The HOMFLY Braidor Algebra

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The Algebra

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

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

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

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


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

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

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

The Equations

We seek to find a "braidor"; an element B of A_2 satisfying:

  • B=(21)+x(12)+(higher order terms).
  • B(\Delta B)B=(\Delta B)B(\Delta B) in A_3.

With the vector space identification of A_n with B_n 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

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 f_{12} and f_{21} of the variables t_1, t_2 and x, whose Taylor expansions begin with

f_{12}=x+\frac{x t_2}{3}-\frac{x t_1}{3}

-\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
-\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
-\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
-\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}
-\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}
   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
-\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}
-\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}
+\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
-\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


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

-\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
-\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}
-\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}?