Theorem 11.2.1904BD04D6C676

Proof. Step 1: the Hessian formula. Pluriclosedness and Stokes' theorem give $\int _ X\omega \wedge (\mathrm{dd}^{\mathrm{c}}\varphi )^3=\int _ X(\mathrm{dd}^{\mathrm{c}}\varphi )^4=0$. Thus the volume and, by (11.2.18), the energy along a potential line $t\mapsto \omega +t\mathrm{dd}^{\mathrm{c}}\varphi $ are polynomials of degree at most two in $t$. Differentiating twice and integrating by parts, we obtain

For the second equality, apply lemma 11.1.2 once and then use the identity $\mathrm{dd}^{\mathrm{c}}(\varphi ^2/2)=\varphi \mathrm{dd}^{\mathrm{c}}\varphi +\mathrm d\varphi \wedge \mathrm d^{\mathrm c}\varphi $ against the closed form $\mathrm{dd}^{\mathrm{c}}(\omega ^2)$.

Step 2: both indices are infinite if $\mathrm{dd}^{\mathrm{c}}(\omega ^2)\ne 0$. Put $\Psi =\mathrm{dd}^{\mathrm{c}}(\omega ^2)$. At each point, $\xi \mapsto \Psi \wedge \mathrm i\xi \wedge \bar\xi $, divided by a positive volume form, is a Hermitian quadratic form in the $(1,0)$-covector $\xi $, and these quadratic forms determine $\Psi $. Their traces with respect to $\omega $ are positive multiples of the density $\Psi \wedge \omega $, whose integral vanishes by Stokes' theorem and $\mathrm{dd}^{\mathrm{c}}\omega =0$. If the quadratic forms took no strictly positive value, they would be negative semidefinite everywhere, with nonpositive traces of integral zero; the traces, and hence $\Psi $, would vanish identically. The same argument applies with the signs reversed. Therefore, if $\Psi \ne 0$, the quadratic forms take a strictly positive value somewhere and a strictly negative value somewhere.

Fix one of the two signs. Choose holomorphic coordinates on a ball such that, with $x=\operatorname {Re}z_1$, the density $\Psi \wedge \mathrm dx\wedge \mathrm d^{\mathrm c}x$ has that sign on the ball; this is possible because $\mathrm dx\wedge \mathrm d^{\mathrm c}x$ is a positive multiple of $\mathrm i\, \mathrm dz_1\wedge \mathrm d\bar z_1$. For a nonzero real cutoff function $a$ supported in the ball, set $\varphi _ N=a\cos (Nx)$. Formula (11.2.20) gives

where the constant in $O(N)$ depends on $\Psi $, the chart and the cutoff, but not on $N$. By integration by parts in $x$, the integral tends to half the same integral without the factor $\sin ^2(Nx)$. Hence the Hessian has the chosen sign for large $N$. For a positive integer $k$, apply this construction on $k$ disjoint smaller balls in that region. By the local integral formula, the resulting nonzero directions are linearly independent and pairwise orthogonal for the Hessian, so their span is a $k$-dimensional subspace on which $H_\omega $ is definite of the chosen sign. Repeating the argument with the opposite sign proves that both indices are infinite. Each finite-dimensional span admits sufficiently small variations preserving positivity.

Step 3: flatness if $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$. If $\Psi =0$, lemma 11.1.3 and pluriclosedness give $\mathrm{dd}^{\mathrm{c}}(\omega ^3)=3\omega \wedge \Psi =0$. Both derivatives at $t=0$ of the quadratic polynomial $t\mapsto F(\omega +t\mathrm{dd}^{\mathrm{c}}\varphi )$ vanish, so it is constant on the interval of $t$ for which $\omega +t\mathrm{dd}^{\mathrm{c}}\varphi >0$. This interval contains $0$ and $1$ whenever $\omega +\mathrm{dd}^{\mathrm{c}}\varphi >0$. Finally, a local minimum cannot have a potential direction along which the Hessian is negative, so it satisfies $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$ by Step 2. $\square$