1045: Tensors Space w.r.t. Field and k Finite-Dimensional Vectors Spaces over Field and Field Is 'Vectors Spaces - Linear Morphisms' Isomorphic to Tensor Product of k Covectors Spaces

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description/proof of that tensors space w.r.t. field and $k$ finite-dimensional vectors spaces over field and field is 'vectors spaces - linear morphisms' isomorphic to tensor product of $k$ covectors spaces

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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 tensors space with respect to field and $k$ vectors spaces and vectors space over field.
• The reader knows a definition of covectors (dual) space of vectors space.
• The reader knows a definition of tensor product of $k$ vectors spaces over field.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of tensor product of tensors.
• The reader admits the proposition that for any multilinear map from any finite product vectors space, there is the unique linear map from the tensor product of the finite number of vectors spaces such that the multilinear map is the linear map after the canonical map from the product vectors space into the tensor product.
• 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.
• The reader admits the proposition that the tensor product of any $k$ finite-dimensional vectors spaces has the basis that consists of the classes induced by any basis elements.
• The reader admits the proposition that for any linear map between any modules with bases, if its restriction on any domain basis is a bijection onto any codomain basis, the map is a 'modules - linear morphisms' isomorphism.
• The reader admits the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any tensors space w.r.t. field and $k$ finite-dimensional vectors spaces over field and field is 'vectors spaces - linear morphisms' isomorphic to the tensor product of the $k$ covectors spaces.

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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:
$F$: $\in \{\text{ the fields }\}$
$\{V_1,...,V_k\}$: $\subseteq \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$L(V_1,...,V_k:F)$: $=\text{ the tensors space }$
$\{V_1^{\ast },...,V_k^{\ast }\}$: $V_j^{\ast }=\text{ the covectors space of }{V}_j$
$V_1^{\ast }\otimes ...\otimes V_k^{\ast }$: $=\text{ the tensor product }$
$\{B_1^{\ast },...,B_k^{\ast }\}$: $B_j^{\ast }=\{b_j^l\}\in \{\text{ the bases for }{V}_j^{\ast }\}$
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Statements:
$\mathrm{\exists }{f}^{\prime }:V_1^{\ast }\otimes ...\otimes V_k^{\ast }\to L(V_1,...,V_k:F),t_{j_1,...,j_k}[((b_1^{j_1},...,b_k^{j_k}))]\mapsto t_{j_1,...,j_k}{b}_1^{j_1}\otimes ...\otimes b_k^{j_k}\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

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2: Note

Each $V_j$ needs to be finite-dimensional, because Proof uses 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 and the proposition that the tensor product of any $k$ finite-dimensional vectors spaces has the basis that consists of the classes induced by any basis elements, which require $V_j$ s to be finite-dimensional.

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3: Proof

Whole Strategy: Step 1: take a multilinear map, $f:V_1^{\ast }\times ...\times V_k^{\ast }\to L(V_1,...,V_k:F),(v^1,...,v^k)\mapsto v^1\otimes ...\otimes v^k$; Step 2: get the unique linear map, $f^{\prime }:V_1^{\ast }\otimes ...\otimes V_k^{\ast }\to L(V_1,...,V_k:F)$; Step 3: see that $f^{\prime }$ is a 'vectors spaces - linear morphisms' isomorphism.

Step 1:

Let us take a map, $f:V_1^{\ast }\times ...\times V_k^{\ast }\to L(V_1,...,V_k:F),(v^1,...,v^k)\mapsto v^1\otimes ...\otimes v^k$.

Let us see that $f$ is a multilinear map.

$f((v^1,...,r{v}^j+r^{\prime }{v}^{\prime j},...,v^k))=v^1\otimes ...\otimes (r{v}^j+r^{\prime }{v}^{\prime j})\otimes ...\otimes v^k=r{v}^1\otimes ...\otimes v^j\otimes ...\otimes v^k+r^{\prime }{v}^1\otimes ...\otimes v^{\prime j}\otimes ...\otimes v^k$, by the property of tensor product of tensors mentioned in Note for the definition of tensor product of tensors.

$=rf((v^1,...,v^j,...,v^k))+r^{\prime }f((v^1,...,v^{\prime j},...,v^k))$, which means that $f$ is multilinear.

Step 2:

By the proposition that for any multilinear map from any finite product vectors space, there is the unique linear map from the tensor product of the finite number of vectors spaces such that the multilinear map is the linear map after the canonical map from the product vectors space into the tensor product, there is the unique linear map, $f^{\prime }:V_1^{\ast }\otimes ...\otimes V_k^{\ast }\to L(V_1,...,V_k:F)$, that satisfies that $f=f^{\prime }\circ g$, where $g:V_1^{\ast }\times ...\times V_k^{\ast }\to V_1^{\ast }\otimes ...\otimes V_k^{\ast },(v^1,...,v^k)\mapsto [((v^1,...,v^k))]$.

Step 3:

The remaining issue to see is that $f^{\prime }$ is a 'vectors spaces - linear morphisms' isomorphism.

$V_1^{\ast }\otimes ...\otimes V_k^{\ast }$ has the standard basis, $B=\{[((b_1^{j_1},...,b_k^{j_k}))]|b_l^{j_l}\in B_l^{\ast }\}$ where $B_l^{\ast }$ is a basis for $V_l^{\ast }$, by the proposition that the tensor product of any $k$ finite-dimensional vectors spaces has the basis that consists of the classes induced by any basis elements.

$L(V_1,...,V_k:F)$ has the standard basis, $B=\{b_1^{j_1}\otimes ...\otimes b_k^{j_k}|b_l^{j_l}\in B_l^{\ast }\}$, by 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.

$f^{\prime }([((b_1^{j_1},...,b_k^{j_k}))])=b_1^{j_1}\otimes ...\otimes b_k^{j_k}$, because $f^{\prime }\circ g=f$, and $f^{\prime }\circ g((b_1^{j_1},...b_k^{j_k}))=f((b_1^{j_1},...b_k^{j_k}))=b_1^{j_1}\otimes ...\otimes b_k^{j_k}$ while $f^{\prime }\circ g((b_1^{j_1},...b_k^{j_k}))=f^{\prime }([((b_1^{j_1},...,b_k^{j_k}))])$.

That means that $f^{\prime }$ maps the basis of $V_1^{\ast }\otimes ...\otimes V_k^{\ast }$ to the basis of $L(V_1,...,V_k:F)$ bijectively.

By the proposition that for any linear map between any modules with bases, if its restriction on any domain basis is a bijection onto any codomain basis, the map is a 'modules - linear morphisms' isomorphism, $f^{\prime }$ is a 'modules - linear morphisms' isomorphism, which means that $f^{\prime }$ is a bijection, and $f^{\prime }$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

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References

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