3
$\begingroup$

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 complex-valued 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 real-valued function?

$\endgroup$
  • 1
    $\begingroup$ Please do not cross-post to math.SE without saying so. That way, we can avoid duplication of effort. $\endgroup$ – S. Carnahan Sep 24 '13 at 0:14
  • $\begingroup$ sorry about that.should I delete one of them? $\endgroup$ – Vicky Cheung Sep 24 '13 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 '13 at 2:17
  • 1
    $\begingroup$ math.stackexchange.com/questions/498782/… $\endgroup$ – Vicky Cheung Sep 25 '13 at 23:45
  • $\begingroup$ See my answer mathoverflow.net/questions/143575/… $\endgroup$ – user21574 Jul 17 '17 at 0:20
3
$\begingroup$

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 1-eigenfunction 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.