# Fundamental theorem of calculus and pullback bundles

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 $\phi(DF(t))$? –  Florian Jul 12 '11 at 9:56
good point, now it makes more sense –  fran Jul 12 '11 at 10:27
presumably $f:[0,1]\to X$? (You wrote $\to M$). –  Willie Wong Jul 12 '11 at 11:32
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. –  Florian Jul 12 '11 at 11:46
@Florian: the vector bundle is equipped with a metric. (First line of the question; at least that's what I hope the OP means.) –  Willie Wong Jul 12 '11 at 13:02

No. A simple counter example comes when $E$ is not orientable: consider the Möbius strip as a one dimensional bundle over $S^1$. We can give a metric on it and consider the induced Levi-Civita connection.

If you start with the vector $v$ over $0$, when you parallel transport $v$ around the circle you get $-v$. The norm of the derivative is 0, but the total difference is $2|v|$.

In the orientable case it is not better. Just take $E$ to be the tangent bundle $TX$ and treat it as a Riemannian manifold. Then parallel transport around a closed loop (for which by definition your RHS expression vanishes) does not necessarily bring a vector back to itself: you will in general pick up a defect related to the area of the loop and the enclosed curvature.

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From the analysis point of view, you should think of this failure as the following: in a local trivialisation, the covariant derivative is $\nabla v = dv + \omega\wedge v$, where $\omega$ is the connection one form. You have then that in the trivialisation, using the usual fundamental theorem of calculus $v(1) - v(0) = \int_0^1 \dot{\gamma}\cdot dv~ds$. To conclude your inequality you necessarily need strong assumptions on the 0th order term $\omega\wedge v$. –  Willie Wong Jul 12 '11 at 11:50

As Willie says, it's not true. It is true if $E$ is a flat trivial vector bundle over $X$. In general however $f$ and $D$ determine a map $\mathrm{Hol}(f,D): E_{f(0)} \rightarrow E_{f(0)}$ called the holonomy map, and if $F$ is parallel then the RHS vanishes but the LHS is $|\mathrm{Hol}(f,D)(F(0))-F(0)|$ which is not necessarily zero.

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