description/proof of that for permutation, which moves \(j\)-th item to \(l\)-th place, sign is \((-1)^{l - j}\) times sign of corresponding \(j\)-th-item-lacking permutation
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of sign of permutation.
Target Context
- The reader will have a description and a proof of the proposition that for any permutation, which moves the \(j\)-th item to the \(l\)-th place, the sign is \((-1)^{l - j}\) times the sign of the corresponding \(j\)-th-item-lacking permutation.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\((1, ..., n)\):
\(\sigma\): \(: (1, ..., n) \mapsto (\sigma_1, ..., \sigma_n)\), \(\in \{\text{ the permutations }\}\), such that \(\sigma_l = j\)
\((1, ..., \widehat{j}, ..., n)\):
\(\sigma'\): \(: (1, ..., \widehat{j}, ..., n) \mapsto (\sigma_1, ..., \widehat{\sigma_l}, ..., \sigma_n)\), \(\in \{\text{ the permutations }\}\)
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Statements:
\(sgn \sigma = (-1)^{l - j} sgn \sigma'\)
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2: Proof
Whole Strategy: Step 1: suppose that \(j \lt l\), and see that the proposition holds; Step 2: suppose that \(l \lt j\), and see that the proposition holds; Step 3: suppose that \(j = l\), and see that the proposition holds.
Step 1:
Let us suppose that \(j \lt l\).
\(\sigma\) is \((1, ..., j - 1, j, j + 1, ..., l - 1, l, l + 1, ..., n) \mapsto (1, ..., j - 1, l, j + 1, ..., l - 1, j = \sigma_l, l + 1, ..., n) \mapsto (\sigma_1, ..., \sigma_{j - 1}, \sigma_j, \sigma_{j + 1}, ..., \sigma_{l - 1}, j = \sigma_l, \sigma_{l + 1}, ..., \sigma_n)\).
The sign of the 1st permutation is \(-1\), because it is just swapping \(j\) and \(l\).
Let the sign of the 2nd permutation be \((-1)^m\).
\(\sigma'\) is \((1, ..., j - 1, \widehat{j}, j + 1, ..., l - 1, l, l + 1, ..., n) \mapsto (1, ..., j - 1, l, j + 1, ..., l - 1, \widehat{j = \sigma_l}, l + 1, ..., n) \mapsto (\sigma_1, ..., \sigma_{j - 1}, \sigma_j, \sigma_{j + 1}, ..., \sigma_{l - 1}, \widehat{j = \sigma_l}, \sigma_{l + 1}, ..., \sigma_n)\).
The sign of the 1st permutation is \((-1)^{l - 1 - j}\), because \(l\) is 1st swapped with \(l - 1\), then with \(l - 2\), ..., then with \(j + 1\), so, is swapped with \(l - 1, ..., j + 1\), \(l - 1 - j\) times: \(l\) cannot be just swapped with \(\widehat{j}\), because \(\widehat{j}\) does not exist and \(l\) cannot be swapped with something that does not exist.
The sign of the 2nd permutation is \((-1)^m\), because \(\sigma_l\) is not moved anyway.
So, \(sgn \sigma = (-1) (-1)^m\) and \(sgn \sigma' = (-1)^{l - 1 - j} (-1)^m\).
So, \((-1)^m = sgn \sigma' (-1)^{- l + 1 + j}\), and so, \(sgn \sigma = (-1) sgn \sigma' (-1)^{- l + 1 + j} = (-1)^{- l + j + 2} sgn \sigma' = (-1)^{- l + j + 2 l - 2 j} sgn \sigma' = (-1)^{l - j} sgn \sigma'\).
Step 2:
Let us suppose that \(l \lt j\).
The logic is parallel with Step 1 and the result is easily expected, but let us do it diligently.
\(\sigma\) is \((1, ..., l - 1, l, l + 1, ..., j - 1, j, j + 1, ..., n) \mapsto (1, ..., l - 1, j = \sigma_l, l + 1, ..., j - 1, l, j + 1, ..., n) \mapsto (\sigma_1, ..., \sigma_{l - 1}, j = \sigma_l, \sigma_{l + 1}, ..., \sigma_{j - 1}, \sigma_j, \sigma_{j + 1}, ..., \sigma_n)\).
The sign of the 1st permutation is \(-1\), because it is just swapping \(j\) and \(l\).
Let the sign of the 2nd permutation be \((-1)^m\).
\(\sigma'\) is \((1, ..., l - 1, l, l + 1, ..., j - 1, \widehat{j}, j + 1, ..., n) \mapsto (1, ..., l - 1, \widehat{j = \sigma_l}, l + 1, ..., j - 1, l, j + 1, ..., n) \mapsto (\sigma_1, ..., \sigma_{l - 1}, \widehat{j = \sigma_l}, \sigma_{l + 1}, ..., \sigma_{j - 1}, \sigma_j, \sigma_{j + 1}, ..., \sigma_n)\).
The sign of the 1st permutation is \((-1)^{j - 1 - l}\), because \(l\) is 1st swapped with \(l + 1\), then with \(l + 2\), ..., then with \(j - 1\), so, is swapped with \(l + 1, ..., j - 1\), \(j - 1 - l\) times: \(l\) cannot be just swapped with \(\widehat{j}\), because \(\widehat{j}\) does not exist and \(l\) cannot be swapped with something that does not exist.
The sign of the 2nd permutation is \((-1)^m\), because \(\sigma_l\) is not moved anyway.
So, \(sgn \sigma = (-1) (-1)^m\) and \(sgn \sigma' = (-1)^{j - 1 - l} (-1)^m\).
So, \((-1)^m = sgn \sigma' (-1)^{- j + 1 + l}\), and so, \(sgn \sigma = (-1) sgn \sigma' (-1)^{- j + 1 + l} = (-1)^{- j + l + 2} sgn \sigma' = (-1)^{l - j} sgn \sigma'\).
Step 3:
Let us suppose that \(j = l\).
The logic is parallel with Step 1 and the result is easily expected, but let us do it diligently.
\(\sigma\) is \((1, ..., j - 1, j, j + 1, ..., n) \mapsto (\sigma_1, ..., \sigma_{j - 1}, j = \sigma_j, \sigma_{j + 1}, ..., \sigma_n)\).
Let the sign of the permutation be \((-1)^m\).
\(\sigma'\) is \((1, ..., j - 1, \widehat{j}, j + 1, ..., n) \mapsto (\sigma_1, ..., \sigma_{l - 1}, \widehat{j = \sigma_j}, \sigma_{j + 1}, ..., \sigma_n)\).
The sign of the permutation is \((-1)^m\), because \(\sigma_j\) is not moved anyway.
So, \(sgn \sigma = (-1)^m\) and \(sgn \sigma' = (-1)^m\).
So, \(sgn \sigma = sgn \sigma' = sgn \sigma' (-1)^0 = sgn \sigma' (-1)^{l - j}\).
