6
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ I think when you say "linearly independent" you mean "linearly independent at every point". $\endgroup$ – Moishe Kohan Sep 10 '16 at 11:24
  • 1
    $\begingroup$ @Anubhav: The given equality gives $\Gamma_{XY} = 0$ but it's not clear to me why the diagonal Christoffel terms should be zero. $\endgroup$ – Anthony Carapetis Sep 10 '16 at 12:53
  • $\begingroup$ @studiosus: Yes you are correct! $\endgroup$ – Mike F Sep 10 '16 at 16:39
  • $\begingroup$ @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. $\endgroup$ – Ted Shifrin Sep 10 '16 at 18:21
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your vector fields are not tangent to $M$. $\endgroup$ – Moishe Kohan Sep 10 '16 at 20:33
  • $\begingroup$ @studiosus: Of course they are. I'm using $(x,y)$ to parametrize the surface. I'll edit to make this clear. $\endgroup$ – Ted Shifrin Sep 10 '16 at 20:34
  • $\begingroup$ OK, now it works. +1. $\endgroup$ – Moishe Kohan Sep 10 '16 at 20:43
  • $\begingroup$ 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. $\endgroup$ – Mike F Sep 10 '16 at 21:12
  • $\begingroup$ 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. $\endgroup$ – Mike F Sep 10 '16 at 21:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.