1035: Canonical 'Vectors Spaces - Linear Morphisms' Isomorphism Between Tensors Space at Point on C∞ Manifold with Boundary and Tensors Space w.r.t. Real Numbers Field and p Cotangent Vectors Spaces and q Tangent Vectors Spaces and Field

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definition of canonical 'vectors spaces - linear morphisms' isomorphism between tensors space at point on $C^{\mathrm{\infty }}$ manifold with boundary and tensors space w.r.t. real numbers field and $p$ cotangent vectors spaces and $q$ tangent vectors spaces and field

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of $(p,q)$-tensors space at point on $C^{\mathrm{\infty }}$ manifold with boundary.
• 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 %category name% isomorphism.

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Target Context

• The reader will have a definition of canonical 'vectors spaces - linear morphisms' isomorphism between tensors space at point on $C^{\mathrm{\infty }}$ manifold with boundary and tensors space with respect to real numbers field and $p$ cotangent vectors spaces and $q$ tangent vectors spaces and field.

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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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$T_{m}M$: $=\text{ the tangent vectors space at }m$
$B$: $\in \{\text{ the bases for }{T}_{m}M\}$, $=\{b_1,...,b_d\}$
$T_m{M}^{\ast }$: $=\text{ the cotangent vectors space at }m$
$B^{\ast }$: $=\text{ the dual basis of }B\text{ for }{T}_m{M}^{\ast }$, $=\{b^1,...,b^d\}$
${B^{\ast }}^{\ast }$: $=\text{ the dual basis of }{B}^{\ast }\text{ for }{T_m{M}^{\ast }}^{\ast }$, $=\{{\tilde{b}}_1,...,{\tilde{b}}_d\}$
$p$: $\in \mathbb{N}$
$q$: $\in \mathbb{N}$
$T_q^p(T_{m}M)$: $=T_{m}M\otimes ...\otimes T_{m}M\otimes T_m{M}^{\ast }\otimes ...\otimes T_m{M}^{\ast }$
$L(T_m{M}^{\ast },...,T_m{M}^{\ast },T_{m}M,...,T_{m}M:\mathbb{R})$:
$\ast f$: $:T_q^p(T_{m}M)\to L(T_m{M}^{\ast },...,T_m{M}^{\ast },T_{m}M,...,T_{m}M:\mathbb{R}),t_{l_1,...,l_q}^{j_1,...,j_p}[((b_{j_1},...,b_{j_p},b^{l_1},...,b^{l_q}))]\mapsto t_{l_1,...,l_q}^{j_1,...,j_p}{\tilde{b}}_{j_1}\otimes ...\otimes {\tilde{b}}_{j_p}\otimes b^{l_1}\otimes ...\otimes b^{l_q}$
//

Conditions:
//

$f$ does not depend on the choice of $B$ as is seen in Note, which is the reason why $f$ is called "canonical".

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2: Note

$f$ is indeed a 'vectors spaces - linear morphisms' isomorphism, because $\{[((b_{j_1},...,b_{j_p},b^{k_1},...,b^{k_q}))]\}$ is a basis for $T_q^p(T_{m}M)$, 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; $\{{\tilde{b}}_{j_1}\otimes ...\otimes {\tilde{b}}_{j_p}\otimes b^{k_1}\otimes ...\otimes b^{k_q}\}$ is a basis for $L(T_m{M}^{\ast },...,T_m{M}^{\ast },T_{m}M,...,T_{m}M:\mathbb{R})$, 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; and the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism applies.

Let us see that $f$ does not depend on the choice of $B$.

Let $B^{\prime }=\{b_1^{\prime },...,b_d^{\prime }\}$ be any other basis for $T_{m}M$.

Let the map constructed by $B^{\prime }$ be $f^{\prime }$.

$b_j^{\prime }=b_l{M}_j^l$ for an invertible matrix, $M$. So, $b_j=b_l^{\prime }{M^{-1}}_j^l$.

The dual basis of $B^{\prime }$, $B^{\prime \ast }=\{b^{\prime 1},...,b^{\prime d}\}$, is $\{{M^{-1}}_l^j{b}^l\}$, by the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this. So, $b^j=M_l^j{b}^{\prime l}$.

The dual basis of $B^{\prime \ast }$, ${B^{\prime \ast }}^{\ast }=\{{\widetilde{b^{\prime }}}_1,...,{\widetilde{b^{\prime }}}_d\}$, is $\{{\tilde{b}}_l{M}_j^l\}$, by the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this. So, ${\tilde{b}}_j={\widetilde{b^{\prime }}}_l{M^{-1}}_j^l$.

Then, $[((b_{j_1},...,b_{j_p},b^{l_1},...,b^{l_q}))]=[((b_{m_1}^{\prime }{M^{-1}}_{j_1}^{m_1},...,b_{m_p}^{\prime }{M^{-1}}_{j_p}^{m_p},M_{n_1}^{l_1}{b}^{\prime n_1},...,M_{n_q}^{l_q}{b}^{\prime n_q}))]={M^{-1}}_{j_1}^{m_1}...{M^{-1}}_{j_p}^{m_p}{M}_{n_1}^{l_1}...M_{n_q}^{l_q}[((b_{m_1}^{\prime },...,b_{m_p}^{\prime },b^{\prime n_1},...,b^{\prime n_q}))]$, which is mapped by $f^{\prime }$ to ${M^{-1}}_{j_1}^{m_1}...{M^{-1}}_{j_p}^{m_p}{M}_{n_1}^{l_1}...M_{n_q}^{l_q}{\widetilde{b^{\prime }}}_{m_1}\otimes ...\otimes {\widetilde{b^{\prime }}}_{m_p}\otimes b^{\prime n_1}\otimes ...\otimes b^{\prime n_q}={\widetilde{b^{\prime }}}_{m_1}{M^{-1}}_{j_1}^{m_1}\otimes ...\otimes {\widetilde{b^{\prime }}}_{m_p}{M^{-1}}_{j_p}^{m_p}\otimes M_{n_1}^{l_1}{b}^{\prime n_1}\otimes ...\otimes M_{n_q}^{l_q}{b}^{\prime n_q}={\tilde{b}}_{j_1}\otimes ...\otimes {\tilde{b}}_{j_p}\otimes b^{l_1}\otimes ...\otimes b^{l_q}$.

So, $f^{\prime }=f$.

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References

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