【第2章】対称群 \(S_4\)
【2-4】 \(\mathbb{C}[S_4]\) の原始冪等元
表現論の本には、「 \(S_4\) の正則表現は(24x24)の行列は、既約表現に分解すると、下式(4.1)の様に(1x1)+3x(3x3)+2x(2x2)+3x(3x3)+(1x1) の合計10個の小行列に分解される。」と説明されております。
\begin{align} \widetilde{L_{i}}=T^{-1} \cdot L_{i} \cdot T &= \begin{pmatrix} \boxed{1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \boxed{ \begin{matrix} *&*&* \\ *&*&* \\ *&*&* \end{matrix}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \boxed{ \begin{matrix} *&*&* \\ *&*&* \\ *&*&* \end{matrix}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \boxed{ \begin{matrix} *&*&* \\ *&*&* \\ *&*&* \end{matrix}} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \boxed{ \begin{matrix} *&* \\ *&* \end{matrix}} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \boxed{ \begin{matrix} *&* \\ *&* \end{matrix}} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \boxed{ \begin{matrix} *&*&* \\ *&*&* \\ *&*&* \end{matrix} } & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \boxed{ \begin{matrix} *&*&* \\ *&*&* \\ *&*&* \end{matrix} } & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \boxed{ \begin{matrix} *&*&* \\ *&*&* \\ *&*&* \end{matrix} } & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \boxed{\pm 1}\end{pmatrix} \\ \end{align}
その為に重要となる \(\mathbb{C}[S_4]\) における10個の原始冪等元が以下の \(\{e_{1},...,e_{10}\}\) です。
【第1章】と同様に、原始冪等元の導出の仕方は【第3章】で説明することにします。
\begin{align} e_{1}&=\frac{1}{24}\biggl( {g_{24}}+{g_{23}}+{g_{22}}+{g_{21}}+{g_{20}}+{g_{19}}+{g_{18}}+{g_{17}}+{g_{16}}+{g_{15}}+{g_{14}}+{g_{13}}\biggr. \notag \\ &\qquad \qquad \biggl. +{g_{12}}+{g_{11}}+{g_{10}}+{g_9}+{g_8}+{g_7}+{g_6}+{g_5}+{g_4}+{g_3}+{g_2}+{g_1} \biggr) \\ \notag \\ e_{2}&=\frac{1}{8}\biggl( -{g_{24}}-{g_{21}}-{g_{20}}-{g_{15}}-{g_{14}}+{g_{12}}+{g_{10}}-{g_7}+{g_6}+{g_5}+{g_3}+{g_1} \biggr) \\ \notag \\ e_{3}&=\frac{1}{8}\biggl( -{g_{23}}-{g_{19}}-{g_{18}}+{g_{14}}-{g_{13}}-{g_{12}}+{g_{11}}+{g_7}-{g_6}+{g_5}+{g_4}+{g_1} \biggr) \\ \notag \\ e_{4}&=\frac{1}{8}\biggl( -{g_{22}}-{g_{17}}-{g_{16}}+{g_{15}}+{g_{13}}-{g_{11}}-{g_{10}}+{g_7}+{g_6}-{g_5}+{g_2}+{g_1} \biggr) \\ \notag \\ e_{5}&=\frac{1}{12}\biggl( {g_{24}}+{g_{23}}+{g_{22}}+{g_{21}}-{g_{20}}+{g_{19}}-{g_{16}}-{g_{15}} \biggr. \notag \\ &\qquad \qquad \biggl. -{g_{12}}-{g_{11}}-{g_8}-{g_6}+{g_5}-{g_4}+{g_2}+{g_1} \biggr) \\ \notag \\ e_{6}&=\frac{1}{12}\biggl( {g_{24}}+{g_{23}}+{g_{22}}-{g_{21}}+{g_{20}}-{g_{19}}+{g_{16}}-{g_{14}} \biggr. \notag \\ &\qquad \qquad \biggl. -{g_{13}}-{g_{10}}-{g_9}+{g_6}-{g_5}+{g_4}-{g_2}+{g_1} \biggr) \\ \notag \\ e_{7}&=\frac{1}{8}\biggl( -{g_{22}}+{g_{20}}+{g_{18}}+{g_{15}}-{g_{14}}+{g_{13}}-{g_{12}}-{g_7}-{g_6}+{g_5}-{g_2}+{g_1} \biggr) \\ \notag \\ e_{8}&=\frac{1}{8}\biggl( -{g_{23}}+{g_{21}}+{g_{17}}-{g_{15}}+{g_{14}}+{g_{11}}-{g_{10}}-{g_7}+{g_6}-{g_5}-{g_4}+{g_1} \biggr) \\ \notag \\ e_{9}&=\frac{1}{8}\biggl( -{g_{24}}+{g_{19}}+{g_{16}}-{g_{13}}+{g_{12}}-{g_{11}}+{g_{10}}+{g_7}-{g_6}-{g_5}-{g_3}+{g_1} \biggr) \\ \notag \\ e_{10}&=\frac{1}{24}\biggl( {g_{24}}+{g_{23}}+{g_{22}}-{g_{21}}-{g_{20}}-{g_{19}}-{g_{18}}-{g_{17}}-{g_{16}}+{g_{15}}+{g_{14}}+{g_{13}} \biggr. \notag \\ &\qquad \qquad \biggl. +{g_{12}}+{g_{11}}+{g_{10}}+{g_9}+{g_8}-{g_7}-{g_6}-{g_5}-{g_4}-{g_3}-{g_2}+{g_1} \biggr) \\ \end{align}
【2-5】 \(\mathbb{C}[S_4]\) の原始冪等元の右正則表現
既約表現に分解する為の変換行列 \(T\) を生成する為には、まず上記10個の原始冪等元の右正則表現を計算する必要があります。その計算法は(3.2)の \(x\) に \(e_{i}\) を代入してそれそれの \(R_i\) を計算すればよいだけです。一例だけ原始冪等元 \(e_5\) の右正則表現 \(R_5\) の計算結果を以下に示します。
\begin{align} R_{5}=\frac{1}{12}\begin{bmatrix}1 & 1 & 0 & -1 & 1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & -1 & 0 & 0 & 1 & -1 & 1 & 1 & 1 & 1\\ 1 & 1 & -1 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 0 & -1 & -1 & 1 & 0 & 1 & 1 & 1 & 1\\ 0 & 0 & 1 & -1 & 0 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ -1 & -1 & 0 & 1 & -1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & -1 & 1 & -1 & -1 & -1 & -1\\ 1 & 1 & -1 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 0 & -1 & -1 & 1 & 0 & 1 & 1 & 1 & 1\\ -1 & -1 & 0 & 1 & -1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & -1 & 1 & -1 & -1 & -1 & -1\\ 0 & 0 & 1 & -1 & 0 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ -1 & -1 & 1 & 0 & -1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & -1 & 0 & -1 & -1 & -1 & -1\\ 0 & 0 & -1 & 1 & 0 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 1 & 0 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ -1 & -1 & 1 & 0 & -1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & -1 & 0 & -1 & -1 & -1 & -1\\ -1 & -1 & 1 & 0 & -1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & -1 & 0 & -1 & -1 & -1 & -1\\ 0 & 0 & -1 & 1 & 0 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 1 & 0 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ -1 & -1 & 1 & 0 & -1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & -1 & 0 & -1 & -1 & -1 & -1\\ -1 & -1 & 0 & 1 & -1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & -1 & 1 & -1 & -1 & -1 & -1\\ 0 & 0 & 1 & -1 & 0 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & -1 & 0 & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ 1 & 1 & -1 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 0 & -1 & -1 & 1 & 0 & 1 & 1 & 1 & 1\\ -1 & -1 & 0 & 1 & -1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & -1 & 1 & -1 & -1 & -1 & -1\\ 1 & 1 & -1 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 0 & -1 & -1 & 1 & 0 & 1 & 1 & 1 & 1\\ 1 & 1 & 0 & -1 & 1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & -1 & 0 & 0 & 1 & -1 & 1 & 1 & 1 & 1\\ 1 & 1 & 0 & -1 & 1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & -1 & 0 & 0 & 1 & -1 & 1 & 1 & 1 & 1\\ 1 & 1 & 0 & -1 & 1 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & -1 & 0 & -1 & 0 & 0 & 1 & -1 & 1 & 1 & 1 & 1\end{bmatrix} \\ \end{align}
