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In the following comment and references, it is mentioned that gravity can be understood as yang mills squared. What is the Lagrangian of a Yang Mills squared theory? Can anyone provide a quick primer?


In this comment:

Why are string theorist so indifferent to the gauge structure of gravity?

Then in these references:

https://arxiv.org/abs/1602.08267

https://www.youtube.com/watch?v=JTUQw2wxck8

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    $\begingroup$ See also here. $\endgroup$
    – J.G.
    Commented Jun 26, 2022 at 9:33

1 Answer 1

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  1. The statement $$ \text{(super) gravity} = [\text{(super) Yang-Mills]}^2 $$ is the punch line of the Bern-Carrasco-Johansson (BCJ) colour-kinematic duality conjecture that the perturbative on-shell scattering amplitudes for (super) gravity are the double-copy of gluon scattering amplitudes in (super) Yang-Mills theory.

  2. Refs. 1-2 are some recent work on implementing manifest BCJ colour-kinematic duality at the level of the off-shell action/Lagrangian, cf. OP's title question.

References:

  1. L. Borsten, B. Jurco, H. Kim, T. Macrelli, C. Saemann & M. Wolf, BRST-Lagrangian Double Copy of Yang-Mills Theory, arXiv:2007.13803.

  2. L. Borsten, B. Jurco, H. Kim, T. Macrelli, C. Saemann & M. Wolf, Double Copy from Homotopy Algebras, arXiv:2102.11390.

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  • $\begingroup$ Thank you. In the following reference ift.uni.wroc.pl/~mborn35/files/proc/Krasnov.pdf page 23. The author mentions this Lagrangian as a "double yang-mills" version of gravity: $ S=\int (\operatorname{Tr} \sqrt{F \wedge F})^2$, but does not provide a reference. Can you shed more light on this Lagrangian or know of a link where I can read more about it? $\endgroup$
    – Anon21
    Commented Jun 30, 2022 at 13:17
  • $\begingroup$ FWIW, Krasnov's talk seems based on his work: arxiv.org/abs/1503.08640 , arxiv.org/abs/1406.7159 , and a Plebanski-like action. $\endgroup$
    – Qmechanic
    Commented Jun 30, 2022 at 14:12
  • $\begingroup$ Is the Lagrangian of a double copy Yang mills $\mathcal{L}=\mathcal{L}_{\text{copy 1}} + \mathcal{L}_{\text{copy 2}}$ or $\mathcal{L}=\mathcal{L}_{\text{copy 1}} \mathcal{L}_{\text{copy 2}}$? $\endgroup$
    – Anon21
    Commented Jul 7, 2022 at 20:55

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