# When can we factor out the time dimension?

First of all, my knowledge of both GR and differential geometry is quite weak, so forgive me if the physics here doesn't make much sense.

Let $(M, g)$ be a smooth, connected Lorentzian manifold of dimension $n$. Let $f: \mathbb{R}\to M$ be a smooth curve such that the pullback of $g$ through $f$ is everywhere negative (where we've chosen an orientation on $\mathbb{R}$); we say that $f$ is time-like. Say that we can "factor" $f$ out of $M$ if there exists a manifold $S$ of dimension $n-1$ and an isomorphism $M\simeq S\times \mathbb{R}$ so that the map $\mathbb{R}\to M\simeq S\times \mathbb{R}\to S$ is constant and the map $\mathbb{R}\to M\simeq S\times \mathbb{R}\to \mathbb{R}$ is the identity. Intuitively, this factorization exhibits $f$ as "time" in some reference frame, and $S$ as space. My question is:

For which $(M, g)$ can every time-like path be factored out?

Minkowski space seems like an obvious example unless I'm missing something; it seems one can take a tangent vector to $f$ at any point and consider a perpendicular subspace to that vector as $S$. I'd accept as an answer a characterization of all such $(M, g)$ in dimension $4$, or some nice sufficient condition on $M$ for factorization to always work.

If the motivation isn't obvious already, this is supposed to codify the intuition that in my reference frame, I seem to be standing still -- and that the same is true for everyone else, even if they seem to me to be moving. My apologies if I've overloaded terms, or used them incorrectly.

Added: Note that this condition is much stronger than stable causality; indeed, it certainly implies stable causality, as choosing any timelike path $f$ and then considering the given projection to $\mathbb{R}$ gives a global time function. However, I am asking for (1) a product structure on $M$ for each path $f$ and (2) in order to formalize the notion that I seem to be standing still (to myself), the projection of $f$ to $S$ must be constant.

Added: I don't think global hyperbolicity suffices either. The theorem of Geroch (it and other splitting theorems are discussed here, for example) does indeed give a decomposition of $M$ as $\mathbb{R}\times S$. But I don't think this is enough. In particular, I am asking for the following---for every timelike path $f: \mathbb{R}\to M$, there is a product structure $M\simeq \mathbb{R}\times S$ such that the projection to $\mathbb{R}$ is a section of $f$, and that $f$ is constant upon projection to $S$. This is much stronger than Geroch's splitting theorem, as far as I can tell.

Added: As the accepted answerer rightly points out in the comments to his question, I was wrong to claim that my condition is stronger than global hyperbolicity. They are in fact equivalent.

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This sounds to me like you're asking that your spacetime admit a family of Cauchy surfaces (modulo annoyances like having $f$ be closed and acausal). There's a theorem of Geroch which guarantees that this is equivalent to global hyperbolicity.

I don't think that globally hyperbolic and stably causal are equivalent conditions, but I'm not an expert on causality conditions in GR, so take this claim with a grain of salt.

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Reading the wikipedia article, it seems to me that this implies that one path can be factored out in the sense I described; does this guarantee it for every path? – Daniel Litt Jul 15 '10 at 20:04
You're worried that $S$ may depend on $f$? It might in the metric category, but in the topological category, any two $S$ are certainly isomorphic. – userN Jul 15 '10 at 20:54
I'm worried that the projection of $f$ to $S$ might not be constant. If you read the question carefully, you'll see we need a different product structure for each $f$. – Daniel Litt Jul 15 '10 at 20:57
I still suspect the two conditions are essentially equivalent, as long as you're not worried about metric structure. We know that $M$ is globally hyperbolic, so let's equate $M = S \times \mathbb{R}$. Let $f_0$ be the "base" path, given by $t \mapsto (f_0(t),t)$. Now suppose we have some other path $f_1$, given by $t \mapsto (f_1(t),t)$. I claim that, as long as $S$ is path-connected, for each $t$, we can find a diffeomorphism $d_t: S \to S$ which moves $f_1(t)$ to $f_0(t)$. Moreover, we can make these diffs depend smoothly on $t$. That should give product structure you want... – userN Jul 15 '10 at 22:30
Take a path connecting the first point to the second point. Thicken the path to a neighborhood. Said neighborhood is diffeomorphic to the ball. It should at least be intuitive that you can always map the ball onto itself in a way that moves any given point to the origin. Just grab the point and drag it where you want it. – userN Jul 15 '10 at 22:42

Edit

The answer below is not correct. Upon further reflection, I believe that the correct causality condition is indeed global hyperbolicity and not the weaker stable causality.

I believe this is a causality condition known as stably causal. You can read about this in Wald's "General Relativity" Chapter 8, in particular his Theorem 8.2.2.

A spacetime $(M,g)$ is stably causal if there exists a continuous nowhere vanishing timelike vector $t$ such that the spacetime $(M,\tilde g)$ with $$\tilde g = g - t^\flat \otimes t^\flat,$$ where $t^\flat$ is the dual one-form to $t$ relative to $g$, has no closed timelike curves.

Wald's Theorem 8.2.2 says that this is equivalent to the existence of a global time function on $M$.

Theorem 8.2.2 A spacetime $(M,g)$ is stably causal if and only if there is a differentiable function $f$ on $M$ such that its gradient is a past directed timelike vector field.

There is a whole hierarchy of causality conditions in GR. Global hyperbolicity implies stable causality and this in turn implies strong causality. Global hyperbolicity might be too strong.

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According to Wikipedia "stably causal" is something different, or at least not obviously the same. (en.wikipedia.org/wiki/Causality_conditions#Stably_causal) The book doesn't seem to be on google books; can you elaborate? – Daniel Litt Jul 15 '10 at 19:58