Proposition 24.4.13.3D0A0BF3A375 The form $\Gamma $ is strictly positive and $\mathrm{dd}^{\mathrm{c}}$-closed exactly when $a$ is a positive constant and $g>0$. For such a form put $\bar g=T^{-1}\int _0^ Tg(s)\, \mathrm ds$ and $d_\Gamma =\int _ X\omega \wedge \Gamma $. Its positive square root and volume are
\[ \gamma =\sqrt a\, \eta +\frac{g(t)}{2\sqrt a}D, \qquad \int _ X\gamma ^3=\frac32TA\sqrt a\, \bar g \leq \frac{d_\Gamma ^{3/2}}{\sqrt V}. \]
Equality holds exactly when $\bar g=2a$, without requiring $g$ to be constant. In particular every degree-one test in this family has root volume at most $V^{-1/2}$.
Proof.
Since $\eta \wedge c=0$, the form $g\eta \wedge D$ is closed. The chain rule gives
\[ \mathrm{dd}^{\mathrm{c}}(a\eta ^2)=a''D\wedge \eta ^2, \]
the $a'c\wedge \eta ^2$ term having excess base degree. Thus $\mathrm{dd}^{\mathrm{c}}\Gamma =0$ forces $a''=0$, and periodicity makes $a$ constant. In a positive coframe with $\eta =\sigma _1+\sigma _2$, $D=\sigma _3$, the complementary coefficient matrix of $\Gamma $ is $\operatorname {diag}(g,g,2a)$. Hence positivity is exactly $a,g>0$. The displayed $\gamma $ is positive and squares to $\Gamma $. Multiplication and fiber integration give
\[ d_\Gamma =TA(a+\bar g),\qquad \gamma ^3=\tfrac 32\sqrt a\, g\eta ^2\wedge D. \]
For fixed $d_\Gamma >0$ set $s=d_\Gamma /(TA)$. Then $0<a<s$, $\bar g=s-a$, and $\sqrt a(s-a)$ has its unique maximum at $a=s/3$. Substitution proves the bound and its equality condition.
$\square$
Pome
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