1
$\begingroup$

(I believe that) de Sitter space is the only maximally symmetric Lorentzian spacetime, and that for $n$ spacetime dimensions, it has the hypercylindrical topology $\mathbb{R} \times S^{n-1}$.

This is already a little counterintuitive to me, because I would have thought that for a maximally symmetric spacetime, space and time would be on a quite equal footing, so it seems a bit strange that the spatial dimensions are compact but the time dimension isn't. I understand how it's logically possible though; the space and time directions are mathematically distinct because of their different signs in the metric signature, and moreover differential geometry is full of non-obvious constraints that purely local requirements can impose on the global topology of manifolds (e.g. the Gauss-Bonnet theorem).

My question is how this generalizes to pseudo-Riemannian manifolds with arbitrary metric signature. Specifically, my question is twofold:

  1. If we consider the $n$-dimensional pseudo-Riemannian manifolds with some fixed (not necessarily Lorentzian) metric signature, is there a unique maximally symmetric manifold with positive scalar curvature? (By maximally symmetric manifold, I mean one with the maximum possible number $\frac{1}{2} n (n+1)$ of linear independent Killing fields on the whole manifold.)

  2. If so, what is its global topology? A guess is that if the metric signature is $(p,m)$, then the global topology is $R^{\min(p,m)} \times S^{\max(p,m)}$, but I have no idea if that's correct.

$\endgroup$
3
  • $\begingroup$ Leonard Susskind said in his class, answering a question, that he does not know what to make of more than one time dimension, how to even make sense of it. I also read somewhere that a second time dimension will result in ultra hyperbolic equations that have no predictive power. It is just hearsay at the moment for me. $\endgroup$
    – user338734
    Commented Sep 3, 2022 at 1:14
  • $\begingroup$ @CarlosGauss Multiple time dimensions might be physically nonsensical, but at the mathematical level this question is perfectly well-posed. $\endgroup$
    – tparker
    Commented Sep 3, 2022 at 15:40
  • $\begingroup$ @A.V.S. Thanks for catching that typo - corrected. $\endgroup$
    – tparker
    Commented Sep 4, 2022 at 20:12

1 Answer 1

1
$\begingroup$

$(n{-}1){+}1$ dimensional de Sitter space can be constructed as a sphere of spacelike radius in $n{+}1$ dimensional Minkowski space. The sign of the squared radius breaks the symmetry.

The sign of the scalar curvature is ambiguous for mixed-signature spaces, since the Ricci scalar inherits the sign convention of the metric. It's probably better to say that de Sitter space has spacelike curvature (spacelike radius of curvature).

Anti de Sitter spaces are also maximally symmetric and can be constructed as spheres of timelike radius in a flat space of $(n{-}1){+}2$ signature. They are topologically $\def\R{\mathbb R} \R^{n-1}\times S^1$ (or $\R^n$ if you take the universal cover, as is typically done so that time isn't cyclic).

The flat spaces $\R^{p+q}$ are also maximally symmetric.

That probably generalizes to: a sphere in flat $p{+}(q{+}1)$ space whose squared radius has the same sign as the $q{+}1$ dimensions is topologically $\R^p S^q$, and moreover, those and $\R^{p+q}$ are probably the only maximally symmetric $p{+}q$ dimensional manifolds up to covers and quotients by $\mathbb Z_2$. But it's not obvious to me that it's true, and I haven't found a source online that confirms it in a quick search.

The answer to your first question is no, since, for instance, $1{+}1$ dimensional de Sitter space isn't homeomorphic to its universal cover or its quotient by $\mathbb Z_2$.

$\endgroup$
1
  • $\begingroup$ The third paragraph of your excellent answer also answers my question at physics.stackexchange.com/questions/726324/…. Please feel free to copy it into that question, and I'll mark that one correct as well. $\endgroup$
    – tparker
    Commented Sep 5, 2022 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.