In the review of cosmological perturbations by Mukhanov et al linked:
https://doi.org/10.1016/0370-1573(92)90044-Z
on page 212 of the pdf, they introduce a symmetric, trace-free, rank-2 tensor perturbation $h_{ij}$ with constraints $h^i_i = 0$ and $\nabla^jh_{ij} = 0$. He claims that the rationale behind these constraints is that this does away with the scalar and vector components, thus making it truly tensorial.
I do not understand how the latter condition reconciles with the 'vector' component of rank-2 tensors usually being interpreted as the anti-symmetric component.
Any rank-2 tensor under the action of rotations (as is our interest in cosmology) can be decomposed into three irreducible components: an anti-symmetric part, and a symmetric part that is further broken down into a trace-free and scalar (trace) part:
$$X_{ab} = X_{[ab]} + \frac{1}{n}X\delta_{ab} + (X_{(ab)} - \frac{1}{n}X\delta_{ab}).$$
The trace transforms as a scalar under rotations; while the 3 independent components of the anti-symmetric part transform like a vector. Finally, the five independent components of the symmetric, trace-free part transform like a tensor.
How does setting $\nabla^jh_{ij} = 0$ reconcile with this interpretation?
In response to the answer by knzhou:
This makes sense, in that you're decomposing the symmetric perturbation tensor into its scalar, purely vector and purely tensor components. The first term accounts for the trace; the remaining are trace-free. The claim is then that the trace-free symmetric tensor can be broken down into components that transform like a vector and a tensor. The former is then broken down into the gradient of a scalar and a pure vector (the second and third terms respectively), with derivatives taken appropriately to tally indices. I just don't see how this exclusion of vector components happens via the derivative condition as imposed by Mukhanov et al.