**Background/Notation:**


Given the Iwahori-Hecke algebra $H\_{n}$ (over some ring commutative ring $R$ with identity) with generators $\{T\_{1},\ldots T\_{n-1}\}$, we know it has basis 
$\{T\_{w}\}\_{w\in S\_{n}}$, where $S\_{n}$ is the permutation group and for some reduced expression $w=s\_{i\_{1}}\cdots s\_{i\_{r}}$, we have $T\_{w}=T\_{i\_{1}}\cdots T\_{i\_{r}}$. This basis can be pictured as the set of all positive braids corresponding to each $w\in S\_{n}$. *E.g.* for $w=s\_{i\_{1}}\cdots s\_{i\_{r}}$, we write $\sigma\_{w}=\sigma\_{i\_{1}}\cdots \sigma\_{i\_{r}}$. 


That is, if we consider the isomorphism $\varphi:R[B\_{n}]\backslash{I}\to H\_{n}$ (where $I$ is the ideal generated by the Hecke relation $(T\_{i}-q\_{1})(T\_{i}-q\_{2}) \ \forall i$ with units $q\_{1}, q\_{2}\in R$), have $\varphi(\sigma\_{w})=T\_{w}$ whence we may identify the two.


We obtain the annular Skein algebra $V$ by taking the braid closure of all elements in $H\_{n}$ (so $V$ is the vector space consisting of $R$-linear combinations of closed $n$-braids *modulo* Reidemeister-II/III moves and the Skein relation (i.e. Hecke relation)).


**Problem:**


*What is a basis for $V$*? Consider the linear map $\pi:H\_{n}\to V$ (which we can think of as braid closure). $V$ should have the same basis as $H\_{n}$ "modulo closure" i.e. the problem reduces to *when does $\pi(T\_{w})=\pi(T\_{w'})$?* By Markov's theorem, this should be *iff* $\sigma\_{w}$ is conjugate to $\sigma\_{w'}$ (since only Markov-I moves are relevant here). So this appears to be equivalent to finding the conjugacy classes amongst the positive braids $\{\sigma\_{w}\}\_{w\in S\_{n}}$. Note that $V\cong H\_{n}\backslash{[.,.]}$ (since $\pi(T\_{w\_{1}}T\_{w\_{2}})=\pi(T\_{w\_{2}}T\_{w\_{1}})$).


**Pursuing an 'Algebraic' Solution:**


In [[Big05]](https://arxiv.org/abs/math/0505064), section 5 is devoted to determining such a basis. This is shown to be the image of braids $b\_{\lambda}$ under $\pi\circ\varphi$ where $b\_{\lambda}=b\_{(\lambda\_{1})}\sqcup\cdots\sqcup b\_{(\lambda\_{k})}$ with $b\_{(m)}=\sigma\_{m-1}\cdots\sigma\_{1}$ given a partition $\lambda=(\lambda\_{1},\ldots,\lambda\_{k})$ of $n$. This is done through geometrical/topological arguments. *Can we do this algebraically instead?*


**Sketch:**


From Bigelow's result, we can conclude that dim $V=p(n)$ (the $n^{th}$ partition no.) and $\pi$ is a linear projector. This tells us that the basis of $V$ is $\{T\_{\lambda}\}\_{\lambda\in\mathcal{P}}\subseteq \{T\_{w}\}\_{w\in S\_{n}}$, where $\mathcal{P}$ is the set of conjugacy classes in $S\_{n}$ (which we know to be the classes of different $\lambda$-cycles), and the chosen representative $\lambda\in S\_{n}$ is the 'canonical' permutation corresponding to the $\lambda=(\lambda\_{1},\ldots,\lambda\_{k})$-cycle: that is, in cycle-notation, we fill out $k$ brackets from left-to-right with the numbers $1$ to $n$, where the $i^{th}$ bracket has length $\lambda\_{i}$. 


Essentially, his result tells us that $\pi(T\_{w})=\pi(T\_{w'})$ *iff $w$ is conjugate to $w'$ in $S\_{n}$*. By our above development of the problem, does this mean that the algebraic route to Bigelow's result is to show that:


*$\sigma\_{w}$ is conjugate to $\sigma\_{w'}$ in $B\_{n}$ iff $w$ is conjugate to $w'$ in $S\_{n}$ ?*


I don't know if this final statement about conjugacy of positive braids is true, but if it is and the above reasoning is fine, then surely this is a valid route. If this isn't true, what's the best algebraic route to the solution?



Basis for Annular Skein Algebra
The outlined proof almost works: only need to consider the weaker statement "*$\sigma\_{w}$ is conjugate to $\sigma\_{w'}$ in $B\_{n} \implies w$ is conjugate to $w'$ in $S\_{n}$*". (A useful discussion with Wade Bloomquist sent me in the right direction regarding conjugacy). One can easily show this:


Use the homomorphism $\phi:B\_{n}\to S\_{n}$ (where $ker \ \phi=PB\_{n}$ (pure braid group)). Clearly, $\phi(\sigma\_{w})=w$ for $w$ reduced. Thus, $\sigma\_{w}=b\sigma\_{w'}b^{-1} \implies w=\phi(b)w'(\phi(b))^{-1}$. $\square$


So $\pi(T\_{w})=\pi(T\_{w'})\implies w$ conjugate to $w'$. Since the $T\_{i}$ are indexed by all words in $S\_{n}$, the basis of $V$ corresponds to the conjugacy classes of $S\_{n}$, whence we can arrive at the same conclusion as Bigelow (we write the basis in this way as it is the most convenient representative of the conjugacy class).


