description/proof of that for \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary that is expressed as multiplication of function by wedge product of \(q\) exterior derivatives of functions in 2 ways, expressions with heading functions replaced by its exterior derivatives represent same object
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of exterior derivative of \(C^\infty\) function over \(C^\infty\) manifold with boundary.
- The reader knows a definition of wedge product of multicovectors.
- The reader admits the proposition that the exterior derivation of \(C^\infty\) function over \(C^\infty\) manifold with boundary satisfies the Leibniz rule.
Target Context
- The reader will have a description and a proof of the proposition that for any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary that is expressed as multiplication of function by wedge product of \(q\) exterior derivatives of functions in any 2 ways, the expressions with the heading functions replaced by its exterior derivatives (and the multiplications replaced by wedge products) represent the same object.
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:
\(M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(t\): \(\in \Omega_q (TM)\)
//
Statements:
\(t = f_1 d f_2 \wedge ... \wedge d f_q = f'_1 d f'_2 \wedge ... \wedge d f'_q\)
\(\implies\)
\(d f_1 \wedge d f_2 \wedge ... \wedge d f_q = d f'_1 \wedge d f'_2 \wedge ... \wedge d f'_q\)
//
2: Note
While it is said like "the expressions with the heading functions replaced by its exterior derivatives (and the multiplications replaced by wedge products)", is it not just an exterior derivative of \(t\)? In fact, it is, but 'exterior derivative of \(C^\infty\) \(q\)-form' has not been defined (as far as we are concerned), and in fact, this proposition is used for defining 'exterior derivative of \(C^\infty\) \(q\)-form'.
3: Proof
Whole Strategy: Step 1: for each \(m \in M\), take any chart, \((U_m \subseteq M, \phi_m)\), and define a chart-dependent operation, \(p: \Omega_r (T U_m) \to \Omega_{r + 1} (T U_m)\), and see its 2 properties; Step 2: see that \(p (t) = d f_1 \wedge d f_2 \wedge ... \wedge d f_q\) and \(p (t) = d f'_1 \wedge d f'_2 \wedge ... \wedge d f'_q\).
Step 1:
Let \(m \in M\) be any.
Let \((U_m \subseteq M, \phi_m)\) be any chart around \(m\).
Let us define a chart-dependent operation, \(p: \Omega_r (T U_m) \to \Omega_{r + 1} (T U_m)\), like this: for any \(u \in \Omega_r (TM)\) such that \(0 \lt r\), express \(u\) with the standard basis for \(\Lambda_r (T U_m)\) as \(u = u_{j_1, ...j_r} d x^{j_1} \wedge ... \wedge d x^{j_r}\), and let \(p (u) = d u_{j_1, ...j_r} \wedge d x^{j_1} \wedge ... \wedge d x^{j_r}\); for any \(u \in \Omega_0 (TM)\), let \(p (u) = d u\).
There is no ambiguity for the definition, because although it may depend on the chart, we have chosen the chart: the standard basis is uniquely determined, the expression of \(u\) is unique, and \(p (u)\) is uniquely specified.
And it is really into \(\Omega_{r + 1} (T U_m)\).
So, the definition is well-defined.
Let us see some 2 properties of \(p\): 1) \(p (u_1 \wedge u_2) = p (u_1) \wedge u_2 + (-1)^r u_1 \wedge p (u_2)\) where \(u_1 \in \Omega_r (T U_m)\) and \(u_2 \in \Omega_s (T U_m)\); 2) \(p (d u) = 0\) where \(u \in \Omega_0 (T U_m)\).
Let us see 1).
Let \(0 \lt r\).
\(p (u_1 \wedge u_2) = p (u_{1, j_1, ..., j_r} d x^{j_1} \wedge ... \wedge d x^{j_r} \wedge u_{2, l_1, ..., l_s} d x^{l_1} \wedge ... \wedge d x^{l_s}) = p (u_{1, j_1, ..., j_r} u_{2, l_1, ..., l_s} d x^{j_1} \wedge ... \wedge d x^{j_r} \wedge d x^{l_1} \wedge ... \wedge d x^{l_s})\).
Note here that although some of \(d x^{j_1} \wedge ... \wedge d x^{j_r} \wedge d x^{l_1} \wedge ... \wedge d x^{l_s}\) s may not be any elements of the standard basis (some may be \(0\); some may be the negatives of a basis element) and there may be some duplications of a same basis element, \(= d (u_{1, j_1, ..., j_r} u_{2, l_1, ..., l_s}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_r} \wedge d x^{l_1} \wedge ... \wedge d x^{l_s})\) holds, because the \(0\) terms vanish anyway and the negatives and the duplications of each same basis element are 1st added up and then operated by \(p\), but as \(d\) is linear, it is the same with 1st each term operated by \(p\) and then added up.
\(= (u_{2, l_1, ..., l_s} d u_{1, j_1, ..., j_r} + u_{1, j_1, ..., j_r} d u_{2, l_1, ..., l_s}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_r} \wedge d x^{l_1} \wedge ... \wedge d x^{l_s}\), by the proposition that the exterior derivation of \(C^\infty\) function over \(C^\infty\) manifold with boundary satisfies the Leibniz rule.
\(= (d u_{1, j_1, ..., j_r} \wedge d x^{j_1} \wedge ... \wedge d x^{j_r}) \wedge (u_{2, l_1, ..., l_s} d x^{l_1} \wedge ... \wedge d x^{l_s}) + (-1)^r (u_{1, j_1, ..., j_r} d x^{j_1} \wedge ... \wedge d x^{j_r}) \wedge (d u_{2, l_1, ..., l_s} \wedge d x^{l_1} \wedge ... \wedge d x^{l_s})\), where \((-1)^r\) appears because moving \(d u_{2, l_1, ..., l_s}\) there is switching \(d u_{2, l_1, ..., l_s}\) and \(d x^{j_1}\), then, \(d u_{2, l_1, ..., l_s}\) and \(d x^{j_2}\), ..., and then, \(d u_{2, l_1, ..., l_s}\) and \(d x^{j_r}\); on the other hand, \(u_{2, l_1, ..., l_s}\) can be just moved.
\(= p (u_1) \wedge u_2 + (-1)^r u_1 \wedge p (u_2)\).
Let \(0 = r\).
\(p (u_1 \wedge u_2) = p (u_1 u_{2, l_1, ..., l_s} d x^{l_1} \wedge ... \wedge d x^{l_s}) = d (u_1 u_{2, l_1, ..., l_s}) \wedge d x^{l_1} \wedge ... \wedge d x^{l_s}\).
\(= (u_{2, l_1, ..., l_s} d u_1 + u_1 d u_{2, l_1, ..., l_s}) \wedge d x^{l_1} \wedge ... \wedge d x^{l_s}\), by the proposition that the exterior derivation of \(C^\infty\) function over \(C^\infty\) manifold with boundary satisfies the Leibniz rule.
\(= d u_1 \wedge (u_{2, l_1, ..., l_s} d x^{l_1} \wedge ... \wedge d x^{l_s}) + u_1 \wedge (d u_{2, l_1, ..., l_s} \wedge d x^{l_1} \wedge ... \wedge d x^{l_s})\).
\(= p (u_1) \wedge u_2 + (-1)^r u_1 \wedge p (u_2)\).
Let us see 2).
\(p (d u) = p (\partial u / \partial x^j d x^j) = d (\partial u / \partial x^j) \wedge d x^j = \partial (\partial u / \partial x^j) / \partial x^l d x^l \wedge d x^j\), but for each \(d x^l \wedge d x^j\), there is the corresponding \(d x^j \wedge d x^l = - d x^l \wedge d x^j\), and \(\partial (\partial u / \partial x^j) / \partial x^l = \partial (\partial u / \partial x^l) / \partial x^j\), so, the pair annihilates each other, and \(p (d u) = 0\).
Step 2:
\(p (t) = p (f_1 d f_2 \wedge ... \wedge d f_q) = p (f_1 \wedge (d f_2 \wedge ... \wedge d f_q)) = p (f_1) \wedge (d f_2 \wedge ... \wedge d f_q) + f_1 \wedge p (d f_2 \wedge ... \wedge d f_q)\).
But \(p (d f_2 \wedge ... \wedge d f_q) = 0\), because while \(p (d f_j) = 0\), \(p (d f_{q - 1} \wedge d f_q) = p (d f_{q - 1}) \wedge d f_q - d f_{q - 1} \wedge p (d f_q) = 0 \wedge d f_q - d f_{q - 1} \wedge 0 = 0\), \(p (d f_{q - 2} \wedge d f_{q - 1} \wedge d f_q) = p ((d f_{q - 2} \wedge d f_{q - 1}) \wedge d f_q) = p (d f_{q - 2} \wedge d f_{q - 1}) \wedge d f_q + (d f_{q - 2} \wedge d f_{q - 1}) \wedge p (d f_q) = 0 \wedge d f_q + (d f_{q - 2} \wedge d f_{q - 1}) \wedge 0 = 0\), and so on.
So, \(p (t) = p (f_1) \wedge (d f_2 \wedge ... \wedge d f_q) + 0 = d f_1 \wedge d f_2 \wedge ... \wedge d f_q\).
Likewise, \(p (t) = d f'_1 \wedge d f'_2 \wedge ... \wedge d f'_q\).
So, \(d f_1 \wedge d f_2 \wedge ... \wedge d f_q = p (t) = d f'_1 \wedge d f'_2 \wedge ... \wedge d f'_q\).
As \(d f_1 \wedge d f_2 \wedge ... \wedge d f_q = d f'_1 \wedge d f'_2 \wedge ... \wedge d f'_q\) around any point, \(m \in M\), it is so all over \(M\).
