# I don't understand holonomy well

I'm just trying to understand how a vector can rotate around a smooth loop $\gamma$ on some manifold $M$. By Picard's theorem, the differential equation $\nabla_{\frac{\partial}{\partial t}} W =0$ with initial condition $W_{\gamma(t_0)} = v$ for this given connection just have one solution and ,since W is a vector field, $W_{\gamma(t_f)} = v$. I think it's very idiot, but I'm not getting it. Thanks in advance.

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Think of a smooth loop in $M$ as a smooth map $$\gamma: [0, 1] \longrightarrow M$$ such that $$\gamma(0) = \gamma(1).$$ When you parallel transport a tangent vector $v \in T_{\gamma(0)} M$ along $\gamma$, you're not actually defining a vector field on the manifold $M$ but a section of the pullback bundle $\gamma^\ast TM$. To put this in simpler terms, you can interpret this as a map $$W: [0,1] \longrightarrow TM$$ such that $$W(t) \in T_{\gamma(t)} M.$$ Therefore $W(0)$ and $W(1)$ lie in the same tangent space, but are not necessarily equal.
So you can think of a solution to the parallel transport equation $$\begin{cases} \nabla_{\dot{\gamma}} W = 0, \\ W(0) = v \end{cases}$$ as a family $W(t)$ of tangent vectors to $M$ parametrized by $t \in [0,1]$ such that $W(t) \in T_{\gamma(t)} M$. Your error was in thinking that $W$ is a vector field on $M$, when in reality it is a "vector field along $\gamma$," which is the notion I defined in the first paragraph.
$W$ is not a section of $TM$, it is a section of the pullback bundle $\gamma^\ast TM$. The first problem you have in calling $W$ a section of $TM$ is that it is only defined on the image of $\gamma$ in $M$, instead of on all of $M$. The second problem is that at points where $\gamma$ intersects itself (i.e. $\gamma(t_1) = \gamma(t_2)$), we have that $W(t_1)$ and $W(t_2)$ lie in the same tangent space but may not be equal. This isn't a problem when we consider $W$ as a section of $\gamma^\ast TM$ though. –  Henry T. Horton Dec 29 '12 at 21:30