1043: Pullback of (0,q)-Tensors at Point by C∞ Map Between C∞ Manifolds with Boundary

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definition of pullback of $(0,q)$-tensors at point by $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary

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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 differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.

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Target Context

• The reader will have a definition of pullback of $(0,q)$-tensors at point by $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary.

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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:
$M_1$: $\in \{\text{ the }{d}_1\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_2$: $\in \{\text{ the }{d}_2\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$m$: $\in M_1$
$q$: $\in \mathbb{N}$
$\ast f_m^{\ast }$: $:T_q^0(T_{f(m)}{M}_2)\to T_q^0(T_m{M}_1)$, $\in \{\text{ the linear maps }\}$
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Conditions:
when $q=0$, $\mathrm{\forall }t\in T_q^0(T_{f(m)}{M}_2)(f_m^{\ast }(t)=t\circ f(m))$
when $0<q$, $\mathrm{\forall }t\in T_q^0(T_{f(m)}{M}_2),\mathrm{\forall }{v}_1,...,v_q\in T_m{M}_1(f_m^{\ast }(t)(v_1,...,v_q)=t(d{f}_m(v_1),...,d{f}_m(v_q)))$
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2: Note

Let us see that $f_m^{\ast }$ is indeed into $T_q^0(T_m{M}_1)$.

When $q=0$, $f_m^{\ast }(t)=t\circ f(m)\in \mathbb{R}$.

Let us suppose that $0<q$.

$f_m^{\ast }(t)(v_1,...,r{v}_j+r^{\prime }{v}_j^{\prime },...,v_q)=t(d{f}_m(v_1),...,d{f}_m(r{v}_j+r^{\prime }{v}_j^{\prime }),...,d{f}_m(v_q))=t(d{f}_m(v_1),...,rd{f}_m(v_j)+r^{\prime }d{f}_m(v_j^{\prime }),...,d{f}_m(v_q))$, because $d{f}_m$ is linear, $=rt(d{f}_m(v_1),...,d{f}_m(v_j),...,d{f}_m(v_q))+r^{\prime }t(d{f}_m(v_1),...,d{f}_m(v_j^{\prime }),...,d{f}_m(v_q))$, because $t$ is multilinear, $=r{f}_m^{\ast }(t)(v_1,...,v_j,...,v_q)+r^{\prime }{f}_m^{\ast }(t)(v_1,...,v_j^{\prime },...,v_q)$.

Let us see that $f_m^{\ast }$ is indeed linear.

When $q=0$, $f_m^{\ast }(rt+r^{\prime }{t}^{\prime })=r{f}_m^{\ast }(t)+r^{\prime }{f}_m^{\ast }(t^{\prime })$, because $f_m^{\ast }(rt+r^{\prime }{t}^{\prime })=(rt+r^{\prime }{t}^{\prime })\circ f(m)=rt\circ f(m)+r^{\prime }{t}^{\prime }\circ f(m)=r{f}_m^{\ast }(t)+r^{\prime }{f}_m^{\ast }(t^{\prime })$.

Let us suppose that $0<q$.

$f_m^{\ast }(rt+r^{\prime }{t}^{\prime })=r{f}_m^{\ast }(t)+r^{\prime }{f}_m^{\ast }(t^{\prime })$, because $f_m^{\ast }(rt+r^{\prime }{t}^{\prime })(v_1,...,v_q)=(rt+r^{\prime }{t}^{\prime })(d{f}_m(v_1),...,d{f}_m(v_q))=rt(d{f}_m(v_1),...,d{f}_m(v_q))+r^{\prime }{t}^{\prime }(d{f}_m(v_1),...,d{f}_m(v_q))=r{f}_m^{\ast }(t)(v_1,...,v_q)+r^{\prime }{f}_m^{\ast }(t^{\prime })(v_1,...,v_q)=(r{f}_m^{\ast }(t)+r^{\prime }{f}_m^{\ast }(t^{\prime }))(v_1,...,v_q)$.

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References

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