The classical Yang-Mills action is of the form

$$S=\frac{1}{2g^2}\int_{\mathcal{M}}\text{tr}\left[F\wedge\star F\right]\\ =\frac{1}{4g^2}\int\mathrm{d}^dx\sqrt{g}\,\text{tr}\left[F^{\mu\nu}F_{\mu\nu}\right],$$

where $F=dA+A\wedge A$ is the Yang-Mills field strength $2$-form and $g$ is a coupling constant that is irrelevant in the classical description.

Now, the classical Einstein-Hilbert action takes the form


where $\mathcal{R}=\text{tr}_g(\text{tr}(R))=R^{\mu\nu}_{\,\,\,\,\mu\nu}$ is the Ricci scalar curvature of the manifold and $R$ represents the Riemann curvature tensor, and $\kappa=8\pi G$ is again a coupling constant.

My question is this: why is the Einstein-Hilbert action linear in the curvature tensor while the Yang-Mills action is quadratic in the gauge-field curvature tensor? Both theories are required to be invariant under a specific set of transformations (Gauge transformations for Yang-Mills and diffeomorphisms for Einstein gravity), so it seems like their actions should be of a similar form (I know that gravity isn't exactly the gauge theory of diffeomorphisms, but this still seems odd to me). For instance, is there an a priori reason why we don't write down

$$S=\frac{1}{2\kappa'}\int\text{tr}\left[R\wedge\star R\right]=\frac{1}{4\kappa'}\int\mathrm{d}^dx\sqrt{g}\,g^{\alpha\mu}g^{\beta\nu}R^{\rho}_{\,\,\alpha\sigma\beta}R^{\sigma}_{\,\,\mu\rho\nu}$$

instead of the obvious reason that it doesn't give the Einstein equations?

Any insight into this would be excellent. Thanks!

Note: This question is essentially the last part of Is GR the Gauge Theory of Diffeomorphisms? Why is the EH action linear in the Curvature?. However, this question was never answered and was closed since the first few parts were essentially duplicates.

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    $\begingroup$ 1. Diffeomorphisms and gauge transformations are of very different nature (though there are some similarities, of course). 2. I don't get your question, why do you expect them to be of the same order in curvature? Gravity and gauge interactions are different. 3. Note that the Hodge star is external in the Yang-Mills theory, while in gravity it also depends on the dynamical field $g_{\mu \nu}$. Thus, even your action doesn't fully resemble the Yang-Mills action. $\endgroup$ Commented Jun 27, 2017 at 3:54
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    $\begingroup$ Related: physics.stackexchange.com/a/340466/92058. Weinberg says "The analogy breaks down in one important respect: in general relativity the affine connection is itself constructed from first derivatives of the metric tensor, while in gauge theories the gauge fields are not expressed in terms of any more fundamental fields ...". BTW, what equations of motion do you get from your proposed Lagrangian $R^2$ instead of $R$? $\endgroup$
    – tparker
    Commented Jul 12, 2017 at 15:10
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    $\begingroup$ It's worth noting here that the Ricci tensor is symmetric and can in principle be decomposed as the product of two antisymmetric matrices, through a variation of the cholesky decomposition. Then tracing either one to a lagrangian scalar more closely resembles what you are seeking $\endgroup$
    – R. Rankin
    Commented Feb 24, 2018 at 4:45

3 Answers 3


In Yang-Mills, the gauge connection plays the role of a potential and the curvature form plays the role of a "field strength".

In GR, the metric tensor plays the role of a potential, and the connection plays the role of a field strength.

This is why, in particular, the gravitational force is not a real force, as the connection is not gauge-covariant. Of course, we say that there is nonzero gravity somewhere, if the curvature is nonzero there, but that's because the curvature tensor obstructs the trivialization of the connection.

Furthermore, you should look into Ostrogradskij-instability. Because in YM, the gauge connection is the potential, and the curvature is the field strength, a lagrangian containing any function of the field strength will procude second order field equations at most.

But for gravity, the curvature tensor contains second derivatives of the metric, so if you don't restrict the curvature expression's form, Ostrogradskij instability will kick in. A lagrangian that is linear in the second metric derivatives avoids this issue.


Your are raising a legitimate question, as from an effective field theory point of view, all symmetry-permitting action terms should be included.

However, there is a catch when it comes to an action linear in Yang-Mills curvature $F$: you can't possibly contract the Lorentz indices of antisymmetric $F$ with symmetric metric $g$.

For special unitary group related YM fields there is another hindrance: $Tr\langle F\rangle $ is identically zero since special unitary groups are traceless, while the YM action (with two Fs) does not experience the said issue.

The action you mentioned with two Riemann curvature $R$ tensors is perfectly allowed, only that this term is highly suppressed at low energies so that it is not relevant under normal circumstances. Note that these high order $R$ terms play an important role at the embryonic stage of our universe, which is usually overlooked for the prevalent cosmological models.


Not so fast, there! You're jumping to conclusions. Who said you couldn't write the action as quadratic in a gauge field strength?

More precisely, though you may not necessarily be able to do that for the Einstein-Hilbert action without a cosmological coefficient, that's not believed to be physically relevant anymore: there's a non-zero cosmological coefficient present. You can do it, if the cosmological coefficient is not zero, which is the physically relevant case.

In that case, the requirement that the action be quadratic in the field strengths carries, with it, the prediction (after the fact, so "retrodiction") that the cosmological coefficient be non-zero.

An Einstein-Hilbert action with a cosmological coefficient has the form: $$S = \int ε_{abcd} \left(K Ω^{ab} + L θ^aθ^b\right)θ^cθ^d,$$ for some coefficients $K$ and $L$, where $θ^a = h^a_μ dx^μ$ are the frame one-forms, $Ω^{ab} = ½ {R^{ab}}_{μν} dx^μ dx^ν$ the curvature two-forms, indexes are raised and lowered with the frame metric $η_{ab}$ and $ε_{abcd}$ is completely anti-symmetric with $ε_{0123} = 1$.

For convenience I'm writing Grassmann products as juxtaposition without the wedge here and below; e.g. $θ^aθ^b$, instead of $θ^a ∧ θ^b$; i.e. as anti-symmetric products $θ^aθ^b = -θ^bθ^a$, in the case of odd-degree forms.

The Cartan structure equations, with upstairs indexes, are: $$dθ^a + ω^{ab}η_{bc}θ^c = Θ^a, \hspace 1em dω^{ab} + ω^{ac}η_{cd}ω^{db} = Ω^{ab},$$ where $$ω^{ab} = Γ^α_{μc}η^{cb}dx^μ, \hspace 1em Θ^a = ½ T^a_{μν} dx^μdx^ν$$ are, respectively, the connection one-forms and and torsion two-forms.

In Riemann-Cartan geometries, the frame metric $η_{ab}$(which is constant) is non-degenerate, with inverse $η^{ab}$ and is constrained to have zero covariant derivatives, which entails the following equations: $$Η^{ab} ≡ dη^{ab} + ω^{ab} + ω^{ba} = 0 \hspace 1em⇒\hspace 1em 0 = dH^{ab} + ω^{ac}η_{cd}H^{db} + ω^{bc}η_{cd}H^{ad} = Ω^{ab} + Ω^{ba},$$ and makes the connection one-form and curvature two-form anti-symmetric in their frame indices: $ω^{ab} = -ω^{ba}$ and $Ω^{ab} = -Ω^{ba}$. This also entails zero covariant derivative on the space-time metric $g_{μν} = η_{ab}h^a_μh^b_ν$.

You can either constrain the torsion to be zero, in which case you have the Palatini action with a cosmological coefficient - which is just Einstein-Hilbert, itself, (with cosmological coefficient) transcribed from Riemannian geometry to Riemann-Cartan geometry, or else you can leave it unconstrained, in which case you have the Einstein-Cartan action with cosmological coefficient ... where zero torsion will follow dynamically, at least outside of matter sources.

The gauge potential, here, is $$A = θ^aP_a + ½ω^{ab}S_{ab},$$ where we'll adopt the convention of treating products of the underlying Lie basis $P_a$ and $S_{ab}$ as freely interspersed with $dx^μ$'s, for convenience, and will work within an algebra where the Lie bracket can be expressed as $[u,v] = uv - vu$. Then, we may write $$A^2 = ½θ^aθ^c\left[P_a,P_c\right] + ⅓θ^aω^{cd}\left[P_a,S_{cd}\right] + ¼ω^{ab}θ^c\left[S_{ab},P_c\right] + ⅛ω^{ab}ω^{cd}\left[S_{ab},S_{cd}\right],$$ with the anti-symmetry of the form products, e.g. $$θ^aP_aθ^cP_c = θ^aθ^cP_aP_c = ½\left(θ^aθ^c - θ^cθ^a\right)P_aP_c = ½θ^aθ^c\left(P_aP_c - P_cP_a\right) = ½θ^aθ^c\left[P_a,P_c\right].$$

Adopting the following Lie brackets $$\begin{align} \left[P_a,P_c\right] &= λS_{ac}, & \left[P_a,S_{cd}\right] &= η_{ac}P_d - η_{ad}P_c, \\ \left[S_{ab},P_c\right] &= η_{bc}P_a - η_{ac}P_b, & \left[S_{ab},S_{cd}\right] &= η_{bc}S_{ad} - η_{bd}S_{ac} - η_{ac}S_{bd} + η_{ad}S_{bc}, \end{align}$$ for some constant $λ$, then for the field strength, we have: $$F = dA + A^2 = Θ^aP_a + ½\left(Ω^{ab} + λθ^aθ^b\right)S_{ab}.$$

Therefore, an action of a form quadratic in the field strengths $$\begin{align} S &= \int M ε_{abcd}\left(Ω^{ab} + λθ^aθ^b\right)\left(Ω^{cd} + λθ^cθ^d\right) \\ &= \int ε_{abcd}\left(M Ω^{ab}Ω^{cd} + 2Mλ Ω^{ab}θ^cθ^d + Mλ^2 θ^aθ^bθ^cθ^d\right) \end{align}$$ reduces equivalently to $$ S = \int ε_{abcd}\left(2Mλ Ω^{ab}θ^cθ^d + Mλ^2 θ^aθ^bθ^cθ^d\right) = \int ε_{abcd}\left(K Ω^{ab}θ^cθ^d + L θ^aθ^bθ^cθ^d\right) $$ provided that $$M = \frac{K^2}{4L}, \hspace 1em λ = \frac{2L}{K}.$$ because $ε_{abcd}Ω^{ab}Ω^{cd}$ is a boundary term and drops out from the analysis.

I'm not the only one to notice this. Just name the Lie algebra, A, whose Lie brackets I just listed above and do a web search on "A-gravity", whatever "A" might be. There's going to be something out there on it.

This also provides room for further expansion. You can just as well consider more fully quadratic actions of the form: $$S = \int \left(M ε_{abcd} + N η_{ac}η_{bd}\right)\left(Ω^{ab} + λθ^aθ^b\right)\left(Ω^{cd} + λθ^cθ^d\right) + O η_{ab}Θ^aΘ^b$$ and see what results from that. Now you have an "Immirzi" coefficient somewhere in there, too.


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