= Solution
{c}
The only nonconstant entry of the <metric tensor> is $g_{\phi\phi}=R^2\sin^2\theta$, whose <partial derivative> is $\partial_\theta g_{\phi\phi}=2R^2\sin\theta\cos\theta$. The nonzero <Christoffel symbols> are
$$
\boxed{\Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta,\qquad
\Gamma^\phi_{\theta\phi}=\Gamma^\phi_{\phi\theta}=\cot\theta.}
$$
Along a fixed latitude with $0<\theta<\pi$, <parallel transport> gives
$$
\frac{dV^\theta}{d\phi}=\sin\theta\cos\theta\,V^\phi,
\qquad
\frac{dV^\phi}{d\phi}=-\cot\theta\,V^\theta.
$$
Set $X=V^\theta$ and $Y=\sin\theta\,V^\phi$. Since $\theta$ is fixed, this <linear ordinary differential equation> becomes
$$
X'=\cos\theta\,Y,\qquad Y'=-\cos\theta\,X.
$$
It is a rotation at constant rate $-\cos\theta$. Its general solution, also valid at the equator where the rate is zero, is
$$
\boxed{\begin{aligned}
V^\theta&=A\cos(\phi\cos\theta)+B\sin(\phi\cos\theta),\\
V^\phi&=\frac{-A\sin(\phi\cos\theta)+B\cos(\phi\cos\theta)}{\sin\theta}.
\end{aligned}}
$$
The initial data give $A=1$ and $B=0$. At $\theta=\pi/3$, a complete circuit has $\phi\cos\theta=\pi$, so
$$
\boxed{(V^\theta,V^\phi)_{\phi=2\pi}=(-1,0).}
$$
The physical squared length uses the <metric tensor>, rather than the Euclidean sum of coordinate components:
$$
\boxed{|\mathbf V|^2=R^2\bigl[(V^\theta)^2+\sin^2\theta(V^\phi)^2\bigr]
=R^2(A^2+B^2).}
$$
It is constant and equals $R^2$ for these data. More generally, <metric compatibility> gives $d\,g(V,V)/d\phi=(\nabla_{\partial_\phi}g)(V,V)+2g(\nabla_{\partial_\phi}V,V)=0$. The final change of direction is <holonomy>, exemplified by <parallel transport around a latitude of the unit sphere>; <parallel transport> preserves length without requiring the direction to return unchanged.
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