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.

  • $\begingroup$ 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. $\endgroup$ May 25, 2016 at 1:44
  • $\begingroup$ Do you have a candidate for a counterexample? $\endgroup$ May 25, 2016 at 12:35
  • $\begingroup$ @AntonPetrunin: No, I don't have a candidate in mind. $\endgroup$ May 25, 2016 at 17:33

2 Answers 2


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

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

See the paper https://projecteuclid.org/download/pdf_1/euclid.jdg/1214448444

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 $M-X_{sing}$

  • 6
    $\begingroup$ 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 $\endgroup$
    – user21574
    Jul 24, 2017 at 7:14
  • 5
    $\begingroup$ 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 $\endgroup$
    – user21574
    Nov 15, 2017 at 9:36
  • 3
    $\begingroup$ 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. $\endgroup$
    – user21574
    Nov 21, 2017 at 17:21
  • 4
    $\begingroup$ 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/…. $\endgroup$
    – user21574
    Nov 21, 2017 at 17:41
  • 4
    $\begingroup$ 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 $\endgroup$
    – user21574
    Nov 22, 2017 at 15:46

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.