description/proof of that for \(C^\infty\) map between \(C^\infty\) manifolds with boundary, pullback of exterior derivative of form is exterior derivative of pullback of form
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of pullback of \(q\)-covectors at point by \(C^\infty\) map between \(C^\infty\) manifolds with boundary.
- The reader knows a definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary.
- The reader admits the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary, the pullback of the wedge product of any multicovectors is the wedge product of the pullbacks of the multicovectors.
- The reader admits the proposition that the exterior derivative of the wedge product of any \(C^\infty\) \(q_1\)-form and any \(C^\infty\) \(q_2\)-form over any \(C^\infty\) manifold with boundary is the wedge product of the exterior derivative of the \(q_1\)-form and the \(q_2\)-form plus \(-1\) to power \(q_1\) the wedge product of the \(q_1\)-form and the exterior derivative of the \(q_2\)-form.
- The reader admits the proposition that the double exterior derivative of any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary is \(0\).
Target Context
- The reader will have a description and a proof of the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary, the pullback of the exterior derivative of any form is the exterior derivative of the pullback of the form.
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:
\(M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\(t\): \(\in \Omega_q (M_2)\)
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Statements:
\(f^* d t = d f^* t\)
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2: Proof
Whole Strategy: Step 1: see it when \(q = 0\); Step 2: when \(0 \lt q\), for each \(m_1 \in M_1\), take any chart around \(f (m_1)\), \((U_{f (m_1)} \subseteq M_2, \phi_{f (m_1)})\), and express \(t\) as \(\sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\), and compute \(f^* d t\) and \(d f^* t\).
Step 1:
Let \(q = 0\).
Let \(m_1 \in M_1\) be any.
Let \(v \in T_{m_1}M_1\) be any.
\(f^* d t (v) = d t (d f_{m_1} (v)) = (d f_{m_1} (v)) (t) = v (t \circ f)\).
\((d f^* t) (v) = v (f^* t) = v (t \circ f)\).
So, \(f^* d t = d f^* t\).
Step 2:
Let \(0 \lt q\).
Let \(m_1 \in M_1\) be any.
Let us take any chart around \(f (m_1)\), \((U_{f (m_1)} \subseteq M_2, \phi_{f (m_1)})\).
\(t = \sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\) there.
\(d t = \sum_{j_1 \lt ... \lt j_q} d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\), by the definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary.
So, \(f^* d t = f^* (\sum_{j_1 \lt ... \lt j_q} d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = \sum_{j_1 \lt ... \lt j_q} f^* (d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = \sum_{j_1 \lt ... \lt j_q} f^* (d t_{j_1, ..., j_q}) \wedge f^* (d x^{j_1}) \wedge ... \wedge f^* (d x^{j_q})\), by the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary, the pullback of the wedge product of any multicovectors is the wedge product of the pullbacks of the multicovectors, \(= \sum_{j_1 \lt ... \lt j_q} d f^* (t_{j_1, ..., j_q}) \wedge d f^* (x^{j_1}) \wedge ... \wedge d f^* (x^{j_q})\), by Step 1.
On the other hand, \(d f^* t = d f^* (\sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}) = d (\sum_{j_1 \lt ... \lt j_q} f^* (t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q})) = d (\sum_{j_1 \lt ... \lt j_q} f^* (t_{j_1, ..., j_q}) \wedge f^* (d x^{j_1}) \wedge ... \wedge f^* (d x^{j_q}))\), by the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary, the pullback of the wedge product of any multicovectors is the wedge product of the pullbacks of the multicovectors, \(= \sum_{j_1 \lt ... \lt j_q} d (f^* (t_{j_1, ..., j_q}) d f^* (x^{j_1}) \wedge ... \wedge d f^* (x^{j_q}))\), by Step 1.
\(= \sum_{j_1 \lt ... \lt j_q} d f^* (t_{j_1, ..., j_q}) \wedge d f^* (x^{j_1}) \wedge ... \wedge d f^* (x^{j_q})\), which is by the proposition that the exterior derivative of the wedge product of any \(C^\infty\) \(q_1\)-form and any \(C^\infty\) \(q_2\)-form over any \(C^\infty\) manifold with boundary is the wedge product of the exterior derivative of the \(q_1\)-form and the \(q_2\)-form plus \(-1\) to power \(q_1\) the wedge product of the \(q_1\)-form and the exterior derivative of the \(q_2\)-form and the proposition that the double exterior derivative of any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary is \(0\): \(d (f^* (t_{j_1, ..., j_q}) d f^* (x^{j_1}) \wedge ... \wedge d f^* (x^{j_q})) = d f^* (t_{j_1, ..., j_q}) \wedge (d f^* (x^{j_1}) \wedge ... \wedge d f^* (x^{j_q})) + (-1)^0 f^* (t_{j_1, ..., j_q}) \wedge d (d f^* (x^{j_1}) \wedge ... \wedge d f^* (x^{j_q}))\), but the 2nd term is \(0\), because by further expanding it, each term has a \(d d\) factor.
So, the both hand sides are equal, so, \(f^* d t = d f^* t\).
