description/proof of that connection depends only on section values on vector curve
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) vectors bundle of rank \(k\).
- The reader knows a definition of \(C^\infty\) vectors bundle connection.
- The reader knows a definition of \(C^\infty\) trivializing open subset and \(C^\infty\) local trivialization.
- The reader knows a definition of section of continuous surjection.
- The reader knows a definition of \(C^\infty\) vectors field over \(C^\infty\) manifold with boundary.
- The reader knows a definition of local \(C^\infty\) frame on \(C^\infty\) vectors bundle.
- The reader admits the proposition that for any \(C^\infty\) vectors bundle, any \(C^\infty\) frame exists over and only over any trivializing open subset.
- The reader admits the proposition that for any \(C^\infty\) vectors bundle, any rough section over any trivializing open subset is \(C^\infty\) if and only if the coefficients of the rough section with respect to any local \(C^\infty\) frame over the trivializing open subset are \(C^\infty\).
Target Context
- The reader will have a description and a proof of the proposition that any \(C^\infty\) vectors bundle connection depends only on the section values on any vector curve.
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:
\((E, M, \pi)\): \(\in \{\text{ the } C^\infty \text{ vectors bundles of rank } k\}\)
\(V\): \(: M \to TM\), \(\in \{\text{ the } C^\infty \text{ vectors fields }\}\)
\(s_1\): \(: M \to E\), \(\in \{\text{ the } C^\infty \text{ sections of } \pi\}\)
\(s_2\): \(: M \to E\), \(\in \{\text{ the } C^\infty \text{ sections of } \pi\}\)
\(\nabla\): \(\in \{\text{ the connections on } E\}\)
\(m\): \(\in M\)
\(U_m\): \(\in \{\text{ the neighborhoods of } m \text{ on } M\}\)
\(I\): \(\in \{\text{ the intervals of } \mathbb{R}\}\)
\(\gamma\): \(: I \to U_m\), \(\in \{\text{ the } C^\infty \text{ maps }\}\), such that \(m = \gamma (0) \land d \gamma / d t \vert_0 = V (m)\)
//
Statements:
\(s_1 \vert_{\gamma (I)} = s_2 \vert_{\gamma (I)}\)
\(\implies\)
\((\nabla_V s_1)_m = (\nabla_V s_2)_m\)
//
2: Proof
Whole Strategy: Step 1: take any trivializing open neighborhood of \(m\) for \(E\) with a local \(C^\infty\) frame, \((e_1, ..., e_k)\), such that \(U^`_m \subseteq U_m\); Step 2: see that \((\nabla_V s_1)_m = (V ({s_1}^j) e_j) (m) + ({s_1}^j \nabla_V e_j) (m)\) and \((\nabla_V s_2)_m = (V ({s_2}^j) e_j) (m) + ({s_2}^j \nabla_V e_j) (m)\), and conclude the proposition.
Step 1:
Let \(U^`_m \subseteq M\) be any trivializing open neighborhood of \(m\) for \(E\) with a local \(C^\infty\) frame, \((e_1, ..., e_k)\), which is possible, by the proposition that for any \(C^\infty\) vectors bundle, any \(C^\infty\) frame exists over and only over any trivializing open subset.
Let \(U^`_m \subseteq U_m\), which is possible, because otherwise, we can take \(U^`_m \cap U_m\) instead of \(U_m\).
Step 2:
Over \(U^`_m\), \(s_1 = {s_1}^j e_j\) and \(s_2 = {s_2}^j e_j\), where \({s_1}^j: U^`_m \to \mathbb{R}\) and \({s_2}^j: U^`_m \to \mathbb{R}\) are \(C^\infty\), by the proposition that for any \(C^\infty\) vectors bundle, any rough section over any trivializing open subset is \(C^\infty\) if and only if the coefficients of the rough section with respect to any local \(C^\infty\) frame over the trivializing open subset are \(C^\infty\).
\((\nabla_V s_1)_m = (V ({s_1}^j) e_j) (m) + ({s_1}^j \nabla_V e_j) (m)\), by the Leibniz rule; \((\nabla_V s_2)_m = (V ({s_2}^j) e_j) (m) + ({s_2}^j \nabla_V e_j) (m)\), likewise.
As \(V\) is represented by \(\gamma\), \(V ({s_1}^j)\) and \(V ({s_2}^j)\) depend only on the values of \({s_1}^j\) and \({s_2}^j\) on \(\gamma (I)\), but as \(s_1 \vert_{\gamma (I)} = s_2 \vert_{\gamma (I)}\), \(V ({s_1}^j) = V ({s_2}^j)\) at \(m\).
As \(s_1 \vert_{\gamma (I)} = s_2 \vert_{\gamma (I)}\), \(({s_1}^j \nabla_V e_j) (m) = ({s_2}^j \nabla_V e_j) (m)\).
So, \((\nabla_V s_1)_m = (V ({s_1}^j) e_j) (m) + ({s_1}^j \nabla_V e_j) (m) = (V ({s_2}^j) e_j) (m) + ({s_2}^j \nabla_V e_j) (m) = (\nabla_V s_2)_m\).
