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It is well known, how to construct Einstein gravity as gauge theory of Poincare algebra. See for example General relativity as a gauge theory of the Poincaré algebra.

There are

  1. Construction of covariant derivative:

$$ \nabla_m = \partial_m -i e_m^{\;a}P_a -\frac{i}{2}\omega_m^{\;\;\;cd}M_{cd}.$$

  1. Impose covariant constraint on geometry: $$ [\nabla_m, \nabla_n] = -i R_{mn}^{\;\;\;a}P_a -\frac{i}{2}R_{mn}^{\;\;\;ab}M_{ab} $$ $$ R_{mn}^{\;\;\;a} = 0. $$ From this equation, spin connection $ω^{\;\;\;cd}_m$ is expressed in terms of veilbein $e^{\;\;a}_m$.

  2. Now, one can easily construct Einstein-Hilbert action: $$ S_{EH} = \int d^d x e \;R_{mn}^{\;\;\;ab} e_a^{\;m}e_b^{\;n} $$ $e_a^{\;m}$ is inverse veilbein $e_a^{\;m} e_m^{\;b}= \delta_a^b $. Metric tensor: $$ g_{mn} = e_m^{\;a}e_n^{\;b} \eta_{ab}. $$

But one can modify second step and obtain another actions, with additional dynamical spin connection:

  1. $$ S_{EH} = \int d^d x e \;R_{mn}^{\;\;\;ab} e_a^{\;m}e_b^{\;n}. $$

  2. $$ S_{YM} = \int d^d x e \left(\;R_{mn}^{\;\;\;ab} R_{kl}^{\;\;\;cd}g^{mk}g^{nl}\eta_{ad}\eta_{bc} + R_{mn}^{\;\;\;a} R_{kl}^{\;\;\;b}g^{mk}g^{nl}\eta_{ab}\right). $$

So I have few questions:

What will standard Einstein-Hilbert action describe in this case?

What is Yang-Mills theory for Poincare group? Which properties have such theory?

Why Einstein action is not Yang-Mills theory for Poincare group?

Qmechanic
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Nikita
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    Related question here. – knzhou Jul 18 '20 at 19:57
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    Minor complaint: it is better to use different alphabets (e.g. Greek vs. Latin or uppercase vs. lowercase) rather than different parts of the same alphabet to distinguish between different types of indices, it is less work for the eyes to sort out indices in complicated expressions. – A.V.S. Jul 19 '20 at 06:10

1 Answers1

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The YM action for the Poincare group as you write down is perfectly allowable in the effective field theory framework, as long as you double check that pathological tachyons are absent. There are tons of papers devoted to the so called $f(R)$ and $f(T)$ theories with higher-order Lagrangian terms (like $R^2$, $T^2$).

The catch is that, comparing with the EH term, the YM term is suppressed by a factor of $O(p^2/M_p^2)$, where $M_p$ is the Planck mass. Therefore, the YM term is negligible, except in extreme situations, e.g. shortly after the Big Bang.

Frederic Thomas
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MadMax
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  • Is some papers where spin connection is independent field? – Nikita Jul 20 '20 at 21:36
  • @Nikita, for Einstein–Cartan theory, see wiki (and refs therein) here: https://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory – MadMax Jul 20 '20 at 23:32
  • Independent spin connection and the related dynamical torsion tensor would lead to Big Bounce, rather than Bid Bang. See here: https://en.wikipedia.org/wiki/Big_Bounce – MadMax Jul 20 '20 at 23:39
  • In Einstein–Cartan theory there's independent torsion, not spin connection, as I understand. But what if one will consider independent spin connection?? – Nikita Jul 21 '20 at 07:49
  • Unfortunately, in Big Bounce article I didn't found any actions and equations. Could you explain your remark? – Nikita Jul 21 '20 at 07:51
  • Please see Nikodem Poplawski's original Big Bounce paper here: https://arxiv.org/abs/1111.4595 – MadMax Jul 21 '20 at 14:31
  • For independent spin connection, please see the exposition in section 2.5.2 of https://arxiv.org/abs/0911.0334. – MadMax Jul 21 '20 at 14:38