description/proof of that for \(C^\infty\) map between \(C^\infty\) manifolds with boundary and corresponding charts, components function of pullback of \((0, q)\)-tensors w.r.t. standard bases is this
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of pullback of \((0, q)\)-tensors at point by \(C^\infty\) map between \(C^\infty\) manifolds with boundary.
- The reader admits the proposition that for any field and any \(k\) finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.
- The reader admits the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary and any corresponding charts, the components function of the differential of the map with respect to the standard bases is this.
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 and any corresponding charts, the components function of the pullback of the \((0, q)\)-tensors with respect to the standard bases is this.
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:
\(M_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } d_2 \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\((U_1 \subseteq M_1, \phi_1)\): \(\in \{\text{ the charts for } M_1\}\)
\((U_2 \subseteq M_2, \phi_2)\): \(\in \{\text{ the charts for } M_2\}\), such that \(f (U_1) \subseteq U_2\)
\(m\): \(\in U_1\)
\(f^*_m\): \(: T^0_q (T_{f (m)}M_2) \to T^0_q (T_mM_1)\), \(= \text{ the pullback }\)
\(\{d x^{j_1} \otimes ... \otimes d x^{j_q} \vert j_l \in \{1, ..., d_1\}\}\): \(= \text{ the standard basis for } T^0_q (T_mM_1)\)
\(\{d y^{j_1} \otimes ... \otimes d y^{j_q} \vert j_l \in \{1, ..., d_2\}\}\): \(= \text{ the standard basis for } T^0_q (T_{f (m)}M_2)\)
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Statements:
\(\forall t = t_{j_1, ..., j_q} d y^{j_1} \otimes ... \otimes d y^{j_q} \in T^0_q (T_{f (m)}M_2) (f^*_m (t) = u_{j_1, ..., j_q} d x^{j_1} \otimes ... \otimes d x^{j_q} \text{ where } u_{j_1, ..., j_q} = t_{l_1, ..., l_q} \partial \hat{f}^{l_1} / \partial x^{j_1} ... \partial \hat{f}^{l_q} / \partial x^{j_q} \text{ where } \hat{f} = \phi_2 \circ f \circ {\phi_1}^{-1})\)
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\(\{d x^{j_1} \otimes ... \otimes d x^{j_q}\}\) and \(\{d y^{j_1} \otimes ... \otimes d y^{j_q}\}\) are indeed some bases, by the proposition that for any field and any \(k\) finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces: \(\{d x^j \vert j \in \{1, ..., d_1\}\}\) and \(\{d y^j \vert j \in \{1, ..., d_2\}\}\) are the dual bases of \(\{\partial / \partial x^j \vert j \in \{1, ..., d_1\}\}\) and \(\{\partial / \partial y^j \vert j \in \{1, ..., d_2\}\}\).
In other words, \(\hat{f}\) is the components function of \(f\) with respect to \((U_1 \subseteq M_1, \phi_1)\) and \((U_2 \subseteq M_2, \phi_2)\).
2: Proof
Whole Strategy: Step 1: compute \(f^*_m (t) ((\partial / \partial x^{l_1}, ..., \partial / \partial x^{l_q}))\) as \(t (d f_m (\partial / \partial x^{l_1}), ..., d f_m (\partial / \partial x^{l_q}))\); Step 2: compare the result of Step 1 with \(f^*_m (t) ((\partial / \partial x^{l_1}, ..., \partial / \partial x^{l_q})) = u_{j_1, ..., j_l} d x^{j_1} \otimes ... \otimes d x^{j_q} ((\partial / \partial x^{l_1}, ..., \partial / \partial x^{l_q}))\).
Step 1:
We know that \(d f_m (\partial / \partial x^{l_n}) = \partial \hat{f}^j / \partial x^{l_n} \partial / \partial y^j\), by the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary and any corresponding charts, the components function of the differential of the map with respect to the standard bases is this: \(\partial / \partial x^{l_n}\) has only the \(l_n\)-th component, \(1\).
\(f^*_m (t) ((\partial / \partial x^{l_1}, ..., \partial / \partial x^{l_q})) = t (d f_m (\partial / \partial x^{l_1}), ..., d f_m (\partial / \partial x^{l_q})) = t_{j_1, ..., j_l} d y^{j_1} \otimes ... \otimes d y^{j_q} ((\partial \hat{f}^{m_1} / \partial x^{l_1} \partial / \partial y^{m_1}, ..., \partial \hat{f}^{m_q} / \partial x^{l_q} \partial / \partial y^{m_q})) = t_{j_1, ..., j_q} d y^{j_1} (\partial \hat{f}^{m_1} / \partial x^{l_1} \partial / \partial y^{m_1}) ... \otimes d y^{j_q} (\partial \hat{f}^{m_q} / \partial x^{l_q} \partial / \partial y^{m_q}) = t_{j_1, ..., j_q} \delta^{j_1}_{m_1} \partial \hat{f}^{m_1} / \partial x^{l_1} ... \delta^{j_q}_{m_q} \partial \hat{f}^{m_q} / \partial x^{l_q} = t_{j_1, ..., j_q} \partial \hat{f}^{j_1} / \partial x^{l_1} ... \partial \hat{f}^{j_q} / \partial x^{l_q}\).
Step 2:
On the other hand, \(f^*_m (t) ((\partial / \partial x^{l_1}, ..., \partial / \partial x^{l_q})) = u_{j_1, ..., j_q} d x^{j_1} \otimes ... \otimes d x^{j_q} ((\partial / \partial x^{l_1}, ..., \partial / \partial x^{l_q})) = u_{j_1, ..., j_q} d x^{j_1} (\partial / \partial x^{l_1}) ... d x^{j_q} (\partial / \partial x^{l_q}) = u_{j_1, ..., j_q} \delta^{j_1}_{l_1} ... \delta^{j_q}_{l_q} = u_{l_1, ..., l_q}\).
So, \(u_{l_1, ..., l_q} = t_{j_1, ..., j_q} \partial \hat{f}^{j_1} / \partial x^{l_1} ... \partial \hat{f}^{j_q} / \partial x^{l_q}\), which is nothing but \(u_{j_1, ..., j_q} = t_{l_1, ..., l_q} \partial \hat{f}^{l_1} / \partial x^{j_1} ... \partial \hat{f}^{l_q} / \partial x^{j_q}\).
