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?

  • 3
    $\begingroup$ Related question here. $\endgroup$
    – knzhou
    Commented Jul 18, 2020 at 19:57
  • 1
    $\begingroup$ 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. $\endgroup$
    – A.V.S.
    Commented Jul 19, 2020 at 6:10

1 Answer 1


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.

  • $\begingroup$ Is some papers where spin connection is independent field? $\endgroup$
    – Nikita
    Commented Jul 20, 2020 at 21:36
  • $\begingroup$ @Nikita, for Einstein–Cartan theory, see wiki (and refs therein) here: en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory $\endgroup$
    – MadMax
    Commented Jul 20, 2020 at 23:32
  • $\begingroup$ Independent spin connection and the related dynamical torsion tensor would lead to Big Bounce, rather than Bid Bang. See here: en.wikipedia.org/wiki/Big_Bounce $\endgroup$
    – MadMax
    Commented Jul 20, 2020 at 23:39
  • $\begingroup$ In Einstein–Cartan theory there's independent torsion, not spin connection, as I understand. But what if one will consider independent spin connection?? $\endgroup$
    – Nikita
    Commented Jul 21, 2020 at 7:49
  • $\begingroup$ Unfortunately, in Big Bounce article I didn't found any actions and equations. Could you explain your remark? $\endgroup$
    – Nikita
    Commented Jul 21, 2020 at 7:51

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