Tangent fields spanning the distribution of principal directions on a surface Suppose $S$ is an orientable regular surface in $\mathbb R^3$ without umbilical points (not necessarily compact, and with no boundary). There are two well-defined smooth $1$-dimensional tangent distributions on $S$ corresponding to the directions of curvature.

Can each of these be generated by a single unitary vector field?

Of course this is possible locally and a standard argument using Čech cocycles gets one from there to simply connected surfaces, which is not very much. This argument is generic —does not use the fact that these are the directions of curvature of the surface— so this is to be expected. It may well be the case that the cocycles one gets from this are special in some way which I cannot see, so that one does not need the vanishing of $\pi_1$.
 A: Let me add a small comment to the answer of Professor Bryant: There is the notion of constant mean curvature (CMC) surfaces which are defined as critical points of the area functional with the constrained of fixed enclosed volume. These surfaces are characterized (in space-forms) by the fact that the complex bilinear part (the so-called Hopf differential $Q$) of the second fundamental form is a holomorphic quadratic differential with respect to the induced Riemann surface structure. There do exists CMC surfaces whose Hopf differentials have simple zeros (e.g. the Lawson surface $\xi_{2,2}$ or its conjugate cousin (here is an image showing the curvature lines)), i.e., with respect to an appopriate holomorphic coordinate $z$ the Hopf differential  is given by $Q=z(dz)^2,$ and after removing the zeros you cannot find a vector field $X$ such that $Q(X,X)>0$ everywhere. But this inequality would hold for a vector field which generates a principal curvature distribution.
Globally on a CMC surface whose Hopf differentials have simple zeros the obstruction to the existence of a vector field $X$ which generates a principal curvature distribution is that the Hopf differential $Q$ admits a square root $\omega$ (which is a holomorphic 1-form).
A: No.  Consider the case of an ellipsoid with three distinct axes, and remove the four umbilic points.  Then you cannot find such vector fields on a (punctured) neighborhood of the deleted umbilics.  Have a look at this reference on umbilics and try drawing the vector field on such a punctured neighborhood, and you'll see why.
