Let $M$ be a compact Fano Kähler–Einstein manifold, and $V$ a holomorphic $(1,0)$ vector field on $M$. The Fano conditions say that $V = \nabla^{1,0} f$ for some smooth complexvalued function. By Matsushima's theorem, the Kähler–Einstein condition implies that $\operatorname{div} V = \Delta f$ is an eigenfunction of the complex Laplacian $\Delta$. My question is: would it be possible to take the potential function $f$ to be a realvalued function?

1$\begingroup$ Please do not crosspost to math.SE without saying so. That way, we can avoid duplication of effort. $\endgroup$– S. Carnahan ♦Sep 24, 2013 at 0:14

$\begingroup$ sorry about that.should I delete one of them? $\endgroup$– Vicky CheungSep 24, 2013 at 20:01

$\begingroup$ Just put up a link, so people can check to see if there is additional information or answers at the other site. $\endgroup$– S. Carnahan ♦Sep 25, 2013 at 2:17

1$\begingroup$ math.stackexchange.com/questions/498782/… $\endgroup$– Vicky CheungSep 25, 2013 at 23:45

$\begingroup$ See my answer mathoverflow.net/questions/143575/… $\endgroup$– user21574Jul 17, 2017 at 0:20
1 Answer
actually, I just figure out an argument. Take any hol'c (1.0) vector field, say X, then Fano implies $X=\nabla^{1,0}f$, with complex valued f. And $div X=\Delta f$. Now let f=u+iv, then since div X is 1eigenfunction of complex Laplacian, so is \Delta u and \Delta v, then by matsushima theorem, one knows that $\nabla^{1,0}(\Delta u)$ is also a hol'c v.f. Which has real potential $\delta u$. since the laplacian is an real operator in the kahler case.

$\begingroup$ See my answer mathoverflow.net/questions/143575/… $\endgroup$– user21574Jul 17, 2017 at 0:20