Lemma 16.16.1C8D21E3D949A

Proof. Step 1 reduces to a metric defined by $\kappa $ and collects the bounds for $a_\beta $ and its gradient. Steps 2 and 3 apply the maximum principle to the logarithm of the largest eigenvalue of $\nabla ^2a_\beta $, corrected by functions of the gradient and of $a_\beta $, and Step 4 absorbs the negative term with third derivatives.

Step 1: reductions and first bounds. Any two smooth Riemannian metrics on $X$ are uniformly equivalent, and, as observed before the lemma, the Hessians of a function for the two metrics differ by a bounded tensor applied to its gradient. Changing $g_0$ therefore changes $B$ and the constants only by factors depending on the two metrics, and we may assume that $g_0$ is the following Riemannian metric defined by $\kappa $. In local holomorphic coordinates write $\kappa =\frac{\mathrm i}{2\pi }k_{i\bar j}\, \mathrm dz_ i\wedge \mathrm d\bar z_ j$, so that $k$ is the coefficient matrix of $2\pi \kappa $ as in section 16.15, and let $g_0=2\operatorname{Re}(k_{i\bar j}\, \mathrm dz_ i\otimes \mathrm d\bar z_ j)$; this is $2\pi $ times the Riemannian metric $\kappa (\cdot ,J\cdot )$, where $J$ is the complex structure. At the center of normal coordinates for $\kappa $, given by lemma 16.15.3, we have $k_{i\bar j}=\delta _{ij}$ and $\mathrm dk=0$; hence the Christoffel symbols of $g_0$ vanish and $g_0=2\sum _ i(\mathrm dx_ i^2+\mathrm dy_ i^2)$. Comparing coordinate derivatives at such a point, we obtain for every smooth function $b$

In particular the upper bound for $\nabla ^2f$ gives $D_ kf\leq nB$.

Write $a=a_\beta $, $G=(k_{i\bar j}+a_{i\bar j})$ and $D_ G=G^{i\bar j}\partial _ i\partial _{\bar j}$; thus $G$ is the matrix denoted $g$ in section 16.15. We apply lemma 16.15.7 with $\kappa $ also as the measuring metric. It bounds $|a|$ by $\| f\| _\infty $ and, for all sufficiently large $\beta $, bounds $\operatorname {tr}_ kG$ in terms of $B$ and the fixed geometry. Since $G>0$ and $\operatorname {tr}_ kG$ is bounded, all coefficients of $G$, and hence of the complex Hessian of $a$, are bounded; in particular the metric Laplacian $La$ for the fixed Kähler form $\omega $, which is a constant multiple of the trace of $\mathrm{dd}^{\mathrm{c}}a$ with respect to $\omega $ (lemma 10.2.12), is uniformly bounded. Fix $p>2n$. By proposition 16.14.9, $\| a\| _{W^{2,p}(X)}\leq C(\| La\| _{L^ p}+\| a\| _{L^ p})$ is uniformly bounded, and Morrey's inequality (lemma 2.9.16), applied in the charts defining the norm to the first derivatives of the products of $a$ with the cutoff functions, bounds $\nabla a$ uniformly. From now on $C$ denotes a constant depending only on these bounds, on $B$ and on the fixed geometry; it may increase from line to line, but it never depends on $\beta $.

Step 2: the auxiliary function. Let $\ell $ be the largest eigenvalue of the real covariant Hessian $\nabla ^2a$ with respect to $g_0$. By (16.16.2), the sum of all eigenvalues is $2D_ ka=2\operatorname {tr}_ kG-2n\geq -2n$. Since each eigenvalue is at most $\ell $, it is at least $-2n-(2n-1)\ell $; hence every eigenvalue of $\nabla ^2a$ has absolute value at most $2n(\max \{ \ell ,0\} +1)$. It therefore suffices to bound $\ell $ from above, and we do so by the maximum principle. Put

Then $S\leq C$, the number $\tau \in (0,1]$ has a uniform positive lower bound, and since $0\leq s\leq S$,

For a constant $A>0$ to be chosen later, consider the function $Q=\log \ell +\eta (s)-Aa$ on the open set $\{ \ell >0\} $. If this set is empty, then $\ell \leq 0$ and there is nothing to prove. Otherwise, since $a$ and $\eta (s)$ are bounded, $Q$ tends to $-\infty $ where $\ell $ tends to zero, so by compactness of $X$ it attains its maximum at some point $x_0$ with $\ell (x_0)>0$. For the same reason a uniform bound for $\ell (x_0)$ bounds $\ell $ on all of $X$. We may therefore assume that $\ell (x_0)\geq 1$, and that $\ell (x_0)$ exceeds any uniform threshold used below.

Choose normal coordinates for $\kappa $ at $x_0$ in which $G(x_0)$ is diagonal, with diagonal entries $r_ i>0$: by lemma 16.15.3, applied to $k$ in one of finitely many fixed charts that contains $x_0$ in a fixed compact subset, and followed by a unitary change of coordinates that diagonalizes $G(x_0)$, there are such coordinates in which the derivatives of $k$ of every fixed order at $x_0$ are bounded independently of $x_0$. Since $\sum _ ir_ i=\operatorname {tr}_ kG\leq C$, we have $r_ i\leq C$, hence $r_ i^{-1}\geq c$ for a uniform $c>0$. Write $T=\sum _ ir_ i^{-1}$; thus $T\geq c$. Take a $g_0$-orthonormal basis $V_1,\ldots ,V_{2n}$ of the real tangent space at $x_0$ consisting of eigenvectors of $\nabla ^2a(x_0)$, with $V_1$ belonging to the eigenvalue $\ell (x_0)$, and extend these vectors to the chart as vector fields with constant real coordinate coefficients.

The largest eigenvalue $\ell $ may be multiple and need not be smooth, so we perturb the Hessian. Let $B_0$ be the symmetric bilinear form with constant coefficients in this chart which at $x_0$ is the $g_0$-orthogonal projection onto $V_1^\perp $, that is, $B_0(Y,Z)=g_0(PY,PZ)$ at $x_0$ for this projection $P$, and set

The form $B_0$ is positive semidefinite everywhere in the chart. Hence the largest eigenvalue $\lambda _1(\Phi )$ of $\Phi $ is at most $\ell $, and it equals $\ell $ at $x_0$. Denote the eigenvalues of $\Phi $ at $x_0$ by $\lambda =\ell (x_0),\lambda _2,\ldots ,\lambda _{2n}$, with eigenvectors $V_1,\ldots ,V_{2n}$. For $\alpha >1$, $\lambda _\alpha $ is the Hessian eigenvalue belonging to $V_\alpha $ minus one, so $\lambda -\lambda _\alpha \geq 1$. In particular $\lambda _1(\Phi )$ is a smooth function near $x_0$. Indeed, $\lambda $ is a simple root of the characteristic polynomial of $\Phi (x_0)$, so by the implicit function theorem the characteristic polynomial of $\Phi $ has, near $x_0$, exactly one root close to $\lambda $, and this root depends smoothly on the point; it equals $\lambda _1(\Phi )$, which is continuous as the maximum of $(\nabla ^2a-B_0)(Y,Y)$ over $g_0$-unit vectors $Y$, and it is a simple eigenvalue. The smooth function $\widehat Q=\log \lambda _1(\Phi )+\eta (s)-Aa$ satisfies $\widehat Q\leq Q\leq Q(x_0)=\widehat Q(x_0)$, so it has a local maximum at $x_0$.

Step 3: the inequality at the maximum point. From now on all quantities are evaluated at $x_0$. Write $a_{V_\alpha V_\gamma }=\nabla ^2a(V_\alpha ,V_\gamma )$, a smooth function near $x_0$, and define the nonnegative quantities

We first compute $D_ G\log \lambda _1(\Phi )$, keeping the third-derivative terms. Let $M(t)$ be a smooth family of real symmetric matrices, and let $\lambda (t)$ be a simple eigenvalue with unit eigenvector $V_1$ at $t=0$, the other eigenvalues being $\lambda _\alpha $ with orthonormal eigenvectors $V_\alpha $. Then

Indeed, near $t=0$ the simple eigenvalue $\lambda (t)$ is a smooth function of $t$, by the implicit function theorem applied to the characteristic polynomial, and the columns of the adjugate matrix of $\lambda (t)I-M(t)$, which has rank one, span the eigenspace; normalizing a column that does not vanish at $t=0$ gives a smooth unit eigenvector $y(t)$ with $y(0)=\pm V_1$, and we may take the sign $+$. Differentiate the eigenvector equation $M(t)y(t)=\lambda (t)y(t)$. Pairing with $V_1$ gives the first formula, and pairing with $V_\alpha $ shows that the derivative of the eigenvector has component $\langle M'V_1,V_\alpha \rangle /(\lambda -\lambda _\alpha )$ along $V_\alpha $, and it has no component along $V_1$, because $|y(t)|=1$. Differentiating the identity $\lambda '(t)=\langle M'(t)y(t),y(t)\rangle $, where $y(t)$ is the normalized eigenvector with $y(0)=V_1$, and substituting these components gives the second formula.

We apply these formulas to the symmetric matrix $M$ with entries $(\nabla ^2a-B_0)(e_\alpha ,e_\gamma )$, where $e_1,\ldots ,e_{2n}$ is the $g_0$-orthonormal frame obtained from the constant vector fields $V_1,\ldots ,V_{2n}$ by the Gram–Schmidt process; $M$ has the same eigenvalues as $\Phi $. The frame depends smoothly on the point, equals $V_1,\ldots ,V_{2n}$ at $x_0$, and its first derivatives vanish at $x_0$, because those of $g_0$ do. We differentiate along the real coordinate directions and use $\partial _ i\partial _{\bar i}=\frac14(\partial _{x_ i}^2+\partial _{y_ i}^2)$. Since $B_0$ is constant, the first derivatives of $M$ at $x_0$ are those of the functions $\nabla ^2a(V_\alpha ,V_\gamma )$. In the second derivatives, the second derivatives of the frame multiply $\nabla ^2a-B_0$, which is bounded by $C(\lambda +1)$; after weighting with $r_ i^{-1}$ and summing they contribute at most $C\lambda T$. Since $|\partial _ ib|^2=\frac14(|\partial _{x_ i}b|^2+|\partial _{y_ i}b|^2)$ for a real function $b$, we obtain

Here the middle term comes from $\sum _ ir_ i^{-1}\partial _ i\partial _{\bar i}a_{V_1V_1}$ after commuting derivatives, and we check that this commutation costs at most $C\lambda T$. In real coordinates $a_{V_1V_1}=V_1V_1a-\Gamma ^ t_{uv}V_1^ uV_1^ v\partial _ ta$, where the Christoffel symbols $\Gamma $ of $g_0$ vanish at $x_0$. When we apply $\partial _ i\partial _{\bar i}$, the term containing third derivatives of $a$ is multiplied by $\Gamma (x_0)=0$; the remaining terms are a first derivative of $\Gamma $ times a second derivative of $a$, or a second derivative of $\Gamma $ times a first derivative of $a$. By the Hessian and gradient bounds above they are at most $C(\lambda +1)$. Moreover $\partial _ i\partial _{\bar i}V_1V_1a=V_1V_1a_{i\bar i}=V_1V_1G_{i\bar i}-V_1V_1k_{i\bar i}$, and the derivatives of the fixed matrix $k$ are bounded. Summing with the weights $r_ i^{-1}$ gives an error at most $C\lambda T$, and no third derivative of $a$ enters it.

Next we use the equation. Taking logarithms in (16.15.8) gives $\log \det G=\log \det k+\beta (a-f)$. Differentiating twice along $V_1$ and using $\partial (G^{-1})=-G^{-1}(\partial G)G^{-1}$, we obtain at $x_0$

Here we used that at the center of normal coordinates directional and covariant second derivatives agree, so that $V_1V_1a=\lambda $ and $V_1V_1f=f_{V_1V_1}$. If $\lambda \geq 2\max \{ 1,\sup _{X,|Y|_{g_0}=1}f_{YY}\} $, which is a uniform threshold because $\nabla ^2f\leq Bg_0$, the last term is at least $\beta \lambda /2$. The term $V_1V_1\log \det k$ is bounded, hence at least $-CT$ because $T\geq c$. By the first variation formula, $\partial _ i\lambda _1(\Phi )=\partial _ i a_{V_1V_1}$ at $x_0$. Since $D_ G\log \lambda _1=\lambda ^{-1}D_ G\lambda _1-\sum _ ir_ i^{-1}|\partial _ i\lambda _1|^2/\lambda ^2$, we conclude that

Next we differentiate the logarithmic equation once and apply the product rule to $s=k^{p\bar q}a_ pa_{\bar q}$. At $x_0$,

The last term involves only the bounded gradient. In the middle term the third derivatives occur only in the contraction $\sum _ ir_ i^{-1}\partial _ pa_{i\bar i}=\partial _ p\log \det G=\beta (a_ p-f_ p)$, because $\mathrm dk(x_0)=0$; so we need no estimate for third derivatives. Weighting with $r_ i^{-1}$ and summing, we get

Also $D_ Ga=\sum _ ir_ i^{-1}(r_ i-1)=n-T$. At the local maximum $x_0$ we have $D_ G\widehat Q\leq 0$, where

Since $0<\eta '\leq \frac12$ and $\eta '|\partial f|_ k^2\leq \frac\tau 2\sup _ X|\partial f|_ k^2\leq \frac14$, the three preceding estimates yield

where $C_0$ is a uniform constant that does not depend on $A$.

Step 4: absorption of the third-order term. It remains to absorb the negative term $-N$ completely. Fix $0<\varepsilon <1/2$, to be chosen below, and assume that $\lambda \geq C/\varepsilon ^2$. Let $J$ be the complex structure of $X$, and define a $(1,0)$-vector $W$ and real numbers $\mu _\alpha $ by

Since $k(x_0)$ is the identity, $\sum _ q|\nu _ q|^2=g_0(V_1,V_1)=1$; since $JV_1$ is a $g_0$-unit vector orthogonal to $V_1$, also $\sum _{\alpha >1}\mu _\alpha ^2=1$. The coefficients $\nu _ q,\mu _\alpha $, and $J$ itself, are constant in our coordinate system.

Writing $V_1=\sqrt2\, \overline W-\mathrm iJV_1$ in the outer derivative of $\partial _ ia_{V_1V_1}$, we obtain

Indeed, the third coordinate derivatives of the scalar function $a$ commute, and $\partial _ i\overline W=\sum _ q\overline\nu _ q\partial _ i\partial _{\bar q}$. Replacing coordinate second derivatives by covariant ones before the last differentiation introduces only first derivatives of $\Gamma $ times the bounded gradient of $a$, since $\Gamma (x_0)=0$; and the derivatives of the fixed matrix $k$ are bounded. Thus the error $E_ i$ involves only the gradient bound.

We use twice the inequality $|b+c|^2\leq (1+t)|b|^2+(1+t^{-1})|c|^2$, $t>0$. First take $t=\varepsilon /(1-2\varepsilon )$ to separate $E_ i$; then $(1-2\varepsilon )(1+t)=1-\varepsilon $ and $(1-2\varepsilon )(1+t^{-1})\leq \varepsilon ^{-1}$, so the contribution of $E_ i$ to $(1-2\varepsilon )N$ is at most $CT/(\varepsilon \lambda ^2)\leq T$. Then take $t=\varepsilon $ for the two principal terms (with the factor $1+\varepsilon $ on the second one). This gives

By the weighted Cauchy–Schwarz inequality and $\sum _ qr_ q|\nu _ q|^2\leq C$, the first line is at most $CR/(\varepsilon \lambda )\leq R$. For the second line the Cauchy–Schwarz inequality and $\sum _{\alpha >1}\mu _\alpha ^2=1$ give

To bound the first factor we use the positivity of $G$. Since the Hessian eigenvalue of $V_\alpha $ is $\lambda _\alpha +1$ (this is where the perturbation by $B_0$ enters) and $\partial \bar\partial a(W,\overline W)=\frac12\bigl (a_{V_1V_1}+a_{JV_1JV_1}\bigr )$ at $x_0$, we have

Thus $\lambda -\sum _{\alpha >1}\lambda _\alpha \mu _\alpha ^2\leq 2\lambda +3$. If $\lambda \geq 3/(2\varepsilon ^2)$, then $(1-\varepsilon ^2)(2\lambda +3)\leq 2\lambda $, so the second line is at most $U$. Finally, the critical point equation for $\widehat Q$ at $x_0$ reads

Together with $|b+c|^2\leq 2|b|^2+2|c|^2$ and $4\varepsilon <2$, it bounds the remaining part of $N$:

Combining the three bounds, we obtain

Now choose $A=C_0+3$, after enlarging $C_0$ so that it is positive, and $\varepsilon =[4A^2(S+1)]^{-1}<1/2$. Both choices are uniform, and so are the thresholds for $\lambda $ used above. By (16.16.3), the $\eta ''$ term of (16.16.4) absorbs the last term of (16.16.6), and $4\varepsilon A^2\sum _ i|a_ i|^2/r_ i\leq sT/(S+1)\leq T$. Substituting (16.16.6) into (16.16.4) and dropping its nonnegative $\beta $ terms, we get

By the lower bound for $\eta '$ in (16.16.3) and $r_ i^{-1}\geq c$, this bounds every $a_{ip}$ and $a_{i\bar p}$ at $x_0$, and hence every real second derivative of $a$ there; at $x_0$ these coincide with the covariant ones. Thus $\lambda =\ell (x_0)$ is uniformly bounded, either by the thresholds assumed above or by this estimate. As observed at the beginning, the maximum property of $Q$ then bounds $\ell $ everywhere, and the lower bound for the trace bounds the remaining eigenvalues of $\nabla ^2a$. This proves the lemma. $\square$