815: Antisymmetrized After Symmetrized Same-Length Multi-Dimensional Array or Symmetrized After Antisymmetrized Same-Length Multi-Dimensional Array Is 0

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description/proof of that antisymmetrized after symmetrized same-length multi-dimensional array or symmetrized after antisymmetrized same-length multi-dimensional array is 0

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Topics

About: set

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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

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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.

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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.

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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.

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Main Body

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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.

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the same-length }n\text{ -dimensional arrays }\}$, $=\left(\begin{array}{c}{M}_{j_1,...,j_n}\end{array}\right)$
$S$: $\subseteq \{1,...,n\}$, $=\{l_1,...,l_k\}$
$M^{\prime }$: $=M\text{ antisymmetrized after symmetrized with respect to }S$, $=\left(\begin{array}{c}{M}_{j_1,...,j_n}^{\prime }\end{array}\right)$
$M^{″}$: $=M\text{ symmetrized after antisymmetrized with respect to }S$, $=\left(\begin{array}{c}{M}_{j_1,...,j_n}^{″}\end{array}\right)$
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Statements:
$M^{\prime }=0$.
$\wedge$
$M^{″}=0$.
//

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2: Natural Language Description

For any same-length $n$-dimensional array, $M=\left(\begin{array}{c}{M}_{j_1,...,j_n}\end{array}\right)$, any subset, $S\subseteq \{1,...,n\}$, $=\{l_1,...,l_k\}$, the antisymmetrized after symmetrized $M$ with respect to $S$, $M^{\prime }=\left(\begin{array}{c}{M}_{j_1,...,j_n}^{\prime }\end{array}\right)$, and the symmetrized after antisymmetrized $M$ with respect to $S$, $M^{″}=\left(\begin{array}{c}{M}_{j_1,...,j_n}^{″}\end{array}\right)$, $M^{\prime }=0$ and $M^{″}=0$.

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3: Proof

Whole Strategy: Step 1: express $M_{j_1,...,j_n}^{\prime }$ with the symmetrized $M$ with respect to $S$, $N^{\prime }$, and see that $M^{\prime }=0$; 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 $N^{\prime }$ be the symmetrized $M$ with respect to $S$.

$M_{j_1,...,j_n}^{\prime }=1/k!\sum _{\sigma }sgn\sigma N_{\sigma ((j_1,...,j_n){)}_1,...,\sigma ((j_1,...,j_n){)}_n}^{\prime }$, where $\sigma$ is each permutation of $(j_1,...,j_n)$ such that only $(j_{l_1},...,j_{l_k})$ is permutated. But $N_{\sigma ((j_1,...,j_n){)}_1,...,\sigma ((j_1,...,j_n){)}_n}^{\prime }=N_{j_1,...,j_n}^{\prime }$, because that is the purpose of symmetrization. So, $=1/k!\sum _{\sigma }sgn\sigma N_{j_1,...,j_n}^{\prime }=1/k!N_{j_1,...,j_n}^{\prime }\sum _{\sigma }sgn\sigma =0$, 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.

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References

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