Lemma 12.25.1.DAA29135206D Let $X$ be a compact Kähler manifold of dimension $n$, let $\Omega $ be a Kähler form, and let $B$ be a smooth closed semipositive $(1,1)$-form. 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 Hermitian form $h$ and smooth strictly positive probability form $\mu $, there are constants $k,K>0$ such that, for $0<\epsilon \leq 1$,
ED15CA2DA3FC
12.25.2
\begin{equation} k\lambda _\mu (B+\epsilon \Omega ) \leq \lambda _\mu (\theta +\epsilon h) \leq K\lambda _\mu (B+\epsilon \Omega ). \end{equation}
If $B$ has constant rank $R$, there are moreover constants $L,U>0$ with
\[ L\epsilon ^{n-R}\leq \mathcal V_\mu (\theta +\epsilon h) \leq U\epsilon ^{n-R},\qquad 0<\epsilon \leq 1. \]
In particular, $\theta $ then has common order $n-R$.
Proof.
Choose $c_ h,C_ h>0$ with $c_ h\Omega \leq h\leq C_ h\Omega $. For every Gauduchon metric $\gamma $, distributional Stokes gives
\[ \int _ X\mathrm{dd}^{\mathrm{c}}\varphi \wedge \gamma ^{n-1} =\langle \varphi ,\mathrm{dd}^{\mathrm{c}}\gamma ^{n-1}\rangle =0. \]
The lower bound for $\theta +\mathrm{dd}^{\mathrm{c}}\varphi $ therefore gives $\int _ X(\theta +\epsilon h)\wedge \gamma ^{n-1}\geq k\int _ X(B+\epsilon \Omega )\wedge \gamma ^{n-1}$ with $k=\min (c,c_ h)$, and the upper bound for $\theta $ gives the reverse inequality with $K=\max (C,C_ h)$. Dividing by $D_\mu (\gamma )>0$ and taking infima over all Gauduchon metrics proves (12.25.2); no rank hypothesis has been used.
Now suppose that $B$ has constant rank $R$. Its $R$ positive eigenvalues relative to $\Omega $ are then bounded above and away from zero, uniformly on $X$. With $A_\epsilon =B+\epsilon \Omega $, compactness gives constants $m_0,M_0>0$ with
022F404D6957
12.25.3
\begin{equation} m_0\epsilon ^{n-R}\mu \leq A_\epsilon ^ n \leq M_0\epsilon ^{n-R}\mu ,\qquad 0<\epsilon \leq 1. \end{equation}
For every Gauduchon metric $\gamma $, the pointwise arithmetic–geometric mean inequality gives
\[ A_\epsilon \wedge \gamma ^{n-1} \geq \left(\frac{A_\epsilon ^ n}{\gamma ^ n}\right)^{1/n}\gamma ^ n, \]
and hence $\lambda _\mu (A_\epsilon )\geq m_0^{1/n}\epsilon ^{(n-R)/n}$. In the other direction, the positive closed form $\gamma =A_\epsilon $ is itself an admissible test. Writing $A_\epsilon ^ n=f_\epsilon \mu $, we have
\[ \begin{aligned} D_\mu (A_\epsilon )& =\int _ X f_\epsilon ^{(n-1)/n}\mu \geq m_0^{(n-1)/n}\epsilon ^{(n-R)(n-1)/n},\\ \int _ X A_\epsilon ^ n& \leq M_0\epsilon ^{n-R}, \end{aligned} \]
so the quotient of this test is at most $M_0m_0^{-(n-1)/n}\epsilon ^{(n-R)/n}$. Raising both bounds to the $n$th power and using (12.25.2) gives the two-sided power bounds. The argument includes $R=0$; when $n=1$, $D_\mu =1$. Fixed positive factors do not affect the order, since their logarithms are divided by $\log \epsilon $.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.