This is Problem 9.6(1) from the book The Geometry of Physics:
What's wrong with the following argument?
A vector $\mathbf v$ is parallel displaced around a small closed curve $C = \partial{U^2}$ in an $n$-dimensional manifold $M^n$. Then $dv^i = -\omega^i_jv^j$ along $C$. Thus the total change in $v^i$ on going around $C$ is given by
$$\begin{align}\Delta v^i &= \oint_{C}\,dv^i = -\oint\omega^i_jv^j\\ &= -\iint_U\,d(\omega^i_jv^j)\\&= -\iint_U\,d(\omega^i_j)v^j-\omega^i_j\wedge dv^j \\&= -\iint_U [d\omega^i_k + \omega^i_j\wedge\omega^j_k]v^k\\&=-\iint_U\theta^i_kv^k\\&=-\iint_U\frac{1}{2}R^i_{krs}v^k\,dx^r\wedge dx^s\end{align}$$
where $\omega$ is the connection, $\theta$ is the curvature $2$-form. I think this should be independent of $v$, but I am not sure where it is wrong.
If the curve $c$ is a boundary curve and if $\{e_i\}$ is a moving along $c$ and $\omega_i$ is a coframe then we have a connection $$ \omega_{is} (X):=\omega_i (\nabla_X e_s)$$ And $$ v:=v^i e_i,\ \nabla_{c'(t)} v =0 $$
Hence if $c'(t)=Y^ie_i$ then $$ Y(v^i) + v^s \omega_{is} (Y) =0$$
That is $$ \int dv^i =\int -v^s\omega_{is} $$
So $v$ is dependent.