Does a Kähler manifold always admit a complete Kähler metric?

Every smooth manifold admits a complete Riemannian metric. In fact, every Riemannian metric is conformal to a complete Riemannian metric, see this note. What about in the Kähler case?

Does a Kähler manifold always admit a complete Kähler metric?

Of course, every metric on a compact manifold is complete, so the question is only of interest in the non-compact case.

One might hope that the proof in the aforementioned note will still be of use, but as is shown in this question, a metric conformal to a Kähler metric cannot be Kähler (with respect to the same complex structure) except in complex dimension one.

The only condition that I am aware of that ensures the existence of complete Kähler metrics is that the manifold is weakly pseuodoconvex (i.e. it admits a plurisubharmonic exhaustion function), see Demailly's Complex Algebraic and Analytic Geometry, Chapter VIII, Theorem 5.2.

• Is the existence of a complete Kaehler metric enough to endow the cohomology with a mixed Hodge structure? If so, that might be enough to rule out certain open Kaehler manifolds. – Jason Starr May 25 '16 at 1:44
• Do you have a candidate for a counterexample? – Anton Petrunin May 25 '16 at 12:35
• @AntonPetrunin: No, I don't have a candidate in mind. – Michael Albanese May 25 '16 at 17:33

Grauert proved that a relatively compact domain with real analytic boundary in C^n has a complete Kahler metric iff the boundary is pseudoconvex .For any relatively compact domain in C^n which is the interior of its closure ,with a complete Kahler metric, Diederich and Pflug showed that it is locally Stein. See the paper of Demailly in Annales ENS vol 15 1982 page 487 .

I give a related answer for the following non-compact case which we can get complete Kähler metric

Take $\overline M$ be a compact Kähler manifold and $Y\subset \overline M$ be the simple normal crossing divisor and take $M=\bar M\setminus Y$ now we can define complete Kähler metric $\omega_P$ on non-compact manifold $M$ as follows

Since $Y$ is simple normal crossing divisor , so it can be defined by the equation $z_1^{\alpha}\cdots z_{n_\alpha}^\alpha=0$

Take a cover for $\overline M=U_1\cup\cdots\cup U_p\cup \cdots \cup U_q$ such that $\overline{U_{p+1}}\cup\cdots \cup \overline{U_{q}}=\phi$

Let $\{η_i\}_{1≤i≤q}$ be the partition of unity subordinate to the cover $\{U_i\}_{1≤i≤q}$. Let $\omega$ be a Kähler metric on $M$ and let $C$ be a positive constant. Then for $C$ enough large, the following Kähler form is complete Kähler metric

$$\omega_P=C\omega+\sum_{i=1}^p\sqrt{-1}\partial\bar\partial\left(\eta_i\log\log\frac{1}{z_1^{i}\cdots z_{n_i}^i}\right)$$

Moreover Let $X$ be a singular subvariety of the compact Kähler manifold $M$ and let $\omega$ be a Kahler $(1,1)$ form on $M$ then the Saper-form

$$\omega_{Saper}=\omega-\frac{\sqrt{-1}}{2\pi}\partial\bar\partial \log(\log F)^2$$

is a complete Kähler metric on $M-X_{sing}$

• Let $M$ be a compact complex manifold and $S\subset M$ a proper analytic subset Let $\omega$ be a $d$-closed $(1, 1)$-current satisfying the conditions 1) $ω$ is smooth on $M-S$ , where $S$ is some proper analytic subset in $M$ 2)$ω > εσ$ in the sense of currents, where $ε > 0$ is some real number and $σ$ is a fixed positive definite $(1, 1)$-form (not necessarily d-closed) on $M$ , Then $M - S$ admits a complete Kahler metric. See lemma 4. 1 , msp.org/pjm/1993/158-2/pjm-v158-n2-p09-p.pdf – user21574 Jul 24 '17 at 7:14
• Let $V$ be a smooth projective variety and $D$ an ample divisor with simple normal crossings . Then the complement $A = V-D$ is a special affine variety. Griffiths showed that There exists a complete Kahler metric $\varphi$ on $A$ whose associated Ricci form satisfies $$Ric( \varphi)=O(\omega_{Poincare})$$ see (3.5) Proposition. Analytic Cycles and Vector Bundles on Non-Compact Algebraic Varieties. Maurizio Cornalba; Phillip Griffiths Inventiones mathematicae (1975) Volume: 28, page 1-106 ISSN: 0020-9910; 1432-1297/e eudml.org/doc/142315 – user21574 Nov 15 '17 at 9:36
• A Kahler manifold with a $C^\infty$ exhaustive pluri-subharmonic function is a complete Kahler manifold. So, Stein manifolds are complete Kahler manifolds. Let $D$ be a bounded domain with a smooth pseudoconvex boundary in a Kahler manifold. Then, $D$ is a complete jstage.jst.go.jp/article/kyotoms1969/20/1/20_1_21/_pdf . Ohsawa asked an interesting conjecture after Example 3 Kahler manifold. – user21574 Nov 21 '17 at 17:21
• About my previous comment for proof of "Every weakly pseudoconvex Kähler manifold $X$ carries a complete metric"see Proposition 14 math.u-psud.fr/~merker/Enseignement/Geometrie-complexe/Proietti/…. – user21574 Nov 21 '17 at 17:41
• Let $X$ be a compact Kahler Stein manifold and $Z$ be a analytic space in $X$, then $X∖Z$ pocesses a complete Kahler metric, see Theorem 0.2 of numdam.org/article/ASENS_1982_4_15_3_457_0.pdf See Theorem 1.5 also and Proposition of 1.6 which is as same as the result of Ohsawa – user21574 Nov 22 '17 at 15:46