Lorentz invariance violation for manifolds

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  • Thread starter sqljunkey
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I was looking at this video , and I was wondering if a (Riemannian)manifold violates the "lorentz invariance" would it become a discrete manifold?
 

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robphy
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Not necessarily.
It depends on the nature of the violation.

From a glance at the video, it looks like something similar to the “causal set” approach to quantum gravity ( https://en.m.wikipedia.org/wiki/Causal_sets
https://www.einstein-online.info/en/spotlight/causal_sets/
)
There, there is no manifold but a discrete set with a partial order (akin to the causal order). In some classical limit, the causal sets would be expected to resemble the manifold (similar to how many molecules of water might resemble a fluid at some macroscopic scale).
 
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Thanks for the links robphy. What do you mean by not necessarily and by the nature of the violation. Are there more types of violations?

I'm confused about this. Wouldn't a Riemannian manifold be inclined to have many "mini manifolds" because you added curvature to it?

As you go to the infinitesimal close patches in the curved manifold it has to somewhere break to start forming the curve. Making each section have different characteristics than the other .

And the only special case is when you are careful how you add the curvature to the manifold, so every "mini manifold" as it were had the exact metric.

Otherwise you are saying the manifold is equipped with some kind structure that requires the manifold to become lorentz invariant every time it is perturbed.

Or maybe one of the axis's is staying unperturbed while the others are being perturbed?
 

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