HHOMFLY Polynomials of Pretzel Knots
William Qin ∗ Abstract
HOMFLY polynomials are one of the major knot invariants being actively studied. They are difficultto compute in the general case but can be far more easily expressed in certain specific cases. In thispaper, we examine two particular knots, as well as one more general infinite class of knots.From our calculations, we see some apparent patterns in the polynomials for the knots 9 and 9 ,and in particular their F -factors. These properties are of a form that seems conducive to finding a generalformula for them, which would yield a general formula for the HOMFLY polynomials of the two knots.Motivated by these observations, we demonstrate and conjecture some properties both of the F -factors and HOMFLY polynomials of these knots and of the more general class that contains them,namely pretzel knots with 3 odd parameters. We make the first steps toward a matrix-less generalformula for the HOMFLY polynomials of these knots. HOMFLY polynomials are subcategory of knot polynomial, that generalizes the Jones and Alexanderpolynomials. In the fundamental representation, they satisfy the skein relation A H ( L + ) − A H ( L − ) =( q − q − ) H ( L ).HOMFLY polynomials can be generalized to non-fundamental representations of SU (2), creating the“colored” HOMFLY polynomials. While the normal HOMFLY polynomials are relatively easy to com-pute and have been computed for hundreds of knots in repositories like [1], colored HOMFLY polynomialsare difficult to compute in general for any non-(anti)symmetric representations, and computations even inthe simplest such representation, namely that corresponding to the smallest L-shaped Young diagram, aredifficult to do for more than a few knots.The crossing changing formula (skein relation) of the HOMFLY polynomial in the fundamental repre-sentation, or the original HOMFLY polynomial, was initially constructed to concern variables t and ν [2].However, when expanded to other representations, the HOMFLY polynomials are quantum invariants asso-ciated with irreducible representations of the quantum group U q ( sl N ), and it is then possible to express itin terms of variables q and A = q N [2]. This is the formulation we use in this text.Certain methods have been developed in the past that can compute general colored HOMFLY polyno-mials. However, each of these general methods relies on the use of matrix multiplication. The most generalsuch method, using quantum R-matrices and braid diagrams is not feasible, despite working for all knotsand all representations. This is because it involves the multiplication of square matrices with dimensionsincreasing as the product of the size of the representation and of the number of strands. The matrix entriesare sufficiently complex that this multiplication takes on the order of minutes even for very small represen-tations and knots, and increases very quickly from there. Even less general methods which rely on matrices,despite having smaller matrices, become infeasible at relatively low representation size, and often work onlyfor (anti)symmetric representations or rectangular representations. Therefore, it is desirable to find somemethod of computing HOMFLY polynomials that does not involve matrix multiplication so that it is feasibleto compare knots using these polynomials, even if only for specific knots or small classes of knots.Certain classes of knots such as torus and twist knots already have short general formulas for(anti)symmetric representations and even for rectangular representations but a general, simple formula forHOMFLY polynomials in all representations is currently out of reach for most if not all knots. However, thereare easier methods for computing the HOMFLY polynomial of a certain class of knot called an aborescent ∗ [email protected], Millburn High School, Millburn, USA a r X i v : . [ m a t h . G T ] J a n not, which allow us to compute many colored knot polynomials for these knots. Therefore, in this paper weexamine the two smallest aborescent knots that are neither twist nor torus, and attempt to find a formulafor their HOMFLY polynomials. Both of these knots, 9 and 9 in the Rolfsen table [1], are part of a moregeneral class of knot called a pretzel knot, which we define later in this paper. In particular, they are bothpretzel knots with 3 odd parameters, which are (3 , ,
3) and (3 , , −
3) respectively.One of the new methods, the techniques of which we use heavily in this paper, is the Racah matrixapproach, which we use in this text to derive the non-fundamental knot polynomials of these two knots, andwe use the general formula for pretzel knots, derived by this same approach, in the latter half of the paper.We do not yet know a general formula for the HOMFLY polynomials of these knots that does not usematrices, which as mentioned before are slow to the point of computational infeasibility, but we begin tomake progress towards such an explicit formula at least for symmetric and anti-symmetric representations,particularly using the differential expansion of [3].
We use the standard notation { x } = x − x ,D j = { Aq j }{ q } , and [ n ] = { q n }{ q } = q n − + q n − + ... + q − n , which is equal to the character of the n -dimensional representation of SL (2).We also use the intuitive notation [ n ]! = n (cid:89) i =1 [ i ] . Representations of GL ( N ) are parametrized by Young diagrams with at most n rows. Their charactersare symmetric functions in x , ..., x N , which are called Schur functions s λ ( x , ..., x N ). We denote these Youngdiagrams as [ c , c , · · · , c n ], where the c i are the number of boxes in the i th column.Schur functions can be computed by the following determinantal identity: s λ = n det i,j =1 ( h λ i − i + j ) , where h n = s [ n ] can be computed using the identity (cid:88) n h n t n = e (cid:80) n pnn t n . It is often useful, in practice, by way of the Taylor series of e x , to express them more explicitly as h n = (cid:88) x +2 x +3 x + ··· = n p x x ( x !) p x x ( x !) p x x ( x !) · · · . Here p n are the power-sum polynomials p k = (cid:88) i x ki . In order to consider stabilization in the limit n → ∞ , it is convenient to introduce A such that A = q N . A and q , and the dependence on N goes through thisrelation.At the special point p ∗ i = { A i }{ q i } , Schur functions have the hook-length factorization s λ ( p ∗ i ) = (cid:89) ( i,j ) ∈ λ { Aq i − j }{ q hook ij } , Figure 1: A hook of length 4 in the Young diagram [4,4,3,1], beginning at at box (1,1)and this expression coincides with the λ -colored HOMFLY polynomial of the unknot. Throughout the paperwe divide all λ -colored HOMFLY polynomials by this quantity, so the unknot HOMFLY polynomials arealways equal to 1. In this text, we primarily consider pretzel knots (discussed in Section 3) with 3 odd parameters a, b, c . Definition 2.1.
We denote the HOMFLY polynomial of a 3-parameter pretzel knot in some symmetricrepresentation [ r ] as H r ( a, b, c ) = H ( a, b, c, r ) = χ [ r, (cid:80) rx =0 1 S ,x ( S · T a · S ) ,x ( S · T b · S ) ,x ( S · T c · S ) ,x .This formula was derived using Racah matrices in equation 22 of [4] in terms of different Racah matrices,though we change conventions here and therefore modify the formula slightly to fit these new conventions.We denote H r ( a, b, c ) as Q ( c, r ) for some arbitrary fixed a, b . In this text we typically set a = b = 1, butwhen we use the notation Q ( c, r ) our only restriction on a, b is that within a formula, they are always thesame constants. We may also write Q ( a, b, c, r ), when we wish to specify a, b . Definition 2.2.
We recursively define the n th difference of a genus-2 pretzel knot’s HOMFLY polynomial Q n ( c, r ) = Q n − ( c + 2 , r ) − Q n − ( c, r ) for positive integers n , and Q ( c, r ) = Q ( c, r ). We may also write Q n ( a, b, c, r ), when we wish to specify a, b .In particular, c is always odd, and the “next” n th difference, where we talk about it, is actually Q n ( c +2 , r ).We denote by X ( P ( x )) , for all P ( x ) ∈ Z [ x ], the factor of P ( x ) that cannot be factored further with thehighest degree. In this paper we use this only in the context of X ( Q n ( c, r )). Remark 2.1.
We can alternatively define the n th difference as X ( Q n − ( c + 2 , r )) − X ( Q n − ( c, r )). Wedenote this expression Q ( n ) ( c, r ) in this text. When we list the computed differences, we use this definition(occasionally omitting the parentheses), however in our proofs we use the former for simplicity. Thus far itseems that the two definitions are related in that Q n ( c, r ) = Q ( n ) ( c, r ) (cid:81) n − i =1 Q i ( c,r ) X ( Q i ) c,r ) . When we refer to F -factors, we are referring to the factors in the differential expansion of the HOMFLYpolynomial as used in [5]. While it is used there for twist knots, the formula H [ r ] = 1 + r (cid:88) s =1 [ r ]![ s ]![ r − s ]! F s ( A | q ) s − (cid:89) j =0 { Aq r + j }{ Aq j − } (1)applies more generally for defect-zero knots, which was used to derive them for the knots considered here.Both of the knots considered in Section 4 are defect-zero and so we can find and analyze their F -factorsgiven their HOMFLY polynomials, using the methods outlined in [6]. 3lso, we get from [3] that the defect δ K of the differential expansion depends on the degree of theAlexander polynomial in a way that allows us to compute the F -factors examined in Section 4. If we denotethe Alexander polynomial by Al K , then δ K = 12 Power q ( Al K ) − . This is very convenient because many Alexander polynomials are readily available from [1] (where wesubstitute t = q ), and are also easily computed as H K [1] ( A = 1 , q ) from only the HOMFLY polynomial inthe fundamental representation.In particular, from this property, we can conclude that both 9 and 9 have defect 0, which allows usto use Formula ( 1). Example 2.1. Al = 7 q −
13 + q , hence Power q ( Al ) = 2 and δ K = 0. Example 2.2. Al = − q + 5 − q , hence Power q ( Al ) = 2 and δ K = 0. Example 2.3. Al = q − q + q − q + 1 − q + q − q + q , hence Power q ( Al K ) = 8 and δ K = 3.All of the pretzel knots we consider here can be trivially confirmed with the tables in [3] to be defect-zero,however whether this is true in general for genus 2 pretzel knots is not yet known. Pretzel knots of genus g are knots created by connecting pairs of crossing strands as in Figure 2. We dealwith genus 2 pretzel knots in this text (or equivalently, pretzel knots with 3 parameters). Each parameter(the n i ) determines the number of crossings in each pair.Figure 2: An illustration of a genus g pretzel knotFrom [7], we have the following two formulas about S and ¯ S Racah matrices, which are used in thegeneral formula for computing HOMFLY polynomials of genus g knots. S km = min( r + k + m, r ) (cid:88) j =max( r + m,r + k ) σ km ( j ) · (cid:115) [2 m + 1]∆ k [2 k + 1] χ [ r + m,r − m ] · G ( r − m ) G ( j + 1) G ( r + k + 1) G ( j − r − m ) , (2)and¯ S km = [ r + 1]! (cid:81) r − i =0 D i · min( r + k + m, r ) (cid:88) j =max( r + m,r + k ) σ km ( j ) · (cid:115) ∆ k ∆ m [2 k + 1][2 m + 1] · G ( r + 1) G ( j + 1) G ( r + k + 1) G ( r + m + 1) G ( r + k + m − j ) . (3)In these formulas, 4 km ( j ) = ( − r + k + m (cid:112) [2 k + 1][2 m + 1] · ([ k ]![ m ]!) [ r − k ]![ r − m ]![ r + k + 1]![ r + m + 1]! · ( − j [ j + 1]![2 r − j ]!([ j − r − k ]![ j − r − m ]![ r + k + m − j ]!) , for all 0 ≤ k, m ≤ r,G ( n ) = n (cid:89) j =1 { Aq j − }{ q j } , ∆ m = D m − D − G ( m ) and χ [ r + m,r − m ] = G ( r + m + 1) G ( r − m ) D − [2 m + 1] . We also have that ¯ T km = (cid:40) k (cid:54) = m ( − q m − A ) m k = m. (4)Before we begin, we prove a small lemma about genus-2 pretzel knots. Lemma 3.1.
For any genus-2 pretzel knot with parameters a, b, c , the knot is invariant under any permuta-tion of the parameters.Proof.
We can prove that the genus-2 pretzel knots are invariant under rotation of the parameters as well asorder reversal, which will allow us to demonstrate that the knot is invariant under any permutation of theparameters.For rotation of parameters, we use the Figure 3.Figure 3: Parameter rotation with the knot described by (3,2,1)Flipping an intersection does not change the intersection type, which allows us to perform the transfor-mation shown, keeping all of the intersections of the same type, by pulling the first two strands over the restof the knot, and then flipping the formerly first pair of strands.To prove that flipping the order of the parameters does not change the not, we need only to flip the knot,which does not change any of the intersection types but reverses the order of the parameters.For the sake of completeness we list the sequences of these two operations that generate each parameterpermutation, with F representing a flip and R representing a rotation.Permutation Sequence(a,b,c) Identity(a,c,b) RF(b,a,c) RRF(b,c,a) R(c,a,b) RR(c,b,a) F 5herefore, we can permute the parameters of a genus-2 pretzel knot and the knot will remain the same.
Remark 3.1.
This means also that the HOMFLY polynomial is invariant under these changes. Additionally,the components of this proof also hold for arbitrary genus but except in genus 2 do not prove the lemma. and and 9 . Both of these knots are pretzel knots with 3 parameters, for which an explicit F-factor formula is not yet known. F -factors here are parts of the differential expansion formula for HOMFLYpolynomials, as described in [3]. Both of these knots are small enough to explicitly calculate some of theHOMFLY polynomials, and we are therefore able to compute some F -factors to make some conjectures asto their properties. We present limited examples of the conjectures in the text, and provide more F -factorsin the ancillary files and in the GitHub repository for both knots. We can use the formula for HOMFLY polynomials of pretzel knots in [6] to get the following F -factors.We find the F -factors by using Formula 1 after computing the HOMFLY polynomials. 9 is the pretzelknot (3 , , F i ( A | q ) as F i throughout this text.We get F = − A ( A + 3 A + 2 A + 1) ,F = A q ( A q + 3 A q + 3 A q + 5 A q + 5 A q ++ 3 A q + 4 A q + 6 A q + 3 A q + A + 3 A q + 4 A q + 3 A + 2 A q + 2 A + 1) ,F = − A q ( A q + 3 A q + 3 A q + 3 A q + 5 A q +8 A q + 11 A q + 6 A q + 3 A q + 7 A q + 9 A q + 18 A q +16 A q + 12 A q + 3 A q + A q + 6 A q + 10 A q + 19 A q +19 A q + 18 A q + 9 A q + 3 A q + 5 A q + 8 A q + 17 A q + 17 A q +17 A q + 8 A q + 3 A q + 4 A q + 6 A q + 12 A q + 12 A q + 9 A q + 3 A q + A + 3 A q + 4 A q + 7 A q + 4 A q + 3 A + 2 A q + 2 A q + 2 A + 1) . For all positive integers i F (cid:48) i = A q i F i When fully factored, we conjecture that the F (cid:48) i are such that the factor of A and q is the same for F (cid:48) i − and F i , for all positive integer i >
2. It can be verified for all integers 2 ≤ i ≤ A q i − ) | F i + F (cid:48) i − .(See ancillary files) This leads us to the following hypothesis. Conjecture 4.1.
For the knot , and positive integers i > , (1 + A q i − ) | F i + F (cid:48) i − . is the pretzel knot (3 , , − F = A ( A + 1) , = A ( A q + A q + A q + 1) ,F = A q ( A q + 1)( A q + A q + A q − q + q + 1) . We see that both F and F have a factor of 1 + A q i − . This also holds true for F and F , giving usthe following conjecture. Conjecture 4.2.
For the knot , and odd positive integer i > , (1 + A q i − ) | F i . After applying the transformation A = ⇒ Aq to F i , multiplying by A q , and denoting the result F ∗ i , wefind that q ( q − | F A q − F ∗ , because F − F ∗ = A q − A q . In particular, we can also verify thatfor odd integers 3 ≤ i ≤ q ( q i − − | F i A q i − − F ∗ i − . This leads to the following conjecture. Conjecture 4.3.
For the knot , and odd i > , q ( q i − − | F i A q i − − F ∗ i − . If the above two conjectures are true, this could help in deriving a formula for arbitrary F -factors. Forall 3 conjectures in this section, the result after dividing by the stated factor seems to begin similar for all i though they differ after some terms, and we suspect that for high i , the remaining parts of each quantitymay converge to some infinite polynomial. n th Differences
The n th differences for general pretzel knots have many properties making them useful for computingHOMFLY polynomials. In particular, these differences tend to factor nicely as products of quantum numbers,and they are constructed in a similar way to the differential expansion of [3]. As at least low-valued 3-parameter pretzel knots seem to all be defect-zero, as mentioned in 3, we can likely use the n th differencesand the differential expansion formula to generate a relatively simple formula for HOMFLY polynomials inat least symmetric and anti-symmetric representations.In this section, where applicable we fix a, b = (1 , a, b . In these subsections a, b are arbitrary fixed constants.Additionally, in this section we assume that a, b, c are all odd, as this guarantees a knot rather than alink. Here we give some conjectures that are not proven here and while not integral to the main result, couldbe helpful in the future if proven.
Conjecture 5.1. Q ( c,r ) X ( Q ( c,r )) = Q ( c +1 ,r ) X ( Q ( c +1 ,r )) . Remark 5.1.
Conjecture 5.1, if true, can easily be extended to show that every Q ( c,r ) X ( Q ( c,r )) is a constantwith respect to c , by repeated application. Remark 5.2.
There are simple counterexamples for higher differences; in a later section we address whatoccurs when we do not take the largest factor. All that results is that previous factors carry over, but it doesturn out to be useful also to study what these factors are, despite corresponding less directly to F -factors. We begin by proving the following proposition. This provides a motivation for a general result about thefirst differences.
Proposition 5.1.
For the representation [1] , ( A − q )( A + q )( Aq − Aq + 1) | Q ( m, , for all a, b, m. roof. The proposition is equivalent to the statements that Q (2 m + 1 , r ) − Q (2 m − , r ) | A = q = 0 ,Q (2 m + 1 , r )) − Q (2 m − , r )) | A = − q = 0 ,Q (2 m + 1 , r )) − Q (2 m − , r )) | A = q = 0 ,Q (2 m + 1 , r )) − Q (2 m − , r )) | A = − q = 0based on our definition of Q ( m,
1) and the fact that any polynomial is zero exactly when one if its polynomialfactors are. We know that H ( a, b, c, r ) = Q ( c, r ) = χ [1 , r (cid:88) x =0 S ,x ( S · T a · S ) ,x ( S · T b · S ) ,x ( S · T c · S ) ,x . Evaluating Equation (4) from [7], we get that with representation [1] ,S = (cid:113) ( A − q )( A + q )( A − q +1) (cid:113) ( Aq − Aq +1)( A − q +1) (cid:113) ( Aq − Aq +1)( A − q +1) (cid:113) ( A − q )( A + q )( A − q +1) . At A = ± q , S , = 1 and S , = 0, and similarly for A = ± q , S , = 0 and S , = 1. We consider first thecase where A = ± q .If A = ± q , then as S , = S , and S , = S , , this means that S = I .If A = ± q , S becomes the anti-diagonal identity matrix for the same reasoning.By evaluating Equation (5) of [7], we get that S = A ( q − A − q A ( q − (cid:114) ( A − q )( A + q )( Aq − Aq +1) A q − ( A − qA ( q − (cid:114) ( A − q )( A + q )( Aq − Aq +1) A q − ( A − q A ( q − A − q . Clearly S , = S , = 0 at each of the 4 points. At A = ± q , it is trivial to verify that the diagonal entriesare ±
1, and for A = ± q , they are ∓ T = (cid:18) − A (cid:19) = (cid:18) ∓ q (cid:19) , for A = ± q , we find that each of the terms of the form S · T m · S becomes exactly (cid:18) ∓ m q m (cid:19) . As m is odd, this is exactly (cid:18) ∓ q m (cid:19) . Because we are only looking atthe elements (1 , x ) of this matrix, we are left with only a contribution of ∓ x = 2 , S , and S , gives a final value for the entire polynomialof ∓ a, b, c .For A = ± q , by going through the same process with our already computed values, we get that the finalpolynomial at this point is exactly ± a, b, c .Regardless of what the constant value is, in all 4 cases the Q ( c, r ) are constant for all odd integers c ,and so it is also equal to this value for both c = 2 m + 1 and c = 2 m −
1. Therefore, upon subtractingthe two polynomials, we get 0 as the 1 th difference. Recalling that we chose the four points specificallybecause ( A + q )( A − q )( Aq − Aq + 1) | Q ( m,
1) if and only if Q ( m,
1) is zero at all of those points, wecan conclude that in fact ( A + q )( A − q )( Aq − Aq + 1) | Q ( m, Remark 5.3.
Both of ( A − q )( A + q ) and ( Aq + 1)( Aq −
1) are quantum numbers up to a monomial factor.In particular, they are { Aq } and { Aq } , respectively. In particular, each of the Q ( c,r ) X ( Q ( c,r )) seems to be a productof quantum numbers in a simple way outlined in Conjecture 5.1, though proving this is for a future work.This proposition can be easily extended to all representations [ r ], as outlined in Theorem 5.1. We firstintroduce two lemmas. 8 emma 5.1. The first row of the Racah matrix S ,m = (cid:40) m = r m (cid:54) = r at A = ± q and A = ± q , and the firstrow of ¯ S ,m = (cid:40) m = 00 m (cid:54) = 0 at A = ± q and A = ± q .Proof. To prove this, we use Equation 2.It is clear that A = ± q = ⇒ G ( i ) = n (cid:89) j =1 {± q j − }{ q j } = {± }{ q } n (cid:89) j =2 { q j − }{ q j } = ± − ( ± − { q } n (cid:89) j =2 { q j − }{ q j } = 0 , for all i > . We begin by considering S . As {± q } = {± } = ± − ( ± − = 0, D − = 0 . Additionally, [2 k + 1] =[1] = { q }{ q } = 1.At A = ± q in general, D j = [ j + 1], and G ( n ) = n (cid:89) j =1 [ j − j ] = [0][ n ]Also, at k = 0, S ,m = σ ,m ( r + m ) (cid:115) G (0) D m − G ( r + m + 1) G ( r − m ) G ( r − m ) G ( r + m + 1) G ( r + 1)= σ ,m ( r + m ) (cid:112) D m − G ( r − m ) G ( r + m + 1) G ( r + 1) , by applying the above simplifications, without using A = ± q .[ x ] is nonzero exactly when x (cid:54) = 0, so as σ ,m ( r + m ) = ( − r + m (cid:112) [1][2 m + 1] · ([ m ]!) [ r ]![ r − m ]![ r + 1]![ r + m + 1]! · ( − r + m [ r + m + 1]![ r − m ]!([ m ]!) , for all 0 ≤ k, m ≤ r, and r ≥ m > σ ,m ( r + m ) is neither 0 and undefined for 0 < m < r , and so can be ignored withinthese ranges. We examine m = 0 , m = r separately. G ( i ) always has exactly one factor that becomes 0 at A = q for i ∈ Z unless i ≤
0, and otherwise isexactly 1, as nothing is being multiplied together. Additionally, this factor, { q } , is the same regardless of i . This means that at r (cid:54) = m , as r, m ≥
0, there are at least two factors in the numerator that become 0 at A = ± q , and only 1 in the denominator, so the entire expression is zero at m (cid:54) = r .At m = r , σ ,m ( r + m ) = σ ,r (2 r ) = ( − r (cid:112) [2 r + 1] · ([ r ]!) [ r ]![ r + 1]![2 r + 1]! · ( − r [2 r + 1]!([ r ]!) = (cid:112) [2 r + 1][ r + 1] , for all 0 ≤ k, m ≤ r As the [0]s in the G ( i ) cancel out at r = m as G (0) = 1, the remainder of the expression for S k,m simplifiesto [ r +1] √ [2 r +1] , which exactly cancels out with σ ,m ( r + m ), giving that S , = 1. Therefore, S ,m = (cid:40) m = r m (cid:54) = r at A = ± q The other cases (with one or both of A = ± q rather than A = ± q and ¯ S rather than S ) follow in preciselythe same way. Theorem 5.1.
For all symmetric representations [r], ( A − q )( A + q )( Aq − Aq + 1) | Q ( m, r ) , for all a, b, m ∈ Z + . roof. Much like the proof of the proposition, the theorem is equivalent to the statements that Q (2 m + 1 , r ) − Q (2 m − , r ) | A = q = 0 ,Q (2 m + 1 , r ) − Q (2 m − , r ) | A = − q = 0 ,Q (2 m + 1 , r ) − Q (2 m − , r ) | A = q = 0 ,Q (2 m + 1 , r ) − Q (2 m − , r ) | A = − q = 0from the definition of Q ( m, r ). We know that H ( a, b, c, r ) = Q ( c, r ) = χ [ r, r (cid:88) x =0 S ,x ( S · T a · S ) ,x ( S · T b · S ) ,x ( S · T c · S ) ,x . We now recall Lemma 5.1. In particular, the formula only involves the first row of ¯ S · ¯ T n S , so weprove a simple form for this. By formula 4, ¯ T n is a diagonal matrix with ¯ T , = 1. We consider each of A = ± q , A = ± q together using the lemma. Therefore, ¯ S · ¯ T n has a first row equal to ( ¯ S · ¯ T n ) ,m = (cid:40) ¯ S , ¯ T n , m = 00 m (cid:54) = 0 = (cid:40) m = 00 m (cid:54) = 0 , from the lemma. Then, again by the lemma, (( ¯ S · ¯ T n ) · S ) ,m = S ,m .Therefore, Q ( c, r ) = χ [ r, r (cid:88) x =0 S ,x S ,x = χ [ r, r (cid:88) x =0 S ,x = χ [ r, , (5)which is constant. Therefore, Q ( m, r ) = Q (2 m + 1 , r ) − Q (2 m − , r ) = 0 , for all m ∈ Z + , regardless of a, b , as expected. Remark 5.4.
This proof, with minimal modification, also works for any genus g pretzel knot, becauseregardless of the power of S ,x in 5, so long as it is at least 1 (which it is because it is one more than thegenus), it is still a constant for constant genus, and so the first differences are always zero at A = ± q, ± q .In particular, for higher genus, we define differences such that all but the last parameter are constant. In this section, we present some conjectured properties. Each property has been verified up to r = [5], c = 2, a = b = 1, unless otherwise specified. Some are also verified for higher c for lower r . Conjecture 5.2. Q r ( c, r ) · A r q r )( r − = Q r ( c +2 , r ) for all odd integers a, b, c and symmetric representations [ r ] . Remark 5.5.
This conjecture, in addition to Lemma 3.1, would allow for computing infinitely many HOM-FLY polynomials with only a finite number computed using the Racah matrices, which would result in drastictime savings. In particular, as shown in Figure 4 for r = [2], only r + 1 HOMFLY polynomials would needto be computed using Racah matrices to be able to compute recursively any HOMFLY polynomial with thesame a, b . H (1 , , , H (1 , , , H (1 , , , Q (1 , Q (3 , Q (1 , Q (3 , Q (5 , H (1 , , , Figure 4: Assuming that the red values are known, we can compute all of the black values from top tobottom with very few computations using Conjecture 5.2 10 emark 5.6.
This would also imply that Q r ( a, b, c, r ) | Q r ( a + 2 , b, c, r ) and Q r ( a, b, c, r ) | Q r ( a, b + 2 , c, r ),for all representations [ r ] and a, b, c odd integers. This would be very powerful as it would allow one togenerate the r th difference for every odd integer a, b, c , from only r +1 Racah matrix computations, relativelycomputationally inexpensively. Acknowledgements
WQ is grateful to Yakov Kononov for advising him during this project.WQ is grateful to the MIT PRIMES program for facilitating this research.
Data Availability Statement
Python code and more explicitly computed values have been deposited at https://github.com/MathTauAthogen/KnotTheory . Supporting Files
In the arXiv submission of this paper, we include as ancillary files some explicitly computed values ofdifferences.
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