1011: Tensors Space w.r.t. Field and k Finite-Dimensional Vectors Spaces over Field and Field Has Basis That Consists of Tensor Products of Elements of Dual Bases

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description/proof of that tensors space w.r.t. field and $k$ finite-dimensional vectors spaces over field and field has basis that consists of tensor products of elements of dual bases

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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 w.r.t. 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.

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Target Context

• The reader will have a description and a proof of 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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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:
$F$: $\in \{\text{ the fields }\}$
$\{V_1,...,V_k\}$: $\subseteq \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$L(V_1,...,V_k:F)$: $=\text{ the tensors space }$
$\{B_1,...,B_k\}$: $B_j\in \{\text{ the bases of }{V}_j\}=\{{b_j}_l|1\le l\le dim{V}_j\}$
$\{B_1^{\ast },...,B_k^{\ast }\}$: $B_j^{\ast }=\text{ the dual basis of }{B}_j=\{{b_j}^l|1\le l\le dim{V}_j\}$
$B^{\ast }$: $=\{{b_1}^{j_1}\otimes ...\otimes {b_k}^{j_k}|\mathrm{\forall }l\in \{1,...,k\}(1\le j_l\le dim{V}_l)\}$
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Statements:
$B^{\ast }\in \{\text{ the bases for }L(V_1,...,V_k:F)\}$
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Let us call $B^{\ast }$ "the standard basis with respect to $\{B_1,...,B_k\}$": it is not determined unless $\{B_1,...,B_k\}$ is specified.

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2: Note 1

$L(V_1,...,V_k:F)$ cannot be more general $L(V_1,...,V_k:W)$, because ${b_1}^{j_1}\otimes ...\otimes {b_k}^{j_k}\in L(V_1,...,V_k:F)$.

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3: Proof

Whole Strategy: Step 1: see that $B^{\ast }$ spans $L(V_1,...,V_k:F)$; Step 2: see that $B^{\ast }$ is linearly independent.

Step 1:

Let $f\in L(V_1,...,V_k:F)$ be any.

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

$v_j=v_j^l{b_j}_l$.

$f(v)=f((v_1,...,v_k))=f((v_1^{l_1}{b_1}_{l_1},...,v_k^{l_k}{b_k}_{l_k}))=v_1^{l_1}{v}_k^{l_k}f(({b_1}_{l_1},...,{b_k}_{l_k}))$.

Let us think of $f(({b_1}_{l_1},...,{b_k}_{l_k})){b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}\in L(V_1,...,V_k:F)$.

$f(({b_1}_{l_1},...,{b_k}_{l_k})){b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}(v)=f(({b_1}_{l_1},...,{b_k}_{l_k})){b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}((v_1,...,v_k))=f(({b_1}_{l_1},...,{b_k}_{l_k})){b_1}^{l_1}(v_1^{m_1}{b_1}_{m_1})...{b_k}^{l_k}(v_k^{m_k}{b_k}_{m_k})=f(({b_1}_{l_1},...,{b_k}_{l_k}))v_1^{m_1}{\delta }_{m_1}^{l_1}...v_k^{m_k}{\delta }_{m_k}^{l_k}=f(({b_1}_{l_1},...,{b_k}_{l_k}))v_1^{l_1}...v_k^{l_k}$.

That means that $f=f(({b_1}_{l_1},...,{b_k}_{l_k})){b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}$, so, $B^{\ast }$ spans $L(V_1,...,V_k:F)$.

Step 2:

Let $c_{l_1,...,l_k}{b_1}^{l_1}\otimes ...\otimes b_k{l}_k=0$.

Let it operate on $({b_1}_{m_1},...,{b_k}_{m_k})$. Then, $(c_{l_1,...,l_k}{b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k})(({b_1}_{m_1},...,{b_k}_{m_k}))=0(({b_1}_{m_1},...,{b_k}_{m_k}))=0$, but $(c_{l_1,...,l_k}{b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k})(({b_1}_{m_1},...,{b_k}_{m_k}))=c_{l_1,...,l_k}{b_1}^{l_1}({b_1}_{m_1})...{b_k}^{l_k})({b_k}_{m_k})=c_{l_1,...,l_k}{\delta }_{m_1}^{l_1}...{\delta }_{m_k}^{l_k}=c_{m_1,...,m_k}$, so, each $c_{m_1,...,m_k}=0$.

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4: Note 2

Once $B^{\ast }$ as in this proposition is fixed, $f=f(({b_1}_{l_1},...,{b_k}_{l_k})){b_1}^{l_1}\otimes ...\otimes {b_k}^{l_k}$, and $f$ can be uniquely represented by the components, $\{f(({b_1}_{l_1},...,{b_k}_{l_k}))\}$.

While this proposition admits that $V_j$ s are some different $F$ vectors spaces, a typical case is $T_q^p(V):=L(V^{\ast },...,V^{\ast },V,...,V:F)$, where there are $p$ $V^{\ast }$ s and $q$ $V$ s. Then, for any basis for $V$, $\{b_j|1\le j\le n\}$, letting the dual basis for $V^{\ast }$ as $\{b^j|1\le j\le n\}$, $B^{\ast }$ is $\{b_{j_1}\otimes ...\otimes b_{j_p}\otimes b^{j_{p+1}}\otimes ...\otimes b^{j_{p+q}}|1\le j_l\le n\}$: in fact, $b_{j_l}$ there is not exactly it but the one that corresponds to it in ${V^{\ast }}^{\ast }$ by the proposition that the double dual of any finite dimensional real vectors space is 'vectors spaces - linear morphisms' isomorphic to the original vectors space: as the one that corresponds to $b_l$ operates on $b^m$ as $b^m(b_l)={\delta }_l^m$, indeed, the basis that corresponds to $\{b_j\}$ is the dual basis of $\{b^j\}$. Then, the components of $f$ are denoted as $f_{j_{p+1},...,j_{p+q}}^{j_1,...,j_p}:=f((b^{j_1},...,b^{j_p},b_{j_{p+1}},...,b_{j_{p+q}}))$.

Although some people call $\{f_{j_{p+1},...,j_{p+q}}^{j_1,...,j_p}\}$ "tensor", they are really the components of the tensor, $f$, with respect to the basis, $B^{\ast }$, not "tensor".

While $B^{\ast }$ as in this proposition is a basis, another basis does not need to be of that form: for an $L(V_1,V_2:F)$ where $B_1=\{{b_1}_1,{b_1}_2\}$ and $B_2=\{{b_2}_1,{b_2}_2\}$, $\{{b_1}^1\otimes {b_2}^1,{b_1}^1\otimes {b_2}^2+{b_1}^2\otimes {b_2}^1,{b_1}^1\otimes {b_2}^2-{b_1}^2\otimes {b_2}^1,{b_1}^2\otimes {b_2}^2\}$ is a basis: it spans $L(V_1,V_2:F)$, because ${b_1}^1\otimes {b_2}^2$ and ${b_1}^2\otimes {b_2}^1$ are realized as ${b_1}^1\otimes {b_2}^2=1/2(({b_1}^1\otimes {b_2}^2+{b_1}^2\otimes {b_2}^1)+({b_1}^1\otimes {b_2}^2-{b_1}^2\otimes {b_2}^1))$ and ${b_1}^2\otimes {b_2}^1=1/2(({b_1}^1\otimes {b_2}^2+{b_1}^2\otimes {b_2}^1)-({b_1}^1\otimes {b_2}^2-{b_1}^2\otimes {b_2}^1))$; it is linearly independent, because $c^1{b_1}^1\otimes {b_2}^1+c^2({b_1}^1\otimes {b_2}^2+{b_1}^2\otimes {b_2}^1)+c^3({b_1}^1\otimes {b_2}^2-{b_1}^2\otimes {b_2}^1)+c^4{b_1}^2\otimes {b_2}^2=0$ implies that $c^1=c^2=c^3=c^4=0$: $c^1{b_1}^1\otimes {b_2}^1+c^2({b_1}^1\otimes {b_2}^2+{b_1}^2\otimes {b_2}^1)+c^3({b_1}^1\otimes {b_2}^2-{b_1}^2\otimes {b_2}^1)+c^4{b_1}^2\otimes {b_2}^2=c^1{b_1}^1\otimes {b_2}^1+(c^2+c^3){b_1}^1\otimes {b_2}^2+(c^2-c^3){b_1}^2\otimes {b_2}^1+c^4{b_1}^2\otimes {b_2}^2$, which implies that $c^1=c^2+c^3=c^2-c^3=c^4=0$, which implies that $c^2=c^3=0$.

Then, the components of $f$ cannot be denoted like $f_{j_{p+1},...,j_{p+q}}^{j_1,...,j_p}$, which would not make sense for that basis. So, the usual components expression, $f_{j_{p+1},...,j_{p+q}}^{j_1,...,j_p}$, is valid only with respect to the specific type of bases used in this proposition.

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References

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