description/proof of that for \(C^\infty\) map between \(C^\infty\) manifolds with boundary, pullback of wedge product of multicovectors is wedge product of pullbacks of multicovectors
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
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 wedge product of any multicovectors is the wedge product of the pullbacks of the multicovectors.
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:
\(M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\(m_1\): \(\in M_1\)
\(t\): \(\in \Lambda_q (T_{f (m_1)}M_2)\)
\(t'\): \(\in \Lambda_{q'} (T_{f (m_1)}M_2)\)
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Statements:
\(f^*_{m_1} (t \wedge t') = (f^*_{m_1} t) \wedge (f^*_{m_1} t')\)
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This holds also when \(q = 0\) or \(q' = 0\).
2: Note
As a corollary, \(f^*_{m_1} (t \wedge t' \wedge ... \wedge t'') = (f^*_{m_1} t) \wedge (f^*_{m_1} t') \wedge ... \wedge (f^*_{m_1} t'')\): \(f^*_{m_1} (t \wedge t' \wedge t'') = f^*_{m_1} ((t \wedge t') \wedge t'') = (f^*_{m_1} (t \wedge t')) \wedge (f^*_{m_1} t'') = ((f^*_{m_1} t) \wedge (f^*_{m_1} t')) \wedge (f^*_{m_1} t'') = (f^*_{m_1} t) \wedge (f^*_{m_1} t') \wedge (f^*_{m_1} t'')\), and so on.
3: Proof
Whole Strategy: Step 1: compute \(f^*_{m_1} (t \wedge t') (v_1, ..., v_{q + q'})\); Step 2: compute \((f^*_{m_1} t) \wedge (f^*_{m_1} t') (v_1, ..., v_{q + q'})\); Step 3: conclude the proposition.
Step 1:
Let \(q = 0\).
\(f^*_{m_1} (t \wedge t') = f^*_{m_1} (t t') = (t \circ f (m_1)) f^*_{m_1} t'\), because the pullback is linear.
Let \(q' = 0\).
\(f^*_{m_1} (t \wedge t') = f^*_{m_1} (t t') = f^*_{m_1} (t' t) = (t' \circ f (m_1)) f^*_{m_1} t\).
Let \(0 \lt q\) and \(0 \lt q'\).
Let \((v_1, ..., v_{q + q'}) \in T_{m_1}M_1 \times ... \times T_{m_1}M_1\) be any.
\(f^*_{m_1} (t \wedge t') (v_1, ..., v_{q + q'}) = t \wedge t' (d f_{m_1} v_1, ..., d f_{m_1} v_{q + q'}) = (q + q')! / (q! q'!) Asym (t \otimes t') (d f_{m_1} v_1, ..., d f_{m_1} v_{q + q'}) = (q + q')! / (q! q'!) 1 / (q + q')! \sum_{\sigma \in S_{q + q'}} sgn \sigma t \otimes t' (d f_{m_1} v_{\sigma_1}, ..., d f_{m_1} v_{\sigma_{q + q'}}) = 1 / (q! q'!) \sum_{\sigma \in S_{q + q'}} sgn \sigma t (d f_{m_1} v_{\sigma_1}, ..., d f_{m_1} v_{\sigma_q}) t' (d f_{m_1} v_{\sigma_{q + 1}}, ..., d f_{m_1} v_{\sigma_{q + q'}})\).
Step 2:
Let \(q = 0\).
\((f^*_{m_1} t) \wedge (f^*_{m_1} t') = (t \circ f (m_1)) (f^*_{m_1} t')\).
Let \(q' = 0\).
\((f^*_{m_1} t) \wedge (f^*_{m_1} t') = (f^*_{m_1} t) (t' \circ f (m_1)) = (t' \circ f (m_1)) f^*_{m_1} t\).
Let \(0 \lt q\) and \(0 \lt q'\).
\((f^*_{m_1} t) \wedge (f^*_{m_1} t') (v_1, ..., v_{q + q'}) = (q + q')! / (q! q'!) Asym ((f^*_{m_1} t) \otimes (f^*_{m_1} t')) (v_1, ..., v_{q + q'}) = (q + q')! / (q! q'!) 1 / (q + q')! \sum_{\sigma \in S_{q + q'}} sgn \sigma (f^*_{m_1} t) \otimes (f^*_{m_1} t') (v_{\sigma_1}, ..., v_{\sigma_{q + q'}}) = 1 / (q! q'!) \sum_{\sigma \in S_{q + q'}} sgn \sigma (f^*_{m_1} t) (v_{\sigma_1}, ..., v_{\sigma_q}) (f^*_{m_1} t') (v_{\sigma_{q + 1}}, ..., v_{\sigma_{q + q'}}) = 1 / (q! q'!) \sum_{\sigma \in S_{q + q'}} sgn \sigma t (d f_{m_1} v_{\sigma_1}, ..., d f_{m_1} v_{\sigma_q}) t' (d f_{m_1} v_{\sigma_{q + 1}}, ..., d f_{m_1} v_{\sigma_{q + q'}})\).
Step 3:
The result of Step 1 and the result of Step 2 are the same.
So, \(f^*_{m_1} (t \wedge t') = (f^*_{m_1} t) \wedge (f^*_{m_1} t')\).
