Example 8.13.5.0DB9D80DE230 Let $X=(\mathbb C/(\mathbb Z+\mathrm i\mathbb Z))^ n$, with $z_ j=x_ j+\mathrm i y_ j$, and put
\[ \beta _ j=\frac{\mathrm i}{2\pi }\, \mathrm dz_ j\wedge \mathrm d\bar z_ j, \qquad b=2+\cos (2\pi x_1), \qquad \omega =\beta _1+b\beta _2+\sum _{j=3}^ n\beta _ j. \]
For $|t|<\pi ^{-2}$ the potential $u_ t=t\cos (2\pi x_1)$ is smooth strictly admissible, and, with $B_ n=\int _ X\beta _1\wedge \cdots \wedge \beta _ n$,
\[ \int _ X H_ m(u_ t)=n!\left(2-\frac{m\pi ^2t}{2n}\right)B_ n. \]
Thus every $-\pi ^{-2}<t<0$, together with a constant $C>|t|$, gives a pair $u=0$, $v=u_ t+C$ that violates the unmodified global comparison inequality.
Proof.
As in example 11.1.12, $\mathrm{dd}^{\mathrm{c}}u_ t=-\pi ^2t\cos (2\pi x_1)\beta _1$. The coefficient of $\beta _1$ in $\omega +\mathrm{dd}^{\mathrm{c}}u_ t$ is strictly positive in the stated range, and all the other coefficients are positive. Since $(\mathrm{dd}^{\mathrm{c}}u_ t)^2=0$, expansion gives
\[ H_ m(u_ t)=\omega ^ n+m\mathrm{dd}^{\mathrm{c}}u_ t\wedge \omega ^{n-1} =n!b\left(1-\frac mn\pi ^2t\cos (2\pi x_1)\right) \beta _1\wedge \cdots \wedge \beta _ n. \]
The period averages of $b$ and $b\cos (2\pi x_1)$ are respectively $2$ and $1/2$, proving the mass formula. For $t<0$ this mass is strictly larger than $\int _ XH_ m(0)=2n!B_ n$. Meanwhile $C>|t|$ gives $\{ 0<u_ t+C\} =X$, so the comparison inequality for this pair would assert the opposite ordering of these masses.
$\square$
Pome
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