description/proof of that antisymmetrized after symmetrized same-length multi-dimensional array or symmetrized after antisymmetrized same-length multi-dimensional array is 0
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 0: Note 1
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of same-length multi-dimensional array.
- The reader knows a definition of same-length multi-dimensional array symmetrized with respect to set of indexes.
- The reader knows a definition of same-length multi-dimensional array antisymmetrized with respect to set of indexes.
- The reader admits the proposition that for any sequence of any finite elements, the set of the permutations has the same number of the even permutations and the odd permutations.
Target Context
- The reader will have a description and a proof of the proposition that for any same-length multi-dimensional array and any set of indexes, the antisymmetrized after symmetrized array or the symmetrized after antisymmetrized array with respect to the set of indexes is 0.
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
0: Note 1
Any \(n\)-length 2-dimensional array is an \(n \times n\) matrix.
The components of any \((p-q)\) tensor with respect to any bases is a vectors-space-dimension-length (p + q)-dimensional array, but a same-length multi-dimensional array is not necessarily the components of a tensor: a same-length multi-dimensional array is in general just a collection of numbers not necessarily related to any tensor.
The dimensions of the same-length multi-dimensional array have to have the same length for our purpose, because otherwise, a permutation of indexes would not make sense.
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the same-length } n \text{ -dimensional arrays }\}\), \(= \begin{pmatrix} M_{j_1, ..., j_n} \end{pmatrix}\)
\(S\): \(\subseteq \{1, ..., n\}\), \(= \{l_1, ..., l_k\}\)
\(M'\): \(= M \text{ antisymmetrized after symmetrized with respect to } S\), \(= \begin{pmatrix} M'_{j_1, ..., j_n} \end{pmatrix}\)
\(M''\): \(= M \text{ symmetrized after antisymmetrized with respect to } S\), \(= \begin{pmatrix} M''_{j_1, ..., j_n} \end{pmatrix}\)
//
Statements:
\(M' = 0\).
\(\land\)
\(M'' = 0\).
//
2: Natural Language Description
For any same-length \(n\)-dimensional array, \(M = \begin{pmatrix} M_{j_1, ..., j_n} \end{pmatrix}\), any subset, \(S \subseteq \{1, ..., n\}\), \(= \{l_1, ..., l_k\}\), the antisymmetrized after symmetrized \(M\) with respect to \(S\), \(M' = \begin{pmatrix} M'_{j_1, ..., j_n} \end{pmatrix}\), and the symmetrized after antisymmetrized \(M\) with respect to \(S\), \(M'' = \begin{pmatrix} M''_{j_1, ..., j_n} \end{pmatrix}\), \(M' = 0\) and \(M'' = 0\).
3: Proof
Whole Strategy: Step 1: express \(M'_{j_1, ..., j_n}\) with the symmetrized \(M\) with respect to \(S\), \(N'\), and see that \(M' = 0\); Step 2: express \(M''_{j_1, ..., j_n}\) with the antisymmetrized \(M\) with respect to \(S\), \(N''\), and see that \(M'' = 0\).
Step 1:
Let \(N'\) be the symmetrized \(M\) with respect to \(S\).
\(M'_{j_1, ..., j_n} = 1 / k! \sum_\sigma sgn \sigma N'_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n}\), where \(\sigma\) is each permutation of \((j_1, ..., j_n)\) such that only \((j_{l_1}, ..., j_{l_k})\) is permutated. But \(N'_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} = N'_{j_1, ..., j_n}\), because that is the purpose of symmetrization. So, \(= 1 / k! \sum_\sigma sgn \sigma N'_{j_1, ..., j_n} = 1 / k! N'_{j_1, ..., j_n} \sum_\sigma sgn \sigma = 0\), by the proposition that for any sequence of any finite elements, the set of the permutations has the same number of the even permutations and the odd permutations.
Step 2:
Let \(N''\) be the antisymmetrized \(M\) with respect to \(S\).
\(M''_{j_1, ..., j_n} = 1 / k! \sum_\sigma N''_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n}\), where \(\sigma\) is each permutation of \((j_1, ..., j_n)\) such that only \((j_{l_1}, ..., j_{l_k})\) is permutated. But \(N''_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} = sgn \sigma N''_{j_1, ..., j_n}\), because that is the purpose of antisymmetrization. So, \(= 1 / k! \sum_\sigma sgn \sigma N''_{j_1, ..., j_n} = 1 / k! N''_{j_1, ..., j_n} \sum_\sigma sgn \sigma = 0\), by the proposition that for any sequence of any finite elements, the set of the permutations has the same number of the even permutations and the odd permutations.
