4: Connection Depends Only on Section Values on Vector Curve

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A description and a proof of that any vector bundle connection depends only on the section values on any vector curve

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Topics

About: differential geometry
About: vector bundle connection

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ vector bundle.
• The reader knows a definition of vector bundle connection.
• The reader knows a definition of trivializing neighborhood of point with respect to vector bundle.
The reader knows a definition of $C^{\mathrm{\infty }}$ section.

The reader knows a definition of $C^{\mathrm{\infty }}$ tangent vector field.


• The reader knows a definition of local operator.
• The reader knows a definition of $C^{\mathrm{\infty }}$ frame.
• The reader admits the proposition that any vector bundle connection ($\mathrm{\Gamma }(TM)\times \mathrm{\Gamma }(E)\to \mathrm{\Gamma }(E)$) is a local operator with respect to its any argument.
• The reader admits the proposition that there is a $C^{\mathrm{\infty }}$ frame over any trivializing neighborhood of point with respect to any vector bundle.

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Target Context

• The reader will have a description and a proof of the proposition that any vector bundle connection depends only on the section values on any vector curve.

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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.

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Main Body

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1: Description

For any $C^{\mathrm{\infty }}$ vector bundle, $(E,M,\pi )$, any $C^{\mathrm{\infty }}$ tangent vector field on M, $X\in \mathrm{\Gamma }(TM)$, any $C^{\mathrm{\infty }}$ section on E, $s\in \mathrm{\Gamma }(E)$, any $C^{\mathrm{\infty }}$ section on E, $t\in \mathrm{\Gamma }(E)$, any connection on E, $\mathrm{\nabla }$, and any point on M, $p\in M$, if there are any neighborhood, $U_p$, of p on M and any curve on $U_p$ through p that (the curve) represents X such that $s=t$ there, then $({\mathrm{\nabla }}_{X}s{)}_p=({\mathrm{\nabla }}_{X}t{)}_p$.

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2: Proof

As any connection is a local operator with respect to any argument, its result can be evaluated on a trivializing neighborhood, $U_p$, of p. There is a $C^{\mathrm{\infty }}$ frame, {$e_1,\dots ,e_r$}, over the trivializing neighborhood of p. Over the trivializing neighborhood of p, $s=s^i{e}_i$ and $t=t^i{e}_i$, where $s^i,t^i\in C^{\mathrm{\infty }}$. $({\mathrm{\nabla }}_{X}s{)}_p=((X{s}_i)e_i)(p)+(s_i{\mathrm{\nabla }}_X{e}_i)(p)$ and $({\mathrm{\nabla }}_{X}t{)}_p=((X{t}_i)e_i)(p)+(t_i{\mathrm{\nabla }}_X{e}_i)(p)$ by the Leibniz rule, but as X is a directional derivative, $X{s}_i$ or $X{t}_i$ depends only on the values of $s_i$ or $t_i$ on any curve that represents X, but as $s_i=t_i$ on the curve, $X{s}_i=X{t}_i$ at p. As also $s_i(p)=t_i(p)$, $({\mathrm{\nabla }}_{X}s{)}_p=({\mathrm{\nabla }}_{X}t{)}_p$.

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References

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