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.
$\square$
Pome
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