2025-03-09

1030: For \(C^\infty\) Manifold with Boundary and Tangent Vectors Space at Point, Transition of Standard Bases w.r.t. Charts Is This

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description/proof of that for \(C^\infty\) manifold with boundary and tangent vectors space at point, transition of standard bases w.r.t. charts is this

Topics


About: \(C^\infty\) 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 \(C^\infty\) manifold with boundary and the tangent vectors space at any point, the transition of the standard bases with respect to any charts 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:
\(M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(m\): \(\in M\)
\(T_mM\): \(= \text{ the tangent vectors space at } m\)
\((U_m \subseteq M, \phi_m)\): \(\in \{\text{ the charts for } M \text{ around } m\}\)
\((U'_m \subseteq M, \phi'_m)\): \(\in \{\text{ the charts for } M \text{ around } m\}\)
\(B\): \(= \text{ the standard basis for } T_mM\) with respect to \((U_m \subseteq M, \phi_m)\), \(= \{\partial / \partial x^1, ..., \partial / \partial x^d\}\)
\(B'\): \(= \text{ the standard basis for } T_mM\) with respect to \((U'_m \subseteq M, \phi'_m)\), \(= \{\partial / \partial x'^1, ..., \partial / \partial x'^d\}\)
//

Statements:
\(\partial / \partial x'^j = \partial x^k / \partial x'^j \partial / \partial x^k\)
//

\(x\) as a function of \(x'\) is \(\phi_m \circ {\phi'_m}^{-1} \vert_{\phi'_m (U_m \cap U'_m)}: \phi'_m (U_m \cap U'_m) \to \phi_m (U_m \cap U'_m)\).


2: Proof


Whole Strategy: Step 1: let \(\partial / \partial x'^j = M^k_j \partial / \partial x^k\); Step 2: make the both sides of it operate on \(x^l: U_m \cap U'_m \to \mathbb{R}, p \mapsto {\phi_m (p)}^l\), and see that \(M^k_j = \partial x^k / \partial x'^j\).

Step 1:

Let \(\partial / \partial x'^j = M^k_j \partial / \partial x^k\), which is possible because \(B\) is a basis for \(T_mM\) and \(\partial / \partial x'^j \in T_mM\).

Step 2:

\(x^l: U_m \cap U'_m \to \mathbb{R}, p \mapsto {\phi_m (p)}^l\) is a \(C^\infty\) function.

\(\partial / \partial x'^j (x^l) = M^k_j \partial / \partial x^k (x^l) = M^k_j \partial_k (x^l \circ {\phi_m}^{-1}) = M^k_j \delta^l_k = M^l_j\). So, \(M^k_j = \partial x^k / \partial x'^j\).


References


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