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], 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?

up vote 1 down vote accepted

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).

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