The invariant cones above have complete unit-volume hypersurfaces. On the product of a compact Kähler manifold with a curve, a nonconstant coefficient in the fiber direction gives cones of nonclosed backgrounds whose unit-volume hypersurfaces are flat and complete instead to a strip or a half-plane. The volume is the reciprocal of an integral of the reciprocal fiber coefficient, and the variance of that reciprocal coefficient measures the distance to the boundary. The probability measure in this variance is a trace of a minimizing balanced metric.
Let $(B,\kappa )$ be a connected compact Kähler manifold of dimension $m\geq 2$, with $\int _ B\kappa ^ m=1$. Let $C$ be a connected compact complex curve with a smooth positive $(1,1)$-form $w$ of integral one. Fix a smooth strictly positive probability form $\sigma $ on $B$ and a smooth nonconstant real function $a$ on $B$ with $\min a=0$ and $\max a=M>0$. On $X=B\times C$, of dimension $n=m+1$, put $\mu =\sigma \wedge w$; we omit pullbacks from the notation. For real parameters $b,c,e$, the backgrounds in the real span of $\kappa ,aw,w$ are
\[ \theta _{b,c,e}=b\kappa +Lw,\qquad L=ca+e. \]
Theorem 12.6.32.40DF235D735C This span contains no nonzero smooth $\mathrm{dd}^{\mathrm{c}}$ direction, and its intersection with $\mathscr P$ is the cone
\[ \mathscr C=\{ b>0,\ e>0,\ e+cM>0\} . \]
For $\theta =\theta _{b,c,e}\in \mathscr C$, let $H(c,e)=\int _ B\sigma /L$. Then
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\begin{equation} \mathcal V_\mu (\theta )=\frac{nb^ m}{H(c,e)}. \end{equation}
A minimizing Gauduchon metric is
\[ \gamma =L^{-1/m}(A+Lw),\qquad A=b\kappa +\mathrm{dd}^{\mathrm{c}}\phi >0,\qquad A^ m=\frac{b^ m}{H}\frac{\sigma }{L}, \]
where $\phi $ is a smooth real function on $B$. This metric is balanced. Backgrounds with $c\ne 0$ are nonclosed.
Proof.
Pairing a smooth positive representative $\theta +\mathrm{dd}^{\mathrm{c}}u$ with the closed semipositive form $\kappa ^{m-1}\wedge w$ gives $b>0$, and its integral over the compact fiber $\{ y\} \times C$ is $L(y)>0$, by Stokes' theorem. These are exactly the stated inequalities; conversely, they make $\theta $ itself positive. The same tests applied to a smooth $\mathrm{dd}^{\mathrm{c}}$ direction give $b=0$ and $ca+e=0$ everywhere, and since $a$ is nonconstant, $c=e=0$. Also $\mathrm{d}\theta =c\, \mathrm{d}a\wedge w$, which is not identically zero when $c\ne 0$.
Next we apply theorem 14.2.1 on $B$ to the closed positive background $b\kappa $ and the density $(b^ m/H)\sigma /L$. Since this density has integral $b^ m=\int _ B(b\kappa )^ m$, Stokes' theorem forces the constant in the equation to be one, and we obtain $A$. Put $\Omega =A+Lw$. Since $A^{m+1}=0$ and $(Lw)^2=0$, the binomial expansion gives
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\begin{equation} \Omega ^ n=\frac{nb^ m}{H}\mu ,\qquad \gamma ^ m=L^{-1}A^ m+mA^{m-1}\wedge w. \end{equation}
The first summand is the pullback of a top-degree form on $B$, and the second is closed. Hence $\gamma $ is balanced, and in particular Gauduchon. For every Gauduchon metric $\eta $ on $X$, Stokes' theorem and the pointwise arithmetic–geometric mean inequality give
\[ \int _ X\theta \wedge \eta ^ m =\int _ X\Omega \wedge \eta ^ m \geq \left(\frac{nb^ m}{H}\right)^{1/n}D_\mu (\eta ). \]
For $\eta =\gamma $ the numerator is $nb^ m$, while $D_\mu (\gamma )=(nb^ m/H)^{m/n}H$. The quotient of $\gamma $ therefore attains the lower bound, which proves (12.6.33) and the minimizing property of $\gamma $.
$\square$
The minimizing metric gives the probability identity
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\begin{equation} \frac{Lw\wedge \gamma ^ m}{b^ m} =\frac{\sigma \wedge w}{HL}. \end{equation}
Its marginal $\sigma /(HL)$ on $B$ is the probability measure that governs the Hessian metric. To compute that metric we use the ratio $v=c/e$ and set
\begin{align*} v& =c/e\in (-1/M,\infty ), & Q(v)& =\int _ B\frac{\sigma }{1+va},\\ \mathrm{d}\nu _ v& =\frac{\sigma }{Q(v)(1+va)}, & \ell _ v& =\frac{a}{1+va}. \end{align*}
Proposition 12.6.36.02AE1B3B4027 On the unit-volume hypersurface $\mathscr S=\{ \theta \in \mathscr C:\mathcal V_\mu (\theta )=1\} $, the functions $U=\log b$ and $v$ are global coordinates, in which
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\begin{equation} \begin{aligned} g_{\mathscr S}& =m\, \mathrm{d}U^2+\rho (v)\, \mathrm{d}v^2,\\ \rho (v)& =\frac{Q''/Q-2(Q'/Q)^2}{n} =\frac2n\operatorname {Var}_{\nu _ v}(\ell _ v)>0. \end{aligned} \end{equation}
For any $v_*\in (-1/M,\infty )$, the coordinates
\[ Y=\sqrt m\, U,\qquad R(v)=\int _{v_*}^ v\sqrt{\rho (t)}\, \mathrm{d}t \]
define a global isometry onto $\mathbb R\times I$, where $I$ is an open interval with the Euclidean metric. The intrinsic distance is the Euclidean distance in $(Y,R)$.
Proof.
Consider an affine variation $\delta =(\delta b,\delta c,\delta e)$ of the coefficients, write $F=\delta L/L$, and note that $\sigma /(HL)=\mathrm{d}\nu _ v$ because $HL=Q(v)(1+va)$. Differentiating under the integral sign gives
\[ D^2\log H(\delta ,\delta ) =2\mathbb E_{\nu _ v} F^2-(\mathbb E_{\nu _ v} F)^2, \]
where $\mathbb E_{\nu _ v}$ denotes expectation with respect to $\nu _ v$. This is positive unless $\delta L=0$ everywhere, which forces $\delta c=\delta e=0$. Since $\log \lambda _\mu =(\log n+m\log b-\log H)/n$,
\[ g=-D^2\log \lambda _\mu =\frac mn\left(\frac{\mathrm{d}b}{b}\right)^2 +\frac1nD^2\log H \]
is positive definite. On the unit-volume hypersurface, the relation $H(c,e)=Q(v)/e=nb^ m$ gives
\[ e=\frac{Q(v)}{n\exp (mU)},\qquad c=ve. \]
To pull back the Hessian, write $t=\mathrm{d}e/e$ and $z=Q'/Q$. Along an affine path of coefficients one has $v''=-2tv'$, so differentiating $\log H=-\log e+\log Q(v)$ twice gives
\[ D^2\log H=t^2-2zt\, \mathrm{d}v +(Q''/Q-z^2)\, \mathrm{d}v^2. \]
On $\mathscr S$ we have $t=z\, \mathrm{d}v-m\, \mathrm{d}U$; substituting this and using $n=m+1$ gives (12.6.37). Since $Q'/Q=-\mathbb E_{\nu _ v}\ell _ v$ and $Q''/Q=2\mathbb E_{\nu _ v}\ell _ v^2$, the variance formula follows, and nonconstancy of $a$ makes this variance positive. In the coordinates $(Y,R)$ the metric is Euclidean on the convex set $\mathbb R\times I$, so straight segments prove the distance assertion.
$\square$
Pome
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