Let $M$ be a Riemann surface and let $\nabla$ be its Levi-Cevita connection. In particular, $\nabla$ is torsion free, i.e. $\nabla_X Y - \nabla_Y X = [X,Y]$ for vector fields $X$ and $Y$.
Question: Suppose there exist linearly independent vector fields $X$ and $Y$ on $M$ such that $\nabla_XY = \nabla_YX = 0$, i.e. each is parallel along the other. Note that, because $\nabla$ is torsion-free, this implies the vector fields commute.
Does this imply that the connection $\nabla$ is flat?
By ad hoc arguments, I've mostly convinced myself such $X$ and $Y$ cannot exist on any small open subset of the 2-sphere, at least in the case where they are also orthogonal (but not orthonormal, which would make their nonexistence obvious). However, I'm not satisfied with my argument. I suspect the real reason this is not possible is that it would force a flat geometry, but I can't see how to prove this....
Trivial comment: the angle between such $X$ and $Y$ should be constant (at least locally, and it is really the local question I am interested in).
The answer is, in fact, no. We can take $M\subset\Bbb R^3$ to be the graph of a function $f(x,y)=g(x)+h(y)$, parametrized by $\mathbf r(x,y)=\big(x,y,f(x,y)\big)$. We take $X=\partial/\partial x=\mathbf r_x$ and $Y=\partial/\partial y=\mathbf r_y$. Then $\nabla_Y X = \nabla_X Y = \mathbf r_{xy} =\mathbf 0$. On the other hand, if we take $f$ to be nonlinear (e.g., $f(x,y)=x^2+y^2$), the surface will have nonzero curvature.
By the way, the angle between $X$ and $Y$ is far from constant.