11.26 Conformal changes and directional curvature positivityA4E01A8B6F30

Proof. The trace of $\alpha _{h,v}$ with respect to $\omega $ is $h(K_{\omega ,h}v,v)\geq 0$. A nonzero Hermitian form with nonnegative trace has a positive eigenvalue, since if all its eigenvalues were nonpositive, their nonnegative sum would force all of them to vanish. This gives the required tangent direction. Conversely, a tangent vector with positive evaluation shows that $\alpha _{h,v}\ne 0$. $\square$

Proof. By theorem 11.4.2, choose a conformal Gauduchon form $\gamma =\mathrm e^ a\omega $. Then $K_{\gamma ,h}=\mathrm e^{-a}K_{\omega ,h}\geq 0$. Choose a closed coordinate ball $Q\subset U$ and a smooth function $\chi \geq 0$ supported in its interior, with $\int _ X\chi \gamma ^ n>0$, and set

Thus $\rho =1$ on $X\setminus Q$ and $\int _ X\rho \gamma ^ n=0$. Since $\gamma $ is Gauduchon, proposition 11.23.2, applied with $\gamma $ in place of $\omega $ and $u=0$, gives a smooth real solution $f$ of

Indeed, with $L_\gamma f=n\mathrm{dd}^{\mathrm{c}}f\wedge \gamma ^{n-1}/\gamma ^ n$ as in (11.23.1), we have $P_\gamma =2\pi L_\gamma $, and the adjoint probability of $L_\gamma $ is $\gamma ^ n/\int _ X\gamma ^ n$.

The curvature transforms by

On $X\setminus Q$, the $\gamma $-trace of the expression in parentheses is at least $t|v|_ h^2\rho =t|v|_ h^2>0$ for $v\ne 0$, so it has a positive eigenvalue there.

Over $Q$, the unit sphere bundle $\{ |v|_ h=1\} $ is compact. The largest $\gamma $-eigenvalue of $\alpha _{h,v}$ is continuous and strictly positive on it, so it has a positive minimum $\delta $. Let $C$ be the maximum on $X$ of the operator norm of $\mathrm i\partial \bar\partial f$ relative to $\gamma $. For every unit vector over $Q$, the largest eigenvalue of the expression in parentheses in (11.26.5) is at least $\delta -tC$, which is positive for small $t>0$. Homogeneity in $v$ and the positive factor $\mathrm e^{-tf}$ prove the assertion on $Q$ as well. $\square$

Proof. The two eigenvalues of $H(z)$ are $1-\tfrac 12|z|^2$ and $1+\tfrac 12|z|^2$, so the metric is positive on a small ball. Its first derivatives vanish at the origin, where

The sum $F_{1\bar1}+F_{2\bar2}$ is zero. For each nonzero $v$, taking $u=v$ gives a positive evaluation. For each nonzero $u$, taking $v\perp u$ gives a negative evaluation, so no single $u$ works for every $v$. $\square$

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. $\square$