Any rank-2 tensor can be decomposed in the following way

$$ \phi_{\mu\nu} =\phi_{\mu\nu}^{TT} + \partial_{(\mu}\xi_{\nu)} +\frac{1}{4}T_{\mu\nu}s+\frac{1}{4}L_{\mu\nu}(w-3s) $$

where $\phi_{\mu\nu}^{TT}$ is transverse and traceless (i.e. ${\phi_{\mu}^{\mu}}^{TT} = 0$ and $\partial^\mu\phi_{\mu\nu}^{TT} = 0$), $\xi_{\nu}$ is transverse $\partial_\mu \xi^{\mu}=0$, $ s$ and $w$ are two scalar functions and the operators $T_{\mu\nu}\,,L_{\mu\nu}$ are defined as projectors

$$ T_{\mu\nu} = \eta_{\mu\nu}-\frac{\partial_\mu\partial_\nu}{\square}\,,\qquad L_{\mu\nu} = \frac{\partial_\mu\partial_\nu}{\square}. $$

Let us suppose we write an action for this rank-2 field whose free part implies the following equation of motion $\square s=0$.

The following interaction in terms of $\phi_{\mu\nu}$ looks local

$$ \partial_\mu\phi_{\nu\rho}\partial^\mu\phi^{\nu\rho}\phi^2\,,\qquad\qquad\qquad \text{Eq.}\,(1) $$

where $\phi^2 = \phi^{\alpha}_{\alpha} \phi^{\beta}_{\beta} = w^2/16$. If instead I look at interactions involving $s$ and $w$ I get terms like

$$ \left(\eta_{\nu\rho}\partial_\mu s - 4\frac{\partial_\mu\partial_\nu\partial_\rho}{\square}s\right)^2w^2 =-4(\partial s)^2w^2 +16\frac{(\partial_{\mu\nu\rho}s\partial^{\mu\nu\rho}s )}{\square^2}w^2\,,\qquad \text{Eq.}\,(2) $$

which looks non-local. However, if we use the leading equation of motion for $s$ (i.e. $\square s=0$) this interaction looks local; Indeed

$$ \partial_{\mu}s\partial^{\mu}s = \frac{1}{2}\square(s^2) = \partial_{\mu}s\partial^{\mu}s + 2s\square s = \partial_{\mu}s\partial^{\mu}s. $$ and the $\square^2$ in the denominator of the second term in Eq.(2) cancels.

This operation is allowed if we look at on-shell scattering amplitudes of the scalar modes $s$, in which case $s$ always do not propagate as an internal leg. However, this is not always the case: e.g. a loop of $s$ would contribute to the $ww\rightarrow ww$ amplitude.

So, my question is the following: Eq.(1) does not manifest non-local interactions, whereas Eq.(2) apparently does. How do I solve this contradiction? What am I missing? Probably I am making confusion on some basic concept.


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