07-1352/Class Notes for January 23

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In Preparation

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

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 are

$f_{12}=x+{\frac {xt_{2}}{3}}-{\frac {xt_{1}}{3}}$

$-{\frac {1}{5}}t_{1}x^{3}+{\frac {t_{2}x^{3}}{5}}+{\frac {t_{1}^{3}x}{45}}-{\frac {t_{2}^{3}x}{45}}+{\frac {1}{15}}t_{1}t_{2}^{2}x-{\frac {1}{15}}t_{1}^{2}t_{2}x$

$-{\frac {1}{7}}t_{1}x^{5}+{\frac {t_{2}x^{5}}{7}}+{\frac {11}{315}}t_{1}^{3}x^{3}-{\frac {11}{315}}t_{2}^{3}x^{3}+{\frac {11}{105}}t_{1}t_{2}^{2}x^{3}-{\frac {11}{105}}t_{1}^{2}t_{2}x^{3}$

$-{\frac {2t_{1}^{5}x}{945}}+{\frac {2t_{2}^{5}x}{945}}-{\frac {2}{189}}t_{1}t_{2}^{4}x+{\frac {4}{189}}t_{1}^{2}t_{2}^{3}x-{\frac {4}{189}}t_{1}^{3}t_{2}^{2}x+{\frac {2}{189}}t_{1}^{4}t_{2}x$

$-{\frac {1}{9}}t_{1}x^{7}+{\frac {t_{2}x^{7}}{9}}+{\frac {598t_{1}^{3}x^{5}}{14175}}-{\frac {598t_{2}^{3}x^{5}}{14175}}+{\frac {1619t_{1}t_{2}^{2}x^{5}}{14175}}-{\frac {1619t_{1}^{2}t_{2}x^{5}}{14175}}$

$-{\frac {74t_{1}^{5}x^{3}}{14175}}+{\frac {74t_{2}^{5}x^{3}}{14175}}-{\frac {74t_{1}t_{2}^{4}x^{3}}{2835}}+{\frac {148t_{1}^{2}t_{2}^{3}x^{3}}{2835}}-{\frac {148t_{1}^{3}t_{2}^{2}x^{3}}{2835}}+{\frac {74t_{1}^{4}t_{2}x^{3}}{2835}}$

$+{\frac {t_{1}^{7}x}{4725}}-{\frac {t_{2}^{7}x}{4725}}+{\frac {1}{675}}t_{1}t_{2}^{6}x-{\frac {1}{225}}t_{1}^{2}t_{2}^{5}x+{\frac {1}{135}}t_{1}^{3}t_{2}^{4}x-{\frac {1}{135}}t_{1}^{4}t_{2}^{3}x+{\frac {1}{225}}t_{1}^{5}t_{2}^{2}x-{\frac {1}{675}}t_{1}^{6}t_{2}x$

and

$f_{21}=1+{\frac {1}{9}}x^{2}t_{1}t_{2}-{\frac {1}{9}}x^{2}t_{1}^{2}-{\frac {13}{135}}t_{1}^{2}x^{4}+{\frac {13}{135}}t_{1}t_{2}x^{4}+{\frac {2}{135}}t_{1}^{4}x^{2}+{\frac {2}{45}}t_{1}^{2}t_{2}^{2}x^{2}-{\frac {8}{135}}t_{1}^{3}t_{2}x^{2}$

$-{\frac {1147t_{1}^{2}x^{6}}{14175}}+{\frac {1147t_{1}t_{2}x^{6}}{14175}}+{\frac {13}{525}}t_{1}^{4}x^{4}+{\frac {878t_{1}^{2}t_{2}^{2}x^{4}}{14175}}-{\frac {1229t_{1}^{3}t_{2}x^{4}}{14175}}-{\frac {1}{525}}t_{1}^{6}x^{2}+{\frac {2}{105}}t_{1}^{3}t_{2}^{3}x^{2}-{\frac {1}{35}}t_{1}^{4}t_{2}^{2}x^{2}+{\frac {2}{175}}t_{1}^{5}t_{2}x^{2}$?

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. $[t_{i},t_{j}]=x\sigma _{ij}(t_{i}-t_{j})$, 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 $A_{n}^{1}$.

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 $B_{n}$.

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.

A Solution

The first few terms of a solution can be computed using a computer, as shown above and 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.