Eventually the people in the universe of Greg Egan's "Orthogonal" series figure out the topology of their cosmos.

They decide it can't be a 4-torus because the curvature needs to be positive everywhere. It therefore needs to be a 4-sphere.

But then they say the opposite, that

the curvature will be negative in many places

because of the presence of matter. So what gives?

Also, why can't the curvature be negative everywhere, resulting in a hyperbolic universe?

  • 1
    As a physicist: the presence of mass distorts the shape of spacetime on a local scale. The book probably refers to the global topology of spacetime.
    – Adamant
    Jul 17, 2016 at 20:04
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    Per wikipedia: "Another way of saying this is that if all forms of dark energy are ignored, then the curvature of the universe can be determined by measuring the average density of matter within it, assuming that all matter is evenly distributed (rather than the distortions caused by 'dense' objects such as galaxies). This assumption is justified by the observations that, while the universe is "weakly" inhomogeneous and anisotropic (see the large-scale structure of the cosmos), it is on average homogeneous and isotropic."
    – Adamant
    Jul 17, 2016 at 20:09

1 Answer 1


They realise that if their universe was a 4-torus, that would result in extra modes for fermionic vacuum energy that led to an overall negative energy density (fermionic vacuum energy is negative) which, in this kind of universe, would require space to be positively curved everywhere. But you can't have a space with the topology of a 4-torus that is positively curved everywhere. So what they conclude is that the universe being a 4-torus is self-contradictory.

However ...

They have also known for a long time that the universe must be finite in all directions, to avoid exponentially growing solutions to the wave equation. So the simplest alternative topology to a 4-torus is a 4-sphere. In that case, the topology doesn't have the extra fermionic modes, and the overall vacuum energy is positive, which, in this kind of universe, requires space to be negatively curved.

So ...

It's not that they conclude that the curvature actually is positive everywhere, and hence the universe can't be a 4-torus. It's that they see why a 4-torus both implies positive curvature and at the same time is ruled out by positive curvature, which eliminates the whole possibility.

You ask:

Also, why can't the curvature be negative everywhere, resulting in a hyperbolic universe?

This universe ...

can't be an infinite hyperbolic universe, which is what is usually meant by that phrase. However, you can have a finite universe with the topology of a 4-sphere but negative curvature. It just can't be uniform negative curvature, it has to vary in magnitude from place to place.

More details at http://www.gregegan.net/ORTHOGONAL/06/GRExtra.html

  • 1
    I am not sure so much of this needs to be in spoiler markup.
    – Adamant
    Jul 18, 2016 at 1:02
  • 2
    @Adamant It does if you enjoy working out the physics puzzles in Egan's work, which is at least half the fun for some sf fans.
    – Kyle Jones
    Jul 18, 2016 at 2:34
  • Straight from the horse's mouth! (apparently)
    – Sam Sippe
    Aug 24, 2016 at 1:36
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    @SamSippe -- Yeah I was wondering about this. It's very unlikely he would have seen my question at all unless he had been lurking on Stack Exchange, but if he's been lurking, it's very unlikely he would make an account just to answer my question. He's had no other activity since. It's more likely that this is an impostor. But then... this is about as cogent an answer as one could hope for. Puzzling. Oct 4, 2016 at 19:32
  • 1
    @MackTuesday If it's an imposter, it's a very convincing one.
    – Sam Sippe
    Oct 6, 2016 at 6:39

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