14.14 The largest-eigenvalue estimate148E6041EC3F

The matrix structure now gives a bound for the complex Hessian in terms of the squared gradient. We apply the maximum principle to the largest eigenvalue of $g$, corrected by functions of the gradient and of the potential. The missing torsion indices allow the concavity square to absorb the third derivatives. If the smallest eigenvalue is not too negative, the positive background closes the estimate; otherwise that eigenvalue produces a large positive Hessian square.

We keep the notation of section 14.13. Plain subscripts on scalar functions and subscripts after a comma on matrix coefficients denote coordinate derivatives, and the coefficients of the Chern connection of $\gamma $ are written $\Gamma ^ q_{ik}$. Choose $\tau >0$ with $Q_0\geq \tau \gamma $.

14.14.2 Differentiating the largest eigenvalue.9994CF39B2B0

Since $\lambda _1$ need not be smooth where it is a multiple eigenvalue, we first perturb $g$. Choose a smooth local unitary frame that is parallel for the Chern connection at the point, with first vector an eigenvector for $\lambda $. Let $B$ be the tensor that is diagonal in this frame, with eigenvalues $0,\epsilon _0,\ldots ,\epsilon _0$, where $\epsilon _0>0$ is fixed and small. Then $B\geq 0$, $\nabla B=0$ at the point, and its second derivatives are bounded by fixed constants. The largest eigenvalue $\widetilde\lambda $ of $g-B$ is simple near the point, is at most $\lambda _1(g)$, and equals $\lambda $ at the point. Hence the smooth function $\widetilde H$, defined like $H$ with $\widetilde\lambda $ in place of $\lambda _1$, also has a local maximum there. Only the largest eigenvalue needs to be simple.

For a family of self-adjoint matrices with a simple largest eigenvalue, differentiate the equation for the normalized eigenvector. Its component in the $p$th eigenspace is the corresponding derivative of the matrix divided by the eigenvalue gap. Differentiating the eigenvalue once more therefore gives

This formula remains valid if some of the lower eigenvalues coincide, with the sum taken over any orthonormal basis of their eigenspaces. The gaps are positive and at most $n\lambda +\epsilon _0$. In holomorphic coordinates, $a_ k=g_{1\bar1,k}+O(\lambda )$, while converting the second covariant derivative to partial derivatives introduces an error of at most $C\sum _ p|g_{1\bar p,k}|+C\lambda $. We also convert the off-diagonal covariant derivatives in the positive squares, with errors $O(\lambda )$, and absorb the linear off-diagonal terms by Young's inequality. Thus, after decreasing a fixed positive coefficient,

The precise positive constant will not be used.

Write $V_ k=\partial _ k(|\partial v|_\gamma ^2)$. The first derivative of $\widetilde H$ gives

In the second derivative, the derivatives of the fixed metric in $|\partial v|_\gamma ^2$ cost only a small fraction of the positive Hessian squares and a constant. Therefore

The third and fourth terms are obtained by differentiating $\sum \gamma ^{p\bar q}v_ pv_{\bar q}$ twice and applying $ab\leq \varepsilon a^2+C_\varepsilon b^2$ to each term containing a derivative of the metric.

14.14.6 The third derivatives and the torsion term.D5E11D27FAAA

We now estimate the third derivatives in this inequality. Define

The first differentiated equation is

Since $P_\gamma (W)=Z$ and the connection preserves $\gamma $, one has the exact contractions

The weights in these identities are evaluated at the point; they are not differentiated.

The large weights are controlled as follows. By (14.13.6), the only weight $t_ i$ not bounded by $(n-1)f_1$ is $t_1$. In $Z_{1\bar1}$, every gradient index differs from $1$, by (14.13.4). Thus a pure second derivative in $\nabla _ pZ_{1\bar1}$ has an index $q>1$, and it can be weighted by $f_ q$, which is comparable to $\mathcal F$; here we use that pure derivatives commute in these coordinates. A mixed second derivative can similarly be placed in column $q$ by conjugate symmetry. All other $t_ i$ are bounded by a constant times every $f_ k$, and derivatives of coefficients cost $C\sqrt K\mathcal F$. Applying these observations to the first identity in (14.14.8), and then to (14.14.7), we obtain, for each $p$,

The connection terms in the differentiated equation contribute weighted mixed second derivatives and terms bounded by $C\sqrt K\mathcal F$, which are already on the right. The bounded derivatives of $h$ are absorbed using the positive lower bound for $\mathcal F$ in lemma 14.13.5.

Next we insert the second derivative of the equation into the eigenvalue estimate. Since partial derivatives of $v$ commute,

Differentiating the equation twice covariantly gives

The conversion to partial derivatives uses

We rewrite the derivative of $g$ in this term using $g_{k\bar q,1}=g_{1\bar q,k}+W_{k\bar q,1}-W_{1\bar q,k}+O(1)$ and its conjugate. The first part is absorbed by the off-diagonal squares in (14.14.3), at the cost of $C\lambda \mathcal F+C\sum _ k f_ k|g_{1\bar1,k}|$. The part involving $W_{1\bar q,k}$ is bounded by $C\lambda \mathcal F+C\sum _{k,p}f_ k|v_{pk}|$. For the remaining part we use

This cancels the partial second derivative $-W_{k\bar k,1\bar1}$ above, and we obtain

The second contraction in (14.14.8) bounds the negative torsion term here. In the part with weight $t_1$, the possible third derivatives are $v_{q1\bar1}$ and their conjugates with $q>1$; all other parts have weights bounded by $Cf_1$. We commute partial derivatives and use

The exclusions in lemma 14.13.2 ensure that any remaining pure derivative $v_{11}$ carries only the small weight $Cf_1$, and derivatives of the connection contribute second derivatives with the same weighted bounds. It follows that

For the positive torsion term, write $W_{1\bar1}=\sum _ p(c^ pv_ p+\overline{c^ p}v_{\bar p})$ with fixed smooth coefficients $c^ p$. Two derivatives give $c^ p\sum _ k f_ kv_{pk\bar k}$ and its conjugate, plus terms bounded by $C(S+\sqrt K\mathcal F)$. The former are bounded by (14.14.9), and in $S$ the mixed second derivatives can be bounded by $C\lambda $, giving

Substitute these estimates in (14.14.10), multiply (14.14.5) by $f_ k$ and sum over $k$. The third derivatives in the latter are bounded by (14.14.9) and $\phi '|\partial v|\leq C/\sqrt K$, and the resulting linear terms in the second derivatives are absorbed by a small part of $E/K$, at the cost of $C\mathcal F$. Since $\widetilde H$ has a local maximum, $\sum _ kf_ k\widetilde H_{k\bar k}\leq 0$, and thus

14.14.12 Cancellation when the smallest eigenvalue is controlled.DA6F05F8F203

Fix $0<\delta <1/2$, to be chosen after $D_1,D_2$. First suppose that $\lambda _ n\geq -\delta \lambda $. Put $I=\{ k:f_ k>\delta ^{-1}f_1\} $, so that $1\notin I$. Equation (14.14.4) and $\phi ''=2(\phi ')^2$ give

Choose $\delta $ small enough that $4\delta (\psi ')^2\leq \psi ''/2$ throughout $[-M,0]$. By (14.13.8),

Here we used $(1-\delta )/(1+\delta )\geq 1-2\delta $.

We need the following comparison of the two derivatives $\nabla _1g_{k\bar1}$ and $a_ k=\nabla _ kg_{1\bar1}$. For every $\varepsilon >0$, provided $\lambda /K$ is sufficiently large depending on the fixed function $\psi $,

To prove this, let $U=\sum _{p>1,q}|v_{pq}|$. Commuting partial derivatives in $g$ gives, for $k>1$,

The second inequality uses the absence of $v_1$ in $Z_{k\bar1}$, so that differentiation in the direction $1$ introduces no pure derivative $v_{11}$. All mixed second derivatives are bounded by $C\lambda $, and every other pure second derivative occurs in $U$.

Use $|a+e|^2-|a|^2\geq -2|a||e|$, and then (14.14.4). For $k>1$, the pure second derivatives in $V_ k$ all occur in $U$. With $H_ k=\sum _ p|v_{p\bar k}|$, the resulting negative terms, divided by $\lambda ^2$, are bounded by a constant times

Here we used $H_ k\leq C\lambda $ and $\lambda \geq K$ to simplify the cross terms. The weight comparison (14.13.6), the symmetry of pure derivatives and the Cauchy–Schwarz inequality give $\mathcal F U^2\leq C E$. Consequently, after multiplication by $f_ k$ and summation, the first two terms are bounded by any fixed small fraction of $E/K$ plus $C\mathcal F$. The third is at most $CE/(\lambda \sqrt K)$, which is absorbed by another such fraction when $\lambda /K$ is large. For the last two terms we use

The contribution of $U^2$ is at most $C|\psi '|E/\lambda $, which can again be absorbed when $\lambda /K$ is large. Choosing the fractions with sum $1/12$ proves (14.14.13).

Insert this cancellation and the preceding inequalities into (14.14.11). The remaining derivative error satisfies, by (14.14.4),

after Young's inequality and an adjustment of constants. The torsion contraction also satisfies

The first inequality uses again the missing gradient index in $Z_{1\bar1}$. Since $\psi '<0$ and $v_{k\bar k}=\lambda _ k-\chi _{k\bar k}-W_{k\bar k}$, we conclude, after decreasing the remaining coefficient of $E$ if necessary, that

For the first term we used $|v_{1\bar1}|^2\geq \lambda ^2/2-CK$, which follows from $\chi +W=O(\sqrt K)$, and $E\geq f_1|v_{1\bar1}|^2$.

For large $\lambda $, lemma 14.13.5 bounds the last sum below by $\tau \mathcal F/2$. Choose first $\varepsilon $ with $\varepsilon C_0<\tau /4$. Next choose $D_2$ so that $\psi ''/2>C_\varepsilon |\psi '|$, and then choose $D_1$ so that $|\psi '|\tau /4>C_0$ on $[-M,0]$. Finally choose $\delta $ as above. All these choices depend only on the fixed data. Inequality (14.14.14) then forces $\lambda ^2\leq 80\, \delta ^{-1}\sup _{[-M,0]}|\psi '|^2K^2$, which proves the desired bound in the first case.

14.14.15 The remaining eigenvalue configuration.57C7A82680F3

Suppose now that $\lambda _ n<-\delta \lambda $, keeping the same constants. Since $f_ n\geq \mathcal F/n$ and $|\chi +W|\leq C\sqrt K$,

In (14.14.11), discard $\mathcal C$ and the term with $\psi ''$. Since $\psi $ is now fixed, $|\psi '\sum _ k f_ kv_{k\bar k}|\leq C\lambda \mathcal F$. Using $g_{1\bar1,k}=a_ k+O(\lambda )$ and Young's inequality, we bound the last error term by $\frac12\sum _ k f_ k|a_ k|^2/\lambda ^2+C\mathcal F$. This leaves

By (14.14.4) and the scalar inequality $\tfrac 32|a+b|^2\leq 2|a|^2+6|b|^2$,

Since $\lambda \geq K$, combining the last two inequalities and dividing by $\lambda \mathcal F>0$ gives $\lambda /K\leq C$. This finishes both cases. Since $H$ attains its maximum at the chosen point, the bound for $\lambda _1/K$ there gives a bound at all points, and $|\lambda _ i|\leq (n-1)\lambda _1$ then bounds every entry of the complex Hessian, as asserted.