I've heard it speculated that the spatial dimensions of the universe is a 3-sphere. Or a 3-torus. But usually, I guess, it's assumed that the "time" dimension just has its own geometry, like a line, in Cartesian product with the geometry of the spatial dimensions.

I don't know much about topology, nor the constraints on topology placed by geometry. So I don't know if the shape of the manifold "cares" that the geometry treats one of those dimensions differently (particularly because the dimensions are symmetric in a sphere). Basically, can a manifold whose metric has the Lorentz signature $-+++$ be a 4-sphere? Or more generally, can a manifold with a $(1,n-1)$ signature metric be an $n$-sphere?

Also, let me know if I'm using imprecise, bad language here.


1 Answer 1


No, Lorentzian manifolds can be spheres only in odd dimensions. This is because the Euler characteristic of compact Lorentzian manifolds must vanish, which it doesn't for $S^n$ for even $n$, cf. e.g. this MathOverflow question.

  • $\begingroup$ Thank you. I trust the conclusion, even though I don't understand the explanation in the linked answer. $\endgroup$ Oct 9, 2019 at 16:29
  • $\begingroup$ In physics, this answer doesn't seem to be consistent with Nikodem Poplawski's torsion-based cosmological model (described in many 2010-2021 preprints that can be found by his name on Cornell University's Arxiv site), comprised of local universes each expanding "indefinitely", each of whose shapes he has analogized to "the skin of a basketball". Might the incompatibility be due to his use of 1929's Einstein-Cartan Theory (a collaboration by Einstein with the mathematician Ely Cartan), which assigns a tiny spatial extent to fermions, rather than 1915's GR? $\endgroup$
    – Edouard
    May 29 at 8:47

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