Proposition 21.14.8F23340CADBE0

Proof. On the covering space, $k\alpha +\mathrm{dd}^{\mathrm{c}}u=\mathrm{dd}^{\mathrm{c}}F$. In the coordinates $z_ i$, the Hermitian coefficient matrix of this form is a positive constant times

This matrix is congruent to the real Hessian $(F_{ij})$ through the invertible diagonal matrix with entries $1/z_ i$, which proves the first equivalence. Differentiating (21.14.7) gives $\sum _ i F_ i=k$ and $D^2F\, \mathbf1=0$.

Since $v$ is bounded on the compact base, there is a constant $C$ such that

Fix $x$ and a coordinate vector $e_ i$. Since $n\geq 2$, the convex function $t\mapsto F(x+te_ i)$ is bounded above as $t\to -\infty $. Its derivative cannot be negative anywhere: otherwise the supporting line at such a point would tend to $+\infty $ as $t\to -\infty $. Thus $F_ i\geq 0$, which proves $\nabla F\in \Sigma _ k$.

Now assume that $k\sigma +\mathrm{dd}^{\mathrm{c}}v$ is positive definite. The kernel of the differential of the projection $\mathbb C^ n\setminus \{ 0\} \to \mathbb P^{n-1}$ is exactly the complex scalar direction, so the Hessian of $F$ is positive definite transverse to $\mathbf1$. In particular $F$ is strictly convex on $H=\{ x:\sum _ i x_ i=0\} $. Along each line $x+te_ i$, the derivative of $F$ is strictly increasing and nonnegative everywhere, hence positive. Thus every gradient lies in the relative interior of $\Sigma _ k$.

Conversely, take $p\in \mathbb R^ n$ with $p_ i>0$ and $\sum _ i p_ i=k$. The function $F(x)-p\cdot x$ is invariant under translation along $\mathbf1$, and we claim that it tends to $+\infty $ on $H$ as $|x|\to \infty $. Indeed,

is strictly positive at every nonzero vector of $H$, and hence has a positive minimum on the unit sphere of $H$. Since $F$ differs from $k\log \sum _ i \mathrm{e}^{x_ i}$ by a bounded function and $\log \sum _ i \mathrm{e}^{x_ i}\geq \max _ i x_ i$, the claim follows. Hence $F(x)-p\cdot x$ has a unique minimizer on $H$. At this point $\nabla F-p$ is a multiple of $\mathbf1$; both vectors have coordinate sum $k$, so the multiple is zero. This proves surjectivity, and strict convexity on $H$ proves injectivity. $\square$