Acta Math. 142 (1979), 123–155.
  • [12] T. Ito, Braid ordering and knot genus, J. Knot Theory Ramification, 20, (2011), 1311-1323.
  • [13] T. Ito, A note on knot fertility, arXiv:2010.04371.
  • [14] R. Kirby, Problems in low-dimensional topology In Geometric Topology (Athens, GA, 1993), 35–473. AMS/IP Stud. Adv. Math. 2(2). Providence, RI: American Mathematical Society, 1997.
  • [15] M. Lackenby, The crossing number of composite knots. J. Topol. 2 (2009), no. 4, 747–768.
  • [16] M. Lackenby, The crossing number of satellite knots. Algebr. Geom. Topol. 14 (2014), no. 4, 2379–2409.
  • [17] D. LaFountain, and W. Menasco, Braid foliations in low-dimensional topology. Graduate Studies in Mathematics, 185. American Mathematical Society, Providence, RI, 2017. xi+289 pp.
  • [18] A. Mayutin, On the question of genericity of hyperbolic knots, IMRN Int. Math. Res. Not. IMRN. to appear.
  • [19] R. Williams, The braid index of generalized cables, Pacific J. Math. 155 (1992), no. 2, 369–375