Pome

Example 11.1.12F336FEED9D87

Example 11.1.12.F336FEED9D87 Let $X=(\mathbb C/(\mathbb Z+\mathrm i\mathbb Z))^2$. For $j=1,2$, let $z_ j=x_ j+\mathrm i y_ j$ be the standard coordinates on the universal cover $\mathbb C^2$, with $x_ j,y_ j$ real, and put

\[ \beta _ j=\frac{\mathrm i}{2\pi }\, \mathrm{d}z_ j\wedge \mathrm{d}\bar z_ j. \]

The forms $\beta _ j$ descend to $X$. For a real parameter $t$, define

\[ b(x_1)=2+\cos (2\pi x_1),\qquad \omega =\beta _1+b(x_1)\beta _2,\qquad u_ t=t\cos (2\pi x_1). \]

Since $1\leq b\leq 3$, the form $\omega $ is Hermitian. It is not closed, because $\mathrm{d}\omega =\mathrm{d}b\wedge \beta _2$ is not identically zero.

For a function $f$ of $x_1$ alone, one has $\mathrm{dd}^{\mathrm{c}}f=(f''/4)\beta _1$, where the prime denotes differentiation in $x_1$. Hence

\[ \omega _{u_ t} =\bigl (1-\pi ^2t\cos (2\pi x_1)\bigr )\beta _1+b\beta _2>0 \quad \text{if}\quad |t|<\pi ^{-2}. \]

Writing $B=\int _ X\beta _1\wedge \beta _2>0$, we have

914CB776FFB8 11.1.13
\begin{equation} \mathrm M_\omega (u_ t)=(4-\pi ^2t)B. \end{equation}

Indeed,

\[ \omega _{u_ t}^2 =2\bigl (1-\pi ^2t\cos (2\pi x_1)\bigr ) \bigl (2+\cos (2\pi x_1)\bigr )\beta _1\wedge \beta _2, \]

and over a period the average of $\cos (2\pi x_1)$ is zero while that of its square is $1/2$.

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