3567ADAA7365
If $\theta $ is nef and $h$ is Hermitian, the volumes $\mathcal V_\mu (\theta +\epsilon h)$ are positive and, by theorem 12.11.1, decrease to $\mathcal V_\mu (\theta )$ as $\epsilon \downarrow 0$. When this limit is zero, the rate of decay measures how degenerate $\theta $ is at the nef boundary. For a closed nef class on a compact Kähler manifold, choosing $h$ Kähler makes the volume of the perturbation the intersection polynomial $\int _ X(\theta +\epsilon h)^ n$, so it decays like an integral power of $\epsilon $ determined by the numerical dimension. For a nonclosed nef background the logarithmic rate of decay can be fractional, and it need not converge. We therefore define lower and upper orders of degeneration as lower and upper limits of this rate. Changing the probability measure or the perturbing metric only introduces bounded factors, so the orders are invariants of the background.
Let $X$ again be a connected compact complex manifold of dimension $n\geq 1$, and $\mu $ a smooth strictly positive probability form on $X$. Let $\theta $ be nef and let $h$ be Hermitian. For $0<\epsilon <1$, put
\[ a_{\mu ,h}(\theta ;\epsilon ) =\frac{\log \mathcal V_\mu (\theta +\epsilon h)}{\log \epsilon }. \]
The logarithm is finite because $\theta +\epsilon h$ belongs to $\mathscr P$, so that its volume is positive.
Proposition 12.20.1.EBFCF76A5A51 The limits
\[ \underline a(\theta )=\liminf _{\epsilon \downarrow 0} a_{\mu ,h}(\theta ;\epsilon ), \qquad \overline a(\theta )=\limsup _{\epsilon \downarrow 0} a_{\mu ,h}(\theta ;\epsilon ) \]
are independent of $\mu $ and $h$, and satisfy
\[ 0\leq \underline a(\theta )\leq \overline a(\theta )\leq n. \]
They are unchanged when $\theta $ is replaced by $\theta +\mathrm{dd}^{\mathrm{c}}f$, with $f$ smooth, or by $r\theta $, with $r>0$. If $\mathcal V_\mu (\theta )>0$, both limits are zero.
Proof.
Since $\theta $ is pseudoeffective, monotonicity and homogeneity give
\[ \epsilon \lambda _\mu (h) \leq \lambda _\mu (\theta +\epsilon h) \leq \lambda _\mu (\theta +h), \qquad 0<\epsilon \leq 1. \]
Thus there are constants $c,C>0$ such that
\[ c\epsilon ^ n\leq \mathcal V_\mu (\theta +\epsilon h)\leq C, \qquad 0<\epsilon \leq 1. \]
Taking logarithms and dividing by the negative number $\log \epsilon $ shows that the quotients $a_{\mu ,h}(\theta ;\epsilon )$ stay bounded as $\epsilon \downarrow 0$, and that their lower and upper limits lie in $[0,n]$.
By proposition 12.15.1, changing the smooth probability measure multiplies the volume by a factor between two fixed positive constants. The resulting error in the logarithm is bounded, and after division by $\log \epsilon $ it tends to zero.
For another Hermitian metric $h'$, choose constants $c_2\geq c_1>0$ with $c_1h\leq h'\leq c_2h$. Then
\[ \mathcal V_\mu (\theta +c_1\epsilon h) \leq \mathcal V_\mu (\theta +\epsilon h') \leq \mathcal V_\mu (\theta +c_2\epsilon h). \]
For any fixed $t>0$ and small $\epsilon $,
\[ \frac{\log \mathcal V_\mu (\theta +t\epsilon h)}{\log \epsilon } =a_{\mu ,h}(\theta ;t\epsilon ) \frac{\log (t\epsilon )}{\log \epsilon }. \]
The last factor tends to one and the first is bounded, while the rescaling $\epsilon \mapsto t\epsilon $ does not change lower and upper limits as $\epsilon \downarrow 0$. Hence the left side has the same lower and upper limits as $a_{\mu ,h}(\theta ;\epsilon )$. Dividing the preceding inequalities by $\log \epsilon <0$ reverses them, and taking lower and upper limits proves independence of the metric.
Invariance under smooth $\mathrm{dd}^{\mathrm{c}}$ changes follows from that of $\mathcal V_\mu $. For scaling, homogeneity gives
\[ \mathcal V_\mu (r\theta +\epsilon h) =r^ n\mathcal V_\mu (\theta +(\epsilon /r)h). \]
The bounded error $n\log r$ and the same rescaling argument prove scale invariance. Finally, if $\mathcal V_\mu (\theta )>0$, then by theorem 12.11.1 the perturbed volumes lie between $\mathcal V_\mu (\theta )$ and $\mathcal V_\mu (\theta +h)$ for $0<\epsilon \leq 1$, so both limits are zero.
$\square$
We call $\underline a(\theta )$ and $\overline a(\theta )$ the lower and upper orders of degeneration of the nef background $\theta $. When they agree, their common value is the common order, and we then also say that $\theta $ has this order. The two orders can differ, and a common order can be irrational. Even when both are zero, the volume can tend to zero, at a logarithmic rate. The elliptic examples in section 12.21 exhibit each of these behaviors. Closed nef classes on Kähler manifolds, in contrast, degenerate polynomially.
Proposition 12.20.2.883742E77C19 Suppose $X$ is Kähler, $\theta $ is closed and nef, and $h$ is Kähler. Set
\[ r=\max \left\{ 0\leq k\leq n: \int _ X\theta ^ k\wedge h^{n-k}>0\right\} . \]
Then
\[ \underline a(\theta )=\overline a(\theta )=n-r. \]
Thus $n$ minus the common order is the numerical dimension of the closed nef class of $\theta $.
Proof.
All the intersection numbers in the definition of $r$ are nonnegative. Indeed, the mixed integrals against $h$ of smooth positive representatives of $\theta +\delta h$ are positive, agree with the cohomological integrals by closed Stokes, and converge to the stated numbers as $\delta \downarrow 0$. The number for $k=0$ is $\int _ Xh^ n>0$, so $r$ is well defined.
By the closed nef formula following theorem 12.11.1,
\[ \mathcal V_\mu (\theta +\epsilon h) =\int _ X(\theta +\epsilon h)^ n =\sum _{k=0}^ n\binom nk\epsilon ^{n-k} \int _ X\theta ^ k\wedge h^{n-k}. \]
As $\epsilon \downarrow 0$, the leading nonzero term is a positive constant times $\epsilon ^{n-r}$, and taking the logarithmic quotient proves the assertion. By proposition 12.20.1, the result holds for every Hermitian perturbation, not only for Kähler ones.
$\square$
The orders can also be computed on the non-Kähler Hopf threefold $X$ of section 23.1. Keep its forms $\omega =\alpha +\beta $, where $\alpha \geq 0$ is closed of rank two, $\beta \geq 0$ has rank one, $\alpha ^3=\beta ^2=0$, and $\int _ X\omega ^3=6$. Take $\theta =\alpha $, $h=\omega $, and $\mu _0=\omega ^3/6$. The metric
\[ \Omega _\epsilon =\alpha +\epsilon \omega =(1+\epsilon )\alpha +\epsilon \beta \]
is strictly positive. Its volume is
\[ \Omega _\epsilon ^3=\epsilon (1+\epsilon )^2\omega ^3 =6\epsilon (1+\epsilon )^2\mu _0. \]
Hence the zero potential solves the smooth normalized equation, and
\[ \mathcal V_{\mu _0}(\alpha +\epsilon \omega ) =6\epsilon (1+\epsilon )^2, \qquad \underline a(\alpha )=\overline a(\alpha )=1. \]
By proposition 12.20.1, the same orders are obtained for every smooth positive probability measure and every Hermitian perturbation.
Each perturbed metric nevertheless has infinite upper mass. If $V_ j$ are the smooth potentials of theorem 23.1.1, then
\[ \Omega _\epsilon +\mathrm{dd}^{\mathrm{c}}(\epsilon V_ j) =\alpha +\epsilon (\omega +\mathrm{dd}^{\mathrm{c}}V_ j)>0, \]
and monotonicity of the determinant of positive matrices gives
\[ \int _ X\bigl (\Omega _\epsilon +\mathrm{dd}^{\mathrm{c}}(\epsilon V_ j)\bigr )^3 \geq \epsilon ^3\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}V_ j)^3\longrightarrow +\infty . \]
Thus a finite order of degeneration at the boundary coexists with infinite upper mass at every positive perturbation.
Pome
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