The HOMFLY Braidor Algebra

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In Preparation

The information below is preliminary and cannot be trusted! (v)

This paperlet is about yet another construction of the HOMFLY polynomial, this time using "braidor equations". Though at the moment the term "braidor equations", the relationship with HOMFLY and the rationale for the whole plan is not yet described here. If you know what this is about, good. If not, bummer.

The Algebra

Let [math]\displaystyle{ A^0_n=\langle S_n, x, t_1,\ldots t_n\rangle }[/math] be the free associative (but non-commutative) algebra generated by the elements of the symmetric group [math]\displaystyle{ S_n }[/math] on [math]\displaystyle{ \{1,\ldots,n\} }[/math] and by formal variables [math]\displaystyle{ x }[/math] and [math]\displaystyle{ t_1\ldots t_n }[/math], and let [math]\displaystyle{ A^1_n }[/math] be the quotient of [math]\displaystyle{ A^0_n }[/math] by the following "HOMFLY" relations:

1. [math]\displaystyle{ x }[/math] commutes with everything else.
2. The product of permutations is as in the symmetric group [math]\displaystyle{ S_n }[/math].
3. If [math]\displaystyle{ \sigma }[/math] is a permutation then [math]\displaystyle{ t_i\sigma=\sigma t_{\sigma i} }[/math].
4. [math]\displaystyle{ [t_i,t_j]=x\sigma_{ij}(t_i-t_j) }[/math], where [math]\displaystyle{ \sigma_{ij} }[/math] is the transposition of [math]\displaystyle{ i }[/math] and [math]\displaystyle{ j }[/math].

Finally, declare that [math]\displaystyle{ \deg x=\deg t_i=1 }[/math] while [math]\displaystyle{ \deg\sigma=0 }[/math] for every [math]\displaystyle{ 1\leq i\leq n }[/math] and every [math]\displaystyle{ \sigma\in S_n }[/math], and let [math]\displaystyle{ A_n }[/math] be the graded completion of [math]\displaystyle{ A^1_n }[/math].

We say that an element of [math]\displaystyle{ A_n }[/math] is "sorted" if it is written in the form [math]\displaystyle{ x^k\cdot\sigma t_1^{k_1}t_2^{k_2}\cdots t_n^{k_n} }[/math] where [math]\displaystyle{ \sigma }[/math] is a permutation and [math]\displaystyle{ k }[/math] and the [math]\displaystyle{ k_i }[/math]'s are all non-negative integer. The HOMFLY relations imply that every element of [math]\displaystyle{ A_n }[/math] is a linear combinations of sorted elements. Thus as a vector space, [math]\displaystyle{ A_n }[/math] can be identified with the ring [math]\displaystyle{ B_n }[/math] of power series in the variables [math]\displaystyle{ x,t_1,\ldots,t_n }[/math] tensored with the group ring of [math]\displaystyle{ S_n }[/math]. The product of [math]\displaystyle{ A_n }[/math] is of course very different than that of [math]\displaystyle{ B_n }[/math].

Examples.

1. The general element of [math]\displaystyle{ A_1 }[/math] is [math]\displaystyle{ (1)f(x,t_1) }[/math] where [math]\displaystyle{ (1) }[/math] denotes the identity permutation and [math]\displaystyle{ f(x,t_1) }[/math] is a power series in two variables [math]\displaystyle{ x }[/math] and [math]\displaystyle{ t_1 }[/math]. [math]\displaystyle{ A_1 }[/math] is commutative.
2. The general element of [math]\displaystyle{ A_2 }[/math] is [math]\displaystyle{ (12)f(x,t_1,t_2)+(21)g(x,t_1,t_2) }[/math] where [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math] are power series in three variables and [math]\displaystyle{ (12) }[/math] and [math]\displaystyle{ (21) }[/math] are the two elements of [math]\displaystyle{ S_2 }[/math]. [math]\displaystyle{ A_2 }[/math] is not commutative and its product is non-trivial to describe.
3. The general element of [math]\displaystyle{ A_3 }[/math] is described using [math]\displaystyle{ 3!=6 }[/math] power series in 4 variables. The general element of [math]\displaystyle{ A_n }[/math] is described using n! power series in [math]\displaystyle{ n+1 }[/math] variables.

The algebra [math]\displaystyle{ A_n }[/math] embeds in [math]\displaystyle{ A_{n+1} }[/math] in a trivial way by regarding [math]\displaystyle{ \{1,\ldots,n\} }[/math] as a subset of [math]\displaystyle{ \{1,\ldots,n+1\} }[/math] in the obvious manner; thus when given an element of [math]\displaystyle{ A_n }[/math] we are free to think of it also as an element of [math]\displaystyle{ A_{n+1} }[/math]. There is also a non-trivial map [math]\displaystyle{ \Delta:A_n\to A_{n+1} }[/math] defined as follows:

1. [math]\displaystyle{ \Delta(x)=x }[/math].
2. [math]\displaystyle{ \Delta(t_i)=t_{i+1}+x\sigma_{1,i+1} }[/math].
3. [math]\displaystyle{ \Delta }[/math] acts on permutations by "shifting them one unit to the right", i.e., by identifying [math]\displaystyle{ \{1,\ldots,n\} }[/math] with [math]\displaystyle{ \{2,\ldots,n+1\}\subset\{1,\ldots,n+1\} }[/math].

The Equations

We seek to find a "braidor"; an element [math]\displaystyle{ B }[/math] of [math]\displaystyle{ A_2 }[/math] satisfying:

• [math]\displaystyle{ B=(21)+x(12)+ }[/math](higher order terms).
• [math]\displaystyle{ B(\Delta B)B=(\Delta B)B(\Delta B) }[/math] in [math]\displaystyle{ A_3 }[/math].

With the vector space identification of [math]\displaystyle{ A_n }[/math] with [math]\displaystyle{ B_n }[/math] in mind, we are seeking two power series of three variables each, whose low order behaviour is specified and which are required to satisfy 6 functional equations written in terms of 4 variables.

The Equations in Functional Form

Lemma. The following identities hold in [math]\displaystyle{ A_n }[/math]:

1. [math]\displaystyle{ [t_i^k, t_j] = x\sigma_{ij}(t_i^k-t_j^k) }[/math] and therefore [math]\displaystyle{ [e^{\alpha t_i}, t_j] = x\sigma_{ij}(e^{\alpha t_i}-e^{\alpha t_j}) }[/math].
2. [math]\displaystyle{ [t_i^k, t_j^l] = x\sigma_{ij}\left(\frac{t_i^{k+l}+t_j^{k+l}-t_i^kt_j^l-t_i^lt_j^k}{t_i-t_j}\right) }[/math] and therefore [math]\displaystyle{ [e^{\alpha t_i}, e^{\beta t_j}] = x\sigma_{ij}\left(\frac{e^{(\alpha+\beta)t_i}+e^{(\alpha+\beta)t_j}-e^{\alpha t_i+\beta t_j}-e^{\beta t_i+\alpha t_j}}{t_i-t_j}\right) }[/math]
   (The right hand sides of these expressions should be interpreted as polynomials / power series in commuting variables [math]\displaystyle{ x }[/math], [math]\displaystyle{ t_i }[/math] and [math]\displaystyle{ t_j }[/math], and then the "true" [math]\displaystyle{ x }[/math], [math]\displaystyle{ t_i }[/math] and [math]\displaystyle{ t_j }[/math] are to be substituted in, in "normal order" - in every monomial the variables are written so that every [math]\displaystyle{ t_i }[/math] occurs before any [math]\displaystyle{ t_j }[/math]).
3. [math]\displaystyle{ \Delta(t_i^k) = t_{i+1}^k + \frac{x}{2} \left(\frac{(t_{i+1}+x)^k-t_1^k}{t_{i+1}+x-t_1} - \frac{(t_{i+1}-x)^k-t_1^k}{t_{i+1}-x-t_1}\right) + \sigma_{1,i+1}\frac{x}{2} \left(\frac{(t_{i+1}+x)^k-t_1^k}{t_{i+1}+x-t_1} + \frac{(t_{i+1}-x)^k-t_1^k}{t_{i+1}-x-t_1}\right) }[/math] and therefore [math]\displaystyle{ \Delta(e^{t_i}) = e^{t_{i+1}} + \frac{x}{2} \left(\frac{e^{t_{i+1}+x}-e^{t_1}}{t_{i+1}+x-t_1} - \frac{e^{t_{i+1}-x}-e^{t_1}}{t_{i+1}-x-t_1}\right) + \sigma_{1,i+1}\frac{x}{2} \left(\frac{e^{t_{i+1}+x}-e^{t_1}}{t_{i+1}+x-t_1} + \frac{e^{t_{i+1}-x}-e^{t_1}}{t_{i+1}-x-t_1}\right) }[/math]. (The right hand sides of these expressions should be interpreted as polynomials / power series in commuting variables [math]\displaystyle{ x }[/math], [math]\displaystyle{ t_1 }[/math] and [math]\displaystyle{ t_{i+1} }[/math], and then the "true" [math]\displaystyle{ x }[/math], [math]\displaystyle{ t_1 }[/math] and [math]\displaystyle{ t_{i+1} }[/math] are to be substituted in, in "normal order" - in every monomial the variables are written so that their subscripts form a non-decreasing sequence).

A Solution

The first few terms of a solution can be computed using a computer, as shown below. But a true solution, written in a functional form, is still missing.

Computer Games

A primitive mathematica program to play with these objects is here.

A Numerology Problem

Question. Can you find nice formulas for the functions [math]\displaystyle{ f_{12} }[/math] and [math]\displaystyle{ f_{21} }[/math] of the variables [math]\displaystyle{ t_1 }[/math], [math]\displaystyle{ t_2 }[/math] and [math]\displaystyle{ x }[/math], whose Taylor expansions begin with

[math]\displaystyle{ f_{12}=x+\frac{x t_2}{3}-\frac{x t_1}{3} }[/math]

[math]\displaystyle{ -\frac{1}{5} t_1 x^3+\frac{t_2 x^3}{5}+\frac{t_1^3 x}{45}-\frac{t_2^3 x}{45}+\frac{1}{15} t_1 t_2^2 x-\frac{1}{15} t_1^2 t_2 x }[/math]

[math]\displaystyle{ -\frac{1}{7} t_1 x^5+\frac{t_2 x^5}{7}+\frac{11}{315} t_1^3 x^3-\frac{11}{315} t_2^3 x^3+\frac{11}{105} t_1 t_2^2 x^3-\frac{11}{105} t_1^2 t_2 x^3 }[/math]

[math]\displaystyle{ -\frac{2 t_1^5 x}{945}+\frac{2 t_2^5 x}{945}-\frac{2}{189} t_1 t_2^4 x+\frac{4}{189} t_1^2 t_2^3 x-\frac{4}{189} t_1^3 t_2^2 x+\frac{2}{189} t_1^4 t_2 x }[/math]

[math]\displaystyle{ -\frac{1}{9} t_1 x^7+\frac{t_2 x^7}{9}+\frac{598 t_1^3 x^5}{14175}-\frac{598 t_2^3 x^5}{14175}+\frac{1619 t_1 t_2^2 x^5}{14175}-\frac{1619 t_1^2 t_2 x^5}{14175} }[/math]

[math]\displaystyle{ -\frac{74 t_1^5 x^3}{14175}+\frac{74 t_2^5 x^3}{14175}-\frac{74 t_1 t_2^4 x^3}{2835}+\frac{148 t_1^2 t_2^3 x^3}{2835}-\frac{148 t_1^3 t_2^2 x^3}{2835}+\frac{74 t_1^4 t_2 x^3}{2835} }[/math]

[math]\displaystyle{ +\frac{t_1^7 x}{4725}-\frac{t_2^7 x}{4725}+\frac{1}{675} t_1 t_2^6 x-\frac{1}{225} t_1^2 t_2^5 x+\frac{1}{135} t_1^3 t_2^4 x-\frac{1}{135} t_1^4 t_2^3 x+\frac{1}{225} t_1^5 t_2^2 x-\frac{1}{675} t_1^6 t_2 x }[/math]

[math]\displaystyle{ -\frac{1}{11} t_1 x^9+\frac{t_2 x^9}{11}+\frac{2414 t_1^3 x^7}{51975}-\frac{2414 t_2^3 x^7}{51975}+\frac{53243 t_1 t_2^2 x^7}{467775}-\frac{53243 t_1^2 t_2 x^7}{467775} }[/math]

[math]\displaystyle{ -\frac{4058 t_1^5 x^5}{467775}+\frac{4058 t_2^5 x^5}{467775}-\frac{3904 t_1 t_2^4 x^5}{93555}+\frac{782 t_1^2 t_2^3 x^5}{10395}-\frac{782 t_1^3 t_2^2 x^5}{10395}+\frac{3904 t_1^4 t_2 x^5}{93555} }[/math]

[math]\displaystyle{ +\frac{331 t_1^7 x^3}{467775}-\frac{331 t_2^7 x^3}{467775}+\frac{331 t_1 t_2^6 x^3}{66825}-\frac{331 t_1^2 t_2^5 x^3}{22275}+\frac{331 t_1^3 t_2^4 x^3}{13365}-\frac{331 t_1^4 t_2^3 x^3}{13365}+\frac{331 t_1^5 t_2^2 x^3}{22275}-\frac{331 t_1^6 t_2 x^3}{66825} }[/math]

[math]\displaystyle{ -\frac{2 t_1^9 x}{93555}+\frac{2 t_2^9 x}{93555}-\frac{2 t_1 t_2^8 x}{10395}+\frac{8 t_1^2 t_2^7 x}{10395}-\frac{8 t_1^3 t_2^6 x}{4455}+\frac{4 t_1^4 t_2^5 x}{1485}-\frac{4 t_1^5 t_2^4 x}{1485}+\frac{8 t_1^6 t_2^3 x}{4455}-\frac{8 t_1^7 t_2^2 x}{10395}+\frac{2 t_1^8 t_2 x}{10395} }[/math]

[math]\displaystyle{ -\frac{1}{13} t_1 x^{11}+\frac{t_2 x^{11}}{13}+\frac{231523 t_1^3 x^9}{4729725}-\frac{231523 t_2^3 x^9}{4729725}+\frac{14046661 t_1 t_2^2 x^9}{127702575}-\frac{14046661 t_1^2 t_2 x^9}{127702575} }[/math]

[math]\displaystyle{ -\frac{2589746 t_1^5 x^7}{212837625}+\frac{2589746 t_2^5 x^7}{212837625}-\frac{285224 t_1 t_2^4 x^7}{5108103}+\frac{462340 t_1^2 t_2^3 x^7}{5108103}-\frac{462340 t_1^3 t_2^2 x^7}{5108103}+\frac{285224 t_1^4 t_2 x^7}{5108103} }[/math]

[math]\displaystyle{ +\frac{1304 t_1^7 x^5}{875875}-\frac{1304 t_2^7 x^5}{875875}+\frac{34493 t_1 t_2^6 x^5}{3378375}-\frac{891986 t_1^2 t_2^5 x^5}{30405375}+\frac{114577 t_1^3 t_2^4 x^5}{2606175}-\frac{114577 t_1^4 t_2^3 x^5}{2606175} }[/math]

[math]\displaystyle{ +\frac{891986 t_1^5 t_2^2 x^5}{30405375}-\frac{34493 t_1^6 t_2 x^5}{3378375} }[/math]

[math]\displaystyle{ -\frac{19178 t_1^9 x^3}{212837625}+\frac{19178 t_2^9 x^3}{212837625}-\frac{19178 t_1 t_2^8 x^3}{23648625}+\frac{76712 t_1^2 t_2^7 x^3}{23648625}-\frac{76712 t_1^3 t_2^6 x^3}{10135125}+\frac{38356 t_1^4 t_2^5 x^3}{3378375}-\frac{38356 t_1^5 t_2^4 x^3}{3378375} }[/math]

[math]\displaystyle{ +\frac{76712 t_1^6 t_2^3 x^3}{10135125}-\frac{76712 t_1^7 t_2^2 x^3}{23648625}+\frac{19178 t_1^8 t_2 x^3}{23648625} }[/math]

[math]\displaystyle{ +\frac{1382 t_1^{11} x}{638512875}-\frac{1382 t_2^{11} x}{638512875}+\frac{1382 t_1 t_2^{10} x}{58046625}-\frac{1382 t_1^2 t_2^9 x}{11609325}+\frac{1382 t_1^3 t_2^8 x}{3869775}-\frac{2764 t_1^4 t_2^7 x}{3869775}+\frac{2764 t_1^5 t_2^6 x}{2764125}-\frac{2764 t_1^6 t_2^5 x}{2764125} }[/math]

[math]\displaystyle{ +\frac{2764 t_1^7 t_2^4 x}{3869775}-\frac{1382 t_1^8 t_2^3 x}{3869775}+\frac{1382 t_1^9 t_2^2 x}{11609325}-\frac{1382 t_1^{10} t_2 x}{58046625} }[/math]

and

[math]\displaystyle{ f_{21}=1+\frac{1}{9} x^2 t_1 t_2-\frac{1}{9} x^2 t_1^2 -\frac{13}{135} t_1^2 x^4+\frac{13}{135} t_1 t_2 x^4+\frac{2}{135} t_1^4 x^2+\frac{2}{45} t_1^2 t_2^2 x^2-\frac{8}{135} t_1^3 t_2 x^2 }[/math]

[math]\displaystyle{ -\frac{1147 t_1^2 x^6}{14175}+\frac{1147 t_1 t_2 x^6}{14175}+\frac{13}{525} t_1^4 x^4+\frac{878 t_1^2 t_2^2 x^4}{14175}-\frac{1229 t_1^3 t_2 x^4}{14175}-\frac{1}{525} t_1^6 x^2+\frac{2}{105} t_1^3 t_2^3 x^2-\frac{1}{35} t_1^4 t_2^2 x^2+\frac{2}{175} t_1^5 t_2 x^2 }[/math]

[math]\displaystyle{ -\frac{2939 t_1^2 x^8}{42525}+\frac{2939 t_1 t_2 x^8}{42525}+\frac{1327 t_1^4 x^6}{42525}+\frac{2896 t_1^2 t_2^2 x^6}{42525}-\frac{4223 t_1^3 t_2 x^6}{42525} }[/math]

[math]\displaystyle{ -\frac{199 t_1^6 x^4}{42525}+\frac{20}{567} t_1^3 t_2^3 x^4-\frac{97 t_1^4 t_2^2 x^4}{1701}+\frac{1124 t_1^5 t_2 x^4}{42525}+\frac{2 t_1^8 x^2}{8505}+\frac{2}{243} t_1^4 t_2^4 x^2-\frac{16 t_1^5 t_2^3 x^2}{1215}+\frac{8 t_1^6 t_2^2 x^2}{1215}-\frac{16 t_1^7 t_2 x^2}{8505} }[/math]

[math]\displaystyle{ -\frac{2953639 t_1^2 x^{10}}{49116375}+\frac{2953639 t_1 t_2 x^{10}}{49116375}+\frac{1740446 t_1^4 x^8}{49116375}+\frac{3411068 t_1^2 t_2^2 x^8}{49116375}-\frac{5151514 t_1^3 t_2 x^8}{49116375} }[/math]

[math]\displaystyle{ -\frac{382048 t_1^6 x^6}{49116375}+\frac{152648 t_1^3 t_2^3 x^6}{3274425}-\frac{87874 t_1^4 t_2^2 x^6}{1091475}+\frac{2046658 t_1^5 t_2 x^6}{49116375} }[/math]

[math]\displaystyle{ +\frac{7472 t_1^8 x^4}{9823275}+\frac{25931 t_1^4 t_2^4 x^4}{1403325}-\frac{45641 t_1^5 t_2^3 x^4}{1403325}+\frac{5377 t_1^6 t_2^2 x^4}{280665}-\frac{57697 t_1^7 t_2 x^4}{9823275} }[/math]

[math]\displaystyle{ -\frac{1382 t_1^{10} x^2}{49116375}+\frac{2764 t_1^5 t_2^5 x^2}{779625}-\frac{2764 t_1^6 t_2^4 x^2}{467775}+\frac{11056 t_1^7 t_2^3 x^2}{3274425}-\frac{1382 t_1^8 t_2^2 x^2}{1091475}+\frac{2764 t_1^9 t_2 x^2}{9823275} }[/math]?

(These Taylor expansions are also available within the mathematica notebook HOMFLY Braidor - Braidor Computations.nb).