Timeline for Finding $\Omega$ such that the 1-form $\Omega^2 \omega$ is $L^2$ orthogonal to conformal killing vector fields on $S^2$

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Jan 8, 2021 at 19:42	comment	added	Willie Wong		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.
Jan 8, 2021 at 18:42	comment	added	Laithy		Oh thank you! What if $\omega$ is exact?
Jan 8, 2021 at 18:11	comment	added	Robert Bryant		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 $f\in \mathcal{A}$.
Jan 8, 2021 at 18:06	comment	added	Robert Bryant		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.
Jan 8, 2021 at 17:51	history	edited	Laithy		edited tags
Jan 8, 2021 at 17:33	history	asked	Laithy	CC BY-SA 4.0