Theorem 11.27.4.8C2D7F0F7808 If $X$ has a nonflat Ricci-flat Kähler metric, then $T^{1,0}X$ admits an RC-positive Hermitian metric whose associated $(1,1)$-form is nonclosed. It admits neither a uniformly RC-positive Hermitian metric nor a metric with positive definite mean curvature relative to any Hermitian form.
Theorem 11.27.48C2D7F0F7808
Proof. By lemma 11.27.2, lemma 11.27.1, the original tangent metric $h$ is RC-positive on a nonempty open set. Its mean curvature is zero, so theorem 11.26.3 gives a metric $h_ t=\mathrm e^{-tf}h$ that is RC-positive everywhere.
In that construction we may take the Gauduchon form to be $\omega $ itself, and the closed coordinate ball to be a proper subset of $X$. Since $P_\omega f=1$ off that ball, $f$ is not constant. The associated form is $\omega _ t=\mathrm e^{-tf}\omega $, with
On a real four-dimensional symplectic vector space, $\zeta \mapsto \zeta \wedge \omega $ is injective on one-forms, as a unitary real frame shows. Thus $\mathrm d\omega _ t\ne 0$ somewhere.
Finally, the induced determinant metric of $h$ is flat because its Chern curvature is the Ricci form of $\omega $, which vanishes. The two obstructions follow from proposition 11.26.7. $\square$
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