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Given a compact Kähler manifold $M$, let $D$ be an effective divisor on $M$.

  1. Is $M\setminus D$ pseudoconvex? That is, can we find a smooth plurisubharmonic function that exhausts $M\setminus D$ ?
  2. Can we find a complete Kähler metric on $M\setminus D$ ?

If $1$ and $2$ are not true, can we find any obstructions? Or necessary and sufficient conditions?

Note: if $D$ is an ample divisor, we can choose an Hermitian metric on $[D]$ which is the associated divisor bundle to $D$, then we can take $-ln|s_{D}|^{2}_{h}$ to be the strongly smooth exhaust plurisubharmonic function on $M\setminus D$. Surely, this is just a Stein (or Affine) manifold. Given such a plurisubharmonic function on $M\setminus D$ we can easily contruct a compelte Kähler metric.

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    $\begingroup$ For part 2 of your question see my answer and my comments here mathoverflow.net/questions/239677/… $\endgroup$
    – user21574
    Commented Nov 25, 2017 at 22:18
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    $\begingroup$ If you mean $M\setminus D$ is strongly pseudoconvex domain? then such variety must admit complete Kahler Einstein metric due to Cheng-Yau and see Tsuji paper also arxiv.org/pdf/1311.4038.pdf which such $X\setminus D$ does not admit Kahler-Einstein metric in general, so it can not be strongly pseudoconvex domain in general. $\endgroup$
    – user21574
    Commented Nov 25, 2017 at 22:28
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    $\begingroup$ If $M$ is compact and $D\subset M$ is compact and has complex $\text{codim}\geq 2$, then $M\setminus D$ is not Stein, hence it cannot be strongly pseudoconvex domain $\endgroup$
    – user21574
    Commented Nov 25, 2017 at 22:47
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    $\begingroup$ Moreover about your first part of your question: An strong result of Griffiths show that a domain $\Omega\subset \mathbb C^n$which admits a complete Hermitian metric with non-positive holomorphic sectional curvature is pseudoconvex. link.springer.com/article/10.1007/BF01418742 and also see Shiffman paper link.springer.com/article/10.1007/BF01350128 $\endgroup$
    – user21574
    Commented Nov 28, 2017 at 4:50
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    $\begingroup$ Assume $M$ is a Stein manifold and $ D ⊂ M$ is a closed subset. If $M \setminus D$ locally admits complete Kahler metrics induced by bounded plurisubharmonic functions, then there exists a complete Kahler metric on $M \setminus D$ induced by a globally defined plurisubharmonic function on $M$ see Proposition 2.1. of arxiv.org/pdf/1505.06451.pdf $\endgroup$
    – user21574
    Commented Nov 29, 2017 at 5:21

2 Answers 2

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Question 1: Plurisubharmonic functions extend across codimension 2 subvarieties . Let X be the complex projective plane blown up at one point and D be the exceptional divisor then any plurisubharmonic function on the complement of D in X extends to X and is therefore a constant.

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  • $\begingroup$ Hi, Mohan, this is a really nice answer. I would appreciate it if you can provide me the proof for the fact that plurisubharmonic functions extend across codimension 2 singularities. Thank you. $\endgroup$
    – user40184
    Commented Sep 17, 2016 at 12:52
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    $\begingroup$ Thank you. See Grauert Remmert Mathematische Zeitschrift vol 65 page 175-194 or Noguchi and Ochiai 's book Geometric Function Theory of Several Complex Variables page 136 . The idea of the proof is to show that psh functions locally bounded above extend past pluripolar sets and then show psh function are locally bounded above in case of codimension 2 singularities $\endgroup$
    – Mohan Ramachandran
    Commented Sep 18, 2016 at 16:41
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For 1 it is not always the case.

It is well-known, that critical points of a plurisubharmonic Morse function on a complex manifold $X$ have Morse index $\le \dim_\mathbb C X$. Hence, if $X=M\backslash D$ admits a plurisubharmonic exhaustion, its homology vanishes in dimensions $>\dim_\mathbb CM$.

At the same time it is easy to come up with with a pair $(M, D)$ such that $\dim_\mathbb C M=2$ while $H^3(M\backslash D)\neq 0$:

For example, take $M={\rm Bl}_p T^4$ to be a blow-up of a torus in a point and let $D$ be the exceptional divisor. Then $M\backslash D=T^4\backslash p$.

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  • $\begingroup$ Hi, Yury, thanks for ur answer. But what about the case when the plurisubharmonic function is not Morse? $\endgroup$
    – user40184
    Commented Sep 16, 2016 at 2:48
  • $\begingroup$ If $f$ is a psh exhaustion of $X$, then for an appropriately small (and decaying at infinity) function $g$ the sum $f+g$ is also psh. If you choose $g$ generic enough, the sum $f+g$ should be Morse. $\endgroup$
    – Yury Ustinovskiy
    Commented Sep 16, 2016 at 12:35
  • $\begingroup$ Hi, Yury, I am sorry but I do not get it. By being plurisubharmonic, I mean the complex hessian of the function is nonnegative definite which is a closed condition. So I do not see why the perturbation works here. $\endgroup$
    – user40184
    Commented Sep 16, 2016 at 13:51
  • $\begingroup$ Sorry, my bad, i had in mind strongly psh functions, like for Stein manifolds. Well, it might be more subtle in this case, but still i believe that 1 does not hold. $\endgroup$
    – Yury Ustinovskiy
    Commented Sep 16, 2016 at 14:02

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