Second fundamental form without orientability? Let $F$ be a $C^2$-hypersurface (or $n$-manifold) embedded in $\mathbb{R}^{n+1}$. Suppose $F$ is not orientable. Since I cannot choose a continuous global normal field, what consequences does this have for the second fundamental form? Specifically, if I start computing the mean curvature or Gauss curvature as functions on $F$, do I have no hope for continuity for these functions?
Thank you!
 A: I usually deal with surfaces in $\mathbb{R}^3$, so what follows deals specifically with this case, although I believe most of it carries through without issue to hypersurfaces of codimension 1 in $\mathbb{R}^n$:

Although usually explained using the shape operator, the Gaussian curvature is in fact a function only of the metric; this is Gauss's celebrated "egregious theorem." Gaussian curvature is thus well-defined (and continuous) on a non-orientable Riemannian manifold. (For some intuition, you can think of Gaussian curvature at a point as comparing the area of a Euclidean disk to that of a geodesic disk, in the limit as the radius goes to zero. A sphere-like point of positive curvature has area deficit in its neighborhood, whereas a point with negative Gaussian curvature has area surplus.)
Reversing the orientation of a surface negates the mean curvature. To see this you can look at the second fundamental form, or alternatively, notice that reversing the orientation of a curve negates the curvature of that curve, and so reversing the orientation of a surface negates the two principal curvatures $\phi_1, \phi_2$. If a surface is not orientable, you can of course still orient the surface locally; globally, you won't have continuous mean curvature, but you will have continuous mean curvature magnitude $|H|$. Moreover mean curvature squared $H^2$ is well-defined and continuous on a non-orientable surface, which leads to the interesting observation that Wilmore energy $2H^2-K$, which governs the physics of how an elastic plate deforms, is well-defined on the Klein bottle -- so it is perfectly possible to simulate the physical behavior of an elastic Klein bottle as if it were a real object.
Lastly, to expand on John's comment: the Laplace-Beltrami operator $\Delta$ on a manifold requires only a Riemannian metric and not orientability. For an oriented surface in $\mathbb{R}^3$ with the induced metric,
$$(\Delta r)(q) = 2H(q)N(q)$$
where $r(q) = [x(q), y(q), z(q)]$ is the vector-valued coordinate function of the embedding, and $N$ is the surface normal. The left-hand side does not require orientability, whereas the right, of course, does -- so we can extend the notion of mean curvature normal $HN$ to non-orientable surfaces with this identity.
