Orientability of spacetime In many theoretical setups it is implicitly assumed that the underlying manifold (i.e. spacetime) is orientable. Then our analysis depends on this implicit assumption. For example, Stokes' theorem assumes orientability of the chain on which we integrate. However, we accept that time "always" points forwardly.
My question is: doesn't a 1-directional arrow of time provides a hint that spacetime should be modeled, afterall, by a non-orientable manifold?
 A: Orientability is a purely topological property. 
Time-orientability is a mix between topology and geometry: a manifold is time orientable if there exists "an arrow of time" on it. Mathematically this means that there must be a non-vanishing vector field (certain topological criterion must be met for the manifold to admit a non-vanishing vector field) that is time-like (so the Lorentzian metric must be compatible with the vector field). 
The two properties are more or less independent of each other. 


*

*The standard Minkowski space is orientable and time-orientable.

*The two dimensional sphere is orientable. But it cannot be time orientable. (On the two dimensional sphere there cannot be a non-vanishing vector field; in fact, the two dimensional sphere cannot admit a continuous Lorentzian metric.)

*The Mobius strip is not orientable. But it can be equipped with a Lorentzian metric that is time-orientable. It can also be equipped with a Lorentzian metric that is not time-orientable. 

*The cylinder ($\mathbb{R}\times\mathbb{S}^1$) is orientable. It can be equipped with a Lorentzian metric that is time-orientable. It can also be equipped with a Lorentzian metric that is not time-orientable. (Let $x$ be the coordinate in the $\mathbb{R}$, and $\theta$ be the coordinate on $\mathbb{S}^1$ taking value between $[0,2\pi)$. The metric
$$ ds^2 = - dx^2 + d\theta^2 $$
is time orientable, but the metric
$$ ds^2 = \cos \theta dx^2 + 2\sin\theta dxd\theta - \cos\theta d\theta^2 $$
is a non-time-orientable Lorentzian metric.)



Edit Ron Maimon brings up a very good point in the comments: that there is a difference between the following two statements:


*

*The manifold $M$ with a Lorentzian metric $g$ is time orientable

*The manifold $M$ can be equipped with a time-orientable Lorentzian metric $g$. 


The distinction is clear when you consider the third and the fourth examples above the cut. And it is clear that trivially we have a one-way implication between the two above statements. 
As it turns out, for paracompact manifolds, we have the equivalence of the following three statements (see Proposition 5.37, O'Neill, Semi-Riemannian Geometry):


*

*$M$ admits a smooth non-vanishing vector field

*$M$ can be equipped with a smooth Lorentzian metric

*$M$ can be equipped with a time-orientable Lorentzian metric


so if one were to consider time orientability to be a property of the manifold that it admits a time-orientable Lorentzian metric, then Ron correctly argues in the comments that this notion of time orientability "is topological". There is, however, one unsatisfying aspect to that statement, namely that any Lorentzian manifold is "topologically" time-orientable, but the given metric may not be time-orientable (again, I refer to the fourth example above the cut). 
What's interesting is the following: there actually is a topological obstruction to the existence of a non -time-orientable Lorentzian metric, when given a manifold that admits a Lorentzian (and hence can be assumed to be time-orientable) metric. Let me introduce some mathematics notations. For the manifold $M$, I will denote by $TM$ the tangent bundle, and by $SM$ the sphere bundle. The sphere bundle is formed by taking $TM$, removing the zero section, and quotienting out in each fiber by the multiplication by scalars. So topologically the fibers are $\mathbb{S}^{n-1}$. And by $PM$ I will denote the "bundle of lines", obtained by identifying the antipodes in the fibres of $SM$. 
One can easily show that the manifold $M$ admits a Lorentzian metric if and only if one can find a section of the bundle $PM$ (for example, see Choi & Suh, "Remarks on the topology of Lorentzian manifolds", Comm. Korean Math. Soc. 15 (2000)). In particular, given a Lorentzian manifold, the section $\gamma$ of $PM$ can be chosen to be time-like. This section is given by a time-like vector field (thereby showing time-orientability), if it is the image of a section of $SM$ under the projection map. 
Now, $SM$ forms a natural double cover of $PM$ under projection $\pi$. So given a section $\gamma$, its preimage $\pi^{-1}\gamma$ can be treated as a double cover of $M$. Therefore a topological condition that guarantees that there cannot be any non-time-orientable Lorentzian metric would be that the pre-image $\pi^{-1}\gamma$ of any section $\gamma$ of $PM$ has two connected components. In particular, a sufficient condition to guarantee this is that $M$ is simply connected (since any covering map of a simply connected manifold $M$ is trivial). So we have that: a simply connected Lorentzian manifold must be time-orientable. 
