Example 12.6.53.27A22321E298 On $B=\mathbb P^2$, take normalized Fubini–Study $\kappa $, $\sigma =\kappa ^2$, $C=\mathbb P^1$, and
\[ a([Z])=\frac{|Z_1|^2/2+|Z_2|^2}{|Z_0|^2+|Z_1|^2+|Z_2|^2}. \]
Then $M=1$, $r_0=r_ M=2$, and the completion is a closed strip with finite faces
\[ b\kappa +\frac{4\log 2}{3b^2}a w,\qquad b\kappa +\frac{4\log 2}{3b^2}(1-a)w. \]
For $v\ne 0$,
\[ Q(v)=\frac4{v^2}\bigl [(v+1)\log (1+v) -(v+2)\log (1+v/2)\bigr ], \qquad Q(0)=1. \]
The width of the strip is the finite positive integral
\[ \int _{-1}^{\infty } \sqrt{\frac{Q''/Q-2(Q'/Q)^2}{3}}\, \mathrm{d}v. \]
Proof.
The extreme sets are the points $[1:0:0]$ and $[0:0:1]$. In affine coordinates, $a$ and $1-a$ vanish there to exactly second order, so both real codimensions are four and both orders are two. To compute the law of $a$ under $\sigma $, polar integration in the chart $Z_0=1$ gives $2\, \mathrm{d}s_1\mathrm{d}s_2/(1+s_1+s_2)^3$, where $s_ j=|z_ j|^2$. The transformation $u_ j=s_ j/(1+s_1+s_2)$ gives the uniform density two on the triangle $u_1,u_2\geq 0$, $u_1+u_2\leq 1$, and $a=u_1/2+u_2$. For $0\leq s\leq 1/2$, the triangle $u_1/2+u_2\leq s$ has area $s^2$, so the law of $a$ has density $4s$ on $[0,1/2]$. With $u_0=1-u_1-u_2$, exchanging $u_0$ and $u_2$ preserves the uniform measure and replaces $a$ by $1-a$; hence the law has density $4(1-s)$ on $[1/2,1]$. Integrating against these two densities gives $H_ a=H_-=4\log 2$ and the formula for $Q$, and the integral definition of $Q$ shows that $Q(0)=1$ is its continuous extension. Theorem 12.6.41 gives the boundary lines and the finite width, with Euclidean coordinates $Y=\sqrt2\log b$ and $R(v)$.
$\square$
Example 12.6.54.CD9181602A11 Let $E=\mathbb C/(\mathbb Z+i\mathbb Z)$, $B=E^ m$, and write $z_ j=u_ j+i\tau _ j$. Put
\[ \eta _ j=\frac i2\mathrm{d}z_ j\wedge \mathrm{d}\bar z_ j, \qquad s=(m!)^{-1/m},\qquad \kappa =s\sum _ j\eta _ j,\qquad \sigma =\kappa ^ m. \]
For $a=\sin ^2(\pi u_1)$, the unit-volume hypersurface is itself a complete Euclidean plane, with Euclidean coordinates
\[ Y=\sqrt m\log b,\qquad R=\frac{\log (1+v)}{2\sqrt n},\qquad v>-1. \]
Its minimizing balanced metrics can be written down by integrating a periodic function twice.
Proof.
The forms descend to $B$, and $\sigma =\eta _1\wedge \cdots \wedge \eta _ m$ is a probability form. Both extreme sets are real hypersurfaces along which $a$, respectively $1-a$, vanishes to second order, so $r_0=r_ M=1/2$. Splitting the period and substituting $t=\tan (\pi u)$ gives
\begin{align*} Q(v)& =\int _0^1\frac{\mathrm{d}u}{1+v\sin ^2(\pi u)}\\ & =\frac2\pi \int _0^\infty \frac{\mathrm{d}t}{1+(1+v)t^2} =\frac1{\sqrt{1+v}}. \end{align*}
Let $F_ v(u)=[Q(v)(1+v\sin ^2(\pi u))]^{-1}$, which has mean one over a period. Integrating the equation
\[ \phi _ v''=4\pi bs(F_ v-1) \]
twice, whose right side has mean zero, gives a smooth periodic potential $\phi _ v$ of $u=u_1$; the first integration constant is chosen so that $\phi _ v'$ has mean zero. In our normalization $\mathrm{dd}^{\mathrm{c}}\phi _ v=\phi _ v''\eta _1/(4\pi )$, hence
\[ A=bs\left(F_ v\eta _1+\sum _{j=2}^ m\eta _ j\right)>0, \qquad A^ m=b^ mF_ v\sigma =\frac{b^ m}{H}\frac{\sigma }{L}. \]
Thus the minimizer of theorem 12.6.32, with its cofactor formula (12.6.34), is explicit in this example, without an appeal to theorem 14.2.1. Differentiating $Q$ gives
\[ \rho (v)=\frac1{4n(1+v)^2},\qquad g_{\mathscr S}=m\, \mathrm{d}U^2+ \frac1{4n}\left(\frac{\mathrm{d}v}{1+v}\right)^2. \]
As $U$ and $v$ range over $\mathbb R$ and $(-1,\infty )$, the coordinates $(Y,R)$ range over all of $\mathbb R^2$. Both boundary faces of the coefficient cone are therefore at infinite distance on the unit-volume hypersurface.
$\square$
These distances are those of the fixed three-parameter cone with the product probability $\mu =\sigma \wedge w$. The endpoint criterion of lemma 12.6.39 uses the hypothesis that the extreme sets are disjoint unions of submanifolds along which $a$ vanishes to finite power order; intersecting extreme strata or vanishing of infinite order would require further estimates. The variances are computed from the minimizing metrics, but we do not assert that these metrics have a pointwise limit at a finite end.
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