"Jones-Ocneanu trace"의 두 판 사이의 차이

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imported>Pythagoras0
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==introduction==
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* linear functional on the Hecke algebra of type $A_n$
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* Jones related a trace found by Ocneanu with [[HOMFLY polynomial]]
  
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==construction==
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* let  $W=S_n$, the symmetric group on $n$ letters and $S$ the set of transpositions $s_i: = (i,i+1)$
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* Coxeter matrix  : $m_{ii}=1$  and $m_{i,i+1}=3$, 2 otherwise
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Let $A$ be a commutative ring with 1 and fix two invertible elements $u, v \in A$.
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For $n\geq 1$, consider $H_A(S_n)$ associated with $S_n$ over the ring $A$ and with parameters $a_{s_i} = u$, $b_{s_i} = v$ for $1\leq i \leq n - 1$.
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For simplicity, let $H_n: = H_A(S_n)$.
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Regard $H_n$ as a subalgebra of $H_{n+1}$.
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;thm (Jones, Ocneanu)
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There is a unique family $\{\tau_n\}_{n\geq1}$ of $A$-linear maps $\tau_n : H_n \to A$  s.t. the following conditions hold :
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$$
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\begin{array}{ll}
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(M1) & \tau_1(T_e)=1 \\
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(M2) & \tau_{n+1}(hT_{s_n}^{\pm})=\tau_{n}(h) \quad& \text{for $n\geq1$ and $h\in H_n$} \\
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(M3) & \tau_n(hh')=\tau_{n}(h'h) & \text{for $n\geq1$ and $h,h'\in H_n$}
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\end{array}
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$$
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Moreover, $\tau_{n+1}(h)=v^{-1}(1-u)\tau_{n}(h)$ for all $n\geq 1$ and $h\in H_n$.
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;proof
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Let us define $\tau_n$ recursively as follows.
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For $n=1$, set $\tau_1(T_e)=1$.
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Let $n\geq 1$ and assume that $\tau_n$ has been defined. Then we set
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\begin{equation}\label{star}
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\tau_{n+1}(a+b T_{s_n}c):=\frac{1-u}{v}\tau_n(a)+\tau_n(bc), \, a,b,c\in H_n
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\end{equation}
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need to check that $\tau_{n+1}$ is well-defined and satisfies M2, M3.
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we need the isomorphism of $A$-modules for $n\geq 2$:
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$$
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\psi_n: H_n\oplus (H_n\otimes_{H_{n-1}} H_n) \to H_{n+1},\, a\oplus (b\otimes c)\mapsto a+bT_{s_n}c
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$$
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* $v^{-1}$ is used when we define $\tau_n$
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* note that to have $T_{s_n}^{-1}$, we need $u^{-1}$.
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[[분류:Knot theory]]
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[[분류:migrate]]

2020년 11월 13일 (금) 04:10 판

introduction

  • linear functional on the Hecke algebra of type $A_n$
  • Jones related a trace found by Ocneanu with HOMFLY polynomial


construction

  • let $W=S_n$, the symmetric group on $n$ letters and $S$ the set of transpositions $s_i: = (i,i+1)$
  • Coxeter matrix : $m_{ii}=1$ and $m_{i,i+1}=3$, 2 otherwise

Let $A$ be a commutative ring with 1 and fix two invertible elements $u, v \in A$.

For $n\geq 1$, consider $H_A(S_n)$ associated with $S_n$ over the ring $A$ and with parameters $a_{s_i} = u$, $b_{s_i} = v$ for $1\leq i \leq n - 1$.

For simplicity, let $H_n: = H_A(S_n)$.

Regard $H_n$ as a subalgebra of $H_{n+1}$.

thm (Jones, Ocneanu)

There is a unique family $\{\tau_n\}_{n\geq1}$ of $A$-linear maps $\tau_n : H_n \to A$ s.t. the following conditions hold : $$ \begin{array}{ll} (M1) & \tau_1(T_e)=1 \\ (M2) & \tau_{n+1}(hT_{s_n}^{\pm})=\tau_{n}(h) \quad& \text{for $n\geq1$ and $h\in H_n$} \\ (M3) & \tau_n(hh')=\tau_{n}(h'h) & \text{for $n\geq1$ and $h,h'\in H_n$} \end{array} $$ Moreover, $\tau_{n+1}(h)=v^{-1}(1-u)\tau_{n}(h)$ for all $n\geq 1$ and $h\in H_n$.

proof

Let us define $\tau_n$ recursively as follows.

For $n=1$, set $\tau_1(T_e)=1$.

Let $n\geq 1$ and assume that $\tau_n$ has been defined. Then we set \begin{equation}\label{star} \tau_{n+1}(a+b T_{s_n}c):=\frac{1-u}{v}\tau_n(a)+\tau_n(bc), \, a,b,c\in H_n \end{equation}

need to check that $\tau_{n+1}$ is well-defined and satisfies M2, M3.

we need the isomorphism of $A$-modules for $n\geq 2$:

$$ \psi_n: H_n\oplus (H_n\otimes_{H_{n-1}} H_n) \to H_{n+1},\, a\oplus (b\otimes c)\mapsto a+bT_{s_n}c $$ ■

  • $v^{-1}$ is used when we define $\tau_n$
  • note that to have $T_{s_n}^{-1}$, we need $u^{-1}$.