"Jones-Ocneanu trace"의 두 판 사이의 차이

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

2020년 11월 16일 (월) 11:03 기준 최신판

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 \(u^{-1}\).