Proposition 11.26.79EFFDFAE03AF

Proof. If $h$ were uniformly RC-positive, then at each point some $u$ would make the $h$-Hermitian endomorphism $R_ h(u,\bar u)$ positive definite, and its trace would be strictly positive. Since the curvature of the induced metric on $\det E$ is $\operatorname {tr}F_ h$, the determinant metric would be RC-positive as a metric on a line bundle. If instead $K_{\omega ,h}>0$, the $\omega $-trace of the determinant curvature would be $\operatorname {tr}K_{\omega ,h}>0$, which again gives a positive tangent evaluation.

Fix a flat Hermitian metric $h_0$ on $\det E$ and write $\det h=\mathrm e^ b h_0$ for a smooth real function $b$. The determinant curvature is then $-\partial \bar\partial b$. At a global minimum of $b$, this curvature has no positive tangent evaluation, contradicting the RC-positivity of the determinant metric. Finally, a holomorphic trivialization of $\det E$ gives a flat metric, by assigning constant norm to the trivializing frame. $\square$