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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 $W$).

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    $\begingroup$ Actually the map from Möbius transformations to the space $\mathcal{A}$ is not one-to-one: The map has $3$-dimensional fibers because any metric of constant curvature $1$ on the $2$-sphere has a $3$-dimensional space of isometries, which are conformal, and, hence, the orientation preserving ones are Möbius transformations. Thus $\mathcal{A}$ is actually a $3$-disc and not $6$-dimensional. $\endgroup$
    – Robert Bryant
    Commented Jan 8, 2021 at 18:06
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    $\begingroup$ Also, the answer to your question is 'no' because, if $W$ is any nonzero vector field on $S^2$, then there is a $1$-form $\omega$ on $S^2$ such that $\omega(W)\ge0$ and it only vanishes at the zeros of $W$. Then $\int_{S^2}f^2\,\omega(W) >0$ for all positive functions $f$ and, in particular for all $f\in \mathcal{A}$. $\endgroup$
    – Robert Bryant
    Commented Jan 8, 2021 at 18:11
  • $\begingroup$ Oh thank you! What if $\omega$ is exact? $\endgroup$
    – Laithy
    Commented Jan 8, 2021 at 18:42
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    $\begingroup$ In standard Spherical coordinates $(\phi,\theta)$ with $\phi\in [0,\pi]$ and $\theta\in [0,2\pi)$, one of the conformal killing fields of the sphere is $\sin(\phi) \partial_\phi$. On the other hand, you also have $\sin(\phi) d\phi = d(\cos(\phi))$ is a smooth, exact one form. Their contraction is strictly positive except at the poles. $\endgroup$
    – Willie Wong
    Commented Jan 8, 2021 at 19:42

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