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 $A_{n}^{0}=\langle S_{n},x,t_{1},\ldots t_{n}\rangle$ 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 $A_{n}^{1}$ be the quotient of $A_{n}^{0}$ 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 $A_{n}^{1}$.

We say that an element of $A_{n}$ is "sorted" if it is written in the form $x^{k}t_{1}^{k_{1}}t_{2}^{k_{2}}\cdots t_{n}^{k_{n}}\cdot \sigma$ 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 $B_{n}$.