# Compactness of spacetime: experiment and math

It is common to find models built on a compact spacetime. In mathematics, compactness is a very nice property $$-$$ and lot of powerful results depend on it. But

• how safe is assuming compactness of spacetime in physics?

Minkowski space is not compact, but, for instance, the treatment given to gauge theories in terms of bundles, assumes a compact basis $$X$$ for the principal bundle $$G\hookrightarrow P \to X$$ and the associated vector bundles containing matter $$P \times_G \mathfrak{g} \to X$$. This basis $$X$$ is thought of as Minkowski space-time (for sake of concreteness, assume an Euclidean spacetime $$\mathbb{R}^4$$, so one compactifies to $$X=S^4$$). Why can or cannot compactes be assumed?

This one is perhaps model-dependent:

• is there an existing experimental validation of the compactness of spacetime?

For compactified Minkowski space see, e.g., Conformal Infinity.

This is sometimes useful to prove mathematical statements about general relativity and its relatives.

But to reach infinity or to get information from there takes infinitely long, given the finite speed of light. This is why we can never find out whether or not spacetime is compactified.

• So, in some sense, is an Occamian argument. We cannot prove that spacetime is compact because we cannot "reach the infinite", so, since a compact space is simpler, we assume compactness?
– c.p.
Oct 25 '12 at 16:34
• It is usually _not_assumed, as Minkowski space is for most purposes far simpler to handle than its compactification. It is assumed only temporarily if you need it for a mathematical argument, just as in math. Oct 25 '12 at 16:42

One mathematical comment which may be interesting:

The universal cover of simply connected boundaryless time-orientable four dimensional space-time manifold must be non-compact

The argument goes something like this: suppose we are given a compact closed manifold without boundary with 4 space-time dimensions that is simply connected. Its Euler characteristic is at least two (using Poincare duality). But if it is time-orientable it admits a non-vanishing vector field, which requires Euler characteristic 0. A contradiction.

The upside to this is that compactification procedures in GR end up with stuff that are even a bit worse than manifolds with boundaries, and are not closed 4 dimensional manifolds.

Note that there is also quite a big difference between a cosmological universe where the spatial Cauchy hypersurface is a closed compact manifold, and requiring that the entire space-time be compact. For starters, the assumption of a compact space-time would require necessarily that the universe ends after finite time, a rather pessimistic world-view to which I rather not subscribe.

• what if the space-time is $S^1\times M$ for some closed 3-manifold $M$? Sep 25 '20 at 18:42
• @kora2g: what if? (note the word "universal cover") Sep 26 '20 at 13:28
• the universe doesn't end in finite time Sep 26 '20 at 18:44
• @kora2g: ah, you refer to the final paragraph. That is not a cosmological universe: it is not globally hyperbolic. (It admits no Cauchy hypersurface.) Sep 27 '20 at 14:25
• Simply-connected is a HUGE assumption given the possibility of topology change in our universe via black hole mergers and more recently the intriguing ER=EPR line of research. youtube.com/watch?v=OBPpRqxY8Uw May 18 at 22:35