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Can anyone think of a conformal Killing vector field $W$ on $S^2$ with the round metric that is not Killing such that its divergence is $L^2$-orthogonal to the spherical harmonics with $\ell = 1$?

One conformal Killing vector field is $W = \sin(\phi) \partial_{\phi}$, but we have

\begin{align} \int_{S^2} \mathrm{div}(W) \, \, Y^{m=0}_{\ell=1} &= \int_{S^2} \mathrm{div}(W) \cos(\phi)\\ &=-\int_{S^2} W \cdot d(\cos(\phi))\\ &= \int_{S^2}\sin^2(\phi) \neq 0 \end{align}

We know that we can find three orthogonal conformal Killing vector fields $W_1,...,W_3$. Is it the case that we can choose $W_1$,...,$W_3$ so that their divergence coincides with the three spherical harmonics with $\ell = 1$? (if yes, then the answer to my first question is no).

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    $\begingroup$ Any conformal-KVF is $K + a^T$ for $a\in \mathbb{R}^3$ constant and $K$ a KVF. You can check that $\textrm{Div}_{S^2}(K+a^T) = - 2 a \cdot x$ which is a first spherical harmonic. $\endgroup$
    – Otis Chodosh
    Commented Dec 15, 2021 at 18:15
  • $\begingroup$ @OtisChodosh Thank you :) How do we see that every conformal KVF is of the form $K + a^T$? It looks like the other way around is true too: for any $F$ that is a linear combination of $Y^{-1}_{1}, Y^0_1, Y^1_1$, there exists a unique conformal KVF with divergence equal $F$. Is that correct? $\endgroup$
    – Laithy
    Commented Dec 15, 2021 at 23:15
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    $\begingroup$ You know the dimension of the space of conformal KVF is 6, right? Given this, $K+a^T$ generates a $6$ dimensional space of KVF, so the dimensions match up. By $Y^{-1}_1$ etc you mean 1-st spherical harmonics, right? If so, then any 1-st spherical harmonic is $a^T$ for some $a$, so the answer is yes. $\endgroup$
    – Otis Chodosh
    Commented Dec 16, 2021 at 0:46

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