403: For C^\infty Vectors Bundle, Global Connection Can Be Constructed with Local Connections over Open Cover, Using Partition of Unity Subordinate to Open Cover

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A description/proof of that for $C^{\mathrm{\infty }}$ vectors bundle, global connection can be constructed with local connections over open cover, using partition of unity subordinate to open cover

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors bundle.
• The reader knows a definition of connection on $C^{\mathrm{\infty }}$ vectors bundle.
• The reader knows a definition of restricted $C^{\mathrm{\infty }}$ vectors bundle.
• The reader knows a definition of partition of unity subordinate to open cover.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle over any $C^{\mathrm{\infty }}$ manifold, a global connection can be constructed with any local connections over any open cover using any partition of unity subordinate to the open cover.

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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 $C^{\mathrm{\infty }}$ manifold, $M$, any $C^{\mathrm{\infty }}$ vectors bundle, $\pi :E\to M$, any open cover of $M$, $\{U_{\alpha }|\alpha \in A\}$, where $A$ is any possible uncountable indices set, any local connection, ${\mathrm{\nabla }}_{\alpha }$, on each restricted $C^{\mathrm{\infty }}$ vectors bundle, $E{|}_{U_{\alpha }}$, and any partition of unity subordinate to the open cover, $\{{\rho }_{\alpha }|\alpha \in A\}$, $\mathrm{\nabla }:=\sum _{\alpha }{\rho }_{\alpha }{\mathrm{\nabla }}_{\alpha }$, which means that ${\mathrm{\nabla }}_{V}s=\sum _{\alpha }{\rho }_{\alpha }{{\mathrm{\nabla }}_{\alpha }}_{V}s{|}_{U_{\alpha }}$ is a connection on $E$.

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2: Proof

Around each point, $p\in M$, there is an open neighborhood, $U_p$, that intersects only finite elements of $\{supp{\rho }_{\alpha }|\alpha \in A\}$, and on $U_p$, $\sum _{\alpha }{\rho }_{\alpha }{{\mathrm{\nabla }}_{\alpha }}_{V}s{|}_{U_{\alpha }}=\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}$ where $i\in I$ where $I$ is a finite indices set.

${\mathrm{\nabla }}_{V}s$ is a $C^{\mathrm{\infty }}$ section on $E$, because on each $U_p$, $\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}$ is $C^{\mathrm{\infty }}$. ${\mathrm{\nabla }}_{V}s$ is $\mathbb{R}$-linear with respect to $s$, because on each $U_p$, ${\mathrm{\nabla }}_V(rs)=\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_V(rs{|}_{U_i})=r\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}=r{\mathrm{\nabla }}_{V}s$. ${\mathrm{\nabla }}_{V}s$ is $C^{\mathrm{\infty }}(M)$-linear with respect to $V$, because on each $U_p$, ${\mathrm{\nabla }}_{fV}s=\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_{f{|}_{U_i}V}s{|}_{U_i}=\sum _i{\rho }_{i}f{|}_{U_i}{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}=f{|}_{U_i}\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}=f{\mathrm{\nabla }}_{V}s$. ${\mathrm{\nabla }}_{V}s$ satisfies the Leibniz rule, because on each $U_p$, ${\mathrm{\nabla }}_V(fs)=\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_V(f{|}_{U_i}s{|}_{U_i})=\sum _i{\rho }_i((Vf{|}_{U_i})s{|}_{U_i}+f{|}_{U_i}{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i})=(Vf{|}_{U_p})s{|}_{U_p}+f{|}_{U_p}\sum _i({\rho }_i{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i})=(Vf)s+f{\mathrm{\nabla }}_{V}s$.

To elaborate more on the evaluations of a construct like $\sum _i{\rho }_i{x}_i$, let us evaluate it at each point, $p^{\prime }\in U_p$, as $\sum _i{\rho }_i(p^{\prime })x_i(p^{\prime })$; let us factor $x_i(p^{\prime })$ as $x(p^{\prime })x_i^{\prime }(p^{\prime })$ where $x(p)$ does not really depend on $i$; $\sum _i{\rho }_i(p^{\prime })x_i(p^{\prime })=x(p^{\prime })\sum _i{\rho }_i(p^{\prime })x_i^{\prime }(p^{\prime })$. In fact, in the previous $\sum _i{\rho }_i(Vf{|}_{U_i})s{|}_{U_i}$, $x_i(p^{\prime })=(Vf{|}_{U_i})(p^{\prime })s{|}_{U_i}(p^{\prime })=x(p^{\prime })$, which does not really depend on $i$, so, $\sum _i{\rho }_i(p^{\prime })x_i(p^{\prime })=x(p^{\prime })\sum _i{\rho }_i(p^{\prime })=x(p^{\prime })$, so, $\sum _i{\rho }_i(Vf{|}_{U_i})s{|}_{U_i}(p^{\prime })=(Vf{|}_{U_p})s{|}_{U_p}(p^{\prime })$. In the previous $\sum _i{\rho }_{i}f{|}_{U_i}{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}$, $x_i(p^{\prime })=f{|}_{U_i}(p^{\prime }){{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}(p^{\prime })=x(p^{\prime })x_i^{\prime }(p^{\prime })$ where $x(p^{\prime })=f{|}_{U_i}(p^{\prime })$, which does not really depend on $i$, and $x_i^{\prime }(p^{\prime })={{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}(p^{\prime })$, so, $\sum _i{\rho }_i(p^{\prime })x_i(p^{\prime })=x(p^{\prime })\sum _i{\rho }_i{{\mathrm{\nabla }}_i}_{V}s{|}_{U_i}(p^{\prime })=f{|}_{U_p}(p^{\prime })\mathrm{\nabla }{|}_{V}s(p^{\prime })$, which is what we did.

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3: Note

The open cover is usually a trivializing open cover (because any trivializing open cover allows a set of local connections), but is not necessarily so.

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References

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