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?

  • 6
    $\begingroup$ There is a discussion about spacetime orientability here at PSE. It's important to distinguish between spacetime orientability, space orientability and time orientability. $\endgroup$ – twistor59 Aug 31 '11 at 9:21
  • 3
    $\begingroup$ A non-orientable manifold lends itself no more to an "arrow of time" than an orientable one does. $\endgroup$ – leftaroundabout Aug 31 '11 at 9:24
  • $\begingroup$ This is the point which is not clear to me: Does the fact that time points always towards the future has no relevance whatsoever to its orientability? $\endgroup$ – user1999 Aug 31 '11 at 13:31
  • $\begingroup$ I would in fact think the converse of your claim is true--on a Möbius strip, there is no notion of 'up' as distinct from 'down'. Similarly, in a non-time orientable spacetime, there can be no notion of 'past' as distinct from 'future'. That we can sensibly talk about past and future is evidence <b>for</b> the time-orientability of spacetime, not against it. $\endgroup$ – Jerry Schirmer Aug 31 '11 at 16:15
  • $\begingroup$ @Jerry, according to your analogy, it occurred to me that space, rather than time, is non-orientable... In the example of the Mobius strip, the "one-directioness" is perpendicular to the strip itself. Hence, it seems that for any scheme in which spacetime is foliated into space-like slices, the "one-directioness" of time (perpendicular to these slices) implies non-orientability of space... Isn't it? $\endgroup$ – user1999 Aug 31 '11 at 19:26

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):

  1. $M$ admits a smooth non-vanishing vector field
  2. $M$ can be equipped with a smooth Lorentzian metric
  3. $M$ 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.

  • $\begingroup$ The two dimensional Minkowski sphere is singular at the poles, so you avoid the hair comb problem in this way--the manifold you get by excluding the singular polar points is, in fact, time orientable. It has a lot of the qualitative features of a big bang/big crunch cosmology, actually. $\endgroup$ – Jerry Schirmer Aug 31 '11 at 16:17
  • $\begingroup$ @Jerry: right, hence I stated that the metric should be continuous. By allowing the singularity at the poles, you are essentially circumventing the problem with the topology (the sphere having positive Euler characteristic) by, well, changing the topology (the manifold you get by excluding the polar points is not a sphere any more). $\endgroup$ – Willie Wong Aug 31 '11 at 16:23
  • $\begingroup$ @Willie: generally a property of Minkowski manifolds which is preserved under diffeomorphisms is called "topological", so time orientability, strictly speaking is "topological", although it depends on the qualitative choice of Minkowski metric structure. $\endgroup$ – Ron Maimon Aug 31 '11 at 16:36
  • 1
    $\begingroup$ @Ron: You have it sooo wrong. A topological property is one such that if $M$ and $N$ are two manifolds and $\phi:M\to N$ a diffeomorphism, than $M$ having the property implies that $N$ having the property. This is what it means by preservation under diffeomorphisms. Take the example I wrote down for the two separate Lorentzian metrics on $\mathbb{R}\times \mathbb{S}^1$. The identity map is a diffeomorphism, but one manifold is time orientable and the other isn't. $\endgroup$ – Willie Wong Aug 31 '11 at 18:05
  • $\begingroup$ @Ron: for your statement to make sense, it requires the choice of the Lorentzian metric on $M$ and $N$ to be compatible through the diffeomorphism. Then by your definition even the Riemann curvature is a topological property, which I think is just absurd. (Furthermore, how can a property be topological if its very definition requires non topological information?) $\endgroup$ – Willie Wong Aug 31 '11 at 18:07

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.