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12.36 A divisorial threshold on the Hopf threefold7F944D8C8D92

7F944D8C8D92

On the Hopf threefold, the pullback of a hyperplane has an explicit divisorial threshold. The residual metric solves its normalized equation with zero potential, so the residual volume can be computed directly, also at the nef endpoint.

In this section $X$ is the Hopf threefold of section 23.1, with its projection $\pi :X\to \mathbb P^2$, and we write $\zeta $ for its vertical form, so that $\omega =\alpha +\zeta $. These forms satisfy

\[ \alpha ^3=0,\quad \zeta ^2=0,\quad \mathrm{dd}^{\mathrm{c}}\zeta =-\alpha ^2,\quad \int _ X\alpha ^2\wedge \zeta =2. \]

Let $H$ be the inverse image under $\pi $ of a projective line in $\mathbb P^2$. The pulled-back Fubini–Study metric gives $\eta _ H=\alpha $ and a logarithmic potential $\sigma _ H$ with $\mathrm{dd}^{\mathrm{c}}\sigma _ H=[H]-\alpha $.

Proposition 12.36.1.3A78333ED147 The divisorial threshold is $\tau _\omega (H)=1$. For $\mu _0=\omega ^3/6$ and $0\leq a\leq 1$,

\[ \mathcal V_{\mu _0}(\omega ;aH)=6(1-a)^2. \]

For $a<1$ this is the scalar of the equation solved, as in theorem 12.35.4, by $a\sigma _ H$ plus a smooth function. At $a=1$ the residual background is nef, and the logarithmic potential $\sigma _ H$ has zero non-pluripolar Monge–Ampère measure. For $a<1$, every residual background has infinite upper mass.

Proof. The residual background is

\[ R_ a=(1-a)\alpha +\zeta . \]

It is positive for $0\leq a<1$ and semipositive for $a=1$, so every coefficient $a\leq 1$ is feasible. For the reverse inequality, consider the metrics $\gamma _ t=\alpha +t\zeta $, $t>0$. They are Gauduchon, since

\[ \mathrm{dd}^{\mathrm{c}}(\gamma _ t^2)=-2t\alpha ^3=0. \]

Using $\zeta ^2=0$, we obtain

\[ \frac{\int _ X\omega \wedge \gamma _ t^2}{\deg _{\gamma _ t}H} =\frac{1+2t}{2t}\longrightarrow 1\qquad (t\to \infty ), \]

and the threshold formula gives $\tau _\omega (H)=1$.

For $a<1$,

\[ R_ a^3=3(1-a)^2\alpha ^2\wedge \zeta =(1-a)^2\omega ^3=6(1-a)^2\mu _0. \]

Hence the zero potential solves the residual smooth equation, and uniqueness of the scalar gives the volume. At $a=1$, the form $R_1=\zeta $ is nef, because $\zeta +\epsilon \omega >0$ for $\epsilon >0$, and the same computation gives

\[ \mathcal V_{\mu _0}(\zeta +\epsilon \omega ) =6\epsilon ^2(1+\epsilon )\longrightarrow 0. \]

By theorem 12.11.1, the residual volume at $a=1$ is zero. Moreover $\omega +\mathrm{dd}^{\mathrm{c}}\sigma _ H=[H]+\zeta $, so the non-pluripolar measure of $\sigma _ H$ is $\zeta ^3=0$.

Finally, fix $a<1$ and choose $0<\delta \leq \min \{ 1-a,1\} $, so that $R_ a-\delta \omega \geq 0$. For the smooth functions $V_ j$ of theorem 23.1.1,

\[ R_ a+\mathrm{dd}^{\mathrm{c}}(\delta V_ j) =(R_ a-\delta \omega )+\delta (\omega +\mathrm{dd}^{\mathrm{c}}V_ j)>0. \]

Monotonicity of the determinant on positive matrices bounds the mass of this form below by $\delta ^3\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}V_ j)^3$, which tends to infinity. Thus a finite residual volume and an explicit divisorial threshold coexist with unbounded upper mass. $\square$

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