Let $\mathbb H=\{(x,y)\in\mathbb R^2\,|\,y>0\}$ be the Poincaré halfplane with metric $g=\frac{1}{y^2}g_\text{eucl}$ for the standard euclidean metric $g_\text{eucl}$. I know that the horizontal line is not a geodesic, but can not find the error in my reasoning:
I "show" that the vector fields $X,Y$ defined by $X_{(x,y)}=(y,0)$ and $Y_{(x,y)}=(0,y)$ are parallel, that means $\nabla_Z X=\nabla_Z Y=0$ for all $Z$. Therefore their integral curves must be geodesics, and the integral curves of $X$ are the horizontal lines.
By symmetry of second derivatives in $\mathbb H\subset\mathbb R^2$ we have $[X,Y]=0$. Since $\nabla$ is torsion free this implies $\nabla_X Y=\nabla_Y X$. Note that $g(X,Y)=0$ and $g(X,X)=g(Y,Y)=1$ are constant, so their derivatives vanish:
$$0=Xg(Y,Y)=2g(\nabla_X Y,Y)=2g(\nabla_Y X,Y)$$ $$0=Yg(X,X)=2g(\nabla_Y X,X)=2g(\nabla_X Y,X)$$
Therefore $0=\nabla_X Y=\nabla_Y X$ (I know that this is already wrong, but don't see the error). Using this we get $$0=Yg(X,Y)=g(X,\nabla_Y Y)$$ $$0=Xg(X,Y)=g(\nabla_X X,Y)$$
and also $$0=Yg(Y,Y)=2g(\nabla_Y Y,Y)$$ $$0=Xg(X,X)=2g(\nabla_X X,X)$$
The last four equations imply $\nabla_X X=\nabla_Y Y=0$, with the result above then $\nabla_Z X=\nabla_Z Y=0$ for all $Z$, so $X,Y$ are parallel. Where is my error?
My error is right at the start, because $[X,Y]=0$ is wrong. This is only true in $\mathbb R^2$ if the fields $X,Y$ are parallel in $\mathbb R^2$, which they are not. For example let $f(x,y)=x$, then $Yf=0$ so $XYf=0$. But $Xf(x,y)=(y\partial_x)x=y$, so $YXf(x,y)=(y\partial_y)y=y$. Therefore $[X,Y]f(x,y)=-y\ne 0$.