Let $M$ be a Riemannian surface with the metric $g$ defined by $g = dx^2 + a(x,y)^2 dy^2$, where $a$ is some smooth function.
Is it true that the curves given in the coordinates $(x,y)$ by $\gamma(t) = (t,y_0)$, where $y_0$ is fixed, are all geodesic with respect to the metric $g$?
I tried to solve $\nabla_{\dot \gamma} \dot \gamma = 0$, but I'm not sure how to proceed. I think that $\dot \gamma(t) = (1,0)$ for any $t$, so I don't see what it would mean to compute $\nabla_{(1,0)} (1,0)$...
The covariant derivative is given in components by $(\nabla_{\dot{\gamma}}\dot{\gamma})^i = \sum_j\dot{\gamma}^j\partial_j\dot{\gamma}^i + \sum_{jk}\Gamma^i_{jk}\dot{\gamma}^j\dot{\gamma}^k$ where the $\Gamma$ are Christoffel symbols of the second kind. Using $\dot{\gamma} = (1,0)$ we get $(\nabla_{\dot{\gamma}}\dot{\gamma})^i = \Gamma^i_{00}$.
Therefore, we just need to compute $\Gamma^i_{00} = \sum_m\frac{1}{2}g^{im}(2\partial_0 g_{m0} - \partial_m g_{00})$. Substituting the metric: $\Gamma^0_{00} = \Gamma^1_{00} = 0$ (because $g_{00}=1$ is constant) and the geodesic equation is satisfied.
To get some intuition, fix two points $(x_1,y_0)$, $(x_2,y_0)$. Consider the curves joining them. To compute their length you have to do an integral, which you can split as an integral over $x$ plus some non-negative term $T$. The integral over $x$ equals $|x_2-x_1|$ and therefore it is constant among our curves. The geodesics are the ones with minimal length, so they minimize $T$. The minimum of $T$ (which is zero) is achieved when the curve doesn't move in the $y$ direction. Thus, the $y=\text{constant}$ lines are geodesics.
