Context: Past exam for class that I sit my final for tomorrow :).
We are in a $2$-dimensional surface, with coordinates $(\theta,\phi)$. Consider the metric $ds^2 = r_0^2 d\theta^2 + r_0^2 \sin^2(\theta)d\phi^2$
I have shown that the only non-zero Christoffel symbols are: $$\Gamma_{12}^{\,\,\, 2}=\Gamma_{21}^{\,\,\, 2}=\cot(\theta),\qquad \Gamma_{22}^{\,\,\, 1}=-\sin(\theta)\cos(\theta)$$
I want to show that if a contravariant vector $a^i$ undergoes parallel transport around the circle from $\theta = \pi/4$ from $\phi=0$ to $\phi=2\pi$ and $a^1=0$, $a^2=1$ at $\phi=0$ then at $\phi=2\pi$ we have: $$a^1 = \frac{\sqrt{2}}{2}\sin(\sqrt{2}\pi),\qquad a^2=\cos(\sqrt{2}\pi)$$
My attempt:
We have that $$0=\frac{Da^\lambda}{Dt}=\frac{da^\lambda}{dt}+\Gamma_{\mu\nu}^{\,\,\, \lambda}a^\mu \frac{dx^\nu}{dt}$$ (where $x^1=\theta,x^2=\phi$ is convention, and does not refer to cartesian coordinates).
So I have:
$$\dot{a}^1=-\Gamma_{\mu\nu}^{\,\,\, 1}a^\mu \dot{x}^\nu=\sin(\theta)\cos(\theta)a^2 \dot{\phi}$$ $$\dot{a}^2=-\Gamma_{\mu\nu}^{\,\,\,2}a^\mu \dot{x}^\nu=-\cot(\theta)(a^1\dot{\phi}+a^2\dot{\theta})$$
Now since $\theta$ stays constant, we have $\theta=\frac{\pi}{4}$ and $\dot{\theta}=0$, and so the above refines to:
$$\dot{a}^1=-\Gamma_{\mu\nu}^{\,\,\, 1}a^\mu \dot{x}^\nu=\frac12a^2 \dot{\phi}$$ $$\dot{a}^2=-\Gamma_{\mu\nu}^{\,\,\,2}a^\mu \dot{x}^\nu=-(a^1\dot{\phi})$$
What to do with $\dot{\phi}$? Should I parametrize $\phi$?
Say we take $\phi(t) = 2\pi t$ so that this moves along a circle over $t\in [0,1]$, then we have: $\dot{\phi}=2\pi$ and hence:
$$\dot{a}^1=-\Gamma_{\mu\nu}^{\,\,\, 1}a^\mu \dot{x}^\nu=\frac12a^2 2\pi$$ $$\dot{a}^2=-\Gamma_{\mu\nu}^{\,\,\,2}a^\mu \dot{x}^\nu=-(a^12\pi)$$
So we have: $$\begin{bmatrix}\dot{a}^1\\\dot{a}^2\end{bmatrix} = \begin{bmatrix}0&\pi \\-2\pi&0\end{bmatrix}\begin{bmatrix}a^1\\a^2\end{bmatrix}$$
and solve this in the classical way: Eigenvalues are $\pm \sqrt{2}\pi i$ and eigenvectors: $$(1,\sqrt{2}i),(1,-\sqrt{2} i)$$
So we have: $$\begin{bmatrix}a^1\\a^2\end{bmatrix} = e^{\sqrt{2}\pi i t}\begin{bmatrix}1\\\sqrt{2}i\end{bmatrix} + e^{-\sqrt{2}\pi i}\begin{bmatrix}1\\-\sqrt{2}i\end{bmatrix}$$
So $$a^1=\cos(\sqrt{2}\pi t) + i\sin(\sqrt{2}\pi t)+\cos(-\sqrt{2}\pi t) + i\sin(-\sqrt{2}\pi t)$$ $$=2\cos(\sqrt{2}\pi t)$$
Made an error apparently. Is this now the right method? Or something is still wrong?
You are supposed to parallel transpose along a circular path, so why not parameterise $\phi$ thus?
Edit: since it seems your exam is likely over, here are the final details. Your solutions to the coupled ODE are wrong (hence my comment that the boundary conditions are not satisfied). By diagonalising the matrix you can easily solve decoupled ODEs and should get $$\begin{pmatrix}a^1\\a^2\end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}}\sin(\sqrt{2}\pi t) \\ \cos(\sqrt{2}\pi t)\end{pmatrix}$$ From here everything follows. Your diff geom was fine but your linear algebra or calculus got lost!