Find all parallel vector fields along a curve I am given a local representation of a metric $g$ on the real projective space $\mathbb{R}P^2$
$$g^{\varphi} = \frac{1}{(\rho^2+1)^2} d \rho^2 +\frac{\rho^2}{(\rho^2 + 1)} d\theta^2$$
where 
$$\varphi \colon (\rho, \theta) \mapsto [\rho \cos \theta, \rho \sin \theta, 1]$$
and the curve
$$\alpha \colon t \mapsto [r \cos t, r \sin t, 1] \in \mathbb{R}P^2$$
I am asked to find all parallel vector fields along $\alpha$, and I'm a bit lost on how to proceed. If anyone could give some directions on what I should do, I'd be grateful. Thank you in advance.
EDIT
So I am going to try something I am not entirely sure it works. First it seems that we need to do $\Gamma_{ij}^k \circ \alpha$ for $k = 1, 2$. For $k=1$ one gets 
$$\begin{pmatrix}
-\frac{2r\cos t}{r^2\cos^2t + 1} & 0\\
0 & -r\cos t
\end{pmatrix}$$
and for $k = 2$
$$\begin{pmatrix}
\frac{1}{r \cos t(r^2\cos^2 t + 1)} & 0 \\
0 & \frac{1}{r\cos t(r^2 \cos^2 t + 1)}
\end{pmatrix}$$
since $\rho(t) = r\cos t, \theta(t) = r\sin t$. Now I have to multiply dos two matrix above by $\alpha'(t)$ so that the I get a system of odes. But my curve $\alpha(t)$ has three coordinates (the two real and the point at the infinity) so how should I treat that? 
 A: Let $X:\ I\rightarrow TM$ be a tangent vector field along $\alpha:\ I\rightarrow M$ (I abbreviate $\Bbb{R}P^2$ to $M$ and call the domain of $\alpha$ $I$). The vector field tangent to $\alpha$ we call $T=\alpha'$. For $X$ to be parallell along $\alpha$ means that $\nabla_TX=0$, where $\nabla$ is the Levi-Civita connection on $M$. Well, it is actually more complicated, since you need technically to go to the pullback bundle $\alpha^*TM$ and use the pullback connection $\alpha^*\nabla$, but for the computation these technicalities don't matter much I think.
What we are interested in is the local coordinate expression of the covariant derivative $\nabla_TX=0$. Given a local basis of tengent vector fields $\{\partial_\rho,\partial_\theta\}=\{\partial_1,\partial_2\}$, the covariant derivative satisfies
\begin{equation}
\begin{split}
\nabla_{\partial_i}\partial_j&=\Gamma_{ij}^k\partial_k\\
\nabla_{fX}Y&=f\nabla_XY\\
\nabla_X(fY)&=(Xf)Y+f\nabla_XY,
\end{split}
\end{equation}
where $\Gamma_{ij}^k$ are the Christoffel symbols of the connection $\nabla$, and $f$ some function which we can multiply a vector field by. These three rules allow you to calculate the covariant derivative, given local coordinate expressions of a vector field, and knowledge of the Christoffel symbols.
The Christoffel symbols of the Levi-Civita connection (the one you are using) have a particular expression in a basis of coordinate tangent vector fields (like in your case). Then we have
\begin{equation}
\Gamma_{ij}^k=\frac{1}{2}g^{kl}\left(\partial_ig_{jl} + \partial_j g_{il} - \partial_l g_{ij}\right),
\end{equation}
where $g^{kl}$ are the components of the inverse of the matrix representation of the metric in the chosen coordinates.
What needs to be done is then to compute the Christoffel symbols and use the three rules above to compute the covariant derivative of a generic tangent vector field along $\alpha$. Feel free to comment if you could use more clarification and I can amend this answer to be more explicit :)
From the coordinate expression of $\alpha$, which is $\rho(t)=r$, $\theta(t) = t$, we get the expression for the tangent field to $\alpha$, which is
\begin{equation}
T(t) = \alpha'(t) = \frac{d\rho}{dt}\partial_\rho + \frac{d\theta}{dt}\partial_\theta=\partial_\theta.
\end{equation}
Now we want the generic vector field along $\alpha$ to have components depending on the time parameter $t$, i.e. write $X(t) = A(t)\partial_\rho + B(t)\partial_\theta$. Then we get some expression like $\nabla_{T(t)}X(t)=0$. We need to modify the third rule, about differentiating the coefficient function, since the components of $X$ here depend on $t$ and are not functions on $M$ (again maybe a technicality which is not the most important thing when first learning how to compute). Since $T$ is the vector field along $\alpha$, we can take the derivative along $T$ as being a time derivative, roughly speaking, and write this as
\begin{equation}
\nabla_{T(t)}(f(t)\partial_i) = \frac{df}{dt}\partial_i + f(t)\nabla_{T(t)}\partial_i,
\end{equation}
where the second term just comes out as multiplication with Christoffel symbols. As reference you want to end up with something like $A'(t) + A(t)\Gamma+B(t)\Gamma = 0$, I think. Let's see if that helps!
So now you put $X(t) = A(t)\partial_\rho + B(t)\partial_\theta$, $T(t)=\partial_\theta$, and the equation $\nabla_TX=0$ becomes
\begin{equation}
A'(t)\partial_\rho + A(t)\Gamma_{\theta\rho}^\rho\partial_\rho + A(t)\Gamma_{\theta\rho}^\theta\partial_\theta + B'(t)\partial_\theta + B(t)\Gamma_{\theta\theta}^\rho\partial_\rho + B(t)\Gamma_{\theta\theta}^\theta\partial_\theta = 0.
\end{equation}
Now equating coefficients, we see that the coefficients before $\partial_\rho$ and $\partial_\theta$ must both be zero, so we have
\begin{equation}
A'(t) + A(t)\Gamma_{\theta\rho}^\rho + B(t)\Gamma_{\theta\theta}^\rho = 0,
\end{equation}
and another similar equation. Now write the Christoffel symbols as functions of $t$, by composing with $\alpha$. At this point it is a matter of solving the two coupled ODEs for $A$ and $B$. If they look really hard to solve one might want to double check that the Christoffel symbols were calculated correctly.
