Proposition 12.26.1.94DF4FEA70D3 Let $X$ be a compact Kähler manifold of dimension $n$, let $B$ be a smooth closed semipositive $(1,1)$-form, and put $R=\max _ X\operatorname {rank}B$. Suppose a smooth nef background $\theta $ satisfies
\[ \theta \leq CB,\qquad \theta +\mathrm{dd}^{\mathrm{c}}\varphi \geq cB \]
as currents, for constants $c,C>0$ and a real function $\varphi \in L^1(X)$. For every smooth positive probability form $\mu $ and Hermitian form $h$, there are constants $L,U>0$ such that
\[ L\epsilon ^{n-R}\leq \mathcal V_\mu (\theta +\epsilon h) \leq U\epsilon ^{n-R},\qquad 0<\epsilon \leq 1. \]
Thus $\theta $ has common order $n-R$, without any constant-rank hypothesis.
Proof.
Fix a Kähler form $\Omega $. The first part of lemma 12.25.1, which does not use the rank, shows that $\lambda _\mu (\theta +\epsilon h)$ and $\lambda _\mu (B+\epsilon \Omega )$ differ by bounded positive factors. Since $B^{R+1}=0$, the closed nef formula following theorem 12.11.1 gives
\[ \mathcal V_\mu (B+\epsilon \Omega ) =\sum _{k=0}^{R}\binom nk\epsilon ^{n-k} \int _ X B^ k\wedge \Omega ^{n-k}. \]
All coefficients are nonnegative. If $R>0$, then at a point where $B$ has rank $R$, some $R\times R$ minor of its coefficient matrix is positive and stays positive nearby, so $B^ R\wedge \Omega ^{n-R}$ has positive integral. If $R=0$, this integral is the Kähler volume. Hence the coefficient of $\epsilon ^{n-R}$ is positive. For $0<\epsilon \leq 1$, every other power of $\epsilon $ that occurs is at most $\epsilon ^{n-R}$, so the polynomial is bounded below by its leading term and above by the sum of its coefficients times $\epsilon ^{n-R}$. Raising the comparison of quotients to the $n$th power gives the asserted bounds, and taking logarithms gives the order.
$\square$
Pome
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