1113: Antisymmetrization of Tensor Product of Tensors Is Antisymmetrizations Applied Sequentially

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description/proof of that antisymmetrization of tensor product of tensors is antisymmetrizations applied sequentially

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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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.

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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.

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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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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.

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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 ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$, where ${\sigma }^{\prime }$ 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 $({\sigma }_1^{\prime },...,{\sigma }_{k_1}^{\prime })$ or $({\sigma }_{k_1+1}^{\prime },...,{\sigma }_{k_1+k_2}^{\prime })$ 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, ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$ means that any $S^{k_1+k_2}$ element is 1st, permutating $(1,...,k_1+k_2)$ such that $({\sigma }_1^{\prime },...,{\sigma }_{k_1}^{\prime })$ and $({\sigma }_{k_1+1}^{\prime },...,{\sigma }_{k_1+k_2}^{\prime })$ are in the increasing orders, and then, permuting $({\sigma }_1^{\prime },...,{\sigma }_{k_1}^{\prime })$ and $({\sigma }_{k_1+1}^{\prime },...,{\sigma }_{k_1+k_2}^{\prime })$.

$sgn\sigma =sgn{\sigma }_{1}sgn{\sigma }_{2}sgn{\sigma }^{\prime }$.

So, $Asym(t_1\otimes t_2)((v_1,...,v_{k_1+k_2}))=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1\sum _{{\sigma }_2}sgn{\sigma }_2{t}_1\otimes t_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})k_2!/k_2!\sum _{{\sigma }_2}sgn{\sigma }_2{t}_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})=(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})$, because ${\sigma }_2$ does not change the concerned components; likewise, $(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

So, $Asym(t_1\otimes t_2)((v_1,...,v_{k_1+k_2}))=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{k}_1!/k_1!\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})k_2!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{k}_1!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})k_2!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

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 ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$.

So, $Asym(Asym(t_1)\otimes t_2)(v_1,...,v_{k_1+k_2})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_1)\otimes t_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}Asym(t_1)(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})(k_2)!/(k_2)!\sum _{{\sigma }_2}sgn{\sigma }_2{t}_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}Asym(t_1)(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})(k_2)!/(k_2)!\sum _{{\sigma }_2}sgn{\sigma }_2{t}_2(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}Asym(t_1)(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

$Asym(t_1)(v_{{{\sigma }_1\circ {\sigma }^{\prime }}_1},...,v_{{{\sigma }_1\circ {\sigma }^{\prime }}_{k_1}})=sgn{\sigma }_{1}Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})$, because $Asym(t_1)$ is antisymmetric.

So, $=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}sgn{\sigma }_{1}Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

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 ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$.

So, $Asym(t_1\otimes Asym(t_2))(v_1,...,v_{k_1+k_2})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1\sum _{{\sigma }_2}sgn{\sigma }_2{t}_1\otimes Asym(t_2)(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_2)(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!/(k_1)!\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_2)(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_2)(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$Asym(t_2)(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=sgn{\sigma }_{2}Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$, because $Asym(t_2)$ is antisymmetric.

So, $=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}sgn{\sigma }_{2}Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})\sum _{{\sigma }_2}Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

That is the same with the result of Step 2.

So, $Asym(t_1\otimes t_2)=Asym(t_1\otimes Asym(t_2))$.

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References

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