Of course, it wasn't a sake of reputation... Just I wanted to know if my answer was satisfactory to you... Thanks anyway! Prego caro!

Hi Gergely! This should be not extremely difficult but for sure not completely trivial... I'll think about it later, if no one answers you meanwhile! All the best!

You meant that but you wrote "My question was if noncompact Kahler manifolds with zero Ricci curvature possess any interesting properties"... So you want non-compact Kähler manifolds with non identically vanishing Ricci curvature but identically zero scalar curvature, right? Then, you should edit your question consequently...

Why should you expect that the sum $T_S+R_S$ is positive without any hypothesis?

Mmm... I know plenty of positive vector (even line) bundles with no non-zero holomorphic global section... I would say rather the contrary... A globally generated vector bundle carries a hermitian metric such that its curvature is Griffiths positive (and hence the dual bundle is Nakano negative).

Gianni, doesn't my answer satisfy you?

Frankly, Peter, from your originally question it is impossible to understand that you actually wanted to have information about non-compact Kähler manifolds with zero Ricci curvature... You should edit your question, indeed...

I am not sure of your terminology. By Ricci scalar, do you mean the scalar curvature, that is the trace of the Ricci curvature? Anyway, in the Kähler case, for example the Ricci curvature is the (double) average (on the unitary holomorphic tangent bundle) of the holomorphic bisectional curvature and the scalar curvature is the average of the holomorphic sectional curvature. I wasn't sure this could be considered as an "answer" so I posted it as a comment. If this satisfies you, just tell me and I'll re-post it as an answer.

What do you mean by "any vector bundle over a curve has vanishing curvature"?

Sorry Sándor, I don't understand... Is your non-trivial line bundle $\mathcal L$ in $\operatorname{Pic}^0(X)$? In this case then $c_1(L)$ shouldn't be zero?

Definitely right Jorge! I meant adapting the strategy of Cano... And I forgot Panazzolo, because Michael Mc Quillan told me that personally quite long ago... Thanks for having given the precision!

Yemon, I was about to write that... Otherwise $f\equiv 0$ is a VERY easy example...