Let $\gamma$ be a metric on $S^2$. I am trying to solve the following PDE on a $(0,2)$ symmetric traceless tensor $A$: $$div_{\gamma} A = \omega$$ where $\omega$ is a 1-form.
It is known that there exists a unique solution if and only if $\omega$ is $L^2$ orthogonal to conformal killing (CK) vector fields on $(S^2,\gamma)$ (the Lie algebra of the conformal group of $S^2$).
It is also known that if $\gamma$ is the round sphere, then the space of CK vector fields is a 6-dim space.
If $\gamma$ is not the round metric, what can we say about the space of CK vector fields? Will it also be 6-dimensional? I think that if $\gamma$ is of negative curvature, then it won't admit any CK vector fields and so the divergence operator is an isomorphism onto the space of 1-forms. What about if $\gamma$ is of positive curvature?
Using the uniformization theorem, we know there exists an $f$ (in a 3-dimensional space I think?) such that $\gamma = f^2 \gamma_0$ where $\gamma_0$ is the round sphere, and so $div_{\gamma} = \frac{1}{f^2} div_{\gamma_0}$. So we need $f^2 \omega$ to be $L^2$ orthogonal to conformal killing vector fields on the round sphere. Can we utilize the freedom of choosing $f$ to figure out precisely which $\omega$ will work?
It seems to me that for any CK vector field $X$ with respect to $\gamma_0$, we have $\mathcal{L}_X \gamma = f^2 \mathcal{L}_X \gamma_0 + X(f^2) \gamma_0 = (h + \frac{X(f^2)}{f^2})\gamma$ where $h$ satisfies $\mathcal{L}_X \gamma_0 = h \gamma_0$. So $X$ is CK with respect to $\gamma$ also. But that seems to say that the space of CK vector fields doesn't depend on the metric. That must be wrong because $\gamma$ might be of negative curvature, which doesn't admit CK vector fields. What am I doing wrong? (There is probably something fundamental that I am not understanding).
Any help or references is appreciated.
