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Let M be a compact Fano Kähler-Einstein manifold, V is a holomorphic (1,0) vector field. Fano conditions tells that $V=\nabla^{1,0} f$ for some smooth complex valued function.By Matsushima theorem, Kähler-Einstein condition implies that $div V=\Delta f$ is an eigenfunction of complex laplacian $\Delta$. My question is would it be possible to take the potential fucntion $f$ to be a real valued function?

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  • $\begingroup$ Hi Vicky: if at any time you think the physics forum might stand a better chance of answering this, let us know and we can try to migrate it. $\endgroup$
    – rschwieb
    Sep 19, 2013 at 18:00
  • $\begingroup$ yes, please. If there is some idea from physics, that would be great. $\endgroup$ Sep 20, 2013 at 2:41
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    $\begingroup$ If you just figured out a solution, please take a moment to stick it into an answer to your own question (that's allowed!) That's preferable to letting this sit around with no solutions. Thanks! $\endgroup$
    – rschwieb
    Sep 20, 2013 at 12:19
  • $\begingroup$ i add the argument to the answer part $\endgroup$ Sep 20, 2013 at 20:28

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Actually, I just figured out an argument. Take any holomorphic $(1,0)$ vector field, say $X$, then Fano implies $X=\nabla^{1,0}f$, with complex valued $f$. And $\operatorname{div} X=\Delta f$. Now let $f=u+iv$, then since $\operatorname{div} X$ is a $1$-eigenfunction of the complex Laplacian, so are $\Delta u$ and $\Delta v$, then by Matsushima theorem, one knows that $\nabla^{1,0}(\Delta u)$ is also a holomorphic vector field Which has real potential $\delta u$. Since the Laplacian is an real operator in the Kähler case.

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