The Einstein-Palatini action can be written as $$ S = M_{pl}^2\int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge R^{cd}\right), $$ where $e^a={e^a}_\mu\text{dx}^\mu$ is the basis one-form and $R^{ab}=\frac{1}{2}{R^{ab}}_{\mu\nu}\text{dx}^\mu\wedge\text{dx}^\nu$ is the Riemann curvature two-form. The Cartan Structure Equations for the torsion-less and metric-compatible connection of GR are $$ R^{ab} = D\omega^{ab} = d\omega^{ab} + {\omega^a}_c \omega^{bc}, \quad 0 = De^a = de^a + {\omega^a}_be^b, $$ where $\omega^{ab}={\omega^{ab}}_\mu\text{dx}^\mu$ is the (antisymmetric) spin connection one-form.

Now, my confusion comes from the fact that if I apply the first structure equation, integrate by parts (neglecting boundary terms), and apply the structure second equation, the whole action seems to vanish. \begin{align} S &= M_{pl}^2\int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\ &= M_{pl}^2\int\varepsilon_{abcd}\left(-D(e^a\wedge e^b)\wedge\omega^{cd}\right) \\ &= M_{pl}^2\int\varepsilon_{abcd}\left(-De^a\wedge e^b\wedge\omega^{cd} + e^a\wedge De^b\wedge\omega^{cd}\right) \\ &= M_{pl}^2\int\varepsilon_{abcd}\left(0+0\right)=0 \end{align}

This obviously doesn't seem correct, so is there an error in my understanding somewhere? Is it incorrect to use the structure equations and integrate by parts in the action in this way?

  • 2
    $\begingroup$ You assumed that the connection is compatible with the tetrad. $\endgroup$ Jan 6, 2021 at 11:47
  • $\begingroup$ @SounakSinha yes I did, but is this not the case in GR? $\endgroup$
    – JeffK
    Jan 6, 2021 at 13:08
  • 1
    $\begingroup$ In the Palatini action tetrad compatibility is not an on shell statement. $\endgroup$ Jan 6, 2021 at 14:09
  • $\begingroup$ Ok, thank you @SounakSinha. If the structure equations are not true on-shell, then how should they be understood in this context? Any references you can recommend are of course welcome as well. $\endgroup$
    – JeffK
    Jan 6, 2021 at 16:18
  • $\begingroup$ Try the Wikipedia article. $\endgroup$ Jan 6, 2021 at 17:55

2 Answers 2


I found a reliable source that addresses my question (Appendix 4.1 - M. Gasperini, Theory of Gravitational Interactions; DOI 10.1007/978-88-470-2691-9), so I will present the resolution to my confusion here for posterity's sake. The issue is not with using the structure equations on-shell (indeed, this is purely a classical theory) and there is no error in my calculation. The problem is that the Palatini action is defined with a non-zero torsion and only corresponds to GR after the limit of vanishing torsion is taken after computing the equations of motion.

The Palatini formalism requires that we use the structure equation $$ T^a = De^a = de^a + {\omega^a}_be^b , $$ where $T^a = \frac{1}{2}{T^a}_{\mu\nu}\text{dx}^\mu\wedge\text{dx}^\nu \neq 0$ is the torsion two-form. One can then integrate by parts to find \begin{align} S &= M_{pl}^2\int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\ &= M_{pl}^2\int\varepsilon_{abcd}\left(-T^a\wedge e^b\wedge\omega^{cd} + e^a\wedge T^b\wedge\omega^{cd}\right) \neq 0 , \end{align} which is a perfectly valid version of the Palatini action.

To make touch with GR (here with no matter sources for simplicity), we must vary the action with respect to the two independent fields in our theory, $e^a$ and $\omega^{ab}$, which yields the following EOMs. $$\delta_e S = \varepsilon_{abcd}R^{ab}\wedge e^c = 0 , \quad \delta_\omega S = \varepsilon_{abcd}T^a\wedge e^b = 0 $$ These are the "Einstein-Cartan" equations, and after converting back to tensor component notation, is easy to see that the first gives precisely the Einstein equations of GR while the second gives an EOM for the torsion which is of course trivial after taking the limit $T^a \rightarrow 0$.

The short story is, the Palatini formalism does not reproduce GR if we simply set $T^a=0$ at the level of the action. The two theories only coincide once we take the limit of vanishing torsion at level of the EOMs.


This part of yours is incorrect: $$ \begin{align} S &\sim \int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\ &= \int\varepsilon_{abcd}\left(-De^a\wedge e^b\wedge\omega^{cd} + e^a\wedge De^b\wedge\omega^{cd}\right) \end{align} $$ The correct derivation is: $$ \begin{align} S &\sim \int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\ &= \int\varepsilon_{abcd}\left(-(de^a + \frac{1}{2} {\omega^a}_k\wedge e^k)\wedge e^b\wedge\omega^{cd} + e^a\wedge (de^b+ \frac{1}{2} {\omega^b}_k\wedge e^k)\wedge\omega^{cd}\right) \\ \end{align} $$ Obviously: $$ de^a + \frac{1}{2} {\omega^a}_k\wedge e^k \neq De^a= de^a + {\omega^a}_k\wedge e^k $$

Added note after discussion with @JeffK (the person who originally asks the question), since @JeffK opted to stand by the wrong calculation.

Moving covariant derivative $D$ between $\omega$ and $e$ is a bit tricky, because of the unique definition of $D\omega$.

One can arrive at the Einstein action by applying the zero torsion condition $$ 0 = De^a = de^a + {\omega^a}_be^b $$ to the Einstein-Palatini action, via expressing $\omega$ as a function of $e$, effectively eliminating explicit dependence of the Einstein action on $\omega$.

The Einstein action is apparently not zero. I hope @JeffK can see the light and would avoid making similar mistakes in future study/research.

  • $\begingroup$ Thanks for the answer @MadMax. Can you explain where the factor of 1/2 comes from? $\endgroup$
    – JeffK
    Jan 7, 2021 at 0:03
  • $\begingroup$ @JeffK, in short hand notation, you have to rewrite $\omega^2 = \frac{1}{2}\omega^2 + \frac{1}{2}\omega^2$ before moving $\omega$ around to match two $de$. On the other hand, when you do variation on $\omega$ to get equation of motion where $De$ naturally appears, you don't have the $\frac{1}{2}$ issue, since $\delta \omega^2 = \omega\delta\omega + \delta\omega \omega$. $\endgroup$
    – MadMax
    Jan 7, 2021 at 15:59
  • $\begingroup$ @JeffK, your self-answer is also wrong ($S = 0$ in the limit $T\rightarrow0$), the notion of action being zero on-shell (using equation of motion) only applies to certain actions. For example Dirac action is zero on-shell, since all terms in Dirac action are linear in $\psi$ (or $\bar\psi$). But the Einstein-Palatini action is not zero on-shell, since $R$ contains both $\omega$ and $\omega^2$ terms. $\endgroup$
    – MadMax
    Jan 7, 2021 at 16:38
  • $\begingroup$ With regard to your first comment, there is no "moving around" of ω happening here. There is an integration by parts between the first two lines in my original question, where I moved the covariant derivative from ω to $e \wedge e2$ (and dropped the boundary term). For your second comment, this is whole point! If you take the zero torsion limit before computing the EOMs then you get a zero action (as you would would any theory if you integrate out all the fields). One must keep the torsion general in the Palatini action and then if you choose, remove torsion at the end to find GR. $\endgroup$
    – JeffK
    Jan 7, 2021 at 18:23
  • $\begingroup$ "I moved the covariant derivative", wrong, you can't move covariant derivative from $D\omega$ to $De$, as I explained earlier. Integration by parts only works for $d$, not for $D$. You have to move $\omega$ around properly. $\endgroup$
    – MadMax
    Jan 7, 2021 at 18:58

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