Let us consider a noncompact Kähler manifold with vanishing scalar curvature but nonzero Ricci tensor. I'm wondering what can it tell us about the manifold. The example (coming from physics) has the following Kähler form

$$K = \bar{X} X + \bar{Y} Y + \log(\bar{X} X + \bar{Y} Y)$$

e.g. this is a 2D complex manifold. I claim that its Ricci form is nonzero, whereas its scalar curvature is identically zero.

I'm wondering if such manifolds possess any interesting properties and how can we classify them.

**UPD**.

Partly the answer for 4 manifolds (2d complex manifolds) is given in the paper by C Lebrun "Counter-examples to the generalized positive action conjecture'' paper. The author considers vanishing scalar curvature and derives the most generic form of the Kähler potential such that it vanishes. There are several integration constants in the final answer, playing with them we can get different manifolds including the one I was talking above. For that case the Kähler metric is the metric of a standard blow-up in the origin

$$K = \bar{X}X+\bar{Y}Y+a\log(\bar{X}X+\bar{Y}Y)$$

where $a>0$.

Now one can ask the same question about manifolds of higher dimension if they all with vanishing scalar curvature (but nonvanishing Ricci tensor) are described by the blow-ups of $\mathbb{C}^n$'s. In particular, I'm interested in the following Kähler potential

$$K = \sum\limits_{i=1}^N \sum\limits_{i=1}^{\tilde N}|X^i Y^j|^2 + a \log \sum\limits_{i=1}^N|X^i|^2.$$

mathsite, so you really should make an effort to follow the terminology mathematicians use. Otherwise, you're creating unneeded confusion. Your question uses the phrase "Ricci curvature", which for any mathematician means the same as "Ricci curvature tensor" and definitely not "Ricci scalar curvature". And I would avoid "Ricci scalar curvature" here. Just say "scalar curvature". $\endgroup$ – Deane Yang Jan 27 '11 at 15:077more comments