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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^k\cdot\sigmat_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 $B_{n}$.

Examples.

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.