2023-09-17

370: For Euclidean C^\infty Manifold and Its Regular Submanifold, Vectors Field Along Regular Submanifold Is C^\infty iff Its Components w.r.t. Standard Chart Are C^\infty on Regular Submanifold

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that for Euclidean \(C^\infty\) manifold and its regular submanifold, vectors field along regular submanifold is \(C^\infty\) iff its components w.r.t. standard chart are \(C^\infty\) on regular submanifold

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 Euclidean \(C^\infty\) manifold and its any regular submanifold, any vectors filed along the regular submanifold is \(C^\infty\) if and only if the components of the vectors field with respect to the standard chart on the Euclidean \(C^\infty\) manifold are \(C^\infty\) on the regular submanifold.

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: Description


For any Euclidean \(C^\infty\) manifold, \(\mathbb{R}^n\), and its any regular submanifold, \(M \subseteq \mathbb{R}^n\), any vectors field along \(M\), \(V = V^i (p \in M) \partial_i\) where \(\{\partial_i\}\) is the canonical basis of \(T_p\mathbb{R}^n\) by the standard chart on \(\mathbb{R}^n\) is \(C^\infty\) if and only if \(V^i (p \in M)\) is \(C^\infty\) on \(M\).


2: Proof


Let us suppose that \(V\) is \(C^\infty\). For any \(C^\infty\) function, \(f\), on \(\mathbb{R}^n\), \(V f\) is \(C^\infty\) on \(M\), by the definition of \(C^\infty\) vectors field along regular submanifold. \(V f = V^i (p \in M) \frac{\partial f}{\partial_i}\), but \(f\) can be taken to be the coordinate function, \(r^i\), then, \(V r^i = V^i (p \in M)\), which is \(C^\infty\) on \(M\).

Let us suppose that \(V^i (p \in M)\) is \(C^\infty\) on \(M\). Then, \(V f = V^i (p \in M) \frac{\partial f}{\partial_i}\) is \(C^\infty\) on \(M\), because \(\frac{\partial f}{\partial_i}\) is \(C^\infty\) on \(\mathbb{R}^n\), and is \(C^\infty\) on \(M\), by the proposition that any \(C^\infty\) function on any \(C^\infty\) manifold is \(C^\infty\) on any regular submanifold of the \(C^\infty\) manifold.


References


<The previous article in this series | The table of contents of this series | The next article in this series>