Let $\phi$ be a stricly plurisubharmonic function on a domain in ${\Bbb C}^n$, and $S=\phi^{-1}(c)$ its level set. Consider $S$ as a Riemannian manifold equipped with a metric induced by $dd^c\phi$. I am interested in curvature restrictions on the Riemannian structure of $S$. In all examples I could check, its Ricci curvature is positive. Is it always true? What kind of restrictions we get?
Any ideas or reference would be appreciated.


I am thinking by analogy with a strictly convex function on a flat space: its Hessian is a metric, and its level set has (I think) positive curvature, though I don't have a formal proof of this either. I would appreciate a reference, or a refutation, if this is false.



Level set of convex and plurisubharmonic functions (Ricci curvature and other curvature conditions)
Well, I'm at the next airport and have a little time. While I don't have a complete answer worked out about the necessary and sufficient conditions on the curvature, I can definitely say that you don't get positivity of the Ricci curvature.


Here's an example: In the domain $D\subset\mathbb{C}^{n+1}$ defined by $|z\_0|^2>|z\_1|^2+|z\_2|^2+\cdots+|z\_n|^2$, consider the nonnegative function
$$
\phi(z) = \frac{1}{|z\_0|^2-|z\_1|^2-|z\_2|^2-\cdots - |z\_n|^2}.
$$
Then it is easy to compute that $\phi$ is strictly plurisubharmonic in $D$ and that $\mathrm{d}\phi$ is nonvanishing in $D$. The level sets of $\phi$, which are noncompact, are homogeneous under the action of the group $\mathrm{U}(1,n)$, so the Ricci curvature of the (necessarily complete) metric induced on each level set has constant eigenvalues. They cannot all be positive, for the level sets are not compact. (In fact, it is not hard to see that all but at most one of the eigenvalues (the one tangent to the circle action) has to be negative.)


Obviously, this is the same sort of example as I gave for the case of the level sets of a strictly convex function on a domain in affine space, i.e., the function $f$ on the domain $D\subset\mathbb{R}^{n+1}$ defined by ${x\_0}^2> {x\_1}^2 + \cdots + {x\_n}^2$ defined by
$$
f(x) = \frac1{{x\_0}^2- {x\_1}^2 - \cdots - {x\_n}^2},
$$
to show that the desired positivity doesn't work for Hessians either.


