Pome

Corollary 11.28.5550E97762605

Corollary 11.28.5.550E97762605 Suppose that $\operatorname {Ric}(g)\geq 0$. Every Hermitian tangent metric $h$ with $H_ h\leq 0$ satisfies $\operatorname {Ric}(g)=0$ and $H_ h=0$ in every direction. In particular, $X$ admits no Hermitian tangent metric with quasi-negative Chern holomorphic sectional curvature.

Proof. The Ricci assumption gives $I_ g(h)\geq 0$, and lemma 11.28.2 gives $-\mathcal R_ g(h)\geq 0$. All three continuous terms in (11.28.4) are nonnegative, so each vanishes everywhere. A positive definite $h$ pairs strictly positively with every nonzero semipositive Ricci matrix; hence $I_ g(h)=0$ forces $\operatorname {Ric}(g)=0$.

At each point, the Gaussian expectation in the averaging proof is now zero. Its integrand $R^ h(Z,\bar Z,Z,\bar Z)$ is nonpositive. If it were negative in one direction, continuity and the strictly positive Gaussian density would make its expectation negative. It is therefore zero in every direction, which gives $H_ h=0$ and excludes quasi-negativity. $\square$

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