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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

\begin{align*} a_ k:=\widetilde\lambda _ k& =\nabla _ k g_{1\bar1},\\ \widetilde\lambda _{k\bar k} & =\nabla _{\bar k}\nabla _ k g_{1\bar1}+O(1) +\sum _{p>1}\frac{|\nabla _ k g_{1\bar p}|^2+ |\nabla _ k g_{p\bar1}|^2}{\lambda -\lambda _ p+\epsilon _0}. \end{align*}

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,

E4B294710238 14.14.3
\begin{equation} \widetilde\lambda _{k\bar k} \geq g_{1\bar1,k\bar k} +\frac1{8n\lambda }\sum _{p>1} (|g_{1\bar p,k}|^2+|g_{p\bar1,k}|^2) -C(|g_{1\bar1,k}|+\lambda ). \end{equation}

The precise positive constant will not be used.

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

1E20E0CF9A10 14.14.4
\begin{equation} \frac{a_ k}{\lambda }=-\phi ' V_ k-\psi 'v_ k, \qquad |V_ k|\leq C\sqrt K\sum _ p(|v_{pk}|+|v_{p\bar k}|)+CK. \end{equation}

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

9.36
\begin{align} \widetilde H_{k\bar k}\geq {}& \frac{\widetilde\lambda _{k\bar k}}\lambda -\frac{|a_ k|^2}{\lambda ^2} +\phi '\sum _ p(v_ pv_{\bar p k\bar k}+v_{\bar p}v_{pk\bar k}) +\frac1{5K}\sum _ p(|v_{pk}|^2+|v_{p\bar k}|^2)\notag \\ & +\phi ”|V_ k|^2+\psi ”|v_ k|^2+\psi 'v_{k\bar k}-C. \tag{14.14.5} \end{align}

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.

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