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12.19.12 The volume of the positive root92006A543CD2

The two averages contribute complementary blocks of the coefficient matrix of a $(2,2)$-form: the translated forms give a large vertical entry, and the curves give a positive horizontal block. A Schur-complement estimate combines them into a lower bound for the volume of the positive root. In the coframe $(x,y,\zeta )$ put $\mathrm{d}V_0=\sigma _ x\wedge \sigma _ y\wedge \sigma _\zeta $. For a real $(1,1)$-form $\xi $, let $H_\xi $ be its coefficient matrix in this normalization. A real $(2,2)$-form $\Theta $ corresponds to a Hermitian matrix $Q_\Theta $, which we call its cofactor matrix, by

\[ \xi \wedge \Theta =\operatorname {tr}(H_\xi Q_\Theta )\, \mathrm{d}V_0. \]

Strict positivity of $\Theta $ is equivalent to $Q_\Theta >0$.

If $\eta ^2=\Theta $ and $H=H_\eta $, expanding the wedge products gives

\[ Q_\Theta =2(\det H)H^{-1},\qquad \eta ^3=\frac{3}{\sqrt2}(\det Q_\Theta )^{1/2}\, \mathrm{d}V_0. \]

Conversely, these formulas uniquely recover a smooth positive $H$ from $Q_\Theta >0$; this is the three-dimensional case of the root construction used in the proof of theorem 11.6.1. We write $\Theta ^{3/2}=\eta ^3$.

Lemma 12.19.13.BD9C00A32F59 There are smooth strictly positive $\mathrm{dd}^{\mathrm{c}}$-closed $(2,2)$-forms $\Gamma _ j$ such that

\[ \int _ X\omega \wedge \Gamma _ j\leq C_2,\qquad R_ j=\int _ X\Gamma _ j^{3/2}\longrightarrow +\infty . \]

Proof. Write the cofactor matrix of $\widehat\Theta _ j$ in block form, with a $2\times 2$ Hermitian block $A_ j$ in the $(x,y)$-directions and a column $\kappa _ j\in \mathbb C^2$:

\[ Q_ j=\begin{pmatrix} A_ j & \kappa _ j \\ \kappa _ j^* & \widehat F_ j \end{pmatrix}. \]

The cofactor matrix of $T_{j,0}$ has the form $\operatorname {diag}(B_ j,0)$. Let $I_2$ denote the $2\times 2$ identity matrix. The lower bound for $K_ f$ and the definition of $\rho _ j$ give a constant $c_0>0$ such that $B_ j\geq c_0\delta _ j^{-2}I_2$ on $V_ j$. Since $Q_ j>0$, the Schur complement $A_ j-\kappa _ j\kappa _ j^*/\widehat F_ j$ is positive definite, and

\[ \det \bigl (Q_ j+\operatorname {diag}(B_ j,0)\bigr ) =\widehat F_ j\det \left(B_ j+A_ j- \frac{\kappa _ j\kappa _ j^*}{\widehat F_ j}\right) \geq \widehat F_ j\det B_ j, \]

by monotonicity of the determinant on positive Hermitian matrices. Combining the root formula with (12.19.8), we obtain

\[ (\widehat\Theta _ j+T_{j,0})^{3/2} \geq c\delta _ j^{-2}\sqrt{J_ j}\, \mathrm{d}V_0 \quad \hbox{on }V_ j. \]

The smaller region $W_ j=\{ |x|,|y|<1/8,\ |\zeta |<\delta _ j/8\} $ has $\mathrm{d}V_0$-volume equal to a fixed positive constant times $\delta _ j^2$. Consequently

\[ I_ j=\int _{W_ j}(\widehat\Theta _ j+T_{j,0})^{3/2} \geq c_1\sqrt{J_ j}>c_1j^2. \]

For this fixed $j$, lemma 12.19.10 gives smooth convergence of $T_{j,\varepsilon }$ to $T_{j,0}$ on the closure of $W_ j$. The cofactor matrix of $\widehat\Theta _ j+T_{j,0}$ is positive definite there, so the integrand of the root volume depends continuously on these smooth coefficients. We can therefore choose $0<\varepsilon _ j\leq 1/16$ so small that

\[ \int _{W_ j}(\widehat\Theta _ j+T_{j,\varepsilon _ j})^{3/2} \geq I_ j/2. \]

Set $\Gamma _ j=\widehat\Theta _ j+T_{j,\varepsilon _ j}$. It is smooth, strictly positive and $\mathrm{dd}^{\mathrm{c}}$-closed. Its degree is bounded by (12.19.7) and lemma 12.19.10, whereas its root volume is at least $c_1j^2/2$. $\square$

of theorem 12.19.1. Fix a Hermitian form $h$ on $X$, and choose $A>0$ with $h\leq A\omega $. Let $\eta _ j$ be the positive root of $\Gamma _ j$, so that $\eta _ j^2=\Gamma _ j$. These metrics are Gauduchon, and

\[ P_ j=\int _ Xh\wedge \eta _ j^2\leq AC_2,\qquad R_ j=\int _ X\eta _ j^3\longrightarrow +\infty . \]

Set $\mu _ j=\eta _ j^3/R_ j$ and solve the smooth equation in the background $h$:

\[ h_ j=h+\mathrm{dd}^{\mathrm{c}}u_ j>0,\qquad h_ j^3=c_ h(\mu _ j)\mu _ j, \qquad \sup _ Xu_ j=0. \]

Since $D_{\mu _ j}(\eta _ j)=R_ j^{2/3}$, using $\eta _ j$ as a test in theorem 12.2.3 gives

\[ 0<\int _ Xh_ j^3=c_ h(\mu _ j)=\mathcal V_{\mu _ j}(h) \leq \frac{P_ j^3}{R_ j^2}\longrightarrow 0. \]

The potentials $u_ j$ are smooth, hence bounded, so $v_-(h)=0$. $\square$

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