A description of what chart induced basis vector on \(C^\infty\) manifold with boundary is
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
- The reader will have a description of what any chart induced basis vector on any \(C^\infty\) manifold with boundary is.
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: Note
The conclusion is in fact what will be rather prevalently guessed intuitively, but this is about more rigorously confirming that that guess is correct.
2: Description
For any \(C^\infty\) manifold with (possibly empty) boundary, \(M\), and any chart, \((U \subseteq M, \phi)\), what is the chart induced basis vector, \(\partial / \partial x^j \vert_p\), at \(p \in U\)?
This argument is profusely based on a definition of map between arbitrary subsets of \(C^\infty\) manifolds with boundary \(C^k\) at point, where \(k\) excludes \(0\) and includes \(\infty\).
\(\partial / \partial x^j \vert_p\) is defined to be such that for any \(C^\infty\) function, \(f: M \to \mathbb{R}\), \(\partial / \partial x^j \vert_p (f) = \partial (f \circ {\phi}^{-1})' / \partial x^j \vert_{\phi (p)}\), where \((f \circ {\phi}^{-1})': U'_{\phi (p)} \to \mathbb{R}\) is any \(C^\infty\) extension of \(f \circ {\phi}^{-1}\), where \(U'_{\phi (p)} \subseteq \mathbb{R}^d\) is open, such that \((f \circ {\phi}^{-1})' \vert_{U'_{\phi (p)} \cap \phi (U)} = f \circ {\phi}^{-1} \vert_{U'_{\phi (p)} \cap \phi (U)}\).
The result does not really depend on the extension, because while \(\phi (U) \subseteq \mathbb{H}^d\) is an open subset of \(\mathbb{H}^d\), \(\partial (f \circ {\phi}^{-1})' / \partial x^j \vert_{\phi (p)}\) has to equal the one-side partial derivative (when \(p\) is on the boundary and \(j = d\)) or the full partial derivative (otherwise), \(\partial (f \circ {\phi}^{-1}) / \partial x^j \vert_{\phi (p)}\).
Let us confirm that \(\partial / \partial x^j \vert_p\) is indeed a derivation. \(\partial / \partial x^j \vert_p (f g) = \partial ((f g) \circ {\phi}^{-1})' / \partial x^j \vert_{\phi (p)}\). But we can take \(((f g) \circ {\phi}^{-1})' = (f \circ {\phi}^{-1})' (g \circ {\phi}^{-1})'\), because while there are some \(C^\infty\) extensions, \((f \circ {\phi}^{-1})': U'_{f, \phi (p)} \to \mathbb{R}\) and \((g \circ {\phi}^{-1})': U'_{g, \phi (p)} \to \mathbb{R}\), \((f \circ {\phi}^{-1})' \vert_{U'_{f, \phi (p)} \cap U'_{g, \phi (p)}} (g \circ {\phi}^{-1})' \vert_{U'_{f, \phi (p)} \cap U'_{g, \phi (p)}}: U'_{f, \phi (p)} \cap U'_{g, \phi (p)} \to \mathbb{R}\) is a \(C^\infty\) extension of \(f g\). So, \(\partial ((f g) \circ {\phi}^{-1})' / \partial x^j \vert_{\phi (p)} = \partial ((f \circ {\phi}^{-1})' (g \circ {\phi}^{-1})') / \partial x^j \vert_{\phi (p)} = \partial (f \circ {\phi}^{-1})' / \partial x^j \vert_{\phi (p)} (g \circ {\phi}^{-1})' (\phi (p)) + (f \circ {\phi}^{-1})' (\phi (p)) \partial (g \circ {\phi}^{-1})' / \partial x^j \vert_{\phi (p)} = \partial / \partial x^j \vert_p (f) g (p) + f (p) \partial / \partial x^j \vert_p (g)\).
\((\partial / \partial x^1 \vert_p, ..., \partial / \partial x^d \vert_p)\) is linearly independent, because for \(c_1 \partial / \partial x^1 \vert_p + ... + c_d \partial / \partial x^d \vert_p = 0\), \((c_1 \partial / \partial x^1 \vert_p + ... + c_d \partial / \partial x^d \vert_p) x^j = c_j = 0 x^j = 0\).
