## When can we factor out the time dimension?

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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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## 2 Answers

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 2010 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 2010 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 2010 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 2010 at 22:30

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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 2010 at 22:42
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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/…)  The book doesn't seem to be on google books; can you elaborate? – Daniel Litt Jul 15 2010 at 19:58
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