6

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.

  • 2
    If D is smooth (or merely snc, which you can restrict to thanks to Hironaka), you can take for 2. a metric of the form $\omega+dd^c(-\log(-\log |s|_h^2))$ for $\omega$ a background Kähler form on $X$. The resulting metric has Poincaré growth in the direction(s) transverse to the divisor. – Henri Sep 14 '16 at 04:03
  • 5
    For part 2 of your question see my answer and my comments here https://mathoverflow.net/questions/239677/does-a-k%C3%A4hler-manifold-always-admit-a-complete-k%C3%A4hler-metric/271277#comment708770_271277 –  Nov 25 '17 at 22:18
  • 4
    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 https://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. –  Nov 25 '17 at 22:28
  • 4
    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 –  Nov 25 '17 at 22:47
  • 5
    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. https://link.springer.com/article/10.1007/BF01418742 and also see Shiffman paper https://link.springer.com/article/10.1007/BF01350128 –  Nov 28 '17 at 04:50
  • 4
    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 https://arxiv.org/pdf/1505.06451.pdf –  Nov 29 '17 at 05:21
  • 3
    Moreover, Yau and Mok showed that a bounded domain which admits a complete Hermitian metric satisfying $−c≤\text{Ricci curvature} ≤0$ must be pseudoconvex. Which is parallel to result of Shiffman and Griffiths but in Ricci curvature setting instead sectional curvature setting see https://books.google.fr/books?id=9S4ECAAAQBAJ&printsec=frontcover –  Dec 03 '17 at 21:10
  • 3
    Also, by using embedding theorem of Bishop-Narasimhan, there exists a complete Kahler metric on $X-V$ , (by using Grauert construction of complete Kahler metric on $\mathbb C^n-{0}$)where $X$ is Stein manifold and $V$ is closed analytic subvariety of any dimension.I learnt it from the paper of Yau-Mok –  Dec 03 '17 at 21:17

2 Answers2

5

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.

Mohan Ramachandran
  • 4,111
  • 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. –  Sep 17 '16 at 12:52
  • 1
    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 – Mohan Ramachandran Sep 18 '16 at 16:41
1

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 $M\backslash D=T^4\backslash p$.

Yury Ustinovskiy
  • 1,520
  • 9
  • 13
  • Hi, Yury, thanks for ur answer. But what about the case when the plurisubharmonic function is not Morse? –  Sep 16 '16 at 02:48
  • 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. – Yury Ustinovskiy Sep 16 '16 at 12:35
  • 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. –  Sep 16 '16 at 13:51
  • 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. – Yury Ustinovskiy Sep 16 '16 at 14:02