description/proof of that antisymmetrization of tensor product of tensors is antisymmetrizations applied sequentially
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of tensor product of tensors.
- The reader knows a definition of antisymmetrization of tensor with respect to some arguments.
Target Context
- The reader will have a description and a proof of the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(\{V, W\}\): \(\subseteq \{\text{ the } F \text{ vectors spaces }\}\)
\(t_1\): \(\in L (V, ..., V: W)\), where \(V\) appears \(k_1\) times
\(t_2\): \(\in L (V, ..., V: W)\), where \(V\) appears \(k_2\) times
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Statements:
\(Asym (t_1 \otimes t_2) = Asym (Asym (t_1) \otimes t_2) = Asym (t_1 \otimes Asym (t_2))\)
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2: Note
By applying this proposition sequentially, \(Asym (t_1 \otimes ... \otimes t_n)\) can be expressed in many ways.
For example, \(Asym (t_1 \otimes t_2 \otimes t_3) = Asym ((t_1 \otimes t_2) \otimes t_3) = Asym (Asym (t_1 \otimes t_2) \otimes t_3) = Asym (Asym (Asym (t_1) \otimes t_2) \otimes t_3)\).
There can be a more general proposition on partial antisymmetrization, but as it seems to become cumbersome and our immediate necessity requires only full antisymmetrizations, this proposition deals with only full antisymmetrizations.
3: Proof
Whole Strategy: Step 1: let \((v_1, ..., v_{k_1 + k_2}) \in V \times ... \times V\) be any; Step 2: let \(Asym (t_1 \otimes t_2)\) operate on \((v_1, ..., v_{k_1 + k_2})\) and expand the result; Step 3: let \(Asym (Asym (t_1) \otimes t_2)\) operate on \((v_1, ..., v_{k_1 + k_2})\) and expand the result; Step 4: let \(Asym (t_1 \otimes Asym (t_2))\) operate on \((v_1, ..., v_{k_1 + k_2})\) and expand the result.
Step 1:
Let \((v_1, ..., v_{k_1 + k_2}) \in V \times ... \times V\) be any.
If some 2 tensors operate on it with the same result, the 2 tensors will be the same.
Step 2:
Let \(Asym (t_1 \otimes t_2)\) operate on \((v_1, ..., v_{k_1 + k_2})\).
\(Asym (t_1 \otimes t_2) ((v_1, ..., v_{k_1 + k_2})) = 1 / (k_1 + k_2)! \sum_{\sigma \in S^{k_1 + k_2}} sgn \sigma t_1 \otimes t_2 (v_{\sigma_1}, ..., v_{\sigma_{k_1 + k_2}})\).
\(\sigma\) can be expressed as \(\sigma_1 \circ \sigma_2 \circ \sigma'\), where \(\sigma'\) is the permutation that permutates \((\sigma_1, ..., \sigma_{k_1})\) and \((\sigma_{k_1 + 1}, ..., \sigma_{k_1 + k_2})\) into the increasing orders after \(\sigma\), and \(\sigma_1\) or \(\sigma_2\) returns \((\sigma'_1, ..., \sigma'_{k_1})\) or \((\sigma'_{k_1 + 1}, ..., \sigma'_{k_1 + k_2})\) back to \((\sigma_1, ..., \sigma_{k_1})\) or \((\sigma_{k_1 + 1}, ..., \sigma_{k_1 + k_2})\), respectively.
The set of \(\sigma_1\) s is practically the symmetric group, \(S^{k_1}\); the set of \(\sigma_2\) s is practically the symmetric group, \(S^{k_2}\).
So, \(\sigma_1 \circ \sigma_2 \circ \sigma'\) means that any \(S^{k_1 + k_2}\) element is 1st, permutating \((1, ..., k_1 + k_2)\) such that \((\sigma'_1, ..., \sigma'_{k_1})\) and \((\sigma'_{k_1 + 1}, ..., \sigma'_{k_1 + k_2})\) are in the increasing orders, and then, permuting \((\sigma'_1, ..., \sigma'_{k_1})\) and \((\sigma'_{k_1 + 1}, ..., \sigma'_{k_1 + k_2})\).
\(sgn \sigma = sgn \sigma_1 sgn \sigma_2 sgn \sigma'\).
So, \(Asym (t_1 \otimes t_2) ((v_1, ..., v_{k_1 + k_2})) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 \sum_{\sigma_2} sgn \sigma_2 t_1 \otimes t_2 (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 t_1 (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1}}) {k_2}! / {k_2}! \sum_{\sigma_2} sgn \sigma_2 t_2 (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}})\).
\((v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1}}) = (v_{(\sigma_1 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma')_{k_1}})\), because \(\sigma_2\) does not change the concerned components; likewise, \((v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}}) = (v_{(\sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_2 \circ \sigma')_{k_1 + k_2}})\).
So, \(Asym (t_1 \otimes t_2) ((v_1, ..., v_{k_1 + k_2})) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' {k_1}! / {k_1}! \sum_{\sigma_1} sgn \sigma_1 t_1 (v_{(\sigma_1 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma')_{k_1}}) {k_2}! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' {k_1}! Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) {k_2}! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}})\).
Step 3:
Let \(Asym (Asym (t_1) \otimes t_2)\) operate on \((v_1, ..., v_{k_1 + k_2})\).
\(Asym (Asym (t_1) \otimes t_2) (v_1, ..., v_{k_1 + k_2}) = 1 / (k_1 + k_2)! \sum_{\sigma \in S^{k_1 + k_2}} sgn \sigma Asym (t_1) \otimes t_2 (v_{\sigma_1}, ..., v_{\sigma_{k_1 + k_2}})\).
As before, \(\sigma\) can be expressed as \(\sigma_1 \circ \sigma_2 \circ \sigma'\).
So, \(Asym (Asym (t_1) \otimes t_2) (v_1, ..., v_{k_1 + k_2}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 \sum_{\sigma_2} sgn \sigma_2 Asym (t_1) \otimes t_2 (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 Asym (t_1) (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1}}) (k_2)! / (k_2)! \sum_{\sigma_2} sgn \sigma_2 t_2 (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}})\).
\(= 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 Asym (t_1) (v_{(\sigma_1 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma')_{k_1}}) (k_2)! / (k_2)! \sum_{\sigma_2} sgn \sigma_2 t_2 (v_{(\sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_2 \circ \sigma')_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 Asym (t_1) (v_{(\sigma_1 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma')_{k_1}}) (k_2)! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}})\).
\(Asym (t_1) (v_{{\sigma_1 \circ \sigma'}_1}, ..., v_{{\sigma_1 \circ \sigma'}_{k_1}}) = sgn \sigma_1 Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}})\), because \(Asym (t_1)\) is antisymmetric.
So, \(= 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 sgn \sigma_1 Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) (k_2)! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) (k_2)! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' (k_1)! Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) (k_2)! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}})\).
That is the same with the result of Step 2.
So, \(Asym (t_1 \otimes t_2) = Asym (Asym (t_1) \otimes t_2)\).
Step 4:
Let \(Asym (t_1 \otimes Asym (t_2))\) operate on \((v_1, ..., v_{k_1 + k_2})\).
\(Asym (t_1 \otimes Asym (t_2)) (v_1, ..., v_{k_1 + k_2}) = 1 / (k_1 + k_2)! \sum_{\sigma \in S^{k_1 + k_2}} sgn \sigma t_1 \otimes Asym (t_2) (v_{\sigma_1}, ..., v_{\sigma_{k_1 + k_2}})\).
As before, \(\sigma\) can be expressed as \(\sigma_1 \circ \sigma_2 \circ \sigma'\).
So, \(Asym (t_1 \otimes Asym (t_2)) (v_1, ..., v_{k_1 + k_2}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 \sum_{\sigma_2} sgn \sigma_2 t_1 \otimes Asym (t_2) (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' \sum_{\sigma_1} sgn \sigma_1 t_1 (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1}}) \sum_{\sigma_2} sgn \sigma_2 Asym (t_2) (v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_1 \circ \sigma_2 \circ \sigma')_{k_1 + k_2}})\).
\(= 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' (k_1)! / (k_1)! \sum_{\sigma_1} sgn \sigma_1 t_1 (v_{(\sigma_1 \circ \sigma')_1}, ..., v_{(\sigma_1 \circ \sigma')_{k_1}}) \sum_{\sigma_2} sgn \sigma_2 Asym (t_2) (v_{(\sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_2 \circ \sigma')_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' (k_1)! Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) \sum_{\sigma_2} sgn \sigma_2 Asym (t_2) (v_{(\sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_2 \circ \sigma')_{k_1 + k_2}})\).
\(Asym (t_2) (v_{(\sigma_2 \circ \sigma')_{k_1 + 1}}, ..., v_{(\sigma_2 \circ \sigma')_{k_1 + k_2}}) = sgn \sigma_2 Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}})\), because \(Asym (t_2)\) is antisymmetric.
So, \(= 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' (k_1)! Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) \sum_{\sigma_2} sgn \sigma_2 sgn \sigma_2 Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' (k_1)! Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) \sum_{\sigma_2} Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}}) = 1 / (k_1 + k_2)! \sum_{\sigma'} sgn \sigma' (k_1)! Asym (t_1) (v_{{\sigma'}_1}, ..., v_{{\sigma'}_{k_1}}) (k_2)! Asym (t_2) (v_{{\sigma'}_{k_1 + 1}}, ..., v_{{\sigma'}_{k_1 + k_2}})\).
That is the same with the result of Step 2.
So, \(Asym (t_1 \otimes t_2) = Asym (t_1 \otimes Asym (t_2))\).
