Consider the space $\mathcal{A}$ of functions $\Omega$ such that $\Omega^2 \gamma_0$ is isometric to the round sphere, where $\gamma_0$ is the round sphere. (so $\Omega^2 \gamma_0$ is of constant curvature 1). Note that this is related to the lack of uniqueness of the uniformization theorem. Also, there is a natural 1-1 correspondence between $\mathcal{A}$ and Mobius transformations of the sphere (which is a 6-parameter family of conformal diffeomorphisms of the round sphere). That correspondence is defined in the following way: for each Mobius transformation $f$, it corresponds to $\Omega \in \mathcal{A}$ satisfying, $f^*(\gamma_0) = \Omega^2 \gamma_0$.

Consider the space conformal killing vector fields on the round sphere, which make a 6-dimensional vector space.

Given an arbitrary 1-form $\omega$ on $S^2$, does there exist $\Omega \in \mathcal{A}$ such that $\Omega^2 \omega$ is $L^2$ orthogonal to conformal killing vector fields? (which means that $\int_{S^2} \Omega^2 \omega_i W^i = 0$ for all conformal killing vectorfields $W$).