definition of pullback of \(q\)-covectors at point by \(C^\infty\) map between \(C^\infty\) manifolds with boundary
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
- The reader will have a definition of pullback of \(q\)-covectors at point by \(C^\infty\) map between \(C^\infty\) manifolds 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:
\( M_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\( M_2\): \(\in \{\text{ the } d_2 \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\( f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\( m\): \(\in M_1\)
\( q\): \(\in \mathbb{N}\)
\(*f^*_m\): \(: \Lambda_q (T_{f (m)}M_2) \to \Lambda_q (T_mM_1)\), \(\in \{\text{ the linear maps }\}\)
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Conditions:
when \(q = 0\), \(\forall t \in \Lambda_q (T_{f (m)}M_2) (f^*_m (t) = t \circ f (m))\)
when \(0 \lt q\), \(\forall t \in \Lambda_q (T_{f (m)}M_2), \forall v_1, ..., v_q \in T_mM_1 (f^*_m (t) (v_1, ..., v_q) = t (d f_m (v_1), ..., d f_m (v_q)))\)
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2: Note
While any \(q\)-covector is a \((0, q)\)-tensor, the pullback of the \(q\)-covector at any point by \(f\) is the pullback of the \((0, q)\)-tensor at the point by \(f\). The issue is whether the pullback is indeed in \(\Lambda_q (T_mM_1)\).
Let us see that \(f^*_m\) is indeed into \(\Lambda_q (T_mM_1)\).
\(f^*_m\) is into \(T^0_q (T_mM_1)\), as is seen in Note for the definition of pullback of \((0, q)\)-tensors at point by \(C^\infty\) map between \(C^\infty\) manifolds with boundary.
The issue is whether \(f^*_m\) is indeed antisymmetric.
When \(q = 0\), \(f^*_m (t) = t \circ f (m)\) is vacuously antisymmetric.
Let us suppose that \(0 \lt q\).
Let \(\sigma \in S_q\) be any, where \(S_q\) is the q-symmetric group.
\(f^*_m (t) (v_{\sigma_1}, ..., v_{\sigma_q}) = t (d f_m (v_{\sigma_1}), ..., d f_m (v_{\sigma_q})) = sgn \sigma t (d f_m (v_1), ..., d f_m (v_q))\), because \(t\) is antisymmetric, \(= sgn \sigma f^*_m (t) (v_1, ..., v_q)\), which means that \(f^*_m (t)\) is antisymmetric.
