Example 12.35.13.76B0694EBA2F Let $\mu =w_1\wedge w_2$, let $\tau (z)=(|z|^2-1)/(|z|^2+1)$ be the height on the first sphere, and set
\[ \theta _ s=w_1+(2+\tau -s)w_2,\qquad 0\leq s<1. \]
Fix the ordinary type of $U=1-(1-G_ p)^\alpha $ with $1/2\leq \alpha <1$, and write $\sigma _-(s),\sigma _+(s)$ for its lower and upper volumes in $\theta _ s$. Then
AFBB384B6340
12.35.14
\begin{equation} V_ s:=\mathcal V_\mu (\theta _ s)=\frac4{\log ((3-s)/(1-s))},\qquad \sigma _-(s)=2(1-s),\qquad \sigma _+(s)=2(3-s). \end{equation}
The improper subtraction cost is finite and equals $2/\sqrt{\log 3}$, while $\sigma _+(s)\to 4$. The endpoint $\theta _1=w_1+(1+\tau )w_2$ is nef, has zero variational volume, and is not big.
Proof.
Take $c=1$, $b=0$, $a_ s=2+\tau -s$ and $m=1$ in proposition 12.35.9. Radial integration on the first sphere gives the law $\mathrm d\tau /2$ on $[-1,1]$, so
\[ H_ s=\frac12\int _{-1}^1\frac{\mathrm dt}{2+t-s} =\frac12\log \frac{3-s}{1-s}. \]
Thus $V_ s=2/H_ s$. Since $b=0<c$, theorem 20.3.66 gives all rough representatives of the fixed type finite positive mass, with the extrema in (12.35.14). The reference has zero Lelong numbers and trivial multiplier ideals but no analytic representative, by proposition 20.3.71. Its infinite closed first moment in this range of $\alpha $ was computed after (20.3.69); no such moment is needed for either extremum.
For the minimizing tests $T_ s=V_ s(2+\tau -s)^{-2}w_1/2+w_2$, corollary 12.35.11 gives
\[ \frac{\int _ Yw_2\wedge T_ s}{D_\mu (T_ s)} =-\frac{\mathrm d}{\mathrm ds}\sqrt{V_ s} =\frac{2}{(1-s)(3-s)[\log ((3-s)/(1-s))]^{3/2}}. \]
Integrating to $r<1$ gives $\sqrt{V_0}-\sqrt{V_ r}$. Since $V_ r\to 0$, the improper integral is $\sqrt{V_0}=2/\sqrt{\log 3}$. By theorem 12.2.3, every minimizing test is a constant multiple of $T_ s$, so the cost is independent of this choice.
The endpoint is smooth semipositive, hence nef. Monotonicity gives
\[ 0\leq \lambda _\mu (\theta _1)\leq \lambda _\mu (\theta _ s) \longrightarrow 0, \]
so its variational volume vanishes. To exclude bigness, suppose that $\theta _1+\mathrm{dd}^{\mathrm{c}}q\geq \epsilon (w_1+w_2)$ for a quasi-plurisubharmonic $q$ and $\epsilon >0$. Choose a nonzero smooth nonnegative $(1,1)$-form $\zeta $ on the first sphere supported where $1+\tau <\epsilon $. Its pullback is closed, so distributional Stokes gives
\[ 0\leq \int _ Y(\theta _1+\mathrm{dd}^{\mathrm{c}}q-\epsilon (w_1+w_2))\wedge \zeta =\int _{\mathbb P^1}(1+\tau -\epsilon )\zeta <0, \]
a contradiction. Thus the positive segment ends outside the big cone, although its upper ordinary-type volume has a positive limit.
$\square$
Pome
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