definition of exterior derivative of \(C^\infty\) function over \(C^\infty\) manifold with boundary
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of function over \(C^\infty\) manifold with boundary.
- The reader knows a definition of \(q\)-form over \(C^\infty\) manifold with boundary.
- The reader knows a definition of section of continuous surjection.
Target Context
- The reader will have a definition of exterior derivative of \(C^\infty\) function over \(C^\infty\) manifold with boundary.
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:
\( M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\( \mathbb{R}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\( f\): \(: M \to \mathbb{R}\), \(\in \Omega_0 (TM)\)
\(*d f\): \(: M \to \Lambda_1 (T M)\), \(\in \Omega_1 (TM)\)
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Conditions:
\(\forall v \in T_mM (d f (m) (v) = v f)\)
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2: Note
As \(d f\) is a section of \(\pi: \Lambda_1 (T M) \to M\), \(d f (m) \in \Lambda_1 (T_mM)\), and \(d f (m) (v)\) makes sense. But usually, \(d f (m) (v)\) is denoted as \(d f (v)\), because as \(v\) determines \(m\), \(m\) does not need to be specified.
For each chart around \(m\), \((U_m \subseteq M, \phi_m)\), \(d f = \partial f / \partial x^j d x^j\), where \((d x^1, ... d x^d)\) is the dual basis of \((\partial / \partial x^1, ..., \partial / \partial x^d)\), because denoting \(v = v^j \partial / \partial x^j\), \(d f (v) = v f = v^j \partial / \partial x^j (f) = v^j \partial f / \partial x^j\) while \(\partial f / \partial x^j d x^j (v) = \partial f / \partial x^j d x^j (v^l \partial / \partial x^l) = \partial f / \partial x^j v^l \delta^j_l = \partial f / \partial x^j v^j\).
Let us see that \(d f\) is indeed a \(C^\infty\) section of \(\pi\).
For each \(m \in M\), \(d f (m)\) operates on \(T_mM\) and it is multilinear, because \(d f (m) (r v + r' v') = (r v + r' v') f = r v f + r' v' f = r d f (m) (v) + r' d f (m) (v')\), so, \(d f (m) \in \Lambda_1 (T_mM)\). So, \(\pi (d f (m)) = m\).
\(d f\) is continuous, in fact \(C^\infty\), because for each \(m \in M\), there are a chart around \(m\), \((U_m \subseteq M, \phi_m)\), and a chart around \(d f (m)\), \((\pi^{-1} (U_m) \subseteq \Lambda_1 (T M), \widetilde{\phi_m})\), and the components function of \(d f\) with respect to the charts is \(: \phi_m (U_m) \to \widetilde{\phi_m} (\pi^{-1} (U_m)), \phi_m (m) \mapsto (\partial f / \partial x^1, ..., \partial f / \partial x^d, \phi_m (m))\), which is \(C^\infty\) because \(f\) is \(C^\infty\).
\((d x^1, ..., d x^d)\) has been defined to be the dual basis of \((\partial / \partial x^1, ..., \partial / \partial x^d)\), but in fact, \(x^j\) is a function over \(U_m\) and \(d x^j\) as the exterior derivative of the function equals \(d x^j\) as the element of the dual basis: regarding \(d x^j\) as the element of the dual basis, \(d x^j (v) = d x^j (v^l \partial / \partial x^l) = v^l d x^j (\partial / \partial x^l) = v^l \delta^j_l = v^j\), on the other hand, regarding \(d x^j\) as the exterior derivative of \(x^j\), \(d x^j (v) = v (x^j) = v^l \partial / \partial x^l (x^j) = v^l \delta^j_l = v^j\).
