Let $\mathfrak{X}\_{CK}^{\perp}$ be the space of vector fields on $S^2$ that are $L^2$-orthogonal to conformal Killing vector fields. Let $\mathfrak{X}\_{CK}$ be the 6-dimensional space of conformal Killing vector fields on $S^2$.


Can we find a vector field $Y \in \mathfrak{X}\_{CK}^{\perp}$ and a vector field $W \in \mathfrak{X}\_{CK}$ that is not Killing such that


$$\mathrm{div}(Y) = \mathrm{div}(W)$$



Finding vector fields on $S^2$ with equal divergence
I think that this is not possible:


Per my comment on [Divergence of conformal Killing vector fields on $S^2$ and the spherical harmonics](https://mathoverflow.net/questions/410824/divergence-of-conformal-killing-vector-fields-on-s2-and-the-spherical-harmoni?noredirect=1#comment1054264_410824) you want to solve
$$
\textrm{div} (Y) = -2a\cdot x
$$
for $Y$ orthogonal to conformal-KVF's and $a \in \mathbb{R}^3$ fixed (nonzero). Suppose you can do this. Then, we find that
$$
Y = a^T + W
$$
for $W$ divergence free. From this question <https://math.stackexchange.com/questions/1898371/divergence-free-vector-field-on-a-2-sphere> you can see that $W = J \nabla f$ for some function $f \in C^\infty(S^2)$ (for $J$ the complex structure on $S^2$). Now, by assumption, $Y$ is orthogonal to $a^T$:
\begin{align\*}
0 & = \int\_{S^2} Y\cdot a^T \\
& = \int\_{S^2}|a^T|^2 + (J \nabla f) \cdot a^T \\
& = \int\_{S^2}|a^T|^2 - \nabla f \cdot J a^T \\
& = \int\_{S^2}|a^T|^2 + f \textrm{div}(J a^T). 
\end{align\*}
However, note that $Ja^T$ is a KVF and thus divergence free. This implies that $\int\_{S^2}|a^T|^2 = 0$. This is a contradiction.


