07-1352/Class Notes for January 23

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In Preparation

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

A HOMFLY Braidor

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. $[t_i,t_j]=x\sigma_{ij}(t_j-t_i)$, 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 $x^kt_1^{k_1}t_2^{k_2}\cdots t_n^{k_n}\cdot\sigma$ 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 $B_n$.

Examples.

1. The general element of $A_1$ is $f(x,t_1)(1)$ 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 $f(x,t_1,t_2)(12)+g(x,t_1,t_2)(21)$ 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.