I want to find the parallel transport on a surface of revolution along parallels and meridians. I considered the following approach:
take the parametrization of a surface of revolution $$ \mathbf{r}(u,\varphi) = (x(u),\rho(u)\cos\varphi, \rho(u)\sin\varphi) $$ the first fundamental coefficients are: $$ E = \rho'(u)^2+x'(u)^2 \qquad F=0 \qquad G=\rho(u)^2 $$ and we can express the Christoffel symbols in terms of these coefficients. By computing all partial derivatives: \begin{align*} E_u &= 2\left(\rho'(u)\rho''(u)+x'(u)x''(u)\right) & E_v &= 0 \\ F_u &= 0 & F_v &= 0 \\ G_u &= 2\rho(u)\rho'(u) & G_v &= 0 \end{align*} Then we have the following systems of equations: \begin{align*} \Gamma_{11}^1 (\rho'(u)^2+x'(u)^2) &= \rho'(u)\rho''(u)+x'(u)x''(u) \\ \Gamma_{11}^2 \rho(u)^2 &= 0 \\ \Gamma_{12}^1 (\rho'(u)^2+x'(u)^2) &= 0 \\ \Gamma_{12}^2 \rho(u)^2 &= \rho(u)\rho'(u) \\ \Gamma_{22}^1 (\rho'(u)^2+x'(u)^2) &= -\rho(u)\rho'(u) \\ \Gamma_{22}^2 G &= 0 \end{align*} Therefore the Christoffel symbols are \begin{align*} \Gamma_{11}^1 &= \dfrac{\rho'(u)\rho''(u)+x'(u)x''(u)}{\rho'(u)^2+x'(u)^2} \\ \Gamma_{11}^2 &= 0 \\ \Gamma_{12}^1 = \Gamma_{21}^1 &= 0 \\ \Gamma_{12}^2 =\Gamma_{21}^2 &= \dfrac{\rho'(u)}{\rho(u)} \\ \Gamma_{22}^1 &= -\dfrac{\rho(u)\rho'(u)}{\rho'(u)^2+x'(u)^2} \\ \Gamma_{22}^2 &= 0 \end{align*} Then a tangent field $Y(t)=\alpha(t)\mathbf{r}_u + \beta(t)\mathbf{r}_\varphi$ is parallel along a curve $\gamma(t)=\mathbf{r}(u(t),\varphi(t))$ if \begin{align*} \alpha' + \left(\dfrac{\rho'(u)\rho''(u)+x'(u)x''(u)}{\rho'(u)^2+x'(u)^2}\right)u'\alpha - \left(\dfrac{\rho(u)\rho'(u)}{\rho'(u)^2+x'(u)^2}\right)\varphi'\beta &= 0 \\ \beta' + \left(\dfrac{\rho'(u)}{\rho(u)}\right)\varphi'\alpha + \left(\dfrac{\rho'(u)}{\rho(u)} \right)u'\beta = 0 \end{align*}
Then, if the curve is the meridian curve, $\varphi(t)=\varphi$ is constant and $\gamma(t)= \mathbf{r}(u(t),\varphi)$. In the case of a parallel, $u(t)=u$ constant.
By studying the two equations in both cases, we can obtain the parallel vector field along the parallel curve $\gamma$ (moreover, by this approach one finds that meridians are geodesics, and we can deduce conditions under which also parallels are geodesic). However, I was wondering whether there was an easier way to prove the parallel transport then having to solve those equations. Any ideas is much appreciated.