Orientability of a differentiable surface knowing the metric

Question

Let $S$ be a differentiable surface, is it possible to understand if $S$ is orientable from its metric $g$?

Answer 1

The way you posed the question doesn't really make sense -- to "know the metric," you first have to know what the surface is, because the metric is a choice of inner product on each tangent plane of the surface. If you know the surface, then you can determine whether it's orientable or not without reference to a metric.

But it is possible in some circumstances to draw conclusions about orientability based on partial knowledge of the metric, such as knowing that its curvature is positive or negative everywhere. If the definition of surface you're working with is a certain kind of subset of $\mathbb R^3$, then there is a little bit that can be said. If $S$ is a compact surface embedded in $\mathbb R^3$, then it is orientable regardless of what kind of metric it has. If it's not compact, then it could have negative curvature, zero curvature, or mixed positive and negative curvatures and still be either orientable or not.

However, here's one theorem that might be useful.

Theorem. A smooth surface in $\mathbb R^3$ with everywhere positive Gaussian curvature is orientable.

Proof. The fact that the Gaussian curvature is positive everywhere means that the principal curvatures are always nonzero and have the same sign. Their sign depends on a choice of unit normal; replacing the normal by its opposite causes both of the principal curvatures to change sign. Thus we can always choose the unit normal that makes the principal curvatures positive, and this yields a global smooth unit normal field and thus an orientation. $\square$

For abstract Riemannian manifolds in arbitrary dimensions, I know of one interesting global result that relates curvature and orientability:

Synge's Theorem. Suppose $M$ is a compact, connected Riemannian manifold with strictly positive sectional curvature.

• If $M$ is odd-dimensional, then $M$ is orientable.
• If $M$ is even-dimensional and not simply connected, then $M$ is not orientable.

There's a proof in my book Introduction to Riemannian Manifolds, 2nd ed.