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12.35.8 A ruled residual and its subtraction costE45BF47B8568

On the ruled surface of subsection 20.3.63, the Gauduchon variational volume has a different dependence on the data from the two volumes of an ordinary type. Keep $Y=\mathbb P^1\times \mathbb P^1$, $\theta =cw_1+aw_2$, $0\leq b<c$ and $\beta =(c-b)w_1+aw_2$. Let $m>0$ be smooth on the first curve, with $\int mw=1$, and put

\[ \mu =mw_1\wedge w_2,\qquad k=c-b,\qquad H=\int _{\mathbb P^1}\frac{m}{a}w. \]

In dimension two the Gauduchon tests are all smooth positive $(1,1)$-forms $T$ with $\mathrm{dd}^{\mathrm{c}}T=0$, and $D_\mu (T)=\int _ Y\sqrt{T^2/\mu }\, \mu $.

Proposition 12.35.9.E45E9799C813 The all-test infimum is attained and

DA1A40EA1D3A 12.35.10
\begin{equation} \mathcal V_\mu (\theta ;bD)=\mathcal V_\mu (\beta )=\frac{2(c-b)}{H}, \qquad \mathcal V_\mu (\theta )=\frac{2c}{H}. \end{equation}

Writing $V=2k/H$ and $g=Vm/(2a)$, a minimizing test is the closed positive form

\[ T=\frac ga w_1+w_2. \]

Proof. Since $\int gw=k$, solve $\mathrm{dd}^{\mathrm{c}}f=(g-k)w$ on the first sphere. Then $\Omega =\beta +\mathrm{dd}^{\mathrm{c}}f=gw_1+aw_2>0$ and $\Omega ^2=V\mu $. For every Gauduchon test, Stokes and the matrix arithmetic–geometric mean inequality give

\[ \int _ Y\beta \wedge T=\int _ Y\Omega \wedge T \geq \sqrt V\, D_\mu (T). \]

The displayed $T$ is closed, since $g/a$ depends only on the first factor, and $T=a^{-1}\Omega $, so equality holds pointwise. Explicitly,

\[ \int _ Y\beta \wedge T=2k, \qquad D_\mu (T)=\sqrt V\int \frac ma w=\sqrt V H. \]

This proves the infimum over all tests. The curvature of $D$ is $w_1$, so the divisorial residual is indeed $\beta $. Replacing $k$ by $c$ gives the formula for $\theta $. $\square$

Fix a real constant $d$ and a smooth real first-factor function $e$. On an interval where $c_ s=c-sd>b$ and $a_ s=a-se>0$, keep $b,m$ fixed and put

\[ k_ s=c_ s-b,\qquad H_ s=\int \frac{m}{a_ s}w, \qquad V_ s=\frac{2k_ s}{H_ s},\qquad T_ s=\frac{V_ sm}{2a_ s^2}w_1+w_2. \]

Corollary 12.35.11.F402CC8C0A7E For the residual background $\beta _ s=k_ sw_1+a_ sw_2$,

10918993D19D 12.35.12
\begin{equation} \frac{\mathrm d}{\mathrm ds}\sqrt{V_ s} =-\frac{\int _ Y(d w_1+e w_2)\wedge T_ s}{D_\mu (T_ s)}. \end{equation}

Consequently, for every compact segment $[s_0,s_1]$ in that interval,

\[ \sqrt{V_{s_0}}-\sqrt{V_{s_1}} =\int _{s_0}^{s_1} \frac{\int _ Y(d w_1+e w_2)\wedge T_ s}{D_\mu (T_ s)}\, \mathrm ds. \]

Proof. We have $H_ s'=\int em/a_ s^2\, w$, $D_\mu (T_ s)=\sqrt{V_ s}H_ s$ and

\[ \int _ Y(d w_1+e w_2)\wedge T_ s=d+\frac{V_ s}{2}H_ s'. \]

Differentiating $V_ s=2k_ s/H_ s$ gives (12.35.12); integration proves the second identity. This is the explicit first variation of proposition 12.3.2 along the residual family. $\square$

For $c=1$, $b=1/4$, $a=2+\tau $ and $m=1$, the height $\tau $ has distribution $\mathrm d\tau /2$ under $w$. Hence $H=(\log 3)/2$, and the residual and original variational volumes are $3/\log 3$ and $4/\log 3$. The ordinary-type volumes from (20.3.68) are instead $3/2$ and $9/2$, independent of $m$. The subtraction identity differentiates the density-dependent variational volume. Pure-pullback invariance of the ordinary-type masses does not identify an arbitrary smooth positive probability density upstairs with the degenerate pullback of $\mu $.

The distinction persists along a segment ending at a nef background. The following surface example combines the subtraction cost with an imposed nonanalytic ordinary type, including the range in which its closed first moment is infinite.

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. $\square$

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