MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be a complex manifold of complex dimension 2. What do we know about the set all Kähler metrics on $M$ in general and in the case of 4-torus $C^2/Z^4$?

For the case of surfaces ($dim_C=1$), any compatible metric is Kähler and by the uniformization theorem, the answer is that every two such metrics are conformally equivalent and the set all Kähler metrics is nonempty.

share|cite|improve this question
up vote 4 down vote accepted

Let $M$ be a compact complex manifold. The set of Kahler metric on $M$ in a given Kahler class is parametrized by the set of positive volume forms with given integral, because (by Calabi-Yau theorem) any given positive volume form is a volume form of a Kahler metric in a given cohomology class, assuming their integrals agree. This is actually used when they put a structure of an infinite-dimensional symmetric space on the space of all Kahler metrics. See for example here: (THE SPACE OF KAHLER METRICS, by XIUXIONG CHEN).

share|cite|improve this answer
Ziegler's answer and yours complements each other in the following way. If our compact complex manifold admits any Kahler metric ever, then on can find all the metrics by first fixing a Kahler class in the Kahler cone and then changing the metric in the same class by choosing a smooth function. – Asghar Ghorbanpour Aug 14 '14 at 12:41

Compact case (since you mention $\mathbf C^2/\mathbf Z^4$):

  • For $M$ to be Kähler its first Betti number must be even, and conversely every compact complex surface with even $b_1(M)$ is Kähler (Kodaira's conjecture, proved by Siu, Lamari, Buchdahl).

  • Lamari and Buchdahl also describe "how many" Kähler metrics then exist, i.e. the so-called "Kähler cone" of classes in $H^{1,1}_{\mathbf R}(M)$ which can be represented by positive closed $(1,1)$-forms.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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