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Let $X$ be a manifold and $\pi:E\rightarrow X$ a vector bundle over $X$ equipped with a metric $\left\langle \cdot,\cdot\right\rangle $.

Let $f:[0,1]\rightarrow M$ be a smooth map, and consider the pullback bundle $f^{*}E\rightarrow[0,1]$. This is the vector bundle whose fiber over $t\in[0,1]$ is

$$ (f^{*}E)_{t}=\left\{ (t,v)\,:\, v\in E_{f(t)}\right\} . $$

Let $\phi:f^{*}E\rightarrow E$ denote the map $\phi(t,v)=v$.

Let $\nabla$ denote a connection on $E$, and let $D:=f^{*}\nabla$ denote the induced connection on $f^{*}E$.

$D$ is defined as follows: suppose $F\in\Gamma(f^{*}E)$ is a section of $f^{*}E$, so that $\phi(F(t))\in E_{f(t)}$ for all $t\in[0,1]$. Fix $s\in[0,1]$, and let $v$ denote any vector field on $X$ with $v(f(s))=\phi(F(s))$.

Then by definition

$$ (D_{\partial_{t}}F)(s)=\left(s,(\nabla_{\dot{f}(s)}v)(f(s))\right)\mbox{ as elements of }(f^{*}E)_{s}, $$

the points being that the right-hand side of the above expression is independent of the choice of $v$.

Assume now that $f(0)=f(1)$. Then if $F\in\Gamma(f^{*}E)$ then $\phi(F(1))$ and $\phi(F(0))$ both lie in the same vector space $E_{f(0)}$. My question is: does the following generalization of the fundamental theorem of calculus always hold?

$$ |\phi(F(1))-\phi(F(0))| \le \int_{0}^{1}|\phi(D_{\partial_{t}}F)(t)|dt. $$

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    $\phi(F(1))$ and $\phi(F(0))$ lie in the same vector space, but $\phi(DF(t))$ for $t\in(0,1)$ do not. So how do you integrate over $\phi(DF(t))$?2011-07-12
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    good point, now it makes more sense2011-07-12
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    presumably $f:[0,1]\to X$? (You wrote $\to M$).2011-07-12
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    This is still not right. Now you introduced $|\cdot|$ applied to vectors of $E_x$. But plain vector bundles do not come with a norm on their fibers; some additional structure is needed.2011-07-12
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    @Florian: the vector bundle is equipped with a metric. (First line of the question; at least that's what I hope the OP means.)2011-07-12
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    @Willie: Oh, OK, I didn't see this. But I still don't see a compatibility condition between the connection and the metric, so the statement is obviously false. Never mind; you already explained that it's even false for Riemannian manifolds.2011-07-13

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