What does it mean that symmetries do not hold globally?

In summary: It is possible, though, for curvature to prevent a global symmetry, as with the symmetries of the FLRW metric being broken by the presence of mass/energy in the universe.In summary, Lorentz symmetry and time translational symmetry are not global in an expanding universe. This means that other Hubble spheres outside our observable universe could have different symmetries. However, the assumption in our models is that the entire universe looks the same as our observable universe, including the presence or absence of symmetries. Local Lorentz symmetry still holds, but curvature can break global symmetry.
  • #1
Suekdccia
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If symmetries in the universe do not hold globally, what does that mean?
Perhaps this is a stupid question but, if Lorentz symmetry and time translational symmetry are not global in an expanding universe, wouldn't that mean that is possible that other Hubble spheres outside our observable universe could have other symmetries or an absence of the Lorentz symmetry? I mean, does this mean that the Lorentz symmetry could hold in our local region of spacetime (that is, our observable universe) but, as it does not hold globally, there could be other regions where it would not hold?
 
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Suekdccia said:
if Lorentz symmetry and time translational symmetry are not global in an expanding universe
Lorentz symmetry isn't global in any curved spacetime.

Time translation symmetry doesn't even exist at all except in stationary spacetimes (spacetimes with a timelike Killing vector field).

Suekdccia said:
wouldn't that mean that is possible that other Hubble spheres outside our observable universe could have other symmetries
The assumption in our models is that the entire universe looks the same as our observable universe. That would include the presence or absence of symmetries.

Of course we have no way of testing that assumption, but it's the simplest one, and without ay evidence against it there's no point in making things more complicated than they need to be.

Suekdccia said:
or an absence of the Lorentz symmetry?
Not locally; any spacetime that GR can model has local Lorentz symmetry.
 
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