There is a saying that the nonlinear completion of free massless spin-2 action in Minkovski spacetime (that is Fierz-Pauli action) must be Einstein-Hilbert action up to Lovelock invariants.

I find a reference tackling this problem Wald R M. Spin-two fields and general covariance[J]. Physical Review D, 1986, 33(12): 3613. http://journals.aps.org/prd/abstract/10.1103/PhysRevD.33.3613

However in this paper it only proves that the nonlinear completion of free massless spin-2 action must be general covariant. So how to prove it must be the Einstein-Hilbert action.


1 Answer 1


If you have a result stating that the non-linear completion of a free massless spin-2 field must be general covariant, then that means the action must only depend on curvature invariants like $R$, $R_{ab}R^{ab}$, $R_{abcd}R^{abcd}$, ...

Because we started with the action for a free massless spin-2 field, we can narrow down the allowed dependence on the infinitely many curvature invariants to be simply the Einstein-Hilbert term, $S \propto \int d^4 \sqrt{-g} R$. This is because the action for a free spin-2 field has only two derivatives, which means that only 1 power of the Riemann tensor $R_{abcd}$ can enter into the action. And the only invariant that contains one single power of the Riemann tensor is the Ricci scalar. No terms like $R_{abcd}R^{abcd}$ can show up because if they did, the action would be more than second derivative.

Certain combinations of curvature invariants can be shown to be simply boundary terms that do not affect the equations of motion, hence the ability to add Love-lock invariants.

The flip side of this is the linearisation of higher derivative curvature corrections leads to additional massive fields, i.e.

$S = \int d^4 x\sqrt{g} \left( a_1 R + a_2 R^2 + a_3 R_{ab}R^{ab} + a_4 R_{abcd} R^{abcd} \right)$,

can be shown to contain a massless spin-2 (the usual graviton), a massive spin-0, and a massive spin-2, http://link.springer.com/article/10.1007%2FBF00760427.

  • $\begingroup$ The Lovelock terms become relevant if you are working in more than 4 dimensions. For example, in 5 dimensions the Lovelock term with two curvature tensors is no longer topological, and affects the equations of motion. The Lovelock terms are special because they do not lead to additional degrees of freedom: there will still be a single massless spin-2 degree of freedom with any of the Lovelock terms. Adding other curvature invariants will introduce new degrees of freedom, and most of the time they will be ghosts (with the exception of $f(R)$ theories). $\endgroup$
    – asperanz
    Mar 17, 2015 at 0:26

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