Pome11FD5CD723A9For potentials invariant under rotations of the individual coordinates, the complex Hessian becomes the real Hessian of a function of the logarithmic coordinates. Invariance under complex scaling gives this Hessian a null direction; transverse to it, positivity becomes convexity, and boundedness of the potential confines the gradient to a simplex. Logarithmic growth makes every interior point of the simplex a gradient value, and strict positivity on the base makes the gradient injective modulo the null direction.
In this section, $k>0$ is a real number. Let $v$ be a smooth real function on $\mathbb P^{n-1}$ that is invariant under the torus action
and put $u=\pi ^*v$. On the locus where all covering coordinates $z_ i$ are nonzero, set $x_ i=\log |z_ i|^2$. By this invariance, $u$ is a function of $x=(x_1,\ldots ,x_ n)$, which we denote by $u(x)$, and we set
Write $\nabla F=(F_ i)$ and $D^2F=(F_{ij})$ for its Euclidean gradient and Hessian, and let $\mathbf1=(1,\ldots ,1)$. Since $u$ is invariant under complex scaling, for every $t\in \mathbb R$,
Proposition 21.14.8.F23340CADBE0 On this locus, $k\alpha +\mathrm{dd}^{\mathrm{c}}u\geq 0$ if and only if $F$ is convex. If $F$ is convex, its gradient lies in
and $D^2F\, \mathbf1=0$. If $k\sigma +\mathrm{dd}^{\mathrm{c}}v$ is positive definite on $\mathbb P^{n-1}$, the gradient induces a bijection from $\mathbb R^ n/\mathbb R\mathbf1$ onto the relative interior of $\Sigma _ k$.
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. $\square$
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