I read the Complex Analytic and Algebraic Geometry, which says that a weakly pseudoconvex Kähler manifolds $(M,\omega)$, in the sense that it admits a smooth plurisubharmonic exhaustive function $\psi$, carries a complete Kähler metric $\hat{\omega}.$
The author construct $$\hat{\omega}:=\omega+i\partial\bar\partial(\psi)^2$$ and by $$\hat{\omega}=\omega+2i\psi\partial\bar\partial\psi+2i\partial\psi\wedge\bar\partial\psi\ge 2i\partial\psi\wedge\bar\partial\psi$$ to show that $|d\psi|_{\hat{\omega}}\le 1$ concluding $(M,\hat{\omega})$ is complete by the lemma before this content.
My problem is that why not just consider $\hat{\omega}:=i\partial\bar\partial(\psi)^2?$ This is a metric and still implies the inequality above. Does this mean that the Kähler condition at first can be omitted?
Thanks a lot in advance!
Assume $ϕ$ is the strictly plurisubharmonic exhaustion function for a Stein manifold $M$. Without loss of generality, we can assume $ϕ ≥ 0$. (Otherwise consider $\exp ϕ$ instead of $ϕ$. ) then $∂\bar ∂ϕ^2$ gives a complete Kahler metric on $M$. See this lecture note for details and this paper .
See also Lemma 1.4.2