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練習問題メモ 7（置換）

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練習問題メモ 7（置換）#

7.1#

次の \(1, 2\) の置換 \(\sigma, \tau\) に対して、積 \(\sigma \tau\) および \(\tau \sigma\) を求めよ。

  1. \(\sigma=\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2\end{array}\right), \tau=\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right)\)

  2. \(\sigma=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3\end{array}\right), \tau=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2\end{array}\right)\)

  1. \(\sigma=\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2\end{array}\right), \tau=\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right)\)

\[\begin{split} \sigma \tau= \begin{pmatrix} 1 & 2 & 3\\ \sigma(\tau(1)) & \sigma(\tau(2)) & \sigma(\tau(3)) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 3 \end{pmatrix} \end{split}\]
\[\begin{split} \tau \sigma = \begin{pmatrix} 1 & 2 & 3\\ \tau(\sigma(1)) & \tau(\sigma(2)) & \tau(\sigma(3)) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2\\ \end{pmatrix} \end{split}\]

  1. \(\sigma=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3\end{array}\right), \tau=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2\end{array}\right)\)

\[\begin{split} \sigma \tau= \begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{pmatrix} \end{split}\]
\[\begin{split} \tau \sigma = \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3 \end{pmatrix} \end{split}\]

7.2#

次の 1、 2 の置換 \(\sigma\) の符号を求めよ。

  1. \(\sigma=\left(\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 4 & 5 & 1 & 6 & 2 & 7 & 3\end{array}\right)\)

  2. \(\sigma=\left(\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 4 & 1 & 5 & 2 & 6 & 3\end{array}\right)\)

  1. \(\sigma=\left(\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 4 & 5 & 1 & 6 & 2 & 7 & 3\end{array}\right)\)

\[\begin{split} \begin{align} \sigma &= (\begin{array}{lll} 1 & 4 & 6 & 7 & 3 \end{array}) (\begin{array}{lll} 2 & 5 \end{array})\\ &= (\begin{array}{lll} 1 & 3 \end{array}) (\begin{array}{lll} 1 & 7 \end{array}) (\begin{array}{lll} 1 & 6 \end{array}) (\begin{array}{lll} 1 & 4 \end{array}) (\begin{array}{lll} 2 & 5 \end{array})\\ \end{align} \end{split}\]

なので

\[ \text{sgn}(\sigma) = (-1)^5 = -1 \]

  1. \(\sigma=\left(\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 4 & 1 & 5 & 2 & 6 & 3\end{array}\right)\)

\[\begin{split} \begin{align} \sigma &= (\begin{array}{lll} 1 & 7 & 3 \end{array}) (\begin{array}{lll} 2 & 4 & 5 \end{array}) \\ &= (\begin{array}{lll} 1 & 3 \end{array}) (\begin{array}{lll} 1 & 7 \end{array}) (\begin{array}{lll} 2 & 5 \end{array}) (\begin{array}{lll} 2 & 4 \end{array}) \\ \end{align} \end{split}\]

なので

\[ \text{sgn}(\sigma) = (-1)^4 = 1 \]

7.3#

\(\sigma\) が \(n\) 文字の置換であることを \(\sigma \in S_n\) と表す。 \(n\) 変数 \(x_1, x_2, \cdots, x_n\) の多項式 \(f_\sigma\left(x_1, x_2, \cdots, x_n\right)\) および \(\sigma \in S_n\) に対して、多項式 \(f_\sigma\) を

\[ f_\sigma\left(x_1, x_2, \cdots, x_n\right)=f\left(x_{\sigma(1)}, x_{\sigma(2)}, \cdots, x_{\sigma(n)}\right) \]

により定める。 \(f\) および \(\sigma\) が次の \(1 \sim 3\) により与えられるとき、 \(f_\sigma\) を求めよ。

  1. \(f\left(x_1, x_2, x_3\right)=x_1+2 x_2+3 x_3, \quad \sigma=\epsilon \in S_3\)

  2. \(f\left(x_1, x_2, x_3, x_4\right)=\left(x_1-x_2\right)\left(x_3-x_4\right), \quad \sigma=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 2 & 1 & 3\end{array}\right) \in S_4\)

  3. \(f\left(x_1, x_2, x_3, x_4\right)=1+x_1+x_2 x_3+x_4^3, \quad \sigma=\left(\begin{array}{lll}1 & 4 & 2\end{array}\right) \in S_4\)

  1. \(f\left(x_1, x_2, x_3\right)=x_1+2 x_2+3 x_3, \quad \sigma=\epsilon \in S_3\)

\(\epsilon\)は恒等置換だと解釈し、\(x_{\sigma(i)}\)は\(x\)を\(\sigma\)で置換したものと考え、\(\sigma f(\cdot) = f_{\sigma}(\cdot)\)と解釈する

\[ f_{\sigma} \left(x_1, x_2, x_3\right) = x_1+2 x_2+3 x_3 \]

  1. \(f\left(x_1, x_2, x_3, x_4\right)=\left(x_1-x_2\right)\left(x_3-x_4\right), \quad \sigma=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 2 & 1 & 3\end{array}\right) \in S_4\)

巡回置換ではあるが一度しか\(\sigma\)を掛けないものとかんがえて

\[ f_{\sigma}\left(x_1, x_2, x_3, x_4\right) = \left(x_4 - x_2\right)\left(x_1 - x_3\right) \]

  1. \(f\left(x_1, x_2, x_3, x_4\right)=1+x_1+x_2 x_3+x_4^3, \quad \sigma=\left(\begin{array}{lll}1 & 4 & 2\end{array}\right) \in S_4\)

\[ f_{\sigma}\left(x_1, x_2, x_3, x_4\right) = 1 + x_4 + x_1 x_3 + x_2^3 \]
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