4
$\begingroup$

Instead of a 4D hyperbolic geometry (saddle-shaped), is it conceivable a spacetime with negative curvature has a 4D spherical geometry where the perspective is from the "inside" of the sphere rather than the "outside"?

$\endgroup$
  • 1
    $\begingroup$ Because by Cartan-Ambrose-Hicks all simply connected manifolds of constant sectional curvature are isometric in any dimension, so you can't put a hyperbolic metric on a topological sphere. $\endgroup$ – Conifold Oct 4 '16 at 19:45
  • 1
    $\begingroup$ @Conifold Meaning it's arbitrary - we'd still see a positive curvature (converging lines) even if we live on the "inside" of a 4D sphere? $\endgroup$ – RobertF Oct 4 '16 at 19:52
  • $\begingroup$ I am not sure what is arbitrary? Spheres simply do not admit hyperbolic metrics, curvature and topology are intrinsic, so "inside" and "outside" do not matter. $\endgroup$ – Conifold Oct 4 '16 at 20:04
  • 2
    $\begingroup$ @RobertF It is not at all arbitrary. The inside of a sphere has positive curvature just like the inside. The curvature is intrinsic to the surface itself. $\endgroup$ – Thriveth Oct 4 '16 at 20:04
  • 1
    $\begingroup$ @Conifold Yea, I think I'm taking simplistic illustrations of positive/negative spacetime curvature showing people walking on hills or inside bowls too literally. Since we're stuck in rather than on the spacetime surface it doesn't make sense to change perspective from inside to outside. $\endgroup$ – RobertF Oct 4 '16 at 20:43
10
$\begingroup$

The curvature of a surface is defined as the product of its principal curvatures. Whether they are both positive (on the outside if a sphere) or both negative (on the inside), the product will still be positive.

The curvature is intrinsic to the surface, as @Cornifold mentions above. It doesn't matter from which side you see it. This is a fundamental mathematical theorem, look e.g. on its Wikipedia page

As you point out in your own comment, the distance and angular measures on a sphere are the same whether you look at it from the inside or the outside. This is basically a different way of saying the same thing - that the curvature is unchanged. This is, in some sense, exactly what curvature means.

$\endgroup$
6
$\begingroup$

Your question should dissolve away when you understand the meaning of curvature.

When we talk about spheres and saddles, we imagine them being in three-dimensional space. We can use a Cartesian 3D system and describe the surfaces completely by some function $F(x,y,z) = 0$. For example, a unit sphere would be described by $$ x^2 + y^2 + z^2 - 1 = 0$$

Mathematically, what we are doing is called embedding. We embedded a two dimensional surface in three dimensional space. It's a mathematical fact that we can embed such surfaces, i.e. manifolds in higher-dimensional ordinary Euclidean spaces.

But here's the thing. We don't have to think about surfaces as being embedded in some higher-dimensional space. The sphere, for example, as a two-dimensional manifold is what it is whether you embed it or not, it's the surface that matters. The curvature of a manifold does not depend on its embedding, it is an intrinsic property of the manifold itself.

So, you see, there is no outside and inside, not really. One way to measure curvature of the sphere would be to draw a circle on it and compare its circumference to its area. The ratio would, of course, be smaller than on a sheet of paper, which is flat space. Similarly, in three-dimensional space, you would compare the volume within a sphere to its surface area.

Therefore, when talking about curvature of 4D spacetime, don't imagine it as being a hypersurface in a higher-dimensional space, it's not necessary.

$\endgroup$
3
$\begingroup$

On a sphere, if you take it as a 2D representation of curved 3D space, it is possible to move straight forward and end up where you started from (which would also be the case in a nonexpanding closed universe; in reality the universe expands too fast for even a light ray to catch up with the increasing curvature circumference) but in an open universe you can't, so the sphere would be the wrong representation for an open universe, but the right one for a closed.

$\endgroup$
  • 1
    $\begingroup$ Negative curvature doesn't imply that geodesics go on forever. There exist 3-D spaces of constant negative curvature that have the property that (at least some) geodesics return to their starting point. See, for example, the Seifert-Weber space (no relation), which is basically a hyperbolic analog of a 3-torus. $\endgroup$ – Michael Seifert Oct 4 '16 at 20:40
  • $\begingroup$ If so I limit my answear to the cosmological perspective. $\endgroup$ – Yukterez Oct 4 '16 at 20:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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