description/proof of that contraction of 2 same-length multi-dimensional arraysame-length multi-dimensional arrays one of which is symmetrized or antisymmetrized w.r.t. set of indexes is contraction with also other array symmetrized or antisymmetrized accordingly
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
- 4: Note 2
Starting Context
- 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 the set of all the sequences for any fixed domain and any fixed codomain, any permutation bijectively maps the set onto the set.
Target Context
- The reader will have a description and a proof of the proposition that the contraction of any 2 same-length multi-dimensional arrays one of which is symmetrized or antisymmetrized with respect to any set of indexes is the contraction with also the other array symmetrized or antisymmetrized accordingly.
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:
\(n\): \(\in \mathbb{N}\), \(2 \le n \lt \infty\)
\(M\): \(\in \{\text{ the same-length } n \text{ -dimensional arrays }\}\), \(= \begin{pmatrix} M_{j_1, ..., j_n} \end{pmatrix}\)
\(N\): \(\in \{\text{ the same-length } n \text{ -dimensional arrays }\}\), \(= \begin{pmatrix} N_{j_1, ..., j_n} \end{pmatrix}\)
\(S\): \(= \{l_1, ..., l_k\} \subseteq \{1, ..., n\}\)
\(M'\): \(= M \text{ symmetrized with respect to } S\), \(= \begin{pmatrix} M'_{j_1, ..., j_n} \end{pmatrix}\)
\(M''\): \(= M \text{ antisymmetrized with respect to } S\), \(= \begin{pmatrix} M''_{j_1, ..., j_n} \end{pmatrix}\)
\(N'\): \(= N \text{ symmetrized with respect to } S\), \(= \begin{pmatrix} N'_{j_1, ..., j_n} \end{pmatrix}\)
\(N''\): \(= N \text{ antisymmetrized with respect to } S\), \(= \begin{pmatrix} N''_{j_1, ..., j_n} \end{pmatrix}\)
//
Statements:
\(\sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M'_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M'_{j_1, ..., j_n}\).
\(\land\)
\(\sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M''_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M''_{j_1, ..., j_n}\).
//
2: Natural Language Description
For any same-length \(n\)-dimensional arrays, \(M = \begin{pmatrix} M_{j_1, ..., j_n} \end{pmatrix}, N = \begin{pmatrix} N_{j_1, ..., j_n} \end{pmatrix}\), where \(2 \le n \lt \infty\), any subset, \(S = \{l_1, ..., l_k\} \subseteq \{1, ..., n\}\), \(M\) symmetrized with respect to \(S\), \(M' = \begin{pmatrix} M'_{j_1, ..., j_n} \end{pmatrix}\), \(M\) antisymmetrized with respect to \(S\), \(M'' = \begin{pmatrix} M''_{j_1, ..., j_n} \end{pmatrix}\), \(N\) symmetrized with respect to \(S\), \(N' = \begin{pmatrix} N'_{j_1, ..., j_n} \end{pmatrix}\), and \(N\) antisymmetrized with respect to \(S\), \(N'' = \begin{pmatrix} N''_{j_1, ..., j_n} \end{pmatrix}\), \(\sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M'_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M'_{j_1, ..., j_n}\) and \(\sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M''_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M''_{j_1, ..., j_n}\).
3: Proof
Whole Strategy: Step 1: see that for \(X := \sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M'_{j_1, ..., j_n}\) and each permutation of \((j_1, ..., j_n)\) such that only \((j_{l_1}, ..., j_{l_k})\) is permutated, \(\sigma\), \(X = \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M'_{j_1, ..., j_n}\); Step 2: take \(X = 1 / k! \sum_{\sigma} X\) and see that \(X = \sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M'_{j_1, ..., j_n}\); Step 3: see that for \(X := \sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M''_{j_1, ..., j_n}\) and each permutation of \((j_1, ..., j_n)\) such that only \((j_{l_1}, ..., j_{l_k})\) is permutated, \(\sigma\), \(X = \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} sgn \sigma M''_{j_1, ..., j_n}\); Step 4: take \(X = 1 / k! \sum_{\sigma} X\) and see that \(X = \sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M''_{j_1, ..., j_n}\).
Step 1:
Let \(\sigma\) be any permutation of \((j_1, ..., j_n)\) such that only \((j_{l_1}, ..., j_{l_k})\) is permutated.
Let \(X := \sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M'_{j_1, ..., j_n}\).
For any fixed \(\sigma\), \(X = \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M'_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n}\), by the proposition that for the set of all the sequences for any fixed domain and any fixed codomain, any permutation bijectively maps the set onto the set.
\(= \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M'_{j_1, ..., j_n}\), because \(M'\) is symmetrized.
Step 2:
While there are \(k!\) such \(\sigma\) s, \(X = 1 / k! \sum_{\sigma} X = 1 / k! \sum_{\sigma} \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M'_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} 1 / k! \sum_{\sigma} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M'_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M'_{j_1, ..., j_n}\).
Step 3:
Let \(X := \sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M''_{j_1, ..., j_n}\).
For any fixed \(\sigma\), \(X = \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M''_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n}\), by the proposition that for the set of all the sequences for any fixed domain and any fixed codomain, any permutation bijectively maps the set onto the set.
\(= \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} sgn \sigma M''_{j_1, ..., j_n}\), because \(M''\) is antisymmetrized.
Step 4:
While there are \(k!\) such \(\sigma\) s, \(X = 1 / k! \sum_{\sigma} X = 1 / k! \sum_{\sigma} \sum_{(j_1, ..., j_n)} N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} sgn \sigma M''_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} 1 / k! \sum_{\sigma} sgn \sigma N_{\sigma ((j_1, ..., j_n))_1, ..., \sigma ((j_1, ..., j_n))_n} M''_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M''_{j_1, ..., j_n}\).
4: Note 2
These of course hold.
\(\sum_{(j_1, ..., j_n)} M'_{j_1, ..., j_n} N_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} M'_{j_1, ..., j_n} N'_{j_1, ..., j_n}\).
\(\sum_{(j_1, ..., j_n)} M''_{j_1, ..., j_n} N_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} M''_{j_1, ..., j_n} N''_{j_1, ..., j_n}\).
\(\sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M'_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M_{j_1, ..., j_n}\), because \(\sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N'_{j_1, ..., j_n} M'_{j_1, ..., j_n}\).
\(\sum_{(j_1, ..., j_n)} N_{j_1, ..., j_n} M''_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M_{j_1, ..., j_n}\), because \(\sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M_{j_1, ..., j_n} = \sum_{(j_1, ..., j_n)} N''_{j_1, ..., j_n} M''_{j_1, ..., j_n}\).
