# Variation with respect to a traceless symmetric tensor

Suppose we have an action variation like

$$\delta S[G]=\int \mathfrak{H}^{\mu\nu}\delta G_{\mu\nu} \,\, d^Nx,$$

where $$\mathfrak{H}^{\mu\nu}$$ is a tensor density. If the variation with respect to $$G_{\mu\nu}$$, for example, takes place in the subspace of symmetric tensors, we get

$$\mathfrak{H}^{(\mu\nu)}=0, \tag{1}$$

instead of $$\mathfrak{H}^{\mu\nu}=0$$ (for a general variation in the whole space of 2-covariant tensors).

I am trying to obtain the equations of motion of a tensor $$G_{\mu \nu}$$ which is symmetric but also traceless. Which is the formal way to do this?

----- A guess

I tried something like:

$$\begin{eqnarray} \mathfrak{H}^{\mu\nu}\delta_{st} G_{\mu\nu} &=&\mathfrak{H}^{\mu\nu}(\delta_{s} G_{\mu\nu} - \frac{1}{N}g_{\mu\nu} g^{\rho\sigma} \delta_{s} G_{\rho\sigma}) \\ &=& \mathfrak{H}^{\mu\nu}(\delta^\rho_\mu \delta^\sigma_\nu -\frac{1}{N}g_{\mu\nu} g^{\rho\sigma})\delta_{s} G_{\rho\sigma} \\ &=& (\mathfrak{H}^{\rho\sigma}-\frac{1}{N}\mathfrak{H}^\lambda{}_\lambda g^{\rho\sigma})\delta_{s} G_{\rho\sigma} \\ &=& (\mathfrak{H}^{(\rho\sigma)}-\frac{1}{N}\mathfrak{H}^\lambda{}_\lambda g^{\rho\sigma})\delta G_{\rho\sigma} \end{eqnarray}$$ where I used the index $$s$$ for "symmetric variation" and $$st$$ for "symmetric-traceless variation". So the equation of motion should be:

$$\mathfrak{H}^{(\rho\sigma)}-\frac{1}{N}\mathfrak{H}^\lambda{}_\lambda g^{\rho\sigma} =0.$$

Is this reasoning correct?