1039: Tensor Is Symmetric or Antisymmetric iff Components w.r.t. Standard Basis w.r.t. Bases That Are Same for Concerned Vectors Spaces Are Symmetric or Antisymmetric

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description/proof of that tensor is symmetric or antisymmetric iff components w.r.t. standard basis w.r.t. bases that are same for concerned vectors spaces are symmetric or antisymmetric

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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 1
• 3: Proof
• 4: Note 2

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Starting Context

• The reader knows a definition of tensors space with respect to field and $k$ vectors spaces and vectors space over field.
• The reader knows a definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
• The reader knows a definition of tensor product of tensors.
• The reader admits the proposition that for any field and any $k$ finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.

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Target Context

• The reader will have a description and a proof of the proposition that any tensor is symmetric or antisymmetric if and only if the components with respect to the standard basis with respect to any bases that are same for the concerned vectors spaces are symmetric or antisymmetric.

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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:
$F$: $\in \{\text{ the fields }\}$
$\{V_1,...,V_k\}$: $\subseteq \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$, where $V_{m_1}=...=V_{m_l}:=V$ for some $\{V_{m_1},...,V_{m_l}\}\subseteq \{V_1,...,V_k\}$
$L(V_1,...,V_k:F)$: $=\text{ the tensors space }$
$\{B_1,...,B_k\}$: $B_j\in \{\text{ the bases for }{V}_j\}=\{b_{j,s}|1\le s\le dim{V}_j\}$, where $B_{m_1}=...=B_{m_l}:=B^{\prime }$
$\{B_1^{\ast },...,B_k^{\ast }\}$: $B_j^{\ast }=\text{ the dual basis of }{B}_j=\{b_j^s|1\le s\le dim{V}_j\}$
$B$: $=\{b_1^{j_1}\otimes ...\otimes b_k^{j_k}|b_l^{j_l}\in B_l^{\ast }\}$
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Statements:
$\mathrm{\forall }f=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}\in L(V_1,...,V_k:F)$
(
(
$f\in \{\text{ the tensors symmetric with respect to }\{V_{m_1},...,V_{m_l}\}\}$
$⟺$
$\{f_{j_1,...,j_k}\}$ is symmetric with respect to the $m_1,...,m_l$ indexes
)
$\wedge$
(
$f\in \{\text{ the tensors antisymmetric with respect to }\{V_{m_1},...,V_{m_l}\}\}$
$⟺$
$\{f_{j_1,...,j_k}\}$ is antisymmetric with respect to the $m_1,...,m_l$ indexes
)
)
//

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2: Note 1

$L(V_1,...,V_k:F)$ cannot be more general $L(V_1,...,V_k:W)$, because the proposition that for any field and any $k$ finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces requires so.

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3: Proof

Whole Strategy: Step 1: suppose that $f$ is symmetric with respect to $\{V_{m_1},...,V_{m_l}\}$, make $f$ operate on $(b_{1,s_1},...,b_{k,s_k})$ and $(b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}})$ where $\sigma$ is any permutation of $(1,...,k)$ that moves only $(m_1,...,m_l)$, and see that $f_{s_1,...,s_k}=f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}$; Step 2: suppose that $\{f_{j_1,...,j_k}\}$ is symmetric with respect to the $m_1,...,m_l$ indexes, make $f$ operate on any $(v_1,...,v_k)$ and $(v_{{\sigma }_1},...,v_{{\sigma }_k})$, and see that the results are equal; Step 3: suppose that $f$ is antisymmetric with respect to $\{V_{m_1},...,V_{m_l}\}$, make $f$ operate on $(b_{1,s_1},...,b_{k,s_k})$ and $(b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}})$ where $\sigma$ is any permutation of $(1,...,k)$ that moves only $(m_1,...,m_l)$, and see that $f_{s_1,...,s_k}=sgn\sigma f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}$; Step 4: suppose that $\{f_{j_1,...,j_k}\}$ is antisymmetric with respect to the $m_1,...,m_l$ indexes, make $f$ operate on any $(v_1,...,v_k)$ and $(v_{{\sigma }_1},...,v_{{\sigma }_k})$, and see that the results are different by the $sgn\sigma$ factor.

Step 1:

Let us suppose that $f$ is symmetric with respect to $\{V_{m_1},...,V_{m_l}\}$.

Let $\sigma$ be any permutation of $(1,...,k)$ that moves only $(m_1,...,m_l)$.

Make $f$ operate on $(b_{1,s_1},...,b_{k,s_k})$ and $(b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}})$.

$f((b_{1,s_1},...,b_{k,s_k}))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((b_{1,s_1},...,b_{k,s_k}))=f_{s_1,...,s_k}$.

$f((b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}}))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}}))=f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}$.

As $f$ is symmetric, $f_{s_1,...,s_k}=f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}$, which is nothing but that $\{f_{j_1,...,j_k}\}$ is symmetric with respect to the $m_1,...,m_l$ indexes.

Step 2:

Let us suppose that $\{f_{j_1,...,j_k}\}$ is symmetric with respect to the $m_1,...,m_l$ indexes.

Let $\sigma$ be any permutation of $(1,...,k)$ that moves only $(m_1,...,m_l)$.

Let $(v_1,...,v_k)\in V_1\times ...\times V_k$ be any.

For each $j\in \{1,...,k\}$, $v_j=v_j^s{b}_{j,s}$.

Let $f$ operate on $(v_1,...,v_k)$ and $(v_{{\sigma }_1},...,v_{{\sigma }_k})$.

$f((v_1,...,v_k))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((v_1^{s_1}{b}_{1,s_1},...,v_k^{s_k}{b}_{k,s_k}))=f_{j_1,...,j_k}{v}_1^{s_1}{\delta }_{s_1}^{j_1}...v_k^{s_k}{\delta }_{s_k}^{j_k}=f_{s_1,...,s_k}{v}_1^{s_1}...v_k^{s_k}$.

$f((v_{{\sigma }_1},...,v_{{\sigma }_k}))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((v_{{\sigma }_1}^{s_{{\sigma }_1}}{b}_{{\sigma }_1,s_{{\sigma }_1}},...,v_{{\sigma }_k}^{s_{{\sigma }_k}}{b}_{{\sigma }_k,s_{{\sigma }_k}}))=f_{j_1,...,j_k}{v}_{{\sigma }_1}^{s_{{\sigma }_1}}{\delta }_{s_{{\sigma }_1}}^{j_1}...v_{{\sigma }_k}^{s_{{\sigma }_k}}{\delta }_{s_{{\sigma }_k}}^{j_k}=f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}{v}_{{\sigma }_1}^{s_{{\sigma }_1}}...v_{{\sigma }_k}^{s_{{\sigma }_k}}$, but as $\{f_{j_1,...,j_k}\}$ is symmetric, $f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}=f_{s_1,...,s_k}$, so, $=f_{s_1,...,s_k}{v}_{{\sigma }_1}^{s_{{\sigma }_1}}...v_{{\sigma }_k}^{s_{{\sigma }_k}}$, but as $v_{{\sigma }_1}^{s_{{\sigma }_1}}...v_{{\sigma }_k}^{s_{{\sigma }_k}}=v_1^{s_1}...v_k^{s_k}$ (which is just a reordering), $=f_{s_1,...,s_k}{v}_1^{s_1}...v_k^{s_k}$.

So, $f((v_1,...,v_k))=f((v_{{\sigma }_1},...,v_{{\sigma }_k}))$, which is nothing but that $f$ is symmetric.

Step 3:

Let us suppose that $f$ is antisymmetric with respect to $\{V_{m_1},...,V_{m_l}\}$.

The logic is analogous to Step 1, as expected.

Let $\sigma$ be any permutation of $(1,...,k)$ that moves only $(m_1,...,m_l)$.

Make $f$ operate on $(b_{1,s_1},...,b_{k,s_k})$ and $(b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}})$.

$f((b_{1,s_1},...,b_{k,s_k}))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((b_{1,s_1},...,b_{k,s_k}))=f_{s_1,...,s_k}$.

$f((b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}}))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((b_{{\sigma }_1,s_{{\sigma }_1}},...,b_{{\sigma }_k,s_{{\sigma }_k}}))=f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}$.

As $f$ is antisymmetric, $f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}=sgn\sigma f_{s_1,...,s_k}$, which is nothing but that $\{f_{j_1,...,j_k}\}$ is antisymmetric with respect to the $m_1,...,m_l$ indexes.

Step 4:

Let us suppose that $\{f_{j_1,...,j_k}\}$ is antisymmetric with respect to the $m_1,...,m_l$ indexes.

The logic is analogous to Step 2, as expected.

Let $\sigma$ be any permutation of $(1,...,k)$ that moves only $(m_1,...,m_l)$.

Let $(v_1,...,v_k)\in V_1\times ...\times V_k$ be any.

For each $j\in \{1,...,k\}$, $v_j=v_j^s{b}_{j,s}$.

Let $f$ operate on $(v_1,...,v_k)$ and $(v_{{\sigma }_1},...,v_{{\sigma }_k})$.

$f((v_1,...,v_k))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((v_1^{s_1}{b}_{1,s_1},...,v_k^{s_k}{b}_{k,s_k}))=f_{j_1,...,j_k}{v}_1^{s_1}{\delta }_{s_1}^{j_1}...v_k^{s_k}{\delta }_{s_k}^{j_k}=f_{s_1,...,s_k}{v}_1^{s_1}...v_k^{s_k}$.

$f((v_{{\sigma }_1},...,v_{{\sigma }_k}))=f_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}((v_{{\sigma }_1}^{s_{{\sigma }_1}}{b}_{{\sigma }_1,s_{{\sigma }_1}},...,v_{{\sigma }_k}^{s_{{\sigma }_k}}{b}_{{\sigma }_k,s_{{\sigma }_k}}))=f_{j_1,...,j_k}{v}_{{\sigma }_1}^{s_{{\sigma }_1}}{\delta }_{s_{{\sigma }_1}}^{j_1}...v_{{\sigma }_k}^{s_{{\sigma }_k}}{\delta }_{s_{{\sigma }_k}}^{j_k}=f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}{v}_{{\sigma }_1}^{s_{{\sigma }_1}}...v_{{\sigma }_k}^{s_{{\sigma }_k}}$, but as $\{f_{j_1,...,j_k}\}$ is antisymmetric, $f_{s_{{\sigma }_1},...,s_{{\sigma }_k}}=sgn\sigma f_{s_1,...,s_k}$, so, $=sgn\sigma f_{s_1,...,s_k}{v}_{{\sigma }_1}^{s_{{\sigma }_1}}...v_{{\sigma }_k}^{s_{{\sigma }_k}}$, but as $v_{{\sigma }_1}^{s_{{\sigma }_1}}...v_{{\sigma }_k}^{s_{{\sigma }_k}}=v_1^{s_1}...v_k^{s_k}$ (which is just a reordering), $=sgn\sigma f_{s_1,...,s_k}{v}_1^{s_1}...v_k^{s_k}$.

So, $f((v_{{\sigma }_1},...,v_{{\sigma }_k}))=sgn\sigma f((v_1,...,v_k))$, which is nothing but that $f$ is antisymmetric.

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4: Note 2

It is crucial that $B$ is a standard basis, because otherwise, the components of $f$ would not have the form, $\{f_{j_1,...,j_k}\}$, at all: see Note for the proposition that for any field and any $k$ finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.

It is crucial that $B_{m_1}=...=B_{m_l}$, because otherwise, $b_{m_j}^{s_{m_j}}(b_{m_n,s_{m_n}})$ cannot be claimed to equal ${\delta }_{s_{m_n}}^{s_{m_j}}$, which is used in Proof.

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References

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