The Jones Polynomial and Khovanov Homology of Weaving Knots W(3,n)

# The Jones Polynomial and Khovanov Homology of Weaving Knots $W(3,n)$

## Abstract

In this paper we compute the signature for a family of knots , the weaving knots of type . By work of E. S. Lee the signature calculation implies a vanishing theorem for the Khovanov homology of weaving knots. Specializing to knots , we develop recursion relations that enable us to compute the Jones polynomial of . Using additional results of Lee, we compute the ranks of the Khovanov Homology of these knots. At the end we provide evidence for our conjecture that, asymptotically, the ranks of Khovanov Homology of are normally distributed.

## 1 Introduction

Weaving knots originally attracted interest, because it was conjectured that their complements would have the largest hyperbolic volume for a fixed crossing number. Here is the weaving knot .

Enumerating strands from the outside inward, our example is the closure of the braid on three strands. Thus, is a righthand twist involving strands 1 and 2, and is a righthand twist involving strands 2 and 3, and so on.

In words, the weaving knot is obtained from the torus knot by making the standard diagram of the torus knot alternating. Symbolically, is the closure of the braid , and is the closure of the braid . Obviously, the parity of is important. If the greatest common divisor , then and are both links with components. In general we are interested only in the cases when is an actual knot.

In [1] the main result is the following theorem.

###### Theorem 1.1 (Theorem 1.1, [1]).

If and , then

 voct(p−2)q(1−(2π)2q2)3/2≤vol(W(p,q))<(voct(p−3)+4vtet)q.

Champanerkar, Kofman, and Purcell call these bounds asymptotically sharp because their ratio approaches 1, as and tend to infinity. Since the crossing number of is known to be , the volume bounds in the theorem imply

 limp,q→∞vol(W(p,q))c(W(p,q))=voct≈3.66

Their study raised the question of examining the asymptotic behavior of other invariants of weaving knots. In this paper we start a study of the asymptotic behavior of Khovanov homology of weaving knots.

Briefly, since weaving knots are alternating knots by definition, we may specialize certain properties of the Khovanov homology of alternating knots to get started. This is accomplished in section 2, where we explain how to calculate the signature of weaving knots. The second main ingredient in our analysis is the fact that for alternating knots knowing the Jones polynomial is equivalent to knowing the Khovanov homology. How this works explicitly in our examples is explained in section 5. In section 3 we prepare to follow the development of the Jones polynomial in [2], starting from representations of braid groups into Hecke algebras. For weaving knots , which are naturally represented as the closures of braids on three strands, we develop recursive formulas for their representations in the Hecke algebras. These formulas are used in computer calculations of the Jones polynomials we need. Section 4 builds on the recursion formulas to develop information about the Jones polynomials . After we explain how to obtain the two-variable Poincaré polynomial for Khovanov homology in section 5, we present the results of calculations in a few relatively small examples. Our observation is that the distributions of dimensions in Khovanov homology resemble normal distributions. We explore this further in section 7, where we present tables displaying summaries of calculations for weaving knots for selected values of satisfying and ranging up to . The standard deviation of the normal distribution we attach to the Khovanov homology of a weaving knot is a significant parameter. The geometric significance of this number is an open question.

## 2 Generalities on Weaving knots

We have already mentioned that weaving knots are alternating by definition. Various facts about alternating knots facilitate our calculations of the Khovanov homology of weaving knots . For example, we appeal first to the following theorem of Lee.

###### Theorem 2.1 (Theorem 1.2, [4]).

For any alternating knot the Khovanov invariants are supported in two lines

 j=2i−σ(L)±1.\qed

We will see that this result also has several practical implications. For example, to obtain a vanishing result for a particular alternating knot, it suffices to compute the signature. Indeed, it turns out that there is a combinatorial formula for the signature of oriented non-split alternating links. To state the formula requires the following terminology.

###### Definition 2.2.

For a link diagram let be the number of crossings of , let be number of negative crossings, and let be the number of positive crossings. For an oriented link diagram, let be the number of components of , the diagram obtained by -smoothing every crossing.

In words, -regions in a neighborhood of a crossing are the regions swept out as the upper strand sweeps counter-clockwise toward the lower strand. An -smoothing removes the crossing to connect these regions. With these definitions, we may cite the following proposition.

###### Proposition 2.3 (Proposition 3.11, [4]).

For an oriented non-split alternating link and a reduced alternating diagram of , . ∎

We now use this result to compute the signatures of weaving knots. For a knot or link drawn in the usual way, the number of crossings . In particular, for , ; for , . Evaluating the other quantities in definition 2.2, we calculate the signatures of weaving knots.

###### Proposition 2.4.

For a weaving knot , , and for , .

###### Proof.

Consider first the example , illustrated by figures 3 and 4 for drawn below. After -smoothing the diagram, the outer ring of crossings produces a circle bounding the rest of the smoothed diagram. On the inner ring of crossings the -smoothings produce circles in a cyclic arrangement. Therefore . The outer ring of crossings consists of positive crossings and the inner ring of crossings consists of negative crossings, so . Applying the formula of theorem 2.3, we obtain the result .

For the general case , we have the following considerations. The crossings are organized into rings. Reading from the outside toward the center, we have first a ring of positive crossings, then a ring of negative crossings, and so on, alternating positive and negative. Thus . Considering the -smoothing of the diagram of , as in the special case, a bounding circle appears from the smoothing of the outer ring. A chain of disjoint smaller circles appears inside the second ring. No circles appear in the third ring, nor in any odd-numbered ring thereafter. On the other hand, chains of disjoint smaller circles appear in each even-numbered ring. Since there are even-numbered rings, we have . Applying the formula of proposition 2.3

 σ(W(2k+1,n))=o(D)−y(D)−1=(1+kn)−kn−1=0.

These figures illustrate the main points of the -cases, and, as explained above, the main points of the -cases.

For the case , we show below in figures 5 and 6 as an example. Our standard diagram may be organized into rings of crossings. In each ring there are crossings, so the total number of crossings is . In our standard representation, there is an outer ring of positive crossings, next a ring of negative crossings, alternating until we end with an innermost ring of positive crossings. There are thus rings of positive crossings and rings of negative crossings. Therefore, and . Considering the -smoothing of the diagram, a bounding circle appears from the smoothing of the outer ring. As before, a chain of disjoint smaller circles appears inside the second ring and in each successive even-numbered ring. As previously noted, there are of these rings. No circles appear in odd-numbered rings, until we reach the last ring, where an inner bounding circle appears. Thus, . Consequently,

 σ(W(2k,n))=o(D)−y(D)−1=((k−1)n+2)−kn−1=−n+1.\qed
###### Theorem 2.5.

For a weaving knot the non-vanishing Khovanov homology lies on the lines

 j=2i±1.

For a weaving knot the non-vanishing Khovanov homology lies on the lines

 j=2i+n−1±1
###### Proof.

Substitute the calculations made in lemma 2.4 into the formula of theorem 2.1. ∎

## 3 Recursion in the Hecke algebra

We review briefly the definition of the Hecke algebra on generators through , and we define the representation of the braid group on three strands in . Theorem 3.2 sets up recursion relations for the coefficients in the expansion of the image in of the braid , whose closure is the weaving knot . The recursion relations are essential for automating the calculation of the Jones polynomial for the knots . Proposition 3.4 uses these relations developed in theorem 3.2 to prove a vanishing result for one of the coefficients. Being able to ignore one of the coefficients speeds up the computations slightly.

###### Definition 3.1.

Working over the ground field containing an element , the Hecke algebra is the associative algebra with on generators , …, satisfying these relations.

 TiTj =TjTi,whenever |i−j|≥2, (3.1) TiTi+1Ti =Ti+1TiTi+1,for 1≤i≤N−1, (3.2) and, finally, T2i =(q−1)Ti+q,for all i. (3.3)

It is well-known [2] that is the dimension of over .

Recasting the relation in the form shows that is invertible in with . Consequently, the specification , combined with relations (3.1) and (3.2), defines a homomorphism from , the group of braids on strands, into the multiplicative monoid of .

For work in , choose the ordered basis . The word in the Hecke algebra corresponding to the knot is formally

 ρ((T1T−12)n)=q−n(Cn,0+Cn,1⋅T1+Cn,2⋅T2+Cn,12⋅T1T2+Cn,21⋅T2T1+Cn,121⋅T1T2T1), (3.4)

where the coefficients of the monomials in and are polynomials in . For ,

 ρ(σ1σ−12)=T1T−12=q−1⋅(T1(−(q−1)+T2))=q−1(−(q−1)⋅T1+T1T2),

so we have initial values

 C1,0(q)=0,C1,1(q)=−(q−1),C1,2(q)=0,C1,12(q)=1,C1,21(q)=0,andC1,121(q)=0. (3.5)
###### Theorem 3.2.

These polynomials satisfy the following recursion relations.

 Cn,0(q) =q2⋅Cn−1,21(q)−q(q−1)⋅Cn−1,1(q) (3.6) Cn,1(q) =−(q−1)2⋅Cn−1,1(q)−(q−1)⋅Cn−1,0(q)+q2⋅Cn−1,121(q) (3.7) Cn,2(q) =q⋅Cn−1,1(q) (3.8) Cn,12(q) =(q−1)⋅Cn−1,1(q)+Cn−1,0(q) (3.9) Cn,21(q)=−(q−1)⋅C(n−1),2(q)+q⋅Cn−1,12(q)−(q−1)2⋅Cn−1,21(q)+q(q−1)⋅Cn−1,121(q) (3.10) Cn,121(q) =Cn−1,2(q)+(q−1)⋅Cn−1,21(q) (3.11)
###### Proof.

We have

 ρ(T1T−12)n=ρ(T1T−12)n−1⋅ρ(T1T−12)=q−n(Cn−1,0+Cn−1,1⋅T1+Cn−1,2⋅T2+Cn−1,12⋅T1T2+Cn−1,21⋅T2T1+Cn−1,121⋅T1T2T1)⋅(−(q−1)⋅T1+T1T2)=q−n(−(q−1)Cn−1,0⋅T1−(q−1)Cn−1,1⋅T21−(q−1)Cn−1,2⋅T2T1−(q−1)Cn−1,12⋅T1T2T1−(q−1)Cn−1,21⋅T2T21−(q−1)Cn−1,121⋅T1T2T21+Cn−1,0⋅T1T2+Cn−1,1⋅T21T2+Cn−1,2⋅T2T1T2+Cn−1,12⋅T1T2T1T2+Cn−1,21⋅T2T21T2+Cn−1,121⋅T1T2T21T2)=q−n((−(q−1)Cn−1,0⋅T1−(q−1)Cn−1,2⋅T2T1−(q−1)Cn−1,12⋅T1T2T1+Cn−1,0⋅T1T2)+{−(q−1)Cn−1,1⋅T21−(q−1)Cn−1,21⋅T2T21−(q−1)Cn−1,121⋅T1T2T21+Cn−1,1⋅T21T2+Cn−1,2⋅T2T1T2+Cn−1,12⋅T1T2T1T2+Cn−1,21⋅T2T21T2+Cn−1,121⋅T1T2T21T2}) (3.12)

after collecting powers of and expanding. In the last grouping, the first four terms inside the parentheses involve only elements of the preferred basis; the second eight terms in the pair of braces all require further expansion, as follows.

 −(q−1)Cn−1,1⋅T21=−(q−1)Cn−1,1⋅((q−1)T1+q)=−(q−1)2Cn−1,1⋅T1−q(q−1)Cn−1,1 (3.13) −(q−1)Cn−1,21⋅T2T21=−(q−1)Cn−1,21⋅T2((q−1)T1+q)=−(q−1)2Cn−1,21⋅T2T1−q(q−1)Cn−1,21⋅T2 (3.14) −(q−1)Cn−1,121⋅T1T2T21=−(q−1)Cn−1,121⋅T1T2((q−1)T1+q)=−(q−1)2Cn−1,121⋅T1T2T1−q(q−1)Cn−1,121⋅T1T2 (3.15) Cn−1,1⋅T21T2=Cn−1,1⋅((q−1)T1+q)T2=(q−1)Cn−1,1⋅T1T2+qCn−1,1⋅T2 (3.16) Cn−1,2⋅T2T1T2 =Cn−1,2⋅T1T2T1 (3.17) Cn−1,12⋅T1T2T1T2=Cn−1,12⋅T21T2T1=Cn−1,12((q−1)T1+q)T2T1=(q−1)Cn−1,12⋅T1T2T1+qCn−1,12⋅T2T1 (3.18) Cn−1,21⋅T2T21T2=Cn−1,21⋅T2((q−1)T1+q)T2=(q−1)Cn−1,21⋅T2T1T2+qCn−1,21⋅T22=(q−1)Cn−1,21⋅T1T2T1+qCn−1,21⋅((q−1)T2+q)=(q−1)Cn−1,21⋅T1T2T1+q(q−1)Cn−1,21⋅T2+q2Cn−1,21) (3.19) Cn−1,121⋅T1T2T21T2=Cn−1,121⋅T1T2((q−1)T1+q)T2=(q−1)Cn−1,121⋅T1T2T1T2+qCn−1,121⋅T1T22=(q−1)Cn−1,121⋅T21T2T1+qCn−1,121⋅T1((q−1)T2+q)=(q−1)Cn−1,121⋅((q−1)T1+q)T2T1+qCn−1,121⋅T1((q−1)T2+q)=(q−1)2Cn−1,121⋅T1T2T1+q(q−1)Cn−1,121⋅T2T1+q(q−1)Cn−1,121⋅T1T2+q2Cn−1,121⋅T1 (3.20)

Collecting the constant terms from (3.13) and (3.19), we get

 Cn,0 =−q(q−1)Cn−1,1+q2Cn−1,21. Collecting coefficients of T1 from (3.12), (3.13), (3.20), we get Cn,1 =−(q−1)Cn−1,0−(q−1)2Cn−1,1+q2Cn−1,121. Collecting coefficients of T2 from (3.14), (3.16), and (3.19), we get Cn,2 =−q(q−1)Cn−1,21+qCn−1,1+q(q−1)Cn−1,21=qCn−1,1. Collecting coefficients of T1T2 from (3.12), (3.15), (3.16), and (3.20), we get Cn,12 =Cn−1,0−q(q−1)Cn−1,121+(q−1)Cn−1,1+q(q−1)Cn−1,121=Cn−1,0+(q−1)Cn−1,1 Collecting coefficients of T2T1 from (3.12), (3.14), (3.18), and (3.20), we get Cn,21 =−(q−1)Cn−1,2−(q−1)2Cn−1,21+qCn−1,12+q(q−1)Cn−1,121. Collecting coefficients of T1T2T1 from (3.12), (3.15), (3.17), (3.18), (3.19), and (3.20), we get Cn,121 =−(q−1)Cn−1,12−(q−1)2Cn−1,121+Cn−1,2 +(q−1)Cn−1,12+(q−1)Cn−1,21+(q−1)2Cn−1,121 =Cn−1,2+(q−1)Cn−1,21

Up to simple rearrangements and expansion of notation, these are formulas (3.6) through  (3.11). ∎

###### Example 3.3.

Applying the recursion formulas just proved to the table of initial polynomials, or by computing directly from the definitions, we find

 C2,0(q) =q2⋅C1,21(q)−q(q−1)⋅C1,1(q)=q(q−1)2, (3.21) C2,1(q) =−(q−1)2⋅C1,1(q)−(q−1)⋅C1,0(q)=(q−1)3, (3.22) C2,2(q) =q⋅C1,1(q)=−q(q−1), (3.23) C2,12(q) =(q−1)⋅C1,1(q)+C1,0(q)=−(q−1)2, (3.24) C2,21(q) =−(q−1)⋅C1,2(q)+q⋅C1,12(q)−(q−1)2⋅C1,21(q)=q, (3.25) C2,121(q) =0. (3.26)

As a first application, we have the following vanishing result.

For all , .

###### Proof.

For , we claim . Make the inductive assumption that