1055: For Finite-Product C∞ Manifold with Boundary and 'Vectors Spaces - Linear Morphisms' Isomorphism from Tangent Vectors Space onto Direct Sum of Tangent Vectors Spaces, Tangent Vector Operates on Function as Sum of Vectors on Projected Functions

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description/proof of that for finite-product $C^{\mathrm{\infty }}$ manifold with boundary and 'vectors spaces - linear morphisms' isomorphism from tangent vectors space onto direct sum of tangent vectors spaces, tangent vector operates on function as sum of vectors on projected functions

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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
• 3: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ map projected from $C^{\mathrm{\infty }}$ map from finite-product $C^{\mathrm{\infty }}$ manifold with boundary by fixing domain components except $j$-th based on point.
• The reader admits the proposition that for any finite-product $C^{\mathrm{\infty }}$ manifold with boundary, at each point of the manifold with boundary, there is a 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at the point onto the direct sum of the tangent vectors spaces of the constituents at the corresponding points.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-product $C^{\mathrm{\infty }}$ manifold with boundary and the 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at each point onto the direct sum of the corresponding tangent vectors spaces of the constituents, any tangent vector operates on any function as the sum of the corresponding vectors on the projected functions.

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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:
$\{M_1,...,M_{n-1}\}$: $\subseteq \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$M_n$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_1\times ...\times M_n$: $=\text{ the finite-product }{C}^{\mathrm{\infty }}\text{ manifold with boundary }$
$m$: $=(m^1,...,m^n)\in M_1\times ...\times M_n$
$g$: $:T_m(M_1\times ...\times M_n)\to T_{m^1}{M}_1\oplus ...\oplus T_{m^n}{M}_n,v_m\mapsto (d{\pi }_1{v}_m,...,d{\pi }_n{v}_m)$, $=\text{ the canonical 'vectors spaces - linear morphisms' isomorphism }$
$v_m$: $\in T_m(M_1\times ...\times M_n)$
$f$: $\in C^{\mathrm{\infty }}(M_1\times ...\times M_n)$
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Statements:
$v_{m}f=\sum _{j\in \{1,...,n\}}(d{\pi }_j{v}_m)f_{j,m}$
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2: Note

We know that $v_m$ corresponds to $(d{\pi }_1{v}_m,...,d{\pi }_n{v}_m)$, but so what? I mean, how can we get $v_{m}f$ with respect to $(d{\pi }_1{v}_m,...,d{\pi }_n{v}_m)$? Certainly, $d{\pi }_j{v}_m$ cannot operate on $f$, because $f$ is not any function on $M_j$. So, what?, which is the motivation of this proposition.

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3: Proof

Whole Strategy: Step 1: take any chart around $m$, $(U_m:=U_{m^1}\times ...\times U_{m^n}\subseteq M_1\times ...\times M_n,{\varphi }_m:={\varphi }_{m^1}\times ...\times {\varphi }_{m^n})$, and let $v_m=v_m^{1,j_1}\partial /\partial x^{1,j_1}+...+v_m^{n,j_n}\partial /\partial x^{n,j_n}$; Step 2: see that $v_{m}f=\sum _{j\in \{1,...,n\}}(d{\pi }_j{v}_m)f_{j,m}$.

Step 1:

Let us take any chart around $m$, $(U_m:=U_{m^1}\times ...\times U_{m^n}\subseteq M_1\times ...\times M_n,{\varphi }_m:={\varphi }_{m^1}\times ...\times {\varphi }_{m^n})$, where $(U_{m^j}\subseteq M_j,{\varphi }_{m^j})$ is a chart around $m^j$ for $M_j$, which is possible by the definition of finite-product $C^{\mathrm{\infty }}$ manifold with boundary.

$v_m=v_m^{1,j_1}\partial /\partial x^{1,j_1}+...+v_m^{n,j_n}\partial /\partial x^{n,j_n}$.

Step 2:

$v_{m}f=v_m^{1,j_1}\partial f/\partial x^{1,j_1}+...+v_m^{n,j_n}\partial f/\partial x^{n,j_n}$.

$(d{\pi }_l{v}_m)f_{l,m}=v_m(f_{l,m}\circ {\pi }_l)=v_m^{1,j_1}\partial (f_{l,m}\circ {\pi }_l)/\partial x^{1,j_1}+...+v_m^{n,j_n}(\partial f_{l,m}\circ {\pi }_l)/\partial x^{n,j_n}$.

$v_m^{o,j_o}\partial (f_{l,m}\circ {\pi }_l)/\partial x^{o,j_o}=v_m^{o,j_o}{\partial }_{o,j_o}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1})$, where ${\partial }_{o,j_o}$ is the partial derivative by the $o,j_o$ component.

$f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1}$ is $(x^{\prime 1,1},...,x^{\prime 1,d_1},...,x^{\prime l,1},...,x^{\prime l,d_l},...,x^{\prime n,1},...,x^{\prime n,d_n})\mapsto (m^{\prime 1},...,m^{\prime l},...,m^{\prime n})\mapsto m^{\prime l}\mapsto f_{l,m}(m^{\prime l})=f(m^1,...,m^{\prime l},...,m^n)$.

So, when $o\ne l$, $v_m^{o,j_o}{\partial }_{o,j_o}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1})=0$, because $f(m^1,...,m^{\prime l},...,m^n)$ does not depend on $(x^{\prime o,1},...,x^{\prime o,d_o})$.

So, $(d{\pi }_l{v}_m)f_{l,m}=v_m^{l,j_l}{\partial }_{l,j_l}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1})$.

On the other hand, $v_m^{l,j_l}\partial f/\partial x^{l,j_l}=v_m^{l,j_l}{\partial }_{l,j_l}(f\circ {{\varphi }_m}^{-1})$.

$f\circ {{\varphi }_m}^{-1}$ is $(x^{\prime 1,1},...,x^{\prime 1,d_1},...,x^{\prime l,1},...,x^{\prime l,d_l},...,x^{\prime n,1},...,x^{\prime n,d_n})\mapsto (m^{\prime 1},...,m^{\prime l},...,m^{\prime n})\mapsto f(m^{\prime 1},...,m^{\prime l},...,m^{\prime n})$.

While the difference between $f(m^1,...,m^{\prime l},...,m^n)$ and $f(m^{\prime 1},...,m^{\prime l},...,m^{\prime n})$ is whether $(m^{\prime 1},...,\widehat{m^{\prime l}},...,m^{\prime n})$ are fixed, $v_m^{l,j_l}{\partial }_{l,j_l}(f\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}$ equals $v_m^{l,j_l}{\partial }_{l,j_l}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}$, because ${\partial }_{l,j_l}$ moves only $m^{\prime l}$ anyway.

So, $(d{\pi }_l{v}_m)f_{l,m}=v_m^{l,j_l}\partial f/\partial x^{l,j_l}$.

So, $v_{m}f=\sum _{j\in \{1,...,n\}}(d{\pi }_j{v}_m)f_{j,m}$.

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References

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