description/proof of that double exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary is \(0\)
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary.
- The reader knows a definition of wedge product of multicovectors.
- The reader admits the proposition that the \(q\)-covectors space of any vectors space has the basis that consists of the wedge products of the increasing elements of the dual basis of the vectors space.
Target Context
- The reader will have a description and a proof of the proposition that the double exterior derivative of any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary is \(0\).
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:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(t\): \(: M \to \Lambda_q (T M)\), \(\in \Omega_q (T M)\)
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Statements:
\(d d t = 0\)
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2: Proof
Whole Strategy: Step 1: around each \(m \in M\), take a chart, \((U_m \subseteq M, \phi_m)\), and express \(t\) as \(\sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\); Step 2: compute \(d d t\) and see that it is \(0\).
Step 1:
Let \(m \in M\) be any.
Let us take a chart, \((U_m \subseteq M, \phi_m)\).
With respect to the chart, \(t = \sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\), by the proposition that the \(q\)-covectors space of any vectors space has the basis that consists of the wedge products of the increasing elements of the dual basis of the vectors space.
Step 2:
\(d t = \sum_{j_1 \lt ... \lt j_q} d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = \sum_{j_1 \lt ... \lt j_q} \partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\).
\(d d t = d (\sum_{j_1 \lt ... \lt j_q} \partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = \sum_{j_1 \lt ... \lt j_q} d (\partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q})\), because \(d\) is \(\mathbb{R}\)-linear.
For each fixed \((j_1, ..., j_q)\), the terms for \(l \in \{j_1, ..., j_q\}\) are \(0\), by a property mentioned in Note for the definition of wedge product of multicovectors.
For each fixed \((j_1, ..., j_q)\), the term for each \(l \notin \{j_1, ..., j_q\}\) is \(d (\partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = \partial (\partial t_{j_1, ..., j_q} / \partial x^l) / \partial x^m d x^m \wedge d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q})\), because while the definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary requires \(d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\) in the increasing order, reordering it in the increasing order yields a sign, but reordering the result back in the original order eliminate the whatever sign.
For each fixed \((j_1, ..., j_q)\), the terms for \(l = m\) are \(0\), as before, and for each \((l, m)\) pair such that \(l \neq m\), there is the corresponding \((m, l)\) pair, and \(\partial (\partial t_{j_1, ..., j_q} / \partial x^l) / \partial x^m d x^m \wedge d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) + \partial (\partial t_{j_1, ..., j_q} / \partial x^m) / \partial x^l d x^l \wedge d x^m \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = 0\), because while \(\partial (\partial t_{j_1, ..., j_q} / \partial x^l) / \partial x^m = \partial (\partial t_{j_1, ..., j_q} / \partial x^m) / \partial x^l\), \(d x^m \wedge d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = - d x^l \wedge d x^m \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\), by a property mentioned in Note for the definition of wedge product of multicovectors.
So, \(d d t = 0\).
