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1 vote
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Yang-Mills-Dirac Lagrangian and Gravity

Let $M$ be a Lorentzian spin $4$ manifold, i.e. admits a spin structure $Spin^+(M)\rightarrow M$, which is just a principal $Spin^+(1,3)$ bundle over $M$, which is compatible with the bundle of ...
Chris's user avatar
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2 votes
1 answer
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Gravity = Yang-Mills squared? What is the Lagrangian?

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? ...
Anon21's user avatar
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