Reference for when a metric on a four-manifold is Kahler?

Question

In a paper of Derdzinski¹ (Proposition 5), he proved that if $\delta W^+=0$ and $W^+$ has at most two distinct eigenvalues, then the metric is (locally) conformally Kahler, and if in addition the scalar curvature or $|W^+|^2$ is constant, then the metric itself is Kahler.

I was wondering are there further characterization of Kahler metrics on four-manifolds? Thank you very much.

¹Derdziński, Andrzej: Self-dual Kähler manifolds and Einstein manifolds of dimension four, Compositio Mathematica, Volume 49 (1983) no. 3 , p. 405-433.

Answer 1

I reformulate you question as follows: suppose we a given a Riemannian metric. How can one decide whether there exists a complex structure $J$ such that the metric is Kähler w.r.t. this complex structure?

Below I will explain how to answer this question in practice, i.e., if say a Kähler metric is explicitly given. Actually, if the metric is given by explicit formulas, sometimes one can relatively easy give a negative answer, and in small dimensions as the rule one can also find the complex structure $J$ or prove its existence.

Easy version of the answer: Kähler metrics have the property that the curvatures commute with the complex structure: for any two vectors $X,Y$ we have $$ [R(X,Y), J]=0. \ \ \ (\ast) $$ Let us view this as as a system of linear equations on $J$ (the vectors $X$ and $Y$ are the one we choose freely, and we know the curvature since we know the metric).
It is a very overdetermined system, in the case it has no nontrivial (i.e., $\ne Id$) solution (which is a generic case) the metric is definitely not Kähler. In the case the space of solution is two-dimensional, one need to understant whether the solution has only two eigevalues and both eigenvalues are not real valued. This is again an algorithmically doable procedure. If not, then the metric is not Kähler. If yes, we have the only candidate for the complex structure and one simply need to check whether Nijenhuis is zero. In the case we have many solutions, one could try to enlarge the system $(\ast)$ to get only one solution by adding to it the equations $[(\nabla_XR)(Y,Z), J]=0$ and so on.

Formally correct but in practice not applyable version of the aswer. One need to consider the holonomy group of the metric, and look whether it preserves a complex structure.

Relation between formally correct and easy versions of the answer. By Ambrose-Singer, the holonomy group is generated by $R(X,Y)$ and also by $ (\tau^*R)(X,Y) $ over all parallel translations $\tau$ along all possible curves. Since $ (\tau^*R)(X,Y) $ is almost impossible to calculate in practice, in the easy version of the answer it is replaced by its derviative at the initial point, which is $(\nabla_XR)(Y,Z)$.

Answer 2

Apostolov, Davidov, Muskarov and Gauduchon (et al.) looked at questions of this sort a few years back.

You might find this paper interesting to you "Compact self-dual Hermitian surfaces" by V. Apostolov, J. Davidov and O. Muskarov, Trans. Amer. Math. Soc. 348 (1996), 3051-3063

Abstract: In this paper, we obtain a classification (up to conformal equivalence) of the compact self-dual Hermitian surfaces. As an application, we prove that every compact Hermitian surface of pointwise constant holomorphic sectional curvature with respect to either the Riemannian or the Hermitian connection is Kähler. - See more at: http://www.ams.org/journals/tran/1996-348-08/S0002-9947-96-01585-1/#sthash.HQ8Zy1yx.dpuf