Lemma 8.4.4.C357698AC1AF At a point with $\lambda _1>1$, the largest eigenvalue has a smooth local lower support $\widetilde\lambda $, equal to $\lambda _1$ at the point, for which
\[ L\log \widetilde\lambda \geq \frac{f_{1\bar1}}{\lambda _1}+N -\sum _ iF_ i|Z_ i|^2-C\mathcal F -\frac{C}{\lambda _1^2}\sum _ iF_ i|H_{1\bar1;i}|, \]
where
\[ N=-\frac{D^2F_ W(\nabla _1W,\nabla _{\bar1}W)}{\lambda _1}, \qquad Z_ i=\frac{H_{i\bar1;1}}{\lambda _1}. \]
Here $C$ depends only on the fixed metric. Derivatives of $F$ are evaluated at $W$.
Proof.
Choose a smooth unitary frame parallel for the Chern connection at the point and diagonalizing $H$ there. The positive semidefinite tensor $B$, diagonal in this frame with entries $0,\beta ,\ldots ,\beta $, satisfies $\nabla B=0$ at the point and $|\nabla ^2B|\leq C\beta $. For $0<\beta <1$, the largest eigenvalue of $W-B$ is simple, lies below that of $W$, and agrees with it at the point. This gives the required smooth support. Only its largest eigenvalue is perturbed; the equation still uses $F(W)$. Differentiating the normalized eigenvector equation gives
\[ \widetilde\lambda _{;k}=H_{1\bar1;k},\qquad \widetilde\lambda _{;k\bar k}\geq H_{1\bar1;k\bar k} +\sum _{p>1}\frac{|H_{p\bar1;k}|^2+|H_{1\bar p;k}|^2}{\lambda _1-\lambda _ p+\beta }-C\beta . \]
The denominators are at most $(n+1)\lambda _1$. Let $U=\partial \bar\partial u$, and take $T^ p_{ki}=\Gamma ^ p_{ki}-\Gamma ^ p_{ik}$. The first commutations are
\[ U_{i\bar j;k}=U_{k\bar j;i}-T^ p_{ki}U_{p\bar j},\qquad U_{i\bar j;\bar k}=U_{i\bar k;\bar j} -\overline{T^ p_{kj}}U_{i\bar p}. \]
Apply the first identity, commute one mixed pair of covariant derivatives, and apply the conjugate identity. The result has the form
\[ U_{1\bar1;k\bar k}=U_{k\bar k;1\bar1} -2\operatorname {Re}\sum _ p U_{k\bar p;1}\overline{T^ p_{k1}}+E_ k, \qquad |E_ k|\leq C|U|. \]
Indeed the mixed commutator contracts curvature with $U$, differentiating the first torsion coefficient contracts $\nabla T$ with $U$, and the second use of the first identities produces contractions of two torsion coefficients with $U$. Thus all the omitted terms satisfy the displayed bound. Since $g$ is parallel, the same derivative relations apply to $H$ apart from the explicit $g$ terms in the first commutation. At the diagonal point $|U|\leq C\lambda _1$. For $p>1$, rewrite $U_{k\bar p;1}=U_{1\bar p;k}+T^ j_{k1}U_{j\bar p}$. Young's inequality absorbs its third derivative into the positive eigenvalue square, at the cost of $C\lambda _1$. Consequently
\[ \widetilde\lambda _{;k\bar k}\geq H_{k\bar k;1\bar1} -2\operatorname {Re}(H_{k\bar1;1}\overline{T^1_{k1}})-C\lambda _1. \]
Differentiate $F(W)=f$ twice and sum this inequality with weights $F_ k$. Write $D_ k=H_{1\bar1;k}$ and $t_ k=T^1_{k1}$. The first commutation, using $U_{1\bar1}=\lambda _1-1$, gives
\[ D_ k=\lambda _1Z_ k-(\lambda _1-1)t_ k. \]
The negative logarithmic square and the remaining torsion term therefore satisfy
\[ -\frac{|D_ k|^2}{\lambda _1^2}-2\operatorname {Re}(Z_ k\bar t_ k) \geq -|Z_ k|^2-C-C\lambda _1^{-2}|D_ k|. \]
For example, expanding the left side leaves $-|Z_ k|^2-2\lambda _1^{-1}\operatorname {Re}(Z_ k\bar t_ k)-(1-\lambda _1^{-1})^2|t_ k|^2$, and $\lambda _1|Z_ k|\leq |D_ k|+C\lambda _1$. This proves the claim, uniformly for small $\beta $.
$\square$
Pome
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