53E844FE515C
The Hodge star converts the positive $(n-1,n-1)$-form $\Gamma _ u$ into a Hermitian matrix whose logarithmic determinant is a concave function of the complex Hessian. Two features will control the largest eigenvalue in the second-order estimate: the positive background gives coercivity, and the torsion coefficients omit exactly those gradient indices that would otherwise carry the largest inverse weight. We derive both properties before differentiating the equation.
For the matrix and derivative calculations, set $v=u/(2\pi )$, so that $\mathrm{dd}^{\mathrm{c}}u=\mathrm i\partial \bar\partial v$. All covariant derivatives are taken with respect to the Chern connection $\nabla $ of $\gamma $. In holomorphic coordinates $z^1,\ldots ,z^ n$, write $v_ i=\partial _ i v$ and $v_{\bar j}=\partial _{\bar j}v$; indices range from $1$ to $n$, and $\delta _{ij}$ is the Kronecker symbol. Write $\gamma =\mathrm i\gamma _{a\bar b}\, \mathrm{d}z^ a\wedge \mathrm{d}\bar z^ b$, with repeated indices summed. We denote by $\star _\gamma $ the Hodge star of $\gamma $, and by $\det _\gamma $ the determinant of the Hermitian matrix of a real $(1,1)$-form relative to $\gamma $. Write the torsion of $\nabla $ as $T_{ab}^{c}$ and lower the last index with $\gamma $. Thus
\[ T_{ab\bar c}=\partial _ a\gamma _{b\bar c}-\partial _ b\gamma _{a\bar c}, \qquad T_{ab\bar c}=-T_{ba\bar c}. \]
Let $Q_0=\star _\gamma \gamma _0^{n-1}/(n-1)!$ and
\[ Z=\frac1{(n-1)!}\star _\gamma \operatorname {Re}\bigl (\mathrm i\partial v\wedge \bar\partial \gamma ^{n-2}\bigr ). \]
Here the real $(1,1)$-forms on the right are identified with their Hermitian coefficient matrices relative to $\gamma $. Set
\[ \chi =(\operatorname {tr}_\gamma Q_0)\gamma -(n-1)Q_0, \quad W=(\operatorname {tr}_\gamma Z)\gamma -(n-1)Z, \quad g=\chi +\mathrm i\partial \bar\partial v+W. \]
The linear map $P_\gamma (A)=((\operatorname {tr}_\gamma A)\gamma -A)/(n-1)$ on real $(1,1)$-forms, or Hermitian matrices, is invertible, and the equation of theorem 14.10.1 becomes
D281466AD7B7
14.13.1
\begin{equation} \begin{aligned} Q_ u& :=P_\gamma (g)=Q_0+P_\gamma (\mathrm i\partial \bar\partial v)+Z>0,\\ \log \det _\gamma Q_ u& =h=(n-1)(F+b). \end{aligned} \end{equation}
Here $Q_ u=\star _\gamma \Gamma _ u/(n-1)!$ is the coefficient matrix of $\Gamma _ u$. The tensor $g$ need not be positive. To check the Hessian term, diagonalize $\mathrm i\partial \bar\partial v$ in a unitary frame: the coefficient of $\star _\gamma (\mathrm i\partial \bar\partial v\wedge \gamma ^{n-2})/(n-1)!$ in each diagonal position is the sum of the other diagonal entries divided by $n-1$. If $\eta _ u$ has eigenvalues $a_ i$ relative to $\gamma $, then $Q_ u$ has eigenvalues $\prod _{j\ne i}a_ j$, whose product is $(\prod _ i a_ i)^{n-1}$. This proves the determinant identity.
Lemma 14.13.2.6168D5D4EF96 At a point in a unitary frame,
9.29
\begin{align} 2(n-1)Z_{i\bar j} ={}& \delta _{ij}\sum _{p,k}v_ p\overline{T_{pk\bar k}} -v_ i\sum _ k\overline{T_{jk\bar k}} -\sum _ kv_ k\overline{T_{kj\bar i}}\notag \\ & +\delta _{ij}\sum _{p,k}v_{\bar p}T_{pk\bar k} -v_{\bar j}\sum _ kT_{ik\bar k} -\sum _ kv_{\bar k}T_{ki\bar j}. \tag{14.13.3} \end{align}
In particular $Z_{i\bar j}$ contains no $v_ j$ or $v_{\bar i}$. Moreover
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14.13.4
\begin{equation} 2(n-1)Z_{i\bar i} =\sum _{p\ne i,\ k\ne i} \bigl (v_ p\overline{T_{pk\bar k}}+v_{\bar p}T_{pk\bar k}\bigr ). \end{equation}
Covariant differentiation preserves these coefficient identities. For example $\nabla _1 Z_{1\bar1}$ contains no pure Hessian $v_{11}$ or its conjugate, and $\nabla _{\bar1}\nabla _1 Z_{1\bar1}$ contains no $v_{11\bar1}$ or its conjugate, nor a pure $v_{11}$ or its conjugate.
Proof.
For $n=2$, the defining expression of $Z$ is zero, and so are the formulas below, because the generalized Kronecker symbol with three indices vanishes in two dimensions. Suppose $n\geq 3$ and use coordinates that are unitary at the point. Direct antisymmetrization gives
\[ \bar\partial \gamma =-\frac{\mathrm i}{2}\sum _{a,b,c} \overline{T_{bc\bar a}}\, \mathrm{d}z^ a\wedge \mathrm{d}\bar z^ b\wedge \mathrm{d}\bar z^ c. \]
We use $\bar\partial \gamma ^{n-2}=(n-2)\bar\partial \gamma \wedge \gamma ^{n-3}$. To write the contraction explicitly, put
\[ \delta ^{paj}_{bci} =\det \begin{pmatrix} \delta _{pb}
& \delta _{pc}
& \delta _{pi}
\\ \delta _{ab}
& \delta _{ac}
& \delta _{ai}
\\ \delta _{jb}
& \delta _{jc}
& \delta _{ji}
\end{pmatrix}. \]
Let $A=\star _\gamma (\mathrm i\partial v\wedge \bar\partial \gamma ^{n-2})/(n-1)!$. Expanding the $n-3$ remaining copies of $\gamma $ leaves the three complementary indices and gives
\[ A_{i\bar j}=\frac1{2(n-1)} \sum _{p,a,b,c}v_ p\overline{T_{bc\bar a}}\, \delta ^{paj}_{bci}. \]
Indeed, the factor is $(n-2)(n-3)!/[2(n-1)!]$, and the determinant records the signs of the remaining holomorphic and antiholomorphic indices. Expanding its six terms and using $T_{bc\bar a}=-T_{cb\bar a}$, we obtain
\[ A_{i\bar j}=\frac1{n-1}\left( \delta _{ij}\sum _{p,k}v_ p\overline{T_{pk\bar k}} -v_ i\sum _ k\overline{T_{jk\bar k}} -\sum _ kv_ k\overline{T_{kj\bar i}}\right). \]
Since $Z=(A+A^*)/2$, where $A^*$ denotes the Hermitian adjoint matrix, this gives (14.13.3).
For $i\ne j$, the coefficient of $v_ j$ in this expression vanishes because $T_{jj\bar i}=0$, and the coefficient of $v_{\bar i}$ vanishes in the conjugate expression. For $i=j$, canceling the terms with $p=i$ and $k=i$ gives (14.13.4). These are tensor identities valid in every unitary frame. Since the Chern connection preserves the metric, they can be differentiated in a unitary frame that is parallel at the point. For instance, the first derivative of a summand in (14.13.4) is
\[ (\nabla _1v_ p)\overline{T_{pk\bar k}} +v_ p\nabla _1\overline{T_{pk\bar k}},\qquad p,k\ne 1, \]
together with its conjugate counterpart. A further derivative in the direction $\bar1$ still has $p\ne 1$. This gives the stated exclusions, including the terms in which a derivative falls on a coefficient.
$\square$
The second-order estimate uses these exclusions; an arbitrary first-order perturbation could not be substituted for $Z$.
Lemma 14.13.5.DBC2334A4FC0 Diagonalize $g$ at a point and order its eigenvalues as $\lambda _1\geq \cdots \geq \lambda _ n$. Put
\[ \mu _ i=\frac{\sum _{j\ne i}\lambda _ j}{n-1}>0, \quad t_ i=\mu _ i^{-1},\quad f_ i=\frac1{n-1}\sum _{j\ne i}t_ j, \quad \mathcal F=\sum _ i f_ i=\sum _ i t_ i. \]
Then $\mu _1\leq \cdots \leq \mu _ n$, $f_1\leq \cdots \leq f_ n$, and
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14.13.6
\begin{gather} \sum _ i f_ i\lambda _ i=n,\qquad \frac{\mathcal F}{n(n-1)}\leq f_ k\leq \frac{\mathcal F}{n-1} \quad (k>1),\qquad t_ i\leq (n-1)f_1\quad (i>1), \\ f_ n\geq \mathcal F/n. \notag \end{gather}
Fix $\tau ,D>0$ and assume $Q_0\geq \tau \gamma $ and $\det P_\gamma (g)\leq D$. For $\lambda _1$ larger than a constant depending only on $n,\tau ,D$,
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14.13.7
\begin{equation} \sum _ i f_ i(\chi _{i\bar i}-\lambda _ i) \geq \frac\tau 2\mathcal F. \end{equation}
There is also a uniform positive lower bound for $\mathcal F$ depending only on $n,D$.
Proof.
The ordering follows from the formula for $\mu _ i$, and then from that for $f_ i$. The identity $\sum _ if_ i\lambda _ i=n$, which reflects the relation $\log \det P_\gamma (sg)=\log \det P_\gamma (g)+n\log s$, can be checked directly:
\[ \sum _ i f_ i\lambda _ i =\sum _ j\frac{t_ j}{n-1}\sum _{i\ne j}\lambda _ i =\sum _ jt_ j\mu _ j=n. \]
For $k>1$, the sum defining $f_ k$ contains the largest weight $t_1\geq \mathcal F/n$. The upper bound for $f_ k$ and the estimate for $t_ i$ are immediate, and the largest weight $f_ n$ is at least the average $\mathcal F/n$.
The map $P_\gamma $ is self-adjoint for the matrix trace pairing. Consequently
\[ \sum _ i f_ i\chi _{i\bar i} =\sum _ i t_ i(Q_0)_{i\bar i}\geq \tau \mathcal F. \]
Moreover $\sum _ i\lambda _ i=\sum _ i\mu _ i>\lambda _1>0$, where the strict inequality follows from $\mu _1>0$. Hence $\mu _ n>\lambda _1/n$. Since the other $n-1$ numbers $\mu _ i$ are at least $\mu _1$,
\[ \mu _1^{n-1}\mu _ n\leq \prod _ i\mu _ i\leq D, \qquad \mathcal F\geq \mu _1^{-1} \geq \left(\frac{\lambda _1}{nD}\right)^{1/(n-1)}. \]
For $\lambda _1$ sufficiently large this is at least $2n/\tau $, so $n\leq \tau \mathcal F/2$. Subtracting $\sum _ i f_ i\lambda _ i=n$ therefore proves (14.13.7). Finally, the arithmetic–geometric mean inequality applied to the $t_ i$ gives $\mathcal F\geq nD^{-1/n}$.
$\square$
The concavity of the logarithmic determinant will supply a positive term with an explicit formula. For the matrix function $\Phi (g)=\log \det P_\gamma (g)$, the negative second differential in a complex matrix direction $B$ and its adjoint is
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14.13.8
\begin{equation} -D^2\Phi _ g(B,B^*) =\operatorname {tr}\bigl (P_\gamma (g)^{-1}P_\gamma (B) P_\gamma (g)^{-1}P_\gamma (B)^*\bigr )\geq 0. \end{equation}
This follows by differentiating the inverse matrix. In a unitary frame diagonalizing $g$, the off-diagonal terms of this expression include
\[ \frac{|B_{k\bar1}|^2}{(n-1)^2\mu _ k\mu _1}. \]
When $\lambda _ k<\lambda _1$, that coefficient is exactly
\[ \frac{f_ k-f_1}{\lambda _1-\lambda _ k} =\frac1{(n-1)^2\mu _ k\mu _1}. \]
The expression on the right also makes sense when the two eigenvalues coincide. This positive square and the weights $f_ i$ enter the largest-eigenvalue estimate in the next section.
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