This may be overtly obvious or simple and I'm being very dense, but it's something that has been bothering me. I am confused about how correlation functions of generic spin operators work in 2-d CFTs. There is a formula that is quoted in many texts (e.g. diFrancesco's text, see section 5.1-5.2) for the four point function of operators with weights $(h_i, \bar{h}_i)$:

\(\langle \phi_1(z_1,\bar{z}_1) \phi_2(z_2,\bar{z}_2) \phi_3(z_3,\bar{z}_3) \phi_4(z_4,\bar{z}_4)\rangle = f(\eta,\bar{\eta}) \prod z_{ij}^{h/3-h_i-h_j} \bar{z}_{ij}^{\bar{h}/3-\bar{h}_i - \bar{h}_j}\)

where the product is supposed to be for $i<j$ up to 4 and $h = \sum_i h_i$. The function $f$ is allowed to depend arbitrarily on the conformally invariant cross ratio $\eta$ and its conjugate. Here lies the crux of my problem:

If we consider any tensor; for example, $D^{\mu \nu \sigma}$, which denotes the four point function of a spin-3 current and three scalars, then it is clear that under the change to the complex coordinates, we will have a tensor $D^{abc}$ where $a,b,c$ may denote either $z$ or $\bar{z}$. If the current is symmetric, then of course we may eliminate all but 4 degrees of freedom. The issue now is the following: how does the above four point function capture all of these independent components? Setting $(h_1,\bar{h}_1) = (h_1, h_1-3)$, and the rest scalars so that $\bar{h}=h$, we clearly only have an expression for one of the components. Which one is it? What happens to the other degrees of freedom? Must we assume tracelessness?