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If there are no localized observables in quantum gravity, does spacetime really exist, or might spacetime really be an illusion?

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    $\begingroup$ If there are no observables in quantum gravity, then it means that the theory of quantum gravity is useless. It does not imply that space time does not exist or is an illusion. I vote to close this question because the question doesn't make sense. $\endgroup$ – QEntanglement Jun 20 '11 at 8:53
  • $\begingroup$ @QEntanglement: partially true but I feel like the question could make sense. More or less everyone agrees that the classical notion of space-time should break down at Planck scale and get replaced by some non-commutative version. There is lot to be said on this topic. $\endgroup$ – Marek Jun 20 '11 at 9:14
  • $\begingroup$ I think the question should be restated to be more specific because "there is a lot to be said on this topic." $\endgroup$ – QEntanglement Jun 20 '11 at 9:54
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    $\begingroup$ He did not say "no observables", he said "no localised observables" $\endgroup$ – Philip Gibbs - inactive Jun 21 '11 at 6:36
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    $\begingroup$ This question should be reformulated as: how do you reconstruct a space-time from the observables admitted in quantum gravity, namely the S-matrix in flat space, or the AdS boundary theory in asymptotically AdS spaces? This is a topic of very active research. $\endgroup$ – Ron Maimon Sep 4 '11 at 0:07
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Andrew Strominger thinks spacetime is an illusion. It's all a Cosmic Hologram at the future boundary of spacetime at the end of time. It projects out the illusion we see around us.

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The answer is divided more on the lines of "TOE vs. non-TOE" than "quantum gravity vs. non-quantum gravity". In string theory, the metric tensor is an approximation that arises as an effective description of the graviton field (see also this post on TRF). But in non-unification theories of quantum gravity, like LQG and its other discrete-spacetime cronies, spacetime is considered fundamental.

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