# Parallel vector fields imply a flat connection?

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).

• I think when you say "linearly independent" you mean "linearly independent at every point". – Moishe Kohan Sep 10 '16 at 11:24
• @Anubhav: The given equality gives $\Gamma_{XY} = 0$ but it's not clear to me why the diagonal Christoffel terms should be zero. – Anthony Carapetis Sep 10 '16 at 12:53
• @studiosus: Yes you are correct! – Mike F Sep 10 '16 at 16:39
• @Anthony: Note that we're not requiring all the Christoffel symbols to vanish. We're only requiring the curvature to vanish. If $X$ and $Y$ can be chosen to form an orthonormal frame, then of course the surface is flat, but I don't have an answer quite yet. – Ted Shifrin Sep 10 '16 at 18:21

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.

• Your vector fields are not tangent to $M$. – Moishe Kohan Sep 10 '16 at 20:33
• @studiosus: Of course they are. I'm using $(x,y)$ to parametrize the surface. I'll edit to make this clear. – Ted Shifrin Sep 10 '16 at 20:34
• OK, now it works. +1. – Moishe Kohan Sep 10 '16 at 20:43
• This is a great example! It's very easy to see $X$ and $Y$ are parallel along one another since the derivative of, say, $Y$ along $X$ is already zero in $\mathbb{R}^3$, with no need to project back into the tangent plane of $M$. You are right that there is no need for the angle to stay constant. I confused myself into thinking that the angle should stay the same as you flow along, say, $X$, but this isn't true because $X$ doesn't need to be parallel along itself. – Mike F Sep 10 '16 at 21:12
• I had come up other examples just by writing things out in coordinates, but they are not so clean or conceptual as this answer, so I will not put those calculations here. I still wonder about the (non)possibility of such fields $X$ and $Y$ for constant curvature geometries (I now see that my argument for $S^2$ was wrong). I think I will be posting this as another question, which will at least give me a chance to recyclethe aforementioned calculations. – Mike F Sep 10 '16 at 21:16