# Does space-time have a projection to time?

I get confused when I see expressions like "the universe is $x$ years old" or "$10^{-2}$ seconds after the big bang" since it seems to me that relativity shows such statements don't have meaning. Is it assumed or experimentally verified or proved that space-time is equipped with a projection to the real numbers whose differential is non-zero on tangent vectors with a non-zero time component? If assumed, why is this a reasonable assumption? and if proved, what are the initial axioms (causality)? Looking at a similar question, perhaps the point is that all geodesic given by an initial point and negative time-pointing tangent vector must converge to some given point in finite (backwards) time? And to get completely cranky, are there good scientific reasons to assume no closed geodesics?

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Relativity certainly doesn't imply that such statements have no meaning. It's probably because relativity isn't theory of relativity at all, it's a theory of invariants (already Einstein made this observation, but it was too late, the current name had caught on). Some (and actually many) things in relativity are absolute. In particular all scalar products and that means all geometrical measurements by any observer whatsoever because measurements are just projections to the local frames carried by those observers and these are defined by scalar products. Therefore, one just needs (cont.) – Marek Jun 16 '11 at 15:26
(cont.) well-defined observer to be able to form such absolutely valid statements. These observers are indeed present in the standard cosmological models: because universe is homogenous and isotropic, it carries a natural comoving frame. – Marek Jun 16 '11 at 15:28
@Marek: By "natural comoving frame" you mean parallel transport WRT the Lorentz metric, right? If so, this would presumably mean that time is well defined to any particular observer. What I'm asking is that relativity says that there does not exist a function from space-time to the reals whose value at two different points is the change in time, since this value depends on how the observers travel between these points. But is the universe believed to be (homeomorphic to) $M X [0,\infty)$ for a 3-manifold $M$? – paul Jun 16 '11 at 21:37
I meant comoving with the homogenous matter that fills the universe. Or, if you prefer different terminology, the time-like Killing vector field. As for the latter part: yes, we assume the space-time to be homogeneous because it's natural but more importantly, it's consistent with observations. As for the last sentence, being homeomorphic is a very weak requirement. The space-time in these models actually is (by construction) $M$ at every $t = {\rm const}$ space-like slice and moreover, this $M$ is homogeneous and isotropic, so it can be e.g. 3-sphere. – Marek Jun 16 '11 at 21:51

@Bjorn. The wikipedia entry you linked says "The FLRW metric starts with the assumption of homogeneity and isotropy of space. It also assumes that the spatial component of the metric can be time-dependent." and concludes from GR that it is $-c^2 d\tau^2=-c^2 dt^2 + a(t)d\Sigma^2$. What's not clear (to me) is if the space-time is assumed to be (homeo or diffeomorphic to) a product $\Sigma \times R$ or if this is a consequence of solving the GR equations. (or, perhaps, the solution is only valid locally). – paul Jun 16 '11 at 21:51
@paul: GR equations don't enforce topology, they are completely local. E.g. there's no problem in rolling up Minkowski space-time into a torus (both in time and space; although rolling up time has obvious problems with causal structure). Also, such questions are very hard to decide by observation. Therefore basically in all of GR one indeed assumes that one works on some very concrete manifold, like $\Sigma \times {\mathbb R}$. – Marek Jun 16 '11 at 22:34
@Marek. Thanks, that helps. Am I correct then in concluding that when theorizing that universe starts with a big bang the assumption is made that space time does have the form $\Sigma\times I$ ($I$ some interval)? And if so, what if one doesn't assume this? (I.e. might ST be a smooth manifold with geodesics defined for all backwards time?) Maybe my question is pure geometry rather than physics: Does the existence of a Lorenz metric on a smooth 4-manifold with no closed geodesics imply that it has a product structure, or must it be incomplete or its completion singular? – paul Jun 16 '11 at 23:21
@paul: IIRC FRLW models do not imply Big Bang. There are lots of solutions (depending on the matter content), including a static model which is defined for all $\mathbb R$. So we input some observed facts, like Hubble constant, speed of acceleration, etc. and recover the matter distribution that should produce that. That is where number on dark energy and dark matter come from. As for the latter part, I am pretty sure it doesn't need to have a product structure. E.g. a manifold resembling a pair of pants should be all right (given a correct matter content of course). – Marek Jun 17 '11 at 6:57