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One of the (many different, somewhat independent) routes to gauge theory is to start from a global symmetry of some kind and "gauge" it, which involves promoting it to a local symmetry and then introducing one or many massless gauge boson(s) to enforce consistency of the theory. This procedure can be done for the case of a phase symmetry of a fermion field to yield QED, more general invariance of a collection of fermions under a unitary group to yield Yang-Mills, and for translational symmetries to give general relativity. In general the gauge bosons need to couple to the conserved (Noether) currents of the global symmetry that is being gauged, which tells us that the photon couples to charge and that the graviton couples to stress-energy.

Focusing particularly on the last part-- it seems as though if GR is what you get when you gauge one part of the Poincaré group (translations), it would be natural to gauge the other part (boosts and rotations). Following this chain of logic, this would yield a gauge boson that would need to couple to (relativistic) angular momentum. To the best of my (limited) knowledge there exists no theory of such a boson. My question is-- what (if any) obstruction is there that prevents us from "gauging" the Lorentz group in this way?

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  • $\begingroup$ You statement is very misleading. The first ever gauge theory of gravity by Utiyama in 1956 is gauging Lorentz symmetry. See the paper here: doi.org/10.1103/PhysRev.101.1597 $\endgroup$
    – MadMax
    Commented Sep 21, 2023 at 17:49
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    $\begingroup$ " it seems as though if GR is what you get when you gauge one part of the Poincaré group (translations)" I'm not sure what this means but if we cast GR in gauge theory terms it's usually either $\mathrm{GL}(n)$ or $\mathrm{SO}(1,3)$ as the gauge group. Where did you get this idea about translations from? For more on GR as a gauge theory see physics.stackexchange.com/q/46324/50583, physics.stackexchange.com/q/346793/50583 and their linked questions. $\endgroup$
    – ACuriousMind
    Commented Sep 21, 2023 at 17:50
  • $\begingroup$ It's entirely possible I'm misguided about this however I've seen this statement in a few different places. Offhand at reddit.com/r/Physics/comments/71k5u2/… some of the commenters make the point that it is inappropriate to regard GR as the gauging of the full Poincaré group as opposed to strictly the group of translations ("Gauging the Poincaré group creates what is known as Poincaré Gauge Theory, which has GR as a subset"). I'm not really equipped to know if this comment is wrong which is part of why I'm asking this question. $\endgroup$
    – Panopticon
    Commented Sep 21, 2023 at 18:16
  • $\begingroup$ Maybe the core of my difficulty here is that normally your gauge bosons couple to conserved currents under the global symmetry that you are gauging. The stress-energy tensor can be obtained as the Noether current associated to translations; the Noether charge associated to SO(1,3) is the relativistic angular momentum which is distinct from the stress energy and does not per se couple to gravitons. So perhaps the more interesting question from my point of view, leaving aside the question of where GR comes from exactly, is why is there no gauge boson that couples to the angular momentum. $\endgroup$
    – Panopticon
    Commented Sep 21, 2023 at 18:26

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