2025-03-23

1044: For C Map Between C Manifolds with Boundary and Corresponding Charts, Components Function of Pullback of (0,q)-Tensors w.r.t. Standard Bases Is This

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description/proof of that for C map between C manifolds with boundary and corresponding charts, components function of pullback of (0,q)-tensors w.r.t. standard bases is this

Topics


About: C manifold

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any C map between any C manifolds with boundary and any corresponding charts, the components function of the pullback of the (0,q)-tensors with respect to the standard bases is this.

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.


Main Body


1: Structured Description


Here is the rules of Structured Description.

Entities:
M1: { the d1 -dimensional C manifolds with boundary }
M2: { the d2 -dimensional C manifolds with boundary }
f: :M1M2, { the C maps }
(U1M1,ϕ1): { the charts for M1}
(U2M2,ϕ2): { the charts for M2}, such that f(U1)U2
m: U1
fm: :Tq0(Tf(m)M2)Tq0(TmM1), = the pullback 
{dxj1...dxjq|jl{1,...,d1}}: = the standard basis for Tq0(TmM1)
{dyj1...dyjq|jl{1,...,d2}}: = the standard basis for Tq0(Tf(m)M2)
//

Statements:
t=tj1,...,jqdyj1...dyjqTq0(Tf(m)M2)(fm(t)=uj1,...,jqdxj1...dxjq where uj1,...,jq=tl1,...,lqf^l1/xj1...f^lq/xjq where f^=ϕ2fϕ11)
//

{dxj1...dxjq} and {dyj1...dyjq} are indeed some bases, 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: {dxj|j{1,...,d1}} and {dyj|j{1,...,d2}} are the dual bases of {/xj|j{1,...,d1}} and {/yj|j{1,...,d2}}.

In other words, f^ is the components function of f with respect to (U1M1,ϕ1) and (U2M2,ϕ2).


2: Proof


Whole Strategy: Step 1: compute fm(t)((/xl1,...,/xlq)) as t(dfm(/xl1),...,dfm(/xlq)); Step 2: compare the result of Step 1 with fm(t)((/xl1,...,/xlq))=uj1,...,jldxj1...dxjq((/xl1,...,/xlq)).

Step 1:

We know that dfm(/xln)=f^j/xln/yj, by the proposition that for any C map between any C manifolds with boundary and any corresponding charts, the components function of the differential of the map with respect to the standard bases is this: /xln has only the ln-th component, 1.

fm(t)((/xl1,...,/xlq))=t(dfm(/xl1),...,dfm(/xlq))=tj1,...,jldyj1...dyjq((f^m1/xl1/ym1,...,f^mq/xlq/ymq))=tj1,...,jqdyj1(f^m1/xl1/ym1)...dyjq(f^mq/xlq/ymq)=tj1,...,jqδm1j1f^m1/xl1...δmqjqf^mq/xlq=tj1,...,jqf^j1/xl1...f^jq/xlq.

Step 2:

On the other hand, fm(t)((/xl1,...,/xlq))=uj1,...,jqdxj1...dxjq((/xl1,...,/xlq))=uj1,...,jqdxj1(/xl1)...dxjq(/xlq)=uj1,...,jqδl1j1...δlqjq=ul1,...,lq.

So, ul1,...,lq=tj1,...,jqf^j1/xl1...f^jq/xlq, which is nothing but uj1,...,jq=tl1,...,lqf^l1/xj1...f^lq/xjq.


References


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